121 questions — 86 objective (1 mark), 18 short (5 marks), 17 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) — 86
📖 §7.2 Range, Approach & Effectiveness
1. In a cooling tower, the 'Approach' is correctly defined as the difference between:
Hot water inlet temperature and cold water outlet temperature
Cold water outlet temperature and the wet bulb temperature (WBT)
Hot water inlet temperature and the dry bulb temperature (DBT)
Dry bulb temperature and wet bulb temperature
Answer: B) Cold water outlet temperature and the wet bulb temperature (WBT)
Confirmed vs Book-3 §7.2 — 'Approach is the difference between the cooling tower outlet cold water temperature and ambient wet bulb temperature.'
Approach = Cold water out − WBT; it is the better indicator of tower performance because a lower approach means the tower has pushed the water closer to its theoretical limit (the WBT).
Option a) is Range (hot in − cold out), which is set by the process heat load, not by the tower; DBT never appears in either definition.
Source: AI practice
📖 §7.2 Range, Approach & Effectiveness
2. A cooling tower receives water at 42 °C and cools it to 33 °C, with a wet bulb temperature of 28 °C. The Range and Approach respectively are:
Range = 9 °C, Approach = 5 °C
Range = 5 °C, Approach = 9 °C
Range = 14 °C, Approach = 4 °C
Range = 9 °C, Approach = 14 °C
Answer: A) Range = 9 °C, Approach = 5 °C
Confirmed vs Book-3 §7.2 — Range = hot water in − cold water out = 42 − 33 = 9 °C. Approach = cold water out − WBT = 33 − 28 = 5 °C.
Range uses water temperatures only; Approach pairs the COLD water with the wet bulb.
Option d) (Approach = 14 °C) comes from using the hot water temperature (42 − 28), the standard trap.
Source: AI practice
📖 §7.2 Range, Approach & Effectiveness
3. If a cooling tower has a Range of 6 °C and an Approach of 4 °C, its effectiveness is:
40%
60%
66.7%
150%
Answer: B) 60%
Confirmed vs Book-3 §7.2 — Effectiveness % = Range / (Range + Approach) × 100 = 6 / (6 + 4) × 100 = 60%.
The denominator is the IDEAL range, i.e. hot water in − WBT, which equals Range + Approach.
Option c) 66.7% is Range/Approach × ... i.e. dividing by the wrong quantity; effectiveness can never exceed 100%, which rules out option d).
Source: AI practice
📖 §7.2 Cycles of Concentration (C.O.C.)
4. Cycles of Concentration (COC) in a cooling tower is defined as:
TDS in make-up water / TDS in blowdown water
TDS in circulating (blowdown) water / TDS in make-up water
Blowdown rate / Evaporation rate
Make-up rate / Evaporation rate
Answer: B) TDS in circulating (blowdown) water / TDS in make-up water
Confirmed vs Book-3 §7.2 — 'Cycles of concentration (C.O.C) is the ratio of dissolved solids in circulating water to the dissolved solids in make up water.'
By water balance this also equals Make-up / Blow down, since the salts entering with make-up must leave with blowdown.
Option a) is the same ratio inverted, which would give a value below 1 — impossible for COC.
Source: AI practice
📖 §7.2 Blow Down Losses
5. As the Cycles of Concentration (COC) of a cooling tower is increased, the blowdown requirement:
Increases
Decreases
Remains unchanged
First increases then becomes constant
Answer: B) Decreases
Confirmed vs Book-3 §7.2 — Blow Down = Evaporation Loss / (C.O.C. − 1). Raising COC increases the denominator, so blowdown falls.
Less blowdown also means less make-up water, which is why raising COC (within the limits of water chemistry) is a standard water- and energy-saving measure.
At COC = 1 the denominator is zero: all the water left after evaporation would have to be blown down — the book's extreme case.
Source: AI practice
📖 §7.2 Evaporation Loss
6. A cooling tower circulates 1000 m³/hr of water over a Range of 8 °C. Using the book formula, the evaporation loss is approximately:
6.12 m³/hr
12.24 m³/hr
1.53 m³/hr
24.48 m³/hr
Answer: B) 12.24 m³/hr
Confirmed vs Book-3 §7.2 — Evaporation Loss (m³/hr) = 0.00085 × 1.8 × circulation rate (m³/hr) × (T₁ − T₂) = 0.00085 × 1.8 × 1000 × 8 = 12.24 m³/hr.
That is about 1.2% of the circulation rate, the sanity check the book implies (roughly 1% per 5.5 °C of range).
Option a) 6.12 m³/hr is the result of dropping the 1.8 factor — memorise the constant as the pair 0.00085 × 1.8.
Source: AI practice
📖 §7.2 Blow Down Losses
7. If evaporation loss is 12 m³/hr and the tower operates at a COC of 3, the blowdown is:
4 m³/hr
6 m³/hr
3 m³/hr
36 m³/hr
Answer: B) 6 m³/hr
Confirmed vs Book-3 §7.2 — Blow Down = Evaporation Loss / (C.O.C. − 1) = 12 / (3 − 1) = 6 m³/hr.
The denominator is (COC − 1), not COC — that single term is what the question is testing.
Option a) 4 m³/hr is 12/3, i.e. dividing by COC itself.
Source: AI practice
📖 §7.1 Components of a Cooling Tower — Fill
8. Which component of a cooling tower has the MAXIMUM effect on its thermal performance?
Fill media
Drift eliminators
Louvers
Casing
Answer: A) Fill media
Confirmed vs Book-3 §7.1 — the book states that most towers employ fills 'to facilitate heat transfer by maximising water and air contact', and the end-of-chapter objective question confirms fill media has the maximum effect on cooling tower performance.
Fill increases surface area, contact time and turbulence between water and air; film fill gives the same heat transfer in a smaller volume than splash fill.
Drift eliminators only capture entrained droplets (a water-loss item), and louvers and casing are structural — none of them drive thermal performance.
Source: AI practice
📖 §7.1 Cooling Tower Types — Natural Draft
9. Natural draft (hyperbolic) cooling towers are generally used in:
Small commercial air-conditioning plants
Power and utility stations
Domestic rooftop cooling units
Small batch process industries
Answer: B) Power and utility stations
Confirmed vs Book-3 §7.1 — natural draft (hyperbolic) towers rely on a very tall concrete chimney to create the draft, and 'because of the large size of these towers, they are generally used for water flow rates above 45,000 m³/hr'.
That scale exists almost only in power and utility stations, which is also the answer given in the book's own objective question.
Small commercial, domestic and small-process duties use mechanical draft (forced or induced) towers instead.
(Options were rewritten: the flow figures previously attached to individual options signposted the answer, so the >45,000 m³/hr threshold now sits here in the explanation.)
10. The coldest (closest) approach that a cooling tower manufacturer will normally guarantee is about:
0.5 °C
1.4 °C
2.8 °C
5.6 °C
Answer: C) 2.8 °C
Confirmed vs Book-3 §7.3 — 'Usually a 2.8 °C approach to the design wet bulb is the coldest water temperature that cooling tower manufacturers will guarantee.'
The closer the approach to the wet bulb, the larger and more expensive the tower, because the driving thermal potential shrinks.
Option a) 0.5 °C would mean cooling almost exactly to the WBT, which the book says is impossible under heat load.
Source: AI practice
📖 §7.6 Case Study — VFD for Cooling Tower Fan
11. Because cooling tower fan power varies with the cube of air flow, the most effective energy-saving retrofit on the fan is to:
Increase the fan motor supply voltage
Install a Variable Frequency Drive to vary the fan air flow with load
Keep the fan at full speed and control the cold water temperature by blowdown
Increase the water circulation rate through the tower
Answer: B) Install a Variable Frequency Drive to vary the fan air flow with load
Confirmed vs Book-3 §7.6 — the book's case study notes that cooling tower fan power follows 'a cube law' with air flow, so reducing air flow when ambient WBT or load is low gives large savings; the recommended implementation is a VFD with closed-loop control on the cold water outlet temperature.
Option c) is the near-miss: blowdown controls water chemistry (COC), not the cold water temperature, and wastes treated water.
Raising the circulation rate or the supply voltage increases power draw, not savings.
Source: AI practice
📖 §7.3 Factors that affect cooling tower SIZE
12. Holding heat load, approach and WBT constant, the physical SIZE (and cost) of a cooling tower varies with the Range as follows:
Directly with range
Inversely with range
Independent of range
Directly with the square of range
Answer: B) Inversely with range
Confirmed vs Book-3 §7.3 — with the other three of {heat load, range, approach, WBT} held constant, tower size varies DIRECTLY with heat load and INVERSELY with the range, the approach and the entering WBT.
The book's own illustration: a tower cooling 4540 m³/hr through a 13.9 °C range is larger than one cooling 4540 m³/hr through a 19.5 °C range.
Option a) is the trap — a bigger range sounds like more duty, but for a FIXED heat load a bigger range means less water to handle, hence a smaller tower.
13. A cooling tower reduces the temperature of water from 40°C to 30°C. If the mass flow rate of water is 5 kg/s, what is the heat removed by the cooling tower?
500 kW
209 kW
2 kW
5 kW
Answer: B) 209 kW
Confirmed vs Book-3 §7.2 (iv) — Cooling capacity = mass flow × specific heat × temperature difference (Range). Q = 5 kg/s × 4.186 kJ/kg°C × (40−30)°C = 209.3 kW → option (b). Option (a) 500 kW ignores Cp; (c)/(d) are dimensionally meaningless. In book units this is 18,000 kg/h × 1 kcal/kg°C × 10°C = 1,80,000 kcal/h ≈ 209 kW.
Source: Sep 2024
📖 §7.2 Cooling Tower Performance (i) Range
14. The approach temperature of a cooling tower is 5°C, and the range is 10°C. If the inlet water temperature is 40°C, what is the outlet water temperature?
25°C
30°C
35°C
45°C
Answer: B) 30°C
Confirmed vs Book-3 §7.2 (i) — Range = hot water inlet − cold water outlet, so outlet = 40 − 10 = 30°C → (b). The 5°C approach (cold water − WBT) is a distractor: it would only be needed to find the WBT (30 − 5 = 25°C, which is option (a), a common trap).
Source: Sep 2024
📖 §7.5 Energy Saving Opportunities in Cooling Towers
15. How can the performance of a cooling tower be improved?
Proper water treatment
Regular maintenance
Optimizing air and water flow
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §7.5 — The book's energy-saving list covers all three: water treatment (algae/scale block nozzles and cut ΔT – §7.3), periodic cleaning of nozzles/fill and fan balance, and matching water and air flow (monitor L/G, blade angle, VFD). Hence (d) all of the above.
Source: Sep 2024
📖 §7.2 Factors Affecting Performance – Fill Media Effects
16. The most influential component for cooling tower performance is:
Fill media
Drift eliminator
Casing
Fan motor
Answer: A) Fill media
Confirmed vs Book-3 §7.2 Fill Media Effects — Book: fill media is responsible for surface area, contact time and turbulence of heat exchange – i.e. the whole heat/mass transfer – and efficient fill directly cuts fan power and pumping head. Drift eliminator, casing and fan motor do not set the transfer area. Same as Book EOC Q8.
Source: Sep 2025
📖 §7.2 Cooling Tower Performance (vii) Blow down
17. A cooling tower has an evaporation loss of 12 m³/hr and COC of 2.5. What will be the blowdown loss in m³/hr?
5.2
8.0
9.6
10.2
Answer: B) 8.0
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation Loss / (COC − 1) = 12 / (2.5 − 1) = 12 / 1.5 = 8.0 m³/hr → (b). Option (a) 5.2 would be 12/2.3, (c) 9.6 is 12/1.25 – both mis-apply the (COC − 1) denominator.
Source: Sep 2025
📖 §7.1 Components of Cooling Tower – Fill
18. The main function of fill media in a cooling tower is to:
Reduce drift losses
Increase water–air contact
Reduce fan noise
Filter suspended solids
Answer: B) Increase water–air contact
Confirmed vs Book-3 §7.1 Components – Fill — Book: fills 'facilitate heat transfer by maximising water and air contact' (splash or film type). Reducing drift is the drift eliminator's job, and fill neither reduces fan noise nor filters solids (side-stream filters do that).
Source: Sep 2025
📖 §7.5 Energy Saving Opportunities in Cooling Towers
19. Energy-saving opportunities in cooling towers include:
Optimizing fan blade angle seasonally
Maintaining correct water chemistry
Cleaning fill media regularly
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §7.5 — All three appear in the book's §7.5 list / §7.3: 'optimise cooling tower fan blade angle on a seasonal and/or load basis', correct water chemistry (cooling water treatment prevents scale/algae that block nozzles) and cleaning fill/nozzles. Hence (d).
Source: Sep 2025
📖 §7.2 Cooling Tower Performance (viii) L/G ratio
20. The L/G ratio in a cooling tower is:
Ratio of liquid water mass flow to gas (air) mass flow
Ratio of drift loss to make-up water
Ratio of cooling load to fan power
Ratio of TDS in blowdown to TDS in make-up water
Answer: A) Ratio of liquid water mass flow to gas (air) mass flow
Confirmed vs Book-3 §7.2 (viii) — Book: 'Liquid/Gas (L/G) ratio of a cooling tower is the ratio between the water and the air mass flow rates' (kg/kg), from L(T1−T2) = G(h2−h1). Option (d) describes COC, (b) and (c) are not book parameters.
21. In a cooling tower, if any three of the four parameters: heat load, range, approach, and wet-bulb temperature, are kept constant, the required tower size will vary
Directly with the heat load
Inversely with the range
Inversely with the approach
All the above
Answer: D) All the above
Confirmed vs Book-3 §7.2 Factors that affect cooling tower size — Book sidebar: when three of heat load, range, approach and WBT are held constant, tower size varies directly with heat load, inversely with range, inversely with approach and inversely with entering WBT. All three listed relations are book statements → (d).
22. The ratio of dissolved solids in circulating water to the dissolved solids in make up water is termed as
liquid gas ratio
cycles of concentration
cooling tower effectiveness
none of the above
Answer: B) cycles of concentration
Confirmed vs Book-3 §7.2 (vi) — Book definition: 'Cycles of concentration (C.O.C) is the ratio of dissolved solids in circulating water to the dissolved solids in make up water' → (b). L/G ratio is a mass-flow ratio (water/air) and effectiveness is Range/(Range + Approach) – neither involves dissolved solids.
Source: Book EOC
📖 §7.2 Cooling Tower Performance (ii) Approach (Book EOC Q2)
23. If inlet and outlet water temperatures of a cooling tower are 40 C and 32 C respectively and atmospheric DBT and WBT are 35 C and 28 C respectively, then the approach of the cooling tower is
3 C
4 C
5 C
7 C
Answer: B) 4 C
Confirmed vs Book-3 §7.2 (ii) — Approach = cold (outlet) water temperature − ambient WET bulb temperature = 32 − 28 = 4°C → (b). The DBT (35°C) is a distractor: 35 − 32 = 3°C (option a) is wrong; 40 − 32 = 8°C would be the Range.
24. Higher the COC in a cooling tower, the blow down quantity will
increases
decreases
no change
none of the above
Answer: B) decreases
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation Loss / (COC − 1). A higher COC enlarges the denominator, so blowdown (and make-up) decreases → (b). This is why the book lists 'COC improvement measures for water savings' in §7.5.
25. The L/G ratio of a cooling tower does not depend on
range
enthalpy of inlet air
outlet wet bulb temperature
dry bulb temperature
Answer: D) dry bulb temperature
Confirmed vs Book-3 §7.2 (viii) — From the heat balance L(T1−T2) = G(h2−h1): L/G = (h2 − h1)/(T1 − T2), where h1, h2 are air enthalpies at inlet and exhaust WET-bulb temperatures and T1−T2 is the range. Dry bulb temperature does not enter the relation → (d).
26. Which of the following ambient conditions will evaporate the minimum amount of water in a cooling tower?
35 C DBT and 30 C WBT
38 C DBT and 31 C WBT
38 C DBT and 37 C WBT
35 C DBT and 29 C WBT
Answer: C) 38 C DBT and 37 C WBT
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Evaporation depends on how much moisture the air can still absorb, i.e. the wet-bulb depression (DBT − WBT). Option (c) has only 1°C depression (38/37 – air almost saturated), so it evaporates the least; (d) with 6°C depression evaporates the most.
27. Natural draft cooling towers are mainly used in
steel industry
alumina industry
fertilizer industry
power stations
Answer: D) power stations
Confirmed vs Book-3 §7.1 — Book: natural draft towers use very large concrete chimneys, are used for flows above 45,000 m³/hr and 'are used only by utility power stations' → (d). Steel, alumina and fertilizer plants use mechanical draft towers.
Confirmed vs Book-3 §7.2 (iii) — Book: effectiveness = Range / ideal range, where ideal range = inlet water temp − WBT = Range + Approach; i.e. Range/(Range + Approach) → (a). Option (c) Range/Approach can exceed 1 and is not the book definition.
Source: Book EOC
📖 §7.2 Factors Affecting Performance – Fill Media Effects (Book EOC Q8)
29. Which one of the following has the maximum effect on cooling tower performance?
fill media
drift
louvers
casing
Answer: A) fill media
Confirmed vs Book-3 §7.2 Fill Media Effects — Book: heat exchange is governed by surface area, contact time and turbulence, and 'fill media in a cooling tower is responsible to achieve all of above' → (a). Drift eliminators, louvers and casing only contain water/air; they do not create transfer area.
30. Normally the guaranteed best approach a cooling tower can achieve is
5 C
8 C
12 C
2.8 C
Answer: D) 2.8 C
Confirmed vs Book-3 §7.2 Range/Approach — Book: 'Usually a 2.8°C approach to the design wet bulb is the coldest water temperature that cooling tower manufacturers will guarantee' → (d). A closer approach needs a disproportionately larger, costlier tower (Table 7.1).
Source: Book EOC
📖 §7.2 Cooling Tower Performance (ii) Approach (Book EOC Q10)
31. Select the wrong statement.
for a given heat rejection duty, a higher range will reduce the circulating water flow rate
when the cycle of concentration is left at one, all water left in the cooling tower after evaporation needs to be removed as blowdown
a better indicator for cooling tower performance is Range
a cooling tower size will be greater for 20 C Wet bulb temperature (WBT) than for a 30 C WBT, for the same circulation, range and approach
Answer: C) a better indicator for cooling tower performance is Range
Confirmed vs Book-3 §7.2 (ii) — Book: 'the Approach is a better indicator of cooling tower performance' – so statement (c) naming Range is the wrong one. (a) is true (Range = heat load/flow), (b) is true (at COC = 1 nothing may concentrate, so all remaining water is blowdown), (d) is true (size varies inversely with WBT).
32. A plant wants to replace the existing 100 TR water cooled vapour compression refrigeration system with a waste heat driven vapour absorption chiller. The capacity of the existing cooling tower
needs no change
is to be doubled
is to be raised to 1.2 times
none of the above
Answer: B) is to be doubled
Confirmed vs Book-3 §7.2 Heat Load — Book heat-rejection table: Refrigeration, Compression = 63 kcal/min/TR; Refrigeration, Absorption = 127 kcal/min/TR – almost exactly double. So replacing a 100 TR VCR with a VAM of the same capacity roughly doubles the heat rejected to the cooling tower → (b). '1.2 times' (c) understates it badly.
33. If water is flowing through a cooling tower at 120 m3/h with 5oC range, the load on cooling tower at an ambient wet bulb temperature of 33 oC is
198.4 TR
357 TR
158 TR
none of the above
Answer: A) 198.4 TR
Confirmed vs Book-3 §7.2 (iv) — Load = 120 m³/h × 1000 kg/m³ × 1 kcal/kg°C × 5°C = 6,00,000 kcal/h; ÷ 3024 kcal/h per TR = 198.4 TR → (a). This is Book EOC S-1. The 33°C WBT only affects approach, not the heat load.
Source: 15th Exam
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
34. The wet bulb temperature normally chosen for designing of cooling tower is
average maximum wet bulb for rainy months
average maximum wet bulb for summer months
average minimum wet bulb for summer months
average maximum wet bulb for winter months
Answer: B) average maximum wet bulb for summer months
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Book: 'The temperature selected is generally close to the average maximum wet bulb for the summer months' (a value not exceeded more than 5% of the time) → (b). Designing for winter or minimum WBT would undersize the tower for the worst case.
Source: 15th Exam
📖 §7.2 Choosing a Cooling Tower (Table 7.4)
35. Which one of the following types of cooling towers consumes least power for the same operating conditions?
counter flow film fill cooling tower
cross-flow splash fill cooling tower
counter flow splash fill cooling tower
none of the above
Answer: A) counter flow film fill cooling tower
Confirmed vs Book-3 §7.2 Choosing a Cooling Tower — Book Table 7.4 (16,000 m³/hr, 41.5→32.5°C, WBT 27.6°C): power at motor terminal per tower = 253 kW (counter-flow film fill), 310 kW (counter-flow splash) and 330 kW (cross-flow splash); 'the power consumption is least in Counter Flow Film Fill' → (a). Film fill needs less air and lower pumping head (Table 7.3).
Source: 15th Exam
📖 §7.2 Cooling Tower Performance (i) Range
36. The range of a cooling tower with inlet and outlet temperature 41°C and 32°C respectively and wet bulb temperature 29°C is
9 °C
3 °C
29 °C
12 °C
Answer: A) 9 °C
Confirmed vs Book-3 §7.2 (i) — Range = cooling tower water inlet − outlet = 41 − 32 = 9°C → (a). 3°C (b) is the Approach (32 − 29) and 12°C (d) is the ideal range (41 − 29) – both are distractors.
37. Find the correct equation, if M = makeup water, E = evaporation losses, B = blow-down losses and D = drift losses of a cooling tower:
M = E + B + D
M = E + B - D
M = E - B + D
M = E - B - D
Answer: A) M = E + B + D
Confirmed vs Book-3 §7.2 (v)–(vii) — Make-up water must replace every water loss from the circuit: evaporation (E), blow-down (B) and drift (D). Book: make-up = evaporation + blow down (+ drift) → M = E + B + D → (a). Any minus sign would imply a loss adds water.
Source: 14th Exam
📖 §7.2 Cooling Tower Performance (vii) Blow down
38. For a cooling tower, if blowdown is 10 m3/hour and Cycles of Concentration (CoC) is 2.5, the evaporation loss is equal to:
25 m3/hour
15 m3/hour
0.25 m3/hour
6.67 m3/hour
Answer: B) 15 m3/hour
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1) → Evaporation = Blowdown × (COC − 1) = 10 × (2.5 − 1) = 15 m³/hr → (b). 25 (a) would be 10 × 2.5 (forgetting the −1); 6.67 (d) inverts the relation.
Source: 14th Exam
📖 §7.2 Cooling Tower Performance (ii) Approach
39. Which of the following is incorrect in the case of cooling towers?
'Range' is the difference between the cooling tower water inlet and outlet temperature
'Approach' is the difference between the outlet cold water temperature and ambient wet bulb temperature
'Range' is a better indicator of cooling tower performance
Cooling capacity is the heat rejected in kCal/hr or TR
Answer: C) 'Range' is a better indicator of cooling tower performance
Confirmed vs Book-3 §7.2 (ii) — Book: both range and approach should be monitored, but 'the Approach is a better indicator of cooling tower performance' – so (c), which credits Range, is the incorrect statement. (a), (b) and (d) are the book's own definitions of range, approach and cooling capacity (kcal/hr or TR).
Source: Set-A
📖 §7.2 Factors Affecting Performance – Fill Media Effects
40. In a cooling tower, Statement A: Surface of heat exchange is the surface area of the water droplets in contact with air. Statement B: Area of heat exchange is the surface area of the fill sheets in contact with air. Judgement:
statements A & B are false
statement A is True & B is false
statements A & B are True
statement A is false & B is True
Answer: C) statements A & B are True
Confirmed vs Book-3 §7.2 Fill Media Effects — Book: with splash fill 'surface of heat exchange is the surface area of the water droplets, which is in contact with air' (Statement A) and in film fill 'area of heat exchange is the surface area of the fill sheets, which is in contact with air' (Statement B). Both are book sentences → (c).
Source: Set-A
📖 §7.2 Cooling Tower Performance (vii) Blow down
41. If the evaporation loss is 16 m3/hr per cell and Cycles of Concentration is 3, the blow down requirement in m3/hr per cell of a cooling tower is:
8
5.33
4
2
Answer: A) 8
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1) = 16/(3 − 1) = 8 m³/hr per cell → (a). 5.33 (b) is 16/3 (forgetting the −1); 4 (c) is 16/4.
Source: Set-A
📖 §7.2 Cooling Tower Performance (vi) Cycles of Concentration
42. Cycles of Concentration (C.O.C) of a cooling tower will depend on
TDS in circulating water
TDS in make-up water
both a & b
none of the above
Answer: C) both a & b
Confirmed vs Book-3 §7.2 (vi) — Book: COC is the ratio of dissolved solids in circulating water to dissolved solids in make-up water – a ratio of two TDS values, so it depends on both → (c).
Source: Set-A
📖 §7.2 Cooling Tower Performance (i) Range
43. The term "cooling range" in a cooling tower refers to the difference in the temperature of:
dry bulb and wet bulb
hot water entering the tower and the wet bulb temperature of the surrounding air
cold water leaving the tower and the wet bulb temperature of the surrounding air
hot water entering the tower and the cooled water leaving the tower
Answer: D) hot water entering the tower and the cooled water leaving the tower
Confirmed vs Book-3 §7.2 (i) — Book: 'Range is the difference between the cooling tower water inlet and outlet temperature' → (d). Option (c) is the Approach; (b) is the ideal range used in the effectiveness formula; (a) is the wet-bulb depression of air.
Source: 17th Sep-2016
📖 §7.2 Cooling Tower Performance (ii) Approach
44. A better indicator for cooling tower performance is:
Heat load in tower
Range
RH of air leaving cooling tower
Approach
Answer: D) Approach
Confirmed vs Book-3 §7.2 (ii) — Book: 'Although both range and approach should be monitored, the Approach is a better indicator of cooling tower performance' → (d). Range and heat load are fixed by the process, not by the tower (§7.2 Range).
Source: 17th Sep-2016
📖 §7.2 Cooling Tower Performance (vii) Blow down
45. The blow down requirement in m3/hr of a cooling tower with evaporation rate of 16 m3/hr and CoC of 3 is:
4
2
8
16
Answer: C) 8
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1) = 16/(3 − 1) = 8 m³/hr → (c). 16 (d) ignores COC; 4 (a) divides by COC + 1; 2 (b) has no basis.
Source: 17th Sep-2016
📖 §7.1 Components of Cooling Tower – Drift eliminators
46. What is the function of drift eliminators in cooling towers?
maximize water and air contact
capture water droplets escaping with air stream
enables entry of air to the cooling tower
eliminates uneven distribution of water into the cooling tower
Answer: B) capture water droplets escaping with air stream
Confirmed vs Book-3 §7.1 Components – Drift eliminators — Book: drift eliminators 'capture water droplets entrapped in the air stream that otherwise would be lost to the atmosphere' → (b). Maximising contact is the fill (a); air entry/equalising is the air inlet/louvers (c); uniform water distribution is the nozzles (d).
Source: Sep 2019
📖 §7.2 Cooling Tower Performance (ii) Approach
47. The indicator of cooling tower performance is best assessed by:
wet bulb temperature
dry bulb temperature
range
approach
Answer: D) approach
Confirmed vs Book-3 §7.2 (ii) — Book: Approach (cold water outlet − ambient WBT) 'is a better indicator of cooling tower performance' → (d). WBT is an ambient input, DBT is irrelevant, and range is fixed by process heat load and flow.
48. The cooling tower size is _____ with the entering WBT when heat load, range and approach are constant:
directly proportional
inversely proportional
constant
none of above
Answer: B) inversely proportional
Confirmed vs Book-3 §7.2 Factors that affect cooling tower size — Book sidebar: with heat load, range and approach constant, tower size varies 'inversely with the entering WBT' → (b). Air at a higher WBT can pick up more heat (§7.2 Approach & WBT), so a smaller tower suffices.
Source: 16th Exam
📖 §7.2 Cooling Tower Performance (viii) L/G ratio
49. L / G ratio in cooling tower is the ratio of:
length and girth
length and gradient of temperature
water mass flow rate and air mass flow rate
water volume flow rate and air volume flow rate
Answer: C) water mass flow rate and air mass flow rate
Confirmed vs Book-3 §7.2 (viii) — Book: L/G 'is the ratio between the water and the air mass flow rates' (kg/kg) → (c). It is a mass ratio, not a volume ratio (d) – the air volume must be multiplied by density (1.08 kg/m³ in the book trial).
Source: 16th Exam
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
50. Which among the following inlet air conditions would result in the best cooling tower performance?
air with lowest wet bulb temperature and high relative humidity
air with lowest wet bulb temperature and low relative humidity
air with same dry bulb and wet bulb temperature
air with high dry bulb temperature and high moisture.
Answer: B) air with lowest wet bulb temperature and low relative humidity
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Book: WBT of entering air 'is a controlling factor from the aspect of minimum cold water temperature'; the lower the WBT (and the drier the air), the greater the evaporative driving force and the colder the water → (b). Air with DBT = WBT (c) is saturated and can evaporate almost nothing.
51. As the 'approach' increases while other parameters remain constant, the effectiveness of a cooling tower:
increases
remains unchanged
decreases
none of the above
Answer: C) decreases
Confirmed vs Book-3 §7.2 (iii) — Effectiveness = Range/(Range + Approach). With range constant, a larger approach increases the denominator, so effectiveness falls → (c). A low approach is what makes a tower 'effective'.
Source: 18th Exam
📖 §7.1 Components of Cooling Tower – Fans
52. Which among the following types of fans is predominantly used in cooling towers?
centrifugal fan
axial fan
radial fan
all the above
Answer: B) axial fan
Confirmed vs Book-3 §7.1 Components – Fans — Book: 'Both axial (propeller type) and centrifugal fans are used… Generally, propeller fans are used in induced draft towers' (the dominant type), while centrifugal fans appear only in some forced-draft units. Hence axial (propeller) fans predominate → (b).
Source: 18th Exam
📖 §7.2 Cooling Tower Performance (vii) Blow down
53. The blow down requirement in m3/hr of a cooling tower for site Cycle of Concentration of 2.5 and approach of 4°C is:
10
0.63
1.6
Data not sufficient to calculate
Answer: D) Data not sufficient to calculate
Confirmed vs Book-3 §7.2 (v),(vii) — Blow Down = Evaporation/(COC − 1) and evaporation = 0.00085 × 1.8 × circulation × range. Only COC (2.5) and approach (4°C) are given – neither circulation rate nor range – so blowdown cannot be computed → (d). Approach never enters the blowdown formula.
54. The cooling tower size is _____ to the entering Wet Bulb Temperature (WBT), when the heat load, range and approach are constant.
Directly proportional
Inversely proportional
Constant
None of above
Answer: B) Inversely proportional
Confirmed vs Book-3 §7.2 Factors that affect cooling tower size — Book sidebar: when heat load, range and approach are constant, tower size varies 'inversely with the entering WBT' → (b). Book example: a tower for 4.45°C approach to 21.11°C WBT is larger than the same duty at 26.67°C WBT.
Source: 19th Exam
📖 §7.2 Cooling Tower Performance (viii) L/G ratio
55. L/G ratio in a cooling tower is the ratio of _________________.
Length and girth
Length and Temperature gradient
Water flow rate and air mass flow rate
Air mass flow rate and water flow rate
Answer: C) Water flow rate and air mass flow rate
Confirmed vs Book-3 §7.2 (viii) — Book: L/G = ratio between the water (liquid) and air (gas) mass flow rates, L over G → (c). Option (d) inverts it (G/L); (a) and (b) are nonsense expansions.
Source: 19th Exam
📖 §7.2 Cooling Tower Performance (vii) Blow down
56. Increasing the Cycles of Concentration (C.O.C) of circulating water in a cooling tower, the blow down quantity will
Increase
Decrease
Not change
None of the above
Answer: B) Decrease
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1): raising COC increases the denominator, so blowdown decreases → (b). Book §7.5: 'Consider COC improvement measures for water savings'.
Source: 19th Exam
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
57. Which of the following ambient conditions will evaporate minimum amount of water in a cooling tower?
35 °C DBT and 30 °C WBT
38 °C DBT and 31 °C WBT
38 °C DBT and 37 °C WBT
35 °C DBT and 29 °C WBT
Answer: C) 38 °C DBT and 37 °C WBT
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Evaporation is limited by the moisture-absorbing capacity of the air, indicated by DBT − WBT. Option (c) has the smallest depression (1°C – near-saturated air) so the least water evaporates; (d) with 6°C depression evaporates the most. Same as Book EOC Q5.
58. A two stage air compressor drawing 75 kW has heat rejection of 862 kCal/kWh. The required capacity of the cooling tower when the operating temperature difference is 5 °C will be ________ TR.
21.55
107.5
22.93
57.4
Answer: A) 21.55
Confirmed vs Book-3 §7.2 Heat Load — Heat rejected = 75 kW × 862 kcal/kWh = 64,650 kcal/hr (book table: two-stage compressor with intercooler and after cooler = 862 kcal/kW/hr). TR = 64,650/3000 = 21.55 TR (exam key uses 3000 kcal/hr per TR; with 3024 it is 21.4 TR – option (a) either way). The 5°C ΔT is only needed for the cooling-water flow, not the TR load.
Source: 19th Exam
📖 §7.2 Factors Affecting Performance – Fill Media Effects
59. Which of the following component has maximum effect on cooling tower performance?
Fill media
drift
louvers
casing
Answer: A) Fill media
Confirmed vs Book-3 §7.2 Fill Media Effects — Book: fill media provides the surface area, contact time and turbulence that govern heat exchange, and efficient fill reduces both fan power and pumping head → (a). Drift eliminators, louvers and casing are containment components. Same as Book EOC Q8.
Source: Sep 2019
📖 §7.2 Cooling Tower Performance (i) Range
60. Which one of the following is true to estimate the range of cooling tower?
Range = Cooling water inlet temperature - Wet bulb temperature
Range = Cooling water outlet temperature - Wet bulb temperature
Range = Wet Bulb Temperature - Cooling Water Inlet Temperature
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §7.2 (i) — Range = cooling water inlet (hot) temperature − outlet (cold) temperature; WBT does not appear. (b) outlet − WBT is the Approach; (a) inlet − WBT is the ideal range; (c) is negative. So none of the listed expressions gives the range → (d).
61. The difference between wet bulb temperature and cooling water inlet temperature in a cooling tower is called _______________.
Approach
Range
Effectiveness
None of the above
Answer: D) None of the above
Corrected (was a) — Book-3 §7.2: Approach is defined as the difference between the cooling tower OUTLET (cold) water temperature and the ambient WBT, and Range is inlet − outlet water temperature. The difference between the INLET (hot) water temperature and the WBT is what the book calls the 'ideal range' (denominator of effectiveness = Range + Approach) – it is neither approach, range nor effectiveness → (d). Tempting option (a) is right only if 'inlet' is misread as the cold water leaving the tower. ⚠ Answer changed a→d on book definitions (inlet − WBT = ideal range, not approach). If the examiner's key treats 'cooling water inlet' as the cold water entering the plant, (a) would be intended – learn both readings.
62. If inlet and outlet water temperatures of a cooling tower are 44°C and 38°C respectively and atmospheric DBT and WBT are 40°C and 35°C respectively, then the effectiveness of cooling tower is
54.5%
66.6%
75%
none of the above
Answer: B) 66.6%
Corrected (was a) — Book-3 §7.2 (iii): Effectiveness = Range/(Range + Approach) × 100. Range = 44 − 38 = 6°C; Approach = 38 − 35 = 3°C (cold water − WBT, DBT is a distractor); Effectiveness = 6/(6 + 3) = 66.6% → (b). 54.5% (a) = 6/11 would only result if the WBT were 33°C (approach 5°C) – it does not follow from the figures printed here. ⚠ Answer changed a→b: with the printed data (WBT 35°C) the book formula gives 66.6%. The old key value 54.5% corresponds to a 33°C WBT variant of this question.
Source: 9th Dec-2009
📖 §7.2 Cooling Tower Performance (vii) Blow down
63. Increasing the Cycles of Concentration (C.O.C) in circulating water in a cooling tower, the blow down quantity will
increase
decrease
not change
none of the above
Answer: B) decrease
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1); a higher COC means a larger denominator and therefore a smaller blowdown → (b). This is the basis of the book's 'COC improvement for water savings'.
Source: 9th Dec-2009
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
64. Which of the following ambient conditions will not evaporate maximum amount of water in a cooling tower
41°C DBT and 38°C WBT
38°C DBT and 37°C WBT
36°C DBT and 30°C WBT
36°C DBT and 31°C WBT
Answer: B) 38°C DBT and 37°C WBT
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Evaporation is driven by the wet-bulb depression (DBT − WBT). Options (c) and (d) have 6 and 5°C depression and evaporate a lot; (b) 38/37°C has only 1°C – near-saturated air – so it is the condition that will NOT evaporate the maximum → (b).
Source: 9th Dec-2009
📖 §7.2 Cooling Tower Performance (ii) Approach
65. If inlet and outlet water temperatures of a cooling tower are 39°C and 33°C respectively and atmospheric DBT and WBT are 35°C and 28°C respectively then the approach of cooling tower is
3°C
4°C
5°C
6°C
Answer: C) 5°C
Confirmed vs Book-3 §7.2 (ii) — Approach = outlet cold water temperature − ambient WBT = 33 − 28 = 5°C → (c). Using DBT (35 − 33 = 2°C) or range (39 − 33 = 6°C, option d) are the common traps.
📖 §7.2 Cooling Tower Performance (ii) Approach; §7.3 IV Performance Assessment
68. When you do a walk through energy audit of a cooling tower which salient parameter will you quickly spot check for its water cooling performance?
makeup water tap is on or off
hot water entry temperature to the cooling tower
cooling tower fan is on or off
cold well and ambient wet bulb temperature
Answer: D) cold well and ambient wet bulb temperature
Confirmed vs Book-3 §7.2 (ii) — Approach (cold-well water temperature − ambient WBT) is the book's best indicator of performance, so those two readings are the quickest spot check → (d). Hot water inlet (b) only reflects process load; fan/make-up status (a, c) says nothing about cooling performance.
69. Which of the following is correct statement in the case of cooling towers:
Range is the difference between the cooling tower water inlet and ambient wet bulb temperature.
Approach is the difference between the cooling tower outlet cold water temperature and hot inlet water temperature.
Range is the only indicator of cooling tower performance.
Cooling tower capacity is expressed as heat rejected in Ton of Refrigeration (TR)
Answer: D) Cooling tower capacity is expressed as heat rejected in Ton of Refrigeration (TR)
Confirmed vs Book-3 §7.2 (i)–(iv) — Book: Range = inlet − outlet water temp (so (a) is wrong); Approach = outlet cold water − WBT (so (b) is wrong); Approach, not Range, is the better indicator (so (c) is wrong); 'Cooling capacity is the heat rejected in kcal/hr or TR' → (d).
Source: 10th Jul-2010
📖 §7.1 Components of Cooling Tower – Fill; §7.2 Fill Media Effects
70. The surface area of heat exchange in a cooling tower is enhanced by
fill media
louvers
drift eliminator
cold water basin
Answer: A) fill media
Confirmed vs Book-3 §7.1/§7.2 — Fill maximises water–air contact; splash fill creates droplet surface and film fill provides sheet surface (30–45 vs 150 m²/m³, Table 7.3) → (a). Louvers equalise air flow, drift eliminators trap droplets, the basin only collects water.
Source: 10th Jul-2010
📖 §7.2 Cooling Tower Performance (vii) Blow down
71. The blowdown quantity required in cooling towers is given by
evaporation loss/ (cycle of concentration -1)
(cycle of concentration -1)/ evaporation loss
evaporation loss/ (1 - cycle of concentration)
evaporation loss/ (cycle of concentration +1)
Answer: A) evaporation loss/ (cycle of concentration -1)
Confirmed vs Book-3 §7.2 (vii) — Book: 'Blow Down = Evaporation Loss / (C.O.C. − 1)' → (a). (b) inverts the ratio, (c) gives a negative value for COC > 1, and (d) uses +1, all contradicting the book relation.
Source: 11th Feb-2011
📖 §7.2 Cooling Tower Performance (vii) Blow down
72. What will be the blowdown loss of a cooling tower if evaporation loss is 15.32 m3/hr and COC is 2.7 ?
9.01 m3/hr
5.67 m3/hr
41.3 m3/hr
0.17 m3/hr
Answer: A) 9.01 m3/hr
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1) = 15.32/(2.7 − 1) = 15.32/1.7 = 9.01 m³/hr → (a) – exactly the book's 3 × 200 MW trial figures. 5.67 (b) is 15.32/2.7 (forgetting −1).
Source: Mar 2021 (Set B)
📖 Not a Ch-7 topic (thermal power plant heat rate); η = 3600 kJ/kWh ÷ heat rate
73. One of the thermal power plants operating with 2 numbers of 500 MW units has reported the operating heat rate of 11250 kJ/kWh. The Plant Load Factor (PLF) of the power plant is 73 %. The operating efficiency of the power plant will be ___________
38 %
35 %
30 %
32 %
Answer: D) 32 %
Confirmed by calculation (not Book-3 Ch-7 content) — Plant efficiency = 3600 kJ/kWh / 11,250 kJ/kWh = 0.32 = 32% → (d). PLF (73%) and unit count are distractors; they do not enter the heat-rate → efficiency conversion.
Source: Mar 2021 (Set B)
📖 §7.2 Cooling Tower Performance (vii) Blow down
74. Increasing the cycles of concentration of circulating water in a cooling tower will
increase blow down quantity
decrease blow down quantity
increase drift losses
decrease fan power consumption
Answer: B) decrease blow down quantity
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1); a higher COC lowers the blowdown quantity → (b). Drift loss is set by the eliminators and air velocity, and fan power by air flow – neither depends on COC. ⚠ Option d repaired: garbled 'decrease bar create power consumption' → 'decrease fan power consumption'.
75. If water is flowing through a cooling tower at 120 m3/h with 5 oC range, the load on cooling tower at ambient wet bulb temperature of 33 oC is
198.4 TR
357 TR
158 TR
none of the above
Answer: A) 198.4 TR
Confirmed vs Book-3 §7.2 (iv) — Heat load = 120 × 1000 × 1 × 5 = 6,00,000 kcal/h; ÷ 3024 = 198.4 TR → (a) (Book EOC S-1). The WBT affects only the approach, not the load.
Source: Mar 2021
📖 §7.2 Factors Affecting Performance – Fill Media Effects
76. Which one has the maximum effect on cooling tower performance ?
fill media
drift
louvers
casing
Answer: A) fill media
Confirmed vs Book-3 §7.2 Fill Media Effects — Book: fill media is responsible for the surface area, contact time and turbulence of heat exchange → (a). Drift, louvers and casing are secondary components (Book EOC Q8).
77. If water is flowing through a cooling tower at 120 m3/h with 5 degC range, the load on cooling tower at ambient wet bulb temperature of 33 degC is ___________
198.4 TR
357 TR
158 TR
none of the above
Answer: A) 198.4 TR
Confirmed vs Book-3 §7.2 (iv) — Heat load = 120,000 kg/h × 1 × 5°C = 6,00,000 kcal/h; 6,00,000/3024 = 198.4 TR → (a). Same as Book EOC S-1.
78. The blow down loss in a cooling tower depends on
TDS in circulating water
TDS in make-up water
evaporation loss
all the above
Answer: D) all the above
Confirmed vs Book-3 §7.2 (vi)–(vii) — Blow Down = Evaporation/(COC − 1) and COC = TDS in circulating water / TDS in make-up water, so blowdown depends on all three quantities → (d).
Source: Jul 2022
📖 §7.2 Cooling Tower Performance (vii) Blow down
79. For a Cooling Tower, if evaporation loss is 15 m³/hour and Cycles of Concentration is 2.5, the blowdown is equal to
6 m³/hour
10 m³/hour
22.5 m³/hour
37.5 m³/hour
Answer: B) 10 m³/hour
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1) = 15/(2.5 − 1) = 15/1.5 = 10 m³/hour → (b). 6 (a) is 15/2.5 (forgetting −1); 22.5 (c) multiplies instead of divides.
Source: Jul 2022
📖 §7.2 Factors Affecting Performance – Range (Range = Heat Load / Water Circulation Rate)
80. The heat load of the cooling tower depends on __________.
Range
Approach
Cooling water temp
Wet Bulb Temperature
Answer: A) Range
Confirmed vs Book-3 §7.2 Range — Book: 'Range °C = Heat Load in kcal/hour / Water Circulation Rate' – heat load = flow × Cp × Range, so the heat load is tied to the Range → (a). Approach and WBT relate to tower performance/ambient, not to the process heat load.
81. A Cooling Tower is operating with the following parameters: Wet Bulb Temp. = 27°C, Cooling Water in Temperature = 45°C, Cooling Water out Temperature = 35°C. What is the cooling tower effectiveness?
50%
44.44%
55.55%
None of the above
Answer: C) 55.55%
Confirmed vs Book-3 §7.2 (iii) — Range = 45 − 35 = 10°C; Approach = 35 − 27 = 8°C; Effectiveness = 10/(10 + 8) = 55.55% → (c). 44.44% (b) is Approach/(Range + Approach), the inverted ratio.
Source: Mar 2023
📖 §7.1 Components of Cooling Tower – Drift eliminators
82. __________ is used to capture water droplets in the air stream leaving the cooling tower.
Splash fill
Film fill
Drift eliminator
Any of the above
Answer: C) Drift eliminator
Confirmed vs Book-3 §7.1 Components — Book: drift eliminators 'capture water droplets entrapped in the air stream' → (c). Splash and film fill are heat-transfer media, not droplet catchers.
84. Find the correct example, if M = makeup water (from the mains water supply), E = losses due to evaporation, B = losses due to blow-down and D = drift losses of a cooling tower:
M = E + B + D
M = E − B + D
M = E + B − D
M = E − B − D
Answer: A) M = E + B + D
Confirmed vs Book-3 §7.2 (v)–(vii) — Make-up water must replace all water leaving the circuit – evaporation, blow-down and drift – so M = E + B + D → (a). Any subtraction would mean a loss returns water to the basin. ⚠ Options c and d were duplicates ('M = E - B - D'); c repaired to 'M = E + B − D' so exactly one option is correct.
85. If temperature of air increases, the amount of water vapor needed to become saturated __________.
Increases
Decreases
not change
Can't say
Answer: A) Increases
Confirmed vs Book-3 §7.2 Approach & WBT — Warmer air can hold more moisture (the book's enthalpy example: air at 26.67°C WBT holds 24.17 kcal/kg, at 37.8°C WBT 39.67 kcal/kg), so more water vapour is needed to saturate it → (a). This is why hotter, drier air evaporates more water.
86. The wet bulb temperature entering the cooling tower is 38°C. Range is twice the approach, the effectiveness of the cooling tower is __________.
33.3%
50%
66.7%
Insufficient data
Answer: C) 66.7%
Confirmed vs Book-3 §7.2 (iii) — Effectiveness = Range/(Range + Approach). With Range = 2 × Approach: 2A/(2A + A) = 2/3 = 66.7% → (c). The 38°C WBT is not needed – the ratio alone fixes effectiveness, so (d) 'insufficient data' is a trap.
Source: Mar 2023
Short questions (5 marks) — 18
📖 §7.2 Evaporation Loss, Blow Down and Make-up
1. A cooling tower circulates 1250 m³/hr of water over a Range of 7 °C and operates at a COC of 3.5. Calculate the evaporation loss, blowdown and make-up water requirement (ignore drift).
Model answer: Evaporation loss = 13.39 m³/hr; Blowdown = 5.36 m³/hr; Make-up = 18.75 m³/hr.
Step 1 — Evaporation loss (§7.2) = 0.00085 × 1.8 × circulation rate × (T₁ − T₂) = 0.00085 × 1.8 × 1250 × 7 = 13.39 m³/hr (≈ 1.07% of circulation).
Step 2 — Blow down (§7.2) = Evaporation Loss / (C.O.C. − 1) = 13.39 / (3.5 − 1) = 13.39 / 2.5 = 5.36 m³/hr.
Step 3 — Make-up = Evaporation + Blowdown + Drift = 13.39 + 5.36 + 0 = 18.75 m³/hr (drift ignored as stated).
Cross-check: Make-up / Blowdown = 18.75 / 5.36 = 3.5 = C.O.C., which is the water-balance form of the COC definition.
Confirmed vs Book-3 §7.2 — the three formulas run in a fixed chain: evaporation first, then blowdown from COC, then make-up as the sum.
The usual errors are dividing by COC instead of (COC − 1), and forgetting that make-up must cover BOTH evaporation and blowdown.
2. Estimate the cooling tower capacity (TR) with: water flow rate 120 m3/h, sp. heat 1 kcal/kg C, inlet water temperature 37 C, outlet water temperature 32 C, ambient WBT 29 C.
Model answer: Heat rejected = mass flow × sp. heat × Range. Water flow = 120 m³/h = 120,000 kg/h. Range = 37 − 32 = 5°C. Heat load = 120,000 × 1 × 5 = 6,00,000 kcal/h. 1 TR = 3024 kcal/h, so capacity = 6,00,000 / 3024 = 198.4 TR (≈ 200 TR). (Approach = 32 − 29 = 3°C; the WBT is not needed for the TR figure.)
Heat load = 120000 kg/h x 1 x 5 C = 6.0 lakh kcal/h; dividing by 3024 kcal/h per TR gives about 198 TR.
Source: Book EOC
📖 §7.2 Cooling Tower Performance (vi) Cycles of Concentration; (vii) Blow down (Book EOC S-2)
3. What do you mean by the term Cycles of Concentration and how is it related to cooling tower blow down?
Model answer: Cycles of Concentration (COC) is the ratio of the concentration of dissolved solids (measured as TDS or conductivity, or as the ratio of chlorides) in the circulating water to that in the make-up water. As pure water evaporates, dissolved solids concentrate in the recirculating water, raising the COC. Blowdown is the deliberate removal of a portion of the concentrated circulating water to keep the COC (and hence scaling/corrosion) within limits, and is related by: Blowdown = Evaporation loss / (COC - 1). A higher COC means a smaller blowdown and lower make-up water consumption.
COC = TDS in circulating water / TDS in make-up water; Blowdown = Evaporation / (COC - 1), so higher COC reduces blowdown.
4. In a cooling tower the cooling water circulation rate is 1250 m3/hr and the operating range is 7 C. If the blowdown rate is 1% of the circulation rate, calculate the evaporation loss and COC.
Model answer: Evaporation loss (m3/hr) = approx 0.00085 x 1.8 x circulation rate x range = 0.00085 x 1.8 x 1250 x 7 = 13.39 m3/hr (about 1.07% of circulation). Blowdown = 1% of 1250 = 12.5 m3/hr. COC = (Evaporation + Blowdown) / Blowdown = (13.39 + 12.5)/12.5 = 25.89/12.5 = 2.07. So evaporation loss is about 13.4 m3/hr and COC is about 2.0.
Evaporation = 0.00085 x 1.8 x 1250 x 7 = 13.4 m3/hr; COC = (Evap + Blowdown)/Blowdown = (13.4 + 12.5)/12.5 = approx 2.07.
Model answer: The L/G ratio is the ratio of the mass flow rate of water (Liquid, L) to the mass flow rate of air (Gas, G) through a cooling tower. From a heat/mass balance, the heat lost by the water equals the heat gained by the air, giving L/G = (h2 - h1)/(T1 - T2), where h1 and h2 are the enthalpies of air at inlet and outlet wet bulb temperatures and (T1 - T2) is the water range. It is an important thermal design parameter: for fixed range and approach, an optimum L/G gives the required performance; an excessively high L/G overloads the tower (poor cooling) while a low L/G means more fan power per unit water.
L/G = water mass flow / air mass flow = (h2 - h1)/(T1 - T2); a key cooling tower design parameter relating water and air loading.
Source: Book EOC
📖 §7.1 Components of Cooling Tower – Fill; §7.2 Fill Media Effects (Book EOC S-5)
6. What is the function of fill media in a cooling tower?
Model answer: The fill (packing) is the heart of the cooling tower. Its function is to maximise the contact area and contact time between the falling water and the rising air, so that heat and mass (evaporation) transfer between water and air is maximised. By breaking the water into droplets (splash fill) or spreading it into thin films over a large surface (film fill), the fill greatly increases the water surface exposed to air, improving cooling efficiency for a given tower size and air flow.
Fill maximises air-water contact area and time, enabling efficient heat and mass transfer (evaporative cooling); splash type and film type are the two kinds.
7. Define Range, approach and effectiveness in cooling tower operation.
Model answer: i) Range = difference between cooling tower water inlet and outlet temperature. ii) Approach = difference between the cooling tower outlet cold water temperature and the ambient wet bulb temperature (approach is a better indicator of cooling tower performance). iii) Effectiveness (%) = Range / ideal range = Range / (Range + Approach), where ideal range = inlet water temp - ambient WBT.
8. An induced draft cooling tower is designed for a range of 7°C. An energy manager finds the operating range as 4°C. What could be the reasons for this situation?
Model answer: 1. Excess cooling water flow rate. 2. Reduced heat load from the process. 3. Some cooling tower cell fans switched off. 4. Poor approach due to high humid conditions. 5. Nozzles may be blocked.
Lower-than-design range implies either reduced heat load, excess water flow or degraded heat-transfer.
Source: 15th Exam
📖 §7.2 Factors Affecting Performance – Fill Media Effects (surface area, time of contact, turbulence/relative velocity)
9. List any five factors that affect the rate of evaporation of water in cooling towers.
Model answer: 1. Amount of water surface area exposed. 2. The time of exposure. 3. The relative velocity of air passing over the droplets. 4. The relative humidity of air. 5. The direction of airflow relative to water.
Standard factors governing evaporative heat/mass transfer in cooling towers.
10. Estimate the cooling tower capacity (TR) and approach: water flow 2 m³/min, specific heat 1 kcal/kg°C, inlet water 43°C, outlet water 35°C, ambient WBT 30°C.
Model answer: Capacity (TR) = (flow x density x sp.heat x ΔT)/3024 = (2x60) x 1000 x 1.0 x (43-35)/3024 = 317.5 TR. Approach = 35 - 30 = 5°C.
TR from heat-load formula; approach = cold water out - ambient WBT.
Source: 14th Exam
📖 §7.2 Factors Affecting Performance – Heat Load (heat-rejection table); Range = Heat load / Water circulation rate
11. An 18 MW cogeneration plant: max condenser load 7 MW; extraction steam 57 TPH for process and VAM. Condenser heat load 550 kcal/kg steam; steam rate 5 kg/kW for condenser power; VAM heat load 127 kcal/min/TR with VAM capacity 1100 TR. Estimate cooling tower heat load (kcal/hr). For 6°C range, calculate cooling water flow. Design approach 5°C.
12. S-1: Operating data of an induced-draft cooling tower: range 8 °C, cooling water flow 12,500 m3/hr, drift loss 0.1% of circulation, WBT 27 °C, ambient DBT 35 °C, effectiveness 67%, COC 3. Estimate evaporation loss, make-up water requirement and TR load.
Model answer: Evaporation loss = 0.00085 x 1.8 x 12500 x 8 = 153 m3/hr. Blowdown = 153/(3-1) = 76.5 m3/hr. Drift = 12500 x 0.001 = 12.5 m3/hr. Make-up = evaporation + blowdown + drift = 153 + 76.5 + 12.5 = 242 m3/hr. TR load = (12500 x 1000 x 8)/3024 = 33,069 TR.
Use book evaporation formula (0.00085 x 1.8 x circulation x range), blowdown = evap/(COC-1), make-up = evap+blowdown+drift, heat-load TR = m·Cp·ΔT/3024.
13. S-4: In a thermal power station, steam input to a turbine on fully condensing mode is 100 TPH. Condenser heat rejection = 555 kcal/kg of steam condensed; cooling water inlet/outlet = 27 °C / 37 °C. Find the circulating cooling water flow.
Model answer: Heat rejected = 100,000 kg/hr x 555 kcal/kg = 55.5 million kcal/hr. Cooling water flow = heat rejected/(ΔT x Cp) = 55,500,000/((37-27) x 1) = 5,550,000 kg/hr = 5550 m3/hr.
Equate condenser heat rejection to heat gained by cooling water (m·Cp·ΔT) and solve for flow.
Source: 19th Exam
📖 §7.6 Case Study – VFD for CT fan (cube law); §7.4 Flow Control Strategies (two-speed fans)
14. A 4-cell cooling tower has 45 kW CT fans per cell operating at 40 kW at 1450 rpm. Fans are replaced with two-speed motors at 1450 rpm and 740 rpm. High-speed mode 5300 hours, low-speed mode 1800 hours/year. Estimate annual energy savings vs continuous fixed-speed 1450 rpm operation.
Model answer: Present (all at 1450 rpm) = 4 x 40 x (5300+1800) = 11,36,000 kWh. After two-speed: high speed = 4 x 40 x 5300 = 8,48,000 kWh; low speed = (740/1450)^3 x 40 x 4 x 1800 = 38,281 kWh. Annual savings = 11,36,000 - (8,48,000 + 38,281) = 2,49,719 kWh.
Printed solution using fan affinity law (P proportional to N^3): savings 2,49,719 kWh/year.
15. S-6: Power plant cooling tower audit: generation 785 MW, circulation rate 107000 m3/hr, range 10.5 C, design COC 3.8. Find (a) total water consumption/hr, (b) specific water consumption (m3/MW). If COC raised to 7.0, (c) potential water savings in m3/hr and m3/MW.
Model answer: Evaporation loss = 0.00085 x 107000 x 10.5 x 1.8 = 1719 m3/hr. Blowdown = 1719/(3.8-1) = 614 m3/hr. (a) Total = 1719 + 614 = 2333 m3/hr. (b) Specific = 2333/785 = 2.97 m3/MW. At COC 7.0: blowdown = 1719/(7-1) = 286.5 m3/hr; total = 1719 + 286.5 = 2005.5 m3/hr; specific = 2005.5/785 = 2.56 m3/MW. (c) Water saving = 2333 - 2005.5 = 327.5 m3/hr; per MW = 327.5/785 = 0.417 m3/MW.
Printed solution table: total 2333 m3/hr, specific 2.97 m3/MW, savings 327.5 m3/hr and 0.417 m3/MW.
16. Estimate the cooling tower capacity (TR) and approach with the following parameters: Water flow rate = 120 m3/hr, Specific heat of water = 1 kCal/kg°C, Inlet water temperature = 42°C, Outlet water temperature = 36°C, Ambient WBT = 32°C.
17. Find the blow down rate of a cooling tower: cooling water flow 600 m3/hr; operating range 8°C; TDS in circulating water 1500 ppm; TDS in make-up water 300 ppm.
Model answer: Evaporation loss = 0.00085 × 1.8 × circulation rate × range = 0.00085 × 1.8 × 600 × 8 = 7.344 m³/hr. COC = TDS in circulating water / TDS in make-up water = 1500/300 = 5. Blow down = Evaporation loss/(COC − 1) = 7.344/(5 − 1) = 1.836 m³/hr.
Evaporation loss = 0.00085×1.8×600×8 = 7.344 m3/hr. COC = 1500/300 = 5. Blowdown = Evap/(COC−1) = 7.344/(5−1) = 1.836 m3/hr. ⚠ answerText expanded from bare result to full book-method working.
18. S-8: An induced draft-cooling tower is designed for a range of 8 C. The energy auditor finds the operating range as 2 C during the conduct of energy audit. In your opinion what could be the reasons for this situation?
Model answer: 1. There may be excess cooling water flow rate; 2. There may be reduced heat load from the process; 3. Some of the cooling tower cells fan are switched off; 4. Approach may be poor because of high humid condition; 5. Cooling tower nozzles may be blocked.
Low range (cooling) caused by excess water flow, reduced heat load, or operational/maintenance issues.
Source: 11th Feb-2011
Long questions (10 marks) — 17
📖 §7.2 Range, Approach & Effectiveness
1. Define cooling tower Range and Approach, and state why Approach is considered the better indicator of performance.
Model answer: Range (Book-3 §7.2): Range = cooling tower water INLET temperature − OUTLET temperature, i.e. the temperature drop achieved across the tower. The book also gives Range °C = Heat Load (kcal/hr) / Water Circulation Rate (LPH), showing that range is fixed by the process.
Approach (Book-3 §7.2): Approach = cooling tower outlet COLD water temperature − ambient WET BULB temperature. Example from the book: a tower cooling 4540 m³/hr from 48.9 °C to 32.2 °C at 26.7 °C WBT has Range = 16.7 °C and Approach = 32.2 − 26.7 = 5.5 °C.
Why Approach is the better indicator: the book states that although both should be monitored, 'the Approach is a better indicator of cooling tower performance'. Range is 'determined not by the cooling tower, but by the process it is serving' — it is set entirely by the heat load and the circulation rate. Approach, by contrast, measures how close the tower can drive the cold water towards the WBT, which is the theoretical minimum achievable by evaporative cooling. A smaller approach means better tower performance; the closest approach manufacturers normally guarantee is about 2.8 °C, and in ranking the four sizing parameters the book puts approach FIRST in importance.
Effectiveness ties the two together: % Effectiveness = Range / (Range + Approach) × 100.
Confirmed vs Book-3 §7.2 and §7.3 — Range uses water temperatures only; Approach uses cold water against the WET bulb (never the dry bulb).
Quote the book's reason directly: range is imposed by the process, approach reflects the tower's own capability.
Source: AI practice
📖 §7.1 Components — Fill; §7.3 Fill Media Effects
2. State the function of fill media in a cooling tower and explain why film fill is considered more efficient than splash fill.
Model answer: Function of fill (Book-3 §7.1): Most towers employ fills, made of plastic or wood, to facilitate heat transfer by MAXIMISING WATER AND AIR CONTACT. The fill does this by increasing the heat-transfer surface area, increasing the contact time between water and air, and increasing the turbulence/intermixing of the two streams. It is the single component with the maximum effect on cooling tower thermal performance.
The two types: with SPLASH fill the water falls over successive layers of horizontal splash bars, continuously breaking into smaller droplets while also wetting the fill surface (plastic splash fill promotes better heat transfer than wood splash fill). With FILM fill the water spreads as a thin film over thin, closely spaced plastic sheets.
Why film fill is more efficient: the book states that 'the film type of fill is the more efficient and provides same heat transfer in a smaller volume than the splash fill'. The thin film exposes far more water surface per unit volume, so for the same duty the tower needs less air flow and a shorter fill height, which is directly reflected as savings in fan power consumption and in pumping head. Film fill is therefore the fill of choice wherever the circulating water is free of debris; where the water carries debris that could plug the narrow passages, splash fill must still be used.
Confirmed vs Book-3 §7.1 — the exam wants the three contact mechanisms (surface area, contact time, turbulence) plus the film-vs-splash comparison.
Also state the consequence the book highlights: efficient fill design shows up as lower fan power and lower pump head, not just better cooling.
3. a) In a large-scale steel manufacturing facility, a cooling tower is used to reject heat from continuous casting operations. The circulating water flow rate is 2000 m³/hr. The cooling tower is currently operating at a Cycles of Concentration (COC) of 3. The evaporation loss is estimated at 1.0% of the circulating flow, and the drift loss is 0.1% of the circulating flow. The facility is planning to improve the COC from 3 to 6 through advanced water treatment. Evaluate the following:
i. Calculate the make-up water requirement at the current COC of 3.
ii. Calculate the revised make-up water requirement if the COC is increased to 6.
iii. Estimate the total water savings per day.
iv. Discuss one limitation or risk associated with increasing the COC.
b) True or False:
i. Cooling towers primarily reject heat through evaporative cooling.
ii. The approach temperature in a cooling tower is the difference between the hot water temperature and the ambient dry bulb temperature.
iii. Blowdown in a cooling tower is required to prevent the build-up of dissolved solids.
iv. Drift losses in a cooling tower refer to water carried away with the exhaust air.
v. Cooling tower effectiveness improves with higher approach temperatures.
vi. Cycles of concentration in a cooling tower relate to how many times the water is reused before discharge.
Model answer: a) Given: Circulating Water Flow (CWF) = 2000 m³/hr; Initial COC = 3; Final COC = 6; Evaporation Loss (E) = 1% of 2000 = 20 m³/hr; Drift Loss (D) = 0.1% of 2000 = 2 m³/hr
i) Current Make-up at COC = 3: B = 20 / (3 − 1) = 10 m³/hr; Make-up = E + D + B = 20 + 2 + 10 = 32 m³/hr
ii) Revised Make-up at COC = 6: B = 20 / (6 − 1) = 4 m³/hr; Make-up = 20 + 2 + 4 = 26 m³/hr
iii) Hourly Savings = 32 − 26 = 6 m³/hr; Daily Savings = 6 × 24 = 144 m³/day
iv) Increasing COC can lead to higher concentrations of dissolved solids in the water, which may cause scaling, corrosion, and microbiological fouling in the system. Effective water treatment and frequent monitoring are necessary to avoid operational issues.
b) i) True; ii) False (It is the difference between the cold-water temperature and the wet bulb temperature.); iii) True; iv) True; v) False (Lower approach means better effectiveness.); vi) True
Blowdown B = E/(COC−1); make-up = E + D + B; higher COC reduces blowdown and make-up.
4. A steel industry has 100 MW of captive power plant with 2 nos. of identical extraction condensing steam turbine. Power demand is 80 MW. The turbine specific condensing load is 3.20 kg/kWh and heat rejection in condenser is 560 kCal/kg.
Cold cooling water temperature: 32.0 °C
Hot cooling water temperature: 39.2 °C
Calculate the following:
a) The Cooling Water circulation flow (m³/hr) through both condensers, if only one cooling tower supplies water to condenser of both the steam turbines.
b) Make-up water flow rate (kg/hr) to basin, assuming blowdown loss is 1.0 % of circulation flow.
📖 §7.3 Efficient System Operation – IV Performance Assessment of Cooling Towers (Book EOC L-1 – identical to book trial)
5. A cooling tower cools 1565 m3/hr of water from 44 C to 37.6 C at 29.3 C wet bulb temperature. The fan air flow rate is 989544 m3/hr (air density 1.08 kg/m3) and it operates at 2.7 cycles of concentration. Find: a) Range, b) Approach, c) % CT Effectiveness, d) L/G ratio in kg/kg, e) Cooling duty in TR, f) Evaporation losses in m3/hr, g) Blowdown in m3/hr, h) Make-up water in m3/hr.
Model answer: a) Range = 44 − 37.6 = 6.4°C. b) Approach = 37.6 − 29.3 = 8.3°C. c) Effectiveness = Range/(Range + Approach) × 100 = 6.4/(6.4 + 8.3) × 100 = 43.53%. d) Water mass L = 1565 × 1000 = 15,65,000 kg/hr; air mass G = 989544 × 1.08 = 10,68,708 kg/hr; L/G = 15,65,000/10,68,708 = 1.46 kg/kg. e) Cooling duty = 1565 × 1000 × 1 × 6.4 = 1,00,16,000 kcal/hr = 10016 × 10³ kcal/hr; in TR = 1,00,16,000/3024 ≈ 3312 TR. f) Evaporation loss = 0.00085 × 1.8 × 1565 × 6.4 = 15.32 m³/hr (≈ 0.97% of circulation). g) Blow down = Evaporation/(COC − 1) = 15.32/(2.7 − 1) = 9.01 m³/hr. h) Make-up water = Evaporation + Blow down = 15.32 + 9.01 = 24.33 m³/hr (book's printed '2433' is a typo for 24.33).
6. Explain the significance of the following on cooling tower performance: a) Range, b) Approach, c) Effectiveness, d) L/G ratio, e) WBT.
Model answer: a) Range (hot water temp - cold water temp) is set by the heat load and water flow (Range = Heat load / (flow x sp.heat)); a higher range for a given duty means a lower water flow rate and less pumping. b) Approach (cold water temp - WBT) reflects how well the tower performs; a smaller approach means better performance but needs a larger tower, and it cannot be reduced below the WBT. c) Effectiveness = Range/(Range + Approach); higher effectiveness indicates the cold water gets closer to the WBT, i.e. a better-performing tower. d) L/G ratio (water to air mass flow) is a key design/operating parameter; air flow must be matched to water load for good cooling, and increasing air flow (within limits) improves cooling at the cost of fan power. e) Wet Bulb Temperature (WBT) is the theoretical lower limit to which water can be cooled; cooling tower capacity is rated at a design WBT and a lower ambient WBT improves achievable cold water temperature, while a higher WBT worsens it.
Range depends on load and flow; Approach indicates performance (limited by WBT); Effectiveness = Range/(Range+Approach); L/G balances air and water flow; WBT is the theoretical cooling limit.
Source: Book EOC
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down (part b is transformer loss – not Ch-7)
7. a) Cooling tower circulation rate 1200 m³/hr, water enters at 38°C, ambient WBT 26°C, approach 4°C, blowdown 1% of circulation. Calculate evaporation loss and COC. b) An industry with 450 kVA load has two 500 kVA transformers (no-load loss 760 W, full-load copper loss 5400 W each). Should it run one transformer at full load or two sharing the load? Recommend with reasons.
Model answer: a) Cold water out = 26+4 = 30°C; range = 38-30 = 8°C. Evaporation loss = 0.00085 x 1.8 x 1200 x 8 = 14.69 m³/hr. Blowdown = 1% x 1200 = 12 m³/hr. COC = Evap/Blowdown + 1: 12 = 14.69/(COC-1) → COC = 2.224. b) One 500 kVA at 450 load: loss = 760 + (450/500)² x 5400 = 760 + 4374 = 5134 W. Two 500 kVA at 50% (225 each): 2 x [760 + (225/500)² x 5400] = 2 x (760 + 1093.5) = 3707 W. Two transformers are better - losses are least (3707 W < 5134 W).
Evaporation loss formula; transformer loss = no-load + (load ratio)² x copper loss.
8. Power plant cooling tower audit: Generation 785 MW; Circulation 107000 m3/hr; Range 10.5°C; COC 3.8. Find a) total water consumption per hour, b) specific water consumption in m3/MW. If COC is increased to 7.0, c) potential water savings in m3/hr and m3/MW.
Model answer: Evaporation loss = 0.00085×107000×10.5×1.8 = 1719 m3/hr. Blowdown = 1719/(3.8−1) = 614 m3/hr. Total = 1719 + 614 = 2333 m3/hr. Specific = 2333/785 = 2.97 m3/MW. At COC 7.0: blowdown = 1719/6 = 286.5; total = 1719 + 286.5 = 2005.5 m3/hr; specific = 2.56 m3/MW. Water saving = 2333 − 2005.5 = 327.5 m3/hr and 327.5/785 = 0.417 m3/MW.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity; (v) Evaporation loss (part b is fan efficiency – Fans chapter)
9. a) Two streams of cooling water 9000 m3/hr at 41°C and 6000 m3/hr at 52°C are mixed and fed to one cooling tower; measured heat rejection 45,000 TR at 31°C WBT. Calculate effectiveness and evaporation loss. b) In a 0.5 m × 0.5 m AC duct, average air velocity 28 m/s; suction static −20 mmWC, discharge 30 mmWC; 3-ph motor draws 10.8 A at 415 V, 0.9 PF; motor efficiency 88%. Find fan efficiency (neglect density correction).
Model answer: a) Mixed flow = 15000 m3/hr; mixed hot temp = (9000×41+6000×52)/15000 = 45.4°C. Range = 45000×3024/(15000×1000) = 9.072°C. Cold water temp = 45.4 − 9.072 = 36.33°C. Approach = 36.33 − 31 = 5.33°C. Effectiveness = Range/(Range+Approach) = 9.072/(9.072+5.33) = 63%. Evaporation loss = 0.00085×1.8×15000×9.072 = 208.2 m3/hr. b) Duct area = 0.25 m2; flow = 28×0.25 = 7 m3/s. Motor power = √3×415×10.8×0.9/1000 = 6.99 kW. Air power = 7×(30−(−20))/102 = 3.43 kW. Shaft power = 6.99×0.88 = 6.15 kW. Fan static efficiency = 3.43/6.15×100 = 55.76%.
Range from TR; effectiveness=Range/(Range+Approach); fan eff=air power/shaft power.
Source: 16th Exam
📖 §7.2 Cooling Tower Performance (viii) L/G ratio; §7.2 Fill Media Effects (enthalpy values from Approach & WBT example)
10. L-6: (a) What is L/G ratio and how is it useful in cooling tower operation? (b) Functions of fill media. (c) Calculate L/G ratio: water flow 4540 m3/hr, approach 4.45 C, air entering enthalpy 24.17 kcal/kg at 26.67 C, air leaving enthalpy 39.67 kcal/kg at 37.8 C, hot water 47.77 C, cold water 31.11 C.
Model answer: (a) L/G ratio = ratio of water (liquid) to air (gas) mass flow rates. By energy balance L(T1-T2) = G(h2-h1), so L/G = (h2-h1)/(T1-T2). Against design, seasonal variation requires tuning of water/air flow (water box loading, blade angle) for best effectiveness. (b) Fill media increases the air-water contact surface and contact time, promoting heat and mass transfer (evaporative cooling). (c) L/G = (39.67-24.17)/(47.77-31.11) = 15.5/16.66 = 0.93.
Printed solution: L/G = 0.93; fill media function referenced to Page 209.
Source: Sep 2019
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity; (v) Evaporation loss (part b is fan efficiency – Fans chapter)
11. L-4: (a) Cooling water 9000 m3/hr at 41 C and 6000 m3/hr at 52 C mixed and fed to one cooling tower. Measured heat rejection 45,000 TR; WBT 31 C. Calculate effectiveness and evaporation loss. (b) AC duct 0.5 x 0.5 m, air velocity 28 m/s, suction static -20 mmWC, discharge 30 mmWC, motor draws 10.8 A at 415 V, 0.9 PF; motor eff 88%. Find fan efficiency (neglect density correction).
Model answer: (a) Mixed flow = 15000 m3/hr; mixed hot water temp = (9000x41 + 6000x52)/15000 = 45.4 C. Range = (45000 x 3024)/(15000 x 1000) = 9.072 C. Cold water temp = 45.4 - 9.072 = 36.328 C; approach = 36.328 - 31 = 5.328 C. Effectiveness = range/(range+approach) = 9.072/(9.072+5.328) = 63%. Evaporation loss = 0.00085 x 1.8 x 15000 x 9.072 = 208.2 m3/hr. (b) Area = 0.25 m2; airflow = 0.25 x 28 = 7 m3/s. Motor power = 1.732 x 415 x 10.8 x 0.9/1000 = 6.99 kW. Air power = 7 x (30-(-20))/102 = 3.43 kW. Shaft power = 6.99 x 0.88 = 6.15 kW. Fan static efficiency = 3.43/6.15 x 100 = 55.76%.
📖 §7.2 Factors Affecting Performance – Heat Load (heat-rejection table); Range = Heat load / Water circulation rate
12. L-5: An 18 MW cogeneration plant: max condenser load 7 MW, extraction steam 57 TPH for process and VAM. Condenser heat load 550 kcal/kg of steam, steam rate 5 kg/kW for condenser power. VAM heat load 127 kcal/min/TR, VAM capacity 1100 TR. Estimate cooling tower heat load (kcal/hr); for 6 C range and 5 C approach calculate the cooling water flow.
Model answer: Steam for condenser power = 7000 x 5 = 35,000 kg/hr. Condenser heat load = 35000 x 550 = 1,92,50,000 kcal/hr. VAM heat load = 1100 x 127 x 60 = 83,82,000 kcal/hr. Total CT heat load = 1,92,50,000 + 83,82,000 = 2,76,32,000 kcal/hr. Cooling water flow = total heat load/range = 27632000/6 = 46,05,333 litres/hr = 4605 m3/hr.
Printed solution: total heat load 2,76,32,000 kcal/hr; water flow 4605 m3/hr.
13. a) Define Range, approach and effectiveness in cooling tower operation. b) An induced draft cooling tower is designed for a range of 8°C. The energy auditor finds the operating range as 2°C. What could be the reasons for such a situation?
Model answer: a) Range = difference between cooling tower water inlet and outlet temperature. Approach = difference between cooling tower outlet cold water temperature and ambient wet bulb temperature (a better indicator of performance). Effectiveness (%) = ratio of range to the ideal range = Range/(Range + Approach). b) Possible reasons: excess cooling water flow rate; reduced heat load from the process; some cooling tower cell fans switched off; poor approach due to high humidity; nozzles blocked.
Standard cooling-tower definitions plus typical causes of a much lower-than-design operating range.
Source: 9th Dec-2009
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity (part b is DG set – not Ch-7)
14. (a) Estimate the cooling tower approach and capacity (TR): water flow 150 m3/hr; sp.heat 1 kCal/kg°C; inlet water 42°C; outlet water 34°C; ambient WBT 30°C. (b) A 180 kVA, 0.80 PF rated DG set has a diesel engine rating of 240 BHP. What is the maximum power factor maintainable at full load on the alternator without overloading the diesel engine? (alternator losses + exciter power = 5.44 kW; no derating.)
Model answer: (a) Approach = outlet cold water − ambient WBT = 34 − 30 = 4°C. Range = 42 − 34 = 8°C. Capacity = flow × density × Cp × range / 3024 = 150 × 1000 × 1 × 8 / 3024 = 12,00,000/3024 = 396.8 TR. (b) Engine output = 240 BHP × 0.746 = 179.04 kW; power available to alternator output = 179.04 − 5.44 = 173.6 kW; maximum PF = kW/kVA = 173.6/180 = 0.964.
(a) Approach = outlet − WBT = 34−30 = 4°C; capacity = 150×1000×1×(42−34)/3024 = 396.8 TR. (b) Engine power = 240×0.746 = 179.04 kW; power available for alternator = 179.04−5.44 = 173.6 kW; max PF = 173.6/180 = 0.964. ⚠ answerText expanded from bare results to full working.
Source: 10th Jul-2010
📖 §7.3 Efficient System Operation – IV Performance Assessment of Cooling Towers (book trial method)
15. L-6: A cooling tower cools 1450 m3/hr of water from 43 C to 36.6 C at 30 C wet bulb temperature. The cooling tower fan air flow rate is 9,50,000 m3/hr (air density = 1.08 kg/m3) and operates at 2.7 cycles of concentration. Find (a) Range, (b) Approach, (c) % CT Effectiveness, (d) L/G Ratio in kg/kg, (e) Cooling Duty Handled in TR, (f) Evaporation Losses in m3/hr, (g) Blow down requirement in m3/hr, (h) Make up water requirement in m3/hr.
Model answer: CT water flow = 1450 m3/hr = 1450000 kg/hr; CT fan flow = 950000 m3/hr; fan flow mass @1.08 kg/m3 = 1026000 kg/hr. (d) L/G Ratio = 1450000/1026000 = 1.41325 kg/kg. (a) Range = 43 - 36.6 = 6.4 C. (b) Approach = 36.6 - 30 = 6.6 C. (c) % CT Effectiveness = 100 x Range/(Range + Approach) = 100 x 6.4/(6.4+6.6) = 49.23%. (e) Cooling duty = 1450 x 6.4 x 10^3 = 9280 x 10^3 kCal/hr; /3024 = 3068 TR. (f) Evaporation losses = 0.00085 x 1.8 x 1450 x 6.4 = 14.1984 m3/hr (% evaporation loss = 14.1984/1450 x 100 = 0.98%). (g) Blow down = Evaporation losses/(COC - 1) = 14.198/(2.7-1) = 8.352 m3/hr. (h) Make up water = Evaporation loss + Blow down loss = 14.198 + 8.352 = 22.55 m3/hr.
Standard CT formulae: range, approach, effectiveness, L/G, duty in TR, evaporation (0.00085 x 1.8 x flow x range), blowdown = evap/(COC-1), make-up = evap + blowdown.
Source: 11th Feb-2011
📖 §7.2 Factors Affecting Performance – Heat Load (heat-rejection table) (condenser 555 kcal/kg); Range = Heat load / Water circulation rate (pump power is Pumps chapter)
16. In a Thermal Power Station, the steam input to a turbine operating on a fully condensing mode is 100 Tonnes/hr. The heat rejection requirement of the steam turbine condenser is 555 kcals/kg of steam condensed. The head developed by the cooling water pump is 2.5 kg/cm2. During 4500 hours of normal operation per year, the cooling water temperatures at the inlet and outlet of turbine condenser are measured to be 27degC and 35degC and during the winter period operation of 3000 hours per year the cooling water temperatures at the inlet and outlet of turbine condenser are measured to be 15degC and 25degC. Find out: i. The circulating cooling water flow for normal operation as well as for winter operation (5 marks). ii. Calculate the annual energy reduction during winter operation if the combined efficiency of the pump and motor is 70% (5 marks).
Model answer: i. Normal operation (27/35degC, 8degC rise, 4500 hrs): Heat rejected = 100000 x 555 = 55.5 million kcal/hr. Cooling water flow = 55.5x10^6/(35-27) x sp.heat 1 x (1/1000) = 6937.5 m3/hr. Head = 2.5 kg/cm2. Hydraulic power = (6937.5/3600)*25*9.81 = 472.62 kW. Input power = 472.62/0.7 = 675.16 kW. Winter operation (15/25degC, 10degC rise, 3000 hrs): Cooling water flow = 55.5x10^6/(25-15) x 1 x (1/1000) = 5550 m3/hr. Hydraulic power = (5550/3600)*25*9.81 = 378.09 kW. Input power = 378.09/0.7 = 540.13 kW. ii. Energy reduction during winter = (675.16 - 540.13) x 3000 = 405090 kWh.
Cooling water flow = heat rejected/(temperature rise x sp.heat x 1000); pump hydraulic power = (flow/3600) x head(m) x 9.81; input = hydraulic/combined efficiency; annual saving = power difference x winter hours.
17. In a steel industry, cooling water of 7500 m³/hr and 4200 m³/hr from two different sections with temperatures of 38 °C and 55 °C respectively, are fed to cooling tower after proper mixing. If the measured heat rejection by the cooling tower is 38,000 TR, calculate the effectiveness and evaporation loss of the cooling tower at 28 °C WBT.
Model answer: Mixed Hot Water Temp, °C = [(Flow1 × Temp1) + (Flow2 × Temp2)] / (Total Flow) = [(7500 × 38) + (4200 × 55)] / 11700 = 44.1 °C. Range of Cooling Tower, °C = Heat Rejection / (Flow × Density × Sp. Heat) = (38000 × 3024) / (11700 × 1000 × 1) ≈ 9.82 °C [book: 9.82]. Cold Water Temp, °C = Hot Water Temp - Range = 44.1 - 9.82 = 34.28 °C. Approach, °C = Cold Water Temp - WBT of Air = 34.28 - 28 = 6.28 °C. Effectiveness = Range / (Range + Approach) = 9.82 / (9.82 + 6.28) = 60.99% or 61%. Evaporation Loss (m³/hr) = 0.00085 × 1.8 × circulation rate (m³/hr) × Range = 0.00085 × 1.8 × 11700 × 9.82 = 175.8 m³/hr.
Mixed temp by flow-weighted average; range from heat rejection; effectiveness = range/(range+approach); evaporation = 0.00085×1.8×flow×range. ⚠ Question stem repaired: garbled 'heat rejection ... is at 1700 m³/hr' → '38,000 TR' (the value used in the printed solution: 38000 × 3024 / 11,700,000 = 9.82 °C range).