108 questions — 73 objective (1 mark), 21 short (5 marks), 14 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) — 73
📖 §5.3.4 Fan Laws (Affinity Laws)
1. According to the fan affinity laws, fan power (kW) varies with which power of the rotational speed N?
N
N²
N³
√N
Answer: C) N³
Confirmed vs Book-3 §5.3.4 — the book states Flow ∝ Speed, Pressure ∝ (Speed)², Power ∝ (Speed)³. So fan shaft power follows N³.
This cube law is exactly why a VFD is the most efficient capacity control: a small speed cut gives a large power cut.
The tempting wrong option is b) N², which is the PRESSURE law, not the power law.
Source: AI practice
📖 §5.3.4 Fan Laws (Affinity Laws)
2. A fan running at 800 rpm draws 16 kW. If the speed is reduced to 600 rpm, the new power input will be approximately:
12.0 kW
9.0 kW
6.75 kW
4.5 kW
Answer: C) 6.75 kW
Confirmed vs Book-3 §5.3.4 — kW₂ = kW₁ × (N₂/N₁)³ = 16 × (600/800)³ = 16 × (0.75)³ = 16 × 0.4219 = 6.75 kW.
Speed ratio 0.75; cubing it (not squaring it) is the whole point of the fan law.
Option b) 9.0 kW is what you get from the wrong law kW ∝ N² (16 × 0.5625) — the classic exam trap.
Source: AI practice
📖 §5.1 Table 5.1 — Difference between Fans, Blowers and Compressors
3. Per the ASME definition, the specific ratio that classifies fans, blowers and compressors is defined as:
suction pressure ÷ discharge pressure
discharge pressure ÷ suction pressure
discharge flow ÷ suction flow
static pressure ÷ total pressure
Answer: B) discharge pressure ÷ suction pressure
Confirmed vs Book-3 §5.1 — ASME defines the specific ratio as discharge pressure ÷ suction pressure.
Book values: Fan up to 1.11, Blower 1.11 to 1.20, Compressor more than 1.20; hence Compressor > Blower > Fan.
Option a) inverts the ratio and would make the fan value greater than 1 only if pressure fell across the fan, which is not the book definition.
Source: AI practice
📖 §5.2 Types of Fans — Centrifugal vs Axial Flow
4. In a centrifugal (radial-flow) fan, how many times does the air change direction as it passes through the fan?
None
Once
Twice
Three times
Answer: C) Twice
Confirmed vs Book-3 §5.2 — the book states: 'In centrifugal flow, airflow changes direction twice' — once entering and once leaving the impeller.
By contrast, in axial flow the air enters and leaves the fan with no change in direction (propeller, tube-axial, vane-axial).
Option a) None describes the axial fan, not the centrifugal fan.
Source: AI practice
📖 §5.2.1 Centrifugal Fan Types / Table 5.3
5. Which centrifugal fan type is described as 'non-overloading' because its power draw drops off so static-pressure changes do not overload the motor?
Radial (paddle blade)
Forward-curved (multi-vane)
Backward-curved / backward-inclined
Propeller (axial flow)
Answer: C) Backward-curved / backward-inclined
Confirmed vs Book-3 §5.2.1 — 'Backward-inclined fans are known as "non-overloading" because changes in static pressure do not overload the motor'; they reach peak power and then power demand drops off within the useable airflow range.
Radial and forward-curved fans are the trap: Table 5.3 says their power rises continuously with flow, so they CAN overload the motor.
(Option d) was originally 'Airfoil', which Table 5.3 defines as 'same as backward curved type' — that made two options correct, so it has been replaced.)
Source: AI practice
📖 §5.6.8 Fan Efficiency (static & mechanical efficiency)
6. In the fan static-efficiency formula η = [Q(m³/s) × SP(mmWC)] / [constant × Shaft kW] × 100, the fixed constant is:
75
102
367
1000
Answer: B) 102
Confirmed vs Book-3 §5.6.8 — Fan Static Efficiency % = [Volume (m³/s) × Static pressure (mmWC)] / [102 × Power input to fan shaft (kW)] × 100.
The same 102 appears in the mechanical (total) efficiency formula, with total pressure instead of static pressure.
Option c) 367 belongs to the PUMP hydraulic-power formula (Q m³/hr × H × ρ / 367), not to fans.
Source: AI practice
📖 §5.3.4 Fan Laws (Affinity Laws)
7. A fan delivers 3000 Nm³/hr at 600 mmWC static pressure at 900 rpm. If the speed is reduced to 600 rpm, the new static pressure is about:
400 mmWC
266.7 mmWC
200 mmWC
600 mmWC
Answer: B) 266.7 mmWC
Confirmed vs Book-3 §5.3.4 — SP ∝ N², so SP₂ = 600 × (600/900)² = 600 × 0.4444 = 266.7 mmWC.
Flow would fall linearly: 3000 × 600/900 = 2000 Nm³/hr.
Option a) 400 mmWC is 600 × (600/900), i.e. the FLOW law wrongly applied to pressure.
Source: AI practice
📖 §5.3.4 Fan Laws — the book's '10% change' facts
8. Reducing fan speed by 10% changes the air delivery (flow) by approximately:
19%
27%
10%
33%
Answer: C) 10%
Confirmed vs Book-3 §5.3.4 — Flow ∝ Speed, so a 10% speed reduction gives a 10% flow reduction.
The book's own recall set: 10% rpm reduction → flow down 10%, static pressure down 19% (0.9² = 0.81) and power down 27% (0.9³ = 0.729).
Options a) 19% and b) 27% are those pressure and power figures planted as distractors for the flow question.
Source: AI practice
📖 §5.6.8 Fan Efficiency
9. For a given duty, the power drawn by a fan is:
directly proportional to its efficiency
inversely proportional to its efficiency
independent of its efficiency
proportional to the square of its efficiency
Answer: B) inversely proportional to its efficiency
Confirmed vs Book-3 §5.6.8 — rearranging η = (Q × SP)/(102 × kW) gives kW = (Q × SP)/(102 × η), so for a fixed duty (fixed Q and SP) power is inversely proportional to efficiency.
A lower-efficiency fan therefore draws MORE power for the same air delivery and pressure.
Option a) is the trap: efficiency and power move in opposite directions, not together.
Source: AI practice
📖 §5.7 End-of-chapter L-2 — ventilation by air changes per hour
10. An engine room measures 30 × 20 × 5 m and requires 20 air changes per hour (ACH). The required ventilation flow is:
6,000 m³/hr
30,000 m³/hr
60,000 m³/hr
120,000 m³/hr
Answer: C) 60,000 m³/hr
Confirmed vs Book-3 §5.7 (L-2 data) — Required flow = room volume × air changes per hour.
Volume = 30 × 20 × 5 = 3000 m³; flow = 3000 × 20 = 60,000 m³/hr (= 16.67 m³/s, the value you feed into the 102 formula).
Option b) 30,000 m³/hr comes from halving, and a) 6,000 m³/hr from using only 2 ACH — neither uses the full 3000 m³ volume.
Source: AI practice
📖 §5.5 Flow Control Strategies (§5.5.2 Dampers, §5.5.5 VSD)
11. Which fan flow-control method is the LEAST energy-efficient?
Variable Speed Drive (VFD)
Inlet Guide Vanes
Variable pitch blades
Damper control
Answer: D) Damper control
Confirmed vs Book-3 §5.5.2 — dampers change air volume 'by adding or removing system resistance', forcing the fan up or down its own curve; the book says they 'are not particularly energy efficient'.
Inlet guide vanes are efficient down to about 80% flow, variable-pitch blades give 'dramatically higher energy efficiency', and the VSD is the best because power varies as the cube of speed.
So damper control is the least efficient of the four listed.
Source: AI practice
📖 §5.3.1 System Characteristics (System Resistance)
12. In a fan-duct system, the system resistance varies with flow as:
directly proportional to flow
the square of flow
the cube of flow
independent of flow
Answer: B) the square of flow
Confirmed vs Book-3 §5.3.1 — 'The system resistance varies with the square of the volume of air flowing through the system.'
System resistance is the sum of static pressure losses in ducts, elbows, pickups and equipment; the operating point is where this system curve cuts the fan curve.
Option c) the cube of flow is the POWER law (kW ∝ N³), not the resistance law.
Source: AI practice
📖 §5.3 Fan Laws
13. What is the effect of decreasing the RPM of a fan by 10% on its power requirement?
Decreases the power requirement by 27%
Decreases the power requirement by 19%
Increases the power requirement by 10%
No significant effect
Answer: A) Decreases the power requirement by 27%
Confirmed vs Book-3 §5.3 — Fan law: Power ∝ (Speed)³, so at 90% speed power = 0.9³ = 0.729, a 27% reduction; the book states this figure verbatim. Option (b) 19% is the STATIC-PRESSURE reduction (0.9² = 0.81), not power — the classic distractor.
Source: Sep 2024
📖 §5.3 System characteristics & fan curves
14. How does an increase in system resistance affect the operation of a centrifugal fan?
Increases the airflow
Reduces the airflow
Reduces the static pressure
No effect on fan performance
Answer: B) Reduces the airflow
Confirmed vs Book-3 §5.3 — Raising system resistance (e.g. partly closing a damper) creates a steeper system curve SC₂; the operating point moves from A to B on the same fan curve: LOWER flow Q₂ against HIGHER pressure P₂. Option (c) is wrong because static pressure rises, not falls; system resistance varies as (flow)².
Source: Sep 2024
📖 §5.3 Fan Laws
15. The power measured in an Induced Draft (I D) fan operating at 49 Hz is 52 kW. A Variable Frequency Drive (VFD) is installed and the fan was operated at 34 Hz, The estimated Power saving will be_________
35.7 kW
17.3 kW
34.6 kW
36 kW
Answer: C) 34.6 kW
Confirmed vs Book-3 §5.3 — Speed ∝ frequency and Power ∝ (Speed)³: P₂ = 52 × (34/49)³ = 52 × 0.334 = 17.4 kW. Saving = 52 − 17.4 = 34.6 kW. Option (b) 17.3 kW is the NEW power, not the saving — a common trap.
Source: Sep 2024
📖 §5.3 Fan Laws
16. A fan is drawing 16 kW at 800 RPM. If its speed is reduced to 600 RPM, the power drawn by the fan will be____________
6.75 kW
9 kW
12 kW
None of the above
Answer: A) 6.75 kW
Confirmed vs Book-3 §5.3 — kW₂/kW₁ = (N₂/N₁)³ → 16 × (600/800)³ = 16 × 0.4219 = 6.75 kW. Option (c) 12 kW assumes power ∝ speed (linear) — that law applies to FLOW, not power.
Source: Sep 2024
📖 §5.3 System characteristics & fan curves
17. Larger diameter ducts in fans:
Increase system resistance
Reduce system resistance
No effect
Increase static pressure
Answer: B) Reduce system resistance
Confirmed vs Book-3 §5.3 — The book states a fan 'in a system with narrow ducts and multiple short-radius elbows' must work harder than 'in a system with larger ducts'; larger ducts lower velocity and friction losses, so system resistance falls. Option (d) is wrong: lower resistance means the fan operates at LOWER static pressure (and higher flow) on its curve.
Source: Sep 2025
📖 §5.5 Flow control strategies
18. A system resistance curve of a fan changes with:
Inlet guide vanes
Discharge dampers
Speed change with VFD
Any of the above
Answer: B) Discharge dampers
Corrected (was d) — Book-3 §5.5/§5.3: Only dampers 'change air volume by adding or removing SYSTEM RESISTANCE', i.e. they move the system resistance curve (SC₁→SC₂, Fig 5.7). Inlet guide vanes 'change the characteristics of the FAN curve', and a VFD speed change moves the fan onto a new fan curve (N₁→N₂) while the system curve stays the same (point C in Fig 5.7). Hence 'any of the above' contradicts the book; the 18th-exam and Sep-2019 keys on the same point also give 'discharge damper'.
Source: Sep 2025
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
19. The specific ratio as defined by ASME and used in differentiating fans, blowers and compressors, is given by
Confirmed vs Book-3 §5.1 — ASME specific ratio = discharge pressure / suction pressure (Table 5.1): fans up to 1.11, blowers 1.11–1.20, compressors > 1.20. Option (b) inverts the ratio and would give values below 1 — meaningless for classification.
Source: Jul 2022
📖 §5.2 Fan types (Tables 5.2/5.3)
20. The efficiency of backward-inclined fans compared to forward-curved fans is
lower
higher
same
none of the above
Answer: B) higher
Confirmed vs Book-3 §5.2 — 'Backward-inclined fans are more efficient than forward-curved fans' (Table 5.2: backward 79–83% vs forward-curved 60–65%). They are also 'non-overloading' — power peaks and then drops within the usable flow range, unlike forward-curved fans whose power rises continuously.
Source: Book EOC
📖 §5.3 Fan Laws
21. For centrifugal fans, the relation between shaft input Power (kW) and Speed (N) is given by
kW2/kW1 = N2/N1
kW2/kW1 = (N2/N1)^2
kW2/kW1 = (N2/N1)^3
none of the above
Answer: C) kW2/kW1 = (N2/N1)^3
Confirmed vs Book-3 §5.3 — Fan laws: Q ∝ N, SP ∝ N², kW ∝ N³, so kW₂/kW₁ = (N₂/N₁)³. Option (b) (square law) is the PRESSURE relation; option (a) (linear) is the FLOW relation.
Source: Book EOC
📖 §5.5 Series and parallel operation
22. Parallel operation of two identical fans in a ducted system.
will double the flow
will double the fan static pressure
will increase flow by more than two times
will not double the flow
Answer: D) will not double the flow
Confirmed vs Book-3 §5.5 — Two fans in parallel double the volume 'only at free delivery'; with a ducted (resisting) system, 'the higher the system resistance, the less increase in flow'. So flow does NOT double (d). Doubling static pressure (b) is the ideal for SERIES operation, not parallel.
Source: Mar 2023
📖 §5.3 Fan Laws
23. A fan is operating at 970 RPM developing a flow of 3000 Nm3/hr at a static pressure of 650 mmWC. If the speed is reduced to 700 RPM, the static pressure (mmWC) developed will be
338.5 (printed in book as 388.5 — typo)
244.3
469
none of the above
Answer: A) 338.5 mmWC (book prints option (a) as 388.5 — a typo for 338.5)
Confirmed vs Book-3 §5.3 — SP ∝ N²: SP₂ = 650 × (700/970)² = 650 × 0.5207 = 338.5 mmWC. The intended book key is option (a); the printed value 388.5 is a misprint of 338.5 (the Master Notes key also maps Q6 to (a)). Option (c) 469 = 650 × (700/970) is the LINEAR (flow-law) error; (b) 244.3 = 650 × (700/970)³ is the cube (power-law) error. Note: the 19th-exam re-use of this question omits 388.5 and its key is 'none of the above' (338.5).
Source: Book EOC
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
24. Identify the correct statement:
the specific ratio of compressors is higher than blowers
the specific ratio of fans is higher than blowers
the specific ratio of compressors is lower than fans
the specific ratio of blowers is higher than compressors
Answer: A) the specific ratio of compressors is higher than blowers
Confirmed vs Book-3 §5.1 Table 5.1 — Specific ratio: fans up to 1.11 < blowers 1.11–1.20 < compressors > 1.20. Hence compressors > blowers (a). Options (b), (c), (d) each reverse this ordering.
Source: Book EOC
📖 §5.3 Fan Laws
25. Decreasing the rpm of a fan at partial loading by 10% results in:
decrease of 10% in flow rate and decrease of 27% in power requirement
decrease of 10% in flow rate and decrease of 19% in power requirement
decrease of 10% in flow rate and increase of 10% in power requirement
increase of 10% in flow rate and no appreciable change in power requirement
Answer: A) decrease of 10% in flow rate and decrease of 27% in power requirement
Confirmed vs Book-3 §5.3 — 'Varying the RPM by 10% decreases or increases air delivery by 10%'; 'reducing the RPM by 10% decreases the power requirement by 27%' (0.9³ = 0.729). Option (b) 19% is the static-pressure drop (0.9² = 0.81), not the power drop.
Source: Book EOC
📖 §5.6 Fan efficiency formula
26. The power drawn by a centrifugal fan is
inversely proportional to fan efficiency
directly proportional to fan efficiency
inversely proportional to static pressure
inversely proportional to flow rate
Answer: A) inversely proportional to fan efficiency
Confirmed vs Book-3 §5.6 — Fan static efficiency = (Q × SP)/(102 × shaft kW), rearranged: shaft power = (Q × SP)/(102 × η). Power is therefore INVERSELY proportional to fan efficiency and DIRECTLY proportional to flow and static pressure. Options (c) and (d) reverse the direct proportionality to pressure and flow.
Source: Jul 2022
📖 §5.5 Variable speed drives; §5.3 Fan laws
27. The main reason for using a Variable Frequency Drive (VFD) for capacity control in electrical motor-driven centrifugal fans with fluctuating load is:
improved power quality
fan capacity is proportional to its speed whereas the power drawn by the fan is proportional to the cube of its speed
improved power factor
precise closed-loop process control
Answer: B) fan capacity is proportional to its speed whereas the power drawn by the fan is proportional to the cube of its speed
Confirmed vs Book-3 §5.5 — 'Since power input to the fan changes as the cube of the flow, this will usually be the most efficient form of capacity control.' Flow ∝ N while power ∝ N³, so a small speed cut for a fluctuating load gives a large power saving. Power quality, power factor and closed-loop control (a, c, d) are side benefits, not the energy reason.
Source: Book EOC
📖 §5.5 Pulley change
28. A fan with 25 cm pulley diameter is driven by a 2940 rpm motor through a V-belt system. If the motor pulley is reduced from 20 cm to 15 cm keeping the motor rpm and fan pulley diameter the same, the fan speed will reduce by
1176 rpm
1764 rpm
588 rpm
none of the above
Answer: C) 588 rpm
Confirmed vs Book-3 §5.5 — Belt drive: fan rpm = motor rpm × (motor pulley ÷ fan pulley). Before: 2940 × 20/25 = 2352 rpm; after: 2940 × 15/25 = 1764 rpm; reduction = 588 rpm (c). Option (b) 1764 is the NEW speed, not the reduction; (a) 1176 = 2940 × 10/25 is a mis-subtraction.
Source: 15th Exam
📖 §5.5 Series and parallel operation
29. In series operation of identical centrifugal fans, ideally
flow doubles
static pressure doubles
static pressure goes up by four times
flow goes up by four times
Answer: B) static pressure doubles
Confirmed vs Book-3 §5.5 — Series (push-pull) staging raises 'the static pressure capability at a given airflow' — ideally doubling it for two identical fans (though 'not double at every flow point'). Doubling FLOW (a) is the ideal for PARALLEL operation; four-times (c, d) has no basis in the fan laws.
Source: 15th Exam
📖 §5.6 Measurement by pitot tube
30. The inner tube of an L-type pitot tube is used to measure …… in the air duct
total pressure
static pressure
velocity pressure
dynamic pressure
Answer: A) total pressure
Confirmed vs Book-3 §5.6 — 'Total pressure is measured using the inner tube of pitot tube and static pressure is measured using the outer tube.' Connecting both to one manometer gives velocity pressure (TP − SP). Options (c)/(d) (velocity = dynamic pressure) are the manometer DIFFERENCE, not what the inner tube alone senses.
Source: 15th Exam
📖 §5.3 Fan Laws
31. If the speed of a centrifugal fan is reduced to 80% of its rated speed then the power drawn will be _______% of its rated power:
80%
51.2 %
40 %
64 %
Answer: B) 51.2 %
Confirmed vs Book-3 §5.3 — Power ∝ N³: (0.8)³ = 0.512 → 51.2% of rated power. Option (d) 64% = 0.8² is the static-pressure fraction; (a) 80% is the flow fraction.
Source: 15th Exam
📖 §5.2 Fan types (Tables 5.2/5.3)
32. In which of the following fans does the air not change flow direction from suction to discharge?
tube axial fan
vane axial fan
propeller fan
all of the above
Answer: D) all of the above
Confirmed vs Book-3 §5.2 — 'In axial flow, air enters and leaves the fan with no change in direction (propeller, tubeaxial, vaneaxial)'. All three listed are axial types → (d). Centrifugal fans, by contrast, change airflow direction twice.
Source: 14th Exam
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
33. The parameter used by ASME to classify fans, blowers and compressors is
volume ratio
specific ratio
blade ratio
impeller ratio
Answer: B) specific ratio
Confirmed vs Book-3 §5.1 — 'As per ASME the specific ratio — the ratio of the discharge pressure over the suction pressure — is used for defining the fans, blowers and compressors' (Table 5.1). Volume, blade and impeller ratios (a, c, d) are not ASME classification parameters.
Source: 14th Exam
📖 §5.6 Fan efficiency / solved example
34. The pressure to be considered for calculating the power required for centrifugal fans is
vapour pressure
dynamic pressure
total static pressure
velocity pressure
Answer: C) total static pressure
Confirmed vs Book-3 §5.6 — Fan static efficiency (and hence power) uses 'Volume in m³/s × total static pressure in mmWC / (102 × shaft kW)'; in the solved example total static pressure = outlet SP − inlet SP = 185 − (−20) = 205 mmWC. Velocity/dynamic pressure (b, d) is added only for the mechanical (total) efficiency; vapour pressure (a) belongs to pump cavitation.
Source: 14th Exam
📖 §5.6 Measurement by pitot tube
35. The inclined manometer connected to a pitot tube is used for measuring which pressure in a gas stream?
velocity
static
total
all of the above
Answer: A) velocity
Confirmed vs Book-3 §5.6 — 'When the inner and outer tube ends are connected to a manometer, we get the velocity pressure. For measuring low velocities, it is preferable to use an inclined tube manometer.' The manometer reads the DIFFERENCE (TP − SP) = velocity pressure, not static or total alone (b, c).
Source: 14th Exam
📖 §5.7 Energy savings (item 4: hollow FRP impeller)
36. FRP fans consume less energy than aluminium fans because
they are lighter
they have better efficiencies
they encounter less system resistance
they deliver less air flow
Answer: A) they are lighter
Confirmed vs Book-3 §5.7 — ECO item 4: replace metallic/GRP impellers with 'the more energy efficient hollow FRP impeller with aerofoil design' (axial fans, e.g. cooling towers). The hollow FRP blade is much lighter, so less driving power is needed — the standard BEE key is 'they are lighter'. FRP does not change the system resistance (c) or reduce the air delivered (d).
Source: Set-A
📖 §5.5 Variable speed drives; §5.3 Fan laws
37. The main reason for using a Variable Frequency Drive (VFD) for capacity control in electric motor driven centrifugal fans with fluctuating load is:
improved power quality
fan capacity is proportional to its speed whereas the power drawn is proportional to the cube of its speed
improved power factor
precise closed loop process control
Answer: B) fan capacity is proportional to its speed whereas the power drawn is proportional to the cube of its speed
Confirmed vs Book-3 §5.5 — Flow ∝ N but power ∝ N³, so matching a fluctuating load by speed reduction gives cube-law savings; the book calls VSD 'usually the most efficient form of capacity control'. Power quality/PF/closed-loop control (a, c, d) are not the energy justification.
Source: Set-A
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
38. The distinction between fans and blowers is based on:
impeller diameter
specific ratio
speed
volume delivered
Answer: B) specific ratio
Confirmed vs Book-3 §5.1 — Fans, blowers and compressors are distinguished by the ASME specific ratio (discharge/suction pressure): fan up to 1.11, blower 1.11–1.20, compressor > 1.20. Impeller diameter, speed or volume (a, c, d) are not the classification basis.
Source: 17th Sep-2016
📖 §5.6 Calculation of gas density (γ = PM/RT)
39. Find the air density at 35°C temperature at one atmospheric pressure. It is given that at one atmospheric pressure the air density at 20°C is 1.2041 kg/m3.
1.1455
1.2657
1.2024
none of the above
Answer: A) 1.1455
Confirmed vs Book-3 §5.6 — From γ = PM/RT at constant pressure, density ∝ 1/T(K): ρ₃₅ = 1.2041 × (273+20)/(273+35) = 1.2041 × 293/308 = 1.1455 kg/m³. Option (b) 1.2657 inverts the temperature ratio (density cannot rise with heating at constant pressure).
Source: 17th Sep-2016
📖 §5.3 System characteristics & fan curves
40. The fan system resistance is predominately due to:
more bends used in the duct
more equipments in the system
volume of air handled
density of air
Answer: C) volume of air handled
Confirmed vs Book-3 §5.3 — 'The system resistance varies with the square of the volume of air flowing through the system … the system resistance increases substantially as the volume of air increases.' Bends, equipment and density (a, b, d) set the resistance coefficient, but for a given system the resistance is governed predominantly by the volume handled (∝ Q²).
Source: 16th Exam (alt set)
📖 §5.6 Calculation of gas density (γ = PM/RT)
41. Calculate the density of air at 11400 mmWC absolute pressure and 65°C. (Molecular weight of air: 28.92 kg/kg mole and Gas constant: 847.84 mmWC m3/kg mole K)
1.2 kg/m3
1.5 kg/m3
1.15 kg/m3
none of the above
Answer: C) 1.15 kg/m3
Confirmed vs Book-3 §5.6 — γ = P×M/(R×T) with P in mmWC, R = 847.84 mmWC·m³/kg-mole·K: (11400 × 28.92)/(847.84 × 338) = 329,688/286,570 = 1.15 kg/m³ (c). Forgetting to convert 65 °C to 338 K is the usual error.
Source: 16th Exam
📖 §5.3 Fan laws (variable-torque loads); cf. Book-3 Ch2 motors
42. Which of the following is an example of variable torque equipment?
centrifugal pump
reciprocating compressor
screw compressor
roots blower
Answer: A) centrifugal pump
Confirmed vs Book-3 §5.3 — Centrifugal machines (pumps, fans) follow the affinity laws: power ∝ N³, so torque ∝ N² — a VARIABLE-torque load, which is why VFDs save so much on them. Reciprocating and screw compressors and roots blowers (b, c, d) are positive-displacement, essentially constant-torque loads.
Source: 18th Exam
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
43. Which among the following is one of the parameters used to classify fans, blowers & compressors?
air flow
speed RPM
specific ratio
none of the above
Answer: C) specific ratio
Confirmed vs Book-3 §5.1 — ASME specific ratio (discharge/suction pressure) classifies fans (≤1.11), blowers (1.11–1.20) and compressors (>1.20). Air flow or rpm (a, b) are not the classification parameters.
Source: 18th Exam
📖 §5.6 Measurement by pitot tube
44. The inner tube of a L-type Pitot tube facing the flow measures ________ in the fan system
static pressure
velocity pressure
total pressure
all of the above
Answer: C) total pressure
Confirmed vs Book-3 §5.6 — The inner (impact) tube facing the flow measures TOTAL pressure; the outer tube with side holes measures STATIC pressure; the two together on one manometer give VELOCITY pressure. So (c), not (a) or (b); 'all of the above' (d) is wrong since one tube senses only one pressure.
Source: 18th Exam
📖 §5.3 Fan laws; §5.6 gas density
45. State which of the following statements is true?
for a given fan operating at a constant temperature, the power input to fan increases by 4 times when the fan speed becomes double
for a given fan operating at a constant temperature, the power input to fan increases by 8 times when the fan speed becomes double
for a given fan operating at a constant flow rate, the power input increases as the air temperature increases
for a given fan operating at a constant static pressure rise, the flow rate reduces as the air temperature increases
Answer: B) for a given fan operating at a constant temperature, the power input to fan increases by 8 times when the fan speed becomes double
Confirmed vs Book-3 §5.3 — Power ∝ N³, so doubling speed gives 2³ = 8 times the power (b); (a) 4× would be the PRESSURE rise. (c) is wrong: at constant flow, hotter (less dense) air needs LESS power; (d) is wrong: volume flow at a given speed is set by the fan laws, not by temperature.
Source: 18th Exam
📖 §5.5 Flow control strategies
46. Which of the following flow controls in a fan system will change the system resistance curve:
Inlet guide vane
speed change with variable frequency drive
speed change with hydraulic coupling
discharge damper
Answer: D) discharge damper
Confirmed vs Book-3 §5.5 — Dampers change volume 'by adding or removing system resistance', i.e. they shift the SYSTEM resistance curve (SC₁→SC₂ in Fig 5.7). Inlet guide vanes change the FAN curve characteristics; speed changes by VFD or hydraulic coupling (b, c) move the fan to a new fan curve while the system curve is unchanged.
Source: 18th Exam
📖 §5.7 Energy savings (item 4: hollow FRP impeller)
47. Fiberglass Reinforced Plastic (FRP) fans consume less energy than aluminum fans because
They are lighter
They have better efficiencies
They encounter less system resistance
They deliver less air flow
Answer: A) They are lighter
Confirmed vs Book-3 §5.7 — The book lists replacing metallic/GRP impellers with 'hollow FRP impeller with aerofoil design' as an ECO for axial fans 'where significant savings have been reported'. The hollow FRP blade is lighter, so less shaft power is needed (standard BEE key: 'they are lighter'). System resistance (c) is a duct-system property, unaffected by blade material.
Source: 19th Exam
📖 §5.3 Fan Laws
48. A fan is drawing 16 kW at 800 RPM. If the speed is reduced to 600 RPM then the power drawn by the fan would be
12 kW
1.38
6.75 kW
none of the above
Answer: C) 6.75 kW
Confirmed vs Book-3 §5.3 — Power ∝ N³: 16 × (600/800)³ = 16 × 0.4219 = 6.75 kW (c). (a) 12 kW is the linear (flow-law) error; 16 × 0.75² = 9 kW would be the pressure-law error.
Source: Mar 2021
📖 §5.2 Fan types (Tables 5.2/5.3)
49. In which of the following fans does air enter and leave the fan with no change in direction?
Forward curved
Backward curved
Radial
Propeller
Answer: D) Propeller
Confirmed vs Book-3 §5.2 — 'In axial flow, air enters and leaves the fan with no change in direction (propeller, tubeaxial, vaneaxial)'. Propeller is the only axial type listed → (d). Forward-curved, backward-curved and radial (a, b, c) are centrifugal designs, where 'airflow changes direction twice'.
Source: 19th Exam
📖 §5.3 Fan Laws
50. A fan operating at 970 RPM develops a flow of 3000 Nm3/hour at a static pressure of 650 mmWC. If the speed is reduced to 700 RPM, the static pressure (mmWC) developed will be
244.3
650
469
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §5.3 — SP ∝ N²: 650 × (700/970)² = 650 × 0.5207 = 338.5 mmWC, which is not printed → 'None of the above' (d). (c) 469 is the linear-law error, (a) 244.3 the cube-law error, (b) 650 assumes no change.
Source: 19th Exam
📖 §5.3 Fan Laws
51. The power measured in a boiler ID fan is 52 kW operating at 49 Hz. As an energy conservation measure a VFD was installed and the fan was operated at 34 Hz. The estimated power savings will be
36 kW
17.2 kW
34.7 kW
35.7 kW
Answer: C) 34.7 kW
Corrected (was d) — Book-3 §5.3: Power ∝ (Speed)³ and speed ∝ frequency: P₂ = 52 × (34/49)³ = 52 × 0.334 = 17.4 kW; saving = 52 − 17.4 = 34.6 kW → nearest printed option is (c) 34.7 kW, not (d) 35.7 kW (the previous entry's own arithmetic gave 34.6 but picked the wrong option). The Sep-2024 repeat of this question prints 34.6 kW as the key, confirming (c). Option (b) 17.2 kW is roughly the NEW power, not the saving.
Source: 19th Exam
📖 §5.6 Calculation of velocity (pitot)
52. A coal fired boiler primary air fan maintains a velocity pressure of 70 mmWC, air temperature 38 °C, air density 1.135 kg/m3 and pitot tube constant 0.85. The velocity of air in m/sec will be
25.6
29.56
28.67
None of the above
Answer: B) 29.56
Confirmed vs Book-3 §5.6 — V = Cp × √(2 × 9.81 × ΔP/γ) with ΔP in mmWC: 0.85 × √(2 × 9.81 × 70/1.135) = 0.85 × √1210 = 0.85 × 34.78 = 29.56 m/s (b). (a) 25.6 m/s is the book's worked example (47 mmWC, Cp 0.9) — a memory trap; (c) 28.67 results from Cp 0.9 with 47 mmWC and other mixes.
Source: 19th Exam
📖 §5.5 Flow control strategies
53. Flow control with ___________ in a fan system will not change the fan characteristic curve.
Inlet guide vane
speed change with variable frequency drive
speed change with hydraulic coupling
discharge damper
Answer: D) discharge damper
Confirmed vs Book-3 §5.5 — A discharge damper only adds system resistance; the fan stays on the same fan curve and moves along it ('forces the fan to move up or down along its characteristic curve … without changing fan speed'). Inlet guide vanes change the fan-curve characteristics, and speed change (VFD or hydraulic coupling) puts the fan on a different fan curve (a, b, c).
Source: Sep 2019
📖 §5.5 Inlet guide vanes
54. Modest flow variation between 80% to 100%, in a centrifugal fan is achieved more efficiently with ___________________.
Inlet damper
Outlet damper
Inlet guide vanes
Impeller Change
Answer: C) Inlet guide vanes
Confirmed vs Book-3 §5.5 — 'Guide vanes are energy efficient for modest flow reductions — from 100 percent flow to about 80 percent. Below 80 percent flow, energy efficiency drops sharply.' Dampers (a, b) are 'not particularly energy efficient'; impeller change (d) is a permanent derating, not a control method for variation.
Source: Sep 2019
📖 §5.2 Fan types (Tables 5.2/5.3)
55. Backward-inclined fans are known as _____ because change in static pressure does not overload the motor
overloading
non-overloading
radial
axial
Answer: B) non-overloading
Confirmed vs Book-3 §5.2 — 'Backward-inclined fans are known as "non-overloading" because changes in static pressure do not overload the motor' — their power peaks and then falls within the usable flow range. Forward-curved fans, whose power rises continuously with flow, are the overloading type.
Source: 9th Dec-2009
📖 §5.3 System characteristics & fan curves
56. The fan characteristic curve is a plot of
static pressure vs flow
dynamic pressure vs flow
total pressure vs flow
suction pressure vs flow
Answer: A) static pressure vs flow
Confirmed vs Book-3 §5.3 — Among the manufacturer's curves 'the curve static pressure (SP) vs. flow is especially important'; its intersection with the system curve defines the operating point. Total/dynamic/suction pressure vs flow (b, c, d) are not the standard fan characteristic.
Source: 9th Dec-2009
📖 §5.2 Fan types (Tables 5.2/5.3)
57. In which of the following fans air enters and leaves the fan with no change in direction
forward curved
backward curved
radial
propeller
Answer: D) propeller
Confirmed vs Book-3 §5.2 — Axial fans (propeller, tubeaxial, vaneaxial) pass air straight through with no change in direction; propeller (d) is the only axial option. Forward-curved, backward-curved and radial (a, b, c) are centrifugal — airflow turns twice.
Source: 9th Dec-2009
📖 Not in Book-3 Ch5 text (belt drive slip; cf. §5.5 Pulley change)
58. In a "V" belt coupled fan drive, the measured speed at motor end 6" diameter pulley is 1480 rpm and that at fan end 10" diameter pulley is 820 RPM. What is the slippage loss in %?
7.66
8.29
6.67
insufficient data, cannot be worked out
Answer: B) 8.29
Confirmed vs official key (topic not covered in Book-3 Ch5) — Theoretical fan speed = 1480 × 6/10 = 888 rpm; actual 820 rpm, so speed lost to slip = 68 rpm. The official key (b) 8.29% expresses this relative to the ACTUAL fan speed: 68/820 = 8.29%. Relative to the theoretical speed it would be 68/888 = 7.66% (option a) — follow the official key in the exam. The data is sufficient, so (d) is wrong.
Source: 9th Dec-2009
📖 §5.5 Inlet guide vanes
59. Modest flow variation, from 100% to 80%, in a centrifugal fan is achieved more efficiently with which of the following flow control methods
inlet damper
outlet damper
inlet guide vanes
none of the above
Answer: C) inlet guide vanes
Confirmed vs Book-3 §5.5 — Inlet guide vanes pre-swirl the inlet air and change the fan curve; they are 'energy efficient for modest flow reductions — from 100 percent flow to about 80 percent'. Inlet/outlet dampers (a, b) simply add resistance and are the least efficient control.
Source: 10th Jul-2010
📖 §5.5 Pulley change
60. A fan with 30 cm pulley diameter is driven by a 1480 rpm motor through a v-belt. If the motor pulley is reduced from 20 cm to 18 cm at the same motor rpm and fan pulley diameter, the fan speed will reduce by
247 rpm
888 rpm
98 rpm
none of the above
Answer: C) 98 rpm
Confirmed vs Book-3 §5.5 — Fan rpm = motor rpm × motor pulley/fan pulley: 1480 × 20/30 = 986.7 rpm → 1480 × 18/30 = 888 rpm; reduction ≈ 98.7 ≈ 98 rpm (c). (b) 888 is the new speed, not the reduction; (a) 247 would need a much larger pulley change.
Pump raises an incompressible fluid to a higher level of pressure or head.
Compressor raises a compressible fluid to a higher level of pressure.
Blower moves gas volumes with moderate increase of pressure.
Pump raises relatively compressible fluid to a higher level of pressure or head.
Answer: D) Pump raises relatively compressible fluid to a higher level of pressure or head.
Confirmed vs Book-3 — A pump handles INCOMPRESSIBLE liquids (a is correct, d contradicts it); a compressor raises a compressible gas to a higher pressure (b); a blower moves gas volumes with a moderate pressure rise — specific ratio 1.11–1.20 (c). Hence the wrong statement is (d).
Source: 10th Jul-2010
📖 §5.3 Fan Laws
62. As per the fan laws, by reducing the fan RPM by 10%, the fan power requirement:
decreases by 27%
decreases by 19%
does not change
decreases by 73%
Answer: A) decreases by 27%
Confirmed vs Book-3 §5.3 — 'Reducing the RPM by 10% decreases the power requirement by 27%' (0.9³ = 0.729). (b) 19% is the static-pressure reduction; (d) 73% is the REMAINING power fraction, not the reduction.
Source: 10th Jul-2010
📖 §5.3 Fan Laws
63. A centrifugal fan operating at 800 RPM develops a flow of 3000 Nm3/hr at a static pressure of 600 mmWC. If the fan speed is reduced to 600 RPM, the static pressure will become:
450 mmWC
519.6 mmWC
337.5 mmWC
none of the above
Answer: C) 337.5 mmWC
Confirmed vs Book-3 §5.3 — SP ∝ N²: 600 × (600/800)² = 600 × 0.5625 = 337.5 mmWC (c). (a) 450 = 600 × 0.75 is the linear (flow-law) error; (b) 519.6 = 600 × √0.75 is a square-root error.
Source: 10th Jul-2010
📖 §5.2 Fan types (Tables 5.2/5.3)
64. Name the fan which is more suitable for high pressure application
propeller type fan
tube-axial fan
backward curved centrifugal fan
forward curved centrifugal fan
Answer: C) backward curved centrifugal fan
Confirmed vs Book-3 §5.2 Table 5.3 — Backward-curved centrifugal: 'High pressure, high flow, high efficiency' (FD fans etc.); forward-curved: medium pressure, 'best suited for moving large volumes of air against relatively low pressures'. Propeller and tube-axial (a, b) are low/medium-pressure axial fans; 'centrifugal fans are suitable for low to moderate flow at high pressures'.
Source: 10th Jul-2010
📖 Book-3 Ch7 Cooling Towers (misfiled under Ch5)
65. A cooling tower is said to be performing well when:
approach is closer to zero
range is closer to zero
approach is larger than design
range is larger than design
Answer: A) approach is closer to zero
Confirmed vs Book-3 Ch7 — Approach = cold-water outlet temperature − ambient wet-bulb; the closer the approach to zero, the better the tower is performing (a). A larger-than-design approach (c) means poor performance; range depends on heat load, so 'range closer to zero' (b) is not a performance criterion.
Source: Jul 2022
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
66. Specific Ratio is maximum for
backward curved fan
forward curved fan
blowers
Compressors
Answer: D) Compressors
Confirmed vs Book-3 §5.1 Table 5.1 — Specific ratio: fans ≤ 1.11, blowers 1.11–1.20, compressors > 1.20 → maximum for compressors (d). Backward- and forward-curved fans (a, b) both fall in the fan band ≤ 1.11.
Source: Mar 2021
📖 §5.6 Measurement by pitot tube
67. The outer tube connection of the Pitot tube is used to measure ___________ in the fan system
static pressure
total pressure
velocity pressure
none of the above
Answer: A) static pressure
Confirmed vs Book-3 §5.6 — 'Static pressure is measured using the outer tube of pitot tube' (side holes at 90° to flow); the inner tube gives total pressure; both together give velocity pressure. So the outer connection reads static pressure (a), not total (b) or velocity (c).
Source: Mar 2021 (Set B)
📖 Book-3 Ch1 Electrical systems (misfiled under Ch5)
68. Which of the following contributes to increased technical losses ?
lower sized conductors
low power factor
loose connections
all of the above
Answer: D) all of the above
Confirmed vs Book-3 Ch1 — Technical (I²R) losses rise with undersized conductors (higher resistance), low power factor (higher current for the same kW) and loose connections (contact resistance/heating). All three contribute → (d).
Source: Mar 2021
📖 §5.2 Fan types (Tables 5.2/5.3)
69. In which of the following fans the air does not change flow direction from suction to discharge?
tube axial fan
vane axial fan
propeller fan
all the above
Answer: D) all the above
Confirmed vs Book-3 §5.2 — Tube-axial, vane-axial and propeller are all axial-flow fans, in which 'air enters and leaves the fan with no change in direction' → all the above (d). Only centrifugal fans turn the airflow (twice).
Source: Jul 2022
📖 §5.3 Fan Laws
70. For centrifugal fans, relation between Pressure (P) and speed (N) is given by __________.
P1/P2 = N1/N2
P1/P2 = N1³/N2³
P1/P2 = N1²/N2²
None of the above
Answer: C) P1/P2 = N1²/N2²
Confirmed vs Book-3 §5.3 — Fan laws: Pressure ∝ (Speed)² → P₁/P₂ = N₁²/N₂² (c). (a) linear is the FLOW law; (b) cube is the POWER law.
Source: Mar 2023
📖 §5.1 Table 5.1; §5.4 (axial fans produce lower pressure than centrifugal)
71. Which of the following equipment is having least compression ratio?
Compressor
Blower
Axial Fan
Radial Fan
Answer: C) Axial Fan
Confirmed vs Book-3 §5.1/§5.4 — Compression (specific) ratio: compressors > 1.20 > blowers 1.11–1.20 > fans ≤ 1.11; and among fans 'axial-flow fans produce lower pressure than centrifugal fans' (radial fans reach up to 1400 mmWC). So the least ratio is the axial fan (c).
Source: Mar 2023
📖 §5.5 Flow control strategies
72. Capacity control of a cooling tower fan can be achieved by __________.
Changing pulley dimensions
Damper Control
VFD
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §5.5 — 'Various ways to achieve change in flow are: pulley change, damper control, inlet guide vane control, variable speed drive and series and parallel operation of fans.' Pulley change, damper and VFD are all listed → (d).
Source: Mar 2023
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
73. Identify the correct statement:
the Specific Ratio of Compressors is higher than Blowers
the Specific Ratio of Fans is higher than Blowers
the Specific Ratio of Compressors is lower than Fans
the Specific Ratio of Blowers is higher than Compressors
Answer: A) the Specific Ratio of Compressors is higher than Blowers
Confirmed vs Book-3 §5.1 Table 5.1 — ASME specific ratio (discharge/suction pressure): fans up to 1.11, blowers 1.11–1.20, compressors more than 1.20; hence compressors > blowers (a). Options (b), (c) and (d) each reverse the fan < blower < compressor ordering. (Options repaired: 'a)…d)' prefixes were missing; identical to Book-3 EOC objective Q7.)
Source: 13th Sep-2012
Short questions (5 marks) — 21
📖 §5.6.8 Fan Efficiency (end-of-chapter S-2)
1. A fan delivers 18,500 Nm³/hr against a static pressure of 45 mmWC. The total motor input is 8.7 kW and the motor efficiency is 88%. Calculate the fan static efficiency.
Model answer: Fan static efficiency ≈ 29.6%.
Step 1 — Flow: Q = 18,500 / 3600 = 5.14 m³/s.
Step 2 — Shaft power: power input to the shaft = motor input × motor efficiency = 8.7 × 0.88 = 7.656 kW.
Step 3 — Book formula (§5.6.8): Fan Static Efficiency % = [Q (m³/s) × SP (mmWC)] / [102 × shaft kW] × 100 = (5.14 × 45) / (102 × 7.656) × 100 = 231.3 / 780.9 × 100 ≈ 29.6%.
Confirmed vs Book-3 §5.6.8 — the method is: convert Nm³/hr to m³/s by dividing by 3600, convert motor input to SHAFT power by multiplying by motor efficiency (and belt efficiency if given), then apply the 102 constant.
Common errors: leaving flow in Nm³/hr, or dividing by motor efficiency instead of multiplying.
2. A seal-air fan draws 120 kW at 50 Hz with the damper 25% open. If the damper is opened fully and the fan is run at 33 Hz via a VFD, estimate the new power and the saving.
Model answer: New power ≈ 34.5 kW; saving ≈ 85.5 kW.
Speed is proportional to supply frequency, so N₃₃/N₅₀ = 33/50 = 0.66.
Fan law (§5.3.4): Power ∝ (Speed)³, so kW₃₃ = 120 × (33/50)³ = 120 × 0.2875 = 34.5 kW.
Saving = 120 − 34.5 = 85.5 kW (about 71%), assuming motor and fan efficiency stay constant as the book states.
Confirmed vs Book-3 §5.5.5 / L-1 — with the damper fully opened and flow set by speed instead, the cube law applies to frequency (N ∝ f).
This is the book's showcase of why a VFD beats damper throttling: throttling at 25% damper opening still drew the full 120 kW.
Source: AI practice
📖 §5.3.1 System Resistance & §5.4.5 System Resistance and Pressure Drop
3. List four factors that change the system resistance of a fan-duct network, and state how using a larger-diameter duct affects system resistance.
Model answer: Book-3 §5.3.1/§5.4.5 factors that change system resistance (any four):
1. Duct configuration — narrow ducts, long runs and multiple short-radius elbows raise resistance; larger ducts and long-radius bends lower it.
2. Pressure drop across process equipment in the circuit (bag filter, cyclone, economiser, air heater, dust control system).
3. Change of equipment or duct modification — e.g. replacing multi-cyclones with an ESP, or fitting low-pressure-drop cyclones, which drastically shifts the operating point.
4. Coating formation or erosion of the duct lining, which changes the resistance marginally.
5. Volume of air flowing — resistance varies as the SQUARE of flow; it is also a function of gas density and gas velocity (damper position changes it by adding/removing resistance).
Effect of a larger-diameter duct: for the same air volume, the larger cross-section lowers the gas velocity, so the static pressure loss falls and the whole system-resistance curve drops. The fan then delivers more air at the same speed for less power.
Confirmed vs Book-3 §5.3.1 — 'the fan in a system with narrow ducts and multiple short radius elbows is going to have to work harder... than it would in a system with larger ducts and a minimum number of long radius turns.'
Note the book treats system resistance as a property of the DUCT/PROCESS side (and gas density and velocity); guide vanes and speed change alter the FAN curve, not the system curve, so they do not belong in this list.
Source: AI practice
📖 §5.3 Fan laws; §5.6 Volume calculation (Q = V × A)
4. An energy audit in an industrial unit revealed a fan directly coupled with motor was operating at 37 Hz through VFD for 500 hours/month and supplying air through a 150 mm diameter duct. The fan is designed to deliver an air flow of 1300 m³/h with a rated input power of 3 kW at 50 Hz. Calculate the air velocity and annual energy savings ignoring the motor losses.
Model answer: Given: Design air flow = 1300 m³/h at 50 Hz; Operating frequency = 37 Hz; Power at 50 Hz = 3 kW; Operating hours = 500 hours/month; Duct diameter = 150 mm = 0.15 m
Flow rate at 37 Hz: Q2 = 1300 × (37/50) = 962 m³/h = 962/3600 = 0.2672 m³/s
Duct area: A = π/4 × (0.15)² = 0.01767 m²
Air velocity: V = Q/A = 0.2672 / 0.01767 = 15.12 m/s
Power at 37 Hz: P2 = 3 × (37/50)³ = 3 × 0.405 = 1.215 kW
Monthly energy savings: (3 − 1.215) × 500 = 892.5 kWh/month
Annual energy savings: 892.5 × 12 = 10710 kWh/year
Fan laws with speed ∝ frequency: Q₂ = Q₁ × (f₂/f₁), P₂ = P₁ × (f₂/f₁)³; velocity = Q/A with A = (π/4)D²; savings = (P₁ − P₂) × hours. Note the 6000 h version of this question (Sep 2019) gives 10,680 kWh/yr.
Source: Sep 2025
📖 §5.7 Energy savings opportunities
5. List any five energy conservation opportunities in a fan system.
Model answer: Energy conservation opportunities in a fan system (Book-3 §5.7): 1) Minimise demand on the fan — reduce excess air in combustion (cuts FD/ID fan load) and stop air in-leaks in the hot flue-gas path (cold in-leaks raise gas density and choke ID-fan capacity); minimise in-/out-leaks in AC ducting. 2) Replace the impeller with a high-efficiency impeller (with cone), or the whole fan assembly with a higher-efficiency fan. 3) Impeller derating (smaller-diameter impeller) or fan speed reduction by pulley-diameter change where the fan is oversized/throttled. 4) Replace metallic/GRP axial impellers with hollow FRP aerofoil impellers. 5) Two-speed motors or variable-speed drives for variable-duty fans (power ∝ N³). 6) Energy-efficient flat belts or cogged raw-edge V-belts instead of conventional V-belts to cut transmission losses. 7) Inlet guide vanes in place of discharge-damper control. 8) Minimise system resistance and pressure drops by duct-system improvements (e.g. PTFE membrane bags cut bag-house DP 165→115 mmWC, ID fan 51→46 kW).
Any five of the Book-3 §5.7 list score full marks: minimise fan demand (excess air, in-leaks), high-efficiency impeller/fan, impeller derating or pulley speed reduction, hollow FRP impellers, VSD/two-speed motors, efficient belts, IGV instead of dampers, lower system resistance.
Source: Book EOC
📖 §5.6 Fan static efficiency formula
6. An energy audit of a fan showed it delivering 18,500 Nm3/hr of air at a static pressure rise of 45 mm WC. The 3-phase motor recorded 2.9 kW/phase on average with a motor operating efficiency of 88%. What would be the fan static efficiency?
Model answer: Flow Q = 18,500/3600 = 5.139 m³/s; static pressure rise = 45 mmWC.
Motor input power = 3 phases × 2.9 kW = 8.7 kW.
Power input to fan shaft = 8.7 × 0.88 = 7.656 kW.
Fan static efficiency = (Q in m³/s × static pressure in mmWC)/(102 × shaft kW) × 100 = (5.139 × 45)/(102 × 7.656) × 100 = 231.25/780.9 × 100 = 29.6%.
Use the Book-3 §5.6 static-efficiency formula with Q in m³/s, SP in mmWC and the 102 constant; shaft power = motor input × motor efficiency (drive-motor kW × motor efficiency gives shaft power). Answer ≈ 29.6% (≈ 30%).
Source: Book EOC
📖 §5.3 System characteristics; §5.4 System resistance and pressure drop
7. Explain the factors which can change the system resistance.
Model answer: System resistance is the sum of the static-pressure losses (ducts, pickups, elbows and equipment such as bag filters or cyclones) that the fan must overcome; it varies with the SQUARE of the air volume and is a function of gas density and gas velocity. It changes with: 1) Air flow rate — resistance ∝ Q², so any change in volume handled changes it. 2) Duct configuration — length, diameter, number and radius of bends/elbows (narrow ducts and short-radius elbows raise it; larger ducts and long-radius turns lower it), and any duct modification. 3) Process/equipment changes — e.g. replacing multi-cyclones with an ESP or installing low-pressure-drop cyclones drastically shifts the operating point; formation of coatings or erosion of duct lining changes it marginally; fouling of filters/bag houses raises DP. 4) Damper position — closing a damper adds system resistance (new system curve SC₂). 5) Gas density and velocity — temperature, in-leaks of cold air, or change of gas composition alter density and hence pressure drop. Because any of these shifts the system curve and the operating point, the book says system resistance must be checked periodically, especially after modifications.
Book-3 §5.3/§5.4: resistance ∝ (flow)², depends on duct layout and equipment pressure drops, on gas density/velocity, and shifts with dampers, lining coating/erosion and equipment changes (multi-cyclone → ESP); check it periodically after modifications.
Source: Book EOC
📖 §5.3 Fan Laws
8. A fan is operating at 900 RPM developing a flow of 3000 Nm3/hr at a static pressure of 600 mmWC. What will be the flow and static pressure if the speed is reduced to 600 RPM?
Model answer: By the fan laws: Flow varies as speed: Q2 = 3000 x (600/900) = 2000 Nm3/hr. Static pressure varies as speed squared: P2 = 600 x (600/900)^2 = 600 x 0.4444 = 266.7 mmWC. So at 600 RPM the fan delivers about 2000 Nm3/hr at about 266.7 mmWC.
Book-3 §5.3 fan laws: Q₂ = 3000 × (600/900) = 2000 Nm³/hr; SP₂ = 600 × (600/900)² = 266.7 mmWC (power would fall to (2/3)³ = 29.6% of original).
Source: Book EOC
📖 §5.6 Calculation of velocity (worked example) & gas density correction
9. Pitot-tube air flow measurement in the primary air fan of a coal-fired boiler gave: air temperature 38 degC, velocity pressure 47 mmWC, pitot tube constant Cp = 0.9, air density at 0 degC = 1.293 kg/m3. Find the velocity of air in m/sec.
Model answer: Correct the air density to 38 °C: γ = 1.293 × 273/(273 + 38) = 1.135 kg/m³.
Velocity V = Cp × √(2 × 9.81 × ΔP/γ), with ΔP in mmWC: V = 0.9 × √(2 × 9.81 × 47/1.135) = 0.9 × √812.5 = 0.9 × 28.5 = 25.6 m/s (Book-3 worked example answer; 25.7 m/s with unrounded intermediates).
Book-3 §5.6 worked example: density corrected by 273/(273+t), then V = Cp√(2gΔP/γ) with ΔP in mmWC → 25.6 m/s. Forgetting the density correction (using 1.293) gives 24.0 m/s — wrong.
Source: Book EOC
📖 §5.3 Fan laws (Book EOC L-1)
10. A seal air fan for a coal mill operates with the suction damper 25% open, drawing 120 kW at 50 Hz. A VFD is to be installed, eliminating the damper; the damper can then be fully opened and the fan operated at 33 Hz. Calculate the power drawn by the fan motor at 33 Hz, assuming motor and fan efficiency remain constant.
Model answer: By the fan law, power varies as the cube of speed (frequency): P2 = P1 x (f2/f1)^3 = 120 x (33/50)^3 = 120 x (0.66)^3 = 120 x 0.2875 = 34.5 kW. So with the VFD at 33 Hz the fan motor draws about 34.5 kW (a saving of about 85.5 kW versus damper control).
Speed ∝ frequency, so Power ∝ (f₂/f₁)³: 120 × (33/50)³ = 120 × 0.2875 = 34.5 kW; saving ≈ 85.5 kW because the damper throttling is eliminated and the fan runs at the lower speed with damper fully open.
Source: Book EOC
📖 §5.6 Fan static efficiency formula (Book EOC L-2)
11. An engine room of size 30 m x 20 m x 5 m is to be ventilated with 20 air changes per hour. If the static pressure rise across the ventilator fan is 15 mm WC and fan efficiency is 70%, find the motor power drawn at a motor efficiency of 90%.
Model answer: Room volume = 30 × 20 × 5 = 3000 m³; air flow = 3000 × 20 ACH = 60,000 m³/hr = 16.67 m³/s.
Fan shaft power = (Q × ΔP)/(102 × η_fan) = (16.67 × 15)/(102 × 0.70) = 250/71.4 = 3.50 kW.
Motor power drawn = 3.50/0.90 = 3.89 kW ≈ 3.9 kW.
Flow from room volume × air changes; then the Book-3 §5.6 formula Power = Q × ΔP/(102 × η) with Q in m³/s and ΔP in mmWC, divided by motor efficiency → 3.89 kW.
Source: Book EOC
📖 §5.6 Fan static efficiency formula
12. An energy audit of a fan recorded delivery of 18,500 Nm3/hr at static pressure rise of 52 mm WC. The 3-phase motor recorded 3.1 kW/phase average; motor operating efficiency 88%. What is the fan static efficiency?
Model answer: Q = 18,500 Nm3/hr = 5.1388 m3/s; SP = 52 mmWC. Power input to motor = 3.1 x 3 = 9.3 kW. Power input to fan shaft = 9.3 x 0.88 = 8.184 kW. Fan static efficiency = (Q x Pst)/(102 x shaft power) = (5.1388 x 52)/(102 x 8.184) = 0.32 = 32%.
13. Pitot tube air-flow data in a boiler primary air fan: air temperature 38°C, velocity pressure 47 mmWC, pitot constant Cp = 0.9, air density at 0°C = 1.293 kg/m³. Find the air velocity in m/s.
Model answer: Corrected air density = 273 x 1.293/(273+38) = 1.135 kg/m³. Velocity = Cp x √(2 x 9.81 x Δp x γ)/γ = 0.9 x √(2 x 9.81 x 47 x 1.135)/1.135 ≈ 25.6 m/s.
📖 §5.6 Fan static efficiency formula (Book EOC S-2)
14. A fan delivers 18,500 Nm³/hr at static pressure rise 45 mm WC. The 3-phase motor records 2.9 kW/phase; motor operating efficiency 88%. What is the fan static efficiency?
Model answer: Q = 18,500 Nm³/hr = 5.13888 m³/s; SP = 45 mmWC. Power input to motor = 2.9 x 3 = 8.7 kW. Power to fan shaft = 8.7 x 0.88 = 7.656 kW. Fan static efficiency = (5.13888 x 45)/(102 x 7.656) = 0.296 = 29.6%.
15. The input power to a fan is 30 kW for a 2500 Nm3/hr fluid flow. The fan pulley diameter is 300 mm. If the flow is to be reduced by 15% by changing the fan pulley, what should be the diameter of the fan pulley and power input to fan?
Model answer: N2 = 0.85 N1. From N1D1 = N2D2: D2 = D1×(N1/N2) = 300/0.85 = 352 mm. Power: kW2 = (N2/N1)³×kW1 = 0.85³×30 = 18.42 kW ≈ 18.4 kW. So change fan pulley to 352 mm; fan power ≈ 18.4 kW.
Q ∝ N so N₂ = 0.85 N₁; belt ratio N₁D₁ = N₂D₂ → D₂ = 300/0.85 = 352 mm (larger fan pulley = slower fan); kW ∝ N³ → 30 × 0.85³ = 18.4 kW.
Source: 16th Exam
📖 §5.3 Fan laws; §5.5 Variable speed drives
16. An ID fan consumes 35 kW at 100% boiler loading with damper fully open. Estimate daily energy savings if the damper is replaced by a VFD, for the schedule: 80% load, 4 h, 31 kW; 70% load, 12 h, 29 kW; 60% load, 8 h, 26 kW. Air requirement ∝ boiler loading.
Model answer: VFD power = (flow%)³ x 35 kW. At 80%: 0.8³x35 = 17.9 kW, saving 31-17.9 = 13.1 kW x 4 h = 52.32 kWh. At 70%: 0.7³x35 = 12 kW, saving 29-12 = 17 kW x 12 h = 203.94 kWh. At 60%: 0.6³x35 = 7.6 kW, saving 26-7.6 = 18.4 kW x 8 h = 147.52 kWh. Total daily savings ≈ 403.78 kWh.
With VFD, flow ∝ speed so power = 35 × (load fraction)³; damper power is the measured value. Savings: 13.1 kW × 4 h + 17 kW × 12 h + 18.4 kW × 8 h ≈ 403.8 kWh/day.
Source: 18th Exam
📖 §5.3 Fan laws; §5.6 Volume calculation
17. A fan is designed for 1300 m3/hr, 50 Hz, drawing 3 kW. Operated with VFD at 37 Hz for 6000 hours. Calculate the air velocity in a 150 mm diameter duct and the annual energy savings.
Model answer: Flow at 37 Hz = 1300 x (37/50) = 962 m3/hr. Duct area = 0.0177 m2. Velocity = (962/3600)/0.0177 = 15.09 m/s. Power at 37 Hz = (37/50)^3 x 3 = 1.22 kW. Annual savings = 6000 x (3-1.22) = 10,680 kWh.
18. S-4: Input power to a fan is 30 kW for 2500 Nm3/hr flow. Fan pulley diameter 300 mm. If flow is reduced by 15% by changing the fan pulley, what should be the pulley diameter and power input to the fan?
Model answer: Using N1D1 = N2D2 and N2 = 0.85 N1: D2 = D1 x (N1/N2) = 300 x (1/0.85) = 352 mm. Power: KW2 = (N2/N1)^3 x KW1 = 0.85^3 x 30 = 18.42 kW. So pulley should be 352 mm diameter and fan power ~18.4 kW.
19. A fan is delivering 20,000 Nm3/hr of air at static pressure difference of 70 mm WC. If the fan static efficiency is 55%, find out the shaft power of the fan.
Model answer: Q = 20,000/3600 = 5.56 m³/s. Fan static efficiency = (Q × Pst)/(102 × shaft power). 0.55 = (5.56 × 70)/(102 × P), giving shaft power P = 6.94 kW.
📖 §5.6 Pitot velocity, volume & fan static efficiency
20. A V-belt driven centrifugal fan is supplying air to a chemical process. Calculate the fan static efficiency for the following operating parameters: Ambient temperature 40degC; Density of air 1.127 kg/m3; Diameter of discharge air duct 1 meter; Velocity pressure measured by Pitot tube in discharge duct 47 mm WC; Pitot tube coefficient 0.9; Static pressure at fan inlet -22 mm WC; Static pressure at fan outlet 188 mm WC; Power drawn by motor 72 kW; Belt transmission efficiency 95%; Motor efficiency at operating load 90%.
Model answer: Air velocity = Cp x (2 x 9.81 x dP/density)^0.5 = 0.9 x (2 x 9.81 x 47/1.127)^0.5 = 25.7 m/s. Area of discharge duct = 3.14 x 1 x 1/4 = 0.785 m2. Volume = 25.7 x 0.785 = 20.17 m3/s. Power input to fan shaft = 72 x 0.95 x 0.9 = 61.6 kW. Total static pressure = 188 - (-22) = 210 mm WC. Fan static efficiency = (Volume x total static pressure in mm WC) / (102 x shaft power in kW) = (20.17 x 210)/(102 x 61.6) = 67.4%.
Same method as the Book-3 §5.9 solved example: velocity from pitot VP, Q = V × A, shaft kW = motor kW × belt × motor efficiency, total static pressure = outlet − inlet (mind the negative inlet), η_static = (Q × ΔP)/(102 × shaft kW) = 67.4%.
Source: Mar 2021 (Set B)
📖 §5.3 Fan Laws
21. A centrifugal fan drawing 16 kW and operating at 1440 RPM is delivering air at 30000 m³/hr. The head developed by the fan is 400mmWC. If the speed is decreased by 200 rpm, calculate the following: a) Air flow in m³/hr; b) Static Pressure in mmWC; c) Power drawn in kW.
Model answer: New speed = 1440 − 200 = 1240 rpm.
a) Air flow = (1240/1440) × 30,000 = 25,833 m³/hr.
b) Static pressure = (1240/1440)² × 400 = 296.6 mmWC.
c) Power drawn = (1240/1440)³ × 16 = 0.6385 × 16 = 10.22 kW.
(If the fan power in the paper is 54 kW, as in the Sep-2024 repeat of this question, part (c) = 0.6385 × 54 = 34.48 kW.)
Book-3 §5.3 fan laws: Q ∝ N, SP ∝ N², kW ∝ N³ with N₂/N₁ = 1240/1440 = 0.861. The earlier answer text ('34.48 → 9.54 kW') was garbled — 34.48 kW belongs to the 54 kW version; with 16 kW the power is 10.22 kW.
Source: Jul 2022
Long questions (10 marks) — 14
📖 §5.6 Pitot velocity, volume & fan static efficiency; §5.3 Fan laws
1. a) During performance guarantee test of an induced draft cooling tower, it was found that design approach of cooling tower is not achieved. As one of the probable causes, the team decided to check the efficiency of cooling tower fan. If design static efficiency is 70%, estimate the operating static efficiency using following parameters:
Pitot tube coefficient: 0.9
Velocity pressure: 49.0 mmWC
Air Density at operating condition: 1.129 kg/m3
Duct diameter: 2.1 m
Differential pressure across fan: 130.0 mmWC
Motor shaft Power: 190.0 kW
Motor Efficiency: 95.0 %
Gear Box Efficiency: 96.0 %
b) A centrifugal fan drawing 54 kW and operating at 1440 rpm is delivering air at 30,000 m³/hr. The head developed by the fan is 400 mm WC. If the speed is decreased by 200 rpm, calculate the following:
1. Air flow in m³/hr
2. Static pressure in mm WC
3. Power drawn in kW
Model answer: a) Air velocity = Cp × √(2 × 9.81 × ΔP/γ) = 0.9 × √(2 × 9.81 × 49/1.129) ≈ 26.3 m/s
Duct area = π/4 × (2.1)² = 3.46 ≈ 3.5 m²
Volume flow = 26.3 × 3.5 ≈ 92.0 m³/s
Air kW transferred = Q × ΔP/102 = 92 × 130/102 = 117.25 kW
Power input to fan shaft = motor shaft power × gearbox efficiency = 190 × 0.96 = 182.4 kW (190 kW is already the motor OUTPUT, so the 95% motor efficiency is not applied again)
Operating static efficiency = 117.25/182.4 × 100 = 64.28% (design 70% → fan is under-performing)
b) New speed = 1440 − 200 = 1240 rpm
1. Air flow = (1240/1440) × 30,000 = 25,833 m³/hr
2. Static pressure = (1240/1440)² × 400 = 296.6 mmWC
3. Power drawn = (1240/1440)³ × 54 = 34.48 kW
Static efficiency = (Q × ΔP_static)/(102 × shaft kW) per Book-3 §5.6; shaft kW = motor shaft output × gearbox efficiency (the model answer's printed expression '190 × 0.95 × 0.96 = 182.4' is a typo — 182.4 = 190 × 0.96). Part (b) applies Q ∝ N, SP ∝ N², kW ∝ N³ at 1240 rpm.
Source: Sep 2024
📖 §5.6 Fan power formula; §5.3 Fan laws (Book EOC L-1, L-2)
2. a) An engine room 30 m x 20 m x 5 m to be ventilated at 20 air changes per hour. Static pressure rise across ventilator fan 15 mmWC, fan efficiency 70%. Find motor power at 90% motor efficiency. b) A seal air fan for a coal mill operates with suction damper 25% open, drawing 120 kW at 50 Hz. A VFD allows full damper opening at 33 Hz. Find power at 33 Hz (motor and fan efficiency constant).
Model answer: a) Flow = 30 x 20 x 5 x 20 = 60,000 m³/hr. Motor power = (60,000/3600) x 15/(102 x 0.7 x 0.9) = 3.89 kW. b) Power at 33 Hz = 120 x (33/50)³ = 34.5 kW.
📖 §5.3 Fan laws; §5.5 Pulley change; §5.6 Fan power formula
3. A belt-driven centrifugal fan system delivers 12 m3/s. One branch (1.5 m3/s) needs 89 mmWC static; the rest needs only 66 mmWC but the fan runs at 89 mmWC. Auditor proposes reducing fan speed to give 66 mmWC and adding a booster fan (75% eff, motor 85% eff) for the branch. Main fan input = 16.2 kW; initial fan speed 1200 rpm, motor pulley 209 mm, fan pulley 305 mm; 6000 h/yr. Calculate annual energy and cost savings.
Model answer: Revised fan speed = 1200×(66/89)^0.5 = 1031 rpm; new fan pulley = 305×1200/1031 = 355 mm. Initial air kW = 12×89/102 = 10.5; revised air kW = 12×66/102 = 7.8. Revised motor input = 16.2×7.8/10.5 = 12 kW. Main fan saving = (16.2−12)×6000 = 25,200 kWh = Rs 1,76,400/yr. Booster: 1.5 m3/s, ΔP=23 mmWC, air kW=1.5×23/102=0.34; shaft=0.34/0.75=0.45; motor=0.45/0.85=0.53 kW → 0.53×6000=3180 kWh = Rs 22,260/yr. Net saving = 1,76,400 − 22,260 = Rs 1,54,140/yr.
SP ∝ N² gives the reduced main-fan speed (1031 rpm) and the pulley from N₁D₁ = N₂D₂ (355 mm); motor input scaled by the ratio of air powers (Q × ΔP/102); booster fan power = air kW/(η_fan × η_motor). Net saving ≈ Rs 1.54 lakh/yr at Rs 7/kWh.
Source: 17th Sep-2016
📖 §5.5 Flow control strategies
4. Briefly explain any three different methods of flow control for fans.
Model answer: Pulley change: for a permanent flow change, alter fan speed by changing the v-belt pulley diameter (simplest permanent speed change). Damper control: dampers add/remove system resistance, forcing the fan up/down its curve to deliver more or less air without changing speed. Inlet guide vanes: curved vanes pre-swirl the inlet air, changing the fan curve; energy-efficient for modest reductions (100% to ~80% flow), efficiency drops sharply below 80%. Variable speed drive: reduce fan speed to match reduced flow; since power ∝ flow³, this is usually the most efficient capacity control.
Book-3 §5.5 lists pulley change, damper control, inlet guide vanes, variable speed drive (and series/parallel operation, variable-pitch axial blades). Any three with the key points — dampers add resistance (least efficient), IGVs efficient 100→80% flow, VSD most efficient (power ∝ N³).
5. L-3 fill in the blanks: 1) DBT 30°C, WBT 30°C → RH = ___%. 2) Cavitation occurs when local static pressure falls below the ___ pressure of the liquid at actual temperature. 3) As 'Approach' decreases, cooling tower effectiveness will ___. 4) Ratio of luminous flux emitted to power consumed is called ___. 5) A centrifugal pump raises water to 12 m; with brine of SG 1.2, the height raised is ___ m. 6) Harmonics are multiples of the ___ frequency. 7) A motor that can run at lagging as well as leading PF is the ___ motor. 8) Per ECBC, Effective Aperture (EA) = ___ for WWR 0.40 and VLT 0.25. 9) In an amorphous-core transformer, ___ loss is less than conventional. 10) Centrifugal pump impeller is generally trimmed down to about ___% of maximum size.
Model answer: 1) 100%. 2) Vapour. 3) Increases. 4) Luminous efficacy. 5) 12 m (same — a centrifugal pump develops head independent of fluid density). 6) Fundamental (50 Hz). 7) Synchronous. 8) EA = VLT x WWR = 0.25 x 0.40 = 0.10. 9) No-load (iron/core/fixed) loss. 10) 75% (about 75-80%).
Definitional fill-ins across chapters: DBT = WBT → saturated (100% RH); cavitation when pressure < vapour pressure; lower approach → higher effectiveness; lumen/W = luminous efficacy; centrifugal pump head is independent of fluid density (12 m); harmonics = multiples of fundamental; synchronous motor PF control; EA = VLT × WWR; amorphous core → lower no-load loss; impeller trim ≈ 75%.
Source: 18th Exam
📖 §5.3 Fan laws; §5.5 Pulley change; §5.6 Fan power formula
6. L-4: A belt-driven centrifugal fan: air flow 68,400 m3/hr, fan differential static pressure 112 mmWC, pressure drop across main damper 17 mmWC, motor input power 26.8 kW, fan speed 600 rpm, motor speed 1460 rpm, fan pulley 560 mm, motor pulley 230 mm. The auditor recommends opening the main damper fully and reducing fan speed via pulley change. Calculate (a) annual energy savings for 6000 h/yr and (b) the new fan pulley diameter.
Model answer: Flow = 68400/3600 = 19 m3/s. Theoretical air power (partly-closed) WTh1 = 19x112/102 = 20.86 kW. With damper fully open, pressure = 112-17 = 95 mmWC; WTh2 = 19x95/102 = 17.7 kW. Case-2 motor power W2 = W1 x (WTh2/WTh1) = 26.8 x (17.7/20.86) = 22.7 kW. (a) Annual savings = (26.8-22.7) x 6000 = 24,600 kWh. (b) New speed N2 = N1 x (p2/p1)^0.5 = 600 x (95/112)^0.5 = 553 rpm. Using N1D1 = N2D2, new fan pulley D2 = (N1/N2) x D1 = (600/553) x 560 = 608 mm.
Opening the damper removes the 17 mmWC throttling loss, so air power falls from 19 × 112/102 = 20.86 kW to 19 × 95/102 = 17.7 kW; motor power scales in proportion (26.8 → 22.7 kW) → 24,600 kWh/yr. New speed from SP ∝ N²: 600 × √(95/112) = 553 rpm; fan pulley from N₁D₁ = N₂D₂ → 608 mm (larger driven pulley slows the fan).
Source: 18th Exam
📖 §5.6 Gas density (γ = PM/RT) and fan power formula
7. L-3: In a boiler, the forced-draught fan develops a total static pressure of 300 mmWC. Determine the shaft power (kW) to drive the fan if 10,000 kg/hr coal is burnt with 13 kg air/kg coal. Boiler-house temperature 20 °C, static efficiency 80%. Use R = 847.84 mmWC·m3/kg·mole·K and M = 28.92 kg/kg·mole.
Model answer: Mass of air = 10000 x 13/3600 = 36.11 kg/s. Atmospheric pressure P = 1 kg/cm² = 10,000 mmWC; T = 293 K. Density = P·M/(R·T) = (10000 x 28.92)/(847.84 x 293) = 1.164 kg/m3. Volume = 36.11/1.164 = 31.02 m3/s. Shaft power = (volume x total pressure)/(102 x fan efficiency) = (31.02 x 300)/(102 x 0.8) = 114 kW.
8. L-1(a): A fan delivers 24,000 Nm3/hr of air. Suction static pressure -15 mmWC, discharge static pressure 35 mmWC. Motor power 7 kW, motor efficiency 90%. What is the static efficiency of the fan? (b) Match the following: Heat Pump, Compressor, Pumping Pressure, Fan, Pump with NPSHR, Static Head, Static Pressure, Compressor, Free air delivery test.
Model answer: (a) Q = 24,000 Nm3/hr = 6.67 m3/s. Static pressure rise = 35-(-15) = 50 mmWC. Power input to fan shaft = 7 x 0.90 = 6.3 kW. Fan static efficiency = (Q x Pst)/(102 x shaft power) = (6.67 x 50)/(102 x 6.3) = 0.519 = 51.9%. (b) Matches: Heat Pump - Compressor; Compressor - Free air delivery test; Pumping Pressure - Static Head; Fan - Static Pressure; Pump - NPSHR.
9. List down any 5 energy conservation opportunities in fan systems.
Model answer: 1) Minimise excess air in combustion systems to reduce FD/ID fan load; 2) Minimise air in-leaks in hot flue gas path to reduce ID fan load; 3) Avoid cold air in-leaks that choke ID fan capacity; 4) Minimise system resistance/pressure drops via duct improvements; 5) Adopt inlet guide vanes in place of discharge damper control; (also: energy-efficient flat/cogged V-belts; two-speed motors or VSDs; fan speed reduction by pulley dia change; hollow FRP aerofoil impellers; impeller derating; higher-efficiency fan/impeller with cone).
Any five from the Book-3 §5.7 list: minimise excess air and in-leaks, high-efficiency impeller/fan, impeller derating, hollow FRP impellers, pulley speed reduction, VSD/two-speed motors, efficient belts, IGV instead of damper, lower system resistance.
10. (a) How do you calculate the velocity of gas in a duct using the average differential pressure and density of the gas? (b) A V-belt centrifugal fan performance test: density of air at 0°C = 1.293 kg/m3; ambient 40°C; discharge duct dia 0.8 m; velocity pressure 45 mmWC; pitot coefficient 0.9; static pressure at fan inlet −20 mmWC, outlet 185 mmWC; motor power 75 kW; belt transmission efficiency 97%; motor efficiency 93%. Find the static fan efficiency.
Model answer: (a) V (m/s) = Cp × √(2×9.81×Δp×γ)/γ, with Cp = pitot constant (0.85 typical), Δp = average velocity pressure, γ = gas density at test condition. (b) Static fan efficiency ≈ 37.58%.
📖 §5.6 Pitot velocity, volume & fan static efficiency
11. A V-belt driven centrifugal fan is supplying air to a chemical process. Calculate the fan static efficiency for: Ambient temperature 40oC; Density of air 1.127 kg/m3; Diameter of discharge air duct 1 m; Velocity pressure (Pitot) 47 mm WC; Pitot tube coefficient 0.9; Static pressure at fan inlet -22 mm WC; Static pressure at fan outlet 188 mm WC; Power drawn by motor 72 kW; Belt transmission efficiency 95%; Motor efficiency at operating load 90%. (5 Marks)
Model answer: Air velocity = Cp x (2 x 9.81 x dp/rho)^0.5 = 0.9 x (2 x 9.81 x 47/1.127)^0.5 = 25.7 m/s. Area of discharge duct = 3.14 x 1 x 1/4 = 0.785 m2. Volume = 25.7 x 0.785 = 20.17 m3/s. Power input to fan shaft = 72 x 0.95 x 0.9 = 61.6 kW. Fan static efficiency = (Volume m3/s x total static pressure mmWC)/(102 x power input to shaft kW) = [20.17 x (188-(-22))]/(102 x 61.6) = (20.17 x 210)/6283 = 67.4%.
📖 Book-3 Ch4 (AHU/refrigeration) and Ch2 (motor no-load test) — misfiled under Ch5
12. a) Calculate the filter area of an Air Handling Unit (AHU) for Refrigeration Load of 5 TR. The air enthalpy at inlet of AHU is 85 kJ/kg and at outlet is 60 kJ/kg. Air velocity at filter is 1.81 m/sec and air density is 1.26 kg/m3 (5 marks). b) A no load test was conducted on a delta connected 37 kW induction motor. Name plate data: 3 Phase, 415 V, 50 Hz, 55 Amp. Measured data at no load: Voltage V = 415 Volts; Current I = 18 Amps; Frequency F = 50 Hz; Stator phase resistance at 30degC = 0.23 Ohms/phase; No load power = 955 Watts. Calculate: i. The iron loss plus friction loss plus windage loss (2 marks); ii. Stator copper loss at name plate ratings (full load), considering stator temperature as 120degC (2 marks); iii. No load power factor of the motor (1 mark).
Model answer: a) TR of AHU = (Enthalpy difference x density x area x velocity x 3600)/(4.187 x 3024). Filter Area = TR x (4.187 x 3024)/(Enthalpy difference x density x velocity x 3600) = (5 x 4.187 x 3024)/((85-60) x 1.26 x 1.81 x 3600) = 3.08 m2. b) i. Let iron loss plus friction plus windage loss = Pi + fw. Stator copper loss at no load (30degC) = 3 x I^2 x R = 3 x (18/sqrt3)^2 x 0.23 = 3 x 108 x 0.23 = 74.51 Watt. Pi + fw = No load power - no load copper loss = 955 - 74.51 = 880.49 W. ii. Stator resistance at 120degC = 0.23 x [(120+235)/(30+235)] = 0.308 Ohms. Stator copper loss at name plate ratings (full load) = 3 x (55/sqrt3)^2 x 0.308 = 3 x 1008.3 x 0.308 = 931.65 Watt. iii. No load power factor = 955/(1.7321 x 415 x 18) = 0.0738.
(a) Filter area from TR = ṁ × Δh: A = TR × 4.187 × 3024/(Δh × ρ × V × 3600) = 3.08 m². (b) Iron + friction + windage = no-load power − no-load stator Cu loss = 955 − 74.5 = 880.5 W; R at 120 °C = 0.23 × (120 + 235)/(30 + 235) = 0.308 Ω; full-load stator Cu loss = 3 × (55/√3)² × 0.308 = 932 W; no-load PF = 955/(√3 × 415 × 18) = 0.074.
Source: Mar 2021 (Set B)
📖 §5.6 Fan static efficiency formula; §5.6 Volume calculation
13. L-2(A): In a ventilation duct of 0.6 m x 0.6 m size, the average velocity of air measured by vane anemometer is 30 m/s. The static pressure at inlet of the fan is -25 mm WC and at the outlet is 35 mm WC. A 3 phase induction motor coupled with fan through belt drive draws 19 A at 410 V at a power factor of 0.8. Find out the efficiency of the fan. Assume motor efficiency 90% and belt transmission efficiency of 98% (density correction can be neglected). (6 Marks)
Model answer: Volume flow Q = velocity × area = 30 × 0.6 × 0.6 = 10.8 m³/s.
Motor input power = √3 × V × I × PF = 1.732 × 410 × 19 × 0.8 = 10,790 W = 10.79 kW.
Power input to fan shaft = 10.79 × 0.90 (motor) × 0.98 (belt) = 9.52 kW.
Static pressure rise across fan = 35 − (−25) = 60 mmWC.
Fan efficiency = (Q in m³/s × ΔP in mmWC)/(102 × shaft kW) × 100 = (10.8 × 60)/(102 × 9.52) × 100 = 648/971 × 100 = 66.7%.
Book-3 §5.6: air power = Q × ΔP/102 kW; shaft power from electrical input (√3·V·I·PF) × motor × belt efficiency; efficiency = air power/shaft power = 66.7%. Density correction is neglected as instructed.
Source: Mar 2023
📖 §5.9 Computational Fluid Dynamics
14. L-2(B): Explain how Computational Fluid Dynamics (CFD) can be used for enhancing the Energy Efficiency. (4 Marks)
Model answer: Computational Fluid Dynamics (CFD) is a computer-based simulation tool that models the physical system mathematically and solves the mass, momentum and energy equations to predict flow patterns, temperature and pressure profiles and particle movement inside equipment and duct systems. Use for energy efficiency (Book-3 §5.9): (1) Designs and operating parameters can be varied in the model to find the best operating conditions BEFORE any physical change is made in the plant. (2) It is proactive — it identifies the root cause (not just the effect) of plant problems, is scale-independent (based on fundamental physics, so scale-up problems are reduced) and can simulate conditions where measurements are impossible (high temperature, dangerous environments). (3) In fan systems the upstream/downstream ducting matters: turbulent or swirling inflow from sharp bends or abrupt cross-section changes raises pressure drop and degrades fan efficiency — especially for high-efficiency fans (> 80%) that need non-swirling inflow. (4) Case study — double-inlet ID fan: CFD showed sharp-edged transitions causing flow disruption (pressure drops of 260–430 Pa). Smoothing the transitions cut the front inflow-duct drop from 261 Pa to 66 Pa (factor of four) and the rear-duct drop to about one-sixth; the mean saving of 275 Pa at 5,00,000 m³/h reduced fan power by about 49 kW and eliminated swirl-induced vibration.
Model answer built from Book-3 §5.9 (pages 165–166): CFD definition, its proactive/scale-independent advantages, the effect of inflow swirl on fan efficiency, and the ID-fan case study (261 → 66 Pa, 275 Pa mean saving, ≈ 49 kW). The previous answer text was only a page reference.