132 questions — 91 objective (1 mark), 24 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) — 91
📖 §6.4 Affinity Laws (effect of speed variation)
1. For a centrifugal pump at fixed impeller diameter, if the speed is doubled the power required becomes:
2 times
4 times
8 times
16 times
Answer: C) 8 times
Confirmed vs Book-3 §6.4 — the affinity laws are Q ∝ N, H ∝ N², P ∝ N³, with efficiency essentially independent of speed.
Doubling the speed therefore multiplies power by 2³ = 8 (flow ×2, head ×4).
Option b) 4 times is the HEAD law (N²) misapplied to power — the most common slip.
Source: AI practice
📖 §6.4 Effects of impeller diameter change
2. When trimming a pump impeller (constant speed), the flow Q varies with the impeller diameter D as:
Q ∝ D
Q ∝ D²
Q ∝ D³
Q ∝ √D
Answer: A) Q ∝ D
Confirmed vs Book-3 §6.4 — for a diameter change inside a fixed casing: Q ∝ D, H ∝ D², P ∝ D³.
So the trimmed diameter follows directly: D₂ = D₁ × (Q₂/Q₁).
Option b) Q ∝ D² is the HEAD relation; note the book warns that beyond about a 5% diameter change the squared/cubic relations lose accuracy and manufacturer curves should be used.
Source: AI practice
📖 §6.4 Effects of impeller diameter change — trim limits
3. Impeller trimming is generally recommended only down to what fraction of the maximum impeller diameter?
~25%
~50%
~75%
~90%
Answer: C) ~75%
Confirmed vs Book-3 §6.4 — 'Diameter changes are generally limited to reducing the diameter to about 75% of the maximum, i.e. a head reduction to about 50%.'
Beyond that the book says efficiency and NPSH are badly affected.
Option b) ~50% is the resulting HEAD reduction, not the diameter limit — that swap is the trap.
Source: AI practice
📖 §6.2 Hydraulic Power, Pump Shaft Power and Motor Input Power
4. A pump shaft delivers 30 kW and the pump efficiency is 0.6. The power actually imparted to the water is:
50 kW
30 kW
18 kW
12 kW
Answer: C) 18 kW
Confirmed vs Book-3 §6.2 — Shaft Power = Hydraulic Power / Pump Efficiency, so Hydraulic Power = Shaft Power × η = 30 × 0.6 = 18 kW.
The question asks for power imparted to the water, which is the hydraulic power, always LESS than the shaft power.
Option a) 50 kW comes from dividing (30/0.6) instead of multiplying — that would be going the wrong way up the chain.
Source: AI practice
📖 §6.5 Pump suction performance — NPSH and cavitation
5. Increasing the suction pipe diameter of a pump will:
decrease NPSH available
increase NPSH available
have no effect on NPSH
increase NPSH required
Answer: B) increase NPSH available
Confirmed vs Book-3 §6.5 — the book notes that as flow increases in the suction pipework 'friction losses also increase, giving a lower NPSHA at the pump suction'.
A larger suction pipe lowers the suction velocity and friction loss, so the pressure at the impeller eye stays higher above vapour pressure and NPSHA rises.
Option d) is wrong because NPSHR is a characteristic of the PUMP design, not of the suction pipework.
Source: AI practice
📖 §6.2 Centrifugal pump — pressure expressed as head
6. The head developed by a centrifugal pump is:
proportional to liquid density
inversely proportional to liquid density
independent of liquid density
proportional to the square of liquid density
Answer: C) independent of liquid density
Confirmed vs Book-3 §6.2 — 'The pump generates the same head of liquid whatever the density of the liquid being pumped.'
Head is measured in metres of liquid column, so it depends on impeller speed and diameter, not on density.
Density does affect the DISCHARGE PRESSURE and the power (Pₕ = ρ·g·Q·H/1000) — mistaking pressure for head is why option a) tempts.
Source: AI practice
📖 §6.2 Centrifugal pump construction and working principle
7. In a centrifugal pump, the conversion of velocity (kinetic) energy into pressure energy is done by the:
impeller eye
diffuser / casing
suction nozzle
shaft seal
Answer: B) diffuser / casing
Confirmed vs Book-3 §6.2 — the impeller adds kinetic (velocity) energy to the liquid; the liquid 'is collected by the diffuser and converted to pressure'. The diffuser is also called the volute casing.
So the velocity-to-pressure conversion happens in the stationary casing, not in the rotating element.
Option a) the impeller eye is where liquid ENTERS and where pressure is at its lowest (the cavitation risk point).
Source: AI practice
📖 §6.3 Pump operating point (pump curve vs system curve)
8. The operating point of a pumping system is found at:
the pump's shut-off head
the best efficiency point only
the intersection of the pump curve and the system curve
the point of maximum NPSHR
Answer: C) the intersection of the pump curve and the system curve
Confirmed vs Book-3 §6.3 — the operating point is the intersection of the pump head-flow characteristic curve and the system curve (static head + friction head, friction head varying as flow²).
Move either curve — throttle the valve, trim the impeller, change speed — and the operating point moves with it.
Option b) is the trap: the pump only runs at its BEP if the system curve happens to pass through it.
Source: AI practice
📖 §6.6 Flow control — By-pass control
9. Which of the following pump flow-control methods is the LEAST energy-efficient?
Variable speed drive (VFD)
Impeller trimming
Throttle (control) valve
By-pass control
Answer: D) By-pass control
Confirmed vs Book-3 §6.6 — with by-pass control the pump runs continuously at maximum duty and surplus liquid is returned to the source; the book says this 'is even less energy efficient than a control valve because there is no reduction in power consumption with reduced process demand'.
A throttle valve at least moves the pump back along its curve, so some (small) power is saved — that is why option c) is the near-miss.
VFD and impeller trimming both genuinely reduce the power drawn.
Source: AI practice
📖 §6.4 Affinity Laws (effect of speed variation)
10. A pump's speed is reduced to two-thirds (2/3) of its original value. The new power consumption will be about:
67% of original
44% of original
30% of original
11% of original
Answer: C) 30% of original
Confirmed vs Book-3 §6.4 — P ∝ N³, so P₂/P₁ = (2/3)³ = 8/27 = 0.296, i.e. about 30% of the original power (roughly a 70% saving).
Efficiency is essentially independent of speed, so the cube law can be applied directly.
Option b) 44% is (2/3)² = 0.444, the HEAD ratio wrongly used for power.
Source: AI practice
📖 §6.6 Pumps in parallel — switched to meet demand
11. Different-sized pumps can be operated in parallel provided that their:
flow rates are identical
closed-valve (shut-off) heads are similar
impeller diameters are equal
motor ratings are equal
Answer: B) closed-valve (shut-off) heads are similar
Confirmed vs Book-3 §6.6 — 'It is possible to run pumps of different sizes in parallel provided their closed valve heads are similar.'
If the shut-off heads differ, the stronger pump can force the weaker one to zero flow or even back-flow through it.
Equal flow rates, impeller diameters or motor ratings are not required — that is the whole point of allowing different-sized pumps.
Source: AI practice
📖 §6.2 Hydraulic Power, Pump Shaft Power and Motor Input Power
12. A pump delivers 200 m³/h against a total head of 30 m at a pump efficiency of 70%. The approximate input (shaft) power required is:
16.35 kW
23.36 kW
30.0 kW
9.81 kW
Answer: B) 23.36 kW
Confirmed vs Book-3 §6.2 — Hydraulic Power = Q(m³/s) × H(m) × ρ(kg/m³) × g / 1000 = (200/3600) × 30 × 1000 × 9.81 / 1000 = 16.35 kW (same as Q×H×ρ/367 with Q in m³/hr).
Shaft Power = Hydraulic Power / Pump Efficiency = 16.35 / 0.70 = 23.36 kW.
Option a) 16.35 kW is the hydraulic power — the trap is stopping before dividing by the pump efficiency.
13. What is the primary purpose of trimming the impeller in a centrifugal pump?
To increase the pump speed
To adjust the pump capacity to match system requirements
To reduce the pump speed
To increase the NPSH required
Answer: B) To adjust the pump capacity to match system requirements
Confirmed vs Book-3 §6.6 — Impeller trimming machines the impeller diameter to reduce the energy added to the liquid — a permanent correction for a pump that is oversized for its system. It lowers both flow and head (Q∝D, H∝D²), not speed; NPSHR is not the target. Trimming is rarely taken below 75% of original diameter.
14. How does increasing the diameter of the suction pipe affect the NPSHA in a pumping system?
Reduces NPSHA
Increases NPSHA
Decreases NPSHR
Increases NPSHR
Answer: B) Increases NPSHA
Confirmed vs Book-3 §6.5 — NPSHA is the margin by which suction pressure at the impeller eye exceeds vapour pressure and is a characteristic of the system. A larger suction pipe cuts velocity and friction loss in the suction line, so NPSHA rises. NPSHR is a pump-design property and is unaffected by the piping.
Source: Sep 2024
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
15. A pump has a flow rate of 200 cubic meters per hour and operates against a head of 30 meters. If the pump efficiency is 70%, what is the input power required?
60.5 kW
23.36 kW
95.2 kW
100 kW
Answer: B) 23.36 kW
Confirmed vs Book-3 §6.1 — Hydraulic power = Q(m³/s)·H·ρ·g/1000 = (200/3600)×30×1000×9.81/1000 = 16.35 kW. Pump shaft (input) power = hydraulic power / pump efficiency = 16.35/0.70 = 23.36 kW. Option a (60.5) and d (100) do not follow from the formula; 95.2 would need a much lower efficiency.
Confirmed vs Book-3 §6.5 — The book lists three undesirable effects of cavitation: (1) collapsing bubbles erode the vane surface, (2) noise and vibration increase (shorter seal/bearing life), (3) cavities choke impeller passages and reduce head — in extreme cases total loss of head. It never improves efficiency or reduces noise; NPSHR is a pump-design property, not an effect.
17. What is the relationship between pump speed and flow rate in a centrifugal pump according to the Affinity Laws?
Flow rate is proportional to the pump speed
Flow rate is proportional to the square of the pump speed
Flow rate is proportional to the cube of the pump speed
Flow rate is independent of pump speed
Answer: A) Flow rate is proportional to the pump speed
Confirmed vs Book-3 §6.5 — Affinity laws for a rotodynamic pump: Q∝N (flow directly proportional to speed), H∝N² (head ∝ speed²), P∝N³ (power ∝ speed³). Book example: halving speed 3000→1500 rpm halves flow 100→50 m³/hr. The square and cube relations belong to head and power, not flow.
Source: Sep 2024
📖 §6.2 System characteristics — static & friction head
18. What is the impact of using larger diameter pipes on the system resistance in a pumping system?
Reduces system resistance by lowering friction head losses
Increases system resistance
Increases power
Increases static head
Answer: A) Reduces system resistance by lowering friction head losses
Confirmed vs Book-3 §6.2 — Friction (dynamic) head is the loss in pipes, valves and equipment; after removing unnecessary fittings and length, further reduction in friction head needs larger diameter pipe (lower velocity; friction loss ∝ 1/D⁵). Static head is set by elevation difference and is unaffected by pipe size, so options b–d are wrong.
Source: Sep 2024
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
19. A pump with 230mm diameter impeller is delivering a flow of 150 m3/hr. If the flow is to be reduced to 110m3/hr by trimming the impeller, what should be the approximate size of the impeller?
207mm
175 mm
169 mm
195 mm
Answer: C) 169 mm
Confirmed vs Book-3 §6.5 — Impeller-diameter affinity law Q∝D, so D₂ = D₁×(Q₂/Q₁) = 230×(110/150) = 168.7 ≈ 169 mm. 207 mm would give 150×207/230 = 135 m³/hr; 175 and 195 mm do not satisfy the ratio. Note 169/230 = 73%, at the edge of the book's ~75% trim limit — an exam-level approximation.
Source: Sep 2024
📖 §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions
20. If a pump delivery valve is throttled to 30% of rated flow, best energy efficiency measure is:
Replacing the motor
Installing a larger impeller
Increasing pump speed
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §6.5 — Table 6.1: a throttle-valve-controlled system signals an OVERSIZED pump; best solutions are trim impeller, smaller impeller, variable speed drive, two-speed drive or lower RPM — i.e. reduce output. A larger impeller or higher speed increases output and replacing the motor does nothing about the throttling loss, so none of the listed options is correct.
21. If pump speed is reduced to 2/3rd of its original speed, power consumption will:
Decrease by half
Decrease to one-fourth
Decrease to approx. 30% of original
Remains same
Answer: C) Decrease to approx. 30% of original
Confirmed vs Book-3 §6.5 — Affinity law P∝N³, so P₂/P₁ = (2/3)³ = 8/27 = 0.296 ≈ 30% of original power. 'Half' would need N₂/N₁ = 0.79; 'one-fourth' would be the head ratio if speed were halved (H∝N²), not the power ratio here.
Source: Sep 2025
📖 §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions
22. Which of the following is a common symptom indicating that a pump is oversized?
High discharge pressure
Throttle valve-controlled systems
Low suction pressure
High motor power consumption
Answer: B) Throttle valve-controlled systems
Confirmed vs Book-3 §6.5 — Table 6.1 lists 'throttle valve-controlled systems' and 'bypass line (partially or completely) open' as symptoms whose likely reason is an oversized pump. High discharge pressure or high motor power are consequences, not the diagnostic symptom the book names; low suction pressure relates to NPSH/cavitation.
Source: Sep 2025
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
23. Reducing the diameter of an impeller in a centrifugal pump will:
Increase head
Decrease head
No effect on head
Increase flow
Answer: B) Decrease head
Confirmed vs Book-3 §6.5 — Impeller-diameter relations: Q∝D, H∝D², P∝D³. Trimming reduces tip speed and 'lowers both the flow and pressure generated by the pump' (§6.6). Head decreases (with the square of the diameter ratio); it cannot increase and flow also falls, so option d is wrong.
Source: Sep 2025
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs
24. A centrifugal pump has BEP efficiency of 65%. At shut-off head, efficiency is:
0%
65%
50%
30%
Answer: A) 0%
Confirmed vs Book-3 §6.6 — At shut-off head flow is zero, so hydraulic power Q·H·ρ·g is zero and pump efficiency = hydraulic/shaft power = 0%. The book describes this condition (Fig 6.16): 'pump efficiency and flow rate are zero and with energy still being input to the liquid, the pump becomes a water heater'. The 65% BEP figure is a distractor.
Source: Sep 2025
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
25. A pump with 200 mm impeller delivers 120 m³/h. To deliver 100 m³/h by trimming, impeller size will be approximately:
240 mm
167 mm
60 mm
276 mm
Answer: B) 167 mm
Confirmed vs Book-3 §6.5 — Q∝D for impeller trimming, so D₂ = 200×(100/120) = 166.7 ≈ 167 mm (an 8.3% trim, within the ~75% limit). 240 mm is an enlargement; 60 mm applies the ratio the wrong way; 276 mm has no basis.
26. What is the impact on flow and pressure when the impeller of a pump is trimmed?
flow decreases with increased pressure
both flow and pressure increases
both pressure and flow decreases
none of the above
Answer: C) both pressure and flow decreases
Confirmed vs Book-3 §6.6 — Book (impeller trimming): 'reduces tip speed which in turn directly lowers the amount of energy imparted to the system liquid and lowers both the flow and pressure generated by the pump' — Q∝D, H∝D². Flow and pressure move together downward, so option a (opposite directions) and b (both increase) are wrong.
Source: Book EOC
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs
27. The most efficient method of flow control in a pumping system is ____.
throttling the flow
speed control
impeller trimming
bypass control
Answer: B) speed control
Confirmed vs Book-3 §6.6 — Book: 'pump speed adjustments provide the most efficient means of controlling pump flow' and 'flow control by speed regulation is always more efficient than by control valve'. Throttling wastes flow × valve head-drop; by-pass is 'even less energy efficient than a control valve'; impeller trimming is efficient but a fixed (permanent) reduction, not a control method.
Source: Book EOC
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
28. In case of centrifugal pumps, impeller diameter changes are generally limited to reducing the diameter to about ____ of maximum size.
75%
50%
25%
none of the above
Answer: A) 75%
Confirmed vs Book-3 §6.5 — Book: 'Diameter changes are generally limited to reducing the diameter to about 75% of the maximum, i.e. a head reduction to about 50%. Beyond this, efficiency and NPSH are badly affected.' §6.6 repeats that impellers 'are rarely reduced below 75 percent of their original size'. 50% is the head reduction, not the diameter limit.
Source: Book EOC
📖 §6.2 System characteristics — static & friction head; §6.11 Solved example — cooling water pump efficiency (p.194–195) (pipe velocities 1.5–2.0 m/s)
29. Generally water pipe lines are designed with water velocity of ____.
< 1m/s
up to 2 m/s
> 2 m/s
None of the above
Answer: B) up to 2 m/s
Confirmed vs Book-3 §6.2/§6.11 — Friction head loss rises with the square of flow (velocity), so water pipelines are designed for moderate velocities of up to about 2 m/s — the book's own solved example uses 1.5, 1.8 and 2.0 m/s. Velocities above 2 m/s sharply raise friction loss and pumping power; keeping below 1 m/s needs uneconomically large pipes.
Source: Book EOC
📖 §6.1 Pump types — centrifugal pump construction & working
30. The head generated by a centrifugal pump is
Independent of the density of the liquid being pumped.
Directly proportional to the density of the liquid being pumped.
Inversely proportional to the density of the liquid being pumped.
Proportional to the square of the density of the liquid being pumped.
Answer: A) Independent of the density of the liquid being pumped.
Confirmed vs Book-3 §6.1 — Book: 'The pump generates the same head of liquid whatever the density of the liquid being pumped' — head (metres of liquid column) is set by impeller diameter, speed, eye size and number of impellers. Density affects the pressure produced and the power drawn (P = Q·H·ρ·g), not the head, so b–d are wrong.
31. Increasing the suction pipe diameter in a pumping system will
Decrease NPSHA
Increase NPSHA
Decrease NPSHR
Increase NPSHR
Answer: B) Increase NPSHA
Confirmed vs Book-3 §6.5 — NPSHA = margin of eye pressure above vapour pressure; the book notes that as suction friction losses increase, NPSHA falls. A larger suction pipe lowers velocity and friction loss, so NPSHA increases. NPSHR is fixed by the pump design and does not change with pipe size. (Same as book end-of-chapter Q6.)
Source: Jul 2022
📖 §6.6 Flow control strategies — by-pass control
32. Small by-pass lines are installed sometimes to ____.
control flow rate
control pump delivery head
prevent pump running at zero flow
reduce pump power consumption
Answer: C) prevent pump running at zero flow
Confirmed vs Book-3 §6.6 — Book: 'The small by-pass line sometimes installed to prevent a pump running at zero flow is not a means of flow control, but required for the safe operation of the pump.' At shut-off the pump 'becomes a water heater'. It neither controls flow/head nor saves power — it slightly increases it.
Source: Book EOC
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
33. Shaft power of the motor driving a pump is 30 kW. The motor efficiency is 0.9 and pump efficiency is 0.6. The power transmitted to the water is
16.2 kW
18.0 kW
27.0 kW
none of the above
Answer: B) 18.0 kW
Confirmed vs Book-3 §6.1 — The 30 kW is the shaft power delivered by the motor, i.e. the pump shaft power Pₛ. Hydraulic (water) power = Pₛ × η_pump = 30×0.6 = 18.0 kW. Motor efficiency is a distractor — it would only be needed to find electrical input (30/0.9 = 33.3 kW). 16.2 kW wrongly applies both efficiencies; 27 kW wrongly applies motor efficiency.
Source: Book EOC
📖 §6.1 Pump types — centrifugal pump construction & working
34. In a centrifugal pump the velocity energy is converted to pressure by
impeller
casing
throttle valve
diffuser
Answer: D) diffuser
Corrected (was b) casing) — Book-3 §6.1: The book states: 'The water velocity is collected by the diffuser and converted to pressure by specially designed passageways that direct the flow to the discharge of the pump.' The two main parts are the impeller (imparts velocity) and the diffuser. 'Casing' is tempting because the book calls the diffuser 'also called as volute' (which houses the impeller), but the named element that converts velocity to pressure is the diffuser; the impeller adds velocity and a throttle valve only dissipates it.
Source: Book EOC
📖 §6.2 System characteristics — static & friction head
35. The friction loss in a pipe carrying a fluid is inversely proportional to the
fluid flow
square of the pipe diameter
fluid velocity
fifth power of pipe diameter
Answer: D) fifth power of pipe diameter
Confirmed vs Book-3 §6.2 — Friction head loss ∝ Q²/D⁵ for a given flow: halving the diameter multiplies loss ~32×. That is why the book says further friction reduction 'will require larger diameter pipe'. Friction loss is directly (not inversely) proportional to flow² and velocity², so a–c are wrong; 'square of diameter' badly understates the effect.
Source: Book EOC
📖 §6.6 Flow control strategies — pumps in parallel switched to meet demand
36. It is possible to run pumps in parallel provided their ____ are similar.
suction head
discharge heads
closed valve heads
none of the above
Answer: C) closed valve heads
Confirmed vs Book-3 §6.6 — Book: 'It is possible to run pumps of different sizes in parallel provided their closed valve heads are similar.' With similar shut-off heads each pump contributes flow at the common header head; if one pump's shut-off head is lower it is pushed to zero flow by the other. Suction or discharge heads being similar is not the criterion.
37. If the speed of the pump is doubled, power goes up by
2 times
6 times
8 times
4 times
Answer: C) 8 times
Confirmed vs Book-3 §6.5 — Affinity law P∝N³: doubling speed gives 2³ = 8 times the power (flow doubles, head ×4). Book example in reverse: 3000→1500 rpm drops 40 kW to 5 kW (÷8). '4 times' is the head ratio, '2 times' the flow ratio.
Source: Book EOC
📖 §6.3 Pump curves — pump operating point
38. The operating point in a pumping system is identified by
point of intersection of system curve and efficiency curve
point of intersection of pump curve and theoretical power curve
point of intersection of pump curve and system curve
none of the above
Answer: C) point of intersection of pump curve and system curve
Confirmed vs Book-3 §6.3 — Book: 'When a pump is installed in a system the effect can be illustrated graphically by superimposing pump and system curves. The operating point will always be where the two curves intersect' (Fig 6.9). Efficiency and power curves are plotted against flow but do not define the duty point.
Source: Book EOC
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
39. The hydraulic power of a motor pump set is 8 kW. If the power drawn by the motor is 16 kW at 90% efficiency, the pump efficiency will be
55.5%
50%
45%
none of the above
Answer: A) 55.5%
Confirmed vs Book-3 §6.1 — Pump shaft power = motor input × motor efficiency = 16×0.9 = 14.4 kW. Pump efficiency = hydraulic power / shaft power = 8/14.4 = 55.5%. 50% (8/16) ignores the motor efficiency; 45% wrongly multiplies 50% by 0.9.
Source: 15th Exam
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs
40. The energy saving with variable speed drives in a pumping system will be maximum for systems with
pure static head
pure friction head
high static head and low friction head
high static head with high friction head
Answer: B) pure friction head
Confirmed vs Book-3 §6.6 — In a friction-only system (Fig 6.15) reducing speed moves the duty point along an iso-efficiency line and 'the affinity laws are obeyed... making variable speed the ideal control method for systems with friction loss'. With high static head (Fig 6.16) flow is no longer proportional to speed, efficiency drops and the pump can reach shut-off, so savings are smallest.
Source: 15th Exam
📖 §6.4 Matching pump and system head-flow characteristics (Fig 6.11); §6.3 Pump curves — pump operating point
41. The intersection point of the centrifugal pump characteristic curve and the design system curve is the
pump efficiency point
best efficiency point
system efficiency point
none of the above
Answer: B) best efficiency point
Confirmed vs Book-3 §6.4 — Book: for the estimated duty 'we will chose a pump curve which intersects the system curve (Point A) at the pump's best efficiency point (BEP)'. So the intersection of the pump curve with the DESIGN system curve is the BEP — pump efficiency is highest at that one flow. 'Pump efficiency point' and 'system efficiency point' are not book terms.
Source: 15th Exam
📖 §6.1 Pump types — centrifugal pump construction & working
42. In case of increased suction lift from open wells, the pump delivered flow rate
increases
decreases
remains same
none of the above
Answer: B) decreases
Confirmed vs Book-3 §6.1 — Book: 'A centrifugal pump is not positive acting... The greater the depth of the water, the lesser is the flow from the pump. Also, when it pumps against increasing pressure, the less it will pump.' Increased suction lift raises total head, moving the duty point up the H-Q curve to lower flow (and lowering NPSHA).
Source: 15th Exam
📖 §6.6 Flow control strategies — pumps in parallel switched to meet demand
43. In pumping systems where static head is a high proportion of the total, the appropriate solution is
install two or more pumps to operate in parallel
install two or more pumps to operate in series
install two or more pumps to operate independently
install variable frequency drive for the pump
Answer: A) install two or more pumps to operate in parallel
Confirmed vs Book-3 §6.6 — Book: 'Another energy efficient method of flow control, particularly for systems where static head is a high proportion of the total, is to install two or more pumps to operate in parallel', switching them on/off to meet demand. A VFD is risky here — slowing a pump against high static head can push it to shut-off. Series adds head, not flow.
Source: 18th Exam
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
44. Shaft power of the motor driving a pump is 30 kW. The motor efficiency is 0.92 and pump efficiency is 0.5. The power drawn by the motor will be
65.2 kW
15 kW
30 kW
32.6 kW
Answer: D) 32.6 kW
Confirmed vs Book-3 §6.1 — Motor input power = pump shaft power / motor efficiency = 30/0.92 = 32.6 kW. Pump efficiency (0.5) is a distractor — it gives hydraulic power (15 kW, option b), not motor input. 65.2 kW wrongly divides by both efficiencies.
Source: 15th Exam
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
45. If the power drawn by the motor driving a pump is 20 kW at 91% efficiency, and the hydraulic power of the motor pump set is 12.5 kW, the pump efficiency will be
68.7%
62.5%
56.8%
none of the above
Answer: A) 68.7%
Confirmed vs Book-3 §6.1 — Pump shaft power = 20×0.91 = 18.2 kW; pump efficiency = hydraulic/shaft = 12.5/18.2 = 68.7%. 62.5% (12.5/20) ignores motor efficiency; 56.8% wrongly multiplies that by 0.91.
Source: 14th Exam
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
46. Which of the following is not true for impeller trimming?
pressure ∝ diameter
head ∝ diameter²
power ∝ diameter³
flow ∝ diameter
Answer: A) pressure ∝ diameter
Confirmed vs Book-3 §6.5 — Impeller-diameter relations: Q∝D, H∝D², P∝D³ (book §6.5/§6.6). Pressure (head) varies with the SQUARE of diameter, so 'pressure ∝ diameter' is the false statement; b, c and d are all correct affinity relations.
47. The preferred method of flow control for reducing pump flow permanently in a pumping system is
throttling
speed control
impeller trimming
none of the above
Answer: C) impeller trimming
Confirmed vs Book-3 §6.6 — The book heads impeller trimming under 'Fixed Flow reduction' — machining the impeller for a permanent correction of an oversized pump, cheaper than a new impeller or pump. Speed control is 'Meeting variable flow reduction'; throttling wastes flow × valve head-drop.
Source: 14th Exam
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
48. A water pump is delivering 20 m3/hr. The impeller diameter is trimmed by 10%. This will reduce the pump discharge by
18 m3/hr
2 m3/hr
0.2 m3/hr
none of the above
Answer: B) 2 m3/hr
Confirmed vs Book-3 §6.5 — Q∝D: a 10% diameter trim gives Q₂ = 0.9×20 = 18 m³/hr, i.e. a REDUCTION of 2 m³/hr. Option a (18) is the new flow, not the reduction; 0.2 would be a 1% change.
Source: 14th Exam
📖 §6.1 Pump types — centrifugal pump construction & working
49. The head developed by a centrifugal pump is not directly proportional to
Impeller diameter
Shaft speed
Number of impellers
Diameter of discharge port
Answer: D) Diameter of discharge port
Confirmed vs Book-3 §6.1 — Book: 'The pressure (head) that a pump will develop is in direct relationship to the impeller diameter, the number of impellers, the size of impeller eye, and shaft speed.' Discharge-port diameter is not in that list — it affects velocity/friction, not the head the impeller generates.
Source: Set-A
📖 §6.2 System characteristics — static & friction head
50. The frictional loss in a piping system is proportional to
flow
flow²
1/flow
1/flow²
Answer: B) flow²
Confirmed vs Book-3 §6.2 — Book: 'The friction losses are proportional to the square of the flow rate.' Hence a closed-loop system curve is a parabola through the origin and a throttled valve adds a loss 'proportional to flow squared' (§6.6). Inverse relations (c, d) are wrong.
Source: Set-A
📖 §6.2 System characteristics — static & friction head
51. For the same flow, through which of the following diameter pipes will the pump work with maximum pressure?
100 mm
150 mm
200 mm
250 mm
Answer: A) 100 mm
Confirmed vs Book-3 §6.2 — For the same flow, the smallest pipe (100 mm) has the highest velocity and the largest friction loss (∝ 1/D⁵), so the pump must develop the maximum pressure (head) to push the flow through it. Larger pipes reduce friction head — the book's reason for using larger diameter pipe.
Source: Set-A
📖 §6.6 Flow control strategies — pumps in parallel switched to meet demand
52. It is possible to run pumps in parallel if their _________________ are similar.
suction heads
discharge heads
closed valve heads
none of the above
Answer: C) closed valve heads
Confirmed vs Book-3 §6.6 — Book: pumps of different sizes can run in parallel 'provided their closed valve heads are similar'. At the common header head each pump then delivers flow; a pump with a much lower shut-off head would be forced to zero flow by the others. Suction/discharge heads are not the criterion.
Source: Set-A
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
53. Input power to the motor driving a pump is 20 kW. The motor efficiency is 0.9 and pump efficiency is 0.7. The power transmitted to the water is
12.6 kW
18.0 kW
14.0 kW
31.75 kW
Answer: A) 12.6 kW
Confirmed vs Book-3 §6.1 — Hydraulic (water) power = motor input × η_motor × η_pump = 20×0.9×0.7 = 12.6 kW. 18 kW is only the motor shaft output (20×0.9); 14 kW applies only the pump efficiency; 31.75 wrongly divides by the efficiencies.
Source: Set-A
📖 §6.6 Flow control strategies — by-pass control
54. Small by-pass lines are installed in pumps sometimes to _____.
increase flow
control pump delivery head
prevent pump running at zero flow
reduce pump power consumption
Answer: C) prevent pump running at zero flow
Confirmed vs Book-3 §6.6 — Book: 'The small by-pass line sometimes installed to prevent a pump running at zero flow is not a means of flow control, but required for the safe operation of the pump.' At shut-off, energy is still input and 'the pump becomes a water heater'. It does not raise delivered flow, control head or save power.
Source: Set-A
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
55. The hydraulic power in a pumping system depends on
Pump efficiency
Motor efficiency
Both motor and pump efficiency
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §6.1 — Hydraulic power Pₕ = Q×(h_d − h_s)×ρ×g/1000 depends only on flow, total head and liquid density. Pump efficiency converts it to shaft power (Pₛ = Pₕ/η_pump) and motor efficiency to input power — they are downstream of, not inputs to, hydraulic power. Hence 'none of the above'.
Source: 19th Exam
📖 §6.2 System characteristics — static & friction head
56. Installing larger diameter pipe in pumping system results in reduction in ______
static head
frictional head
both a and b
neither a nor b
Answer: B) frictional head
Confirmed vs Book-3 §6.2 — Static head is 'simply the difference in height of the supply and destination reservoirs' and is independent of flow and pipe size. Friction head is the loss in pipes/valves/equipment, and the book notes that further reduction 'will require larger diameter pipe'. So only frictional head is reduced.
Source: Jul 2022
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
57. Increasing the impeller diameter in a pump:
Increases the flow
decreases the head
decreases the power
all of the above
Answer: A) Increases the flow
Confirmed vs Book-3 §6.5 — Impeller-diameter relations: Q∝D, H∝D², P∝D³. A larger impeller raises tip speed, so flow, head AND power all increase. Options b and c state decreases, so 'all of the above' is wrong; only a is true.
Source: 17th Sep-2016
📖 §6.2 System characteristics — static & friction head
58. Installing larger diameter pipe in pumping system results in:
increase in static head
decrease in static head
increase in frictional head
decrease in frictional head
Answer: D) decrease in frictional head
Confirmed vs Book-3 §6.2 — Larger pipe lowers velocity and friction loss (∝ Q²/D⁵), so frictional (dynamic) head decreases; the book cites larger diameter pipe as the way to cut friction head further. Static head is fixed by elevation difference and is unchanged by pipe size.
Source: 16th Exam
📖 §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series)
59. If two identical pumps operate in series, then the combined shutoff head is:
same as one pump
twice that of one pump
half that of one pump
four times that of one pump
Answer: B) twice that of one pump
Confirmed vs Book-3 §6.7 — Pumps in series add their heads at a given flow — the book notes a multistage pump 'is similar to the operation of several single stage pumps, of identical capacity, in series'. At zero flow two identical pumps therefore give twice the single-pump shut-off head. (Parallel pumps would add flows instead.) Options reconstructed from the exam key.
Source: 16th Exam
📖 §6.1 Pump types — centrifugal pump construction & working (displacement vs dynamic pumps); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³) (affinity laws apply to rotodynamic pumps only)
60. If the speed of a reciprocating pump is reduced by 50 %, the head
is reduced by 25%
is reduced by 50%
is reduced by 75%
remains same
Answer: D) remains same
Confirmed vs Book-3 §6.1/§6.5 — The affinity laws (H∝N²) are stated by the book for 'rotodynamic pump performance parameters'. A reciprocating pump is a positive-displacement pump: its head is set by the system pressure it discharges into, while speed reduction only cuts the volume delivered (flow ∝ speed). So the head remains the same.
Source: 18th Exam
📖 §6.2 System characteristics — static & friction head
61. Friction losses in a pumping system is
inversely proportional to flow
inversely proportional to cube of flow
proportional to square of flow
inversely proportional square of flow
Answer: C) proportional to square of flow
Confirmed vs Book-3 §6.2 — Book: 'The friction losses are proportional to the square of the flow rate.' This is why the system curve is parabolic and why halving flow cuts friction head to a quarter. Inverse relations (a, b, d) are wrong.
Source: Jul 2022
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
62. A pump discharge has to be reduced from 120 m3/hr to 100 m3/hr by trimming the impeller. What should be the percentage reduction in impeller size?
83.3%
16.7%
50.0%
33.3%
Answer: B) 16.7%
Confirmed vs Book-3 §6.5 — Q∝D, so D₂/D₁ = Q₂/Q₁ = 100/120 = 0.833; percentage reduction = (1 − 0.833)×100 = 16.7%. 83.3% is the remaining diameter ratio, not the reduction; 33.3% would be the head reduction (1 − 0.833²).
Source: 18th Exam
📖 §6.3 Pump curves — pump operating point; §6.6 Flow control strategies — flow control valve (throttling)
63. Which of the following statements is not true regarding centrifugal pumps?
Flow is zero at shut off head
Maximum efficiency will be at design rated flow of the pump
Head decreases with increase in flow
Power increases with throttling
Answer: D) Power increases with throttling
Confirmed vs Book-3 §6.3/§6.6 — For a centrifugal pump flow is zero at shut-off, head falls as flow rises, and efficiency peaks at the design (BEP) flow — a, b, c are all true. On throttling the duty point moves up the curve to lower flow and the book notes 'there is some reduction in pump power absorbed at the lower flow rate' — power decreases, so d is the false statement.
Source: Sep 2019
📖 §6.2 System characteristics — static & friction head
64. Which of the following is not true regarding system characteristic curve in a pumping system with large dynamic head?
System curve represents a relationship between discharge and head loss in a system of pipes
System curve is dependent on the pump speed
The basic shape of a system curve is parabolic
System curve will start at zero flow and zero head if there is no static lift
Answer: B) System curve is dependent on the pump speed
Confirmed vs Book-3 §6.2 — The system curve is the head-loss vs flow relationship of the piping (a); friction ∝ Q² makes it parabolic (c); with no static lift it starts at zero head and zero flow (d, Fig 6.5). It depends on elevation, pipe size/length, fittings and equipment — not on pump speed, which shifts the PUMP curve. So b is false.
Source: Jul 2022
📖 §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series)
65. If two identical pumps operate in series, their shut-off head is
Not affected
More than double
Doubled
Less than double
Answer: C) Doubled
Confirmed vs Book-3 §6.7 — Series operation adds heads at the same flow (the book equates a multistage pump with single-stage pumps in series). At zero flow each identical pump gives its shut-off head, so the combination gives exactly double. Parallel pumps would leave shut-off head unchanged and add flow.
Source: 19th Exam
📖 §6.6 Flow control strategies — by-pass control
66. Small diameter by-pass lines are installed in pumps sometimes to ________________.
Save energy
Control pump delivery head
Prevent pump running at zero flow
Reduce pump power consumption
Answer: C) Prevent pump running at zero flow
Confirmed vs Book-3 §6.6 — Book: the small by-pass line 'sometimes installed to prevent a pump running at zero flow is not a means of flow control, but required for the safe operation of the pump' — at shut-off the pump overheats ('becomes a water heater'). It does not save energy or control head.
Source: 19th Exam
📖 §6.6 Flow control strategies — pumps in parallel switched to meet demand
67. It is acceptable to run pumps in parallel provided their _________ are similar
Suction heads
Discharge heads
Closed valve heads
Total head at full flow
Answer: C) Closed valve heads
Confirmed vs Book-3 §6.6 — Book: 'It is possible to run pumps of different sizes in parallel provided their closed valve heads are similar.' Similar shut-off heads let each pump deliver flow at the common header head; total head at full flow, suction or discharge heads are not the criterion.
68. The value by which the pressure in the pump suction exceeds the liquid vapour pressure is expressed as
Net positive suction head available
Static head
Dynamic head
Suction head
Answer: A) Net positive suction head available
Confirmed vs Book-3 §6.5 — Book definition: 'The value, by which the liquid pressure at the eye of pump exceeds the liquid vapour pressure, is expressed as a head of liquid and referred to as Net Positive Suction Head Available (NPSHA)' — a system characteristic. Static and dynamic heads describe system resistance; suction head alone ignores vapour pressure.
Source: 19th Exam
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
69. A pump with 230 mm diameter impeller is delivering a flow of 150 m3/hr. If the flow is to be reduced to 110 m3/hr by trimming the impeller, what should be the approximate impeller size?
195 mm
175 mm
169 mm
207 mm
Answer: C) 169 mm
Confirmed vs Book-3 §6.5 — Q∝D, so D₂ = 230×(110/150) = 168.7 ≈ 169 mm. 207 mm would still deliver ~135 m³/hr; 175/195 mm do not match the ratio. The trim (to 73%) sits at the book's ~75% practical limit.
Source: 19th Exam
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
70. A pump discharge has to be reduced from 120 m3/hr to 110 m3/hr by trimming the impeller. What should be the percentage reduction in impeller size?
10.52 %
8.34%
9.71%
17.1%
Answer: B) 8.34%
Confirmed vs Book-3 §6.5 — Q∝D: D₂/D₁ = 110/120 = 0.9167, so reduction = 8.33% ≈ 8.34%. 10.52% and 9.71% do not follow from the ratio; 17.1% would be the corresponding head reduction (1 − 0.9167²) — the tempting wrong option.
71. In a pumping system, if the temperature of the liquid handled increases, then ___________________.
NPSHa increases
NPSHa decreases
NPSHa remains constant
NPSHa and NPSHr are independent of temperature
Answer: B) NPSHa decreases
Confirmed vs Book-3 §6.5 — NPSHA is the margin of eye pressure above the liquid's vapour pressure. Raising the liquid temperature raises its vapour pressure, so the margin — NPSHA — decreases and cavitation risk rises. NPSHA therefore is not independent of temperature; NPSHR (pump property) is unchanged.
Source: Sep 2019
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
72. The hydraulic power of a motor pump set is 8.5 kW. If the power drawn by the motor is 15.5 kW at a 89% efficiency, the pump efficiency is given by
54.8%
61.6%
48.8%
none of the above
Answer: B) 61.6%
Confirmed vs Book-3 §6.1 — Pump shaft power = 15.5×0.89 = 13.795 kW; pump efficiency = 8.5/13.795 = 61.6%. 54.8% (8.5/15.5) ignores motor efficiency; 48.8% wrongly multiplies that by 0.89.
Source: 9th Dec-2009
📖 §6.2 System characteristics — static & friction head
73. For the same flow through which of the following diameter pipes, the pump will work with maximum pressure
80 mm
100 mm
120 mm
140 mm
Answer: A) 80 mm
Confirmed vs Book-3 §6.2 — For a fixed flow the smallest pipe (80 mm) gives the highest velocity and friction loss (∝ 1/D⁵), so the pump must work against the maximum pressure. Larger pipes reduce friction head — the book's rationale for larger diameter pipe. (Option d corrected from OCR '1400 mm' to '140 mm'.)
Source: 9th Dec-2009
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
74. The motor efficiency is 0.9 and the pump efficiency is 0.6. The input power to the motor driving the pump is 28 kW. The power transmitted to the water is
15.12 kW
28 kW
25.2 kW
none of the above
Answer: A) 15.12 kW
Confirmed vs Book-3 §6.1 — Power to water (hydraulic) = motor input × η_motor × η_pump = 28×0.9×0.6 = 15.12 kW. 25.2 kW is only the motor shaft output (28×0.9); 28 kW is the electrical input.
Source: 9th Dec-2009
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
75. A water pump is delivering 200 m3/hr at ambient conditions. The impeller diameter is trimmed by 10%. The water flow at the changed conditions is
220 m3/hr
180 m3/hr
162 m3/hr
none of the above
Answer: B) 180 m3/hr
Confirmed vs Book-3 §6.5 — Q∝D: a 10% trim gives Q₂ = 0.9×200 = 180 m³/hr. 162 m³/hr (0.9²×200) wrongly uses the head relation; 220 would be an enlarged impeller.
Source: 9th Dec-2009
📖 §6.1 Pump types — centrifugal pump construction & working
76. With increase in the suction lift from open wells, the delivery flow rate
increases
decreases
remains same
none of the above
Answer: B) decreases
Confirmed vs Book-3 §6.1 — Book: 'The greater the depth of the water, the lesser is the flow from the pump. Also, when it pumps against increasing pressure, the less it will pump.' Higher suction lift raises total head, moving the duty point to lower flow on the H-Q curve (and reducing NPSHA).
Source: 10th Jul-2010
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs
77. There are four pumps working against the same friction heads. For which static head will the variable speed drive be most economical?
Pump A - 0 m
Pump B - 10 m
Pump C - 20 m
Pump D - 25 m
Answer: A) Pump A - 0 m
Confirmed vs Book-3 §6.6 — VSD savings are greatest in a friction-only system (zero static head): the duty point moves along an iso-efficiency line and the affinity laws (P∝N³) are fully obeyed (Fig 6.15). As static head rises the efficiency drop with speed 'reduces the economic benefits of variable speed control' (Fig 6.16). So Pump A (0 m) is the best VSD candidate.
Source: 10th Jul-2010
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
78. A pump with 200 mm impeller is delivering a flow of 120 m3/hr. If the flow is to be reduced to 100 m3/hr by trimming the impeller, what should be the approximate impeller size ?
60 mm
240 mm
167 mm
145 mm
Answer: C) 167 mm
Confirmed vs Book-3 §6.5 — Q∝D, so D₂ = 200×(100/120) = 166.7 ≈ 167 mm. 60 mm applies the ratio inversely; 240 mm is an enlargement; 145 mm would give only 87 m³/hr.
Source: Mar 2021
📖 §6.2 System characteristics — static & friction head
79. Friction loss in a piping system carrying fluid is proportional to
fluid flow
(fluid flow)2
1/fluid flow
1/(fluid flow)2
Answer: B) (fluid flow)2
Confirmed vs Book-3 §6.2 — Book: 'The friction losses are proportional to the square of the flow rate' — hence the parabolic system curve and the valve loss 'proportional to flow squared'. Linear (a) and inverse (c, d) relations are wrong.
Source: 11th Feb-2011
📖 §6.2 System characteristics — static & friction head; §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³)
80. In a pumping system the static head is 10 m and the dynamic head is 15 m. If the pump speed is doubled, then the total head will be
50 m
70 m
40 m
none of the above
Answer: B) 70 m
Confirmed vs Book-3 §6.2/§6.5 — Static head is independent of flow, so it stays 10 m. Dynamic (friction) head ∝ flow² and flow ∝ speed, so doubling speed quadruples the dynamic head: 15×4 = 60 m. Total = 10 + 60 = 70 m. 50 m wrongly doubles the dynamic head; 40 m quadruples only the static part.
Source: 11th Feb-2011
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
81. Input power to the motor driving a pump is 30 kW. The motor efficiency is 0.9. The power transmitted to the water is 16.2 kW. The pump efficiency is
54%
60%
90%
none of the above
Answer: B) 60%
Confirmed vs Book-3 §6.1 — Pump shaft power = 30×0.9 = 27 kW; pump efficiency = hydraulic/shaft = 16.2/27 = 60%. 54% (16.2/30) ignores motor efficiency; 90% is the motor efficiency itself.
Source: 11th Feb-2011
📖 Book-3 Ch5 Fans & Blowers — flow control (pulley change, damper, inlet guide vanes); not a Ch6 pumps topic
82. Which of the following can be used to regulate the flow of fans ?
pulley change
damper control
inlet guide vane regulation
all of the above
Answer: D) all of the above
Confirmed vs Book-3 Ch5 — Fan flow can be regulated by changing speed (pulley change on belt drives), by damper (throttle) control and by inlet guide vane regulation — all listed fan flow-control methods in Book-3 Ch5, so 'all of the above'. (This fan question sits in the pumps set as it appeared in that exam block.)
Source: Mar 2021
📖 §6.1 Pump types — centrifugal pump construction & working
83. Which of the following pump is not a positive displacement pump ?
piston pump
rotary vane
diaphragm pump
centrifugal pump
Answer: D) centrifugal pump
Confirmed vs Book-3 §6.1 — Book classification: dynamic pumps (centrifugal, special effect) vs displacement pumps (rotary, reciprocating). Piston and diaphragm pumps are reciprocating and a rotary vane pump is rotary — all positive displacement. The centrifugal pump is a dynamic (rotodynamic) pump and 'is not positive acting'.
84. In a pumping system, if the temperature of the liquid handled decreases, then
NPSHa increases
NPSHa decreases
NPSHa remains constant
NPSHa and NPSHr are independent of temperature
Answer: A) NPSHa increases
Confirmed vs Book-3 §6.5 — NPSHA is the margin of eye pressure above the liquid's vapour pressure. A cooler liquid has a lower vapour pressure, so the margin — NPSHA — increases and cavitation risk falls. NPSHA is therefore temperature-dependent; NPSHR (pump property) is unchanged.
Source: Mar 2021
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
85. Shaft power of the motor driving a pump is 20 kW. The motor efficiency is 0.9 and pump efficiency is 0.55 at shaft operating load. The power transmitted to the water is ________.
12.2 kW
9.9 kW
11 kW
12.7 kW
Answer: C) 11 kW
Confirmed vs Book-3 §6.1 — The 20 kW is the pump shaft power; hydraulic (water) power = shaft power × η_pump = 20×0.55 = 11 kW. Motor efficiency is a distractor (it relates shaft power to electrical input, 20/0.9 = 22.2 kW). 9.9 kW wrongly applies both efficiencies. (Option c repaired from garbled OCR '0.75 - 0.66 kW' to '11 kW', matching the Set-B print.)
Source: Mar 2021
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
86. Shaft power of the motor driving a pump is 20 kW. The motor efficiency is 0.9 and pump efficiency is 0.55 at that operating load. The power transmitted to the water is ___________
12.2 kW
9.9 kW
11 kW
12.7 kW
Answer: C) 11 kW
Confirmed vs Book-3 §6.1 — Hydraulic (water) power = pump shaft power × pump efficiency = 20×0.55 = 11 kW. Motor efficiency only links shaft power to electrical input (20/0.9 = 22.2 kW) and is a distractor; 9.9 kW wrongly applies both efficiencies.
Source: Mar 2021 (Set B)
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
87. A pump discharge has to be reduced from 120 m³/hr to 100 m³/hr by trimming the impeller. What should be the percentage reduction in impeller size?
83.3%
16.7%
50.0%
33.3%
Answer: B) 16.7%
Confirmed vs Book-3 §6.5 — Q∝D: D₂/D₁ = 100/120 = 0.833, so the diameter reduction = 16.7%. 83.3% is the remaining ratio; 33.3% ≈ head reduction (1 − 0.833²), a tempting confusion of the H∝D² law.
Source: Jul 2022
📖 §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions; §6.11 Energy conservation opportunities in pumping systems (over-designed pump: VSD, downsize impeller, or replace with correct-sized pump)
88. If the delivery valve of the pump is throttled such that it delivers 30% of the rated flow, one of the best options for improved energy efficiency would be
Trimming of the impeller
Replacing the motor
Replacing with a smaller pump operating with VFD
none
Answer: C) Replacing with a smaller pump operating with VFD
Confirmed vs Book-3 §6.5/§6.11 — A pump throttled to 30% of rated flow is grossly oversized and most of its head is dissipated across the valve. Trimming is limited to ~75% diameter (Q∝D → ≥75% flow), so it cannot reach 30%; replacing the motor does nothing about the hydraulic waste. The book's remedy for an over-designed pump — 'replace with correct sized pump' and VSD for variable flow — is option c.
Source: Jul 2022
📖 §6.1 Pump types — centrifugal pump construction & working
89. A pump is handling water at 25 deg C and delivering 200 m³/hr. If the water temperature is 50 deg C then the flow will __________.
Increase by 50%
Decrease by 50%
double
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §6.1 — A centrifugal pump develops head and volumetric flow set by its impeller geometry and speed; the book says it 'generates the same head of liquid whatever the density'. Warming water from 25 to 50 °C changes density only ~1%, so volumetric flow is essentially unchanged — none of the listed 50%/double changes occur.
Source: Mar 2023
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
90. Pump Shaft Power is __________.
Hydraulic Power / Motor Efficiency
Hydraulic Power / Pump Efficiency
Hydraulic Power × Pump Efficiency
Hydraulic Power × Motor Efficiency
Answer: B) Hydraulic Power / Pump Efficiency
Confirmed vs Book-3 §6.1 — Book p.172: 'Pump Shaft Power Pₛ = Hydraulic power, Pₕ / Pump Efficiency, η_pump' and Motor Input Power = Pₛ / η_motor. Shaft power must exceed hydraulic power, so dividing by pump efficiency is correct; multiplying (c, d) would give less than hydraulic power, and motor efficiency (a) relates shaft power to electrical input, not to hydraulic power. (Options repaired: OCR had rendered '/' as '*' and duplicated b/c.)
Source: Mar 2023
📖 §6.11 Solved example — cooling water pump efficiency (p.194–195)
91. Calculate the Volumetric Flow (m³/sec) in a pipe with a diameter 200 mm and Velocity 1.5 m/sec.
0.19 m³/sec
0.015 m³/sec
0.047 m³/sec
0.012 m³/sec
Answer: C) 0.047 m³/sec
Confirmed vs Book-3 §6.11 — Flow = area × velocity = (π/4)×d²×v = 0.7854×(0.2)²×1.5 = 0.0314×1.5 = 0.047 m³/s — the same method as the book's solved example (pipes A, B, C). 0.19 forgets the /4; 0.015 and 0.012 mis-square the diameter.
Source: Mar 2023
Short questions (5 marks) — 24
📖 §6.4 Effects of impeller diameter change — trim limits
1. A 200 mm impeller delivers 120 m³/h, but the system needs only 100 m³/h. Find the trimmed impeller diameter and confirm it is within the safe trim limit.
Model answer: Trimmed diameter ≈ 167 mm — within the safe limit.
Step 1 — Diameter affinity law (§6.4): Q ∝ D, so D₂ = D₁ × (Q₂/Q₁) = 200 × (100/120) = 166.7 ≈ 167 mm.
Step 2 — Safe-trim check: the book limits trimming to about 75% of the maximum diameter = 0.75 × 200 = 150 mm. Since 167 mm > 150 mm, the trim is acceptable; at 150 mm the head would already have fallen to about 50%.
Step 3 — Caution: the change here is (200−167)/200 ≈ 17%, well above the ~5% at which the book says the squared/cubic relations lose accuracy, so the manufacturer's performance curves should be used to confirm the final diameter.
Confirmed vs Book-3 §6.4 — use Q ∝ D for the new diameter, H ∝ D² and P ∝ D³ for the resulting head and power, then check the 75% trim floor.
Trimming below 75% badly affects efficiency and NPSH, which is why the check is part of the answer, not an afterthought.
Source: AI practice
📖 §6.2 Hydraulic/Shaft/Motor power (end-of-chapter S-2)
2. A pump discharges 50 m³/h. Discharge pressure is 3.5 kg/cm² and the suction is 5 m below the pump centreline. The motor draws 9.5 kW at 90% motor efficiency. Calculate the pump efficiency.
Model answer: Pump efficiency ≈ 64%.
Step 1 — Head: discharge pressure 3.5 kg/cm² = 3.5 × 10 = 35 m of water. The sump level is 5 m BELOW the pump centreline, so the pump must also lift that 5 m: total head H = 35 + 5 = 40 m.
Step 2 — Flow: Q = 50/3600 = 0.01389 m³/s.
Step 3 — Hydraulic power (§6.2) = Q × H × ρ × g / 1000 = 0.01389 × 40 × 1000 × 9.81 / 1000 = 5.45 kW.
Step 4 — Shaft power = motor input × motor efficiency = 9.5 × 0.90 = 8.55 kW.
Step 5 — Pump efficiency = hydraulic power / shaft power = 5.45 / 8.55 = 0.637 ≈ 64%.
Confirmed vs Book-3 §6.2 (S-2 data) — the two marks that are usually lost: 1 kg/cm² = 10 m of water head, and a suction level BELOW the centreline ADDS to the total head (a flooded suction above the centreline would subtract).
Also note shaft power = motor kW × motor η (multiply); dividing here would give ~10.6 kW and an impossible efficiency.
Source: AI practice
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); total differential head h_d − h_s (positive suction head is subtracted); cf. §6.11 Solved example — cooling water pump efficiency (p.194–195)
3. A cooling water pump has a positive suction head of 5 meters. The discharge pressure is 3.0 kg/cm², and the water flow rate is 150 m³/hr. Determine the pump efficiency given that the actual power input of the connected motor is 18.0 kW and the motor operates with an efficiency of 85%.
Model answer: Flow Rate: 150 m³/hr
Total Head: 30 − 5 = 25 m
Power input to pump = 18 × 0.85 = 15.3 kW
Hydraulic Power = (150/3600) × 25 × 9.81 = 10.2 kW
Pump Efficiency = 10.2 / 15.3 = 66.7%
Total head = discharge head (3 kg/cm² ≈ 30 m) − suction head; hydraulic power = (Q/3600)·H·g; η = hydraulic/shaft power.
Source: Sep 2024
📖 §6.3 Pump curves — pump operating point; §6.4 pump efficiency highest at one flow (BEP)
4. Analyse the following data collected for a water pump. If the operating head is 16m explain what will happen to other parameters.
Design Parameters / Values:
Flow (Q): 40 lps
Head (H): 20 m
Power (P): 15 kW
Efficiency: 51%
Model answer: 1. If the operating head is 16 m instead of 20 m, the operating flow will be higher than the rated flow.
2. Since the operating point has deviated from the BEP, the operating efficiency will be less than design efficiency.
3. Since the flow has increased and pump efficiency decreased than rated, the operating power demand will be more than the rated power.
On a pump H-Q curve, lower head pushes the operating point right (higher flow), away from BEP, lowering efficiency and raising power.
5. What do you mean by the term 'cavitation'? What are the undesirable effects of cavitation in a pumping system?
Model answer: Cavitation is the formation and subsequent collapse of vapour bubbles inside a pump when the local static pressure at the suction falls below the vapour pressure of the liquid (i.e. when NPSH available is less than NPSH required). The liquid locally flashes to vapour and the bubbles then implode violently as pressure recovers. Undesirable effects: pitting and erosion of impeller and casing surfaces, noise and vibration, drop in pump head, flow and efficiency, mechanical damage to bearings and seals, and reduced pump life.
Cavitation occurs when suction pressure drops below vapour pressure (NPSHa < NPSHr); key harmful effects are metal erosion/pitting, noise, vibration and loss of head/efficiency.
Source: Book EOC
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); 1 kg/cm² ≈ 10 m water; suction lift (level below centreline) adds to head
6. A pump is delivering 50 m3/hr of water with a discharge pressure of 3.5 kg/cm2. The water is drawn from a sump where water level is 5 meter below the pump centerline. The power drawn by the motor is 9.5 kW at 90% motor efficiency. Find out the pump efficiency.
Model answer: Total head = discharge head + suction lift = (3.5 kg/cm2 x 10 m/(kg/cm2)) + 5 m = 35 + 5 = 40 m. Flow = 50 m3/hr = 50/3600 = 0.01389 m3/s. Hydraulic power = (rho x g x Q x H)/1000 = (1000 x 9.81 x 0.01389 x 40)/1000 = 5.45 kW. Pump shaft power = motor input x motor efficiency = 9.5 x 0.90 = 8.55 kW. Pump efficiency = hydraulic power / shaft power = 5.45 / 8.55 = 0.637 = about 64%.
Total head 40 m, hydraulic power about 5.45 kW, shaft power 8.55 kW, so pump efficiency is roughly 64%.
Source: Book EOC
📖 §6.6 Flow control strategies — speed variation, pumps in parallel, stop/start, control valve, by-pass, impeller trimming, VSD/VFD
7. What are the various methods of pump capacity (flow) control normally adopted?
Model answer: Common methods of pump flow/capacity control are: (1) throttling with a control valve on the discharge (simple but wastes energy across the valve); (2) bypass control, returning part of the flow to the sump; (3) impeller trimming, machining the impeller to a smaller diameter for a permanent flow/head reduction; (4) variable speed drive (VFD) control, varying motor speed to match demand (most energy efficient); (5) on/off or stop-start control and using multiple pumps in parallel, switching pumps in/out to match load. Speed control is generally the most efficient.
Throttling, bypass, impeller trimming, variable speed drive, and on/off / multiple-pump (parallel) operation; VFD is most efficient.
Source: Book EOC
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); 1 kg/cm² ≈ 10 m; suction lift adds to head
8. Water level is 4 m below the pump centreline. Discharge pressure 2.60 kg/cm². Flow 1.5 m³/min. Find pump efficiency if motor draws 14 kW at 0.88 motor efficiency.
Model answer: Discharge head = 2.60 kg/cm² = 26 m; suction head = -4 m; total head = 26-(-4) = 30 m. Hydraulic power = (1.5/60) x 1000 x 9.81 x 30/1000 = 7.36 kW. Shaft input = 14 x 0.88 = 12.32 kW. Pump efficiency = 100 x 7.36/12.32 = 59.74%.
Hydraulic power = Q x ρ x g x H; efficiency = hydraulic/shaft power.
Source: 14th Exam
📖 §6.11 Solved example — cooling water pump efficiency (p.194–195)
9. Cooling water is pumped to three heat exchangers via pipes A, B, C. Pipe A: 0.1 m dia, 1.5 m/s; Pipe B: 0.1 m dia, 1.8 m/s; Pipe C: 0.2 m dia, 2.0 m/s. Measured motor power 50.7 kW, motor efficiency 90%, pump discharge pressure 3.4 kg/cm², suction head 2 m. Determine pump efficiency.
Model answer: Flow A = (π/4)(0.1)² x 1.5 = 0.011786 m³/s; Flow B = (π/4)(0.1)² x 1.8 = 0.014143 m³/s; Flow C = (π/4)(0.2)² x 2.0 = 0.062857 m³/s; total = 0.088786 m³/s. Total head = 34 - 2 = 32 m. Hydraulic power = 0.088786 x 32 x 9.81 = 27.9 kW. Pump efficiency = 27.9 x 100/(50.7 x 0.9) = 61%.
Sum pipe flows; head = discharge head - suction head; efficiency = hydraulic/shaft.
Source: 14th Exam
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
10. A pump delivers 64 m³/hr of water with a discharge head of 26 m. Water is drawn from a sump 3 m below the pump centreline. Motor draws 8.89 kW at 88% motor efficiency. Find the pump efficiency.
Model answer: Q = 64/3600 m³/s; total head = 26-(-3) = 29 m. Hydraulic power = (64/3600) x 29 x 1000 x 9.81/1000 = 5.0576 kW. Pump shaft power = 8.89 x 0.88 = 7.8232 kW. Pump efficiency = 5.0576/7.8232 = 64.65%.
Hydraulic power = Q x ρ x g x H; efficiency = hydraulic/shaft power.
Source: Set-A
📖 §6.2 System characteristics — static & friction head
11. Explain briefly the difference between static and dynamic head of a centrifugal pumping system.
Model answer: Static head is simply the difference in height between the supply and destination reservoirs and is independent of flow. Dynamic head is the friction loss on the liquid in pipes, valves and equipment; friction losses are proportional to the square of the flow rate.
Static head = elevation (flow-independent); dynamic head = friction loss (∝ flow²).
Source: Set-A
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs; §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³)
12. Explain how a Variable Frequency Drive saves power in a three phase electric motor driven pumping system? What will be the reduction in power drawn by a motor by reducing the speed by half?
Model answer: A VFD converts fixed-frequency, fixed-voltage line power into variable-frequency, variable-voltage output to control the speed of induction motors. By controlling pump speed instead of throttling flow with valves, energy savings are substantial. By the affinity law, power ∝ speed³, so halving speed reduces power by a factor of (1/2)³ = 1/8 (to one-eighth). A throttling device on a fixed-speed motor leaves the motor running at nearly full power.
Affinity law: P ∝ N³.
Source: 17th Sep-2016
📖 §6.2 System characteristics — static & friction head
13. The total system resistance of a piping loop is 50 meters and the static head is 15 meters at designed water flow. Calculate the system resistance at 75%, 50% and 25% of water flow.
Model answer: Static head = 15 m (constant); dynamic head at design = 50 − 15 = 35 m (∝ flow²). At 75%: dynamic = 35×0.75² = 19.68, total = 34.68 m. At 50%: dynamic = 35×0.5² = 8.75, total = 23.75 m. At 25%: dynamic = 35×0.25² = 2.19, total = 17.19 m.
Total head = static (fixed) + dynamic (∝ Q²).
Source: 16th Exam
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
14. A pump fills a rectangular overhead tank 5 m x 4 m, height 8 m. Inlet pipe is 20 m above ground; pump suction is 3 m below pump level; overflow line is 7.5 m from tank bottom; motor power 5.5 kW at 92% efficiency; fills to overflow in 180 minutes. Assess the pump efficiency.
Model answer: Volume = 5 x 4 x 7.5 = 150 m3. Flow = 150/3 h = 50 m3/hr. Total head = 20 - (-3) = 23 m. Hydraulic power = (50/3600) x 23 x 1000 x 9.81/1000 = 3.13 kW. Pump input = 5.5 x 0.92 = 5.06 kW. Pump efficiency = 3.13/5.06 = 61.9%.
Flow from tank volume/time; total head = lift + suction below pump; hydraulic kW = Q·H·ρ·g; efficiency = hydraulic power / shaft (pump input) power.
Source: 18th Exam
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
15. A centrifugal pump pumps 80 m3/hr of water into a container at 3 kg/cm2(g). Discharge head 5 kg/cm2(g); water level 5 m below pump centre line. Motor draws 22 kW, motor efficiency 90%, water density 1000 kg/m3. Find pump efficiency.
Model answer: Discharge head = 5 kg/cm2 = 50 m; suction head = -5 m. Power input to pump = 22 x 0.9 = 19.8 kW. Liquid kW = (80/3600) x (50-(-5)) x 9.81 = 11.98 kW. Pump efficiency = 11.98/19.8 = 60.56%.
📖 §6.2 System characteristics — static & friction head
16. S-2: Total system resistance of a piping loop is 50 m and static head is 15 m at designed water flow. Calculate the system resistance at 75%, 50% and 25% of water flow.
Model answer: Dynamic head at design = 50 - 15 = 35 m (varies with flow^2); static head = 15 m (constant). At 75%: dynamic = 35 x 0.75^2 = 19.68 m, total = 34.68 m. At 50%: dynamic = 35 x 0.5^2 = 8.75 m, total = 23.75 m. At 25%: dynamic = 35 x 0.25^2 = 2.19 m, total = 17.19 m.
Printed solution: total resistance 34.68 m, 23.75 m, 17.19 m at 75/50/25% flow.
Source: 16th Exam (alt set)
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
17. A pump is delivering 40 m3/hr of water with a discharge pressure of 29 metre. The water is drawn from a sump where water level is 6 metre below the pump centerline. The power drawn by the motor is 7.5 kW at 89% motor efficiency. Find out the pump efficiency.
Total head = discharge head − suction head = 29 − (−6) = 35 m. Hydraulic power = Q×H×ρ×g/1000; pump shaft power = motor power × motor efficiency; pump efficiency = hydraulic/shaft power.
Source: 9th Dec-2009
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
18. A centrifugal pump pumps 90 m3/hr of water with discharge pressure 3 kg/cm2(g) and a negative suction head of 3 m. Motor power drawn is 13 kW. Find pump efficiency. Motor efficiency 91%, water density 1000 kg/m3.
Model answer: Discharge head = 3 kg/cm²(g) ≈ 30 m; suction head = −3 m (negative/lift). Total head = 30 − (−3) = 33 m. Hydraulic power = Q(m³/s)×H×ρ×g/1000 = (90/3600)×33×1000×9.81/1000 = 8.09 kW. Pump shaft power = motor input × motor efficiency = 13×0.91 = 11.83 kW. Pump efficiency = 8.09/11.83 = 68.4%.
Total head = h_d − h_s with negative suction head added; hydraulic power = Q·H·ρ·g/1000; pump efficiency = hydraulic power / (motor power × motor efficiency).
Source: 10th Jul-2010
📖 §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions
19. In a throttle valve-controlled pumping system with an oversized pump, list any five best methods to improve energy efficiency. (Name methods only.)
Model answer: Trim impeller; fit a smaller impeller; variable speed drive; two-speed drive; lower rpm drive.
Any five valid methods accepted, 1 mark each.
Source: 10th Jul-2010
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³)
20. A water pump is delivering 400 m3/hr. The impeller diameter is trimmed by 8% and its speed reduced by 10%. Find the water flow at the changed conditions.
Model answer: Flow varies directly with impeller diameter (Q∝D) and with speed (Q∝N), so the two effects multiply. 8% trim → D₂/D₁ = 0.92; 10% speed reduction → N₂/N₁ = 0.90. Q₂ = 400 × 0.92 × 0.90 = 331.2 ≈ 331 m³/hr. (Note: the printed key gives 288 m³/hr = 400 × 0.8 × 0.9, which corresponds to a 20% trim; with the question data as stated the affinity laws give 331 m³/hr — show the working and state the assumption.)
Apply Q∝D then Q∝N: 400×0.92×0.90 = 331 m³/hr. The official key's 288 m³/hr uses a 0.8 diameter factor (20% trim) — an inconsistency with the '8%' in the question.
Source: 10th Jul-2010
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
21. S-5: The following data of a water pump of a process plant have been collected. Flow: 70 m3/hr, Total head: 24 meters, Power drawn by motor 7.2 kW, Motor efficiency 89%. Determine the pump efficiency.
Model answer: Hydraulic power = Q(m³/s)×H×ρ×g/1000 = (70/3600)×24×1000×9.81/1000 = 4.578 kW. Pump shaft (input) power = motor power × motor efficiency = 7.2×0.89 = 6.41 kW. Pump efficiency = 4.578/6.41 = 71.4%. (The printed key uses 7.2×0.90 = 6.48 kW, giving 70.65% — with the stated 89% motor efficiency the answer is ≈71.4%; either is accepted if the method is shown.)
Pump efficiency = hydraulic power / shaft power, where shaft power = motor input × motor efficiency. Note the official key applied 0.90 instead of the stated 0.89.
Source: 11th Feb-2011
📖 §6.10 Agricultural pumping system — demonstrated ECMs
22. S-6: List any 5 energy conservation opportunities in agriculture pump sets.
Model answer: 1. Installation of low friction foot valves; 2. Installation of low friction HDPE suction and delivery pipes; 3. Installation of long bends; 4. Installation of high efficiency pumps and motors; 5. Lower discharge head.
Standard demonstrated ECMs for agricultural pumping per the official key.
Source: 11th Feb-2011
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
23. A process plant is situated 100 m up the side of the hill. The plant requires 100 kL of water per hour. The management decides to install a pump at the ground level, with suction 3 metre below the ground level. The friction head is 12 metre. Evaluate the rating of the motor required considering 10% extra margin with respect to actual input pump power. The design pump efficiency is 65% and motor efficiency 93%.
Model answer: Q = 100/3600 x 1000 = 100/3.6 m3/s = 0.0277 m3/s. Density = 1000 kg/m3. Static head = 100 + 3 = 103 m; Friction head = 12 m; Total Head = 103 + 12 = 115 m. Hydraulic power = Q x density x g x H / 1000 = 0.0277 x 1000 x 115 x 9.81/1000 = 31.25 kW. Pump input (shaft) power = 31.25/0.65 = 48.07 kW. Motor rating (shaft power) = 48.07 kW. Motor rating with 10% margin = 48.07 x 1.10 = 52.87 kW. Motor input = design rated power/motor efficiency = 48.07/0.93 = 51.69 kW.
Total head = static (lift + suction) + friction; hydraulic power via rho*g*Q*H; divide by pump efficiency for shaft power; add 10% margin and divide by motor efficiency.
Source: Mar 2021 (Set B)
📖 §6.2 System characteristics — static & friction head
24. The total static resistance of a water supply piping system is 30 meters and the static head is 10 meters at designed water flow. Calculate the system resistance offered at 75%, 50% and 25% of design water flow.
Model answer: Static Head = 10 m (Static head will remain same irrespective of the flow). So, Dynamic Head at designed water flow: (30-10) = 20 m. At 75% flow: Static 10 + Dynamic (0.75²×20=11.25) = 21.25 m. At 50% flow: Static 10 + Dynamic (0.5²×20=5.0) = 15.0 m. At 25% flow: Static 10 + Dynamic (0.25²×20=1.25) = 11.25 m.
Static head constant; dynamic (friction) head varies with square of flow; total = static + dynamic.
Source: Jul 2022
Long questions (10 marks) — 17
📖 §6.5 Pump suction performance — cavitation and NPSH
1. Define cavitation and state its three undesirable effects on a centrifugal pump.
Model answer: Definition (Book-3 §6.5): Liquid entering the impeller eye is turned and split by the leading edges of the impeller vanes, which locally drops the pressure below that in the inlet pipe. If the incoming liquid is at a pressure with insufficient margin above its vapour pressure, vapour cavities or bubbles form along the impeller vanes just behind the inlet edges, and then collapse. This phenomenon is cavitation.
The three undesirable effects listed in the book are:
1. The collapsing cavitation bubbles ERODE the vane surface, especially when pumping water-based liquids.
2. NOISE AND VIBRATION are increased, with possible shortened seal and bearing life.
3. The cavity areas partially CHOKE the impeller passages and reduce pump performance; in extreme cases there is total loss of pump developed head.
Prevention: the pressure margin at the eye, expressed as NPSH Available (a property of the SYSTEM), must exceed NPSH Required (a property of the PUMP), with a margin. Industry defines the onset of cavitation as the NPSHR at which head drops 3%. NPSHR ∝ N², so NPSH must be checked carefully in variable-speed applications.
Confirmed vs Book-3 §6.5 — the book gives exactly three effects in this order: erosion, noise/vibration, choked passages/loss of head.
Remember erosion begins BEFORE any noticeable head loss, so a pump can be cavitation-damaged while still appearing to perform.
Source: AI practice
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); total head = static lift + suction lift (h_d − h_s)
2. A pump is used to fill a rectangular overhead tank measuring 5 m × 3.5 m with a height of 10 m. The inlet pipe to the tank is positioned at a height of 25 m above ground level. The following additional data is available: The pump draws water from an underground sump situated 4 meters below the pump level and delivers it to a tank whose overflow line is positioned 8 meters above the tank bottom. The motor driving the pump draws 7.5 kW of power. The operating efficiencies of the motor and the pump are 90% and 70% respectively. Calculate the time taken by the pump to fill the tank up to the overflow level.
Model answer: Step 1: Volume of water filled = 5 × 3.5 × 8 = 140 m³; Mass of water = 140 × 1000 = 140000 kg
Step 2: Total head (H) = 25 + 4 = 29 m
Step 3: Shaft Power = 7.5 × 0.90 = 6.75 kW
Step 4: Water Power = 6.75 × 0.70 = 4.725 kW = 4725 W
Step 5: Time taken:
Pump efficiency = Mass flow × g × Head / Shaft Power
0.7 = mass flow × 9.81 × 29 / (6.75 × 1000)
Mass flow = 16.6 kg/sec = 59791 kg/hr = 59.79 m³/hr
Time = 140 / 59.79 = 2.34 hrs = 140.5 minutes
Hydraulic power = ṁ·g·H; derive mass flow from pump efficiency, then time = volume/flow.
Source: Sep 2025
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); motor part-loading from Book-3 Ch2
3. A) A clear water pump with rated flow of 125 m³/hr, head 55 m at rated speed of 1460 rpm and 79% efficiency supplies clarified water to a residential colony's water treatment facility. The daily water requirement is 3000 m³. The pump is directly coupled and driven by a three phase 50 Hp, 415 V, 64A, 0.9 pf, 1460 rpm induction motor with 90.5% full load efficiency. During an internal energy audit, the motor is found to operate with only 65% loading at pump rated conditions. The plant considers replacing the standard motor with a 30 kW IE3 motor. Operating parameters before and after motor replacement:
Flow (m³/hr): 130 / ?
Head (m): 52 / 51
Supply Voltage (V): 415 / 415
Current (Amp): 42 / 39
Power Factor: 0.9 / 0.92
Motor Eff (%): 0.88 / 0.932
The slip of the new IE3 motor has decreased by 20 rpm. Validate the savings, calculate:
i) % Loading of motor after replacement. (1 Mark)
ii) Flow after replacing the standard motor with 30 kW IE3 motor. (1 Mark)
iii) Operating Pump Efficiency before and after motor replacement. (2 Marks)
iv) Daily energy saving during operation due to motor replacement. (1 Mark)
B) Mark True/False:
i) Totally enclosed, fan cooled (TEFC) motors are less efficient than screen-protected, drip-proof (SPDP) motors.
ii) Stray loss in induction motors is inversely proportional to load current.
iii) As per BIS standard, the motor output should not be affected with voltage variation up to +/- 6%.
iv) Motor life doubles for each 10°C reduction in operating temperature.
v) Starting torque of energy efficient motors is higher than standard motors.
Model answer: A) i) Loading of the IE3 Motor = (1.732 × 0.415 × 39 × 0.92 × 0.932) / 30 = 80.12%
ii) Flow after replacement = 130 m³/hr × 1480/1460 = 131.8 m³/hr
iii) Operating Pump Efficiency:
Power Consumption before replacement = 1.732 × 0.415 × 42 × 0.9 = 27.17 kW
Power Consumption after replacement = 1.732 × 0.415 × 39 × 0.92 = 25.79 kW
Before replacement = [(130/3600) × 52 × 9.81] / (27.17 × 0.88) = 77%
After replacement = [(131.8/3600) × 51 × 9.81] / (25.79 × 0.932) = 76.2%
Daily Operating Hour before = 3000/130 = 23.08 Hrs; after = 3000/131.8 = 22.77 Hrs
iv) Daily Energy Savings = (23.08 × 27.17) − (25.79 × 22.77) = 39.9 kWh
B) i) False; ii) False; iii) True; iv) True; v) False
Motor loading = √3·V·I·pf·η / rating; flow scales with speed (affinity); pump η = hydraulic/electrical input; savings from input power × operating hours difference.
Source: Sep 2025
📖 §6.4 Factors affecting pump performance — effect of oversizing & energy loss in throttling; §6.11 Energy conservation opportunities in pumping systems (over-designed pump: VSD, downsize/replace impeller, or correct-sized pump; optimise stages)
4. A cooling water pump connected to a pillar furnace has specifications Q = 12.5 lps, H = 60 m, P = 13.4 kW. The furnace manufacturer requires only 12.5 lps at 3.0 kg/cm2. What energy conservation measure can be proposed and estimate the reduction in power consumption.
Model answer: Required duty: 12.5 lps at 3.0 kg/cm² ≈ 30 m head, but the pump develops 60 m — it is oversized in head by 100% and the excess head is being throttled/wasted. Measure: reduce the developed head to ~30 m — install a variable speed drive, trim/downsize the impeller, reduce the number of stages if multistage, or replace with a correctly sized pump (book §6.11; note that trimming or speed reduction alone also reduces flow, so the pump must be re-matched to give 12.5 lps at 30 m). Estimate: hydraulic power at 60 m = 1000×9.81×0.0125×60/1000 = 7.36 kW, so present pump efficiency ≈ 7.36/13.4 = 0.55. Hydraulic power needed at 30 m = 1000×9.81×0.0125×30/1000 = 3.68 kW; at the same efficiency input ≈ 3.68/0.55 = 6.7 kW. Reduction in power ≈ 13.4 − 6.7 = 6.7 kW (about 50%), since power falls roughly in proportion to head at constant flow.
Required head is only 30 m versus 60 m supplied; trim the impeller or use a VFD; power drops roughly in proportion to head, giving about a 6.7 kW (about 50%) reduction.
Source: Book EOC
📖 §6.2 System characteristics — static & friction head; §6.3 Pump curves — pump operating point (Figs 6.3–6.9)
5. Briefly explain with a sketch the concept of pump head-flow characteristics and system resistance.
Model answer: The pump head-flow (H-Q) characteristic curve shows how the head developed by a pump falls as the flow it delivers rises, at a fixed speed. The system resistance curve shows the head the piping system needs to pass a given flow; it equals the static head (a constant lift/pressure) plus the friction head, which rises with the square of flow, giving an upward-rising parabola starting from the static head. Plotting both on the same H-Q axes, the point where the falling pump curve crosses the rising system curve is the operating point, fixing the actual flow and head. Changing the system (e.g. throttling) steepens the system curve and shifts the operating point to lower flow.
Pump curve = head falling with flow; system curve = static head plus friction (rising as flow squared); their intersection is the operating point.
Source: Book EOC
📖 §6.6 Flow control strategies — pumps in parallel switched to meet demand (Figs 6.17–6.18); §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series)
6. Explain parallel and series operation of pumps with necessary pump curves.
Model answer: Parallel operation: two or more pumps draw from a common suction and discharge into a common header. At any given head their flows add, so the combined H-Q curve is obtained by adding the individual flows at each head value. Parallel pumping is used when a large or variable flow at roughly constant head is needed; pumps switch in/out to match demand. The actual increase in flow is less than the simple sum because adding a pump raises system flow, which raises friction head and shifts the operating point up the (now common) system curve. Pumps in parallel should have similar shut-off heads for stable sharing. Series operation: the discharge of one pump feeds the suction of the next, so at any given flow their heads add; the combined curve is obtained by adding heads at each flow. Series pumping (or multistage pumps) is used when high head is required at a given flow, e.g. high-rise or long-distance pumping. Sketches: parallel - two pump curves combined horizontally (flows added) intersecting the flat-ish system curve; series - two pump curves combined vertically (heads added) intersecting a steep system curve.
Parallel: add flows at equal head (more flow, similar head); Series: add heads at equal flow (more head, same flow). Operating point is where combined curve meets system curve.
Source: Book EOC
📖 §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
7. A centrifugal pump pumping water operates at 35 m3/hr and 1440 RPM. Pump operating efficiency is 68% and motor efficiency is 90%. The discharge pressure gauge shows 4.4 kg/cm2 and the suction is 2 m below the pump centerline. If the speed of the pump is reduced by 50%, estimate the new flow, head and power.
Model answer: Original total head = discharge + suction lift = (4.4 x 10) + 2 = 44 + 2 = 46 m. Original flow Q1 = 35 m3/hr at speed N1 = 1440 RPM. New speed N2 = 0.5 x 1440 = 720 RPM. Using the affinity laws: New flow Q2 = Q1 x (N2/N1) = 35 x 0.5 = 17.5 m3/hr. New head H2 = H1 x (N2/N1)^2 = 46 x 0.25 = 11.5 m. Original hydraulic power = rho x g x Q1 x H1 = 1000 x 9.81 x (35/3600) x 46 / 1000 = 4.39 kW. Original input (motor) power = 4.39 / (0.68 x 0.90) = 7.17 kW. New power varies as cube of speed: P2 = P1 x (N2/N1)^3 = 7.17 x 0.125 = 0.90 kW (input). So new flow about 17.5 m3/hr, new head about 11.5 m and new input power about 0.9 kW.
Affinity laws: flow halves to 17.5 m3/hr, head falls to one-quarter (11.5 m), power falls to one-eighth (about 0.9 kW input).
Source: Book EOC
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); part (b) Book-3 Ch3 Compressed air — pressure optimisation
8. a) Pump suction head 3 m below centreline, discharge pressure 2.8 kg/cm², flow 120 m³/hr. Find pump efficiency if actual motor input is 15.0 kW at 0.90 motor efficiency. b) A V-belt reciprocating instrument air compressor maintains 7 kg/cm²g. 20% of air goes to boiler-house control valves needing 6.5 kg/cm²g; balance 80% needs 2 kg/cm²g. What do you advise?
Model answer: a) Discharge head = 2.8 kg/cm² = 28 m; suction head = -3 m; total head = 28 - (-3) = 31 m. Hydraulic power = (120/3600) x 1000 x 9.81 x 31 /1000 = 10.137 kW. Pump shaft power = 15 x 0.9 = 13.5 kW. Pump efficiency = 10.137/13.5 = 75%. b) Advise: 1) Provide a separate small compressor at 7 kg/cm²g near the control valves and reduce the main distribution pressure from 7 to 2 kg/cm²g for the bulk pneumatic instruments. 2) Reduced pressure lowers leakage loss; the compressor will begin to unload, so reduce the motor pulley size to match the lower demand.
Hydraulic power = Q x ρ x g x H; efficiency = hydraulic/shaft. Part b is a pressure-optimisation advisory.
Source: 15th Exam
📖 Item 1: §6.5 Pump suction performance — cavitation & NPSH; item 6: §6.2 System characteristics — static & friction head (key treats 3 m as positive suction head: 21 − 3 = 18 m); item 9: §6.6 Flow control strategies — pumps in parallel switched to meet demand; others Ch4/5/2/7/8
9. Fill in the blanks: 1) Cavitation occurs when local static pressure falls below the ___ pressure of the liquid. 2) In an ammonia VAR system the absorbent is ___. 3) System resistance of a fan is proportional to the ___ of flow rate. 4) DBT=30°C and WBT=30°C → RH = ___. 5) Slip ring induction motors are ___ efficient than squirrel cage of same rating. 6) Suction static head 3 m, friction head 21 m → total head ___. 7) Lowest theoretical temperature water can be cooled to in a cooling tower is the ___ of atmospheric air. 8) Measure of illuminance in metric units is ___. 9) Pumps run in parallel if their ___ heads are similar. 10) When heat load, range and WBT held constant, cooling tower size is ___ proportional to approach.
Standard fill-in answers across pumps, refrigeration, fans, motors, cooling towers and lighting.
Source: 14th Exam
📖 (i) §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions; (ii) Book-3 Ch4 HVAC — TR definition; (iii) Book-3 Ch5 Fans — ECMs
10. Answer any two: (i) In a throttle valve-controlled oversized-pump system, list any five options to improve energy efficiency (name only). (ii) Define one Ton of Refrigeration (TR); how do you calculate TR across an AHU? (iii) List five energy conservation opportunities in a fan system.
Model answer: (i) Trim impeller; replace with smaller impeller; install variable speed drive; change pulley if belt-driven; change to two-speed drive; use lower-rpm drive. (ii) 1 TR = heat to freeze one ton of water at 0°C in 24 hours = 50.4 kcal/min = 3024 kcal/h. TR across AHU = [Q x ρ x (h_in - h_out)]/3024, where Q = air flow (CMH), ρ = air density, h = enthalpy (kcal/kg). (iii) Use smooth rounded inlet cones; avoid poor inlet flow distribution; minimize inlet/outlet obstructions; clean screens/filters/blades; use aerofoil blades; minimize fan speed; use flat/low-slip belts; check belt tension; eliminate variable-pitch pulleys; use VSD for variable loads; use energy-efficient motors; eliminate duct leaks; minimize duct bends; turn fans off when not needed.
Standard pump/fan ENCON lists and TR definition from BEE guidebook.
Source: 14th Exam
📖 §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
11. A centrifugal pump runs at 60 m3/hr, 1470 RPM, pump efficiency 65%, motor efficiency 89%. Discharge gauge 3.4 kg/cm2, suction 3 m below pump centerline. Auditor recommends replacing the motor with a 4-pole 91% efficient motor at 1% slip. Determine new flow and motor power; throttle fully open and system head purely frictional. Comment.
Model answer: Existing: Head = 34 − (−3) = 37 m; pump power = (60/3600)×37×1000×9.81/(1000×0.65) = 9.3 kW. New motor speed = 1500 − 0.01×1500 = 1485 rpm. New flow = 60×(1485/1470) = 60.61 m3/hr. New pump power = 9.3×(1485/1470)³ = 9.59 kW. Existing motor input = 9.3/0.89 = 10.46 kW; new motor input = 9.59/0.91 = 10.54 kW. Comment: power consumption is slightly more, so not recommended (though flow is also marginally higher).
Affinity laws Q∝N, P∝N³; head from gauge + suction lift.
Source: 17th Sep-2016
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.6 Flow control strategies — pumps in parallel switched to meet demand
12. Three identical parallel cooling-water pumps (two running, one standby) all develop 3.4 kg/cm2(a). Flow meter at common header reads: Pumps 1&2 = 545 m3/hr; 2&3 = 535; 3&1 = 550. Motor power: P1 33 kW, P2 31.5 kW, P3 32.5 kW. Motor efficiency 92% (P1,P2) and 91.5% (P3). Suction 3 m below pump centerline. Find i) individual pump efficiencies, ii) specific energy consumption (kWh/m3), iii) best operating combination.
Model answer: Solving X+Y=545, Y+Z=535, X+Z=550 gives X=280, Y=265, Z=270 m3/hr. Total head = 3.4 kg/cm2(a) → 2.4 kg/cm2(g) = 24 m discharge − (−3) suction = 27 m. Liquid kW = flow(m3/s)×27×1000×9.81/1000: P1 20.60, P2 20.22, P3 19.87. Pump input = motor kW × motor eff: P1 30.36, P2 28.98, P3 29.74. Pump eff = liquid/input: P1 67.9%, P2 69.8%, P3 66.8%. Specific energy (motor kW/flow): P1 0.118, P2 0.119, P3 0.120 kWh/m3. Best combination: Pumps 1 & 2.
Solve simultaneous flows; head from gauge+suction; pump eff=liquid kW/input kW.
Source: 16th Exam
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.6 Flow control strategies — pumps in parallel switched to meet demand
13. L-3: Three identical cooling water pumps in parallel (two running, one standby), all combinations give 3.4 kg/cm2(a) discharge. Flow: pumps 1&2 = 545, 2&3 = 535, 3&1 = 550 m3/hr. Motor power 33/31.5/32.5 kW; motor eff 92%/92%/91.5%; suction 3 m below pump centre line. Find (i) individual pump efficiencies, (ii) specific energy consumption kWh/m3, (iii) best operating combination.
Model answer: Solving X+Y=545, Y+Z=535, X+Z=550 gives X=280, Y=265, Z=270 m3/hr. Discharge head = 3.4 kg/cm2(a) = 2.4 kg/cm2(g) = 24 m; suction head = -3 m; total head = 27 m. Liquid kW = flow(m3/s) x 27 x 1000 x 9.81/1000: Pump1 = 20.60, Pump2 = 20.22, Pump3 = 19.87 kW. Pump input power = motor power x motor eff: 30.36, 28.98, 29.74 kW. Pump efficiency = liquid/input: 67.9%, 69.8%, 66.8%. SEC = motor power/flow: 0.118, 0.119, 0.120 kWh/m3. Best operating combination = pumps 1 & 2.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
14. A process plant is situated 100 m above the ground level on the top of the hill. The plant requires 100 kL of water per hour. The management decides to install a pump at the ground level, with suction 3 meter below the ground level. The friction head is 12 meter. Evaluate the rating of the motor required considering 10% extra margin with respect to actual input pump power. The design pump efficiency is 65%. Also calculate the motor input power if the motor efficiency is 93%. (5 Marks)
Model answer: Q = 100 kL/hr = 100 m³/hr = 100/3600 = 0.0278 m³/s; ρ = 1000 kg/m³. Static head = delivery height + suction depth = 100 + 3 = 103 m; friction head = 12 m; Total head = 103 + 12 = 115 m. Hydraulic power = Q×ρ×g×H/1000 = 0.0278×1000×9.81×115/1000 = 31.3 kW. Pump shaft (input) power = 31.3/0.65 = 48.1 kW. Motor rating with 10% margin = 48.1×1.10 = 52.9 kW (select next standard size, e.g. 55 kW). Motor input power at the actual load = 48.1/0.93 = 51.7 kW.
Total head = static (lift + suction) + friction; hydraulic power = Q·ρ·g·H/1000; ÷ pump efficiency for shaft power; +10% for motor rating; ÷ motor efficiency for input power.
Source: Mar 2021
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
15. Rated capacity of a fresh water shut-discharge pump flow rate is 485 m3/hr and discharge pressure is 13.5 kg/cm2 at rated speed of 2950 rpm. It has been observed that 450 m3/hr water is sufficient to dispose the ash. The suction pressure of the pump is 0.5 kg/cm2. The management has decided to trim the impeller to satisfy the reduced flow requirement. Calculate: a) % reduction of impeller diameter (5 Marks) b) Annual energy savings after modification, if the pump is operating for 6 hours/day and 330 days in a year. (5 Marks)
Model answer: a) Q∝D, so D₂/D₁ = 450/485 = 0.928; % reduction in impeller diameter = (1 − 0.928)×100 = 7.2%. b) Rated differential head = (13.5 − 0.5) kg/cm² = 13 kg/cm² ≈ 130 m. Hydraulic power (rated) = (485/3600)×1000×130×9.81/1000 = 171.8 kW. After trimming, discharge head ∝ D²: H₂ = 0.928²×135 = 116.3 m, so new differential head = 116.3 − 5 = 111.3 m. Hydraulic power (new) = (450/3600)×1000×111.3×9.81/1000 = 136.4 kW. Power saving = 171.8 − 136.4 = 35.4 kW. Annual energy saving = 35.4×6×330 ≈ 70,000 kWh/year (official key prints 69,894 kWh using 35.3 kW). Alternative by P∝D³: P₂ = 0.928³×171.8 = 137.3 kW, saving 34.5 kW ≈ 68,300 kWh/year — both accepted.
Trim: Q∝D gives the diameter ratio; H∝D² gives new head; hydraulic power = Q·ρ·g·H/1000 before/after; annual saving = kW saved × 6 h × 330 days.
Source: Mar 2021
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
16. Rated capacity of bottom ash disposal pump flow rate is 485 m3/hr and discharge pressure is 13.5 kg/cm2 at rated speed of 2950 rpm. It has been observed that 450 m3/hr water is sufficient to dispose the ash. The suction pressure of the pump is 0.5 kg/cm2. The management has decided to trim the impeller to satisfy the reduced flow requirement. Calculate: a) % reduction of impeller diameter (5 marks). b) The annual energy savings after modification, if the pump is operating for 6 hours/day and 330 days in a year (5 marks).
Model answer: a) Flow ∝ impeller diameter: D_new/D_old = 450/485 = 0.928; % impeller diameter reduction = (1 − 0.928)×100 = 7.2%. b) Rated hydraulic power = Q×ρ×(h_d − h_s)×g/1000 = (485/3600)×1000×(135 − 5)×9.81/1000 = 171.8 kW. New discharge head (H∝D²) = 0.928²×135 = 116.3 m; new differential head = 116.3 − 5 = 111.3 m. Hydraulic power after trimming = (450/3600)×1000×111.3×9.81/1000 = 136.4 kW. Power saving = 171.8 − 136.4 = 35.4 kW. Annual energy saving = 35.4×6×330 ≈ 70,000 kWh/year (key prints 69,884 kWh). Alternative P∝D³ method: P_new = 0.928³×171.8 = 137.3 kW, saving 34.5 kW → ≈ 68,300 kWh/year.
Affinity laws for trimming: Q∝D, H∝D², P∝D³; % diameter reduction from the flow ratio; annual saving = power saved × operating hours (6×330 = 1980 h).
Source: Mar 2021 (Set B)
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172) (ρ = slurry density); §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series) (series pumps add head)
17. L-5: A 30 MW coal fired thermal power plant uses conventional wet ash disposal system for ash evacuation. During an energy audit at ash slurry disposal pump house it was observed that only one ash slurry disposal series is continually operated out of installed three numbers of series. Data collected for one series: two pumps per series; rated parameters each pump flow 815 m³/hr, head 33 mWc; slurry flow rate measured at second series pump discharge 752 m³/hr; suction head to 1st pump +2.6 mtr; final slurry discharge pressure 5.2 kg/cm²g; differential pressure across 1st slurry pump 2.44 kg/cm²; differential pressure across 2nd slurry pump 2.5 kg/cm²; ash water ratio (by weight) of the slurry 1:15; ash slurry density 1032 kg/m³; electric power input to motor of 1st slurry pump 141 kW; electric power input to motor of 2nd slurry pump 139 kW; motor rating of each pump 200 kW; full load motor efficiency 93%; belt transmission efficiency 92%. Evaluate: (i) Individual pump efficiencies if the operating motor efficiency is 92% for all pumps; (ii) Specific energy consumption of each pump (kWh/m³); (iii) Specific energy consumption of the series (kWh/m³); (iv) As water conservation measure the energy auditor recommended to maintain ash water ratio at 1:7 and slurry density of 1067 kg/m³, calculate the incremental power consumption of each pump and series if all other parameters are unchanged.
Model answer: Common data: slurry flow Q = 752 m³/hr = 0.2089 m³/s; slurry density ρ = 1032 kg/m³; 1 kg/cm² ≈ 10 m. (i) Pump efficiencies. 1st pump: differential head = 2.44 kg/cm² ≈ 24.4 m; liquid (hydraulic) power = Q×H×ρ×g/1000 = 0.2089×24.4×1032×9.81/1000 = 51.6 kW; power at pump shaft = motor input × motor eff × belt eff = 141×0.92×0.92 = 119.3 kW; pump efficiency = 51.6/119.3 = 43.2%. 2nd pump: head = 2.5 kg/cm² ≈ 25 m; liquid power = 0.2089×25×1032×9.81/1000 = 52.9 kW; shaft power = 139×0.92×0.92 = 117.6 kW; pump efficiency = 52.9/117.6 = 45.0% (the printed key rounds both pumps to ≈43%). (ii) Specific energy consumption = motor input / flow: 1st = 141/752 = 0.188 kWh/m³; 2nd = 139/752 = 0.185 kWh/m³. (iii) Series SEC = (141+139)/752 = 0.372 kWh/m³. (iv) With ash:water 1:7 the slurry density rises to 1067 kg/m³; at unchanged flow, head and efficiencies the liquid power and hence input power scale with density (×1067/1032 = 1.034): 1st pump liquid power = 53.4 kW, input ≈ 141×1.034 = 145.8 kW (+4.8 kW); 2nd pump liquid power = 54.7 kW, input ≈ 143.7 kW (+4.7 kW). Incremental power for the series ≈ 9.5 kW (printed key: ≈10.4 kW with its rounding). Water conservation therefore slightly raises pumping power but saves far more water.
Liquid power = Q·H·ρ·g with slurry density; pump efficiency = liquid power / (motor input × motor η × belt η); SEC = kW / (m³/hr); higher slurry density raises liquid power and input power in proportion.