BEE Paper-1 — Chapter 3: Basics of Energy & Its Forms
292 questions — 201 objective (1 mark), 75 short (5 marks), 16 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation. ▶ Practice this chapter interactively (timer, read-aloud, progress saving).
Objective questions (1 mark) — 201
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
1. If wet bulb and dry bulb temperatures read the same, the relative humidity is ____.
0%
50%
100%
none of the above
Answer: C) 100%
Confirmed vs Book-1 §3.4 — When WBT = DBT the air is saturated, so relative humidity is 100%. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2021
📖 §3.5 Energy units and conversions
2. 1 kWh is equivalent to
86000 cal
10000 Wh
3.6 MJ
none of the above
Answer: C) 3.6 MJ
Confirmed vs Book-1 §3.5 — 1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J = 3.6 MJ. Book-1 Ch.3, Energy units and conversions.
Source: Sep 2021
📖 §3.4 Temperature — Celsius, Fahrenheit and Kelvin scales
3. A temperature of -40 deg F will be ____ deg C?
0
-10
-40
none of the above
Answer: C) -40
Confirmed vs Book-1 §3.4 — -40 deg F equals -40 deg C; the two scales coincide at -40. Book-1 Ch.3, Temperature — Celsius, Fahrenheit and Kelvin scales.
Source: Sep 2021
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
4. Unit of maximum demand is ____.
kVAh
kVA
kVAr
kWh
Answer: B) kVA
Confirmed vs Book-1 §3.3 — Maximum demand is billed in kVA (apparent power). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
6. For the purpose of calculating TOE for a designated consumer the calorific value of oil is taken as
10500 kcal/kg
10000 kcal/kg
5000 kcal/kg
8700 kcal/kg
Answer: B) 10000 kcal/kg
Confirmed vs Book-1 §3.5 — 1 tonne of oil equivalent (toe) is based on a calorific value of 10,000 kcal/kg (10^7 kcal/tonne). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2021
📖 §3.5 Energy units and conversions
7. 1 BTU is equal to
252 Joule
252 cal
3600 kcal
3.5 W
Answer: B) 252 cal
Confirmed vs Book-1 §3.5 — 1 BTU ~ 252 calories (~1.055 kJ). Book-1 Ch.3, Energy units and conversions.
Source: Sep 2021
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
8. When the current leads the voltage in an AC electrical circuit, it is caused mainly due to
Inductive load
Resistive load
Capacitive load
none of the above
Answer: C) Capacitive load
Confirmed vs Book-1 §3.3 — In a capacitive load the current leads the voltage. Book-1 Ch.3, Power factor — power triangle kW/kVA/kVAr, PF = cosθ.
Source: Sep 2021
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
9. The power indicated in the name plate of a motor denotes ____.
minimum kW drawn by the motor
maximum kW drawn by the motor
maximum kVA drawn by the motor
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.3 — The motor nameplate kW indicates the rated mechanical (shaft) output power, not input/maximum kW or kVA; hence none of the above. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Source: Sep 2021
📖 §3.4 The laws of thermodynamics
10. The law of conservation of energy is related with
third law of thermodynamics
second law of thermodynamics
first law of thermodynamics
none of the above
Answer: C) first law of thermodynamics
Confirmed vs Book-1 §3.4 — The first law of thermodynamics is the law of conservation of energy. Book-1 Ch.3, The laws of thermodynamics.
Source: Sep 2021
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
11. The producer gas is basically ____.
only CH4
only CO and CH4
CO, H2 and CH4
only CO and H2
Answer: C) CO, H2 and CH4
Confirmed vs Book-1 §3.4 — Producer gas is a mixture of CO, H2 and CH4 (plus N2 and CO2). Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: Sep 2021
📖 §3.4 Steam properties — superheat and dryness fraction (x)
12. The 'superheat' of steam is expressed as ____.
degrees centigrade above saturation temperature
degrees centigrade above critical temperature of the steam
degrees centigrade below the boiling point of water
all of the above
Answer: A) degrees centigrade above saturation temperature
Confirmed vs Book-1 §3.4 — Superheat is the temperature of steam above its saturation temperature at a given pressure. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
Source: Sep 2021
📖 §3.5 Energy units and conversions
13. The electrical power unit GigaWatt (GW) may be expressed as
1,000,000,000 MW
1,000 MW
1,000 kW
10,000 W
Answer: B) 1,000 MW
Confirmed vs Book-1 §3.5 — 1 GW = 1,000 MW = 10^6 kW = 10^9 W. Book-1 Ch.3, Energy units and conversions.
Source: Sep 2021
📖 §3.4 Fuel properties — density, specific gravity, viscosity
14. Which of the following is not true of liquid fuels?
the viscosity of a liquid fuel is a measure of its internal resistance to flow
the viscosity of all liquid fuels decreases with increase in its temperature
higher the viscosity of liquid fuels, higher will be its heating value
viscous fuels need heat tracing
Answer: C) higher the viscosity of liquid fuels, higher will be its heating value
Confirmed vs Book-1 §3.4 — Viscosity has no direct relation to heating value; the statement is false. Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Sep 2021
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
15. 300 litres of water in a tank is heated from 30 deg C to 70 deg C by using a direct steam with an enthalpy of 600 kcal/kg. The mass in kg of steam used is ____.
10
200
40
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.4 — Heat to water = 300 x 1 x (70-30) = 12,000 kcal; steam mass = 12,000/600 = 20 kg, which is not among a/b/c, so none of the above. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Sep 2021
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
16. Which of the following is not a unit of energy?
Joule
Calorie
Watt
BTU
Answer: C) Watt
Confirmed vs Book-1 §3.2 — Watt is a unit of power (energy per unit time), not energy. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: Sep 2021
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
17. What is the heat content of the 200 litres of water at 50 deg C in terms of the basic unit of energy in kilo Joules (kJ)?
3000
4187
1000
41870
Answer: D) 41870
Confirmed vs Book-1 §3.4 — Heat = m c dT = 200 x 4.187 x 50 = 41,870 kJ (taking rise from 0 deg C reference). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Sep 2021
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
18. Power in a 3 phase AC system is
3 x Voltage x Current
Voltage x Current
1.73 x Voltage x Current
None of the above
Answer: D) None of the above
Confirmed vs Book-1 §3.3 — Three-phase active power = sqrt3 x V x I x cos(phi); the listed forms omit power factor, so none of the above. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: Sep 2021
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
19. To maximize the combustion efficiency, it is required to ____ in the flue gas?
maximize O2
maximize CO2
minimize CO2
maximize NOx
Answer: B) maximize CO2
Confirmed vs Book-1 §3.4 — High combustion efficiency corresponds to maximum CO2 (minimum excess air) in flue gas. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: Sep 2021
📖 §3.3 Example 3.6 — resistive load power varies as V²
20. An electric heater of 230 V, 10 kW rating is installed for hot water generation in a hospital. The consumption per hour at 200 V is
10 kWh
8.7 kWh
13.23 kWh
7.56 kWh
Answer: D) 7.56 kWh
Confirmed vs Book-1 §3.3 — P proportional to V^2: P = 10 x (200/230)^2 = 10 x 0.756 = 7.56 kW, so 7.56 kWh in one hour. Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
Source: Sep 2021
📖 §3.4 Fuel properties — density, specific gravity, viscosity
21. The specific gravity of water is expressed as ____.
1
1 kg/m3
1 g/cc
1000 kg/m3
Answer: A) 1
Confirmed vs Book-1 §3.4 — Specific gravity is a dimensionless ratio; for water it is 1. Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Sep 2021
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)
22. The number of moles in 90 kg of water is ____.
5
18
2
none of the above
Answer: A) 5
Confirmed vs Book-1 §3.5 — Moles = mass/molar mass = 90,000 g/18 g/mol... in kmol: 90/18 = 5 kmol. Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
Source: Sep 2021
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy
23. The type of energy possessed by the charged capacitor is
kinetic energy
electrostatic
potential
magnetic
Answer: B) electrostatic
Confirmed vs Book-1 §3.1 — A charged capacitor stores energy in the electrostatic field between its plates. (Answer not marked in source.). Book-1 Ch.3, Energy types & forms — potential (stored) vs kinetic energy.
Source: Apr 2010
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
24. The heat required to change a substance from liquid to vapor state without change of temperature is termed as
latent heat of fusion
latent heat of vaporization
heat capacity
sensible heat
Answer: B) latent heat of vaporization
Confirmed vs Book-1 §3.4 — Liquid-to-vapour phase change at constant temperature absorbs the latent heat of vaporization. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Apr 2010
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
25. The difference between GCV and NCV of coal is
the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapor) in the fuel
the difference in heat released by using theoretical air and allowable excess air
difference in accounting the un-burnt content in the ash
none of the above
Answer: A) the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapor) in the fuel
Confirmed vs Book-1 §3.4 — GCV minus NCV equals the latent heat of the water formed from fuel moisture and hydrogen. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
26. The calorific value of coal is 4,200 kCal/kg. Find out the oil equivalent of 1000 kg of coal if the calorific value of oil is 10,000 kCal/kg
42,000 kg
96 kg
420 kg
128 kg
Answer: C) 420 kg
Confirmed vs Book-1 §3.5 — Oil equivalent = (1000 x 4200)/10000 = 420 kg. (Answer not marked in source.). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Apr 2010
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
27. Name plate kW rating of an induction motor indicates
input kW to the motor
output kW of the motor
minimum input kW to the motor
maximum input kW to the motor
Answer: B) output kW of the motor
Confirmed vs Book-1 §3.3 — Motor nameplate kW is the rated shaft (output) power. (Answer not marked in source.). Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Source: Apr 2010
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
28. Normally the maximum demand is charged for
kWh
kWAh
kVA
kVArh
Answer: C) kVA
Confirmed vs Book-1 §3.3 — Maximum demand is billed in kVA. (Answer not marked in source.). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
Source: Apr 2010
📖 §3.3 Example 3.6 — resistive load power varies as V²
29. For an electric heater, voltage remaining constant, the heat output ___ when resistance decreases.
decreases
increases
first increases then decreases
remains same
Answer: B) increases
Confirmed vs Book-1 §3.3 — P = V^2/R, so at constant V, lower R gives higher heat output. (Answer not marked in source.). Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
Source: Apr 2010
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
30. The active power consumption of AC 3-phase motive drive is determined by using which one of the following relations.
sqrt3 x V x I
sqrt3 x V^2 x I x Cosφ
3 x V x I x Cosφ
sqrt3 x V x I x Cosφ
Answer: D) sqrt3 x V x I x CosO
Confirmed vs Book-1 §3.3 — Three-phase active power = sqrt3 x V x I x cos(phi). (Answer not marked in source.). Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: Apr 2010
📖 §3.4 The laws of thermodynamics
31. The law of conservation of energy states that energy
can be created and destroyed
is destroyed in the process of burning
cannot be converted from one form to another
is neither destroyed nor created
Answer: D) is neither destroyed nor created
Confirmed vs Book-1 §3.4 — Energy is neither created nor destroyed, only transformed. (Answer not marked in source.). Book-1 Ch.3, The laws of thermodynamics.
Source: Apr 2010
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
32. Latent heat is best described as the amount of heat required to cause a change in
both temperature and state
specific heat
state without a change in temperature
temperature without a change in state
Answer: C) state without a change in temperature
Confirmed vs Book-1 §3.4 — Latent heat changes the state at constant temperature. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Pressure: gauge pressure pg = ps - pa, so the absolute (system) pressure ps = gauge pressure + atmospheric pressure. Gauges are calibrated to read zero at atmospheric pressure, hence the atmospheric term must be added back.
Source: 2016
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
34. When heat flows from one place to another by means of liquid or gas, it is being transferred by
radiation
conduction
sublimation
convection
Answer: D) convection
Confirmed vs Book-1 §3.4 — Heat transfer by movement of a fluid is convection. (Answer not marked in source.). Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
Source: Apr 2010
📖 §3.5 Energy units and conversions
35. How many Watts are equivalent to one HP?
760
725
740
746
Answer: D) 746
Confirmed vs Book-1 §3.5 — 1 HP = 746 W. (Answer not marked in source.). Book-1 Ch.3, Energy units and conversions.
Source: Apr 2010
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
36. A process requires 100 kg of fuel with a calorific value of 5000 kCal/kg. If the system efficiency is 80%, then the losses would be
100000 kCal
400000 kCal
50000 kCal
20000 kCal
Answer: A) 100000 kCal
Confirmed vs Book-1 §3.4 — Input = 100 x 5000 = 500,000 kCal; losses = 20% = 100,000 kCal. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Apr 2010
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
37. The lagging reactive power is required for
inductive load
resistive load
capacitive load
all of the above
Answer: A) inductive load
Confirmed vs Book-1 §3.3 — Inductive loads draw lagging reactive power. (Answer not marked in source.). Book-1 Ch.3, Power factor — power triangle kW/kVA/kVAr, PF = cosθ.
Source: Apr 2010
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
38. To maximize the combustion efficiency, which of the following in the flue gas needs to be done?
maximize O2
maximize CO2
minimize CO2
maximize CO
Answer: B) maximize CO2
Confirmed vs Book-1 §3.4 — Complete combustion converts all carbon to CO2; the CO2 in flue gas is therefore highest when combustion is complete and excess air is minimum. Maximising O2 means excess air (dilution and stack loss); CO indicates incomplete combustion.
Source: 2017
📖 §3.1 Chemical energy — fuels store chemical energy
39. Propane is an example of
nuclear energy
radiant energy
chemical energy
thermal energy
Answer: C) chemical energy
Confirmed vs Book-1 §3.1 — Propane (a fuel) stores chemical energy. (Answer not marked in source.). Book-1 Ch.3, Chemical energy — fuels store chemical energy.
Source: Nov 2009
📖 §3.4 Fuel properties — density, specific gravity, viscosity
40. The density of a fuel oil is 0.86. Its specific gravity will be
0.75
0.86
1.75
0.0086
Answer: B) 0.86
Confirmed vs Book-1 §3.4 — Specific gravity equals density relative to water (=0.86). (Answer not marked in source.). Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Nov 2009
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
41. An indication of sensible heat content in air-water vapour mixture is
wet bulb temperature
dew point temperature
density of air
dry bulb temperature
Answer: D) dry bulb temperature
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Dry bulb measures sensible heat content in air-vapour mixtures' and is not influenced by RH. Wet-bulb accounts for RH (latent effect) and dew point is the saturation temperature.
Source: 2017
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy
42. The total mechanical energy of a body free falling in a vacuum
increases
decreases
remains the same
depends on the shape of the body
Answer: C) remains the same
Confirmed vs Book-1 §3.1 — With no air resistance, total mechanical energy is conserved. (Answer not marked in source.). Book-1 Ch.3, Energy types & forms — potential (stored) vs kinetic energy.
Source: Nov 2009
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy
43. Which of the following is false?
electricity is high-grade energy
high grade forms of energy are highly ordered and compact
low grade energy is better used for applications like melting of metals rather than heating water for bath
the molecules of low grade energy are more randomly distributed than the molecules of carbon in coal
Answer: C) low grade energy is better used for applications like melting of metals rather than heating water for bath
Confirmed vs Book-1 §3.1 — Statement (c) is false. Low-grade (disordered, low-temperature) energy is best used for low-temperature duty such as bath-water heating; high-grade energy such as electricity is needed for melting metals. The other three statements are true.
Source: 2017
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ
44. What will be the energy saving if one 1500 W, 25 litre water heater, normally on for 20 minutes/day for 250 days/year, is replaced with a 100 litre solar water heater?
581 units
750 units
125 units
169 units
Answer: C) 125 units
Confirmed vs Book-1 §3.3 — Energy = 1.5 kW x (20/60) h x 250 = 125 kWh (units) saved per year. (Answer not marked in source.). Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
Source: Nov 2009
📖 §3.4 Fuel properties — density, specific gravity, viscosity
45. Which of the following is not applicable to liquid fuels?
the viscosity of a liquid fuel is a measure of its internal resistance to flow.
the viscosity of all liquid fuels decreases with increase in its temperature
higher the viscosity of liquid fuels, higher will be its heating value
viscous fuels need heat tracing
Answer: C) higher the viscosity of liquid fuels, higher will be its heating value
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Fuel properties: viscosity is the internal resistance to flow and falls as temperature rises; heating value correlates with SPECIFIC GRAVITY, not viscosity. So (c) is the statement that does not apply.
Source: 2017
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
46. Which of the following will be true of load factor for a continuous process
higher than batch process plants
comparable to that of a five star hotel with 60% occupancy
less than that of an energy efficient municipal lighting system
closer to the regional grid load factor
Answer: A) higher than batch process plants
Confirmed vs Book-1 §3.3 — Continuous processes run steadily, giving a higher load factor than batch plants. (Answer not marked in source.). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
Source: Nov 2009
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
47. In a heat treatment furnace the material is heated up to 800 deg C from ambient 30 deg C. With specific heat 0.13 kCal/kg deg C, what is the energy content in one kg of material after heating?
700 kCal
250 kCal
350 kCal
100 kCal
Answer: D) 100 kCal
Confirmed vs Book-1 §3.4 — Q = m Cp dT = 1 x 0.13 x (800-30) = 0.13 x 770 = 100.1 kCal ~ 100 kCal. (Answer not marked in source.). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Nov 2009
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
48. Condensation of saturated steam releases
sensible heat
super heat
latent heat
none of the above
Answer: C) latent heat
Confirmed vs Book-1 §3.4 — Saturated steam condensing releases its latent heat. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Nov 2009
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
49. Calorific Value of coal is measured by a device called
bomb calorimeter
calorifier
infrared thermometer
none of these
Answer: A) bomb calorimeter
Confirmed vs Book-1 §3.4 — A bomb calorimeter measures the calorific value of solid fuels. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
53. Energy in one Tonne of Oil Equivalent (toe) corresponds to
4.187 GJ
1.162 MWh
10,000 kcal
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.5 — 1 toe = 10^7 kcal = 4.187 x 10^7 kJ = 41.87 GJ = 11,630 kWh = 11.63 MWh. Option (a) 4.187 GJ, (b) 1.162 MWh and (c) 10,000 kcal (= 1 kg oil equivalent) are all too small, so the answer is 'none of the above'.
Source: 2019
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
54. If 1 kWh of electrical energy is used to heat 10 kg of ice at 0o C, what will be the temperature of water after melting? (Latent heat of fusion of ice is 80 kcal/kg)
0°C
6°C
86°C
none of the above
Answer: B) 6°C
Confirmed vs Book-1 §3.4 — 1 kWh = 860 kcal. Melting 10 kg of ice needs 10 x 80 = 800 kcal, leaving 860 - 800 = 60 kcal. Warming the 10 kg of melt water: dT = 60/(10 x 1) = 6 degC, so the final temperature is 6 degC.
Source: 2019
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
55. The rate of energy transfer from a higher temperature to a lower temperature is measured in
kcal
Watt
Watts per second
none of the above.
Answer: B) Watt
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Heat transfer: 'The energy transferred is measured in Joules. The rate of energy transfer, more commonly called heat transfer, is measured in Watts (J/s).' kcal is a quantity, not a rate; 'Watts per second' is not a unit of rate of heat flow.
56. The theoretical amount of electricity required to heat 500 litres of brine solution with a specific gravity of 1.2 and specific heat of 1 kcal/kg K from 30°C to 70 °C through resistance heating is_________
27.9 kWh
23.3 kWh
20 kWh
none of the above
Answer: A) 27.9 kWh
Confirmed vs Book-1 §3.5 — Mass = 500 litres x 1.2 (sp. gr.) = 600 kg. Q = m·Cp·dT = 600 x 1 x (70-30) = 24,000 kcal. Electricity = 24,000/860 = 27.9 kWh (1 kWh = 860 kcal).
Source: 2019
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
57. SI unit for energy is_____________
Watt
Kilogram
Newton
Joule
Answer: D) Joule
Confirmed vs Book-1 §3.2 — Book-1 §3.2: 'The unit of work or energy is the joule (J), where one joule is one Newton metre.' Watt is power (J/s), kilogram is mass and Newton is force.
58. Which of the following has the lowest energy content in terms of MJ/kg?
LPG
Diesel
Bagasse
Furnace Oil
Answer: C) Bagasse
Confirmed vs Book-1 §3.5 — Bagasse (biomass) has much lower calorific value than LPG, diesel or furnace oil. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2024
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
59. Which of the following industries has the highest Specific Electrical Energy Consumption?
Aluminum
Sugar
Paper & Pulp
Cement
Answer: A) Aluminum
Confirmed vs Book-1 §3.1 — Primary aluminium smelting (Hall-Heroult electrolysis) needs roughly 14,000-16,000 kWh per tonne of metal - far above sugar, paper or cement - so aluminium has the highest specific electrical energy consumption.
Source: 2019
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
60. Which primary energy is used as a feedstock in fertilizer industry?
Steam
Natural gas
Electricity
All of the above
Answer: B) Natural gas
Confirmed vs Book-1 §3.1 — Natural gas is the primary energy source used as FEEDSTOCK (raw material) in fertilizer plants, where it is reformed to hydrogen for ammonia/urea. Steam and electricity are secondary (derived) energy carriers.
Source: 2019
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy
61. Which of the following statements are true? Rice husk is a source of secondary energy ii) nuclear energy is non-renewable energy iii) electricity is basically a convenient form of primary energy iv) steam is a convenient form of secondary energy
(ii) & (iii)
(i) & (iii)
(ii) & (iv)
(ii) & (i)
Answer: C) (ii) & (iv)
Confirmed vs Book-1 §3.1 — (ii) Nuclear energy is non-renewable - TRUE; (iv) steam is a convenient secondary energy form - TRUE. (i) is false because rice husk is a PRIMARY energy source, and (iii) is false because electricity is a SECONDARY (converted) form. Hence (ii) & (iv).
Source: 2019
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
62. An induction motor with 11 kW rating and a rated power factor of 0.9 in its name plate means
it will draw 12.22 kW at full load
it will draw 11 kW at full load
it will draw 9.9 kW at full load
it will deliver 11 kW at full load
Answer: D) it will deliver 11 kW at full load
Confirmed vs Book-1 §3.3 — Book-1 §3.3: the nameplate kW/HP is the motor OUTPUT at full load; the volts, amps and PF are the INPUT conditions. So an 11 kW motor DELIVERS 11 kW at full load and draws more than 11 kW at its input.
Source: 2019
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
63. The sector consuming major share of energy in India
Agriculture Sector
Transport Sector
Industrial Sector
Domestic Sector
Answer: C) Industrial Sector
Confirmed vs Book-1 §3.1 — Industry is the largest energy-consuming sector in India, accounting for roughly half of commercial energy use - ahead of transport, domestic and agriculture.
Source: 2018
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
64. The kW or HP of a motor given on the name plate indicates
The shaft output of the motor at part load
The shaft output of the motor at full load
The input power to the motor at the best efficiency point
The input power to the motor at any load
Answer: B) The shaft output of the motor at full load
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Motor loading calculation: 'The name plate details of motor, kW or HP indicates the output of the motor at full load.' The volt/amp/PF on the plate are the input conditions at that full load.
Source: 2018
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
65. Which of the following has the highest Specific Heat?
Steel
Aluminium
Copper
Water
Answer: D) Water
Confirmed vs Book-1 §3.4 — Book-1 Table 3.1: water 4200 J/kg degC, aluminium 910, iron 470, copper 390. 'The specific heat of water is very high as compared to other common substances.'
Source: 2018
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
66. Heat transfer in an air cooled condenser occurs predominantly by
conduction
convection
radiation
none of the above
Answer: B) convection
Confirmed vs Book-1 §3.4 — In an air-cooled condenser the hot refrigerant/vapour gives up heat to air moving over the finned tubes; the fluid motion carries the heat away, i.e. (forced) convection is the predominant mode.
Source: 2018
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
67. To arrive at the relative humidity at a point we need to know ___________ of air
DBT
WBT
Dew point
Both A and B
Answer: D) Both A and B
Confirmed vs Book-1 §3.4 — Relative humidity is obtained from both dry-bulb (DBT) and wet-bulb (WBT) temperatures. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2024
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
68. What is the heat content of 200 litres of water at 50 °C in terms of the basic unit of energy in kilo Joules (kJ)?
30000
23880
10000
none of the above (BEE awarded 1 mark to every candidate attempting this question)
Answer: D) Note: 1 Mark is awarded to all candidate who have attempted this question.
Confirmed vs Book-1 §3.4 — Q = m·Cp·dT = 200 kg x 4.187 kJ/kg degC x 50 degC = 41,870 kJ, which is not among options (a)-(c). BEE therefore awarded 1 mark to every candidate who attempted this question; the correct value is 'none of the above'.
Source: 2018
📖 §3.4 Temperature — Celsius, Fahrenheit and Kelvin scales
69. Which of the following is used for non-contact measurement of temperature
Thermocouples
Infrared Thermometer
Leaf type contact probe
All of the above
Answer: B) Infrared Thermometer
Confirmed vs Book-1 §3.4 — An infrared (radiation) thermometer senses emitted thermal radiation and therefore needs no contact. Thermocouples and leaf-type contact probes both require physical contact with the surface.
Source: 2018
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
70. The lowest theoretical temperature to which water can be cooled in a cooling tower is
Difference between DBT and WBT of the atmospheric air
Average DBT and WBT of the atmospheric air
DBT of the atmospheric air
WBT of the atmospheric air
Answer: D) WBT of the atmospheric air …….…….
Confirmed vs Book-1 §3.4 — The wet-bulb temperature of the entering air is the theoretical minimum to which evaporative cooling can cool the water; the approach (cold water temp - WBT) can be reduced but never taken to zero.
Source: 2016
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
71. From Voltage, Amps and Power factor given in the name plate of a motor, one can calculate ________.
Rated output power
Shaft power
Rated input power
Both (b) & (c)
Answer: C) Rated input power
Confirmed vs Book-1 §3.3 — Book-1 §3.3: nameplate V, A and PF are the INPUT conditions at full load, so rated input power = sqrt3 x V x I x PF (3-phase). The nameplate kW/HP separately gives the output (shaft) power.
72. Which of the following comes under Capital cost in a project?
Design cost
Installation cost
Commissioning cost
All of the above
Answer: D) All of the above
Confirmed vs Book-1 Ch.3 — Capital cost of an energy project covers the one-time costs of design, supply, installation and commissioning; operating and maintenance costs are recurring (revenue) costs.
Source: 2018
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
73. A three phase induction motor is drawing 10 Ampere at 440 Volts. If the operating power factor of the motor is 0.9 and the efficiency of the motor is 95%, then the mechanical shaft power of the motor is
3.76 KW
4.18 KW
6.51 KW
7.21 KW
Answer: C) 6.51 KW
Confirmed vs Book-1 §3.3 — Input power = sqrt3 x V x I x PF = 1.732 x 440 x 10 x 0.9 = 6859 W = 6.86 kW. Shaft (mechanical) output = input x efficiency = 6.86 x 0.95 = 6.51 kW.
Source: 2018
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
74. The amount of CO2 produced in complete combustion of 18 Kg of Carbon is ______.
50
44
66
792
Answer: C) 66
Confirmed vs Book-1 §3.4 — C + O2 -> CO2: 12 kg carbon gives 44 kg CO2. For 18 kg carbon: CO2 = 18 x 44/12 = 66 kg. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: 2018
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
75. Which mode of heat transfer does not require medium?
Natural convection
Forced convection
Radiation
Conduction
Answer: C) Radiation
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Radiation mode heat transfer requires no medium for the transport of heat.' Conduction needs a solid and convection needs a fluid.
76. The heat rate of a power plant is expressed as
kWh/kg of steam
kCal/kWh
kg of steam / kg of fuel
kWh / kVA
Answer: B) kCal/kWh
Confirmed vs Book-1 §3.5 — Heat rate is the heat input required per unit of electricity generated, expressed in kcal/kWh (or kJ/kWh). It is the inverse of plant efficiency: eta = 860/heat rate.
Source: 2018
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
77. One Silicon cell in a PV module typically produces
0.5 V
1 V
2 V
12 V
Answer: A) 0.5 V
Confirmed vs Book-1 §3.1 — A single crystalline/multi-crystalline silicon solar cell develops an open-circuit voltage of about 0.5-0.6 V; cells are series-connected in a module to reach usable voltages (e.g. 36 cells for a 12 V module).
78. The production factor is defined as the ratio of
current year production to the reference year production
current year production to the reference month production
reference month production to the current month production
reference year production to the current year production
Answer: A) current year production to the reference year production
Confirmed vs Book-1 Ch.3 — Production factor = current year (or period) production / reference year production. It is used to normalise energy consumption to the reference-year output when computing specific energy consumption.
Source: 2018
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
79. To reduce the distribution losses within a plant, the capacitors should be located
Closest to the load
Farthest from the load
In the substation
Before the billing meter
Answer: A) Closest to the load
Confirmed vs Book-1 §3.3 — Capacitors installed closest to the inductive load supply the reactive current locally, so the reactive current no longer flows through the plant cables and transformer - which minimises I²R distribution losses.
Confirmed vs Book-1 §3.4 — Book-1 §3.4: gauge pressure pg = ps - pa, hence absolute pressure ps = gauge + atmospheric. All gas-law calculations must use absolute pressure.
Source: 2018
📖 §3.4 Steam properties — superheat and dryness fraction (x)
81. The dryness (x) fraction of superheated steam is taken as
x= 0
x= 0.9
x= 0.87
x= 1
Answer: D) x= 1
Confirmed vs Book-1 §3.4 — Book-1 §3.4 (T-S diagram): x is the dryness fraction, the mass of steam in 1 kg of the water-steam mixture. Dry saturated and superheated steam contain no moisture, so x = 1 (the region to the right of the x = 1 line is superheated steam).
Source: 2018
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
82. When the evaporation of water from a wet substance is zero, the relative humidity of the air is likely to be
0%
100%
50%
unpredictable
Answer: B) 100%
Confirmed vs Book-1 §3.4 — Evaporation stops when the air can hold no more moisture, i.e. when it is saturated - relative humidity = 100%. At that condition dew-point, wet-bulb and dry-bulb temperatures are equal.
Source: 2018
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
83. Which of the following statements are true? i) reactive current is necessary to build up the flux for the magnetic field of inductive devices ii) some portion of reactive current is converted into work iii) the cosine of angle between kVA and kVAr vector is called power factor iv) the cosine of angle between kW and kVA vector is called power factor
i & iv
ii & iii
i & iii
iii & iv
Answer: A) i & iv
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Power factor: (i) is true - 'the reactive current is necessary to build up the flux for the magnetic field of inductive devices'; (iv) is true - PF = cos of the angle between kW and kVA. (ii) is false (reactive current does no useful work) and (iii) is false (the angle is between kW and kVA, not kVA and kVAr).
Source: 2017
📖 §3.5 Energy units and conversions
84. The electrical power unit Giga Watt (GW) may be written as
1,000,000 MW
1,000 MW
1,000 kW
1,000,000 W
Answer: B) 1,000 MW
Confirmed vs Book-1 §3.5 — 1 GW = 10^9 W = 10^6 kW = 1,000 MW. Book-1 Ch.3, Energy units and conversions.
Source: 2017
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
85. The term missing in the following equation (kVA)² = (kVA cosφ)² + ( ? )² is
cosφ
sinφ
kVA sinφ
kVArh
Answer: C) kVA sin phi
Confirmed vs Book-1 §3.3 — From the power triangle kW = kVA·cosθ and kVAr = kVA·sinθ, so (kVA)² = (kVA cosθ)² + (kVA sinθ)². The missing term is kVA·sinθ (the reactive component).
Source: 2017
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
86. 2000 kJ of heat is supplied to 500 kg of ice at 0°C. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be
1.49
83.75
5.97
None of the above
Answer: C) 5.97
Confirmed vs Book-1 §3.4 — Qₗ = m x h_if, so m = Qₗ/h_if = 2000 kJ / 335 kJ/kg = 5.97 kg. (Only 5.97 kg of the 500 kg of ice melts; the rest stays as ice at 0 degC.)
Source: 2017
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
87. An electric heater draws 5 kW of power for continuous hot water generation in an industry. How much quantity of water in litres per min can be heated from 30°C to 85°C ignoring losses?.
1.3
78.18
275
none of the above
Answer: A) 1.3
Confirmed vs Book-1 §3.4 — Heat available per minute = 5 kW x 60 s = 300 kJ/min = 300/4.187 = 71.6 kcal/min. Water flow = Q/(Cp x dT) = 71.6/(1 x (85-30)) = 1.30 litres/min.
Source: 2017
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
88. An electric heater consumes 1000 Joules of energy in 5 seconds. Its power rating is:
200 W
1000 W
5000W
none of the above
Answer: A) 200 W
Confirmed vs Book-1 §3.2 — Book-1 §3.2: P = W/t = 1000 J / 5 s = 200 J/s = 200 W. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: 2017
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
89. The quantity of heat required to raise the temperature of a given substance by 1 °C is known as:
sensible heat
specific heat
heat capacity
latent heat
Answer: C) heat capacity
Confirmed vs Book-1 §3.4 — Heat capacity is the heat needed to raise the temperature of a GIVEN body/quantity of substance by 1 degC. Specific heat (Book-1 §3.4) is defined per 1 kg of substance, so it is not the right term when no mass is specified.
90. Which of the following parameters is not considered for external Bench Marking?
scale of operation
energy pricing
raw materials and product quality
vintage of technology
Answer: B) energy pricing
Confirmed vs Book-1 Ch.3 — External benchmarking compares plants on technical parameters - scale of operation, vintage of technology, raw material and product quality. Energy PRICE is a commercial/location factor and is excluded because it does not reflect energy performance.
Source: 2017
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)
91. The number of moles of water contained in 36 kg of water is ------------
2
3
4
5
Answer: A) 2
Confirmed vs Book-1 §3.5 — Molar mass of water = 18 g/mol (18 kg/kmol). Moles = 36 kg / 18 kg per kmol = 2 kmol (i.e. 2000 mol). Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
Source: 2017
📖 §3.3 Example 3.6 — resistive load power varies as V²
92. A process electric heater is taking an hour to reach the desired temperature while operating at 440 V. It will take ------- hours to reach the same temperature if the supply voltage is reduced to 220 V.
2
3
4
5
Answer: C) 4
Confirmed vs Book-1 §3.3 — For a fixed resistance, P = V²/R. Halving the voltage from 440 V to 220 V gives one quarter of the power, so the same heat requires four times the time: 1 h x 4 = 4 hours.
Source: 2017
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
93. The component of electric power which yields useful mechanical power output is known as
apparent power
active power
reactive power
none of the above
Answer: B) active power
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Power factor: 'The resistive portion is also known as the active power which is directly converted to useful work.' Reactive power builds flux only; apparent power (kVA) is the vector sum.
Source: 2017
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
94. An oil fired boiler is retrofitted to fire coconut shell chips. Boiler thermal efficiency drops from 82% to 70%. What will be the percentage change in energy consumption to generate the same output
12% increase
14.6% increase
17.1% decrease
17.1% increase
Answer: D) 17.1% increase
Confirmed vs Book-1 §3.4 — For the same useful output, fuel energy is inversely proportional to efficiency. Ratio = 82/70 = 1.171, so the energy consumption rises by 17.1%.
Source: 2017
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
95. A three phase induction motor is drawing 16 Ampere at 440 Volts. If the operating power factor of the motor is 0.90 and the motor efficiency is 92%, then the mechanical shaft power output of the motor is
12.04 kW
10.09 kW
10.97 kW
None of the above
Answer: B) 10.09 kW
Confirmed vs Book-1 §3.3 — Input power = sqrt3 x V x I x PF = 1.732 x 440 x 16 x 0.90 = 10,974 W = 10.97 kW. Shaft output = 10.97 x 0.92 = 10.09 kW. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: 2017
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
96. If a 2 KW immersion heater is used to heat 30litres of water at 30°C, what would be the temperature of water after 15 minutes? Assume no losses in the system
87.3 °C
44.3°C
71.3 °C
none of the above
Answer: B) 44.3°C
Confirmed vs Book-1 §3.4 — Energy = 2 kW x 0.25 h = 0.5 kWh = 0.5 x 860 = 430 kcal. Temperature rise dT = Q/(m·Cp) = 430/(30 x 1) = 14.3 degC. Final temperature = 30 + 14.3 = 44.3 degC.
Source: 2016
📖 §3.4 Fuel properties — density, specific gravity, viscosity
97. Red wood seconds is a measure of
Density
Viscosity
Specific gravity
Flash point
Answer: B) Viscosity
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Viscosity: 'Viscosity is measured in Stokes/Centistokes. Sometimes viscosity is quoted in Engler, Saybolt or Redwood.' Redwood seconds is therefore a viscosity measure.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
99. For every 10°C rise in temperature, the rate of chemical reaction doubles. When the temperature is increased from 30°C to 70°C, the rate of reaction increases __________ times.
8
64
16
none of the above
Answer: C) 16
Confirmed vs Book-1 §3.4 — A rise of 70 - 30 = 40 degC contains 40/10 = 4 doublings, so the rate increases by 2^4 = 16 times. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: 2016
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
100. If the reactive power drawn by a particular load is zero it means the load is operating at
Lagging power factor
Unity power factor
Leading power factor
none of the above
Answer: B) Unity power factor
Confirmed vs Book-1 §3.3 — kVAr = kVA·sinθ. Zero reactive power means sinθ = 0, i.e. θ = 0 and PF = cosθ = 1 - a purely resistive load operating at unity power factor.
Confirmed vs Book-1 Ch.3 — Capital cost is the one-time investment - design, installation and commissioning of the project. Operation and maintenance costs are recurring operating costs, not capital cost.
Source: 2016
📖 §3.5 Energy units and conversions
102. The kilowatt-hour is a unit of
power
work
time
force.
Answer: B) work
Confirmed vs Book-1 §3.5 — The kilowatt-hour is power x time = energy (work). 1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J (Book-1 §3.3). Book-1 Ch.3, Energy units and conversions.
Source: 2016
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
103. Which among the following is a green house gas?
Sulphur Dioxide
Carbon Monoxide
NO2
Methane
Answer: D) Methane
Confirmed vs Book-1 §3.1 — Methane (CH4) is a greenhouse gas and, weight for weight, traps about 21 times more heat than CO2. SO2 and CO are air pollutants but not counted as GHGs; NO2 is a pollutant (N2O is the GHG).
Source: 2016
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
104. The quantity of heat required to raise the temperature of a given substance by 1 o C is known as:
sensible heat
specific heat
heat capacity
latent heat
Answer: C) heat capacity
Confirmed vs Book-1 §3.4 — Heat capacity is the heat required to raise the temperature of a GIVEN substance/body by 1 degC; specific heat is the same quantity referred to 1 kg of substance (Book-1 §3.4).
105. The Metric Tonne of Oil Equivalent (MTOE) value of 125 tonnes of coal having GCV of 4000 kcal/kg is
40
50
100
125
Answer: B) 50
Confirmed vs Book-1 §3.5 — Energy = 125 t x 1000 kg/t x 4000 kcal/kg = 5 x 10^8 kcal. MTOE = 5 x 10^8 / 10^7 = 50 (1 MTOE = 1 x 10^7 kcal). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: 2016
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
106. In a coal fired boiler, hourly consumption of coal is 1300 kg. The ash content in the coal is 6%. Calculate the quantity of ash formed per day. Boiler operates 24 hrs/day.
216 kg
300 kg
1872 kg
none of the above
Answer: C) 1872 kg
Confirmed vs Book-1 §3.4 — Coal fired per day = 1300 kg/h x 24 h = 31,200 kg. Ash = 6% x 31,200 = 1872 kg/day. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: 2016
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
107. A comparison of the trapping of heat by CO2 and CH4 is that
CH4 traps 21 times more heat in the atmosphere than does CO2
CO2 traps 21 times more heat in the atmosphere than does CH4
the same amount of heat is trapped by both CO2 and CH4
none of the above
Answer: A) CH4 traps 21 times more heat in the atmosphere than does CO2
Confirmed vs Book-1 §3.1 — Methane has a global warming potential of about 21 times that of CO2 over 100 years, i.e. CH4 traps 21 times more heat than the same mass of CO2.
108. In a chemical process two reactants A (300 kg) and B (400 kg) are used. If conversion is 50% and A and B react in equal proportions, the mass of the product formed is.
300 kg
350 kg
400 kg
none of the above
Answer: A) 300 kg
Confirmed vs Book-1 Ch.3 — A and B react in equal proportions, so A (300 kg) is limiting: only 300 kg of B can react. At 50% conversion, 150 kg of A reacts with 150 kg of B, giving 150 + 150 = 300 kg of product.
Source: 2016
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
109. Among which of the following fuel is the difference between the GCV and NCV maximum?
coal
furnace oil
natural gas
rice husk
Answer: C) natural gas
Confirmed vs Book-1 §3.4 — The difference between Gross and Net Calorific Value depends on the hydrogen (water-forming) content of the fuel. Natural gas (mainly methane) has the highest hydrogen content, hence forms the most water vapour on combustion and shows the maximum GCV-NCV difference.
Source: Guidebook
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy
110. The type of energy possessed by a charged capacitor is
kinetic energy
electrostatic
potential
magnetic
Answer: B) electrostatic
Confirmed vs Book-1 §3.1 — A charged capacitor stores energy in the electrostatic field between its plates (E = 1/2 CV²) - a form of stored potential energy, but specifically electrostatic energy.
Source: 2013
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
111. What is the heat content of 200 liters of water at 5°C in terms of the basic unit of energy in kilojoules ?
3000
2388
1000
4187
Answer: D) 4187
Confirmed vs Book-1 §3.4 — Q = m·Cp·dT = 200 kg x 4.187 kJ/kg degC x 5 degC = 4187 kJ (referred to 0 degC). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: 2013
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
112. Nameplate kW rating of a motor indicates
input to the motor
rated output of the motor
no-load input to the motor
rated input to the motor
Answer: B) rated output of the motor
Confirmed vs Book-1 §3.3 — Book-1 §3.3: 'The name plate details of motor, kW or HP indicates the output of the motor at full load' - i.e. the rated shaft output, not the input.
Source: 2013
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
113. In inductive and resistive combination circuit, the resultant power factor under AC supply will be
less than unity
more than unity
zero
unity
Answer: A) less than unity
Confirmed vs Book-1 §3.3 — With both resistance and inductance present the current lags the voltage by an angle 0 < θ < 90 deg, so PF = cosθ is less than unity (it is unity only for a purely resistive circuit).
Source: 2013
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ
114. How much carbon dioxide emission will be reduced annually by replacing 60 Watt incandescent lamp with a 15 Watt CFL Lamp, if emission per unit is 1 kg CO2 per kWh and annual burning is 3000 hours?
45 ton
3 ton
0.135 ton
183 ton
Answer: C) 0.135 ton
Confirmed vs Book-1 §3.3 — Saving = (60 - 15) W x 3000 h = 135,000 Wh = 135 kWh per year. CO2 avoided = 135 x 1 kg = 135 kg = 0.135 tonne. Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
115. The annual electricity bill for a plant is Rs 110 lakhs and accounts for 38% of the total energy bill. Furthermore the total energy bill increases by 5% each year. The plant’s annual energy bill at the end of the third year will be about ________
Rs 335 lakhs
Rs 268 lakhs
Rs 386 lakhs
Rs 418 lakhs
Answer: A) Rs 335 lakhs
Confirmed vs Book-1 Ch.3 — Total energy bill now = 110/0.38 = Rs 289.5 lakh. Escalating at 5% p.a. for three years: 289.5 x (1.05)³ = 289.5 x 1.1576 = Rs 335 lakh.
116. A sum of Rs 100,000 is deposited in a bank at the beginning of a year. The bank pays 10% interest annually. How much money will be in the bank account at the end of the fifth year, if no money is withdrawn?
161050
150000
155000
160000
Answer: A) 161050
Confirmed vs Book-1 Ch.3 — Compound interest: A = P(1+i)^n = 1,00,000 x (1.10)^5 = 1,00,000 x 1.61051 = Rs 1,61,051 (approx. 1,61,050). Book-1 Ch.3, Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text).
Source: 2013
📖 Book-1 Ch.4 Energy Audit instruments (outside Ch-3 text)
117. Portable combustion analyzers may have in-built chemical cells for measurement of stack gas components. Which combination of chemical cells for measurement of stack gas components is not possible?
CO, SOx, O2
CO2, O2
O2, NOx, SOx, CO
O2, CO
Answer: B) CO2, O2
Confirmed vs Book-1 Ch.3 — Electrochemical cells are available for O2, CO, NOx and SOx, so combinations (a), (c) and (d) are possible. CO2 cannot be measured by a chemical cell (it needs an infra-red/NDIR analyser), so the O2 + CO2 combination is not possible.
Source: 2013
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
118. The weight (kg) of the water vapour in each kg of dry air(kg/kg) is termed as :
Specific Humidity
relative humidity
humidity
saturation ratio
Answer: A) Specific Humidity
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Specific Humidity or Humidity Ratio - It is the mass (kg) of the water vapour in each kg of dry air (kg/kg).' Relative humidity is a percentage ratio, not kg/kg.
119. The primary energy content of fuels is generally expressed in terms of ton of oil equivalent (toe) and is based on the following conversion factor
1 toe=10x106 kCal
1 toe=11630 kWh
1 toe=41870 MJ
all the above
Answer: D) all the above
Confirmed vs Book-1 §3.5 — 1 toe = 10 x 10^6 kcal = 10^7 kcal (Book-1 §3.5). Also 10^7/860 = 11,630 kWh and 10^7 x 4.187 kJ = 41,870 MJ. All three statements are therefore correct.
Source: 2012
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
120. From rated V, A and PF given in the name-plate of a motor , one can calculate:
rated input Power
rated output Power
both a & b
none of these
Answer: A) rated input Power
Confirmed vs Book-1 §3.3 — Nameplate V, A and PF are the INPUT conditions at full load, so they give the rated INPUT power (sqrt3·V·I·PF for 3-phase). Rated output is separately stamped as the kW/HP rating.
Source: 2012
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
121. One certified emission reduction (CER) is equivalent to:
one kg of carbon
one kg of carbon dioxide
one ton of carbon
one ton of carbon dioxide
Answer: D) one ton of carbon dioxide
Confirmed vs Book-1 §3.1 — One Certified Emission Reduction (CER) under the CDM is a credit for one tonne of CO2-equivalent emission reduction. Book-1 Ch.3, Energy forms — background from Book-1 Ch.1 Energy Scenario.
Source: 2012
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
122. A process requires 10 kg of fuel with a calorific value of 5000 kCal/kg. The system efficiency is 80% The losses then will be
10000 kCal
45000 kCal
40000 kCal
20000 kCal
Answer: A) 10000 kCal
Confirmed vs Book-1 §3.4 — Energy input = 10 kg x 5000 kcal/kg = 50,000 kcal. At 80% system efficiency the useful heat is 40,000 kcal, so the losses = 20% x 50,000 = 10,000 kcal.
Source: 2012
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
123. Ratio of average load (kW) to maximum load (kW) is termed as
load factor
demand factor
form factor
utilization factor
Answer: A) load factor
Confirmed vs Book-1 §3.3 — Load factor = average load / maximum (peak) load over a period = energy consumed / (peak demand x hours). Demand factor is max demand/connected load.
Source: 2012
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
124. The average gross efficiency of thermal power generation on all India bases is about
30 – 34%
36 – 38%
39 - 41%
25 - 28%
Answer: A) 30 – 34%
Confirmed vs Book-1 §3.1 — Coal-based thermal power generation in India has an average gross station efficiency of about 30-34% (station heat rate around 2500-2900 kcal/kWh), the balance being rejected as condenser and flue-gas losses.
Source: 2012
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
125. Assuming total conversion of electrical energy to heat energy, how much heat is produced by a 200 W heater in 5 minutes?
200 kJ
40 kJ
1000 kJ
60 kJ
Answer: D) 60 kJ
Confirmed vs Book-1 §3.2 — Book-1 §3.2: W = P x t = 200 W x (5 x 60) s = 200 x 300 = 60,000 J = 60 kJ. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: 2012
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
126. A motor with 10 kW rating in its name plate, will draw Input power of____
10 kW at full load
more than 10 kW at full load
less than 10 kW at full load
10 kW at 110% of full load
Answer: B) more than 10 kW at full load
Confirmed vs Book-1 §3.3 — The nameplate 10 kW is the OUTPUT at full load. Since input = output/efficiency and efficiency is below 100%, the motor draws MORE than 10 kW at full load.
Source: 2012
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
127. Which of the following statements is not true regarding Maximum Demand Control?
Maximum demand control offers a way of ‘shaving’ the peaks and ‘filling’ the valleys in the consumer load diagram
Maximum demand control is carried out by concerned utility at customer premises
Maximum demand control focuses on critical load for management
All of the above
Answer: B) Maximum demand control is carried out by concerned utility at customer premises
Confirmed vs Book-1 §3.3 — Maximum demand control is done by the CONSUMER at his own premises (load shedding/shifting, staggering, demand controllers); the utility only meters and bills the demand. The other statements are correct.
Source: 2012
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
128. Which of the following statements is false?
reactive current is necessary to build up the flux for the magnetic field of inductive devices
some portion of reactive current is converted into useful work
Cosine of the angle between kVA and kW vector is called power factor
power factor is unity in a pure resistive circuit
Answer: B) some portion of reactive current is converted into useful work
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Power factor: the reactive current builds the magnetic flux but 'otherwise it is non-usable' - none of it is converted into useful work, so statement (b) is false. (a), (c) and (d) are true.
129. Steam leak reduction program can be best achieved through
Small Group Activities
Autonomous Maintenance
TPM
All of the above
Answer: D) All of the above
Confirmed vs Book-1 Ch.3 — A steam-leak reduction programme is a shop-floor housekeeping activity best sustained through small group activities, autonomous maintenance and TPM - all of the listed approaches apply.
130. Consider two competitive projects A and B each entailing investment of Rs.85,000/- . Project A returns Rs.50,000 at the end of each year, but Project B returns Rs.115,000 at the end of Year 2. Which project is superior?
project A since it starts earning by end of first year itself and recovers cost before end of two years
project B since it offers higher return before end of two years
both projects are equal in rank
insufficient information to assess the superiority
Answer: D) insufficient information to assess the superiority
Confirmed vs Book-1 Ch.3 — Project A returns Rs 50,000 per year but the project LIFE is not stated, while B gives Rs 1,15,000 once at year 2. Without the project life (and discount rate) neither NPV nor IRR can be compared - the information is insufficient.
Source: 2012
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
131. Nuclear power development in India is constrained by
low % of Uranium in the ore
inadequate supply of Uranium
constraints in import of Uranium
all of the above
Answer: D) all of the above
Confirmed vs Book-1 §3.1 — India's nuclear programme is constrained by low uranium content in the domestic ore, inadequate indigenous uranium supply and restrictions on uranium imports - all of the listed factors.
132. In a contract when all or part of the savings are guaranteed by contractor, and all or part of the costs of equipment and/or services are paid out of savings as they are achieved, is termed as
traditional contract
guaranteed saving performance contract
shared saving performance contract
extended technical guarantee contract
Answer: B) guaranteed saving performance contract
Confirmed vs Book-1 Ch.3 — In a guaranteed savings performance contract the ESCO guarantees all or part of the savings and the cost of the equipment/services is paid out of the savings as they are realised.
Source: 2012
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
133. If 3350 kJ of heat is supplied to 20 kg of ice at 0o C, how many kg of ice will melt into water at 0o C (latent heat of melting of ice is 335 kJ/kg)
1 kg
4.18 kg
10 kg
29 kg
Answer: C) 10 kg
Confirmed vs Book-1 §3.4 — m = Qₗ/h_if = 3350 kJ / 335 kJ/kg = 10 kg of ice melts (the remaining 10 kg of the 20 kg stays as ice at 0 degC). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: 2012
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
134. If oxygen rich combustion air (25% vol oxygen) is supplied to a furnace instead of normal air (21% vol oxygen), the % CO2 in flue gases will
reduce
increase
remain same
will become zero
Answer: B) increase
Confirmed vs Book-1 §3.4 — Enriching the combustion air with oxygen (21% -> 25%) reduces the nitrogen diluting the flue gas, so the same CO2 appears in a smaller flue-gas volume and the %CO2 increases (stack loss also falls).
135. In project management work breakdown structure defines
temporary endeavour undertaken to create unique product or service
the activities to be completed in the projects
how realistic were the assumptions underlying the project
none of the above
Answer: B) the activities to be completed in the projects
Confirmed vs Book-1 Ch.3 — A work breakdown structure decomposes the project into the activities/deliverables to be completed, forming the basis for scheduling, costing and responsibility assignment.
Source: 2012
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy
136. An example of stored mechanical energy is
water in a reservoir
an arrow in a stretched bow
an air-borne aeroplane
you on top of a mountain
Answer: B) an arrow in a stretched bow
Confirmed vs Book-1 §3.1 — Stored mechanical energy is energy stored in objects by the application of force, such as the elastic potential energy in a stretched bow (or compressed/stretched spring). The other options are examples of gravitational potential or kinetic energy.
Source: Guidebook
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
137. Mega Volt Ampere (MVA) in a three phase electrical circuit could be written as
(Voltage x Ampere) / 1,000
(Voltage x Ampere) / 1,000,000
Voltage x Ampere x 1,000
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.3 — For a three-phase circuit, MVA = (sqrt3 x Voltage x Ampere) / 1,000,000. None of the listed options includes the sqrt3 factor for a three-phase circuit, so the answer is 'none of the above'.
Source: Guidebook
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
138. When the current lags the voltage in an alternating current system, it is caused mainly due to
resistive load
capacitive load
inductive load
none of the above
Answer: C) inductive load
Confirmed vs Book-1 §3.3 — In an inductive load (e.g. motors, transformers), the current lags the voltage. In a capacitive load the current leads, and in a purely resistive load they are in phase.
Source: Guidebook
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy
139. Which energy source is indirect in an overall energy balance in the generation of electricity by a photovoltaic cell?
commercial energy
wave energy
sun light
none of the above
Answer: A) commercial energy
Confirmed vs Book-1 §3.1 — While sunlight is the direct energy input to a PV cell, the commercial (conventional) energy embedded in manufacturing the cells and system is the indirect energy input in the overall energy balance.
140. What is the 'toe' of 125 Ton of coal which has GCV of 4000 kcal/kg
40
50
400
500
Answer: B) 50
Confirmed vs Book-1 §3.5 — Energy = 125 ton x 1000 kg/ton x 4000 kcal/kg = 5 x 10^8 kcal. 1 toe = 10^7 kcal, so toe = 5 x 10^8 / 10^7 = 50. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Guidebook
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
141. The quantity of heat required to raise the temperature of a substance by 1 degree C is known as
sensible heat
specific heat
heat capacity
latent heat
Answer: C) heat capacity
Confirmed vs Book-1 §3.4 — Heat capacity is the quantity of heat required to raise the temperature of a (given) substance by 1 degree C. Specific heat is the heat per unit mass per degree; latent heat involves phase change with no temperature rise.
Source: Guidebook
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
142. Active power in an alternating current (AC) circuit is given by
kVA x power factor
(kVA^2 - kVAr^2)^1/2
[(kVA + kVAr) x (kVA - kVAr)]^1/2
all of the above
Answer: D) all of the above
Confirmed vs Book-1 §3.3 — Active power kW = kVA x power factor. Since kVA^2 = kW^2 + kVAr^2, kW = (kVA^2 - kVAr^2)^1/2 = [(kVA + kVAr)(kVA - kVAr)]^1/2. All three expressions give the active power.
Source: Guidebook
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
143. Nameplate kW or HP rating of a motor indicates
input kW to the motor
output kW of the motor
minimum input kW to the motor
maximum input kW to the motor
Answer: B) output kW of the motor
Confirmed vs Book-1 §3.3 — The nameplate kW/HP rating of a motor indicates its rated shaft output power, not the electrical input power (which is higher due to motor losses).
Source: Guidebook
📖 §3.3 Example 3.6 — resistive load power varies as V²
144. A 230V, 100 W rated Incandescent bulb is operated at a constant voltage of 200V. The power consumption of the bulb is ____.
80W
76W
87W
100W
Answer: B) 76W
Confirmed vs Book-1 §3.3 — Power varies with voltage squared at fixed resistance: P = 100 x (200/230)^2 = 100 x 0.756 = 75.6 ~ 76 W. Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
Source: Mar 2023
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
145. The input current drawn by 3-ph 10 kW induction motor is 20 Amps at 0.8 pf. The input voltage is 410V. The motor efficiency is ____.
86%
90%
88%
None of the above
Answer: C) 88%
Confirmed vs Book-1 §3.3 — Input power = sqrt(3) x 410 x 20 x 0.8 = 11362 W. Efficiency = 10000/11362 ~ 0.88 = 88%. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: Mar 2023
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
146. If 3500 kJ of heat is supplied to 22 kgs of ice at 0 degC, how many kg of ice will melt into water at 0 degC (latent heat of melting is 330 kJ/kg).
10.606 Kg
12 Kg
22 Kg
15 Kg
Answer: A) 10.606 Kg
Confirmed vs Book-1 §3.4 — Mass melted = Heat/Latent heat = 3500/330 = 10.606 kg. Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Mar 2023
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
147. Which among the following fuels has the highest calorific value?
Coal
Diesel
Hydrogen
Natural Gas
Answer: C) Hydrogen
Confirmed vs Book-1 §3.4 — Hydrogen has the highest calorific value per unit mass (~120 MJ/kg) among the listed fuels. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
152. Ten units of electricity are equivalent to ____.
10 ToE
10 kCal
10 KJ
8600 kCal
Answer: D) 8600 kCal
Confirmed vs Book-1 §3.5 — 1 kWh = 860 kcal, so 10 units (kWh) = 8600 kcal. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Mar 2023
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
153. Select the incorrect statement related to energy basics ____.
Superheating is a process of heating vapor above evaporation temperature.
Pump is used to move the fluid in process of natural convection.
Calorific value is a measure of energy content of organic matter of fuel.
Internal resistance of a fluid is measured as viscosity of a fluid.
Answer: B) Pump is used to move the fluid in process of natural convection.
Confirmed vs Book-1 §3.4 — Natural convection occurs due to density differences without a pump; using a pump is forced convection, so statement b is incorrect.
Source: Mar 2023
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
154. Why radiation heat transfer is prominent in applications like boiler and furnace?
It does not require medium
Heat transfer is proportional to T^4
It uses electromagnetic waves to transfer heat
All of above
Answer: D) All of above
Confirmed vs Book-1 §3.4 — Radiation needs no medium, follows the T^4 (Stefan-Boltzmann) law making it dominant at high temperatures, and transfers heat via electromagnetic waves - all true for boilers and furnaces.
Source: Mar 2023
📖 §3.4 Steam properties — superheat and dryness fraction (x)
155. Temperature of steam will be highest in following condition at same pressure ____.
Wet steam
Saturated steam
Superheated steam
At all stages temperature is same
Answer: C) Superheated steam
Confirmed vs Book-1 §3.4 — At a given pressure, superheated steam is heated above saturation temperature, hence has the highest temperature. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
Source: Mar 2023
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
156. The Specific heat is high for ____.
Lead
Water
Mercury
Alcohol
Answer: B) Water
Confirmed vs Book-1 §3.4 — Water has a very high specific heat (~4.187 kJ/kg degC), higher than the other listed substances. Book-1 Ch.3, Specific heat — Table 3.1 Specific heat of common substances.
Source: Mar 2023
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
157. The name plate kW or HP of a motor indicates ____.
Input power drawn
Output power
Max input power
Minimum input power
Answer: B) Output power
Confirmed vs Book-1 §3.3 — The motor nameplate rating (kW or HP) denotes the rated mechanical output power, not the input power. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Source: Mar 2023
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
158. 2000 kJ of heat is supplied to 500 kg of ice at 0 degC. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be ____.
1.49
83.75
5.97
None of the above
Answer: C) 5.97
Confirmed vs Book-1 §3.4 — Mass melted = Heat/Latent heat = 2000/335 = 5.97 kg. Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Jul 2022
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)
159. The number of moles of water contained in 27 kg of water is ____.
5
3
4
1.5
Answer: D) 1.5
Confirmed vs Book-1 §3.5 — Moles = mass/molar mass = 27000 g / 18 g/mol = 1500 mol = 1.5 kmol. Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
Source: Jul 2022
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
160. The amount of energy transfer from a higher temperature to a lower temperature is measured in ____.
kcal
Watt
Watts per second
none of the above
Answer: A) kcal
Confirmed vs Book-1 §3.4 — Heat (energy transferred due to a temperature difference) is measured in kcal (a unit of energy). Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
161. The amount of electricity required to heat 200 litres of water from 30 degC to 70 degC through resistance heating is ____.
0.93 kWh
9.3 kWh
930 kWh
8 kWh
Answer: B) 9.3 kWh
Confirmed vs Book-1 §3.5 — Heat = 200 x 1 x (70-30) = 8000 kcal = 8000/860 = 9.3 kWh. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Jul 2022
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
162. A process requires 100 kg of fuel with a calorific value of 5000 kcal/kg for heating with a system efficiency of 83%. The loss in kcal would be ____.
235,000 kCal
85,000 kCal
103680 kCal
415,000 kCal
Answer: B) 85,000 kCal
Confirmed vs Book-1 §3.4 — Total input = 100 x 5000 = 500,000 kcal. Loss = (1-0.83) x 500000 = 0.17 x 500000 = 85,000 kcal. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Jul 2022
📖 §3.1 Chemical energy — fuels store chemical energy
163. Propane is an example of stored ____ energy.
Nuclear
Radiant
Chemical
Mechanical
Answer: C) Chemical
Confirmed vs Book-1 §3.1 — Propane stores energy in its chemical bonds, i.e. chemical energy. Book-1 Ch.3, Chemical energy — fuels store chemical energy.
Source: Jul 2022
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
164. In a heat treatment furnace the material is heated up to 1053 K from ambient temperature of 303 K. Considering the specific heat of material as 0.125 kCal/kg degC, what is the energy content gained by one kg of material after heating?
94 kCal
250 kCal
350 kCal
100 kCal
Answer: A) 94 kCal
Confirmed vs Book-1 §3.4 — Temperature rise = 1053 - 303 = 750 K (= 750 degC). Heat = 1 x 0.125 x 750 = 93.75 ~ 94 kCal. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Jul 2022
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
165. The quantity of heat required to convert one kg of a liquid into vapour without change of temperature is called ____.
latent heat of fusion
specific heat
sensible heat
Latent heat of Evaporation
Answer: D) Latent heat of Evaporation
Confirmed vs Book-1 §3.4 — The heat needed to convert a liquid to vapour at constant temperature is the latent heat of evaporation (vaporization). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Jul 2022
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
166. The energy consumed by a 55 kW motor loaded at 40 kW over a period of 4 hours is:
220 kW
220 kWh
160 kWh
160 kW
Answer: C) 160 kWh
Confirmed vs Book-1 §3.2 — Energy = Load x time = 40 kW x 4 h = 160 kWh. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
168. Which of the following most closely represents the heat content of 1 kg of LPG:
8000 kilo Calorie
12500 kilo Joule
12500 kilo Calorie
8000 kilo Joule
Answer: C) 12500 kilo Calorie
Confirmed vs Book-1 §3.5 — LPG has a calorific value of approximately 12500 kcal/kg. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Jul 2022
📖 §3.3 Resistance & conductance — Ohm's law R = V/I
169. Resistance of 250 V incandescent lamp drawing 0.5 A:
5,000 Ω
500 Ω
50 Ω
5 Ω
Answer: B) 500 Ω
Confirmed vs Book-1 §3.3 — R = V/I = 250/0.5 = 500 Ω. Book-1 Ch.3, Resistance & conductance — Ohm's law R = V/I.
Source: Sep 2025
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
170. A boiler receives 100 MJ of fuel energy. The steam output is 70 MJ, the flue gas loss is 20 MJ and the radiation plus unaccounted loss is 10 MJ. What is the boiler efficiency?
65%
60%
70%
75%
Answer: C) 70%
Confirmed vs Book-1 §3.2 — Efficiency = useful output/input = 70/100 = 70%. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Confirmed vs Book-1 §3.5 — 1 toe = 10^7 kcal ≈ 41,868 MJ (41.868 GJ). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2025
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
172. Maximum specific heat among the following:
Water
Lead
Mercury
Iron
Answer: A) Water
Confirmed vs Book-1 §3.4 — Water has the highest specific heat (~1 kcal/kg°C) among the listed substances. Book-1 Ch.3, Specific heat — Table 3.1 Specific heat of common substances.
Source: Sep 2025
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
173. Heat required for Cooling 2000 kg of water for ΔT of 10°C ____________
175. 'Daylight harvesting' in lighting systems means:
Collecting solar energy for night lighting
Using flat plate collectors for heating
Adjusting artificial lighting based on natural daylight
Storing energy in battery banks
Answer: C) Adjusting artificial lighting based on natural daylight
Confirmed vs Book-1 Ch.3 — Daylight harvesting dims/switches artificial lighting in response to available natural daylight to save energy. Book-1 Ch.3, Book-3 Ch.8 Lighting System (outside Ch-3 text).
Source: Sep 2025
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
176. Power rating of an electrical heater consuming 12,000 J/min is:
12 W
100 W
200 W
12,000 W
Answer: C) 200 W
Confirmed vs Book-1 §3.2 — Power = 12,000 J / 60 s = 200 W. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: Sep 2025
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ
177. Calculate the energy consumed by a 200-watt appliance used for 5 hours a day over 30 days.
30 kWh
27000 kCal
6500 kJ
30 kJ/h
Answer: A) 30 kWh
Confirmed vs Book-1 §3.3 — Energy = 0.2 kW × 5 h × 30 = 30 kWh. Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
178. Which of the following is not true, equivalent to 1 atm pressure?
1 atm = 101.3 kPa
1 atm = 10332 mmWC
1 atm = 14.7 psi
1 atm = 0.98 kg/cm2
Answer: D) 1 atm = 0.98 kg/cm2
Confirmed vs Book-1 §3.4 — 1 atm ≈ 1.033 kg/cm², not 0.98 kg/cm²; the other equivalences are correct. Book-1 Ch.3, Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar.
Source: Sep 2024
📖 §3.4 Fuel properties — density, specific gravity, viscosity
179. Redwood Seconds is measure of ____________
Density
Viscosity
Specific Gravity
Flash Point
Answer: B) Viscosity
Confirmed vs Book-1 §3.4 — Redwood Seconds is a unit of kinematic viscosity (Redwood viscometer). Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Sep 2024
📖 §3.4 Sensible heat — Q = m · Cp · ΔT
180. What is the heat content of 500 liters of water at 6°C in terms of the basic unit of energy in kilojoules?
12000
3000
500
None of the above
Answer: D) None of the above
Confirmed vs Book-1 §3.4 — Heat content depends on a reference temperature; the figure cannot be determined as stated, so 'None of the above'. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Sep 2024
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
181. Heat transfer in an air-cooled condenser occur predominately by
Conduction
Convection
Radiation
All of the above
Answer: B) Convection
Confirmed vs Book-1 §3.4 — An air-cooled condenser transfers heat predominantly by convection to the cooling air. Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
Source: Sep 2024
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
182. When the evaporation of water from wet substance is zero, the relative humidity of air is likely to be
0%
50%
100%
Unpredictable
Answer: C) 100%
Confirmed vs Book-1 §3.4 — No further evaporation occurs when the air is saturated, i.e. relative humidity = 100%. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2024
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
183. If we heat air without changing absolute humidity, % relative humidity will
Increase
Decrease
No change
Can't Say
Answer: B) Decrease
Confirmed vs Book-1 §3.4 — Heating raises the saturation capacity at constant moisture, so relative humidity decreases. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2024
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
184. Among which of the following fuel the difference between the GCV and NCV is maximum
Coal
Furnace Oil
Natural Gas
Rice Husk
Answer: C) Natural Gas
Confirmed vs Book-1 §3.4 — Natural gas has the highest hydrogen content, producing the most water vapour, so GCV−NCV difference is maximum. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Sep 2024
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
185. An induction motor with 30 kW rating and efficiency of 85% in its name plate means
It will draw 35.29 kW at full load
it will always draw 30 kW at full load
it will draw 25.5 kW at full load
it will draw 28.23 kW at full load
Answer: A) It will draw 35.29 kW at full load
Confirmed vs Book-1 §3.3 — Rated output 30 kW at 85% efficiency → input = 30/0.85 = 35.29 kW at full load. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
186. Energy content in 2500 kgs of coal with a calorific value of 4000 kcal/kg in terms of toe would be
1 toe
10 toe
100 toe
1000 toe
Answer: A) 1 toe
Confirmed vs Book-1 §3.5 — Energy = 2500×4000 = 10^7 kcal = 1 toe (1 toe = 10^7 kcal). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2024
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ
187. An electric iron of power 2000 watts is used for a total of 120 minutes per month. Compute its monthly electricity consumption
2.0 kWh
2.4 kWh
4.0 kWh
24.0 kWh
Answer: C) 4.0 kWh
Confirmed vs Book-1 §3.3 — Energy = 2 kW × 2 h = 4.0 kWh. Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
Source: Sep 2024
📖 §3.4 Steam properties — superheat and dryness fraction (x)
188. The dryness fraction (x) of superheated steam will be
x = 0.8
x = 0.9
x = 1
x = 0
Answer: C) x = 1
Confirmed vs Book-1 §3.4 — Superheated (and dry saturated) steam has a dryness fraction of 1. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
Source: Sep 2024
📖 §3.4 Thermal Energy Basics — Specific Heat (Table 3.1)
189. The specific heat of water is approximately:
470 J/kg°C
910 J/kg°C
2400 J/kg°C
4200 J/kg°C
Answer: D) 4200 J/kg°C
Confirmed vs Book-1 §3.4 — Table 3.1 'Specific Heat of Some Common Substances' gives Water = 4200 J/kg°C, the highest of the listed substances; the book notes "the specific heat of water is very high as compared to other common substances".
4200 J/kg°C is the same as 1 kcal/kg°C (since 1 Calorie = 4.187 J ≈ 4.2 J), which is why Q = m·Cp·ΔT works in either unit set.
The distractors are the book's own Table 3.1 values for other substances: Iron 470, Aluminium 910 and Alcohol 2400 J/kg°C.
Source: AI practice
📖 §3.5 Energy Units and Conversions / Gazette MTOE values
190. One kilowatt-hour (1 kWh) equals:
860 kcal = 3.6 MJ
252 kcal = 1.055 MJ
1000 kcal = 4.2 MJ
11,630 kcal = 41.8 MJ
Answer: A) 860 kcal = 3.6 MJ
Confirmed vs Book-1 §3.5 — the conversion list gives "1 kWh = 3.6 x 10⁶ J" and the Gazette table gives "1 kWh - 860 kilocalories (kcal)"; so 1 kWh = 860 kcal = 3.6 MJ.
This pair is the bridge between electrical and thermal numericals throughout Paper-1.
Option (b) 252 cal / 1055 J is the book's BTU value, and (d) 11,630 (kWh) belongs to 1 toe (Ch-1 §1.2) — both are real book numbers attached to the wrong quantity.
Source: AI practice
📖 §3.3 Electricity Basics — Example 3.6 (heater at reduced voltage)
191. The power drawn by a resistive heater varies with applied voltage as:
P ∝ V
P ∝ V^2
P ∝ 1/V
P ∝ √V
Answer: B) P ∝ V^2
Confirmed vs Book-1 §3.3 — for a fixed resistance R = V/I, power P = V·I = V²/R, so P varies with the SQUARE of the applied voltage.
Example 3.6 applies exactly this: consumption at 200 V = (200/230)² × 5 kW × 1 h = 3.78 kWh.
Option (a) P ∝ V is the trap — it would hold only if the current stayed constant, but in a resistive heater the current also falls in proportion to V.
Source: AI practice
📖 §3.3 Electricity Basics — Example 3.6
192. A 5 kW heater rated at 230 V is operated at 200 V. The approximate power it now draws is:
5.0 kW
4.35 kW
3.78 kW
2.50 kW
Answer: C) 3.78 kW
Confirmed vs Book-1 §3.3, Example 3.6 (identical data: 230 V, 5 kW heater run at 200 V).
P₂ = (V₂/V₁)² × P₁ = (200/230)² × 5 = 0.7561 × 5 = 3.78 kW (so 3.78 kWh in one hour).
Option (b) 4.35 kW is what you get from the linear ratio (200/230) × 5 — the standard mistake of forgetting to square the voltage ratio.
Source: AI practice
📖 §3.3 Electricity Basics — Relationships between Power, Voltage and Current
193. The active power in a balanced three-phase AC system is given by:
P = Vl × Il × cosθ
P = √3 × Vl × Il × cosθ
P = 3 × Vl × Il
P = Vl × Il ÷ √3
Answer: B) P = √3 × Vl × Il × cosθ
Confirmed vs Book-1 §3.3 — "For a balanced three-phase load, Power, Watts = √3 × V_L × I_L × CosΘ" and "For a balanced single-phase load, Power, Watts = V_L × I_L CosΘ", where V_L is line voltage and I_L line current.
Example 3.8 uses it directly: √3 × 0.440 × 25 × 0.90 = 17.15 kW.
Option (a) is the SINGLE-phase formula (it under-reads three-phase power by a factor of √3) and (c) drops the power factor entirely, giving apparent power, not active power.
Source: AI practice
📖 §3.3 Electricity Basics — Example 3.8 (3-phase motor)
194. A 3-phase motor draws 25 A at 440 V with power factor 0.9 for 1 hour. The energy consumed is approximately:
9.9 kWh
12.5 kWh
17.15 kWh
19.8 kWh
Answer: C) 17.15 kWh
Confirmed vs Book-1 §3.3, Example 3.8 (same data: 440 V, 25 A, PF 0.90, one hour).
P = √3 × V_L × I_L × cosΦ = 1.732 × 0.440 kV × 25 A × 0.90 = 17.15 kW, so energy in 1 hour = 17.15 kWh.
Option (d) 19.8 kWh is √3 × 0.44 × 25 without the power factor (i.e. kVA, not kW), and (a) 9.9 kWh omits the √3 — the two classic three-phase slips.
Source: AI practice
📖 §3.3 Electricity Basics — Power Factor and the power triangle
195. Power factor (PF) of an AC load is equal to:
kVAR / kVA
kW / kVA
kVA / kW
kW / kVAR
Answer: B) kW / kVA
Confirmed vs Book-1 §3.3 — the power triangle gives kW = kVA·cosΘ, kVA = kW/cosΘ, kVAR = kVA·sinΘ and "PF = cos Θ"; rearranging kW = kVA·cosΘ gives PF = kW / kVA.
kW is the active (resistive) power actually converted to useful work and kVA the apparent power (the hypotenuse).
Option (c) kVA/kW is the reciprocal (always ≥ 1, so it can never be a power factor) and (a) kVAR/kVA is sinΘ, the reactive fraction.
Source: AI practice
📖 §3.4 Latent Heat of Vaporization / Specific Enthalpy of Saturated Steam
196. The latent heat of vaporization of water at 100°C is approximately:
335 kJ/kg
419 kJ/kg
2257 kJ/kg
4200 kJ/kg
Answer: C) 2257 kJ/kg
Confirmed vs Book-1 §3.4 — "The latent heat of vaporization of water is 2257 KJ/kg", also derived in the book as h_fg = h_g − h_f = 2676 − 419 = 2257 kJ/kg at standard atmosphere (≈540 kcal/kg).
This is the number used in Q = m × h_fg for every evaporation numerical.
Option (b) 419 kJ/kg is the specific enthalpy of SATURATED WATER (h_f) and (a) 335 kJ/kg is the latent heat of FUSION of ice — both are book values for different quantities.
Source: AI practice
📖 §3.4 Energy Content in Fuel — GCV and NCV (chapter objective Q10)
197. Which fuel has the LARGEST difference between its gross (GCV) and net (NCV) calorific value?
Coal
Furnace oil
Natural gas
Bagasse
Answer: C) Natural gas
Confirmed vs Book-1 §3.4 — "The difference between GCV and NCV is the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapour) in the fuel."
The gap is therefore largest for the fuel richest in hydrogen: natural gas is essentially methane (CH₄), so it forms the most water vapour per kg burnt and shows the maximum GCV − NCV difference. This is also the answer to the book's own chapter objective Q10.
Coal and bagasse carry moisture but far less hydrogen, and furnace oil is intermediate (GCV 10,500 vs NCV 9,800 kcal/kg in the book's example) — so all three show a smaller gap.
Source: AI practice
📖 §3.4 Super Heat — T-S steam diagram and dryness fraction
198. In a steam table, a dryness fraction (x) equal to 1 represents:
Saturated water
Wet steam
Dry saturated steam
Superheated steam
Answer: C) Dry saturated steam
Confirmed vs Book-1 §3.4 — "X = dryness factor of steam = in 1 kg of water-steam mixture, x kg is mass of steam and (1-x) kg is mass of water", and "the zone in right side of X = 1.0 line represents the superheated region of steam".
So on the X = 1.0 line itself the mixture is 100% steam with no water: dry saturated steam. At x = 0 the substance is saturated water, and 0 < x < 1 under the dome is wet steam.
Option (d) superheated steam is the trap — that lies BEYOND (to the right of) the x = 1 line, at a temperature above saturation, and is no longer described by a dryness fraction.
Source: AI practice
📖 §3.4 Heat transfer — the three primary modes
199. Which mode of heat transfer requires NO material medium for the transport of heat?
Conduction
Convection
Radiation
Forced convection
Answer: C) Radiation
Corrected (question rewritten) — Book-1 §3.4: the original stem asserted that radiation is "proportional to the fourth power of temperature", a statement the 2014 guidebook never makes; the stem has been re-grounded in the book's own wording while keeping the same topic and the same correct answer.
The book lists three primary modes — "Conduction (Energy transfer in a solid), Convection (Energy transfer in a fluid), Radiation (doesn't need a material to travel through)" — and states "Radiation mode heat transfer requires no medium for the transport of heat".
Conduction needs a solid and both convection options need a fluid; forced convection is only convection with fluid motion induced by a fan or pump, so it still needs a medium.
Source: AI practice
📖 §3.4 Standard Atmospheric Pressure / §3.5 Pressure Units
200. One standard atmosphere (1 atm) is equal to:
1.01325 bar = 760 mmHg
0.5 bar = 380 mmHg
2.0 bar = 1520 mmHg
10 bar = 7600 mmHg
Answer: A) 1.01325 bar = 760 mmHg
Confirmed vs Book-1 §3.4 — "1 atm = 1.01325 bar = 101.3 kPa = 760 mmHg = 10.33 meter H₂O = 1013 mbar = 1.0332 kg/cm²", the Standard Atmospheric Pressure defined at sea level.
The §3.5 pressure table repeats it as 1 atm = 760 mm Hg = 101325 Pa.
Options (b), (c) and (d) are arbitrary multiples that break the fixed 1.01325 bar ↔ 760 mmHg pairing; remember also that absolute pressure = gauge pressure + atmospheric pressure.
Source: AI practice
📖 §3.4 Heat transfer — rate of energy transfer measured in Watts (J/s)
201. The rate of energy transfer from a higher temperature to a lower temperature is measured in
kcal
Watt
Watts per Second
none of the above
Answer: B) Watt
Confirmed vs Book-1 §3.4 — "The energy transferred is measured in Joules. The rate of energy transfer, more commonly called heat transfer, is measured in Watts (J/s)."
Heat always flows from the hotter body to the colder one, and it is the RATE of that flow that the question asks about, so the unit must be a power unit.
Option (a) kcal is a quantity of heat, not a rate, and (c) 'Watts per second' is dimensionally wrong — a Watt is already one Joule per second.
Source: Book EOC
Short questions (5 marks) — 75
📖 §3.1 Energy types & forms
1. List five forms of energy with examples.
Model answer: Energy is broadly potential (stored) and kinetic (working). Five forms: (1) Chemical energy - stored in bonds of atoms/molecules, e.g. coal, petroleum, natural gas, biomass. (2) Nuclear energy - stored in the nucleus, e.g. uranium fission (E = mc2). (3) Gravitational/potential energy - e.g. water in a hydropower reservoir (Ep = mgh). (4) Thermal energy - internal vibration of atoms/molecules, e.g. geothermal energy, steam. (5) Electrical energy - movement of electrons, e.g. lightning, grid electricity. Other forms: radiant (solar, light, X-rays), motion (wind, hydro), sound, stored mechanical (springs).
Memorise the two families first: potential (stored) = chemical, nuclear, stored mechanical, gravitational; kinetic (working) = radiant, thermal, electrical, motion, sound. Pick any five from those lists and you cannot go wrong. Marks are lost when students name a device or a source (solar panel, wind turbine, dam) instead of a FORM of energy — always write the form first, then the example. Hook: stored = potential, moving = kinetic.
Source: Guidebook
📖 §3.1 Potential & Kinetic energy
2. Distinguish between potential energy and kinetic energy with examples and their defining equations.
Model answer: Potential (stored) energy is energy a body possesses due to its position or configuration, e.g. raised pile-driver head, stretched rubber band, water in a dam. Gravitational PE: Ep = m g h. It exists as chemical, nuclear, stored-mechanical and gravitational energy. Kinetic (working) energy is energy possessed by virtue of motion or velocity, e.g. moving vehicle, flowing fluid, machinery. KE: Ek = 1/2 m v2. It exists as radiant, thermal, electrical, motion and sound energy.
Two formulas carry the marks: Ep = m g h and Ek = ½ m v². Both answers come out in joules when m is in kg, h in m and v in m/s. The commonest slip is dropping the ½ in kinetic energy, or forgetting that v is squared — doubling speed gives four times the energy. Hook: potential = position, kinetic = motion.
Source: Guidebook
📖 §3.2 Work, Energy and Power
3. Define work, energy and power and give their SI units and defining relations.
Model answer: Work is done (energy transferred) when a force moves a body: W = F x s, unit joule (J); one joule = work done by a force of 1 newton through 1 metre. Energy is the capacity for doing work, also measured in joules. Power is the rate of doing work or rate of using/converting energy: P = W/t, unit watt (W), where 1 W = 1 joule/second. For a rotating body, W = T x (theta/2pi) and P = T x omega = 2*pi*T*N/60. 1 kWh = 3600 kJ = 3.6 MJ.
One chain to memorise: W = F × s (joule), P = W/t (watt), and 1 W = 1 J/s. For rotation, P = 2πNT/60 — N must be in rpm, which is exactly why you divide by 60. Keep the energy bridge ready: 1 kWh = 3600 kJ = 3.6 MJ. Common mistake: writing the unit of power as joule instead of watt — power always has 'per second' hidden in it.
Source: Guidebook
📖 §3.2 Work & Power (worked example)
4. A portable machine requires a force of 200 N to move it. How much work is done if it is moved 20 m, and what average power is used if the movement takes 25 s?
Model answer: Work done = force x distance = 200 N x 20 m = 4000 Nm = 4 kJ. Average power = work done / time taken = 4000 J / 25 s = 160 J/s = 160 W.
Do it in two clean steps: first work (W = F × s), then power (P = W/t). Never divide force by time. Units look after themselves in SI — N × m = J, and J/s = W. Cross-check: 4000 J spread over 25 s must be a small number of watts; if you get thousands of watts, you multiplied by time instead of dividing.
Source: Guidebook
📖 §3.3 DC vs AC current
5. Differentiate between Direct Current (DC) and Alternating Current (AC).
Model answer: Direct Current (DC) is a non-varying, unidirectional current, e.g. current produced by batteries. Alternating Current (AC) reverses in regularly recurring intervals of time, having alternate positive and negative values occurring a specified number of times per second, e.g. utility supply. In 50 Hz (cycle) AC, current reverses direction 100 times per second, i.e. twice in one cycle. For AC, voltage and current are normally expressed as RMS values so that the same power formulas as DC can be used.
The number the examiner wants is 50 Hz → the current reverses 100 times a second, because there are two reversals in every cycle. Also state that AC voltage and current are quoted as RMS values — that is what lets the DC formulas P = V I and R = V/I be used unchanged for AC. Hook: DC = one direction (battery), AC = to-and-fro (grid supply).
Source: Guidebook
📖 §3.3 Ampere, Volt, Frequency
6. Define ampere, volt and frequency (with units) as used in electricity basics.
Model answer: Ampere (A): current is the rate of flow of charge; ampere is the basic unit of electric current. Volt (V): a measure of electric potential or electromotive force; a potential of one volt appears across a resistance of one ohm when a current of one ampere flows through it. Frequency (Hz): the number of cycles per second at which alternating current changes; unit is cycles/second or hertz. In India the normal utility supply frequency is 50 Hz.
Define each one by the other two: 1 volt across 1 ohm drives 1 ampere. Ampere = rate of flow of charge, volt = electrical pressure, hertz = cycles per second. Do not forget the India-specific fact: utility supply frequency is 50 Hz — it is often a one-mark part of this question. Common mistake: writing hertz as 'cycles' without 'per second'.
Source: Guidebook
📖 §3.3 Ohm's law - Resistance & Conductance
7. State Ohm's law definition of resistance and define conductance, with units.
Model answer: The unit of electric resistance is the ohm (Omega), where one ohm is one volt per ampere - the resistance between two points in a conductor when a constant potential of 1 V applied produces a current of 1 A. Thus R = V/I, where V is the potential difference in volts and I is the current in amperes. The reciprocal of resistance is conductance, measured in siemens (S) or mho: G = 1/R.
The one line to write is R = V/I (ohm = volt per ampere), and its reciprocal G = 1/R in siemens (mho). Common mistake: inverting Ohm's law to R = I/V, and quoting conductance in ohms instead of siemens. Hook: resistance opposes, conductance allows — they are upside-down of each other.
Source: Guidebook
📖 §3.3 Electrical power & energy P=VI
8. Give the relation for electrical power and energy in a DC/AC circuit. An electric heater consumes 1.8 MJ when connected to a 250 V supply for 30 minutes. Find its power rating and the current drawn.
Model answer: Power P = V x I (watts); electrical energy = power x time = V x I x t (joules); the same formulas apply to AC using RMS values. Unit of energy for large amounts is kWh: 1 kWh = 1000 Wh = 3,600,000 J. Solution: Power = energy/time = 1.8x10^6 J / (30x60 s) = 1000 W = 1 kW. Current I = P/V = 1000/250 = 4 A.
Two conversions decide this sum: 30 minutes = 1800 seconds, and 1.8 MJ = 1.8 × 10⁶ J. Then P = energy/time and I = P/V. Marks are lost by leaving time in minutes or energy in MJ. Quick check with 1 kWh = 3.6 MJ: 1.8 MJ is 0.5 kWh used in half an hour, which is 1 kW — the same answer.
Source: Guidebook
📖 §3.3 P=VI, R=V/I (worked example)
9. A 100 W electric light bulb is connected to a 250 V supply. Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.
Model answer: (a) From P = V x I, current I = P/V = 100/250 = 0.4 A. (b) Resistance R = V/I = 250/0.4 = 625 Omega (equivalently R = V2/P = 250^2/100 = 625 Omega).
Use P = V I to get current, then Ohm's law for resistance. The short-cut worth memorising is R = V²/P. Common mistake: writing R = P/V² (upside down) — check by units: 250²/100 = 625 Ω is a sensible bulb resistance, 100/250² is not. Note a bulb draws a small current (0.4 A) but has a high resistance; a heater is the opposite.
Source: Guidebook
📖 §3.3 Ohm's law / power (worked example)
10. An electric kettle has a resistance of 30 ohm. What current flows when connected to a 240 V supply, and what is its power rating?
Model answer: Current I = V/R = 240/30 = 8 A. Power P = V x I = 240 x 8 = 1920 W = 1.92 kW = power rating of the kettle.
Order of work: I = V/R first, then P = V I. The one-step alternative is P = V²/R = 240²/30 = 1920 W — use it to verify. Always convert the final power to kW for a 'rating' answer (1920 W = 1.92 kW). Common mistake: using P = I²R with the wrong current, or forgetting that kettle 'rating' means power, not current.
Source: Guidebook
📖 §3.3 Power proportional to V-squared
11. An electric heater of 230 V, 5 kW rating is used for hot water generation. Find the electricity consumption per hour (a) at rated voltage and (b) at 200 V.
Model answer: Resistive load power varies as the square of voltage. (a) At rated voltage: consumption = 5 kW x 1 h = 5 kWh. (b) At 200 V: consumption = (200/230)^2 x 5 kW x 1 h = 3.78 kWh. So reduced supply voltage reduces a resistive heater's consumption in proportion to (V2/V1)^2.
For a resistive heater the resistance is fixed, so P ∝ V². The line to write is P₂ = (V₂/V₁)² × P₁. The classic error is scaling straight-line: (200/230) × 5 = 4.35 kW is wrong; (200/230)² × 5 = 3.78 kW is right. Hook: square the voltage ratio — a 13% voltage drop cuts power by about 24%.
Source: Guidebook
📖 §3.3 Power triangle, kW/kVA/kVAR
12. Explain the power triangle and define active power (kW), reactive power (kVAR), apparent power (kVA) and power factor.
Model answer: Total power has two components 90 degrees out of phase. Active/resistive power (kW) is directly converted to useful work. Reactive power (kVAR) builds up the magnetic flux for inductive devices but is otherwise non-usable. Apparent power (kVA) is the hypotenuse of the power triangle. Relations: kW = kVA cos(theta); kVA = kW/cos(theta); kVAR = kVA sin(theta); kVA2 = kW2 + kVAR2. Power factor PF = cos(theta) = kW/kVA, the cosine of the angle between kW and kVA.
Draw the right-angled triangle and write kVA² = kW² + kVAR², kW = kVA cosθ, kVAR = kVA sinθ, PF = kW/kVA. Because kVA is the hypotenuse it is always the largest — kVA can never be less than kW. Common mistake: adding kW and kVAR arithmetically; they are 90° apart, so they add vectorially only.
Source: Guidebook
📖 §3.3 Power factor definition
13. Define power factor.
Model answer: Power factor is the cosine of the phase angle between the current and the voltage in an AC circuit, PF = cos(theta). It equals the ratio of active power to apparent power, i.e. PF = kW/kVA. It is also the ratio of the resistive (useful) power to the total apparent power supplied. A low power factor (caused by inductive loads such as motors and transformers) means more current is needed to supply the same real power; it is improved by capacitors.
One line to memorise: PF = cosθ = kW/kVA. Being a cosine it can never exceed 1. Add the practical sentence — low PF means extra current for the same useful kW, so cables and transformers are loaded for nothing, and capacitors correct it. Common mistake: calling kVAR/kVA the power factor; that is sinθ, the reactive fraction.
Source: Guidebook
📖 §3.3 Single-phase & 3-phase power
14. Give the expressions for power in single-phase and three-phase balanced AC loads, and state which loads use single-phase power.
Model answer: For a balanced single-phase load: Power (W) = Vl x Il x cos(theta). For a balanced three-phase load: Power (W) = sqrt(3) x Vl x Il x cos(theta), where Vl = line voltage, Il = line current and cos(theta) is the power factor. Single-phase power is mostly used for lighting, fractional-HP motors and electric heater applications.
P = V I cosφ for single phase, P = √3 × V I cosφ for three phase — the √3 (1.732) is the only thing that separates them, and V and I must be LINE values. Forgetting √3 is the single biggest mark-loser in this chapter. Also remember the second half of the question: single phase is used for lighting, fractional-HP motors and heaters.
Source: Guidebook
📖 §3.3 Single-phase energy (worked example)
15. A 400 W mercury vapour lamp is switched on for 10 hours per day at 230 V (current 2 A, PF 0.8). Find the energy consumption per day.
Model answer: Single-phase energy (kWh) = V x I x cos(phi) x hours = 0.230 x 2 x 0.8 x 10 = 3.7 kWh (units) per day.
Formula: energy (kWh) = V × I × cosφ × hours, with V put in kV (0.230) so the answer lands directly in kWh. The 400 W lamp rating is a distractor — always use the MEASURED V, I and PF, not the nameplate watts. Common mistake: leaving V in volts and reporting watt-hours as kWh, i.e. an answer 1000 times too big.
Source: Guidebook
📖 §3.3 Single-phase energy (worked example)
16. A 250 W sodium vapour street lamp runs at 230 V for 12 hours/day, drawing 2 A at power factor 0.85. Calculate the energy consumption per day.
Model answer: Actual power drawn (single phase) = V x I x PF = 230 x 2 x 0.85 = 391 W = 0.391 kW. Energy per day = 0.391 kW x 12 h = 4.69 kWh/day (about 4.7 units/day).
Two steps: actual power = V × I × PF (single phase), then energy = kW × hours. The 250 W nameplate is a trap — the lamp actually draws 391 W, so use the measured values. Common mistake: quoting 250 W × 12 h = 3 kWh. Write the units 'kWh/day' or 'units/day' to secure the last mark.
Source: Guidebook
📖 §3.3 Motor kVA from HP
17. How is a motor's horsepower rating converted to kVA, and what is the significance of the nameplate rating?
Model answer: Motor loads are specified by horsepower (rated OUTPUT power). kVA = (HP x 0.746)/(eta x PF), where eta = motor efficiency and PF = motor power factor. The nameplate kW or HP indicates the OUTPUT of the motor at full load; the other nameplate parameters (volt, amps, PF) are the input conditions at full load. Smaller motors running partly loaded are the least efficient and have the lowest power factor.
The formula to memorise: kVA = (HP × 0.746)/(η × PF), with 1 HP = 0.746 kW = 745.7 W. The concept mark is for saying nameplate kW/HP is the OUTPUT at full load, while nameplate volts, amps and PF are the INPUT conditions. Common mistake: treating HP as input power — then efficiency never enters the sum.
Source: Guidebook
📖 §3.3 3-phase energy (worked example)
18. A 3-phase AC induction motor (20 kW) is used for pumping. Measured values: 440 V, 25 A, PF 0.90. Find the energy consumption in one hour.
Model answer: Three-phase energy = sqrt(3) x V x I x PF x time = 1.732 x 0.440 x 25 x 0.90 x 1 = 17.15 kWh in one hour.
Three-phase energy = √3 × V × I × PF × hours. Put V in kV (0.440) and the answer comes straight out in kWh. The 20 kW rating is a distractor — the motor is drawing 17.15 kW, not 20 kW. Common mistake: dropping the √3, which would give about 9.9 kWh instead of 17.15 kWh.
Source: Guidebook
📖 §3.3 Motor loading calculation
19. A 3-phase 10 kW motor has nameplate 415 V, 18.2 A, 0.9 PF. Actual measurement shows 415 V, 12 A, 0.7 PF. Find the motor loading and actual input power.
Model answer: Rated input at full load = sqrt(3) x V x I x PF = 1.732 x 0.415 x 18.2 x 0.9 = 11.8 kW. Rated efficiency = output/input = 10/11.8 = 85%. Measured (actual) input power = 1.732 x 0.415 x 12 x 0.7 = 6.0 kW. Motor loading (%) = (measured kW / rated input kW) x 100 = (6.0/11.8) x 100 = 51.2%.
Loading % = measured input kW ÷ rated INPUT kW × 100, with both inputs found from √3 × V × I × PF. The trap is dividing the measured 6.0 kW by the 10 kW OUTPUT rating — you must convert the rating to input first (here 11.8 kW from the nameplate amps and PF). Never judge loading from current alone, because the PF also falls when the motor is lightly loaded.
Source: Guidebook
📖 §3.3 Motor loading (worked example)
20. A 10 kW rated motor has full-load efficiency 85%. Actual input measurement shows 415 V, 10 A, PF 0.68. Find the motor loading in percentage.
Model answer: Measured 3-phase input power = sqrt(3) x V x I x PF = 1.732 x 415 x 10 x 0.68 = 4888 W = 4.89 kW. Rated input at full load = rated output/efficiency = 10/0.85 = 11.76 kW. Motor loading = (measured input / rated input) x 100 = 4.89/11.76 x 100 = 41.6%.
Two lines: measured input = √3 × V × I × PF, and rated input = rated output ÷ efficiency (10/0.85 = 11.76 kW). Loading = measured ÷ rated input. Common mistake: using 10 kW as the rated input — that inflates the loading figure. Note the low measured PF of 0.68 is itself the clue that the motor is lightly loaded.
Source: Guidebook
📖 §3.4 Temperature scales & conversions
21. Define temperature and give the Celsius, Fahrenheit and Kelvin scales with their conversions.
Model answer: Temperature is a physical property quantitatively expressing hot and cold. In the Fahrenheit scale, water freezes at 32 F and boils at 212 F. The Kelvin scale is the scientific standard, with the same increment as Celsius but origin at absolute zero (0 K = -273.15 C). Conversions: F = (C x 1.8) + 32; C = (F - 32)/1.8; K = C + 273.
Three conversions to memorise: °F = (°C × 1.8) + 32, °C = (°F − 32)/1.8, K = °C + 273. The trap is temperature DIFFERENCE: a rise of 1 °C is a rise of 1 K, so in Q = m Cp ΔT you never add 273 to ΔT. Add 273 only when an absolute temperature is needed (gas laws). Hook: 0 K = −273.15 °C is absolute zero — nothing can be colder.
Source: Guidebook
📖 §3.4 Pressure - absolute, gauge, atmospheric
22. Define pressure and explain absolute, gauge and atmospheric pressure with their relationship.
Model answer: Pressure is force per unit area: P = F/A, unit N/m2 or pascal (Pa). Absolute pressure (ps) is the true pressure measured relative to absolute (perfect) vacuum; all gas-law calculations need absolute pressure and temperature in Kelvin. Gauge pressure (pg) is what a gauge reads (gauges are zeroed at atmospheric). Atmospheric pressure (pa) is the surrounding air pressure at the earth's surface, varying with temperature and altitude. Relationship: absolute = gauge + atmospheric, i.e. ps = pg + pa.
The single relation to write: absolute = gauge + atmospheric (ps = pg + pa), with P = F/A in N/m² (pascal). Gauges are zeroed in the atmosphere, so they read zero when they are actually at 1 atm — you must add the atmosphere back. Common mistake: feeding gauge pressure into gas-law or steam-table work; those always need ABSOLUTE pressure and temperature in kelvin.
Source: Guidebook
📖 §3.4 / 3.5 Standard atmosphere units
23. State the value of standard atmospheric pressure in the various units used for pressure measurement.
Model answer: Standard atmospheric pressure is defined at sea level: 1 atm = 1.01325 bar = 101325 Pa (101.3 kPa) = 760 mm Hg = 10.33 metre H2O = 1013 mbar = 1.0332 kg/cm2. Other units: 1 bar = 100000 Pa, 1 kPa = 1000 Pa, 1 N/m2 = 1 Pa, 1 kgf/cm2 = 98066.5 Pa. Four common pressure-measurement units are pascal, kg/cm2, mm of mercury and metre of water column (also pounds/inch2).
Learn one chain: 1 atm = 1.01325 bar = 101325 Pa = 760 mm Hg = 10.33 m water column = 1.0332 kg/cm² = 1013 mbar. Rough working values worth remembering: 1 bar ≈ 1 kg/cm² ≈ 10 m of water. Common mistake: quoting 760 mm of WATER instead of mercury — mercury is 760 mm, water is 10.33 metres.
Source: Guidebook
📖 §3.4 Heat & calorie unit
24. Define heat and the calorie/kilocalorie, and give the relationship between calorie and joule.
Model answer: Heat is energy transferred from one body to another at lower temperature by virtue of temperature difference - it is energy in transition (transitory energy). Calorie is the unit of heat: the quantity of heat that raises the temperature of 1 g of water by 1 C. Kilocalorie (1 kcal = 1000 cal) raises 1 kg of water by 1 C. The internationally accepted unit is the joule: 1 calorie = 4.187 J (approx 4.2 J).
Two definitions and one constant: calorie raises 1 GRAM of water by 1 °C, kilocalorie raises 1 KG of water by 1 °C, and 1 calorie = 4.187 J (≈4.2 J). Say clearly that heat is energy in transition — it exists only while it is flowing because of a temperature difference. Common mistake: mixing cal and kcal, which is a factor-of-1000 error in the final answer.
Source: Guidebook
📖 §3.4 Specific heat
25. Define specific heat and give its units. Why are two specific heats defined for gases?
Model answer: Specific heat is the quantity of heat required to raise the temperature of 1 kg of a substance through 1 C (or 1 K). It is expressed in kcal/kg.C or J/kg.K and varies with temperature. For solids and liquids it is process-independent. For gases, heat can be added in infinitely many processes, so two specific heats are defined: specific heat at constant pressure (Cp) and at constant volume (Cv). Water has a very high specific heat (4200 J/kg.C = 1 kcal/kg.C) compared with other common substances.
Definition to reproduce: heat needed to raise 1 kg of a substance by 1 °C, units kcal/kg·°C or J/kg·K. Water is the number to keep ready: Cp = 1 kcal/kg·°C = 4200 J/kg·°C. Gases get two values (Cp and Cv) because a gas can be heated at constant pressure or constant volume, and heating at constant pressure also does expansion work, so Cp > Cv. Common mistake: using 4200 J/kg·°C with the mass written in grams.
Source: Guidebook
📖 §3.4 Specific heat of common substances
26. Give the specific heat values (J/kg.C) of common substances from the guidebook table.
Model answer: Lead 130, Mercury 140, Copper 390, Iron 470, Aluminium 910, Alcohol 2400, Water 4200 (all in J/kg.C). Water has the highest specific heat among these common substances, so it takes a lot of heat to raise its temperature and it releases a large quantity of heat when cooled.
Learn the table in rising order: Lead 130, Mercury 140, Copper 390, Iron 470, Aluminium 910, Alcohol 2400, Water 4200 J/kg·°C. Add the one-line conclusion: a LOW specific heat means the substance heats up (and cools down) quickly, which is why metals feel hot fast and water does not. Common mistake: quoting water as 4.2 without saying kJ/kg·°C, or as 1 without saying kcal/kg·°C.
Source: Guidebook
📖 §3.4 Sensible vs latent heat
27. Differentiate between sensible heat and latent heat, and give the formula for sensible heat.
Model answer: Sensible heat is the heat which, when added to (or removed from) a substance, causes a change in TEMPERATURE without altering moisture/phase. Sensible heat = mass x specific heat x temperature change: Q = m x Cp x dT (expressed in calories or joules). Latent heat is the change in heat content when a substance changes physical STATE (phase) WITHOUT any change in temperature. So sensible heat changes temperature; latent heat changes phase at constant temperature.
One formula, one distinction: sensible heat Q = m × Cp × ΔT changes TEMPERATURE; latent heat changes PHASE at constant temperature. So if the temperature is changing use Cp; if ice is melting or water is boiling use the latent heat value instead. Hook: 'sensible' = you can sense it on a thermometer; latent = hidden, the thermometer does not move.
Source: Guidebook
📖 §3.4 Fusion, vaporization, condensation
28. Define fusion, vaporization and condensation, and the terms melting point and boiling point.
Model answer: Fusion is the change of state from solid to liquid; the fixed temperature at which a solid changes into a liquid is its melting point. Vaporization is the change from liquid to gaseous state; the fixed temperature at which a liquid changes into vapour is its boiling point. Condensation is the change from gaseous state back to liquid state. Each of these phase changes occurs at constant temperature.
Three words in order of heating: fusion (solid → liquid at the melting point), vaporization (liquid → gas at the boiling point), condensation (gas → liquid). The mark-earning phrase is 'at constant temperature' — all the heat goes into changing state, not into raising temperature. Common mistake: calling fusion 'joining together'; here fusion means melting.
Source: Guidebook
📖 §3.4 Latent heat of fusion
29. Define latent heat of fusion, give its value for ice, and the formula used.
Model answer: Latent heat of fusion is the quantity of heat required to convert 1 kg of solid into liquid state without change of temperature (symbol h_if, unit J/kg or kJ/kg). For ice/water it is 335 kJ/kg. The same quantity is given up when liquid freezes to solid at the fusion temperature. Quantity of latent heat: Ql = m x h_if, where m is mass in kg. Example: 10 kg water at 0 C freezing to ice releases 10 x 335 = 3350 kJ.
The number to memorise: latent heat of fusion of ice = 335 kJ/kg, used as Ql = m × h_if. Freezing gives back exactly the same 335 kJ/kg that melting absorbs — same figure, opposite direction. Common mistake: mixing 335 (fusion, ice↔water) with 2257 (vaporization, water↔steam). Hook: melting is the small number, boiling is the big one.
Source: Guidebook
📖 §3.4 Latent heat of vaporization
30. Define latent heat of vaporization, state its value for water, and explain condensation in heat terms.
Model answer: Latent heat of vaporization is the heat a 1 kg mass of liquid absorbs going from liquid to vapour phase (or gives up vapour to liquid) without change in temperature (symbol h_fg, unit J/kg). For water it is 2257 kJ/kg (540 kcal/kg): when 1 kg of water at 100 C vaporizes to steam at 100 C, it absorbs 2257 kJ. Condensation is the reverse - when 1 kg of steam at 100 C condenses to water at 100 C it gives out about 2260 kJ of heat.
The number to memorise: latent heat of vaporization of water = 2257 kJ/kg = 540 kcal/kg at 100 °C, and condensation releases the same amount. Compare it with sensible heat — heating water 0 to 100 °C takes only about 100 kcal/kg, but boiling it takes 540 kcal/kg. That five-fold gap is why steam is used to carry heat. Common mistake: mixing the kJ and kcal figures in the same sum.
Source: Guidebook
📖 §3.4 Superheat & dryness fraction
31. What is superheating of steam, why is it done, and what is the dryness fraction (x)?
Model answer: Superheating is heating vapour (saturated steam) to a temperature much higher than the boiling/saturation temperature at the existing pressure. It is done in power plants to improve efficiency and avoid condensation in the turbine. The higher the water pressure, the higher the saturation temperature. Dryness fraction x = mass of steam per kg of water-steam mixture: in 1 kg of mixture, x kg is steam and (1-x) kg is water. x = 1 is dry saturated steam; the region right of the x = 1 line is superheated steam.
Two things to state: superheating means heating steam ABOVE its saturation temperature at that pressure (done to raise efficiency and keep the turbine dry), and dryness fraction x = mass of steam ÷ mass of the steam-water mixture. So in 1 kg of wet steam, x kg is steam and (1 − x) kg is water; x = 1 is dry saturated steam. Common mistake: thinking superheating happens at constant temperature — that is evaporation; superheating is exactly the stage where the temperature rises again.
Source: Guidebook
📖 §3.4 Humidity, specific & relative humidity
32. Define humidity, specific humidity (humidity ratio) and relative humidity (RH).
Model answer: Humidity is the moisture contained in air; saturated air holds all the moisture it can at that temperature and pressure. The unit of humidity is kg of moisture per kg of dry air. Specific humidity (humidity ratio) is the mass (kg) of water vapour in each kg of dry air (kg/kg). Relative humidity (RH) is the ratio of the mass of water vapour actually held by air in a given volume to that which the air could hold at the same temperature if saturated, expressed as a percentage. Warmer air holds more water vapour.
Keep the units apart: specific humidity (humidity ratio) is a MASS ratio, kg of moisture per kg of dry air; relative humidity is a PERCENTAGE of what the air could hold at that temperature. Note the 'per kg of DRY air' — that is what makes the humidity ratio constant when only the temperature changes. Hook: warm air holds more moisture, so heating air lowers its RH without adding or removing a single gram of water.
Source: Guidebook
📖 §3.4 Dew point, dry-bulb & wet-bulb temperature
33. Define dew point, dry-bulb temperature and wet-bulb temperature, and state their relationship at 100% RH.
Model answer: Dew point is the temperature at which water vapour in air becomes saturated and starts to condense into droplets; it equals the saturation temperature at the partial pressure of the water vapour. Dry-bulb temperature (DBT) measures sensible heat content and is not influenced by RH (recorded by a dry-bulb thermometer). Wet-bulb temperature (WBT) is recorded with a wick saturated with distilled water; evaporation lowers it, so WBT accounts for RH. If RH is 100%, the dew point, wet-bulb and dry-bulb temperatures are all equal.
The line that earns the mark: at 100% RH, DBT = WBT = dew point — all three read the same. WBT is lower than DBT because water evaporating from the wick takes latent heat away; the drier the air, the bigger the gap. Common mistake: calling dew point the temperature at which water boils off; it is where vapour starts to CONDENSE.
Source: Guidebook
📖 §3.4 Enthalpy of air
34. What is enthalpy of air and how is it determined?
Model answer: Enthalpy of air is the measure of the total heat content of an air and water-vapour mixture, measured from a pre-determined base point. It is expressed as kcal/kg or J/kg. The enthalpy of an air stream can be determined by measuring its dry-bulb and wet-bulb temperatures and referring to the psychrometric chart.
Enthalpy of air is TOTAL heat — sensible plus latent — measured from a chosen base point, in kcal/kg or J/kg of dry air. The practical answer to 'how is it determined' is: measure DBT and WBT, then read the enthalpy off the psychrometric chart. Common mistake: giving only the dry-bulb temperature; DBT alone gives sensible heat, and you need WBT to bring in the moisture.
Source: Guidebook
📖 §3.4 Fuel properties
35. Define fuel density, specific gravity and viscosity, and state the units of viscosity.
Model answer: Density is the ratio of mass of fuel to its volume at a stated temperature, expressed in kg/m3. Specific gravity of fuel is the ratio of the density of the fuel to that of water (specific gravity of water = 1); it has no units/dimensions, and higher specific gravity means higher heating value. Viscosity is a fluid's internal resistance to flow; it decreases with increasing temperature for all liquid fuels. Viscosity is measured in Stokes/Centistokes (also quoted in Engler, Saybolt or Redwood).
Three definitions, one unit trap: density in kg/m³, specific gravity a pure RATIO with NO units (water = 1), viscosity in stokes/centistokes. Add the direction of change — viscosity FALLS as temperature rises, which is exactly why furnace oil is preheated before pumping and atomising. Common mistake: writing units after specific gravity; it is dimensionless.
Source: Guidebook
📖 §3.4 Calorific value GCV vs NCV
36. How is the calorific value of a fuel measured, and what is the difference between GCV and NCV?
Model answer: Calorific (heating) value is the heat released during complete combustion of unit weight of fuel, measured by burning a known mass in a bomb calorimeter (sealed, insulated) and noting the temperature rise. It is expressed as Gross Calorific Value (GCV) or Net Calorific Value (NCV). The difference between GCV and NCV is the heat of vaporization of the moisture and of the atomic hydrogen (converted to water vapour) in the fuel; hence it is maximum for fuels with most hydrogen (e.g. natural gas). Typical heavy fuel oil: GCV approx 10,500 kcal/kg, NCV approx 9,800 kcal/kg.
State the instrument (bomb calorimeter) and then the one key sentence: GCV − NCV = the latent heat of the moisture in the fuel plus the water formed from the fuel's hydrogen. So the gap is largest for hydrogen-rich fuels like natural gas. Typical furnace oil: GCV ≈ 10,500 and NCV ≈ 9,800 kcal/kg. Common mistake: using GCV in an efficiency calculation that is defined on NCV (or the reverse) — always say which one you are using.
Source: Guidebook
📖 §3.4 Modes of heat transfer
37. Name and explain the three primary modes of heat transfer.
Model answer: Heat always flows from hot to cold; transfer rate is measured in watts (J/s). The three modes are: (1) Conduction - energy transfer in a solid, by molecular motion (higher-energy molecules impart energy to adjacent ones) and migration of free electrons (in pure metals). (2) Convection - energy exchange between a fluid and an adjacent solid; forced convection (fluid motion induced by fan/pump) or natural convection (density differences from heating cause hot fluid to rise, cold to sink). (3) Radiation - requires no medium; energy radiated over a range of wavelengths (infrared to ultraviolet) and can be reflected, absorbed or transmitted.
Three modes, one sorting rule: conduction needs contact (solid), convection needs a moving fluid (forced by a fan/pump or natural from density difference), radiation needs NO medium at all. Give the rate unit: heat transfer rate is in watts (J/s), and heat always flows hot → cold. Common mistake: describing convection but forgetting to split it into forced and natural, which is usually a separate mark.
Source: Guidebook
📖 §3.4 Evaporation process & enthalpy of steam
38. Describe the stages of evaporation and define enthalpy and specific enthalpy of steam.
Model answer: Evaporation occurs in stages: the liquid heats up to the evaporation temperature (sensible heat), then evaporates at constant temperature changing from fluid to gas (latent heat of evaporation), then the vapour heats above the evaporation temperature (superheating). The most common vapour is steam. Enthalpy of a system H = m x h, where m = mass (kg) and h = specific enthalpy (kJ/kg). Specific enthalpy h = u + p v, where u = internal energy (kJ/kg), p = absolute pressure (N/m2), v = specific volume (m3/kg).
Three stages in order: sensible heating up to boiling, latent heat at constant temperature, then superheating. The formulas to write: H = m × h, and specific enthalpy h = u + p v with p in N/m² ABSOLUTE and v the specific volume in m³/kg. Common mistake: using gauge pressure in h = u + p v, or forgetting that h is per kg while H is for the whole mass.
Source: Guidebook
📖 §3.4 Enthalpy values at standard atmosphere
39. State the specific enthalpy of saturated water, saturated steam and evaporation for water at standard atmospheric pressure.
Model answer: At standard atmosphere (1 bar, water boils at 100 C): specific enthalpy of saturated water hf = 419 kJ/kg; specific enthalpy of saturated steam hg = 2676 kJ/kg; specific enthalpy of evaporation he = hg - hf = 2676 - 419 = 2257 kJ/kg. hf can be calculated as cw x (tf - t0) with cw = 4.19 kJ/kg.C. For superheated steam, specific heat at constant pressure Cps = 1.860 kJ/kg.C at standard atmosphere.
Three numbers at 1 bar / 100 °C: hf = 419, hg = 2676, hfg = 2257 kJ/kg. They are linked by hfg = hg − hf, so if you remember any two you can produce the third — and 419 + 2257 = 2676 is your built-in check. Note hf ≈ 4.19 × 100, i.e. just water heated from 0 to 100 °C. Common mistake: quoting hfg as the total heat of steam; the total is hg.
Source: Guidebook
📖 §3.4 Laws of thermodynamics
40. State the three laws of thermodynamics.
Model answer: First law (law of conservation of energy): energy in a system can neither be created nor destroyed; it is only converted from one form to another or transferred between systems, so total energy remains constant. Second law: deals with the natural direction of energy processes - heat flows only from a hot to a colder object; it introduces entropy (disorder) and explains why no heat engine can be 100% efficient (some heat is always rejected to surroundings). Third law: it is impossible to reduce the temperature of any system to absolute zero (-273 C).
Sort them by what each one governs: 1st = QUANTITY (energy is conserved), 2nd = DIRECTION (heat flows hot to cold, entropy, no engine is 100% efficient), 3rd = absolute zero (−273 °C) can never be reached. Common mistake: saying the second law forbids heat flowing from hot to cold — it is the reverse (cold to hot on its own) that is forbidden. Hook: quantity, direction, zero.
Source: Guidebook
📖 §3.4 Entropy & second law
41. What is entropy, and what does the second law of thermodynamics tell us about heat-engine efficiency?
Model answer: Entropy, arising from the second law, means disorder; it can be used to quantify the amount of useful work obtainable from a system - the more chaotic/disorderly a system, the more difficult it is to perform useful work. The second law accounts for the fact that a heat engine can never be 100% efficient: some heat energy from the fuel is always rejected to the surroundings and is not converted into mechanical energy. The first law refers to the quantity of energy; the second law governs the direction of flow.
Entropy = disorder, and the more disordered a system, the less useful work you can get out of it. The exam sentence: a heat engine can never be 100% efficient because some heat must always be rejected to the surroundings. Hook: first law counts the energy, second law tells you which way it goes and how much of it is actually useful.
Source: Guidebook
📖 §3.5 SI base & derived units
42. List the SI base units, and give the SI derived units (with symbols) relevant to energy management.
Model answer: SI base units: length-metre (m), time-second (s), electric current-ampere (A), temperature-kelvin (K), amount of substance-mole (mol), luminous intensity-candela (cd). Derived units relevant to energy: frequency-hertz (Hz, s^-1); force-newton (N, m.kg.s^-2); pressure-pascal (Pa, N/m2); energy/work/heat-joule (J, N.m); power-watt (W, J/s); electric potential-volt (V, W/A); electric resistance-ohm (V/A); electric conductance-siemens (S, A/V).
Base units are the six the guidebook lists: metre, second, ampere, kelvin, mole, candela (kilogram is the base unit of mass). Derived: Hz, N, Pa, J, W, V, ohm, S. Each derived unit is best remembered through its definition — Pa = N/m², J = N·m, W = J/s, V = W/A, ohm = V/A, S = A/V. Common mistake: writing °C as the SI unit of temperature; the SI unit is the kelvin.
Source: Guidebook
📖 §3.5 Energy unit conversions
43. Give the key energy unit conversions for kWh, joule, BTU, kcal and HP.
Model answer: 1 joule = 1 watt-second; 1 kW = 1000 W; 1 kWh = 3.6 x 10^6 J = 3.6 million joules; 1 watt-hour = 3600 J; 1 MJ = 278 Wh; 1 BTU = 252 cal = 1055 J; 1 BTU/h = 0.293071 Wh; 1 kcal/h = 1.163 Wh; 1 HP = 745.7 watts (0.746 kW). For energy accounting: 1 kWh = 860 kcal.
Six numbers are worth rote learning: 1 kWh = 3.6 × 10⁶ J = 860 kcal, 1 Wh = 3600 J, 1 BTU = 1055 J = 252 cal, 1 HP = 745.7 W (0.746 kW). Use 3.6 MJ when the answer must be in joules, and 860 kcal when it is fuel/energy accounting. Common mistake: writing 1 kWh = 860 kJ — it is 860 kilo-CALORIES, and confusing the two costs the whole sum.
Source: Guidebook
📖 §3.5 toe / MTOE definition & GCVs
44. Define 1 kg oil equivalent and 1 MTOE, and give the standard calorific values used for fuels in toe accounting.
Model answer: 1 kg of oil equivalent = 10,000 kcal; 1 Metric Tonne of Oil Equivalent (MTOE) = 1 x 10^7 kcal. Energy accounting: 1 kWh = 860 kcal. Standard GCVs (Gazette of India 2007, or supplier certificate): Charcoal 6,900 kcal/kg; Furnace oil/RFO/LSHS/Naphtha 10,050 kcal/kg; HSD 11,840 kcal/kg; Petrol 11,200 kcal/kg; Kerosene 11,110 kcal/kg; LPG 12,500 kcal/kg; Natural gas 8,000-10,500 kcal/m3; coal/coke as per supplier certificate.
The two anchor numbers: 1 kg oil equivalent = 10,000 kcal, so 1 MTOE (a TONNE) = 1000 × 10,000 = 1 × 10⁷ kcal. Add 1 kWh = 860 kcal for the electricity side. These GCVs come from the Gazette of India notification (2007) and apply only when the supplier's certificate is not available. Common mistake: dividing by 10⁴ instead of 10⁷ — always ask yourself whether you are working per kg or per tonne.
Source: Guidebook
📖 §3.5 MTOE conversion formulas
45. Give the formulas used to convert solid, liquid and gaseous fuel consumption to MTOE.
Model answer: For solid fuel: MTOE = (quantity in kg x GCV in kcal/kg) / 10^7. For liquid fuel: MTOE = (quantity in kg or litres x GCV in kcal/kg or kcal/litre) / 10^7. For gaseous fuel: MTOE = (quantity in kg or Nm3 x GCV in kcal/kg or kcal/Nm3) / 10^7. In the absence of a supplier certificate, GCV is taken from a NABL-accredited / State or Government-recognised lab test certificate.
All three formulas are the same shape: MTOE = (quantity × GCV in kcal) ÷ 10⁷. Only the unit of quantity changes — kg for solid, kg or litre for liquid, kg or Nm³ for gas. Just make sure the GCV unit matches the quantity unit (kcal/kg with kg, kcal/litre with litres, kcal/Nm³ with Nm³). Common mistake: leaving the quantity in tonnes; convert to kg first, or the answer is 1000 times too small.
Source: Guidebook
📖 §3.5 toe calculation (worked example)
46. What is the toe of 125 tonnes of coal having GCV 4000 kcal/kg?
Model answer: Mass = 125 tonnes = 125,000 kg. Energy = mass x GCV = 125,000 x 4000 = 5 x 10^8 kcal. toe = energy / 10^7 = 5 x 10^8 / 10^7 = 50 toe.
Two moves only: tonnes → kg (×1000), then toe = (kg × GCV)/10⁷. Common mistake: leaving 125 tonnes as 125 and getting 0.05 toe instead of 50 toe — the tonne-to-kg step is where this question is won or lost. Sanity check: 10⁷ kcal is one toe, so 5 × 10⁸ kcal must be 50 toe.
Source: Guidebook
📖 §3.1 Chemical & nuclear energy
47. Define chemical energy and nuclear energy with their characteristic equations.
Model answer: Chemical energy is energy stored in the bonds of atoms and molecules, released as heat in a chemical reaction; it is specific to each reaction and given per unit mass (kJ/kg) or per mole (kJ/mol). Examples: biomass, petroleum, natural gas, propane, coal. Nuclear energy is energy stored in the nucleus of an atom that holds it together; uranium releases nuclear energy on fission (loss of mass m), given by Einstein's equation E = m c2, where c = 3 x 10^8 m/s.
Chemical energy sits in the BONDS between atoms, nuclear energy sits INSIDE the nucleus — that one word is the difference the examiner looks for. The equation to write is Einstein's E = mc² with c = 3 × 10⁸ m/s, where m is the mass LOST in fission. Common mistake: giving units for chemical energy as kJ only; the book gives it per unit mass (kJ/kg) or per mole (kJ/mol).
Source: Guidebook
📖 §3.3 Lagging current / PF improvement
48. Why does current lag voltage in an AC system, and how is the power factor improved?
Model answer: Current lags voltage mainly due to inductive loads (such as motors and transformers), which draw reactive current (kVAR) to build up the magnetic flux. This reactive component lowers the power factor. Power factor is improved by installing capacitors, which supply leading reactive power (kVAR) to offset the lagging reactive demand of the inductive load, thereby reducing the kVA drawn from the supply for the same useful kW.
Two sentences carry the marks: inductive loads (motors, transformers) draw magnetising kVAR, which makes current lag voltage; capacitors supply leading kVAR and cancel it. After correction the useful kW is unchanged — it is the kVA and the line current that come down. Hook: the coil lags, the capacitor leads, so put them together and they cancel.
Source: Guidebook
📖 §3.3 Contract/Maximum demand, billing, PF
49. A facility has connected load 500 kW and contract demand 500 kVA. Monthly maximum demand approx 350 kW at 0.85 PF. Demand charge is Rs 300 per kVA/month and minimum billing demand is 80% of contract demand. a) Determine the current demand in kVA. b) Calculate excess demand charges above minimum billing demand. c) Find the minimum power factor required to avoid excess demand charges.
Model answer: a) Actual kVA demand = kW/PF = 350/0.85 = 411.76 kVA. b) Minimum billed demand = 500 x 0.8 = 400 kVA; excess demand = 411.76 - 400 = 11.76 kVA; excess charges = 11.76 x 300 = Rs 3528/month. c) Minimum PF to keep demand at 400 kVA = 350/400 = 0.875.
Three relations do the whole sum: kVA = kW/PF, minimum billing demand = 80% of contract demand, and required PF = kW ÷ billed kVA. Common mistake: comparing 350 kW against the 500 kVA contract — demand is billed in kVA, so convert with the power factor first. Note that raising PF from 0.85 to 0.875 alone removes the penalty; that is the answer to part (c).
Source: Sep 2025 (25th NCE)
📖 §3.3 Motor loading from PF & kVAR
50. A 10 HP induction motor (nameplate 415 V, 12 A, PF 0.9) is audited. Monitoring shows reactive power 2 kVAR and power factor 0.758. Calculate the percentage loading of the motor.
Model answer: With PF = kW/kVA and kVA2 = kVAR2 + kW2, using kVAR = 2 and PF = 0.758: tan(theta) = kVAR/kW, and sin(theta) = sqrt(1-0.758^2) = 0.652, so kVA = kVAR/sin = 2/0.652 = 3.07; measured kW = kVA x PF = 3.07 x 0.758 = 2.32 kW. Rated input kW = sqrt(3) x V x I x PF = 1.732 x 0.415 x 12 x 0.9 = 7.76 kW. Percentage loading = 2.32/7.76 x 100 = 29.9%.
When kVAR and PF are known, use sinθ = √(1 − PF²), then kVA = kVAR/sinθ and kW = kVA × PF. Common mistake: dividing kVAR by the power factor instead of by sinθ. Then loading % = measured kW ÷ rated input kW, where rated input = √3 × V × I × PF from the nameplate — not the 10 HP output.
Source: Sep 2024 (24th NCE)
📖 §3.4 Sensible heat / energy balance
51. A drilling machine draws 5 kW input at 50% efficiency to drill a 5 kg aluminium block. A 45 C temperature rise is observed over 100 s (specific heat of aluminium = 900 J/kg.K). What percentage of the machine's output power is lost to the surroundings?
Model answer: Output power = 5 x 0.5 = 2.5 kW. Energy delivered Q = 2.5 x 1000 x 100 = 250,000 J. Energy absorbed by block Q' = m x Cp x dT = 5 x 900 x 45 = 202,500 J. Fraction used for heating = 202,500/250,000 = 81%. Energy lost to surroundings = 100 - 81 = 19%.
Work in joules throughout: output = input × efficiency, energy delivered = output (W) × time (s), and heat absorbed = m × Cp × ΔT. Loss % = 100 − (heat absorbed ÷ energy delivered × 100). Common mistake: using the 5 kW INPUT instead of the 2.5 kW output; the question asks for the loss as a percentage of the machine's OUTPUT.
Source: Sep 2024 (24th NCE)
📖 §3.3 Single-phase R=V/I, P proportional to V2
52. A single-phase electric geyser is rated 2000 W at 230 V. Calculate (a) rated current, (b) resistance in ohms, (c) actual power drawn when the measured supply voltage is 210 V.
Model answer: (a) Rated current I = P/V = 2000/230 = 8.7 A. (b) Resistance R = V/I = 230/8.7 = 26.45 Omega. (c) Actual power at 210 V = V2/R = 210^2/26.45 = 1667 W = 1.67 kW (equivalently (210/230)^2 x 2000 = 1667 W).
Three steps, three formulas: I = P/V, R = V/I, and P = V²/R for the new voltage. The resistance stays the same when the supply voltage falls, so the power drops as the SQUARE of the voltage: (210/230)² × 2000 = 1667 W. Common mistake: assuming the geyser still draws its rated 2000 W at 210 V — it does not, and the water simply takes longer to heat.
Source: Jul 2022
📖 §3.4 Sensible heat balance (Q=mCpdT)
53. A furnace shell (4 tonnes) is to be cooled from 95 C to 45 C. The maximum permissible rise in water temperature is 5 C. Compute the quantity of water required. (Cp shell = 0.122 kcal/kg.C, Cp water = 1 kcal/kg.C)
Model answer: Heat to be removed Q = m x Cp x dT = 4000 x 0.122 x (95-45) = 24,400 kcal. For water: Q = m x Cp x dT, so 24,400 = m x 1 x 5, giving m = 24,400/5 = 4,880 kg of water.
Same formula on both sides: Q = m × Cp × ΔT for the shell gives the heat to be removed, and the same Q for water gives the mass required. Convert first: 4 tonnes = 4000 kg. Cp of water = 1 kcal/kg·°C is what makes the water side easy. Common mistake: using the shell's 50 °C drop for the water — the water is only allowed a 5 °C rise, and that is the ΔT you divide by.
Source: Sep 2021 (21st NCE)
📖 §3.3 Contract demand vs maximum demand
54. Explain the difference between contract demand and maximum demand.
Model answer: Contract Demand is the amount of electric power (in kVA or kW) a customer contracts/agrees to draw from the utility in a specified interval; it is the capacity for which the utility must plan. Maximum Demand is the highest average kVA recorded during any one demand interval within the billing month (the interval is normally 30 minutes, ranging 15-60 minutes), measured by a tri-vector / digital energy meter.
Keep them apart by asking 'agreed' or 'recorded'. Contract demand is the capacity the consumer AGREES to draw; maximum demand is the highest average kVA actually RECORDED in a demand interval during the month. The number to quote is the demand interval: normally 30 minutes (range 15–60 minutes), measured by a tri-vector/digital meter. Common mistake: calling maximum demand the highest instantaneous peak — it is an interval average, not an instant.
Source: Dec 2009 (9th NCE)
📖 §3.3 Power factor from kW & kVAR
55. An induction motor draws 8 kW with a lagging reactive power of 4 kVAR. Calculate the operating power factor.
Model answer: Power factor = kW/kVA = kW/sqrt(kW2 + kVAR2) = 8/sqrt(8^2 + 4^2) = 8/sqrt(80) = 8/8.944 = 0.894 (lagging). Book-1 Ch.3 §3.3 power triangle: kVA² = kW² + kVAr², PF = cos(theta) = kW/kVA. The load is therefore operating at about 0.89 lagging power factor.
One line does it: PF = kW/√(kW² + kVAR²). Here 8/√(64+16) = 8/8.944 = 0.894. Common mistake: writing PF = kVAR/kW (that is tanθ) or adding 8 and 4 straight. Always add the word 'lagging' for an induction motor — the direction of the phase angle carries a mark.
Source: Dec 2009 (9th NCE)
📖 §3.4 Sensible heat (find final temperature)
56. The initial temperature of 150 g of ethanol was 22 C. What is the final temperature if 3240 J is supplied? (Specific heat of ethanol = 2.44 J/g.C)
Model answer: Q = m x C x (Tf - Ti): 3240 = 150 x 2.44 x (Tf - 22) = 366 (Tf - 22). So Tf - 22 = 3240/366 = 8.85, giving Tf = 30.9 C.
Rearranged formula: Tf = Ti + Q/(m × C). The unit trap here is that the specific heat is given per GRAM (2.44 J/g·°C), so keep the mass as 150 g — do not convert to 0.15 kg. Common mistake: forgetting to add the initial 22 °C back at the end; the formula gives the RISE, not the final temperature.
Source: Dec 2009 (9th NCE)
📖 §3.4 Specific heat comparison (iron vs copper)
57. When the same quantity of heat is added to equal masses of iron and copper, the iron's temperature rises by 15 C. Find the rise in copper's temperature. (Cp iron = 470 J/kg.C, Cp copper = 390 J/kg.C)
Model answer: Since masses are equal and heat added is the same: Cp(iron) x 15 = Cp(copper) x dT(copper). So dT(copper) = (470 x 15)/390 = 18.08 C. (The lower specific heat of copper makes it heat up more for the same heat input.)
With equal masses and equal heat, m Cp ΔT is the same for both, so Cp₁ ΔT₁ = Cp₂ ΔT₂ — temperature rise is INVERSELY proportional to specific heat. Copper has the lower Cp (390 vs 470), so it must get hotter: ΔT = 15 × 470/390. Common mistake: multiplying by 390/470 and getting a smaller rise — always sanity-check that the lower-Cp metal heats up more.
Source: Compilation
📖 §3.5 MTOE multi-fuel conversion
58. A textile plant's monthly use: 700,000 kWh electricity, 40 kL furnace oil (sp.gr 0.92, GCV 10,000 kcal/kg), 360 t coal (GCV 3450 kcal/kg), 10 kL HSD (sp.gr 0.885, GCV 10,500 kcal/kg). Compute the monthly energy use in MTOE. (1 kWh = 860 kcal, 1 kg oil equiv = 10,000 kcal)
Model answer: Furnace oil = 40,000 x 0.92 x 10,000 = 36.8 x 10^7 kcal; Coal = 360,000 x 3450 = 124.2 x 10^7 kcal; Electricity = 700,000 x 860 = 60.2 x 10^7 kcal; HSD = 10,000 x 0.885 x 10,500 = 9.29 x 10^7 kcal. Total = 230.5 x 10^7 kcal. MTOE = total / 10^7 = 230.5 MTOE per month (annual approx 2766 MTOE).
Bring everything to kcal first, then divide the total by 10⁷ once at the end. Electricity uses 1 kWh = 860 kcal; oils use litres × specific gravity to get kg, then × GCV. Common mistake: forgetting the specific gravity on furnace oil and HSD (GCV is per kg, not per litre), and forgetting 40 kL = 40,000 litres. Coal is the easy one — tonnes × 1000 × GCV. Add all four streams before the final division.
Source: Compilation
📖 §3.4 Pressure - absolute vs gauge
59. Give the relationship between absolute and gauge pressure, and list four units used for pressure measurement.
Model answer: Absolute pressure is zero-referenced against a perfect vacuum: Absolute = Atmospheric + Gauge. Gauge pressure is zero-referenced against ambient atmospheric pressure: Gauge = Absolute - Atmospheric (gauges read zero at atmospheric). Four pressure-measurement units: Pascal (N/m2), kg/cm2, atmosphere (mm of mercury), and metre of water column (also pounds/inch2).
One relation, both ways round: absolute = atmospheric + gauge, so gauge = absolute − atmospheric. A gauge in open air reads zero even though the true pressure is 1 atm. For the four units, quote pascal (N/m²), kg/cm², mm of mercury and metre of water column. Common mistake: subtracting when you should add — remember absolute pressure is always the BIGGER number.
Source: Compilation
📖 §3.4 GCV/NCV combustion - calorific value
60. A gas-fired water heater heats water flowing at 20 litre/min from 25 C to 85 C. If the GCV of gas is 9200 kcal/kg and heater efficiency 82%, find the gas combustion rate in kg/min.
Model answer: Mass of water = 20 kg/min (density 1 kg/litre). Heat required = m x Cp x dT = 20 x 1 x (85-25) = 1200 kcal/min. Heat from gas x efficiency = heat to water: gas (kg/min) x 9200 x 0.82 = 1200. Gas rate = 1200/(9200 x 0.82) = 0.159 kg/min.
Balance the two sides: fuel (kg) × GCV × efficiency = m × Cp × ΔT. Water makes it easy — 1 litre = 1 kg and Cp = 1 kcal/kg·°C, so the heat load is just 20 × 60 = 1200 kcal/min. Common mistake: multiplying the required heat by 0.82 instead of dividing by it; low efficiency must mean MORE fuel, so the answer has to go up.
Source: Compilation
📖 §3.4 Latent heat / steam saving (heat recovery)
61. In a textile unit, 25,000 kg/hr water is heated from 28 C to 80 C by steam. By recovering effluent heat, water is pre-heated to 45 C before steam raises it to 80 C. Estimate the steam saving (kg/hr), latent heat of steam = 520 kcal/kg.
Model answer: Without recovery: Q1 = m x Cp x dT = 25000 x 1 x (80-28) = 13,00,000 kcal/hr; steam = 1,300,000/520 = 2500 kg/hr. With recovery: Q2 = 25000 x 1 x (80-45) = 8,75,000 kcal/hr; steam = 875,000/520 = 1682.7 kg/hr. Steam saving = 2500 - 1682.7 = 817.3 kg/hr.
Steam required = sensible heat load ÷ latent heat of steam, i.e. (m × Cp × ΔT)/h_fg. Use the latent heat printed in the question (520 kcal/kg here) — do not substitute the standard 540 kcal/kg. Recovery only changes ΔT (52 °C becomes 35 °C); the flow of 25,000 kg/hr stays the same, so the saving comes purely from the smaller temperature rise.
Source: Compilation
📖 §3.4 Latent heat of steam (kerosene heating)
62. A tank with 600 kg kerosene is heated from 10 C to 40 C in 20 minutes using 4 bar(g) steam (latent heat hfg = 2108.1 kJ/kg). Cp of kerosene = 2.0 kJ/kg.C. Heat losses negligible. Determine the steam flow rate in kg/hr.
Model answer: Heat rate Q = m x Cp x dT / time = 600 x 2 x (40-10) / 1200 s = 36,000/1200 = 30 kJ/s. Steam mass flow = Q x 3600 / hfg = 30 x 3600 / 2108.1 = 51.23 kg/hr.
Steam flow = heat rate ÷ latent heat, m = Q/h_fg. The unit work is the whole exam trick: 20 minutes = 1200 seconds gives Q in kJ/s (= kW), then multiply by 3600 to get kg per HOUR. Common mistake: leaving the answer in kg/s or forgetting the ×3600. Note only the latent heat is used, because the steam condenses at constant temperature.
Source: Compilation
📖 §3.4 Heat balance - condensate recovery
63. Boiler feed water is at 70 C. Returning condensate is at 86 C and makeup water at 27 C. Determine the percentage of condensate water that can be recovered (mass/heat balance).
Model answer: Let makeup fraction = x and condensate fraction = (1-x). Heat balance: 27x + 86(1-x) = 70. So 27x + 86 - 86x = 70; -59x = -16; x = 0.27. Thus makeup = 27% and condensate recovered = 1 - 0.27 = 0.73 = 73%.
Set up a 1 kg heat balance: makeup fraction x at 27 °C plus condensate (1 − x) at 86 °C must average to 70 °C, i.e. 27x + 86(1 − x) = 70. Faster form to remember: x = (86 − 70)/(86 − 27) = 16/59 = 0.27, so 73% condensate is recovered. Common mistake: solving for x and then reporting x as the condensate — x is the MAKEUP fraction; the recovery is 1 − x.
Source: Compilation
📖 §3.4 Latent heat of steam (air heating coil)
64. A paint drier needs 75.4 m3/min of air at 93 C heated by a steam coil. How many kg/hr of steam at 4 bar are needed? (air density 1.2 kg/m3, Cp air 0.24 kcal/kg.C, ambient 32 C, latent heat of steam 510 kcal/kg)
Model answer: Air flow = 75.4 x 60 = 4524 m3/hr = 4524 x 1.2 = 5428.8 kg/hr. Sensible heat = m x Cp x dT = 5428.8 x 0.24 x (93-32) = 79,477.6 kcal/hr. Steam required = 79,477.6 / 510 = 156 kg/hr.
Chain of conversions: m³/min × 60 = m³/hr, × density = kg/hr, then Q = m × Cp × ΔT, then steam = Q ÷ latent heat. Common mistake: forgetting the ×60, which makes the answer 60 times too small. Only sensible heat is needed on the air side (air is just being warmed), and only latent heat on the steam side (steam is just condensing).
Source: Compilation
📖 §3.4 Heat (Q=mCpdT) in kcal and kWh
65. Calculate the heat energy required to raise 200 kg of water from 30 C to 80 C (Cp water = 1 kcal/kg.C). Express the answer in kcal and kWh.
Model answer: Q = m x Cp x dT = 200 x 1 x (80-30) = 200 x 50 = 10,000 kcal. Converting: 1 kWh = 860 kcal, so Q = 10,000/860 = 11.63 kWh.
Q = m × Cp × ΔT with Cp of water = 1 kcal/kg·°C, then divide by the constant that links heat and electricity: 1 kWh = 860 kcal. Common mistake: dividing by 3600 or 4.187 — those convert to kJ, not kWh. For kcal → kWh the number is always 860. Hook: 860 for energy accounting, 3.6 MJ when you want joules.
Source: Compilation
📖 §3.2 Power & energy (pump duty)
66. A pump runs at constant head/flow delivering 250 litre/s at 100 m head, drawing 300 kW. Calculate the energy consumption to pump 13,500 kL of water.
Model answer: Time = volume/flow = (13,500 x 10^3 litre) / (250 litre/s x 3600 s/hr) = 13,500,000/900,000 = 15 hours. Energy = power x time = 300 kW x 15 h = 4500 kWh.
Two steps: time = volume ÷ flow rate, then energy = power × time. The unit work is the trap: 13,500 kL = 13,500,000 litres, and 250 litre/s = 900,000 litre/hr. The 100 m head and the flow are given only to confirm the duty is constant — the 300 kW input is what you actually multiply by the hours.
Source: Compilation
📖 §3.1 Mass flow / energy (conveyor)
67. A conveyor delivers coal 1 m wide with a 0.25 m bed height at 0.5 m/s. Determine the coal delivery in tons/hour (coal density 1.1 ton/m3).
Model answer: Volumetric rate = width x height x speed = 1 x 0.25 x 0.5 = 0.125 m3/s = 0.125 x 3600 = 450 m3/hr. Coal delivery = 450 x 1.1 = 495 tonnes/hr.
Volumetric rate = width × bed height × belt speed (m³/s), then × 3600 for m³/hr, then × density for tonnes/hr. Common mistake: forgetting the ×3600, or multiplying by density before converting the time base. Check the answer's units at each step — m × m × m/s really does give m³/s.
Source: Compilation
📖 §3.4 Combustion - unburnt carbon balance
68. A coal sample contains 60% carbon and 23% ash. The combustion refuse contains 7% carbon (rest ash). Compute the percentage of original carbon remaining unburnt in the refuse.
Model answer: Take 100 kg refuse: unburnt carbon = 7 kg, ash = 93 kg. All ash comes from coal (23% of coal): raw coal = 93/0.23 = 404.35 kg. Original carbon in coal = 0.60 x 404.35 = 242.61 kg. Unburnt carbon = 7 kg. Percentage unburnt = (7/242.61) x 100 = 2.89%.
Use ash as the tracer: ash does not burn, so all the ash in the refuse came from the coal. Take 100 kg of refuse as the basis — 7 kg carbon, 93 kg ash. Coal burnt = ash ÷ 0.23, then original carbon = 0.60 × coal, and unburnt % = 7 ÷ that carbon × 100. Common mistake: reporting 7% as the answer — 7% is the carbon in the REFUSE, not the fraction of the coal's original carbon left unburnt.
Source: Compilation
📖 §3.4 Calorific value - DG energy balance
69. A DG set gives 3.5 kWh per litre of diesel. Cooling-water loss is 28% and exhaust loss 32% of fuel input. Diesel CV = 10,200 kcal/kg, sp.gr 0.85. Calculate the unaccounted loss as % of input energy.
Model answer: Heat input per litre = 10,200 x 0.85 = 8670 kcal/litre. Useful output = 3.5 kWh x 860 = 3010 kcal/litre. % output = 3010/8670 = 34.72%. Unaccounted loss = 100 - (34.72 + 28 + 32) = 5.28%.
Two conversions do the whole sum: 1 kWh = 860 kcal (to turn output into kcal) and litres × specific gravity = kg (to turn fuel into kcal). Then unaccounted loss = 100 − (output% + cooling% + exhaust%). Common mistake: multiplying the CV by the litres without the specific gravity of 0.85 — calorific value is per KG, not per litre.
Source: Compilation
📖 §3.5 Fuel substitution cost saving (GCV)
70. A boiler uses 6 t/day coal (GCV 3300 kcal/kg, Rs 4200/t) at 72% efficiency. Coal is replaced by agro-residue (GCV 3100 kcal/kg, Rs 1800/t) at the same 72% efficiency. Calculate the annual cost savings for 300 days.
Model answer: Useful heat from coal = 6000 x 3300 x 0.72 = 1,42,56,000 kcal/day. Agro-residue needed = 14,256,000/(3100 x 0.72) = 6387 kg/day. Daily cost: coal = 6 x 4200 = Rs 25,200; agro = 6.387 x 1800 = Rs 11,497. Daily saving = Rs 13,703. Annual saving = 13,703 x 300 = Rs 41,10,900.
The rule is: the USEFUL heat must stay the same, so quantity × GCV × efficiency is equated for both fuels. Lower GCV therefore means more tonnes needed. Watch the units: 6 t = 6000 kg for the heat calculation, but the price is Rs per TONNE, so convert back to tonnes before costing. Common mistake: comparing the two fuels on price per tonne alone — the agro-residue looks cheaper still, but you must first find how much MORE of it is burnt.
Source: Compilation
📖 §3.4 Gas-fired heater fuel (SI units)
71. A gas-fired water heater heats water at 10 litre/min from 30 C to 95 C. GCV of gas = 4 x 10^4 kJ/kg, efficiency 60%, Cp water = 4.2 kJ/kg.C. Find the gas consumption in kg/min.
Model answer: Mass of water = 10 kg/min. Heat required = m x Cp x dT = 10 x 4.2 x (95-30) = 2730 kJ/min. Gas x GCV x efficiency = heat to water: gas = 2730/(4x10^4 x 0.6) = 2730/24000 = 0.114 kg/min.
Same balance as the kcal version, but in SI: fuel (kg) × GCV (kJ/kg) × efficiency = m × Cp × ΔT with Cp of water = 4.2 kJ/kg·°C. Water again gives 1 litre = 1 kg, so 10 litre/min = 10 kg/min. Common mistake: mixing the two systems — if the GCV is in kJ/kg you must use 4.2 kJ/kg·°C, not 1 kcal/kg·°C.
Source: Compilation
📖 §3.1/3.3/3.4 Objective recall facts
72. State the correct answers to the chapter's key objective recalls: (i) an example of stored mechanical energy; (ii) active power in an AC circuit; (iii) what nameplate kW/HP of a motor indicates; (iv) the fuel with the maximum difference between GCV and NCV.
Model answer: (i) Stored mechanical energy: an arrow in a stretched bow (or a compressed spring) - energy stored by application of force. (ii) Active power in an AC circuit = kVA x power factor (= sqrt(kVA2 - kVAR2)). (iii) Nameplate kW or HP of a motor indicates the OUTPUT power of the motor at full load. (iv) The GCV-NCV difference is maximum for natural gas, because it has the most hydrogen forming water vapour.
Four separate recalls, so give four crisp lines. Active power = kVA × PF (also √(kVA² − kVAR²)). Nameplate kW/HP = OUTPUT at full load, never input. Stored mechanical energy = a stretched bow or compressed spring (force applied and held). GCV − NCV is greatest for natural gas because it has the most hydrogen, and that hydrogen becomes water vapour whose latent heat is lost.
Source: Guidebook
📖 §3.4 Rate of heat transfer unit
73. In what unit is the rate of energy (heat) transfer from a higher to a lower temperature measured, and why?
Model answer: The energy transferred is measured in joules; the RATE of energy transfer (heat transfer) is measured in watts (J/s), since 1 watt = 1 joule per second. Heat always flows from a higher temperature to a lower temperature, independent of the mode (conduction, convection or radiation).
The rate of heat transfer is in WATTS, because 1 watt = 1 joule per second — the joule is the quantity, the watt is the quantity per second. Add that this is true for all three modes (conduction, convection, radiation) and that flow is always hot → cold. Common mistake: answering 'joules' — that is the energy, not the rate.
Source: Guidebook
📖 §3.3 Electricity Basics / §3.4 Thermal Energy Basics — chapter terminology (GCV, NCV, PF, kVA, kVAR) with §3.5 and chapter question S-4 (Load Factor)
74. Expand and briefly explain the common acronyms used in this chapter: GCV, NCV, PF, kVA, kVAR, MD, LF, TOD.
Model answer: GCV = Gross Calorific Value (total heat of combustion including latent heat of water vapour). NCV = Net Calorific Value (excludes that latent heat). PF = Power Factor (cos(theta) = kW/kVA). kVA = kilovolt-ampere (apparent power). kVAR = kilovolt-ampere reactive (reactive power). MD = Maximum Demand (highest average kVA/kW over a demand interval in the billing period). LF = Load Factor (average load / peak load over a period). TOD = Time of Day (tariff with different rates by time-of-day to shift load off-peak).
Group them in threes so nothing drops out under pressure. Fuel pair: GCV minus the latent heat of the water vapour = NCV, so NCV is always the SMALLER number — quoting efficiency on GCV when the question means NCV is the classic mark-loser. Power triangle: kW is real, kVA is apparent, kVAR is reactive, and PF = kW/kVA — write the unit every time, because kVA and kVAR carry no 'h' and are power, not energy. Billing trio: MD is the highest average demand over the demand interval (you are billed on it even if it lasted minutes), LF = average load / peak load (a low LF means a spiky, expensive profile), and TOD charges more at peak hours to push load off-peak. Memory hook: two about the fuel, three about the triangle, three about the bill.
Source: Guidebook
📖 §3.3 Load factor
75. Define load factor and calculate it for a facility that consumed 900,000 kWh over a 30-day billing period with a peak demand of 2000 kW.
Model answer: Load factor is the ratio of the average load (actual energy consumed) to the peak demand over a period, i.e. actual energy / (peak demand x hours). Maximum possible energy = 2000 kW x (30 x 24) h = 2000 x 720 = 1,440,000 kWh. Load factor = 900,000/1,440,000 = 0.625 = 62.5%.
Load factor = actual energy consumed ÷ (peak demand × hours in the period). For 30 days, hours = 30 × 24 = 720. Common mistake: forgetting to convert days to hours, or using the billed kVA instead of the peak kW. Load factor can never exceed 1 (100%); a high load factor means steady, well-spread usage.
1. Define and explain the following, giving the governing relation and its significance in energy management: (a) Specific heat; (b) Power factor. State the specific heat of water and the effect of a low power factor in an industry.
Model answer: (a) SPECIFIC HEAT — the quantity of heat required to raise the temperature of 1 kg of a substance through 1 degC (or 1 K). Governing relation (sensible heat): Q = m x Cp x deltaT. Units: kcal/kg degC or J/kg.K. It measures a material's thermal 'inertia' - for the same heat input a high-specific-heat body heats up less. Water has an exceptionally high specific heat, 4200 J/kg degC (approx 1 kcal/kg degC), the highest among common substances, so it absorbs and gives out large quantities of heat with only a small temperature change - which is why it is the standard heat-transfer and cooling medium. For gases two values are defined: Cp (constant pressure) and Cv (constant volume).
(b) POWER FACTOR — in an AC circuit the total (apparent) power kVA has an active/resistive component kW (does useful work) and a reactive component kVAR (builds the magnetic flux of inductive equipment, otherwise non-usable). From the power triangle: kW = kVA cos(theta); kVAR = kVA sin(theta); kVA^2 = kW^2 + kVAR^2. Power factor PF = cos(theta) = kW/kVA, i.e. the ratio of active power to apparent power. PF = 1 means all power is useful; a low PF (e.g. 0.7 lagging, caused by under-loaded motors and inductive loads) means the utility must supply extra current/kVA for the same kW, causing higher I^2R losses, larger cable and transformer sizing, voltage drop and utility penalties. Industries therefore improve PF, typically with shunt capacitors.
Directly the Ch-3 long question L-1. The two most-tested facts: specific heat of water = 4200 J/kg degC (highest among common substances) and PF = kW/kVA = cos(theta).
2. Define the following psychrometric terms and state the relationship between them: (a) Relative humidity; (b) Wet-bulb temperature; (c) Dew point. Also note the relation to dry-bulb temperature at 100% RH.
Model answer: (a) RELATIVE HUMIDITY (RH) — the ratio of the mass of water vapour actually held by a given volume of air to the maximum mass it could hold at the same temperature if saturated, expressed as a percentage. Warmer air holds more vapour; saturated air (100% RH) can hold no more. RH governs comfort and evaporation; 50% RH means the air holds half the moisture possible at that temperature.
(b) WET-BULB TEMPERATURE (WBT) — the temperature recorded by a thermometer whose bulb is wrapped in a wick saturated with distilled water and exposed to the air stream. Evaporation of water from the wick draws latent heat and lowers the reading, so WBT accounts for RH and is always less than or equal to the dry-bulb temperature; the drier the air, the greater the wet-bulb depression.
(c) DEW POINT — the temperature to which air must be cooled (at constant pressure and moisture content) for the water vapour to become saturated and begin to condense into droplets. It equals the saturation temperature corresponding to the partial pressure of the water vapour in the mixture.
RELATIONSHIP: Dry-bulb temperature (DBT) measures sensible heat and is not influenced by RH. When RH = 100%, the dry-bulb, wet-bulb and dew-point temperatures are all equal.
Ch-3 long question L-2. Key exam fact: at 100% RH, DBT = WBT = dew point; WBT accounts for RH, DBT does not.
Source: unknown
📖 BEE Guidebook-1, Ch.3 §3.3 Motor loading calculation, Example 3.9 (p.68)
3. A 3-phase 10 kW motor has nameplate details 415 V, 18.2 A, 0.9 PF. During running, a power analyser measures 415 V, 12 A and 0.7 PF. Find the rated input power, the rated efficiency, the actual (measured) input power and the percentage motor loading.
Model answer: Given: 3-phase, nameplate OUTPUT = 10 kW; nameplate input parameters 415 V, 18.2 A, 0.9 PF; measured running input 415 V, 12 A, 0.7 PF.
Step 1 — Rated INPUT power at full load = sqrt3 x Vl x Il x PF = 1.732 x 0.415 kV x 18.2 x 0.9 = 11.8 kW.
Step 2 — Rated efficiency = output / input = 10 / 11.8 = 0.85 = 85%.
Step 3 — Measured (actual) INPUT power = sqrt3 x 0.415 x 12 x 0.7 = 6.0 kW.
Step 4 — Motor loading (%) = (Measured input kW / Rated input kW) x 100 = (6.0 / 11.8) x 100 = 51.2%.
Answer: actual input power is about 6.0 kW and the motor is running at about 51% load. Note that the nameplate kW is the OUTPUT; the input at full load is sqrt3 x V x I x PF, and a part-loaded motor shows a lower PF (0.7 vs 0.9), confirming poor loading.
Book worked Example 3.9. Motor loading = measured kW / rated-input kW; rated input = sqrt3 x V x I x PF; nameplate kW = output.
4. State the relationship between power, voltage and current for balanced single-phase and three-phase AC loads. Then solve: (a) a 3-phase AC induction motor draws 440 V, 25 A at 0.90 PF - find the energy consumed in 1 hour; (b) a 400 W mercury-vapour lamp (230 V, 2 A, PF 0.8) runs 10 hours/day - find the energy consumption per day.
Model answer: RELATIONS: For a balanced SINGLE-PHASE load, Power P = Vl x Il x cos(theta). For a balanced THREE-PHASE load, Power P = sqrt3 x Vl x Il x cos(theta), where Vl = line voltage, Il = line current and cos(theta) = power factor. Energy (kWh) = Power (kW) x hours.
(a) 3-phase motor: P = sqrt3 x 0.440 kV x 25 x 0.90 = 1.732 x 0.440 x 25 x 0.90 = 17.15 kW. Energy in 1 hour = 17.15 x 1 = 17.15 kWh (units).
(b) Single-phase lamp: Energy = V x I x cos(theta) x hours = 0.230 x 2 x 0.8 x 10 = 3.68 kWh (approx 3.7 units) per day.
Note: single-phase supply is used mainly for lighting, fractional-HP motors and heaters; the sqrt3 factor is what distinguishes three-phase power from single-phase power.
Book Examples 3.7 & 3.8. Remember the sqrt3 for three-phase power and energy(kWh) = kW x hours; watch the RMS values used for AC.
Source: unknown
📖 BEE Guidebook-1, Ch.3 §3.3, Example 3.6 (p.66)
5. An electric heater rated 230 V, 5 kW is used for hot-water generation in an industry. Find the electricity consumption per hour (a) at the rated voltage and (b) if the supply voltage falls to 200 V. Explain the principle used.
Model answer: PRINCIPLE: For a fixed-resistance heating element the resistance R is constant, so power P = V^2/R varies as the SQUARE of the applied voltage: P2 = (V2/V1)^2 x P1.
(a) At the rated 230 V: consumption = 5 kW x 1 hour = 5 kWh (units).
(b) At 200 V: P2 = (200/230)^2 x 5 = 0.756 x 5 = 3.78 kW, so consumption = 3.78 kWh in one hour.
Thus a roughly 13% drop in voltage (230 to 200 V) cuts the heater output and energy by about 24%, because power falls with the square of the voltage. This square-law behaviour applies to resistive loads such as heaters and incandescent bulbs.
Book Example 3.6. For resistive loads P is proportional to V^2, so P2 = (V2/V1)^2 x P1 - a frequently tested relation.
6. Define latent heat, latent heat of fusion and latent heat of vaporization, and solve: (a) the heat given up by 10 kg of water at 0 degC when it freezes to ice at 0 degC (h_if = 335 kJ/kg); (b) the mass of ice melted when 20 kJ is supplied to ice at 0 degC; (c) the heat required to vaporize 2 m3 of water at 100 degC (h_fg = 2257 kJ/kg).
Model answer: DEFINITIONS: Latent heat is the heat exchanged when a substance changes physical state WITHOUT any change in temperature. Latent heat of fusion (h_if) is the heat needed to convert 1 kg of solid to liquid (or released on freezing) at the melting point - for ice/water = 335 kJ/kg. Latent heat of vaporization (h_fg) is the heat to convert 1 kg of liquid to vapour (or released on condensation) at the boiling point - for water = 2257 kJ/kg (approx 540 kcal/kg) at 100 degC. Governing relation: Q = m x h.
(a) Q = m x h_if = 10 x 335 = 3350 kJ given up on freezing.
(b) m = Q / h_if = 20 / 335 = 0.06 kg of ice melted.
(c) Mass of 2 m3 of water = 2000 kg; Q = m x h_fg = 2000 x 2257 = 4,514,000 kJ (4514 MJ).
Note: the phase change occurs in either direction at the same temperature - freezing releases the same latent heat that melting absorbs, and condensation releases the same as vaporization absorbs.
Book Examples 3.10-3.12. Core numbers: h_if(ice) = 335 kJ/kg, h_fg(water) = 2257 kJ/kg; Q = m x h.
Source: unknown
📖 §3.4 Sensible Heat (Q = m·Cp·ΔT) and Latent Heat of Vaporization (h_fg = 2257 kJ/kg)
7. Determine the total heat required to convert 5 kg of water at 30 degC into dry saturated steam at 100 degC at atmospheric pressure. Take specific heat of water Cp = 4.2 kJ/kg degC and latent heat of vaporization h_fg = 2257 kJ/kg. Express the answer in kJ and in kcal.
Model answer: The process has two stages - a SENSIBLE-heat stage (temperature rise 30 to 100 degC) and a LATENT-heat stage (evaporation at constant 100 degC).
Step 1 — Sensible heat to raise the water to boiling point: Q1 = m x Cp x deltaT = 5 x 4.2 x (100 - 30) = 5 x 4.2 x 70 = 1470 kJ.
Step 2 — Latent heat of vaporization at 100 degC: Q2 = m x h_fg = 5 x 2257 = 11,285 kJ.
Step 3 — Total heat = Q1 + Q2 = 1470 + 11,285 = 12,755 kJ.
Step 4 — Convert to kcal (1 kcal = 4.187 kJ): 12,755 / 4.187 = approx 3046 kcal.
Answer: about 12,755 kJ (approx 3046 kcal). Note that the latent stage needs about 7.7 times the sensible stage; evaporation dominates the heat demand, which is why boiler and steam-system efficiency work focuses on latent heat and on condensate/flash-steam recovery.
Combines Q = m x Cp x deltaT (sensible) with Q = m x h_fg (latent) - the classic two-stage steam-generation numerical. Numbers are composed (not a book worked example) but use only Ch-3 values.
Source: unknown
📖 §3.3 Electricity Basics — Power Factor and the power triangle (kW = kVA·cosΘ, kVAR = kVA·sinΘ)
8. A plant load is 500 kW at 0.75 power factor lagging. It is to be improved to 0.95 lagging by installing shunt capacitors. Calculate (a) the initial apparent power (kVA) and reactive power (kVAR); (b) the reactive power after correction; (c) the capacitor rating (kVAR) required; (d) the new kVA; and (e) the percentage reduction in apparent power (and hence line current). State the benefits of the improvement.
Model answer: Active power kW = 500 (unchanged by PF correction). Use kVAR = kW x tan(theta), where theta = cos-inverse(PF).
(a) Initial: cos(theta1) = 0.75 -> theta1 = 41.41 deg, tan(theta1) = 0.882. kVA1 = kW/cos(theta1) = 500/0.75 = 666.7 kVA; kVAR1 = 500 x 0.882 = 440.9 kVAR.
(b) Target: cos(theta2) = 0.95 -> theta2 = 18.19 deg, tan(theta2) = 0.329. kVAR2 = 500 x 0.329 = 164.3 kVAR.
(c) Capacitor rating = kVAR1 - kVAR2 = 440.9 - 164.3 = 276.6 kVAR (about 277 kVAR of capacitors to install).
(d) New apparent power kVA2 = kW/cos(theta2) = 500/0.95 = 526.3 kVA.
(e) Reduction in kVA = (666.7 - 526.3)/666.7 x 100 = 21.1%. Since line current I is proportional to kVA at fixed voltage, the current also falls by about 21%.
BENEFITS: lower current reduces I^2R distribution and transformer losses, releases system capacity (kVA), improves voltage regulation and avoids the utility low-PF penalty. General rule: required capacitor kVAR = kW x (tan(theta1) - tan(theta2)).
Standard PF-improvement method: capacitor kVAR = kW(tan(theta1) - tan(theta2)). Built on the Ch-3 power triangle; the specific numbers are invented (not a book worked example).
Source: unknown
📖 §3.3 Electricity Basics — chapter short question S-4 (Load Factor); §1.13 for Availability Based Tariff (ABT)
9. Explain the following electricity-tariff terms used in energy-cost management: (a) Maximum (contract) demand; (b) Load factor; (c) Time-of-Day (TOD) tariff; (d) Availability Based Tariff (ABT); (e) power-factor penalty/incentive. Then calculate the load factor of a continuously operating facility that consumed 900,000 kWh during a 30-day billing period with an established peak demand of 2000 kW.
Model answer: (a) MAXIMUM DEMAND — the highest average kVA/kW drawn over a defined interval (usually 30 min) in the billing period; the utility levies a demand charge on it, so controlling the peak reduces the fixed part of the bill.
(b) LOAD FACTOR — the ratio of the average load to the peak (maximum) demand over a period = energy consumed / (maximum demand x hours). A high load factor means capacity is used steadily and the unit cost of energy is lower.
(c) TIME-OF-DAY (TOD) TARIFF — energy is charged at different rates in different time slots (higher in peak hours, lower off-peak), encouraging consumers to shift load to off-peak and flatten the demand curve.
(d) AVAILABILITY BASED TARIFF (ABT) — a frequency-linked tariff for bulk/grid power with three components: a fixed (capacity) charge, an energy charge, and an Unscheduled Interchange (UI) charge priced by grid frequency, rewarding grid discipline.
(e) POWER-FACTOR PENALTY/INCENTIVE — a surcharge when PF falls below a set threshold and a rebate for high PF, because low-PF loads burden the network with reactive current.
CALCULATION — Load factor = energy consumed / (maximum demand x hours). Hours in 30 days = 30 x 24 = 720 h. LF = 900,000 / (2000 x 720) = 900,000 / 1,440,000 = 0.625 = 62.5%.
The load-factor numerical is Ch-3 short question S-4 (answer = 62.5%). The TOD/ABT/maximum-demand descriptions are general BEE tariff knowledge that goes beyond the Ch-3 text, so flagged ai.
Source: unknown
📖 BEE Guidebook-1, Ch.3 §3.4 Energy Content in Fuel (p.75) + Objective Q10 (p.83)
10. Explain how the energy content (calorific value) of a fuel is determined using a bomb calorimeter. Distinguish between Gross Calorific Value (GCV) and Net Calorific Value (NCV), give typical values for heavy fuel oil, and state for which type of fuel the GCV - NCV difference is greatest and why.
Model answer: MEASUREMENT (bomb calorimeter): A sample of KNOWN mass is placed in a bomb calorimeter - a completely sealed, insulated vessel that prevents heat loss. The sample is burned completely and the resulting rise in temperature (read on a thermometer inside, viewed from outside) is measured. From the mass and the temperature rise the heat released per unit weight - the calorific value - is calculated. Calorific value is the heat released during the complete combustion of unit weight of fuel.
GCV vs NCV: The GROSS (higher) calorific value assumes ALL the water vapour formed during combustion (from the fuel's moisture and from the hydrogen in the fuel) is fully condensed, so its latent heat is recovered. The NET (lower) calorific value assumes this water leaves with the flue gases as vapour, so its latent heat is NOT recovered. Hence GCV - NCV = the latent heat of vaporization of the moisture plus the water formed from the atomic hydrogen in the fuel.
TYPICAL VALUES (heavy fuel oil): GCV is about 44,100 kJ/kg (10,500 kcal/kg); NCV is about 41,160 kJ/kg (9,800 kcal/kg).
MAXIMUM DIFFERENCE: The GCV - NCV gap is greatest for the fuel that forms the most water on combustion, i.e. the fuel richest in hydrogen - natural gas - because more hydrogen produces more water vapour whose latent heat separates GCV from NCV.
OCR section 'Energy Content in Fuel'. HFO GCV 10,500 / NCV 9,800 kcal/kg; GCV-NCV is maximum for a high-hydrogen fuel (natural gas) - the answer to Objective Q10.
Source: unknown
📖 BEE Guidebook-1, Ch.3 §3.4 Heat transfer (p.75)
11. Describe the three primary modes of heat transfer, explaining the mechanism and giving an example of each, and distinguish between forced and natural convection.
Model answer: Heat always flows from a hotter body to a colder one, independent of the mode; the rate of heat transfer is measured in watts (J/s). There are three primary modes:
1. CONDUCTION — the primary mode in SOLIDS. Energy passes by (i) molecular motion, in which higher-energy (more vigorously vibrating) molecules impart energy to adjacent lower-energy molecules, and (ii) migration of free electrons, which is dominant in pure metals (so metals are good conductors). Example: heat travelling along a metal rod or through a furnace wall.
2. CONVECTION — occurs when a moving FLUID exchanges energy with an adjacent solid surface; the fluid motion carries the heat away. Two types: (a) FORCED convection - fluid motion is produced by an external device such as a fan or pump; (b) NATURAL (free) convection - motion arises on its own from density differences: heated fluid becomes lighter and rises while the colder, denser fluid sinks, setting up circulation. Example: air heated by a radiator, or a cooling tower.
3. RADIATION — needs NO medium; energy travels as electromagnetic waves and can pass through a vacuum. Thermal radiation spans infrared to ultraviolet, and radiant energy striking a surface can be reflected, absorbed or transmitted. Example: heat from the sun or from a furnace flame. Radiation becomes dominant at high temperatures.
📖 BEE Guidebook-1, Ch.3 §3.4 The laws of thermodynamics (p.79)
12. State and explain the three laws of thermodynamics. Relate the second law to entropy and explain why a heat engine can never be 100% efficient.
Model answer: Thermodynamics is the study of heat, work and the conversion of energy from one form to another; three laws govern it (most of the subject rests on the first two).
FIRST LAW (Law of Conservation of Energy): energy in a system can neither be created nor destroyed - it is only converted from one form to another or transferred from one system to another. Applied to a heat engine (e.g. a gas turbine) converting heat into mechanical energy, the total energy in the system stays constant whatever the intermediate stages. The first law deals with the QUANTITY of energy.
SECOND LAW: while the first law fixes the quantity, it says nothing about DIRECTION. The second law governs the natural direction of energy flow: heat flows on its own only from a hotter body to a colder body, never the reverse. It introduces ENTROPY, a measure of disorder; the more disordered a system, the less useful work can be extracted from it. Crucially, the second law explains why a heat engine can NEVER be 100% efficient - some heat from the fuel must always be rejected to the surroundings (the cold sink) and cannot be turned into mechanical work.
THIRD LAW: concerns absolute zero (-273 degC, i.e. 0 K). It states that it is impossible to reduce the temperature of any system to absolute zero.
Thermal efficiency follows: eta = useful output / input; because of the second law, eta is always less than 100% for any real heat engine.
OCR 'The laws of thermodynamics'. 1st = conservation of energy, 2nd = direction/entropy and no engine is 100% efficient, 3rd = cannot reach absolute zero.
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📖 §3.5 Energy Units and Conversions / MTOE conversions (chapter objective Q6)
13. Working from the standard BEE energy-conversion values, solve: (a) convert 10 kWh (ten units) into kcal and into MJ; (b) convert a 50 HP motor rating into kW; (c) express 2500 kg of coal of GCV 4000 kcal/kg in tonnes of oil equivalent (toe); (d) express 125 tonnes of the same coal in toe.
Model answer: KEY VALUES (BEE / Gazette of India): 1 kWh = 860 kcal = 3.6 MJ; 1 HP = 745.7 W = 0.746 kW; 1 kg of oil equivalent = 10,000 kcal, so 1 toe (tonne of oil equivalent) = 1 x 10^7 kcal; toe = (mass in kg x GCV in kcal/kg) / 10^7.
(a) 10 kWh x 860 = 8600 kcal; and 10 x 3.6 = 36 MJ.
(b) 50 HP x 0.746 = 37.3 kW (i.e. 50 x 745.7 = 37,285 W).
(c) toe = 2500 x 4000 / 10^7 = 10,000,000 / 10^7 = 1.0 toe.
(d) 125 t = 125,000 kg; toe = 125,000 x 4000 / 10^7 = 500,000,000 / 10^7 = 50 toe.
These conversions (1 kWh = 860 kcal, 1 toe = 10^7 kcal, toe = mass x GCV / 10^7) are the backbone of energy-balance and MTOE calculations.
Uses the OCR conversion table plus the Gazette MTOE values. 125 t coal @4000 kcal/kg = 50 toe is the book's own Objective-Q6 figure. Numbers assembled into a fresh multi-part problem, so flagged ai.
14. Explain the enthalpy of steam and the terms specific enthalpy, dryness fraction and superheat. Give the specific enthalpy of saturated water, saturated steam and evaporation at standard atmospheric pressure, and distinguish the sensible and latent heat involved in generating steam.
Model answer: ENTHALPY OF STEAM (H) is the total heat content of the steam = mass x specific enthalpy: H = m x h, where H is in kJ, m in kg and h in kJ/kg. SPECIFIC ENTHALPY h = u + p x v, where u = internal energy (kJ/kg), p = absolute pressure and v = specific volume.
AT STANDARD ATMOSPHERE (1.01325 bar, water boiling at 100 degC):
- specific enthalpy of saturated water, hf = 419 kJ/kg (approx Cw x tf = 4.19 x 100);
- specific enthalpy of saturated steam, hg = 2676 kJ/kg;
- specific enthalpy of evaporation, h_evap = hg - hf = 2676 - 419 = 2257 kJ/kg.
SENSIBLE vs LATENT heat in steam generation: heating the water up to its boiling point adds SENSIBLE heat (hf); evaporating it at constant 100 degC adds the LATENT heat of evaporation (2257 kJ/kg) with no change in temperature.
DRYNESS FRACTION (x): in 1 kg of a water-steam mixture, x kg is steam and (1 - x) kg is water. x = 1 means dry saturated steam; x = 0 means saturated water; the region to the right of the x = 1 line is SUPERHEATED steam.
SUPERHEAT: heating saturated steam above its saturation temperature at the existing pressure; done in power plants to raise efficiency and avoid condensation in the turbine. The higher the pressure of water, the higher the saturation temperature.
OCR 'Steam Properties / Enthalpy of steam'. Remember hf = 419, hg = 2676, h_evap = 2257 kJ/kg at 1 atm; dryness fraction x = 1 is dry saturated steam.
Source: unknown
📖 BEE National Certification Exam, Paper-1 (Nov 2013); concepts per Book-1 Ch.3 §3.4
15. (a) Explain the difference between GCV and NCV of a fuel. (b) A gas-fired water heater heats water flowing at 1.2 m3/hour from 20 degC to 65 degC. If the GCV of the gas is 4 x 10^7 J/kg and the efficiency of the water heater is 80%, find the rate of gas combustion in kg/hr. Take Cp of water = 4.187 kJ/kg degC and density of water = 1000 kg/m3.
Model answer: (a) GCV vs NCV: calorific value is the heat released on complete combustion of unit weight of fuel. The GROSS calorific value (GCV) assumes ALL the water vapour produced during combustion is fully condensed, so the latent heat of that vapour is recovered. The NET calorific value (NCV) assumes the water leaves with the combustion products as vapour without being condensed, so its latent heat is not recovered. The difference between GCV and NCV is therefore the latent heat of condensation of the water vapour (from the fuel moisture and from hydrogen burning to water).
(b) Step 1 — Mass flow of water = 1.2 m3/hr x 1000 = 1200 kg/hr (= 20 kg/min).
Step 2 — Heat gained by water Q = m x Cp x deltaT = 1200 x 4.187 x (65 - 20) = 1200 x 4.187 x 45 = 226,098 kJ/hr.
Step 3 — Heat to be supplied by the gas = Q / efficiency = 226,098 / 0.80 = 282,623 kJ/hr.
Step 4 — GCV = 4 x 10^7 J/kg = 40,000 kJ/kg. Gas rate = 282,623 / 40,000 = 7.07 kg/hr.
Answer: the gas combustion rate is about 7.07 kg/hr.
Past-paper (Nov 2013) combining the GCV/NCV definition with a fuel-firing numerical; fully consistent with Ch-3 method: fuel rate = m x Cp x deltaT / (eta x GCV). Not in the book OCR, so verified=false.
Source: unknown
📖 BEE National Certification Exam Paper-1 Set A (10th NCE, Jul 2010); method per Book-1 Ch.3 Example 3.9
16. A 15 kW, 415 V, 27 A, 4-pole, 50 Hz, 3-phase squirrel-cage induction motor has a full-load efficiency of 90% and PF 0.86. During operation a power analyser reads 406 V, 22 A, PF 0.82. Find (a) the input power in kW and (b) the percentage motor loading.
Model answer: (a) Measured INPUT power = sqrt3 x Vl x Il x PF = 1.732 x 0.406 kV x 22 x 0.82 = 12.68 kW.
(b) Rated INPUT power at full load = rated output / efficiency = 15 / 0.90 = 16.67 kW.
Motor loading (%) = (measured input kW / rated input kW) x 100 = (12.68 / 16.67) x 100 = 76.1% (about 76%).
Note: the nameplate 15 kW is the OUTPUT; the rated input = output / eta (here we use the given efficiency rather than sqrt3 x V x I of the nameplate). Loading compares the actual input power with the full-load input power.
Past-paper motor-loading numerical (10th NCE, Jul 2010). Loading = measured input / rated input; rated input = output / eta. Genuine exam question, not in book OCR, so verified=false.