BEE Paper-1 — Chapter 5: Material & Energy Balance
113 questions — 48 objective (1 mark), 46 short (5 marks), 19 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) — 48
📖 §5.5 Example 5.5 — concentrations (mole fraction)
1. 54 kg of water is mixed with 0.34 moles of salt to make a solution. The mole fraction of the solution is ____.
0.1
18.36
158.8
none of the above
Answer: A) 0.1
Confirmed vs Book-1 §5.5 Ex.5.5: Book computes moles as (mass)/(mol. wt), mol. wt of water = 18. Moles of water = 54/18 = 3. Mole fraction of salt = moles salt/(total moles) = 0.34/(3 + 0.34) = 0.34/3.34 = 0.102 ≈ 0.1. Hence option (a).
Source: Sep 2021
📖 §5.3 Basic principles — element (stoichiometric) balance
2. C2H4 + xO2 ----> 2CO2 + yH2O, what is the value of x + y?
2
3
5
8
Answer: C) 5
Confirmed vs Book-1 §5.3 (mass of each element is conserved): C2H4 + xO2 → 2CO2 + yH2O. Hydrogen: 4 = 2y → y = 2. Oxygen: 2x = (2×2) + 2 = 6 → x = 3. Therefore x + y = 3 + 2 = 5, option (c).
Source: Sep 2021
📖 §5.5 Example 5.5 — weight/weight concentration
3. A solution of common salt is prepared by adding 25 kg of salt to 100 kg of water. The weight fraction of solution is ____.
20%
25%
4%
none of the above
Answer: A) 20%
Confirmed vs Book-1 §5.5 Ex.5.5: weight fraction = weight of solute / total weight of solution = 25/(25 + 100) = 25/125 = 0.20, i.e. % w/w = 20%. (Book's own case: 20/(100+20) = 16.7%.) Option (a).
4. A system uses 100 kg of raw material A, 200 kg of B and 220 kg of C. The mix is heated to 220 deg C. Air carries away on average 60% of A and 30% of B through the chimney. The output product would be
520 kg
400 kg
312 kg
208 kg
Answer: B) 400 kg
Confirmed vs Book-1 §5.3: Total raw material in = 100 + 200 + 220 = 520 kg. Waste carried away by air = 60% of A + 30% of B = (0.60×100) + (0.30×200) = 60 + 60 = 120 kg. With no storage, Products = 520 − 120 = 400 kg. Option (b).
Source: Nov 2009
📖 §5.5 Example 5.6 — evaporator solids (tie-component) balance
5. If feed of 15 tonnes per hour at 6% concentration is fed to an evaporator, the product obtained at 30% concentration is equal to ____ tonnes per hour.
3
9
0.9
4.5
Answer: A) 3
Confirmed vs Book-1 §5.5 Ex.5.6: Solids are conserved. Solids in feed = 15 × 0.06 = 0.9 t/h. Product at 30% solids = 0.9/0.30 = 3 t/h. (Water evaporated = 15 − 3 = 12 t/h.) Option (a).
Source: 2019
📖 §5.5 Material balance — moisture + water formed from hydrogen
6. 1 kg of wood contains 15% moisture and 5% hydrogen by weight. How much water is evaporated during complete combustion of 1kg of wood?
0.6 kg
200 g
0.15 kg
none of the above
Answer: A) 0.6 kg
Confirmed vs Book-1 §5.5 (component mass balance): Free moisture = 1 × 0.15 = 0.15 kg. Hydrogen burns as H2 + ½O2 → H2O, so 2 kg H gives 18 kg water, i.e. 9 kg water per kg of hydrogen: 9 × 0.05 = 0.45 kg. Total water evaporated = 0.15 + 0.45 = 0.60 kg. Option (a).
Source: 2019
📖 §5.5 Material balance procedure — bone-dry solids balance
7. In a drying process product moisture is reduced from 60% to 30%. Inlet weight of the material is 200 kg. Calculate the weight of the outlet product.
80
120.5
114.3
none of the above
Answer: C) 114.3
Confirmed vs Book-1 §5.5 (dry-solids balance, as in Ex.5.11): Bone-dry solids = 200 × (1 − 0.60) = 80 kg and are unchanged. Outlet product at 30% moisture is 70% solids, so outlet = 80/0.70 = 114.3 kg. Option (c).
Source: 2016
📖 §5.2 Components of material and energy balance (Fig 5.1)
8. A mass balance for energy conservation does not consider which of the following
Steam
water
Lubricating oil
Raw material
Answer: C) Lubricating oil
Confirmed vs Book-1 §5.2 Fig 5.1: the streams counted are raw materials, chemicals, water/air, energy/power (inputs) and products, by-products, emissions, wastewater and wastes (outputs). Steam, water and raw material are all such process streams; lubricating oil is a maintenance consumable and is not taken in the mass balance. Option (c).
Source: 2016
📖 §5.8 Energy analysis and the Sankey diagram (Fig 5.9)
9. Diagrammatic representation of input and output energy streams of an equipment or system is known as
mollier diagram
sankey diagram
psychrometric chart
balance diagram
Answer: B) sankey diagram
Confirmed vs Book-1 §5.8: 'The Sankey diagram is a very useful tool to represent an entire input and output energy flow in any energy equipment or system', the width of each arrow being proportional to the flow. Hence it is the diagrammatic representation of input/output energy streams. Option (b).
Source: 2016
📖 §5.1 Purpose of Material and Energy Balance (box)
10. Which of the following tool is made use of to assess the input, conversion efficiency, output, losses, quantification of all material, energy and waste streams in a process or system?
material balance
energy balance
material and energy balance
Sankey diagram
Answer: C) material and energy balance
Confirmed vs Book-1 §5.1 purpose box: material AND energy balance is used 'to assess the input, conversion efficiency, output and losses' and 'to quantify all material, energy and waste streams in a process or a system'. Only the combined material and energy balance covers all of these. Option (c).
Source: 2013
📖 §5.5 Example 5.6 — evaporator solids balance
11. If feed of 100 tonnes per hour at 10% concentration is fed to an evaporator, the product obtained at 25% concentration is equal to ____ tonnes per hour.
25
40
50
62.5
Answer: B) 40
Confirmed vs Book-1 §5.5 Ex.5.6: Solids in feed = 100 × 0.10 = 10 t/h and are conserved. Product at 25% concentration = 10/0.25 = 40 t/h. (Water evaporated = 60 t/h.) Option (b).
Source: 2013
📖 §5.5 Material balance — moisture + water formed from hydrogen
12. 1 kg of wood contains 15% moisture and 7% hydrogen by weight. How much water is evaporated from wood during complete combustion of 1 kg of wood ?
0.78 kg
0.22 kg
0.15 kg
0.63 kg
Answer: A) 0.78 kg
Confirmed vs Book-1 §5.5 (component mass balance): Free moisture = 0.15 kg. Water from hydrogen = 9 × 0.07 = 0.63 kg (9 kg water per kg H, from H2 + ½O2 → H2O). Total water evaporated = 0.15 + 0.63 = 0.78 kg. Option (a).
Source: 2012
📖 §5.1 Purpose of Material and Energy Balance (box)
13. Material and energy balance is used to quantify
material and energy losses
profit
cost of production
all of the above
Answer: A) material and energy losses
Confirmed vs Book-1 §5.1 purpose box: the M&E balance quantifies all material, energy and waste streams and assesses input, conversion efficiency, output and losses — i.e. it quantifies material and energy losses. Profit and cost of production are financial, not balance, outputs. Option (a).
14. The objective of material and energy balance is to assess the
input-output
conversion efficiency
losses
all the above
Answer: D) all the above
Confirmed vs Book-1 §5.1 purpose box: the objective is to assess the input, conversion efficiency, output AND losses; hence all three listed items apply. Option (d) all the above.
15. In the material balance of a process or unit operation, which component will not be considered on the input side?
chemicals
water/air
recycle
by product
Answer: D) by product
Confirmed vs Book-1 §5.2 Fig 5.1: inputs are raw materials, chemicals, water/air, energy/power and the RECYCLE stream ('recycle stream is shown along with input side'). By-products appear on the output side with products, emissions and wastes. Option (d).
16. In material balance of a process, recycle product is always considered as
input to process
output to process
both (a) and (b)
none of them
Answer: A) input to process
Confirmed vs Book-1 §5.2 Fig 5.1 note: 'It may be noted that recycle stream is shown along with input side.' The recycle is fed back into the process, so in the material balance it is always taken as an input. Option (a).
Source: Guidebook
📖 §5.3 Conservation of mass (EOC Objective Q4)
17. In a chemical process two reactants A (200 kg) and B (200 kg) are used as reactants. If conversion is 50% and A and B react in equal proportion, calculate the weight of the product formed.
150 kg
200 kg
250 kg
400 kg
Answer: B) 200 kg
Confirmed vs Book-1 §5.3 (mass is conserved): total reactants charged = 200 + 200 = 400 kg; A and B react in equal proportion so they are consumed together. At 50% conversion the reacted mass = 0.50 × 400 = 200 kg, and by conservation of mass this appears as 200 kg of product. Option (b).
18. In a heat treatment furnace the material is heated up to 800 degC from ambient temperature of 30 degC. Considering the specific heat of material as 0.13 kcal/kg degC, what is the energy content in one kg of material after heating?
50 kcal
250 kcal
350 kcal
100 kcal
Answer: D) 100 kcal
Confirmed vs Book-1 §5.6 heat balance (Q = m·Cp·ΔT, as in Ex.5.10): Q = 1 kg × 0.13 kcal/kg°C × (800 − 30)°C = 0.13 × 770 = 100.1 ≈ 100 kcal per kg. Option (d) 100 kcal.
19. In a utility steam boiler, heat loss due to radiation normally is in the range of
10%
14%
1%
8%
Answer: C) 1%
Confirmed vs Book-1 §5.6: sinks are 'depositories of leakage or rejected energy... usually low-grade heat, as in radiation losses from boilers'. For a large utility steam boiler this radiation and convection loss is small — of the order of 1% of input (it is significant only in small boilers). Option (c).
Source: Guidebook
📖 §5.6 Energy balance — electrical-to-heat conversion (EOC Objective Q7)
20. Energy supplied by electricity, Q in kcal is equal to
kWh x 8.6
kWh x 86
kWh x 860
none
Answer: C) kWh x 860
Confirmed vs Book-1 §5.6: to compare energy streams they are converted to equivalent heat energy. 1 kWh = 860 kcal, so Q (kcal) = kWh × 860. Option (c).
Source: Guidebook
📖 §5.8 Sankey diagram (EOC Objective Q8)
21. Sankey diagram is a useful tool to represent
financial strength of the company
management philosophy
input and output energy flow
human resource strength of the company
Answer: C) input and output energy flow
Confirmed vs Book-1 §5.8: the Sankey diagram 'is very useful tool to represent an entire input and output energy flow in any energy equipment or system', arrow width proportional to the flow. Option (c).
22. Which of the following formula is useful to determine the heat duty in conducting heat balance?
Q = M Cp deltaT
Q = AV
PV = nRT
none of the above
Answer: A) Q = M Cp deltaT
Confirmed vs Book-1 §5.6/Ex.5.10: heat duty (sensible heat) = mass × specific heat × temperature change, Q = M·Cp·ΔT. Q = AV is a flow (continuity) relation and PV = nRT is the gas law used for gas concentrations (Ex.5.7/5.8), not for heat duty. Option (a).
Source: Guidebook
📖 §5.6 Energy balance — electrical energy (EOC Objective Q10)
23. A 230V, 100 W rated incandescent bulb is operated at a constant voltage of 250V. The approximate power consumption of bulb is
100 W
118 W
85 W
none of the above
Answer: B) 118 W
Confirmed vs Book-1 §5.6: for a fixed-resistance (incandescent) load R = V²/P = 230²/100 = 529 Ω, so P ∝ V². P = 250²/529 = 62500/529 = 118.1 W, i.e. P = 100 × (250/230)² ≈ 118 W. Option (b).
Source: Guidebook
📖 §5.5 Example 5.7 — mole fraction of a gas mixture
24. A gaseous mixture contains 7.50 gms of H2 and 3.25 gms of O2 and 5.55gms of N2. The Mole fraction of N2 is ____.
📖 §5.5 Material balance procedure — bone-dry solids balance
25. A dry feed contains 7% moisture was feed to a water spray chamber to increase the moisture content to 35% in the dry feed. The output feed quantity coming from the spray chamber is ____.
0.5 kg/kg of input feed
1.43 kg/kg of input feed
1.48 kg/kg of input feed
2.66 kg/kg of input feed
Answer: B) 1.43 kg/kg of input feed
Confirmed vs Book-1 §5.5 (dry solids are unchanged): per 1 kg of input feed, bone-dry solids = 1 × (1 − 0.07) = 0.93 kg. In the output the moisture is 35%, so solids are 65%: output = 0.93/0.65 = 1.43 kg per kg of input feed. Option (b).
Source: Mar 2023
📖 §5.3 Basic principles — master material balance equation
26. Law of conservation of mass can be expressed by the following equation:
Confirmed vs Book-1 §5.3: 'Raw Materials = Products + Waste Products + Stored Products + Losses', where Losses are the unidentified materials. Input equals the sum of all outputs plus what is stored plus unaccounted losses. Option (d).
Source: Mar 2023
📖 §5.7 Example 5.11 / §5.5 — evaporation of moisture into air
27. When the evaporation of water from a wet substance is zero, the relative humidity of air is likely to be ____.
0%
10%
50%
100%
Answer: D) 100%
Confirmed vs Book-1 §5.5–§5.7 (drying material balance): moisture leaves the wet substance only while the surrounding air can still take up water vapour. When the air is saturated — relative humidity 100% — its moisture-carrying capacity is exhausted and the evaporation rate falls to zero. Option (d).
Source: Mar 2023
📖 §5.3 Basic principles — element (stoichiometric) balance
28. 1 mole of sulphur react with X moles of H2SO4 to form Y moles of H2O and Z moles of SO2 then (X/Y) + Z is ____.
2
4
1
3
Answer: B) 4
Confirmed vs Book-1 §5.3 (element balance): S + 2H2SO4 → 3SO2 + 2H2O. So 1 mole of sulphur reacts with X = 2 moles of H2SO4 giving Y = 2 moles of H2O and Z = 3 moles of SO2. Therefore (X/Y) + Z = (2/2) + 3 = 1 + 3 = 4. Option (b).
Source: Mar 2023
📖 §5.5 Material balance procedure — bone-dry solids balance
29. In a drying process, moisture is reduced from 50% to 30%. Initial weight of the material is 100 kg. Calculate the weight of the final product in kg.
80
86
71.4
74.3
Answer: C) 71.4
Confirmed vs Book-1 §5.5 (dry solids unchanged, as in Ex.5.11): Bone-dry solids = 100 × (1 − 0.50) = 50 kg. Final product at 30% moisture is 70% solids, so final weight = 50/0.70 = 71.4 kg. Option (c).
Source: Jul 2022
📖 §5.5 Example 5.6 — solids balance (crystallizer/evaporator)
30. If feed of 100 tons per hour at 5% concentration is fed to a crystallizer, the rate in tons per hour of the product obtained at 25% concentration is equal to:
40
20
25
100
Answer: B) 20
Confirmed vs Book-1 §5.5 Ex.5.6 method: Solids in feed = 100 × 0.05 = 5 t/h and are conserved. Product at 25% concentration = 5/0.25 = 20 t/h. Option (b).
31. A process receives 1000 kg/hr of raw material. The hourly outputs are 700 kg of product, 200 kg of waste, and 50 kg stored. What is the unaccounted loss?
32. Calculate the quantity of water evaporated when 100 kg of feed containing 6% solids is concentrated to 30% solids.
600 kg
180 kg
80 kg
800 kg
Answer: C) 80 kg
Confirmed vs Book-1 §5.5 Ex.5.6 (book's own numbers): Solids in feed = 100 × 0.06 = 6 kg and are conserved. Output = 6/0.30 = 20 kg. Water evaporated = 100 − 20 = 80 kg. Option (c).
Source: Sep 2025
📖 §5.5 Example 5.5 — moles = mass/molecular weight
33. Moles of water in 54 grams:
3
4
5
6
Answer: A) 3
Confirmed vs Book-1 §5.5 Ex.5.5 (mol. wt of water = 18): moles = 54/18 = 3 moles. Option (a).
Source: Sep 2025
📖 §5.5 Example 5.7(a) — mean molecular weight of air
34. Mean molecular weight of air (77% N2, 23% O2 by weight) is ___________ grams.
26.8
27.8
28.8
29.8
Answer: C) 28.8
Confirmed vs Book-1 §5.5 Ex.5.7(a): basis 100 kg air contains 77/28 = 2.75 moles N2 and 23/32 = 0.72 moles O2; total = 3.47 moles. Mean molecular weight = 100/3.47 = 28.8. Option (c).
Source: Sep 2025
📖 §5.5 Example 5.5 — moles = mass/molecular weight
35. The number of moles of water contained in 72 grams of water is
2
3
4
5
Answer: C) 4
Confirmed vs Book-1 §5.5 Ex.5.5 (mol. wt of water = 18): moles = 72/18 = 4 moles. Option (c).
Source: Sep 2024
📖 §5.5 Example 5.5 — weight/weight concentration
36. A solution of common salt in water is prepared by adding 25 kg of salt to 100 kg of water. The concentration of salt in this solution as a weight fraction shall be
Answer: A) Raw Materials = Products + Waste Products + Stored Products + Losses
Confirmed vs Book-1 §5.3 — the balance box gives Mass In = Mass Out + Mass Stored, i.e. Raw Materials = Products + Wastes + Stored Materials. When part of the input cannot be accounted for, the book writes the full form: Raw Materials = Products + Waste Products + Stored Products + Losses, where the losses are the unidentified materials. Rearranging this equation is what lets you solve for any unknown stream.
Source: AI practice
📖 Book-1 §5.2 — Components of Material and Energy Balance; §5.4 (Figure 5.3)
38. In a process flow diagram, which of the following is NOT on the INPUT side of the process box?
Raw material
Recycle stream
By-product
Combustion air
Answer: C) By-product
Confirmed vs Book-1 §5.2 — the book states explicitly that the RECYCLE stream is shown along with the input side; inputs also include raw materials, water/air, chemicals and energy. Figure 5.3 in §5.4 shows desired products, by-products, wastes and energy coming OUT of the box, so a by-product is an output. This is exactly the chapter's own objective Q2 (which component is not considered on the input side — by-product).
Source: AI practice
📖 Book-1 §5.8 — Energy Analysis and the Sankey Diagram (Figure 5.9)
39. A diagram that visually represents the input and output energy flows of a system, where the width of each arrow is proportional to the magnitude of the flow, is called a:
Gantt chart
Sankey diagram
Pie chart
Fishbone diagram
Answer: B) Sankey diagram
Confirmed vs Book-1 §5.8 — flows in a Sankey diagram are represented by arrows whose width is proportional to the size of the actual flow, so input, useful output and losses can be compared at a glance. Figure 5.9 for an IC engine shows 25% effective power, 5% friction and parasitic losses, 30% coolant and 40% exhaust gas — making exhaust heat recovery the priority.
Source: AI practice
📖 Book-1 §5.5 — Example 5.9 (dust balance) / Short Question S-1
40. Dust-laden gas enters a bag filter at 200,000 m^3/hr carrying 6 g/m^3 of dust. The cleaned gas leaves at 210,000 m^3/hr carrying 0.1 g/m^3. The dust collected in the hopper is:
1179 kg/hr
1200 kg/hr
21 kg/hr
1221 kg/hr
Answer: A) 1179 kg/hr
Confirmed vs Book-1 §5.5 — component (dust) balance: In = Out + hopper. Dust in = 200,000 x 6 / 1000 = 1200 kg/hr; dust out = 210,000 x 0.1 / 1000 = 21 kg/hr; collected in the bin = 1200 - 21 = 1179 kg/hr. Identical method to Example 5.9, where 777.7 - 10.6 = 767.1 kg/hr.
Source: AI practice
📖 Book-1 §5.5 — Example 5.6 (evaporator)
41. A liquor containing 8% solids is concentrated to 40% solids. Starting with 100 kg of feed, the mass of water evaporated is:
32 kg
60 kg
80 kg
20 kg
Answer: C) 80 kg
Confirmed vs Book-1 §5.5 — the dry solids are conserved. Solids = 100 x 0.08 = 8 kg; concentrated output = 8 / 0.40 = 20 kg; water evaporated = 100 - 20 = 80 kg. Example 5.6 works the same case (6% to 30% solids on 100 kg feed) and also gives 80 kg of water evaporated.
Source: AI practice
📖 Book-1 §5.7 — Example 5.10 (furnace cooling water)
42. A furnace shell (mass 2000 kg, specific heat 0.2 kcal/kg°C) is cooled from 90°C to 55°C by water whose temperature rises by 5°C. The mass of cooling water required is (Cp_water = 1 kcal/kg°C):
1400 kg
2800 kg
14000 kg
700 kg
Answer: B) 2800 kg
Confirmed vs Book-1 §5.7 — heat to be removed from the shell Q = m x Cp x dT = 2000 x 0.2 x (90 - 55) = 14,000 kcal. Heat picked up by the water = X x 1 x 5 = 5X kcal. Setting stream 1 = stream 2 gives X = 14,000 / 5 = 2800 kg. This is Example 5.10 verbatim (water enters at 28 degC and may rise only to 33 degC).
Source: AI practice
📖 Book-1 §5.5 — Short Question S-4 (boiler blow-down)
43. Boiler feed water has a TDS of 120 mg/l and the maximum permissible TDS in the boiler drum is 3500 mg/l. The required blow-down (as % of feed) is approximately:
29.2%
0.34%
3.43%
12.0%
Answer: C) 3.43%
Confirmed vs Book-1 §5.5 — a steady-state solids (TDS) balance gives blow-down as % of feed = feed TDS / maximum permissible TDS x 100 = 120 / 3500 x 100 = 3.43%. All dissolved solids entering with the feed water must leave with the blow-down to hold the drum at the IS limit of 3500 mg/l for boilers operating up to 2 MPa.
Source: AI practice
📖 Book-1 §5.5 — Example 5.7 (air composition)
44. Air is approximately 77% N2 and 23% O2 by weight. Taking a 100 kg basis (N2 = 28, O2 = 32), the mean molecular weight of air is about:
30.0
28.8
32.0
28.0
Answer: B) 28.8
Confirmed vs Book-1 §5.5 — on a basis of 100 kg of air: moles N2 = 77/28 = 2.75 and moles O2 = 23/32 = 0.72, total 3.47 moles. Mean molecular weight of air = 100 / 3.47 = 28.8. Cross-check: mole fraction of oxygen = 0.72 / 3.47 = 0.21, the familiar 21% by volume.
Source: AI practice
📖 Book-1 §5.7 — Short Question S-5 (heat rejected to cooling water)
45. Water flowing at 200 m^3/hr is heated through a temperature rise of 7°C. The heat duty is (1 m^3 water = 1000 kg, Cp = 1 kcal/kg°C):
140,000 kcal/hr
1,400,000 kcal/hr
14,000 kcal/hr
1400 kcal/hr
Answer: B) 1,400,000 kcal/hr
Confirmed vs Book-1 §5.7 — 200 m3/hr of water = 200 x 1000 = 200,000 kg/hr (1 m3 of water = 1000 kg). Q = m x Cp x dT = 200,000 x 1 x 7 = 14,00,000 kcal/hr. The usual slip is applying Q = m Cp dT to the volume without converting to mass.
Source: AI practice
📖 Book-1 §5.7 — Example 5.11 (dryer heat balance)
46. A solid is dried from 55% moisture to 10% moisture. For 60 kg/hr of wet feed, the moisture removed is:
27 kg/hr
33 kg/hr
30 kg/hr
45 kg/hr
Answer: C) 30 kg/hr
Confirmed vs Book-1 §5.7 — the bone-dry cloth is the tie component: 60 x (1 - 0.55) = 27 kg/hr, unchanged through the dryer. Final product = 27 / 0.90 = 30 kg/hr, so moisture removed = 60 - 30 = 30 kg/hr. Example 5.11 gets the same 30 kg/hr (33 kg initial moisture less 3 kg final).
47. Electrical energy of 1 kWh is equivalent to how much heat energy?
860 kcal
632 kcal
1000 kcal
427 kcal
Answer: A) 860 kcal
Confirmed vs Book-1 §5.6 — the chapter's own objective question states that energy supplied by electricity, Q in kcal = kWh x 860. The book advises converting every energy stream (oil, gas, coal, steam, electricity) to equivalent heat energy before balancing, and this is the electrical conversion factor. 632 kcal/kg in Example 5.12 is the enthalpy of evaporated moisture, a different quantity.
Source: AI practice
📖 Book-1 §5.5 — Short Question S-3 (unburnt carbon in refuse)
48. Coal contains 67.2% carbon and 22.3% ash. The refuse contains 7.1% carbon (rest is ash). Taking 100 kg of coal as basis, the fraction of original carbon left unburnt in the refuse is about:
7.1%
2.5%
1.7%
24%
Answer: B) 2.5%
Confirmed vs Book-1 §5.5 — ash is inert and is therefore the tie component that fixes the refuse mass: refuse = 22.3 / (1 - 0.071) = 22.3 / 0.929 = 24.0 kg per 100 kg of coal. Carbon in the refuse = 24.0 x 0.071 = 1.70 kg. As a percentage of the original carbon = 1.70 / 67.2 x 100 = 2.5%.
Source: AI practice
Short questions (5 marks) — 46
📖 §5.1 Purpose of Material and Energy Balance (box)
1. State the purpose of carrying out a material and energy (M&E) balance for an industrial process.
Model answer: A material and energy balance accounts for all mass and energy entering and leaving a process. Its purpose is: (i) to assess the input, conversion efficiency, output and losses; (ii) to quantify all material, energy and waste streams in a process or system; and (iii) it is a powerful tool for establishing the basis for improvement and identifying potential savings. It rests on the conservation principle (Mass In = Mass Out + Stored; Energy In = Energy Out + Stored) and exposes losses/leaks that are otherwise invisible.
Guidebook box 'Purpose of Material and Energy Balance' lists exactly these three points. Objective-Q1 answer is 'all the above' (input-output, conversion efficiency, losses).
Source: n/a
📖 §5.1 Introduction — laws of conservation of mass and energy
2. State the law of conservation of mass and the law of conservation of energy on which M&E balances are based.
Model answer: Law of conservation of mass: matter is neither created nor destroyed; matter may flow through a control volume and may react to form another species, but no matter is ever lost or gained. Law of conservation of energy: energy is neither created nor destroyed, it is simply converted from one form into another. Together these laws lead to the mass (material) balance and the energy balance: if there is no accumulation, what goes into a process must come out — true for both batch and continuous operation over any chosen time interval.
3. Write the general material balance equation for a unit operation and explain why a 'Losses' term is included.
Model answer: Treating the unit operation as a box, mass in must balance mass out: Mass In = Mass Out + Mass Stored, i.e. Raw Materials = Products + Waste Products + Stored Products. In practice the measured products, waste and stored material do not fully account for the input; the unidentified shortfall is added as Losses, so the working equation becomes: Raw Materials = Products + Waste Products + Stored Products + Losses. Losses are the unidentified materials (e.g. material chemically changed, or going unnoticed down a drain) that must be tracked down.
§5.3, sugar-plant example: m_Au is the unknown loss; equation extended to include Losses.
Source: n/a
📖 §5.3 Basic principles — energy balance equation; §5.6
4. Write the general energy balance equation for a unit operation and state why energy balances are more complicated than mass balances.
Model answer: Energy In = Energy Out + Energy Stored, where energy entering with raw materials plus energy added in the plant equals energy leaving with products, energy leaving with waste, energy lost to surroundings, plus energy stored. Energy balances are more complicated than mass balances because energy exists in many inter-convertible forms — kinetic, potential, heat, chemical, electrical, mechanical — and these forms can be converted from one to another during processing (e.g. mechanical energy converted by friction into heat). It is the sum total of all forms that is conserved, so the quantities must still balance overall.
§5.3 energy balance + §5.6 conservation of energy.
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📖 §5.2 Components of M&E balance, Fig 5.1 — recycle on input side
5. In the material balance of a process, where are the recycle stream and the by-product placed — on the input side or the output side?
Model answer: The recycle stream is shown on the INPUT side (it re-enters the process and is treated as an input along with raw materials, chemicals, water/air and energy/power). The by-product is an OUTPUT of the process (along with products, gaseous emissions, wastewater, liquid/solid waste). This is why, in the objective questions, recycle is 'always considered as input to process' and the by-product is the component NOT considered on the input side.
Fig 5.1 note states recycle stream is shown along with input side; Objective Q2 (by-product not on input) and Q3 (recycle = input to process).
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📖 §5.4 Classification of processes (steady/unsteady; continuous/batch)
6. Distinguish between a steady-state process and an unsteady-state process, and between a continuous and a batch process.
Model answer: A steady-state process is one in which none of the process variables change with time (flows and quantities held in vessels are constant), so balances can be written per unit time. An unsteady-state process is one in which the process variables change with time. Based on how the process is built to operate: a continuous process has feed and product streams moving into and out of the process all the time (e.g. oil refinery, distillation). A batch process has feed charged in to start, processed through steps with no mass added or removed during the cycle (parameters monitored/controlled), and finished products taken out at specific times.
§5.4 Classification of Processes (A by time, B by operation).
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📖 §5.5 Levels of material balance
7. Describe the different levels at which a material balance can be developed.
Model answer: Material balances can be developed at three levels: (1) Overall material balance — covers input and output streams for the complete plant. (2) Section-wise material balance — made for each section/department/cost centre; helps prioritise focus areas for efficiency improvement. (3) Equipment-wise material balance — made for key equipment; helps assess equipment performance and identify energy and material losses. The choice depends on the reason for the balance; the major factor is cost of materials, so costly materials and products are considered more readily than cheaper ones and waste materials.
§5.5 'Levels of Material Balance'.
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📖 §5.5 Material balance procedure — steps (a), (b), (c)
8. List the typical steps in carrying out a material balance for a process.
Model answer: First identify materials in, materials out and material stored, and decide whether to treat each material as a whole (gross balance) or treat individual constituents separately (e.g. dry solids vs water). Typical steps are: (a) Define basis & units — choose a basis as quantity (mass for batch) or flow rate (mass/hr for continuous) of one process stream, with convenient units (w/w, w/v, molar concentration, mole fraction). (b) Draw a flowchart — establish a boundary so flow streams in/out can be identified, showing inputs, process steps, intermediates, by-products, recycle and outputs with operating parameters and flow rates. (c) Write the material balance equations and solve.
§5.5 'Material Balance Procedure' steps a, b, c.
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📖 §5.5 step (b) Draw a flowchart; Fig 5.4 Typical process flowchart
9. What is a process flow diagram and what should it represent when prepared for a material and energy balance?
Model answer: A process flow (flow chart) is a schematic representation of the production process showing the various input resources, conversion steps, outputs and recycle streams; a boundary is established so flow streams in and out can be identified. It should be drawn stepwise from raw material to finished product, showing intermediates and by-products. It must represent the operating process parameters (temperature, pressure, % concentration, etc.) and the flow rate of each stream in appropriate units (m³/h or kg/h; for batch processes the total cycle time is included). Inputs include raw materials, water, steam and energy; wastes/by-products include solids, water, chemicals and energy. The output is the final product.
§5.5 step (b) Draw a flowchart and the bullets that follow; Fig 5.4 typical arrangement.
10. Define the four ways of expressing material concentration used when choosing units for a material balance.
Model answer: (1) Weight/weight (w/w) concentration — weight of solute divided by total weight of solution (fractional form of % composition by weight). (2) Weight/volume (w/v) concentration — weight of solute in the total volume of the solution. (3) Molar concentration (M) — the number of molecular weights of solute expressed in kg in 1 m³ of solution. (4) Mole fraction — the ratio of the number of moles of solute to the total number of moles of all species present in the solution. (For gases, concentration is measured as weight per unit volume or as partial pressures.)
§5.5 step (a) Define basis & units.
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📖 §5.6 Heat balances — enthalpy and datum
11. What is a heat balance, and why must enthalpy always be referred to a datum?
Model answer: A heat balance is an energy balance that considers only heat energy (enthalpy), ignoring internal energies — the most common and important practical form of energy balance. In heating and drying operations enthalpy (total heat) is conserved, so enthalpy balances can be written around items of equipment, process stages or the whole plant, assuming no appreciable heat is converted to work. Enthalpy (H) is always referred to a reference level or datum so that quantities are relative to that datum; working out the balance is then a matter of considering the masses involved, their specific heats, and their changes in temperature or state (latent heat).
§5.6 'Heat Balances'.
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📖 §5.6 'How can energy be lost in a system?'
12. How can energy be 'lost' in a system, and what are the main causes of this loss of usable energy?
Model answer: Because of inter-conversions it is not easy to isolate the separate forms of energy (heat, kinetic, chemical, potential). When energy is said to be 'lost' it really means it has changed into a form that is not counted — most often work done against friction, which changes other forms into heat and wear. No energy is actually destroyed; only the forms have changed. The causes of loss of usable energy are: in mechanical systems — friction; in electrical systems — resistance; in fluid systems — turbulence, viscosity or mixing. Practically, energy balances take into account only heat balances, ignoring internal energies.
§5.6 'How can energy be lost in a system?'
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📖 §5.6 Energy balance in power plant cycle, Fig 5.6 (ΣQ + ΣW = ΔU)
13. Apply the law of conservation of energy to a power plant (Rankine) cycle and state the cycle energy balance.
Model answer: If a system undergoes a process by heat and work transfer, the net heat supplied ΣQ plus the net work input ΣW equals the change in internal energy ΔU: ΣQ + ΣW = ΔU. For a complete thermodynamic cycle ΔU = 0, so ΣQ + ΣW = 0. For the power plant: heat Q_in is supplied to water in the boiler, feed-pump work W_in is added, the steam drives the turbine producing useful work W_out, and heat Q_out is rejected in the condenser. ΣQ = Q_in − Q_out. The working fluid (feed water) changes state from water to steam and back to condensate.
§5.6 Fig 5.6 Energy Balance in Power Plant Cycle; sign convention: Q supplied (+) / rejected (−), W on system (+) / by system (−).
14. Define the thermodynamic efficiency of a process and explain what 'sinks' represent in an energy balance.
Model answer: The thermodynamic efficiency of a process is the ratio of useful output to input, and is always less than 100% (Efficiency = Output / Input). For example, with input = 15 + 185 = 200 units and useful output = 45, efficiency = 45/200 = 0.225 = 22.5%, so energy lost = 100 − 22.5 = 77.5%. Sinks are the depositories of leakage or rejected energy — usually low-grade heat such as radiation losses from boilers or heat carried away by cooling water. The outputs represent the useful work; when storage does not change, inputs must equal outputs.
§5.6 Efficiency, Fig 5.7; sinks defined just before Fig 5.6.
15. Using the conservation of energy, explain how a change in stored energy is inferred when the input does not equal the output.
Model answer: If the storage in a system does not change, the ingoing and outgoing energy must be equal. If the storage changes, this is reflected in the balance and the energy input need not equal the output. Example: if 75 units enter but only 60 units leave, the first law requires energy be conserved, so the system must have gained (stored) 15 units — storage increased. Conversely, if the input is short by 15 units relative to the output, the system must have lost (released) 15 units of stored energy — storage decreased.
📖 §5.7 Facility as an Energy System, Fig 5.8 Plant Energy Systems
16. Explain the concept of a 'facility as an energy system' and the three areas into which energy/utility systems are classified.
Model answer: In a production facility, primary energy as coal, oil, gas and electricity enters and is converted into more convenient forms such as steam, compressed air and chilled water; the outgoing energy is usually heat and motion. Energy usage splits into: electricity (purchased HT, converted to LT, or self-generated by DG sets/captive plants), fuels (furnace oil, coal converted to steam or electricity), boilers (steam for heating/drying), cooling towers/cooling-water systems (cooling demand) and air compressors (compressed-air needs). For a system approach and energy analysis, all energy/utility systems are classified into three areas: generation, distribution and utilisation.
§5.7 'Facility as an Energy System', Fig 5.8 Plant Energy Systems.
Model answer: A Sankey diagram is a flow diagram that represents the entire input and output energy flow of an energy equipment or system (boiler, fired heater, furnace, etc.) after an energy balance has been carried out. Flows are shown as arrows, and the WIDTH of each arrow is proportional to the magnitude of the actual flow. Better than numbers or tables, it visually shows outputs (benefits) and losses so that energy managers can prioritise improvements. Example: for an internal combustion engine, 100% fuel energy enters and splits into ~25% effective power, ~5% friction/parasitic losses, ~30% coolant loss and ~40% exhaust gas loss. The wide exhaust-gas arrow flags it as the priority area — pointing to a waste-heat-recovery device.
§5.8 + Fig 5.9 (IC engine). Objective Q8: Sankey = input and output energy flow.
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📖 §5.8 Energy analysis and the Sankey diagram
18. What basic data is needed for an energy analysis, and why is the Sankey diagram preferred to present it?
Model answer: The basic data needed for an energy analysis is an energy balance of each process section. The objective is to define in detail the energy input, the energy utilised, and the energy dissipated or wasted. This is best represented by a Sankey diagram because, rather than numbers, tables or descriptions, it visually represents the various outputs (benefits) and losses — with arrow width proportional to flow — so the energy manager can immediately see the largest loss and focus on improvements in a prioritized manner. For an IC engine it makes clear that exhaust flue-gas loss is the key area for attention, justifying a waste-heat boiler.
19. State the heat-duty formula used in a heat balance and the relation converting electrical energy to heat (kcal).
Model answer: The heat duty (sensible heat) is found from Q = m · Cp · ΔT, where m = mass, Cp = specific heat and ΔT = temperature change; for a phase change the latent-heat form Q = m · h_latent is used. Electrical energy supplied, Q in kcal, equals kWh × 860 (since 1 kWh = 860 kcal). These, with PV = nRT for gases, are the core relations for Chapter-5 heat and material balances.
📖 §5.5–§5.8 constants used in Examples 5.5–5.12 and the Solved Example
20. List the key physical constants used in Chapter-5 material and energy balance calculations.
Model answer: Latent heat of evaporation of water ≈ 540 kcal/kg (≈ 2257 kJ/kg). Specific heat of water Cp = 1 kcal/kg°C. Electrical energy: 1 kWh = 860 kcal. Gas law: PV = nRT with R = 0.08206 m³·atm/mol·K; 1 mole of gas at STP occupies 22.4 litres. Common molecular weights: water = 18, NaCl (common salt) = 58.5, CO₂ = 44, N₂ = 28, O₂ = 32; mean molecular weight of air ≈ 28.8. Density of water = 1000 kg/m³.
Drawn from Examples 5.5–5.12 and the Solved Example (water 540 kcal/kg, mol wts, 22.4 L, R value).
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📖 §5.5 Material balance procedure — dry-solids (tie-component) balance
21. Explain the method (tie-component / dry-solids balance) for solving a drying or evaporation material balance.
Model answer: Use a constituent (tie-component) balance: identify a component whose mass is conserved through the process — in drying/evaporation the bone-dry solids do not change. Steps: (1) From the feed, compute the dry-solid mass = feed × (1 − moisture fraction). (2) This same dry-solid mass appears in the product, so product mass = dry solids ÷ (1 − final moisture fraction). (3) Water (or vapour) removed = feed − product. The solids act as the tie component linking inlet and outlet, letting you find the unknown stream without needing the water balance directly.
§5.5 'do material balance for dry solids alone... separating into non-water and water'; method underlies Examples 5.3, 5.6, 5.11, 5.12.
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📖 §5.7 Example 5.10 — furnace shell cooling water
22. Explain how to compute the cooling-water requirement to cool a hot body, using a furnace shell as an example.
Model answer: Use an energy (heat) balance: heat lost by the hot body = heat gained by the cooling water. Heat to be removed = m·Cp·ΔT of the body. Heat gained by water = m_water·Cp_water·ΔT_water (Cp_water = 1 kcal/kg°C). Equate the two and solve for m_water. Example: furnace shell m = 2000 kg, Cp = 0.2 kcal/kg°C, cooled 90→55°C, so heat removed = 2000 × 0.2 × 35 = 14,000 kcal. Water enters 28°C, leaves 33°C (ΔT = 5°C), so 14,000 = m_water × 1 × 5, giving m_water = 2800 kg.
Example 5.10 (furnace); verified numbers.
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📖 §5 EOC Short Q S-1 — dust balance (method of Example 5.9)
23. During an air pollution monitoring study, the inlet gas stream to a bag filter was 200,000 m³/hr and the outlet stream was 210,000 m³/hr. Dust load at the inlet was 6 g/m³ and at the outlet 0.1 g/m³. How much dust (kg/hr) was collected in the bag filter bin?
Model answer: Apply a dust mass balance: Dust in = Dust out + Dust collected in hopper. Inlet dust = 200,000 m³/hr × 6 g/m³ = 1,200,000 g/hr = 1200 kg/hr. Outlet dust = 210,000 m³/hr × 0.1 g/m³ = 21,000 g/hr = 21 kg/hr. Dust collected in bin = 1200 − 21 = 1179 kg/hr.
Short-Q S-1; method from Example 5.9 dust balance. Verified arithmetic.
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📖 §5 EOC Short Q S-3 — unburnt carbon by ash (tie-component) balance
24. A coal sample from the mine contains 67.2% carbon and 22.3% ash. The refuse after combustion contains 7.1% carbon and the rest is ash. Compute the % of the original carbon left unburnt in the refuse.
Model answer: Basis 100 kg coal: carbon = 67.2 kg, ash = 22.3 kg. Ash is inert and is conserved, so ash in refuse = 22.3 kg. Refuse is 7.1% carbon and 92.9% ash, so total refuse = 22.3 / 0.929 = 24.0 kg. Carbon (unburnt) in refuse = 7.1% of 24.0 = 1.70 kg. % of original carbon unburnt = (1.70 / 67.2) × 100 ≈ 2.54%.
Short-Q S-3; ash is the tie (conserved) component. Verified.
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📖 §5 EOC Short Q S-4 — boiler blowdown from TDS balance
25. A boiler is fed with soft water containing 120 mg/l dissolved solids. As per IS standards the dissolved solids in the boiler should not exceed 3500 mg/l (boilers up to 2 MPa). A continuous blowdown system is adopted. Find the percentage of feed water that is blown down.
Model answer: Dissolved solids are conserved: solids entering with feed water must leave with the blowdown (steam carries no solids). At steady state, % blowdown (as fraction of feed water) = Feed-water TDS / Maximum boiler TDS × 100 = 120 / 3500 × 100 = 3.43%. So about 3.4% of the feed water must be blown down to hold the boiler TDS at 3500 mg/l.
Short-Q S-4; rewritten cleanly (existing answer was self-contradictory). Method per master notes: %blowdown = feed TDS / max TDS.
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📖 §5 EOC Short Q S-5 — Q = m·Cp·ΔT for cooling water
26. A shell-and-tube heat exchanger is cooled with a stream of demineralized water. Evaluate the total heat rejected to the cooling water (kcal/hr) if the water flow rate is 200 m³/hr and the temperature rise is 7°C.
Model answer: Mass flow of water = 200 m³/hr × 1000 kg/m³ = 200,000 kg/hr. Specific heat of water Cp = 1 kcal/kg°C. Heat rejected Q = m · Cp · ΔT = 200,000 × 1 × 7 = 1,400,000 kcal/hr (1.4 × 10⁶ kcal/hr).
Short-Q S-5; verified.
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📖 §5 EOC Long Q L-2 — autoclave cooling-water requirement
27. An autoclave holds 1000 cans of pea soup heated to 100°C; the cans must be cooled to 40°C. Each can weighs 60 g (Cp 0.50 kJ/kg°C) and holds 0.45 kg pea soup (Cp 4.1 kJ/kg°C). Heat content of the autoclave walls above 40°C is 1.6×10⁴ kJ. Cooling water enters at 15°C and leaves at 35°C. How much cooling water is required (no wall heat loss)?
Model answer: Heat to be removed (cooling 100→40°C, ΔT = 60°C): Soup = 1000 × 0.45 × 4.1 × 60 = 110,700 kJ. Cans = 1000 × 0.060 × 0.50 × 60 = 1,800 kJ. Walls = 1.6×10⁴ = 16,000 kJ. Total heat removed = 110,700 + 1,800 + 16,000 = 128,500 kJ. Cooling water (Cp = 4.187 kJ/kg°C, ΔT = 35−15 = 20°C) gains this heat: m_w = 128,500 / (4.187 × 20) ≈ 1534 kg of cooling water.
Long-Q L-2 framed as a short worked balance. Numbers from OCR; water Cp 4.187 kJ/kg°C assumed.
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📖 §5.5 Example 5.1 — mixing two solutions (solids balance)
28. A 10% solids solution (entering at 5.0 kg/s) is mixed with a 25% solids solution; a single output of 20% solids is removed. Find the other flow rates (no accumulation).
Model answer: Let B (25% solids) = X kg/s and output C (20% solids) = Y kg/s. Total balance: 5 + X = Y. Solids balance: 0.10×5 + 0.25X = 0.20Y → 0.5 + 0.25X = 0.20(5 + X) → 0.05X = 0.5 → X = 10 kg/s. Then Y = 5 + 10 = 15 kg/s. So the 25% stream is 10 kg/s and the 20% output is 15 kg/s.
Example 5.1; verified.
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📖 §5.5 Example 5.6 — evaporator, water evaporated per 100 kg feed
29. In a textile mill an evaporator concentrates a liquor containing 6% solids (w/w) to an output containing 30% solids (w/w). Calculate the water evaporated per 100 kg of feed.
Model answer: Solids are conserved. Feed = 100 kg → solid content = 100 × 0.06 = 6 kg. Outlet solid content = 6 kg (mass in = mass out). Output (thick liquor) = 6 / 0.30 = 20 kg. Water evaporated = feed − output = 100 − 20 = 80 kg.
Example 5.6; verified.
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📖 §5.5 Example 5.4 — continuous centrifuge (milk) balance
30. In continuous centrifuging, 35,000 kg of whole milk containing 4% fat is separated over 6 hours into skim milk (0.45% fat) and cream (45% fat). Find the flow rates of the two output streams.
Model answer: Feed per hour = 35,000 / 6 = 5833 kg/hr. Let skim milk = Y and cream = Z. Mass balance: 5833 = Y + Z (Z = 5833 − Y). Fat balance: 0.04 × 5833 = 0.0045Y + 0.45Z. Substituting: 233.3 = 0.0045Y + 0.45(5833 − Y) → solving gives Y ≈ 5369 kg/hr (skim milk) and Z = 5833 − 5369 ≈ 464 kg/hr (cream).
Example 5.4; verified.
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📖 §5.5 Example 5.3 — constituent balance, skim vs whole milk
31. Skim milk made by removing fat from whole milk contains 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original whole milk contained 4.5% fat (only fat removed, no losses), find the whole-milk composition.
Model answer: Basis 100 kg skim milk → it contains 0.1 kg fat. Let removed fat = x kg. Original fat = (x + 0.1) kg; original mass = (100 + x) kg. Since original fat was 4.5%: (x + 0.1)/(100 + x) = 0.045 → x = 4.6 kg. So whole milk = 104.6 kg. Composition: fat = 4.5%, water = 90.5/104.6 = 86.5%, protein = 3.5/104.6 = 3.3%, carbohydrate = 5.1/104.6 = 4.9%, ash = 0.8/104.6 = 0.76% (the book prints 0.8%). Whole milk total = 104.6 kg from 100 kg skim milk + 4.6 kg fat.
32. A textile dryer consumes 4 m³/hr of natural gas (CV 800 kJ/mole) and dries 60 kg/hr of wet cloth from 55% to 10% moisture. Estimate its overall thermal efficiency (latent heat of evaporation only).
Model answer: Bone-dry cloth = 60 × (1 − 0.55) = 27 kg; initial moisture = 33 kg. Final product at 10% moisture: dry cloth is 90%, so product = 27/0.9 = 30 kg, final moisture = 3 kg. Moisture removed = 33 − 3 = 30 kg/hr. Heat used = 30 × 2257 = 6.8×10⁴ kJ/hr. Gas at STP: 1 mole = 22.4 L, so 4 m³/hr = 4000/22.4 = 179 mole/hr. Heat available = 179 × 800 = 14.3×10⁴ kJ/hr. Thermal efficiency = 6.8×10⁴ / 14.3×10⁴ ≈ 48%.
Example 5.11; verified.
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📖 §5.7 Example 5.12 — paper machine evaporation rate and steam
33. A paper machine produces 340 TPD; inlet/outlet dryness is 40%/95%; evaporated moisture at 80°C (enthalpy 632 kcal/kg); steam supplied at 35 kg/cm² (latent heat 513 kcal/kg). Find (a) moisture evaporated and (b) steam required per hour.
Model answer: Production = 340 TPD = 14.16 TPH. Paper (bone-dry) in product = 14.16 × 0.95 = 13.45 TPH. Moisture before dryer = [(100−40)/40] × 13.45 = 20.175 TPH. Moisture after dryer = 14.16 × 0.05 = 0.707 TPH. (a) Evaporated moisture = 20.175 − 0.707 = 19.468 TPH. (b) Heat in moisture = 632 × 19,468 = 12,303,776 kcal/hr; steam required = 12,303,776 / 513 = 23,984 kg ≈ 23.98 MT/hour.
Example 5.12; verified.
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📖 §5.8 Solved Example — evaporator vapour and steam (enthalpy balance)
34. An evaporator is fed 10,000 kg/hr of 1% solids solution (feed at 38°C), concentrated to 2% solids. Steam enters at enthalpy 640 kcal/kg, condensate leaves at 100°C; enthalpies: feed 38.1, product 100.8, vapour 640 kcal/kg. Find the vapour formed and the steam used per hour.
Model answer: Solids balance: solids = 10,000 × 0.01 = 100 kg/hr; product = 100/0.02 = 5000 kg/hr; vapour formed = 10,000 − 5000 = 5000 kg/hr. Heat balance: heat in steam + heat in feed = heat in vapour + heat in thick liquor. M×(640−100) + 38.1×10,000 = 640×5000 + 100.8×5000 → 540M + 381,000 = 3,200,000 + 504,000 → 540M = 3,323,000 → M = 6153.7 kg steam/hr.
Solved Example before the Questions section; verified (steam gives 640−100 = 540 kcal/kg).
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📖 §5.5 Example 5.5 — salt solution expressed four ways
35. 20 kg of salt is dissolved in 100 kg of water giving a solution of density 1323 kg/m³. Express the salt concentration as (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molar concentration. (Mol wt: salt 58.5, water 18.)
Model answer: (a) Weight fraction = 20/(100+20) = 0.167 = 16.7% w/w. (b) 1 m³ solution = 1323 kg, of which salt = 20×1323/120 = 220.5 kg/m³; weight/volume fraction = 220.5/1000 = 0.2205 ≈ 22.1%. (c) Moles water = 100/18 = 5.56, moles salt = 20/58.5 = 0.34; mole fraction salt = 0.34/(5.56+0.34) = 0.058. (d) Molar concentration M = 220.5/58.5 = 3.77 mol/m³.
Example 5.5; verified.
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📖 §5 EOC Objective Q4 — conversion and mass conservation
36. In a chemical process two reactants A (200 kg) and B (200 kg) react in equal proportion. If the conversion is 50%, calculate the weight of product formed.
Model answer: Total reactants charged = 200 + 200 = 400 kg. They react in equal proportion, so all of A and B can react together. At 50% conversion, only half of the reactant mass is converted to product: product formed = 0.50 × 400 = 200 kg. (By conservation of mass the product equals the converted reactant mass.)
Objective Q4; correct option = 200 kg.
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📖 §5 EOC Objective Q5 — sensible heat Q = m·Cp·ΔT
37. In a heat-treatment furnace, material is heated from an ambient 30°C to 800°C. Taking the specific heat as 0.13 kcal/kg°C, what is the energy content of one kg of material after heating?
Model answer: Sensible heat Q = m·Cp·ΔT = 1 × 0.13 × (800 − 30) = 0.13 × 770 = 100.1 ≈ 100 kcal per kg.
Objective Q5; correct option = 100 kcal.
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📖 §5.6 Energy balance — milk pasteurizer example (whole and part process)
38. Using the milk pasteurizer as an example, show how an energy (heat) balance can be applied to a whole process and to only a part of a process.
Model answer: In pasteurizing, milk is pumped through a heat exchanger, first heated then cooled; the energy of interest is heat energy in the milk. Whole process: heat energy leaving in the milk = initial heat energy + heat added by pump + heat added in the heating section − heat removed in the cooling section − heat lost to surroundings. Part of a process: considering only the heating section, heat lost by the hot water = heat gained by the milk + heat lost from the heat exchanger to its surroundings. Thus the conservation of energy applies equally to the whole plant or to any chosen sub-section.
§5.6 pasteurizer example; verified.
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📖 §5.1 Introduction — planning → pilot → commissioning → production
39. At what stages of a process is a material balance used, from concept to production?
Model answer: A material balance can be determined from the conceptual stage to the final production stage. Initially it is estimated during the planning stage of a new process or equipment. This estimate is improved while carrying out pilot-scale tests on the new process. The material balance is then verified during the commissioning stage, and finally used as a control measure during actual production. Material balances are fundamental to the control of processing, particularly the control of product yields.
§5.1 Introduction (planning → pilot → commissioning → production).
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📖 §5.3 Component balance with no chemical change — sugar plant
40. If no chemical changes occur in a plant, how does the law of conservation of mass apply to each individual component? Illustrate with a sugar plant.
Model answer: If no chemical changes occur, the conservation of mass applies to each component separately: mass of component A in entering materials = mass of A in exit materials + mass of A stored in the plant. Example (sugar plant): if total sugar (mA) entering is not equalled by purified sugar (mAp) + sugar in waste liquors (mAw) + sugar accumulated (mAs), something is wrong — sugar is either being burned (chemically changed) or going unnoticed down a drain. The unidentified loss mAu must be found, giving: Raw materials = Products + Waste + Stored + Losses.
§5.3 component balance + sugar example.
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📖 §5.5 Example 5.7 (a) and (b) — air composition
41. Air consists of 77% by weight nitrogen and 23% by weight oxygen. Calculate (a) the mean molecular weight of air and (b) the mole fraction of oxygen. (Mol wt: N₂ = 28, O₂ = 32.)
Model answer: Basis 100 kg air: moles N₂ = 77/28 = 2.75, moles O₂ = 23/32 = 0.72. Total moles = 2.75 + 0.72 = 3.47. (a) Mean molecular weight of air = 100/3.47 = 28.8. (b) Mole fraction of oxygen = 0.72/3.47 = 0.21.
42. A coal sample contains 64% carbon and 24% ash. The refuse after combustion contains 8% carbon and the rest ash. Compute the % of the original carbon left unburnt in the refuse.
Model answer: Basis 100 kg coal: carbon = 64 kg, ash = 24 kg. Ash is conserved, so ash in refuse = 24 kg. Refuse is 8% carbon and 92% ash, so total refuse = 24/0.92 = 26.087 kg. Unburnt carbon in refuse = 8% × 26.087 = 2.087 kg. % of original carbon unburnt = (2.087/64) × 100 = 3.26%.
Exam variant of S-3; same ash-tie method. Verified arithmetic.
43. Steam heats 5 kL/hr of furnace oil from 30°C to 90°C. Furnace-oil Cp = 0.22 kcal/kg°C, sp. gravity 0.95. (a) Steam per hour needed if steam latent heat is 510 kcal/kg. (b) If steam costs Rs 3.40/kg and electricity Rs 6/kWh, which is more economical?
📖 §5.4 Classification of processes, Fig 5.3 — process as a control box
44. Explain how a process or unit operation is represented as a 'box', and what must be true of the mass crossing the box.
Model answer: A process can be viewed overall or as a series of units, each a unit operation represented by a box (control volume). Raw materials and energy go into the box, and desired products, by-products, wastes and energy come out. The mass into and out of the control box must be equal (Mass In = Mass Out when there is no accumulation). The equipment within the box makes the required changes with as little waste and energy use as possible. The box may sit between previous unit operations (feeding it) and subsequent unit operations (receiving its product), with wasted energy shown leaving.
§5.4 + Fig 5.3 Representation of Process.
Source: n/a
📖 Book-1 §5.5 — Material Balance Procedure (gross vs constituent balance)
45. Distinguish between a gross (overall) material balance and a constituent (component) material balance.
Model answer: A gross (overall) material balance treats all the material in each stream as a whole — total mass in = total mass out (+ stored). A constituent (component) balance separates the material into individual constituents and balances each one separately; for example, splitting a stream into non-water (dry solids) and water, then writing a balance on the dry solids alone. The constituent approach is essential when one component is conserved (a 'tie component') and is used to find unknown stream rates in drying, evaporation and separation problems.
Synthesised from §5.5 'Material Balance Procedure' (gross vs individual constituents); concept not given as a single Q in OCR.
Source: n/a
📖 Book-1 §5.5 (levels of material balance) & §5.8 (energy analysis); SEC as energy intensity per §4.6
46. What is specific energy consumption (SEC), and how does an energy balance help establish it?
Model answer: Specific energy consumption is the energy used per unit of output — for example kcal (or kWh) per kg or per tonne of product. It normalises energy use against production so that performance can be compared over time or against benchmarks regardless of output level. An energy balance, by quantifying the total energy input and the useful output of a facility, section or equipment, provides exactly the numbers needed to compute SEC (energy input ÷ production). Tracking SEC highlights deteriorating efficiency and the scope for energy-saving measures.
Concept flagged in the task brief; supported by §5.5 levels of balance and §5.8 energy analysis but not defined verbatim in OCR.
Source: n/a
Long questions (10 marks) — 19
📖 §5.8 Solved Example — evaporator (book worked example)
1. An evaporator is fed with 10,000 kg/hr of a solution having 1% solids. The feed is at 38 degC and is to be concentrated to 2% solids. Steam enters at a total enthalpy of 640 kcal/kg and the condensate leaves at 100 degC. Enthalpy of feed = 38.1 kcal/kg, enthalpy of product (thick liquor) = 100.8 kcal/kg and enthalpy of vapour = 640 kcal/kg. Find (i) the mass of vapour formed per hour and (ii) the mass of steam used per hour.
Model answer: STEP 1 - Mass (solids) balance to get product and vapour.
Solids in feed = 10,000 x 1/100 = 100 kg/hr (solids are conserved).
Product (thick liquor) is 2% solids: Product x 2/100 = 100 -> Product = 100/0.02 = 5000 kg/hr.
Vapour formed = Feed - Product = 10,000 - 5000 = 5000 kg/hr.
STEP 2 - Heat (enthalpy) balance to get steam.
Heat in with feed = 10,000 x 38.1 = 3,81,000 kcal/hr.
Heat out in thick liquor = 5000 x 100.8 = 5,04,000 kcal/hr.
Heat out in vapour = 5000 x 640 = 32,00,000 kcal/hr.
Steam gives up (640 - 100) = 540 kcal/kg (enthalpy of steam minus condensate at 100 degC).
Balance: Heat by steam + Heat in feed = Heat in vapour + Heat in thick liquor
M x 540 + 3,81,000 = 32,00,000 + 5,04,000
M x 540 = 37,04,000 - 3,81,000 = 33,23,000
M (steam) = 33,23,000 / 540 = 6153.7 kg/hr.
ANSWER: Vapour formed = 5000 kg/hr; Steam used = 6153.7 kg/hr.
Canonical two-part evaporator problem. Part (a) is a pure solids balance (solids unchanged, water leaves as vapour). Part (b) is an enthalpy balance where steam contributes latent heat = h_steam - h_condensate = 640 - 100 = 540 kcal/kg.
Source: unknown
📖 §5.7 Example 5.12 — paper machine (book worked example)
2. Production rate from a paper machine is 340 tonnes per day (TPD). Inlet and outlet dryness to the paper machine are 40% and 95% respectively. Evaporated moisture temperature is 80 degC. To evaporate the moisture, steam is supplied at 35 kg/cm2 (latent heat = 513 kcal/kg). Assume 24 hours/day operation and enthalpy of evaporated moisture = 632 kcal/kg. Estimate (a) the quantity of moisture to be evaporated per hour and (b) the input steam quantity required for evaporation per hour.
Model answer: Production = 340 TPD = 340/24 = 14.16 TPH (tonnes per hour) of paper.
STEP 1 - Moisture to be evaporated (solids balance; bone-dry paper is conserved).
Bone-dry paper in final product (95% dry) = 14.16 x 0.95 = 13.45 TPH.
Weight of moisture BEFORE dryer (inlet 40% dry means 60% moisture on the dry solids): = [(100-40)/40] x 13.45 = (60/40) x 13.45 = 20.175 TPH.
Weight of moisture AFTER dryer (outlet 95% dry -> 5% moisture) = 14.16 x 0.05 = 0.707 TPH.
Evaporated moisture = 20.175 - 0.707 = 19.468 TPH = 19,468 kg/hr.
STEP 2 - Steam required (heat balance).
Heat to be carried away in the moisture (sensible + latent) = 632 x 19,468 = 1,23,03,776 kcal/hr.
This heat is supplied by the latent heat of the steam (513 kcal/kg):
Steam required = 1,23,03,776 / 513 = 23,984 kg/hr = 23.98 MT/hr.
ANSWER: Moisture evaporated = 19.468 TPH (19,468 kg/hr); Steam required = 23,984 kg/hr (~24 MT/hr).
Drying + steam heat balance. Track bone-dry paper (unchanged). Moisture before/after found from dryness %, evaporated = difference. Steam = heat in evaporated moisture / latent heat of steam.
Source: unknown
📖 §5 EOC Long Q L-1 — wet scrubber make-up water balance
3. A wet scrubber removes fine dust from an inlet gas stream with a spray of water so that the outlet gas meets emission standards. Stream 1 (recirculation liquid back to the scrubber) = 4.54 m3/hr. The liquid withdrawn for treatment and disposal (stream 4) = 0.454 m3/hr. The inlet gas (stream 2) is completely dry and the outlet gas (stream 6) carries 272.16 kg/hr of moisture evaporated in the scrubber. Make-up water is added as stream 5. How much make-up water must be continually added to keep the unit running?
Model answer: Apply an OVERALL WATER BALANCE across the scrubber system boundary: Water IN = Water OUT.
The recirculation stream (stream 1 = 4.54 m3/hr) is internal to the loop and cancels out - it does not cross the system boundary.
Water LEAVES the system by two routes:
(a) Liquid withdrawn for treatment/disposal (stream 4) = 0.454 m3/hr = 454 kg/hr (1 m3 water = 1000 kg).
(b) Moisture evaporated and carried away with the outlet gas (stream 6) = 272.16 kg/hr.
Total water leaving = 454 + 272.16 = 726.16 kg/hr.
Water ENTERS the system only as make-up water (stream 5). By balance:
Make-up water (stream 5) = Water leaving = 726.16 kg/hr ~ 0.726 m3/hr.
ANSWER: Make-up water to be added continually = 726.16 kg/hr (approximately 0.73 m3/hr).
Classic make-up water balance. Key insight: ignore the internal recirculation loop; balance only what crosses the boundary. Make-up replaces water lost to disposal plus water evaporated into the gas.
Source: unknown
📖 §5 EOC Long Q L-2 — autoclave cooling-water requirement
4. An autoclave contains 1000 cans of pea soup heated to an overall temperature of 100 degC. The cans are to be cooled to 40 degC before leaving the autoclave. How much cooling water is required if it enters at 15 degC and leaves at 35 degC? The specific heats of the pea soup and the can metal are 4.1 and 0.50 kJ/kg degC respectively. Each can weighs 60 g and contains 0.45 kg of pea soup. The heat content of the autoclave walls above 40 degC is 1.6 x 10^4 kJ, and there is no heat loss through the walls.
Model answer: HEAT TO BE REMOVED (cooling all hot components from 100 to 40 degC, delta_T = 60 degC).
Use Q = m x Cp x delta_T for each component:
1) Pea soup: m = 1000 x 0.45 = 450 kg; Q = 450 x 4.1 x 60 = 1,10,700 kJ.
2) Can metal: m = 1000 x 0.060 = 60 kg; Q = 60 x 0.50 x 60 = 1,800 kJ.
3) Autoclave walls (heat content above 40 degC, given) = 16,000 kJ.
Total heat to be removed = 1,10,700 + 1,800 + 16,000 = 1,28,500 kJ.
HEAT ABSORBED BY COOLING WATER (15 -> 35 degC, delta_T = 20 degC; Cp water = 4.186 kJ/kg degC).
Heat gained = m_water x 4.186 x 20 = 83.72 x m_water kJ.
ENERGY BALANCE: Heat removed = Heat gained by water
1,28,500 = 83.72 x m_water
m_water = 1,28,500 / 83.72 = 1535 kg.
ANSWER: Cooling water required ~ 1535 kg (about 1530-1540 kg depending on the Cp of water used).
Multi-component sensible-heat balance plus stored wall heat. Total heat released by soup + cans + walls equals heat picked up by cooling water (m Cp delta_T).
Source: unknown
📖 §5.5 Example 5.9 — bag filter dust balance (book worked example)
5. A bag filter is used to remove dust from a gas stream in a cement plant. Inlet gas to the bag filter is 1,69,920 m3/hr with a dust loading of 4577 mg/m3. Outlet gas from the bag filter is 1,85,040 m3/hr with a dust loading of 57 mg/m3. What is the maximum quantity of ash that will have to be removed per hour from the bag filter hopper?
Model answer: Apply a DUST (mass) BALANCE: Mass in = Mass out.
Inlet gas dust = Outlet gas dust + Hopper ash.
STEP 1 - Inlet dust quantity.
= 1,69,920 m3/hr x 4577 mg/m3 x (1/1,000,000) kg/mg
= 7,77,72,... /1,000,000 = 777.7 kg/hr.
STEP 2 - Outlet dust quantity.
= 1,85,040 m3/hr x 57 mg/m3 x (1/1,000,000) kg/mg
= 10.6 kg/hr.
STEP 3 - Hopper ash (removed).
Hopper ash = Inlet dust - Outlet dust = 777.7 - 10.6 = 767.1 kg/hr.
ANSWER: Ash removed from the hopper = 767.1 kg/hr.
Straight dust mass balance: convert loading (mg/m3) x flow (m3/hr) to kg/hr for inlet and outlet; collected ash = inlet - outlet.
Source: unknown
📖 §5.7 Example 5.10 — furnace shell cooling water (book worked example)
6. A furnace shell has to be cooled from 90 degC to 55 degC. The mass of the furnace shell is 2 tonnes and its specific heat is 0.2 kcal/kg degC. Water is available at 28 degC and the maximum allowed increase in water temperature is 5 degC. Calculate the quantity of water required to cool the furnace. Neglect heat loss.
Model answer: ENERGY STREAM 1 - Heat to be removed from the furnace shell.
m = 2 tonnes = 2000 kg; Cp = 0.2 kcal/kg degC; T1 = 90 degC, T2 = 55 degC.
Q = m x Cp x (T1 - T2) = 2000 x 0.2 x (90 - 55) = 2000 x 0.2 x 35 = 14,000 kcal.
ENERGY STREAM 2 - Heat picked up by cooling water.
Cp water = 1 kcal/kg degC; inlet 28 degC, outlet (max) 33 degC, so delta_T = 5 degC.
Heat removed by water = X x 1 x (33 - 28) = 5X kcal, where X = mass of water.
ENERGY BALANCE: Stream 1 = Stream 2
14,000 = 5X
X = 14,000 / 5 = 2800 kg.
ANSWER: Quantity of cooling water required = 2800 kg.
Sensible-heat balance: heat lost by shell (m Cp delta_T) = heat gained by water (m Cp delta_T). Solve for water mass.
Source: unknown
📖 §5.7 Example 5.11 — textile dryer thermal efficiency (book worked example)
7. A textile dryer consumes 4 m3/hr of natural gas with a calorific value of 800 kJ/mole. The throughput of the dryer is 60 kg of wet cloth per hour, drying it from 55% moisture to 10% moisture. Estimate the overall thermal efficiency of the dryer, taking into account the latent heat of evaporation only. (1 mole of gas occupies 22.4 litres at STP; latent heat of evaporation of water = 2257 kJ/kg.)
Model answer: STEP 1 - Moisture removed (solids/moisture balance).
Initial moisture in wet cloth = 60 x 0.55 = 33 kg.
Bone-dry cloth = 60 x (1 - 0.55) = 27 kg (unchanged).
Final product at 10% moisture: dry solids are 90% -> total final mass = 27/0.90 = 30 kg, so final moisture = 30 - 27 = 3 kg.
Moisture removed per hour = 33 - 3 = 30 kg/hr.
STEP 2 - Heat usefully used for drying (latent heat only).
Q_used = 30 kg/hr x 2257 kJ/kg = 67,710 kJ/hr ~ 6.8 x 10^4 kJ/hr.
STEP 3 - Heat available from combustion of the gas.
Gas rate = 4 m3/hr = 4 x 1000 litres/hr; at STP 1 mole = 22.4 litres, so moles = 4000/22.4 = 179 moles/hr.
Q_available = 179 x 800 = 1,43,200 kJ/hr ~ 14.3 x 10^4 kJ/hr.
STEP 4 - Thermal efficiency.
eta = Q_used / Q_available = 6.8 x 10^4 / 14.3 x 10^4 = 0.48 = 48%.
ANSWER: Overall thermal efficiency of the dryer ~ 48%.
Energy balance on a dryer. Efficiency = heat used to evaporate moisture (m x latent heat) divided by heat available from fuel (moles/hr x CV per mole). Sensible heat neglected as latent heat dominates.
Source: unknown
📖 §5.5 Example 5.4 — continuous centrifuge (book worked example)
8. In a continuous centrifuging of milk, 35,000 kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream with 45% fat. What is the flow rate of the two output streams from the continuous centrifuge?
Model answer: Work on a per-hour basis (steady state).
BASIS: Total mass input per hour = 35,000 / 6 = 5833 kg/hr.
Let Y = skim milk (kg/hr), Z = cream (kg/hr).
EQ-1 (Total mass balance): 5833 = Y + Z -> Z = 5833 - Y.
EQ-2 (Fat balance): 0.04 x 5833 = 0.0045 x Y + 0.45 x Z.
Substitute Z from EQ-1 into EQ-2:
0.04 x 5833 = 0.0045 Y + 0.45 (5833 - Y)
233.3 = 0.0045 Y + 2624.85 - 0.45 Y
0.4455 Y = 2624.85 - 233.3 = 2391.55
Y = 2391.55 / 0.4455 = 5369 kg/hr (skim milk).
Z = 5833 - 5369 = 464 kg/hr (cream).
ANSWER: Skim milk = 5369 kg/hr; Cream = 464 kg/hr.
Continuous two-output split. Two equations: total mass balance and the tracked-component (fat) balance. Substitute and solve for the two output flow rates.
Source: unknown
📖 §5.5 Example 5.3 — constituent balance (book worked example)
9. Skim milk is prepared by removing some fat from whole milk. The skim milk contains 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original whole milk contained 4.5% fat, calculate the composition of the whole milk, assuming that only fat was removed and there are no processing losses.
Model answer: BASIS: 100 kg of skim milk (which therefore contains 0.1 kg fat).
Let x = mass of fat removed to make the skim milk.
Total original fat = (x + 0.1) kg.
Total original (whole milk) mass = (100 + x) kg.
The original fat content was 4.5%, so:
(x + 0.1) / (100 + x) = 0.045
x + 0.1 = 0.045 (100 + x) = 4.5 + 0.045 x
x - 0.045 x = 4.5 - 0.1
0.955 x = 4.4 -> x = 4.6 kg.
So total whole milk = 100 + 4.6 = 104.6 kg.
Compositions of the whole milk (each skim-milk component mass now divided by 104.6):
- Fat = 4.5% (given / by construction).
- Water = 90.5 / 104.6 = 86.5%.
- Protein = 3.5 / 104.6 = 3.3%.
- Carbohydrate = 5.1 / 104.6 = 4.9%.
- Ash = 0.8 / 104.6 = 0.8%.
ANSWER: Whole milk = 4.5% fat, 86.5% water, 3.3% protein, 4.9% carbohydrate, 0.8% ash (total 104.6 kg from 100 kg skim).
Constituent balance: fix a 100 kg basis on the known (skim) stream, track the single component that changed (fat), set up the % equation, solve for added/removed mass, then re-express all components on the new total mass.
Source: unknown
📖 §5.5 Example 5.2 — multi-component balance (book worked example)
10. A solution which is 80% oil, 15% usable by-products and 5% impurities enters a refinery. One output is 92% oil and 6% usable by-products. The other output is 60% oil and flows at 1000 lit/hr (assume no accumulation, percentages by volume). Find (a) the flow rate of the input, (b) the percent composition of the 1000 lit/hr output, and (c) what percent of the original impurities are in the 1000 lit/hr output.
Model answer: BASIS: Input A = X lit/hr; Output B = Y lit/hr (92% oil, 6% UBP, 2% impurities); Output C = 1000 lit/hr (60% oil, V fraction UBP, W fraction impurities).
Equations:
EQ-1 Total: X = Y + 1000.
EQ-2 Oil: 0.80 X = 0.92 Y + 0.60 x 1000.
EQ-3 UBP: 0.15 X = 0.06 Y + V x 1000.
EQ-4 Impurities: 0.05 X = 0.02 Y + W x 1000.
(a) Solve EQ-1 and EQ-2. Substitute X = Y + 1000 into EQ-2:
0.80 (Y + 1000) = 0.92 Y + 600
0.80 Y + 800 = 0.92 Y + 600
200 = 0.12 Y -> Y = 1666.7 lit/hr.
X = 1666.7 + 1000 = 2666.7 lit/hr = flow rate of input.
(b) Composition of the 1000 lit/hr stream:
UBP (EQ-3): V x 1000 = 0.15 x 2666.7 - 0.06 x 1666.7 = 400.0 - 100.0 = 300 -> V = 0.30 (30%).
Impurities (EQ-4): W x 1000 = 0.05 x 2666.7 - 0.02 x 1666.7 = 133.3 - 33.3 = 100 -> W = 0.10 (10%).
So the 1000 lit/hr output = 60% oil, 30% usable by-products, 10% impurities. (Check: 60+30+10 = 100.)
(c) Impurities in input = 0.05 x 2666.7 = 133.3 lit/hr; impurities in the 1000 lit/hr stream = 100 lit/hr.
Percent of original impurities in this stream = 100 / 133.3 x 100 = 75%.
ANSWER: (a) Input = 2666.7 lit/hr; (b) 60% oil, 30% UBP, 10% impurities; (c) 75% of the original impurities.
Multi-component (three constituents) material balance with two outputs. Write total + component balances, solve the two-unknown pair first, then back out the unknown output fractions and the impurity ratio.
Source: unknown
📖 §5.8 Sankey diagram + §5.6 energy balance — DG set application
11. Prepare the energy balance of a Diesel Generator and draw a Sankey diagram. Given: calorific value of diesel = 10,000 kcal/litre; average energy generated = 4.07 kWh/litre; alternator efficiency = 96%; stack (flue gas) losses = 33%; coolant losses = 24%; and the balance is radiation losses. (1 kWh = 860 kcal.)
Model answer: Take fuel input = 100% thermal energy (basis: 1 litre diesel = 10,000 kcal).
STEP 1 - Electrical output %.
Energy generated = 4.07 kWh/litre = 4.07 x 860 = 3500.2 kcal/litre.
Electrical output = (3500.2 / 10,000) x 100 = 35%.
STEP 2 - Alternator losses.
Alternator efficiency = 96%, so alternator loss = 100 - 96 = 4% (of the mechanical energy fed to it; taken as ~4% of input for the Sankey split).
STEP 3 - Given loss splits.
Stack (flue gas) loss = 33%; Coolant loss = 24%.
STEP 4 - Radiation (balance) loss.
Radiation loss = 100 - (Electrical 35 + Alternator 4 + Stack 33 + Coolant 24) = 100 - 96 = 4%.
ENERGY BALANCE (of 100% input): 35% electrical output + 4% alternator + 33% stack + 24% coolant + 4% radiation = 100%.
SANKEY DIAGRAM: a single 100% input arrow (fuel energy) branches into a 35% useful electrical output arrow and loss arrows of 33% (stack/flue gas), 24% (coolant), 4% (alternator) and 4% (radiation), the arrow widths drawn proportional to each percentage. Total losses = 65%.
Energy balance expressed as % of fuel input. Electrical output from kWh x 860 / CV. Radiation is the closing (balance) term. Sankey arrow widths are proportional to the magnitude of each stream.
Source: unknown
📖 §5.6 Energy balance — specific energy consumption; EOC Objective Q7 (kWh × 860)
12. A foundry has an induction furnace of 5 TPH with a specific energy consumption of 620 kWh/tonne of liquid metal. The yield of foundry castings is 60%. The castings are heat treated in an oil-fired furnace consuming 75 kg oil/tonne of castings. Find the energy consumption per tonne of finished product in terms of oil equivalent. (GCV of oil = 10,000 kcal/kg; 1 kWh = 860 kcal.)
Model answer: STEP 1 - Melting energy per tonne of finished (heat-treated) product.
SEC of melting = 620 kWh per tonne of liquid metal.
Yield of castings = 60%, so per tonne of castings: 620 / 0.60 = 1033.3 kWh/tonne.
Convert to heat units: 1033.3 x 860 = 8,88,667 kcal/tonne.
STEP 2 - Heat-treatment energy per tonne of castings.
Oil used = 75 kg oil/tonne x 10,000 kcal/kg = 7,50,000 kcal/tonne.
STEP 3 - Total energy per tonne of finished product.
Total = 8,88,667 + 7,50,000 = 16,38,667 kcal/tonne.
STEP 4 - Express as oil equivalent.
Oil equivalent = 16,38,667 / 10,000 = 163.87 kg oil/tonne of finished product.
ANSWER: ~163.9 kg (163.87 kg) oil equivalent per tonne of finished product.
Combine electrical melting energy (adjusted for 60% yield, converted kWh->kcal) with oil heat-treatment energy, then divide the total kcal by the oil GCV to get kg oil equivalent.
Source: unknown
📖 §5.5 Example 5.9 (dust balance) + §5 EOC Short Q S-3 (ash tie-component)
13. (a) A fuel used in a boiler contains 40% carbon and 23% ash. The refuse obtained after combustion is analysed and found to contain 7% carbon and the rest ash. Compute the percentage of the original carbon in the fuel that remains unburnt in the refuse. (5 marks) (b) During an ESP performance study, the inlet gas stream to the ESP is 2,89,920 Nm3/hr with a dust loading of 5500 mg/Nm3, and the outlet gas stream is 3,01,100 Nm3/hr with a dust loading of 110 mg/Nm3. How much fly ash is collected in the system in kg/hr? (5 marks)
Model answer: PART (a) - Unburnt carbon (use ash as the conserved/inert tracer).
Basis: 100 kg of refuse -> unburnt carbon = 7 kg, ash = 93 kg.
All the ash in the fuel reports to the refuse, and ash = 23% of the fuel.
So 93 kg ash corresponds to 23% of the fuel: quantity of raw fuel = 93 / 0.23 = 404.35 kg.
Original carbon in the fuel = 0.40 x 404.35 = 161.74 kg.
Unburnt carbon (in refuse) = 7 kg.
% of original carbon unburnt = (7 / 161.74) x 100 = 4.33%.
PART (b) - Fly ash collected (dust mass balance): Inlet dust = Outlet dust + Fly ash collected.
Inlet dust = 2,89,920 x 5500 / 1,000,000 = 1594.56 kg/hr.
Outlet dust = 3,01,100 x 110 / 1,000,000 = 33.12 kg/hr.
Fly ash collected = 1594.56 - 33.12 = 1561.44 kg/hr.
ANSWER: (a) 4.33% of the original carbon remains unburnt; (b) 1561.44 kg/hr of fly ash collected.
Part (a): ash is inert and conserved, so use it to back-calculate the fuel mass, then compare unburnt carbon to original carbon. Part (b): dust mass balance, collected = inlet loading x flow - outlet loading x flow.
Source: unknown
📖 §5.5 Example 5.6 — evaporator solids balance (improvement case)
14. In a Chlor-Alkali plant, an evaporator was designed to concentrate 500 kg of liquor containing 7% w/w solids to 45% solids w/w in the output. Presently the output from the evaporator has 30% solids w/w. The energy manager suggested overhauling the evaporator to achieve the design solids in the output. Calculate the percentage improvement in water removal in the evaporator after overhauling.
Model answer: Feed = 500 kg with 7% solids.
Solids in feed = 500 x 7/100 = 35 kg (conserved through the evaporator).
PRESENT case (output 30% solids):
Output (thick liquor) = solids / 0.30 = 35 / 0.30 = 116.7 kg.
Water removed = 500 - 116.7 = 383.3 kg.
DESIGN case (output 45% solids, after overhaul):
Output = 35 / 0.45 = 77.8 kg.
Water removed = 500 - 77.8 = 422.2 kg.
IMPROVEMENT:
Incremental water removal = 422.2 - 383.3 = 38.9 kg.
% improvement in water removal = 38.9 / 383.3 x 100 = 10.14%.
ANSWER: About 10.14% improvement in water removal after overhauling.
Solids are conserved; output mass = solids / (solids fraction). Compute water removed = feed - output for both present and design cases, then percentage improvement relative to the present case.
Source: unknown
📖 §5.6 Heat balances — enthalpy balance on a steam mixing point
15. Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 degC and 1 atm is required as feed to a heat exchanger. To produce it, the turbine discharge is mixed with superheated steam at 400 degC, 1 atm (specific volume 3.11 m3/kg). Calculate the amount of superheated steam at 300 degC produced and the volumetric flow rate of the 400 degC steam. (Enthalpies: saturated steam at 1 atm = 2676 kJ/kg; 400 degC steam = 3278 kJ/kg; 300 degC steam = 3074 kJ/kg.)
Model answer: Let m1 = mass flow of 400 degC steam (kg/h), m2 = mass flow of 300 degC product steam (kg/h).
STEP 1 - Mass balance of water:
1200 + m1 = m2 ... (1)
STEP 2 - Energy (enthalpy) balance:
(1200)(2676) + m1(3278) = m2(3074) ... (2)
Solve (1) and (2) simultaneously. Substitute m2 = 1200 + m1 into (2):
32,11,200 + 3278 m1 = (1200 + m1)(3074) = 36,88,800 + 3074 m1
3278 m1 - 3074 m1 = 36,88,800 - 32,11,200
204 m1 = 4,77,600
m1 = 2341.2 kg/h.
m2 = 1200 + 2341.2 = 3541.2 kg/h (superheated steam at 300 degC produced).
STEP 3 - Volumetric flow rate of the 400 degC steam (specific volume 3.11 m3/kg):
= 2341.2 kg/h x 3.11 m3/kg = 7281.1 m3/h.
ANSWER: 300 degC steam produced = 3541.2 kg/h; volumetric flow of 400 degC steam = 7281.1 m3/h.
Mixing problem solved with a mass balance and an enthalpy balance (two equations, two unknowns). Volumetric flow = mass flow x specific volume.
Source: unknown
📖 §5.2 Fig 5.1 / §5.5 — material balance with a recycle stream (dry-solids balance)
16. In a drying operation, the moisture content of the feed to a calciner must be held at 15% (w/w) to prevent lumping and sticking. This is achieved by mixing the fresh feed (30% moisture w/w) with a recycle stream of dried material (3% moisture w/w). What fraction of the dried product must be recycled?
Model answer: Let F = fresh feed, R = recycle, P = product (all mass units). Track SOLIDS (dry matter): feed is 70% solids, recycle and product are 97% solids, mixed stream to dryer is 85% solids (15% moisture).
STEP 1 - Solids balance at the MIXER:
0.70 F + 0.97 R = 0.85 (F + R)
0.70 F + 0.97 R = 0.85 F + 0.85 R
0.12 R = 0.15 F
R = 1.25 F ... (1)
STEP 2 - Solids balance at the DRYER (mixed stream in; product + recycle out, both at 97% solids):
0.85 (F + R) = 0.97 (P + R)
0.85 (F + 1.25 F) = 0.97 P + 0.97 x 1.25 F
0.85 x 2.25 F = 0.97 P + 1.2125 F
1.9125 F = 0.97 P + 1.2125 F
0.70 F = 0.97 P
F = 1.386 P ... (2)
STEP 3 - Recycle in terms of product. Substitute (2) into (1):
R = 1.25 x 1.386 P = 1.7325 P ... (3)
Product + Recycle = P + 1.7325 P = 2.7325 P.
STEP 4 - Recycle as a fraction of dried product stream:
R / (P + R) = 1.7325 P / 2.7325 P = 0.634 = 63.4%.
ANSWER: About 63.4% of the dried product must be recycled.
Recycle problem solved by two solids balances (mixer and dryer), because dry solids are conserved. Express R and F in terms of P, then the recycle fraction of the dried stream.
Source: unknown
📖 §5.7 Example 5.12 — paper drying machine (method)
17. A paper drying machine has a production capacity of 500 TPD and currently operates at an output of 480 TPD. The dryness of the paper is 60% at the inlet and 95% at the outlet. Steam is supplied at 4 kg/cm2 with a latent heat of 510 kcal/kg. The evaporated moisture is at about 100 degC with an enthalpy of 640 kcal/kg. The plant operates 24 hours per day. Assuming only the latent heat of steam is used for drying and neglecting the enthalpy of moisture in the wet paper, estimate (i) the quantity of moisture to be evaporated per hour and (ii) the input steam quantity required per hour.
Model answer: Output = 480 TPD at 95% dryness.
STEP 1 - Bone-dry paper (conserved).
Bone-dry mass at output = 480 x 0.95 = 456 TPD.
STEP 2 - Total wet paper at the inlet (60% dryness = 60% bone-dry).
Total wet paper in = 456 / 0.60 = 760 TPD.
STEP 3 - Moisture evaporated per hour.
Moisture evaporated = (inlet wet - outlet) = (760 - 480) = 280 TPD.
Per hour = 280 / 24 = 11.67 TPH.
STEP 4 - Steam required (heat balance).
Heat to evaporate = moisture x enthalpy of evaporated moisture = 11.67 x 640 kcal (per T basis).
Steam = (11.67 x 640) / 510 = 14.6 TPH.
ANSWER: Moisture evaporated ~ 11.67 TPH; Steam required ~ 14.6 TPH.
Bone-dry paper is conserved; find total inlet wet mass from outlet dry mass and inlet dryness. Moisture evaporated = inlet - outlet. Steam = heat carried by evaporated moisture / latent heat of steam.
Source: unknown
📖 §5.7 Example 5.12 — paper machine evaporation and steam
18. The production through a paper machine is 300 tonnes per day (TPD). The inlet and outlet dryness to the paper machine are 50% and 95% respectively. The evaporated moisture temperature is 80 degC. Steam is supplied at 3.5 kg/cm2 (latent heat 513 kcal/kg). Assuming 24 hours/day operation and enthalpy of evaporated moisture = 632 kcal/kg, estimate (i) the quantity of moisture to be evaporated in kg/hr and (ii) the steam quantity required for evaporation in kg/hr.
Model answer: Production = 300 TPD = 300/24 = 12.5 TPH of paper.
STEP 1 - Moisture to be evaporated (bone-dry paper conserved).
Bone-dry paper in product (95% dry) = 12.5 x 0.95 = 11.875 TPH.
Moisture after dryer (5% of product) = 12.5 - 11.875 = 0.625 TPH.
At the inlet the paper is 50% dry, so on the same 11.875 TPH of dry solids the total inlet wet mass = 11.875 / 0.50 = 23.75 TPH, and moisture before dryer = 23.75 - 11.875 = 11.875 TPH.
Evaporated moisture = 11.875 - 0.625 = 11.25 TPH = 11,250 kg/hr.
STEP 2 - Steam required (heat balance).
Heat carried by evaporated moisture = 632 x 11,250 = 71,10,000 kcal/hr.
Steam required = 71,10,000 / 513 = 13,859.6 kg/hr.
ANSWER: Moisture evaporated = 11,250 kg/hr; Steam required = 13,859.6 kg/hr.
Same method as the book Example 5.12 but with 50% inlet dryness. Track bone-dry paper, find moisture before and after, evaporated = difference; steam = heat in moisture / latent heat.
Source: unknown
📖 §5.8 Solved Example — evaporator mass + enthalpy balance (method)
19. An evaporator is fed with 5000 kg/hr of a solution having 0.5% solids at 38 degC, and is concentrated to 1% solids. Steam enters at total enthalpy 640 kcal/kg and condensate leaves at 100 degC. Enthalpy of feed = 38.1 kcal/kg, product solution = 100.8 kcal/kg, vapour = 640 kcal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour.
Model answer: STEP 1 - Mass (solids) balance.
Solids in feed = 5000 x 0.5/100 = 25 kg/hr (conserved).
Product (thick liquor) at 1% solids: mass x 1/100 = 25 -> product = 25 / 0.01 = 2500 kg/hr.
Vapour formed = 5000 - 2500 = 2500 kg/hr.
STEP 2 - Heat (enthalpy) balance. Steam gives up (640 - 100) = 540 kcal/kg.
Heat by steam + heat in feed = heat in vapour + heat in thick liquor
M x 540 + 38.1 x 5000 = 640 x 2500 + 100.8 x 2500
M x 540 + 1,90,500 = 16,00,000 + 2,52,000 = 18,52,000
M x 540 = 18,52,000 - 1,90,500 = 16,61,500
M (steam) = 16,61,500 / 540 = 3076.8 kg/hr.
ANSWER: Vapour formed = 2500 kg/hr; Steam used = 3076.8 kg/hr.
Same structure as the book Solved Example: solids balance for vapour/product, then enthalpy balance with steam latent heat = 640 - 100 = 540 kcal/kg.