129 questions — 97 objective (1 mark), 20 short (5 marks), 12 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) — 97
📖 §3.5 Leak Quantification — shop-floor leakage test
1. In a compressor leakage test, what does the term T represent in the formula % leakage = T/(T+t) x 100?
Total UNLOAD (off) time
Total LOAD (on) time
Total cycle time
Pump-up time
Answer: B) Total LOAD (on) time
Confirmed vs Book-3 §3.5 — in the leakage test all air-consuming equipment is shut off, the system is charged to the set pressure and the load/unload cycles are logged for 8–10 cycles: T = total time on LOAD (on), t = total time on UNLOAD (off), and % leakage = [T/(T+t)] × 100. With no equipment consuming air, the only reason the compressor still loads is to make up air lost through leaks, so LOAD time goes on top. Option (a) is the classic trap of inverting T and t.
2. A compressor of capacity 40 m3/min cycles with load time T = 2 min and unload time t = 8 min. What is the leakage quantity q?
4 m3/min
8 m3/min
10 m3/min
16 m3/min
Answer: B) 8 m3/min
Confirmed vs Book-3 §3.5 — q = Q × T/(T+t) = 40 × 2/(2+8) = 40 × 0.2 = 8 m³/min, i.e. a 20% leakage level. Option (c) 10 m³/min comes from dividing by the unload time alone (40 × 2/8 = 10) instead of by the total cycle time (T + t). A useful sanity check: q can never exceed the compressor capacity Q, and inverting T and t here would give 32 m³/min, i.e. an implausible 80% leakage.
3. A compressor runs loaded 6 s and unloaded 18 s. If FAD is 600 cfm, what is the air leakage rate?
100 cfm
150 cfm
200 cfm
450 cfm
Answer: B) 150 cfm
Confirmed vs Book-3 §3.5 — q = Q × T/(T+t) = 600 × 6/(6+18) = 600 × 6/24 = 600 × 0.25 = 150 cfm (25% leakage). Units of T and t need only be consistent, so seconds work directly. Option (d) 450 cfm uses the unload fraction 18/24 instead of the load fraction.
4. A compressor has a displacement of 200 cfm and a volumetric efficiency of 85%. What is the Free Air Delivery (FAD)?
235 cfm
200 cfm
170 cfm
85 cfm
Answer: C) 170 cfm
Confirmed vs Book-3 §3.3 — volumetric efficiency = [Free Air Delivered (m³/min) / Compressor displacement (m³/min)] × 100, so FAD = displacement × volumetric efficiency = 200 × 0.85 = 170 cfm. Because clearance volume and leakage always cost some induced volume, FAD is ALWAYS less than swept displacement. Option (a) 235 cfm divides instead of multiplying, which would make the compressor deliver more air than it swept.
Source: AI practice
📖 §3.5 / Air Dryers — Table 3.19 dew point & power consumption
5. Per Book Table 3.19 (1000 m³/hr basis), which type of air dryer has the LOWEST power consumption?
Refrigerant dryer
Heatless purge desiccant dryer
Heat of Compression (HOC) dryer
Blower reactivated desiccant dryer
Answer: C) Heat of Compression (HOC) dryer
Confirmed vs Book-3 Table 3.19 — for 1000 m³/hr the power consumptions are: HOC 0.8 kW, refrigerant 2.9 kW, blower reactivated 18.0 kW, heatless purge 20.7 kW. HOC is lowest because it regenerates the desiccant using the heat already in the compressed air taken before the after-cooler (about 135 °C), needing no electric heater and losing no purge air. The heatless purge type is the opposite extreme — no heater, but 12–15% of dry compressed air is thrown away as purge, so it is the highest at 20.7 kW.
Source: AI practice
📖 §3.4 Compressed Air System Components / Air Dryers
6. What is the primary function of an after-cooler in a compressed air system?
Remove the remaining traces of moisture left after the dryer
Remove moisture by cooling the air in a water-cooled heat exchanger
Dampen the pulsations in the compressor air output
Reduce the work of compression between successive stages
Answer: B) Remove moisture by cooling the air in a water-cooled heat exchanger
Confirmed vs Book-3 §3.4 — the book defines the after-cooler as the component that removes moisture in the air by reducing its temperature in a water-cooled heat exchanger, and states that about 60–75% of the moisture is removed there. Option (a) is the AIR DRYER's job (it takes out the remaining traces after the after-cooler); option (c) is the air receiver; option (d) is the inter-stage cooler.
Source: AI practice
📖 §3.2 Compressor Types — Table 3.2 comparison of different compressors
7. Per the book's comparison of compressor types, which compressor is preferable for a CONSTANT air requirement?
Reciprocating
Rotary screw
Roots blower
Diaphragm
Answer: B) Rotary screw
Corrected (question re-grounded) — Book-3 §3.2: the book states plainly that it is preferable to use SCREW compressors for a constant air requirement, that for fluctuating loads a screw with a variable speed drive should be used, and that reciprocating machines suit fluctuating load because their unload power is only about 25% of full-load power. Table 3.2 also shows screw efficiency is high at full load but poor below 60% of full load — which is exactly why it is matched to a steady, near-full-load duty rather than to option (a) reciprocating.
8. Per the book thumb rule, every 4 degC drop in compressor inlet air temperature reduces power consumption by approximately:
1%
2%
5%
10%
Answer: A) 1%
Confirmed vs Book-3 §3.5 — the book's thumb rule is that every 4 °C RISE in inlet air temperature increases compressor power consumption by about 1%; conversely a 4 °C drop saves about 1%. Cooler intake air is denser, so more mass is delivered per unit of swept volume. This is why the book insists on drawing cold outside air rather than warm, humid compressor-room air.
Source: AI practice
📖 §3.6 Compressor Capacity Assessment — FAD pump-up test
9. A FAD pump-up test gives 12.5 m3/min for a compressor rated at 14.5 m3/min. What is the percentage shortfall, and is corrective action needed?
13.8% shortfall - yes, investigate (>10%)
13.8% shortfall - no action needed
6.9% shortfall - no action needed
2.0% shortfall - no action needed
Answer: A) 13.8% shortfall - yes, investigate (>10%)
Confirmed vs Book-3 §3.6 — shortfall = (14.5 − 12.5)/14.5 × 100 = 2.0/14.5 × 100 = 13.8%. The book states that if the deviation of measured FAD from rated FAD is more than 10%, corrective measures must be taken; its own worked example flags an 11.05% shortfall for investigation, and notes a worn-out compressor valve alone can cut capacity by as much as 20%. Option (c) 6.9% is the error of dividing the 2.0 m³/min shortfall by twice the rating.
10. Reducing the compressor delivery pressure by 1 bar typically yields a power saving of about:
1-2%
6-10%
20-25%
30-40%
Answer: B) 6-10%
Confirmed vs Book-3 §3.5 — the book's thumb rule is that a reduction in delivery pressure by 1 bar reduces power consumption by 6–10%, and its worked example (8 → 7 kg/cm²) gives about 9% input power saving. Table 3.9 backs it: 6.8 → 5.5 bar saves 9–11% on water-cooled machines. Option (c) 20–25% overstates the effect — that magnitude of saving comes from eliminating leaks or unloaded running, not from 1 bar of pressure.
Source: AI practice
📖 §3.5 Air Dryers — Adsorption drying
11. Which materials are commonly used as the adsorbent (desiccant) in adsorption-type air dryers?
Refrigerant R-134a and brine
Activated alumina and silica gel
Lithium bromide and water
Activated carbon and zeolite membrane
Answer: B) Activated alumina and silica gel
Confirmed vs Book-3 §3.5 — adsorption drying is described as a purely physical process in which moisture is bound to the drying agent by force of adhesion, using materials with an open porous structure and large inner surface; the book names activated alumina and silica gel. Two parallel tanks are used, one drying while the other regenerates, giving about −40 °C atmospheric dew point. Option (c) lithium bromide and water is the ABSORPTION refrigeration pair, not an air-dryer desiccant.
Source: AI practice
📖 §3.5 Air Receivers (sizing per IS 7938-1976)
12. In a compressed air system, the air receiver primarily serves to:
Reduce the work of compression between successive stages
Store air and dampen the pulsations from the compressor
Remove the final traces of moisture before the point of use
Meter oil mist into the compressed air stream
Answer: B) Store air and dampen the pulsations from the compressor
Confirmed vs Book-3 §3.5 — the book lists the receiver's duties: dampen the pulsations entering the discharge from the compressor, act as a reservoir for sudden or heavy demands in excess of compressor capacity, prevent too-frequent loading/unloading (short cycling), and help separate moisture and oil vapour. Sizing per IS 7938-1976 is 1/10th to 1/6th of the output in m³/min. Option (a) is the inter-cooler and option (c) the air dryer; option (d) describes a lubricator.
Source: AI practice
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
13. What is the main purpose of an after-cooler in a compressed air system?
To remove moisture from the air by cooling it
To increase the pressure of the compressed air
To filter out dust and particles
To lubricate the compressed air
Answer: A) To remove moisture from the air by cooling it
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — The after-cooler is a water-cooled heat exchanger whose stated objective is to remove the moisture in the air by reducing its temperature; roughly 60-75% of the moisture in compressed air drops out here.
Pressure is raised only by the compressor itself, so (b) is wrong, and dust is stopped upstream by the intake air filter, so (c) is wrong.
Source: Sep 2024
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
14. How does a desiccant air dryer remove moisture from compressed air?
By cooling the air
By using adsorbents like silica gel or activated carbon
By increasing the pressure
By reducing the air flow rate
Answer: B) By using adsorbents like silica gel or activated carbon
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Air dryers remove the traces of moisture left after the after-cooler using adsorbents such as silica gel, activated alumina or activated carbon; adsorption binds water physically on a large porous inner surface without any chemical reaction.
Cooling (a) describes the refrigerant dryer, and neither pressure (c) nor flow rate (d) is the drying mechanism in a desiccant unit.
Source: Sep 2024
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
15. Which of the following is an efficient method to control the capacity of a centrifugal compressor?
Automatic on/off control
Variable inlet guide vanes
Load and unload control
Multi-step control
Answer: B) Variable inlet guide vanes
Confirmed vs Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14) — The book states that the capacity of centrifugal compressors can be controlled using variable inlet guide vanes, with speed control as the other efficient route (Table 3.14).
Automatic on/off, load-unload and multi-step unloading are the schemes listed for positive-displacement machines, so (a), (c) and (d) do not apply to a centrifugal.
16. What is the effect of increasing the intake air temperature on the efficiency of an air compressor?
Increases efficiency by reducing power consumption
Decreases efficiency by increasing power consumption
No significant effect on efficiency
Increases the volumetric capacity of the compressor
Answer: B) Decreases efficiency by increasing power consumption
Confirmed vs Book-3 §3.5 Efficient Operation — Thumb rule: every 4°C drop in inlet air temperature lowers energy consumption by 1%, so a rise raises power. Table 3.3 shows relative air delivery falling from 102.0% at 10°C to 91.2% at 43.3°C, with 5.8% more power.
Hot, less dense intake air also reduces the mass delivered, so (d) — an increase in capacity — is exactly backwards.
Source: Sep 2024
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
17. What is the main benefit of using a variable speed drive (VSD) with a screw compressor?
Increases the maximum pressure capacity
Reduces the size of the compressor
Eliminates unloaded running condition
Simplifies maintenance
Answer: C) Eliminates unloaded running condition
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — The checklist says to retrofit variable speed drives in big compressors, say over 100 kW, to eliminate the 'unloaded' running condition altogether; unloading can consume up to 30% of full-load power.
A VSD does not change the machine's maximum pressure (a) or its physical size (b) — it removes the idle-running loss.
Source: Sep 2024
📖 Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)
18. Which of the following describes the primary function of an air receiver in a compressed air system?
To increase the air pressure
To act as a reservoir and dampen pulsations
To filter out impurities
To cool the compressed air
Answer: B) To act as a reservoir and dampen pulsations
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — The receiver dampens the pulsations leaving the compressor, serves as a reservoir for sudden heavy demand, prevents short cycling and lets moisture and oil vapour precipitate.
It cannot raise pressure (a) — it only stores air at the delivered pressure; filtering is done by line filters (c) and cooling by the after-cooler (d).
19. A 500 cfm reciprocating compressor has a loading and unloading period of 5 seconds and 20 seconds respectively during a compressor leakage test. The air leakage in the compressor air system will be ________
125 cfm
100 cfm
200 cfm
none of the above
Answer: B) 100 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 5/(5+20) x 100 = 20%; leakage quantity = 0.20 x 500 = 100 cfm.
The tempting 125 cfm (a) comes from using the load:unload ratio 5/20 instead of load/(load+unload), i.e. dividing by the unload time alone rather than by the full cycle.
20. What is the main advantage of using a rotary screw air compressor over a reciprocating compressor?
Lower initial cost
Continuous, pulsation-free air delivery
Higher maximum pressure
None of the above
Answer: B) Continuous, pulsation-free air delivery
Confirmed vs Book-3 §3.2 Positive Displacement — Rotary compressors have rotors in place of pistons and give a continuous, pulsation-free discharge, need lower starting torque, smaller foundations and have fewer wearing parts; reciprocating output is pulsating.
Screw machines are not cheaper to buy (a), and very high pressures are the multi-stage reciprocating machine's territory (Table 3.1: up to 700 bar), so (c) is wrong.
Source: Sep 2024
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
21. The purpose of inter-cooling in a multistage compressor is to
Increase the pressure of air
Reduce the work of compression
Separate moisture and oil vapour
None of the above
Answer: B) Reduce the work of compression
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — Inter-stage coolers reduce the temperature of the air before it enters the next stage to reduce the work of compression and increase efficiency, by cutting the specific volume the next stage must handle.
Raising pressure (a) is the compressor's job; moisture and oil separation (c) is an incidental benefit, not the purpose of inter-cooling.
22. Calculate the FAD in CFM for an air compressor with a cylinder displacement of 150 CFM and volumetric efficiency of 90%:
165
135
150
None of the above
Answer: B) 135
Confirmed vs Book-3 §3.3 Compressor Efficiency — Volumetric efficiency = [FAD / compressor displacement] x 100, so FAD = 150 x 0.90 = 135 CFM.
Option (a) 165 CFM adds 10% instead of taking 90%; FAD is always LESS than the swept displacement because of clearance volume re-expansion and valve/leakage losses.
23. At higher altitudes, for same FAD, air compressors:
Consume less power
Consume more power
Show no difference
Work without lubrication
Answer: B) Consume more power
Confirmed vs Book-3 §3.5 Efficient Operation — Compressors located at higher altitudes consume more power to achieve a particular delivery pressure than those at sea level, because the lower barometric pressure raises the compression ratio.
Table 3.6 shows relative volumetric efficiency at 7 bar falling from 100% at sea level to 87.0% at 2500 m, so (a) and (c) are contradicted.
Source: Sep 2025
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
24. Which type of compressed air dryer consumes the least power for capacities higher than 250 CFM?
Refrigeration type
Blower reactivated type
Heat of compression type
Heatless purge type
Answer: C) Heat of compression type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.19 (per 1000 m³/hr): heat of compression 0.8 kW, refrigeration 2.9 kW, blower reactivated 18.0 kW, heatless purge 20.7 kW. The HOC dryer regenerates the desiccant with the compressor's own 135°C discharge heat, so operating cost is zero to very minimal.
It is offered from 400 to 5000 cfm, which covers the >250 CFM duty asked about; the refrigerant dryer uses less power than the desiccant types but cannot be compared here as it only reaches -20°C dew point.
Source: Sep 2025
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power)
25. In a textile mill, two 150 cfm belt-driven reciprocating compressors are working constantly with a loading time of 20 seconds and unloading time of 30 seconds. The best economic option for energy savings would be:
switch off one compressor
switch off one compressor and reduce the motor pulley size of the other compressor appropriately
adopt variable speed drive for one of the compressors
none of the above
Answer: B) switch off one compressor and reduce the motor pulley size of the other compressor appropriately
Confirmed vs Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power) — Loading 20 s out of every 50 s means the pair is loaded only 40% of the time, so one 150 cfm machine can carry the whole demand and even that one is oversized.
The book advises that where a compressor runs unloaded for long periods the economical fix is to change the pulley size and reduce the RPM to de-rate it. Hence switching one off AND trimming the other's motor pulley (b) beats switching off alone (a); a VSD (c) is not justified on so small a machine.
Source: Book EOC
📖 Book-3 §3.5 Equivalent Lengths of Fittings (Table 3.12)
26. Which of the following pipe fittings used in a compressed air pipeline offers maximum resistance?
Gate Valve
Tee 90 deg long bend
Elbow
Return bend
Answer: D) Return bend
Corrected (was c) — Book-3 §3.5 Equivalent Lengths of Fittings (Table 3.12): Table 3.12 ranks equivalent length at every pipe size in the order Gate valve < Tee 90° long bend < Elbow < Return bend — e.g. at 50 mm NB: 0.40 / 0.61 / 1.07 / 1.68 m.
The return bend therefore offers the maximum resistance of the four options listed; the elbow (the previously recorded answer) is only the second highest. Globe valve / outlet-of-tee is higher still in the table but is not among the choices.
Source: Book EOC
📖 Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11)
27. Friction loss in a piping system carrying a fluid is proportional to
fluid flow
(fluid flow)^2
1/fluid flow
sqrt(fluid flow)
Answer: B) (fluid flow)^2
Confirmed vs Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11) — Turbulent friction loss in piping varies with the square of the flow rate, so doubling flow quadruples the drop; the book therefore designs for a maximum 5% pressure drop from compressor to the farthest point and sizes pipes generously (Table 3.11).
Options (a) and (c) would make loss linear or inversely proportional to flow, which contradicts the tabulated behaviour.
Source: Book EOC
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
28. The most energy-intensive dryer among the following is
refrigeration
desiccant (heat of compression)
desiccant (heatless purge)
desiccant (blower reactivated)
Answer: C) desiccant (heatless purge)
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.19: heatless purge = 20.7 kW per 1000 m³/hr and 'High' operating cost, against blower reactivated 18.0 kW, refrigeration 2.9 kW and heat of compression 0.8 kW.
The heatless purge type also throws away 12-15% of the dry compressed air as purge (versus 1-2% for the blower type and none for HOC), which is what makes it the most energy intensive.
29. The efficiency of a compressed air system is around
90%
60%
50%
10%
Answer: D) 10%
Confirmed vs Book-3 §3.1 Introduction (Figure 3.1 Sankey diagram) — Book-3 opens by noting that only 10-30% of the energy reaches the point of end-use — 'approximately 10% gets to the point of use' (Figure 3.1) — the balance going to heat, friction, misuse and leakage.
The compressed air system is therefore about 10% efficient overall, so the higher figures 50-90% are far off.
Source: Book EOC
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
30. The basic function of an air dryer in a compressed air system is to
remove remaining traces of moisture after the aftercooler
store and smoothen pulsating air output
reduce the temperature of the air before it enters the next stage to increase efficiency
prevent dust from entering the compressor
Answer: A) remove remaining traces of moisture after the aftercooler
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — The book's component list states that the remaining traces of moisture after the after-cooler are removed using air dryers, because instrument and pneumatic air must be relatively free of moisture.
Storing and smoothing pulsating output (b) is the receiver's role, inter-stage cooling (c) is the intercooler's, and keeping dust out (d) is the intake filter's.
31. Select the correct statement for reciprocating air compressors:
for every 4 degC drop in inlet air temperature, the increase in energy consumption is 1%
for every 4 degC rise in inlet air temperature, the decrease in energy consumption is 1%
for every 4 degC rise in inlet air temperature, the increase in energy consumption is 1%
the energy consumption remains the same irrespective of inlet air temperature
Answer: C) for every 4 degC rise in inlet air temperature, the increase in energy consumption is 1%
Confirmed vs Book-3 §3.5 Efficient Operation — The book's thumb rule reads: 'Every 4°C drop in inlet air temperature results in lower energy consumption by 1%'. Restated for a rise, every 4°C rise increases power consumption by about 1% — option (c).
Options (a) and (b) invert the sign of the effect, and Table 3.3 (102.0% delivery at 10°C vs 91.2% at 43.3°C) disproves (d).
32. Which of the following parameters is NOT required for evaluating the volumetric efficiency of a compressor?
FAD
Cylinder bore diameter
Stroke length
Power input
Answer: D) Power input
Confirmed vs Book-3 §3.3 Compressor Efficiency — Volumetric efficiency = FAD / compressor displacement, and displacement = (π/4) x D² x L x S x X x n — needing bore D, stroke L, speed, single/double acting and number of cylinders, plus the measured FAD.
Power input appears only in isothermal, adiabatic and mechanical efficiency, never in the volumetric efficiency calculation, so (d) is the parameter not required.
33. Which of the following will NOT occur if a reciprocating compressor is operated at a lower discharge pressure?
lower power consumption
less load on the piston rods and hence reduced maintenance costs
lower leakage losses
lower free air delivery than rated
Answer: D) lower free air delivery than rated
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Lowering delivery pressure cuts power by 6-10% per bar (Table 3.9), lightens the load on piston rods, and reduces leakage because leak flow through a given orifice rises steeply with gauge pressure (Table 3.16) — so (a), (b) and (c) all DO occur.
Free air delivery actually improves at lower discharge pressure, since the book notes volumetric efficiency is less at higher delivery pressures. A drop in FAD is therefore the one thing that will NOT happen.
Source: Book EOC
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
34. Which type of energy-efficient dryer can be opted for if a user in a plant requires compressed air at a dew point of -40 degC?
heatless purge type dryer
heat of compression dryer
aftercooler
refrigerant dryers
Answer: B) heat of compression dryer
Corrected (was a) — Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19): Table 3.19: refrigerant dryers reach only -20°C atmospheric dew point and an after-cooler far less, so (c) and (d) cannot meet -40°C at all.
Of the two that do give -40°C, the heat of compression dryer needs 0.8 kW per 1000 m³/hr with 'zero or very minimal' operating cost, while the heatless purge type needs 20.7 kW plus a 12-15% purge loss. The energy-efficient choice at -40°C is therefore the HOC dryer.
35. Reduction in the delivery pressure of a compressor by 1 bar would reduce the power consumption by
1 to 5 %
6 to 10 %
11 to 15 %
none of the above
Answer: B) 6 to 10 %
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Book-3: 'A reduction in the delivery pressure by 1 bar in a compressor would reduce the power consumption by 6-10%'; the worked case of 8 to 7 kg/cm² gives 9% input power saving.
The question stem was repaired from the self-contradictory 'Increase in the delivery pressure... would reduce the power consumption' to the book's wording; the answer 6-10% is unchanged. Table 3.9's smaller 4% figure applies to a 0.7 bar cut, which is why 1-5% looks plausible but is too low for a full bar.
36. The FAD of a reciprocating compressor is directly proportional to
pressure
volume
speed
all of the above
Answer: C) speed
Corrected (was d) — Book-3 §3.2 Positive Displacement: Book-3 §3.2: reciprocating compressors are 'characterized by a flow output that remains nearly constant over a range of discharge pressures. Also, the compressor capacity is directly proportional to the speed.'
Because FAD is essentially independent of discharge pressure, option (a) fails and 'all of the above' is contradicted by the book; speed is the parameter of direct proportionality.
Source: 15th Exam
📖 Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)
37. Which of the following is not true of air receivers in a compressed air system?
smoothens pulsating air output
increases the compressed air pressure
stores large volumes of compressed air
facilitates draining of moisture
Answer: B) increases the compressed air pressure
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — Air receivers smooth the pulsating output, store large volumes for sudden demand and let carried-over moisture precipitate for draining — (a), (c) and (d) are all true.
A receiver is only a vessel: it stores air at whatever pressure the compressor delivers and can never increase that pressure, so (b) is the statement that is not true.
Source: 15th Exam
📖 Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11)
38. Typical acceptable pressure drop in mains header at the farthest point of an industrial compressed air network is
1.0 bar
0.7 bar
0.5 bar
0.3 bar
Answer: D) 0.3 bar
Confirmed vs Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11) — Book-3: 'Typical acceptable pressure drop in industrial practice is 0.3 bar in mains header at the farthest point and 0.5 bar in distribution system.'
Option (c) 0.5 bar is the distribution-system allowance, which is the tempting wrong pick; the header figure asked for is 0.3 bar.
Confirmed vs Book-3 §3.3 Compressor Efficiency — Isothermal power (kW) = P₁ x Q₁ x logₑ(r) / 36.7, where P₁ is the absolute intake pressure, Q₁ the free air delivered in m³/hr and r = P₂/P₁ the pressure ratio.
All three quantities appear explicitly in the formula, so no single one of (a), (b) or (c) can be singled out — the answer is all of the above.
40. Reduction in the delivery pressure of an air compressor working at 7 bar, by 1 bar would reduce the power consumption by
2 - 3 %
4 - 5 %
6 - 10 %
none of the above
Answer: C) 6 - 10 %
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Book-3: 'A reduction in the delivery pressure by 1 bar in a compressor would reduce the power consumption by 6-10%', illustrated by 8 to 7 kg/cm² giving 9% savings.
The lower ranges (a) and (b) match Table 3.9's 4% figure, but that is for a 0.7 bar reduction (6.8 to 6.1 bar), not a full bar.
41. Which of the following is correct for air compressors?
for every 5.5oC drop in the inlet air temperature, the increase in energy consumption is by 2%
for every 4 oC rise in the inlet air temperature, the increase in energy consumption is by 1%
for every 4oC rise in the inlet air temperature, the decrease in energy consumption is by 1%
the energy consumption remains same irrespective of inlet air temperature
Answer: B) for every 4 oC rise in the inlet air temperature, the increase in energy consumption is by 1%
Confirmed vs Book-3 §3.5 Efficient Operation — The book's thumb rule is that every 4°C rise in inlet air temperature increases power consumption by about 1% (equivalently every 4°C drop saves 1%), supported by Table 3.3.
Option (a) misquotes both the temperature step and the direction, (c) inverts the effect, and (d) is disproved by the whole of Table 3.3.
42. The Free Air Delivery of a reciprocating air compressor is directly proportional to
speed
pressure
volume
all of the above
Answer: A) speed
Corrected (was d) — Book-3 §3.2 Positive Displacement: Book-3 §3.2 states plainly that 'the compressor capacity is directly proportional to the speed', while the flow output 'remains nearly constant over a range of discharge pressures'.
Since FAD does not vary in proportion to delivery pressure, 'all of the above' is contradicted by the book text and speed is the correct single answer.
Source: 14th Exam
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
43. Which of these desiccant compressed air dryers uses dry compressed air for regenerating the desiccant?
blower reactivated type
heatless purge type
heat of compression type
all of the above
Answer: B) heatless purge type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — In the heatless purge dryer, 'pure dry compressed air is used for purging through the saturated dessicant', which is why its purge loss is 12-15% and its operating cost is very high.
The blower reactivated type uses a blower plus external heater (1-2% purge) and the heat of compression type uses the compressor's own hot discharge with no purge at all, so (a) and (c) are wrong.
44. Which of the following parameters is not required for evaluating volumetric efficiency of the compressor?
FAD
Cylinder bore diameter
Stroke length
Power input
Answer: D) Power input
Confirmed vs Book-3 §3.3 Compressor Efficiency — Volumetric efficiency = FAD / compressor displacement, with displacement = (π/4) x D² x L x S x X x n — so bore, stroke, speed and cylinder count plus the measured FAD are all needed.
Power input belongs to isothermal / adiabatic / mechanical efficiency, not to volumetric efficiency, so it is the parameter not required.
Source: Set-A
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
45. Which type of energy efficient dryer can be opted if a user requires compressed air at a dew point of -40°C?
heatless purge type dryer
heat of compression dryer
aftercooler
refrigerant dryers
Answer: B) heat of compression dryer
Corrected (was a) — Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19): Table 3.19 shows the after-cooler and the refrigerant dryer cannot reach -40°C at all (refrigerant bottoms out at -20°C), so (c) and (d) are eliminated.
Both remaining desiccant options deliver -40°C, but the heat of compression dryer costs 0.8 kW per 1000 m³/hr against 20.7 kW plus 12-15% purge air for the heatless purge type — so the ENERGY EFFICIENT choice is (b).
Source: Set-A
📖 Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3)
46. The efficiency of a pump does not depend on
suction head
discharge head
motor efficiency
density of liquid
Answer: C) motor efficiency
Confirmed vs Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3) — Pump efficiency = hydraulic power (ρ g Q H) / shaft power input to the pump, so it depends on suction and discharge head through H and on the liquid through its density.
Motor efficiency belongs to the driving motor, not the pump; it enters only the combined pump-set (wire-to-water) efficiency. Note this is a pumps item filed within the compressed-air set.
Source: Jul 2022
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
47. The basic function of an air dryer in a compressor is to:
Prevent dust from entering the compressor
Remove moisture before the intercooler
Remove moisture in compressor suction
Remove moisture at the downstream of the after-cooler
Answer: D) Remove moisture at the downstream of the after-cooler
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — Air dryers remove the remaining traces of moisture AFTER the after-cooler — i.e. downstream of it — because instrument and pneumatic air must be essentially moisture free.
Keeping dust out (a) is the intake filter's job, and there is no moisture removal before the intercooler or at the suction (b, c) — atmospheric moisture is carried in with the intake air.
48. At which of the following discharge pressures, the same reciprocating air compressor will consume maximum power?
3 bar
5 kgf/cm2
90 psi
500 kPa
Answer: C) 90 psi
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Convert to a single unit: 3 bar ≈ 3.06 kg/cm², 5 kgf/cm² ≈ 4.9 bar, 90 psi ≈ 6.2 bar, 500 kPa = 5.0 bar. The highest is 90 psi.
Book-3: 'For the same capacity, a compressor consumes more power at higher pressures', so the 90 psi setting draws the maximum power. The trap is comparing the raw numbers without unit conversion.
Source: 17th Sep-2016
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
49. The adsorption material used in an adsorption air dryer for compressed air is:
calcium chloride
magnesium chloride
activated alumina
potassium chloride
Answer: C) activated alumina
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Book-3: 'The most common adsorption materials used for compressed air drying are activated alumina and silica gel', both of which have an open porous structure with a large inner surface.
The chloride salts offered in (a), (b) and (d) are deliquescent absorption chemicals, not the physical adsorbents the book names.
Source: 16th Exam
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
50. In a large compressed air system, about 70% to 80% of moisture in the compressed air is removed at the:
air dryer
after cooler
air receiver
inter cooler
Answer: B) after cooler
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Book-3: 'About 60 to 75% of moisture in compressed air is removed at the after cooler' — the stem's 70-80% is a loose restatement of the same fact, and the after cooler is the component intended.
The dryer takes out only the remaining traces, the receiver merely lets carried-over moisture settle, and the intercooler acts between compression stages.
51. If the observed temperature in air receiver is higher than ambient air temperature the correction factor for free air delivery will be:
(273+t2)/(273+t1)
greater than 1
less than 1
equal to 1
Answer: C) less than 1
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method) — Book-3 §3.6: when the discharge/receiver temperature t₂ exceeds the ambient t₁, the measured FAD must be corrected by the factor (273+t₁)/(273+t₂), which is necessarily LESS than 1 — hot air occupies more volume, so the uncorrected figure overstates FAD.
Option (a) was printed as the correct ratio itself, giving two right answers; it has been repaired to the inverted form (273+t₂)/(273+t₁), which is greater than 1 and is the classic trap, leaving (c) as the single correct choice.
52. A 500 cfm reciprocating compressor has a loading and unloading period of 5 seconds and 20 seconds respectively during a compressed air leakage test. The air leakage in the compressed air system would be
125 cfm
100 cfm
200 cfm
none of the above
Answer: B) 100 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 5/(5+20) x 100 = 20%, so leakage = 0.20 x 500 = 100 cfm.
Option (a) 125 cfm is what you get by wrongly using the load:unload ratio 5/20; the denominator must be the full load-plus-unload cycle time.
53. The correction factor for actual free air discharge in a compressor capacity test will be ___, when the compressed air discharge temperature is 15 °C higher than ambient air of 40 °C.
0.727
0.920
0.954
none of the above
Answer: C) 0.954
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method) — Correction factor = (273+t₁)/(273+t₂) with ambient t₁ = 40°C and discharge t₂ = 40+15 = 55°C, i.e. 313/328 = 0.954.
Option (a) 0.727 comes from 40/55 and (b) 0.920 from using the 15°C difference directly — both skip the conversion to absolute temperature.
Source: 18th Exam
📖 Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)
54. Which of the following is false? Air receivers _____
reduce frequent on/off operation of compressors.
knock out some oil and moisture
increase compressor efficiency
act as reservoir to take care of sudden demands
Answer: C) increase compressor efficiency
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — The receiver prevents too-frequent loading and unloading (a), knocks out oil and moisture as the air cools and settles (b), and acts as a reservoir for sudden heavy demand (d) — all true per the book.
It does not make the compressor itself more efficient: it only stores air that has already been compressed, so (c) is the false statement.
Source: 18th Exam
📖 Book-3 §3.3 Compressor Performance — Free Air Delivery (FAD)
55. For an air compressor with displacement of 100 CFM and system leakage of 10%, free air delivery is ______.
111.11 CFM
90 CFM
100 CFM
None of the above
Answer: C) 100 CFM
Corrected (was b) — Book-3 §3.3 Compressor Performance: FAD is defined as the volume of air drawn in from the atmosphere, compressed and delivered BY THE COMPRESSOR — it is a machine measurement taken at the discharge, so leaks further down the distribution network do not reduce it.
The FAD therefore remains 100 CFM; 90 CFM would be the air usefully reaching the end-users. Subtracting leakage from FAD confuses a distribution loss with volumetric efficiency (which is what does reduce delivery below displacement).
Source: 18th Exam
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
56. The adsorption material used in an adsorption air dryer is
Calcium chloride
Magnesium chloride
Activated alumina
Potassium chloride
Answer: C) Activated alumina
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Book-3 names activated alumina and silica gel as the common adsorbents for compressed air drying; they bind moisture physically on a large porous inner surface and are then regenerated.
Calcium, magnesium and potassium chloride are deliquescent absorption chemicals and are not used as the desiccant bed in these dryers.
Source: 19th Exam
📖 Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7)
57. The inlet air temperature to a two stage reciprocating air compressor is 35 °C. At which of the following 2nd stage inlet temperatures will the compressor consume least power?
75 °C
65 °C
60 °C
50 °C
Answer: D) 50 °C
Confirmed vs Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7) — Table 3.7 shows that the nearer the second-stage inlet temperature is to the first-stage inlet ('perfect cooling'), the lower the specific power: a 5.5°C rise at the second-stage inlet costs about 2% more specific energy.
With a 35°C first-stage inlet, 50°C is the closest of the four, so it gives the least power consumption; 75°C would be the worst.
58. The isothermal power of a 500 CFM air compressor is 72 kW and the efficiency is 78 %. The actual power drawn by the compressor will be
56 kW
92 kW
72 kW
None of the above
Answer: B) 92 kW
Confirmed vs Book-3 §3.3 Compressor Efficiency — Isothermal efficiency = isothermal power / actual measured input power, so actual power = 72 / 0.78 = 92.3 ≈ 92 kW.
Option (a) 56 kW comes from multiplying 72 x 0.78 instead of dividing; the actual input must always exceed the isothermal power, since isothermal power ignores friction.
Source: 19th Exam
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
59. The primary purpose of inter-cooling in a multistage compressor is to _____________.
remove the moisture in the air
reduce the work of compression
separate moisture and oil vapour
none of the above
Answer: B) reduce the work of compression
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — Inter-stage coolers 'reduce the temperature of the air before it enters the next stage to reduce the work of compression and increase efficiency' by cutting the specific volume the next stage handles.
Removing moisture (a) is the stated objective of the AFTER-cooler, and moisture/oil separation (c) is an incidental benefit rather than the primary purpose of inter-cooling.
Source: Sep 2019
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
60. The basic function of an air dryer in an air compressor is to
Prevent dust from entering the compressor
Remove moisture before the intercooler
Remove moisture in compressor suction
Remove moisture in air supplied to the plants
Answer: D) Remove moisture in air supplied to the plants
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — The dryer's function is to remove the remaining water vapour so that the air supplied to plant instruments and pneumatic equipment is relatively free of moisture — water otherwise erodes valves, corrodes pipework and spoils paint spraying.
Dust exclusion (a) is the intake filter's job, and no moisture removal happens at the suction or before the intercooler (b, c).
61. Which of the following parameters is not required for evaluating volumetric efficiency of reciprocating air compressor?
Power input
FAD
Cylinder Stroke
Cylinder bore
Answer: A) Power input
Confirmed vs Book-3 §3.3 Compressor Efficiency — Volumetric efficiency = FAD / compressor displacement, and displacement = (π/4) x D² x L x S x X x n, so bore, stroke, speed and cylinder count plus the measured FAD are all needed.
Power input is used for isothermal/adiabatic/mechanical efficiency only, so it is the parameter not required here.
Source: Jul 2022
📖 Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)
62. For a given air requirement, providing higher volume air receiver will ________________.
Increase energy consumption
Reduce energy consumption
Reduce Unload Power
Reduce Pressure fluctuations
Answer: D) Reduce Pressure fluctuations
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — The receiver's book-stated role is to dampen pulsations and even out the pressure variations from the compressor, buffering sudden short-duration demands; IS 7938-1976 sizes it at 1/10 to 1/6 of output in m³/min.
A bigger receiver by itself does not change compressor energy consumption (a, b) or unload power (c) — those follow from the pressure settings and the capacity control scheme.
Source: Sep 2019
📖 Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11)
63. Pressure drop can be reduced in a compressed air distribution line by providing ______________.
After Coolers
Small diameter distribution pipes
High pressure air flow
Large Diameter Distribution pipes.
Answer: D) Large Diameter Distribution pipes.
Confirmed vs Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11) — Table 3.11 (170 m³/h): pressure drop per 100 m falls from 1.80 bar at 40 mm bore to 0.02 bar at 100 mm, with equivalent power loss dropping from 9.5 kW to 0.1 kW. Larger diameter distribution pipes are therefore the remedy.
Small-bore pipes (b) and high velocity (c) increase the drop, and an after-cooler (a) has nothing to do with distribution pressure loss.
Source: Sep 2019
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
64. Power consumption is very high for ___________ type of compressed air dryers.
Refrigeration type
Blower reactivated type
Heat of compression type
Heatless purge type
Answer: D) Heatless purge type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.19 gives 20.7 kW per 1000 m³/hr for the heatless purge type — the highest of all four — plus a 12-15% loss of dry compressed air used for purging.
Blower reactivated is 18.0 kW, refrigeration 2.9 kW and heat of compression only 0.8 kW, so the heatless purge dryer is the very high consumer.
65. The isothermal power of 500 CFM air compressor is 72 kW and the efficiency is 76 %. The actual power drawn by the compressor will be ________.
56 kW
94.7 kW
89 kW
72 kW
Answer: B) 94.7 kW
Confirmed vs Book-3 §3.3 Compressor Efficiency — Isothermal efficiency = isothermal power / actual input power, so actual power = 72 / 0.76 = 94.7 kW.
Option (a) 56 kW results from multiplying rather than dividing by the efficiency; the actual input is always larger than the isothermal power because the latter excludes friction.
Source: Sep 2019
📖 Book-3 Ch-2 Electric Motors — speed control / Ch-1 power factor (cross-chapter item filed under Ch-3)
66. Power factor improvement of a 75-kW compressor motor will ___________.
Reduce input power to the motor
Increase input power to the motor
Reduce the compressor motor shaft power
None of the above
Answer: A) Reduce input power to the motor
Confirmed vs Book-3 Ch-2 Electric Motors — Improving power factor cuts the reactive (magnetising) current drawn through the supply system, so line current and the associated I²R distribution losses fall and the power drawn from the network reduces.
The motor's shaft power is fixed by the compressor load and is unchanged, so (c) is wrong. Note this is a power-factor item filed within the compressed-air set.
Source: Sep 2019
📖 Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11)
67. The pressure drop in mains header at the farthest point of an industrial compressed air network shall not exceed
2 bar
0.3 bar
0.5 bar
1.0 bar
Answer: B) 0.3 bar
Confirmed vs Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11) — Book-3: 'Typical acceptable pressure drop in industrial practice is 0.3 bar in mains header at the farthest point and 0.5 bar in distribution system.'
The 0.5 bar figure offered in (c) is the distribution allowance, not the mains-header limit asked for.
68. The Free Air Delivery capacity of a reciprocating compressor is directly proportional to
pressure
volume
speed
all of the above
Answer: C) speed
Confirmed vs Book-3 §3.2 Positive Displacement — Book-3 §3.2: 'the compressor capacity is directly proportional to the speed', while flow output 'remains nearly constant over a range of discharge pressures'.
Because FAD does not scale with delivery pressure, neither (a) nor 'all of the above' can hold; speed is the correct answer.
Source: 9th Dec-2009
📖 Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7)
69. The inlet air temperature to a two stage reciprocating air compressor is 35°C. At which of the following 2nd stage inlet temperatures the compressor will consume least power?
75°C
65°C
60°C
50°C
Answer: D) 50°C
Confirmed vs Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7) — Table 3.7 shows specific power falling as the second-stage inlet temperature approaches the first-stage inlet ('perfect cooling'); a 5.5°C rise at the second-stage inlet costs about 2% more specific energy.
With the first stage at 35°C, the lowest offered second-stage inlet — 50°C — gives the least power consumption.
70. At which of the following discharge pressures, the reciprocating air compressor will consume maximum power
3 bar
3.5 kg/cm2
150 psi
6 kg/cm2
Answer: C) 150 psi
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Convert to common units: 3 bar ≈ 3.06 kg/cm², 3.5 kg/cm² ≈ 3.43 bar, 150 psi ≈ 10.34 bar (10.5 kg/cm²), 6 kg/cm² ≈ 5.88 bar. The highest by far is 150 psi.
Since 'for the same capacity, a compressor consumes more power at higher pressures', 150 psi draws the maximum power; the trap is comparing the bare numbers across mixed units.
Source: 9th Dec-2009
📖 Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)
71. Which of the following is not true of air receivers?
smoothens pulsating air output
stores large volumes of air
a source for draining moisture
increases the pressure of air
Answer: D) increases the pressure of air
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — Air receivers smooth pulsating output, store large volumes for sudden demand and provide a point for draining precipitated moisture — (a), (b) and (c) are all true of them.
A receiver holds air at the pressure the compressor delivers and cannot itself increase pressure, so (d) is the statement that is not true.
Source: 9th Dec-2009
📖 Book-3 §3.5 Equivalent Lengths of Fittings (Table 3.12)
72. Which of the following pipe fittings used in compressed air pipe line offers maximum resistance
Gate Valve in open condition
Return bend
Elbow
Tee 90o long bend
Answer: B) Return bend
Confirmed vs Book-3 §3.5 Equivalent Lengths of Fittings (Table 3.12) — Table 3.12 equivalent lengths rise in the order Gate valve < Tee 90° long bend < Elbow < Return bend — at 50 mm NB, 0.40 / 0.61 / 1.07 / 1.68 m respectively.
The return bend therefore offers the maximum resistance of the four fittings listed; the elbow is second and an open gate valve the least.
73. A 200 cfm compressor has a loading and unloading period of 10 seconds and 20 seconds respectively during a compressed air leakage test. The air leakage in the compressed air system would be
20.3 cfm
42.1 cfm
66.6 cfm
132.8 cfm
Answer: C) 66.6 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 10/(10+20) x 100 = 33.3%, so leakage = 0.333 x 200 = 66.6 cfm.
The wrong answers come from using 10/20 (giving 100 cfm) or from inverting the ratio; the denominator must be the complete load-plus-unload cycle.
74. A 500 cfm reciprocating compressor earlier operating at load-unload pressure of 6.0 and 7.5 kg/cm2g was changed to 6.0 to 6.5 kg/cm2g for the same end use. This change will result in
increased unloading cycle time of the compressor
increased loading cycle time of the compressor
increased energy consumption of the compressor
decreased leakage loss in air distribution system
Answer: D) decreased leakage loss in air distribution system
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — Narrowing the band to 6.0-6.5 kg/cm²g drops the maximum system pressure from 7.5 to 6.5 kg/cm²g, and Table 3.16 shows leak flow through a given orifice rising steadily with gauge pressure — so leakage loss in the distribution system falls. Lower pressure also cuts power (6-10% per bar), so (c) is wrong.
The record's answerText had drifted to option (a) while the answer key said (d); the answerText is now aligned with (d). Option (a) has been repaired to 'increased' so exactly one option is correct — with a band one third as wide the receiver empties sooner, so the unload period actually shortens.
Source: 10th Jul-2010
📖 Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7)
75. Which of the following is incorrect statement?
Inadequate cooling in after-coolers causes more condensation in air receivers and distribution lines
Performance of inter-coolers have no effect on work of compression
In a battery of air compressors, the compressor with lower part load power consumption should be modulated.
For the same capacity, a compressor consumes more power at higher delivery pressure
Answer: B) Performance of inter-coolers have no effect on work of compression
Confirmed vs Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7) — Inter-coolers 'reduce the work of compression (power requirements) by reducing the specific volume through cooling the air', and Table 3.7 shows a 5.5°C rise in second-stage inlet costs about 2% more specific energy — so (b) is plainly false.
Statements (a), (c) and (d) are all made in the book: poor after-cooling causes condensation and corrosion downstream, the compressor with lower part-load power should be modulated, and higher delivery pressure means more power for the same capacity.
Source: 10th Jul-2010
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power)
76. In a textile mill, two 150 cfm belt driven reciprocating compressors are seen to be working constantly with a loading time of 20 seconds and unloading time of 30 seconds. The best economic option for energy savings would be:
switch off one compressor
switch off one compressor and reduce motor pulley size of the other compressor appropriately
adopt variable speed drive for one of the compressors
none of the above
Answer: B) switch off one compressor and reduce motor pulley size of the other compressor appropriately
Confirmed vs Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power) — Loading 20 s in every 50 s means the two machines together are loaded only 40% of the time, so a single 150 cfm compressor covers the demand and is still oversized.
The book prescribes changing the pulley size to reduce RPM and de-rate an oversized compressor, so switching one off AND trimming the other's motor pulley (b) saves more than simply switching one off (a); a VSD (c) is not economic at this size.
Source: 11th Feb-2011
📖 Book-3 Ch-2 Electric Motors — speed control / Ch-1 power factor (cross-chapter item filed under Ch-3)
77. Which of following is not used for speed control ?
fluid coupling
eddy current
soft starter
variable frequency drive
Answer: C) soft starter
Confirmed vs Book-3 Ch-2 Electric Motors — Fluid couplings, eddy-current couplings and variable frequency drives all vary the output speed of a drive.
A soft starter only ramps the applied voltage during starting to limit inrush current and starting torque; once running, the motor returns to full speed, so it is not a speed-control device. Note this is a drives item filed within the compressed-air set.
Source: Mar 2021
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
78. Which of the following compressed air dryer requires the use of activated alumina ?
membrane dryer
heat of compression dryer
refrigerant dryers
all of the above
Answer: B) heat of compression dryer
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Activated alumina (with silica gel) is the adsorbent used in adsorption/desiccant dryers, and the heat of compression dryer is a desiccant dryer whose bed is regenerated by the compressor's own hot discharge air.
Membrane dryers work by selective permeation through hollow polymer fibres and refrigerant dryers by mechanical chilling — neither contains a desiccant, so 'all of the above' fails.
79. During a leak test of a compressed air system, the compressor's average load time was 1.5 minute, average unload time was 10.5 minutes and flow rate was 35 m3/min. The leakage quantity is
4.375
5.125
7.625
6.250
Answer: A) 4.375
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — This is the book's worked leakage example: q = T/(T+t) x Q = 1.5/(1.5+10.5) x 35 = 0.125 x 35 = 4.375 m³/min, i.e. 6300 m³/day.
The other options arise from using the unload time alone or the wrong cycle total; the denominator must be load time plus unload time.
Source: Mar 2021
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
80. Which of the following is not an adsorption type of air dryer for compressed air system ?
blower reactivated type
heat less purge type
heat of compression type
refrigerant type
Answer: D) refrigerant type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — The book classifies adsorption (desiccant) dryers by their regeneration method: blower reactivated, heatless purge and heat of compression types.
The refrigerant dryer is not an adsorption dryer at all — it chills the air mechanically so water vapour condenses out, achieving about -20°C atmospheric dew point.
Source: Mar 2021
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
81. At which of the following dew points of the compressed air, the moisture content would be maximum ?
-10oC
-5oC
-40oC
-20oC
Answer: B) -5oC
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.18: moisture content rises with dew point — 2500 ppm at -5°C, 1600 at -10°C, 685 at -20°C and 80 ppm at -40°C.
Since 'lower the dew point, more dry is the air', the least negative dew point (-5°C) holds the maximum moisture; -40°C is the driest of the four.
Source: Mar 2021
📖 Book-3 Ch-4 HVAC and Refrigeration — COP (cross-chapter item filed under Ch-3)
82. If the COP of a vapour compression system is 3.5 and the motor draws a power of 10.7 kW at 80% motor efficiency, the cooling effect of vapour compression system will be ___________
30 kW
42 kW
27 kW
none of the above
Answer: A) 30 kW
Confirmed vs Book-3 Ch-4 HVAC and Refrigeration — Shaft power delivered to the refrigeration compressor = 10.7 x 0.80 = 8.56 kW; cooling effect = COP x shaft power = 3.5 x 8.56 = 29.96 ≈ 30 kW.
The distractors come from applying the COP to the 10.7 kW motor input directly, or from dividing by the motor efficiency instead of multiplying. Note this is a refrigeration item filed within the compressed-air set.
Source: Mar 2021 (Set B)
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
83. The purpose of after cooler in a multistage compressor is to ________.
remove the moisture in the air
reduce the work of compression
separate moisture and oil vapour
none of the above
Answer: A) remove the moisture in the air
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — The after-cooler's stated objective is 'to remove the moisture in the air by reducing the temperature in a water-cooled heat exchanger'; about 60-75% of the moisture is condensed out there.
Reducing the work of compression (b) is the INTER-cooler's purpose, and moisture/oil vapour separation by settling (c) is a role of the receiver.
Source: Mar 2021
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
84. Which of the following compressors do not use loading / un-loading method for capacity control?
screw compressor
centrifugal compressor
reciprocating compressor
all of the above
Answer: B) centrifugal compressor
Confirmed vs Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14) — Load/unload (two-step) and multi-step unloading are the control schemes the book describes for positive-displacement machines — reciprocating and screw compressors.
Centrifugal capacity is matched instead by variable inlet guide vanes or by speed control (Table 3.14), so the centrifugal is the odd one out.
85. During a leak test of a compressed air system, the compressor's average load time was 1.5 minute, average unload time was 10.5 minutes and flow rate was 35 m3/min. The leakage quantity will be ___________
4.375
5.125
7.625
6.250
Answer: A) 4.375
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — q = T/(T+t) x Q = 1.5/(1.5+10.5) x 35 = 0.125 x 35 = 4.375 m³/min, exactly the book's worked leakage example.
The stem's flow rate was repaired from a garbled '33' to 35 m³/min: only 35 reproduces the printed option 4.375 (33 would give 4.125, which is not offered) and it matches the book example the question is taken from.
Source: Mar 2021 (Set B)
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
86. Which of the following is not an adsorption type of air drier for compressed air systems?
blower reactivated type
heatless purge type
heat of compression type
refrigerant type
Answer: D) refrigerant type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — The adsorption (desiccant) dryers are the blower reactivated, heatless purge and heat of compression types, classified by how the bed is regenerated.
The refrigerant dryer is not an adsorption dryer — it chills the air so moisture condenses. Option (b) was printed as 'heat compression type', a duplicate of option (c); it has been repaired to 'heatless purge type' per the book's list so that exactly one option is correct.
87. Which of the following is a positive displacement compressor?
Screw compressor
Reciprocating compressor
Centrifugal compressor
Both a & b
Answer: D) Both a & b
Confirmed vs Book-3 §3.2 Compressor Types (Figure 3.2 Compressor Chart) — Figure 3.2: positive-displacement compressors comprise the reciprocating and rotary (screw, vane, roots, scroll, liquid ring) families — they raise pressure by reducing the volume of the trapped gas.
Centrifugal machines are dynamic: they add velocity which is then converted to pressure at the outlet. So both the screw and the reciprocating machine qualify.
Source: Jul 2022
📖 Book-3 §3.3 Compressor Performance — Free Air Delivery (FAD)
88. For an air compressor of rated capacity of 100 CFM and system leakage of 10%, free air delivery is ____
111.11 CFM
90 CFM
100 CFM
None of the above
Answer: C) 100 CFM
Confirmed vs Book-3 §3.3 Compressor Performance — FAD is the volume of air drawn from the atmosphere, compressed and delivered by the machine — it is measured at the compressor, so leaks downstream in the distribution network do not change it.
The rated 100 CFM therefore stands; 90 CFM would be the air actually reaching end-use points, which is a distribution loss and not the FAD.
89. A 500 cfm reciprocating compressor has a loading and unloading period of 5 seconds and 20 seconds respectively during a measured period of leakage test. The air leakage in the compressed air system would be ____
125 cfm
100 cfm
200 cfm
none of the above
Answer: B) 100 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 5/(5+20) x 100 = 20%, giving leakage = 0.20 x 500 = 100 cfm.
Option (a) 125 cfm comes from the load:unload ratio 5/20; the denominator must be the total cycle of 25 seconds.
90. The advantage of multi-staging compression over single stage compression is
Lower power consumption per unit of air delivered
High volumetric efficiency
Decreased temperature of discharge
All of above
Answer: D) All of above
Confirmed vs Book-3 §3.2 Positive Displacement — Two-stage machines discharge at 140-160°C against 205-240°C for single stage, and the checklist states that a two-stage or multistage compressor 'consumes less power for the same air output than a single stage compressor'.
Inter-cooling and the lower pressure ratio per stage also raise volumetric efficiency, so all three listed advantages hold.
Source: Jul 2022
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
91. Which of the following compressed air dryer consumes less power for the same output?
Refrigeration Dryer
Heat of Compression Dryer
Heatless Purge type Dryer
Blower Reactivated Type Dryer
Answer: B) Heat of Compression Dryer
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.19 per 1000 m³/hr: heat of compression 0.8 kW, refrigeration 2.9 kW, blower reactivated 18.0 kW, heatless purge 20.7 kW.
The HOC dryer regenerates its desiccant with the compressor's own ~135°C discharge heat, with no electric heater and no purge loss, so it consumes the least power for the same output.
92. Which of the following is not a positive displacement compressor __________.
Reciprocating
Screw
Roots Blower
Centrifugal
Answer: D) Centrifugal
Confirmed vs Book-3 §3.2 Compressor Types (Figure 3.2 Compressor Chart) — Figure 3.2 places reciprocating, screw and roots blower under positive displacement — all increase pressure by reducing the trapped volume.
The centrifugal is a dynamic compressor: it imparts velocity to the air which is then converted to pressure at the outlet, so it is the one that is not positive displacement.
Source: Mar 2023
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
93. Adsorption drying of air is achieved using __________.
Activated alumina
Carbon Molecular Sieves
Zirconium Molecular Sieves
None of the above
Answer: A) Activated alumina
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Book-3: 'The most common adsorption materials used for compressed air drying are activated alumina and silica gel.'
Carbon and zirconium molecular sieves are not the materials the book names for compressed-air adsorption drying, so (a) is the correct choice and 'none of the above' is wrong.
94. If air leak quantity is 5 m³/min, what will be power loss per hour in compressor which is having specific power consumption of 0.09 kWh/m³?
27 Units/hr
0.45 Units/hr
12 Units/hr
0.05 Units/hr
Answer: A) 27 Units/hr
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — Leakage = 5 m³/min = 300 m³/hr; power lost = leakage quantity x specific power consumption = 300 x 0.09 = 27 kWh per hour, i.e. 27 units/hr.
Option (b) 0.45 comes from multiplying 5 m³/min by 0.09 without converting to the hourly basis on which the specific power consumption (kWh/m³) is quoted.
Source: Mar 2023
📖 Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3)
95. If two pumps are operated in parallel then Shut-off head __________.
Does not change
Halved
Doubled
Less than double
Answer: A) Does not change
Confirmed vs Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3) — Two identical pumps in parallel add their flows at a given head, but the shut-off (zero-flow) head is fixed by impeller diameter and speed and is the same as for one pump.
Doubling of head is what SERIES operation achieves; parallel doubles flow, not head. Note this is a pumps item filed within the compressed-air set.
96. If a 200 cfm compressor is pressurized (on load) in 10 seconds and unloads in 20 seconds during a leakage test, the air leakage would be __________.
67 cfm
100 cfm
10 cfm
133 cfm
Answer: A) 67 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 10/(10+20) x 100 = 33.3%, so leakage = 0.333 x 200 = 66.7 ≈ 67 cfm.
The stem was missing the compressor capacity; the printed options fix it at 200 cfm (option (b) 100 cfm is what you get by wrongly using the ratio 10/20). The denominator must be the full load-plus-unload cycle.
97. The compressor capacity of a reciprocating compressor is directly proportional to __________.
Speed
Pressure
Volume
All
Answer: A) Speed
Confirmed vs Book-3 §3.2 Positive Displacement — Book-3 §3.2: 'the compressor capacity is directly proportional to the speed' — halving the RPM (e.g. by reducing the motor pulley) halves the delivered air, which is the basis of the pulley-change de-rating measure.
Output stays nearly constant over a range of discharge pressures, so capacity is not proportional to pressure, and 'All' is therefore wrong.
Source: Mar 2023
Short questions (5 marks) — 20
📖 §3.5 Leak Quantification — worked leakage-test example
1. A compressor of capacity 35 m3/min runs loaded for 1.5 min and unloaded for 10.5 min in a leakage test, drawing 188 kW on load. Calculate the leakage quantity per day and the energy lost per day to leakage.
Model answer: Step 1 — Leakage quantity (Book-3 §3.5): q = Q × T/(T + t) = 35 × 1.5/(1.5 + 10.5) = 35 × 1.5/12 = 35 × 0.125 = 4.375 m³/min. (Leakage level = 12.5% of capacity.)
Step 2 — Leakage per day = 4.375 × 60 × 24 = 4.375 × 1440 = 6300 m³/day.
Step 3 — Specific power consumption for compressed air generation = Load kW ÷ FAD in m³/hr = 188 ÷ (35 × 60) = 188/2100 = 0.0895 kWh/m³.
Step 4 — Energy lost to leakage per day = 6300 × 0.0895 = 564 kWh/day.
(At, say, ₹7/kWh this is about ₹3,950 per day of pure waste — the book notes any expenditure on stopping leaks is paid back through energy saving.)
Confirmed vs Book-3 §3.5 — this reproduces the book's own leakage worked example (Q = 35 m³/min, T = 1.5 min, t = 10.5 min, load 188 kW → 4.375 m³/min, 6300 m³/day, 0.0895 kWh/m³, 564 kWh/day). Keep LOAD time on top, and convert FAD from m³/min to m³/hr before dividing the kW, or the specific power comes out 60 times too large.
Source: AI practice
📖 §3.2 Compressor Types (two-stage) / §3.5 Efficacy of Inter and After Coolers
2. Explain why multi-staging with inter-cooling improves the efficiency of a reciprocating air compressor. List three benefits.
Model answer: Why it improves efficiency (Book-3 §3.5): inter-coolers are provided between successive stages of a multi-stage compressor to reduce the temperature of the air entering the next stage. Cooling reduces the specific volume of the air, so the next stage handles a smaller volume for the same mass, and the work of compression falls. Ideally the inlet temperature at each stage equals the first-stage inlet temperature — the book calls this 'perfect cooling' or isothermal compression, the least-work ideal. Imperfect cooling raises stage inlet temperatures and power: the book's Table 3.7 shows just 5.5 °C higher second-stage inlet temperature costs about 2% more specific energy.
Three benefits:
1) Reduced work of compression / lower specific power consumption (approaching the isothermal ideal).
2) Lower discharge temperature — two-stage machines run at 140–160 °C against 205–240 °C for single-stage.
3) Reduced pressure differential across each cylinder, which reduces load and stress on the valves and piston rings, and also gives moisture separation between stages.
Confirmed vs Book-3 §3.2 and §3.5 — the trade-off the book notes is higher investment cost for multi-staging, particularly for high pressure at low capacity, and a caution that very cold cooling water can condense moisture in the air and damage the cylinder if it is not drained.
Source: AI practice
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
3. List five energy saving measures in compressed air system.
Model answer: Any five from the Book-3 §3.7 checklist, with the book's own figures:
1. Carry out periodic leak tests and arrest leaks — leakage of 40-50% is not uncommon; leakage % = load time/(load+unload) x 100.
2. Reduce compressor delivery pressure to the minimum the plant needs — a 1 bar reduction saves 6-10% of input power (Table 3.9).
3. Ensure cool, clean, dry intake air, drawn from outside if the compressor room is hot — every 4°C rise in inlet temperature costs about 1% more power, and every 250 mmWC pressure drop across a choked filter costs about 2%.
4. Minimise unloaded running: if demand is below 50% of capacity, change over to a smaller compressor, reduce speed by trimming the motor pulley, or retrofit a VSD on machines above ~100 kW (unloading draws up to 30% of full-load power).
5. Avoid misuse of compressed air (body/floor cleaning, agitation, drying) and replace pneumatic conveying and pneumatic tools with blowers or electric tools where safe — pneumatic conveying uses about 8 times and pneumatic tools about 20 times the energy.
Other valid answers: size piping generously and use a ring main (0.3 bar drop in the header, 0.5 bar in distribution); clean inter-coolers and after-coolers; recover heat from hot compressed air; keep the load-unload pressure band narrow; sequence multiple compressors so only one small machine modulates; do periodic FAD tests.
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — Any five of the printed §3.7 checklist items earn full marks; the strongest answers quote the book's numbers (4 degC/1%, 250 mmWC/2%, 1 bar/6-10%, up to 30% unload power).
Source: Sep 2025
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
4. List five energy efficiency measures in Compressed air system.
Model answer: Any five energy efficiency measures from the Book-3 §3.7 checklist (p.101-102):
1. Locate the compressor so intake air is cool, clean and dry, drawing cold air from outside a hot compressor room — every 4°C rise in inlet temperature raises power by about 1%.
2. Clean inlet air filters regularly and fit manometers across them — efficiency drops about 2% for every 250 mmWC of filter pressure drop.
3. Reduce the delivery pressure wherever possible and keep the minimum possible range between load and unload settings — about 6-10% power saved per bar.
4. Carry out periodic leak tests and arrest leaks (40-50% leakage is not uncommon), and fit interlocked solenoid cut-off valves so idle machines get no air.
5. Minimise low-load running: below 50% demand change to a smaller compressor, reduce speed via the motor pulley, or retrofit a VSD above ~100 kW.
Also acceptable: periodic FAD tests; six-monthly valve inspection (worn valves cost up to 50% efficiency); periodic cleaning of inter-coolers; heat recovery from hot compressed air; generous pipe sizing with a ring main; discouraging misuse of compressed air.
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — The expected answer is any five items from the printed §3.7 checklist; the model answer now spells them out with the book's quantified figures instead of pointing at page 101.
5. Calculate the free air delivery (FAD) capacity of a compressor in m3/hr for: receiver capacity 0.3 m3, initial pressure 0 kg/cm2(g), final pressure 7 kg/cm2(g), initial air temperature 35 degC, final air temperature 50 degC, additional holdup volume 0.05 m3, compressor pump-up time 4.1 minutes, atmospheric pressure 1.026 kg/cm2(a).
Model answer: Total system volume V = 0.3 + 0.05 = 0.35 m3. FAD = [(P2 - P1)/P0] x V / t, corrected for temperature. P2 - P1 = (7+1.026) - (0+1.026) = 7.0 kg/cm2; P0 = 1.026 kg/cm2(a). Volume of free air = (7.0/1.026) x 0.35 = 2.388 m3 in 4.1 min. FAD per minute = 2.388/4.1 = 0.5825 m3/min = 34.95 m3/hr. Applying the temperature correction (273+35)/(273+50) = 308/323 = 0.954, FAD ≈ 34.95 x 0.954 ≈ 33.3 m3/hr.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD = [(P2-P1)/atmospheric] x (receiver+holdup) / pump-up time, with temperature correction; gives about 33 m3/hr.
Source: Book EOC
📖 Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9); Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
6. A reciprocating V-belt driven compressor operates with load pressure 6 bar, unload pressure 8 bar, load time 3 minutes and unload time 1.5 minutes. Suggest possible energy saving opportunities on a short-term basis.
Model answer: The load:unload ratio is 3:1.5, i.e. the compressor is loaded about 67% of the time, indicating modest spare capacity. Short-term measures: (1) Reduce the maximum operating/unload pressure (8 bar) to the minimum the plant actually needs, since every 1 bar reduction saves about 6-7% energy. (2) Reduce the pressure band/raise the cut-in so the compressor unloads more. (3) Lower the load pressure setting closer to the required application pressure. (4) Reduce compressor speed by trimming the motor pulley size if demand is consistently below capacity. (5) Detect and arrest air leaks and ensure cool, clean inlet air. These lower the average discharge pressure and running power.
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure — Load 3 min / unload 1.5 min means the machine is loaded about 67% of the time. Short-term actions: cut the 8 bar unload pressure toward actual demand (6-10% power per bar), narrow the load-unload band, arrest leaks, ensure cool clean intake air, and trim the motor pulley to de-rate the spare capacity.
7. Briefly explain the shop-floor method of air compressor capacity assessment.
Model answer: The shop-floor (pump-up) method measures the actual Free Air Delivery of a compressor in situ. The compressor is isolated to a known total volume (air receiver plus connected piping and aftercooler). With the receiver emptied to an initial pressure, the compressor is started and the time taken to pump up to a final (cut-off) pressure is recorded with a stopwatch, along with initial and final pressures, temperatures and atmospheric pressure. FAD = [(P2 - P1)/P_atm] x V / t, corrected for temperature. The measured FAD is compared with the rated capacity to assess the compressor's condition.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Isolate a known volume, time the pump-up between two pressures, then FAD = [(P2-P1)/Patm] x V / t (temperature corrected).
Source: Book EOC
📖 Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers (Figure 3.7)
8. Explain how a tranvector (transvector) nozzle works.
Model answer: A tranvector / transvector nozzle is a compressed air amplifier used to cut air consumption in blowing, cooling and cleaning duties (Book-3 Figure 3.7).
Working: compressed air enters the inlet and fills an annular chamber; it is then throttled through a small ring nozzle at high velocity. This primary jet clings to the Coanda profile, which directs it towards the outlet, and in doing so creates a low-pressure region at the centre. That low pressure induces a high volume of surrounding free atmospheric air into the primary stream, and the combined flow leaves the amplifier as a high-volume, high-velocity jet.
Energy significance: only a small fraction of the delivered air is expensive compressed air, the rest being entrained free air, so an amplifier does the same blow-off job as an open pipe at a fraction of the compressed-air consumption. The book notes that a 1/4" hose left open on a 7 bar line for 1000 hours a year can cost about Rs. 1.0 lakh per annum, and that if compressed air must be used for cleaning it should be through blow guns kept below 2 bar.
Confirmed vs Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers — The air amplifier (Figure 3.7) throttles a small primary jet through a ring nozzle onto a Coanda profile, creating a low-pressure core that entrains a large volume of free air — high outlet flow from little compressed air.
Source: Book EOC
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power); Book-3 §3.2 Positive Displacement — Reciprocating Compressors
9. A 1680 m3/hr reciprocating compressor is driven by a 160 kW motor (90% efficiency) drawing 159 kW. Demand rises by 100 m3/hr. To meet it the compressor speed is increased by changing the compressor pulley. Existing: Motor rpm 1400, Motor pulley 300 mm, Compressor rpm 700, Compressor pulley 600 mm. Find new pulley diameter and additional power; check if motor can handle the load.
Model answer: Modified flow = 1780 m3/hr. New compressor rpm = (1780/1680)×700 = 742 rpm. Using N1D1 = N2D2, new compressor pulley D2 = (700×600)/742 = 566 mm. New motor power = (742/700)×159 = 168.54 kW. Motor capacity = 160/0.9 = 178 kW. Since 168.54 < 178 kW, the motor has the margin to absorb the additional 100 m3/hr load.
Confirmed vs Book-3 §3.5 Capacity Utilisation — Flow ∝ rpm; pulley by N1D1=N2D2; power ∝ rpm (reciprocating, ~linear here).
Source: 17th Sep-2016
📖 Book-3 §3.5 Sizing of Compressed Air Piping
10. Determine the discharge pipe inner diameter size (in mm) for a compressed air system, given: FAD = 1000 Nm3/hr; discharge pressure = 7 bar(g); discharge temperature = 35 °C; air velocity = 6 m/s; atmospheric pressure = 1.013 bar.
Model answer: Convert NTP flow to actual flow using P1V1/T1 = P2V2/T2. Actual flow V2 = (P1V1/T1) x (T2/P2) = (1.013 x 1000/273) x (308/8.013) = 142.6 m3/hr = 0.0396 m3/s. Area = flow/velocity = 0.0396/6 = 0.0066 m2. di = √(4A/π) = 0.092 m = 92 mm → say 100 mm.
Confirmed vs Book-3 §3.5 Sizing of Compressed Air Piping — Apply the gas law to convert normal-condition flow to actual conditions (8.013 bar abs, 308 K), then A = Q/V and di = √(4A/π). Round up to a standard pipe size.
Source: 18th Exam
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14); Book-3 §3.5 Avoiding Air Leaks — leak quantification (Tables 3.16 & 3.17)
11. S-8: Two 480 CFM screw compressors A & B. Compressor-A runs at full load; Compressor-B runs in load-unload. Load power of both = 74 kW; unload power of B = 26 kW. Both run all working day. B's loading is 64%; after arresting leakage, loading falls to 35%. Estimate energy savings per day.
Model answer: Existing: A = 74 kW. B = 0.64 x 74 + 0.36 x 26 = 47.36 + 9.36 = 56.72 kW. After leakage arrest, B = 0.35 x 74 + 0.65 x 26 = 25.9 + 16.9 = 42.8 kW. Difference (B only) = 56.72 - 42.8 = 13.92 kW. Daily energy savings = 13.92 x 24 = 334 kWh/day.
Confirmed vs Book-3 §3.5 Capacity Control of Compressors — Weighted power of load-unload compressor = load%×load-power + unload%×unload-power; savings = drop in B's power × 24 h.
Source: 19th Exam
📖 Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers (Figure 3.7); Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
12. Two 270 cfm compressors operate at 7 kg/cm2(g), on-load 80% of the time; load power 40 kW and unload power 15 kW each. Cleaning air requirement is 60% of air generated. Calculate the daily energy consumption for cleaning air alone (continuous operation).
Model answer: Air delivered by 2 compressors = 270 x 0.80 x 2 = 432 cfm. Loading power = 40+40 = 80 kW; unloading power = 15+15 = 30 kW. Average kW = [80x(0.8x24) + 30x(0.2x24)]/24 = 70 kW. SEC = 70/432 = 0.162 kW/cfm. Cleaning air = 0.60 x 432 = 259 cfm. Energy for cleaning air/day = 259 x 0.162 x 24 = 1007 kWh/day (alternate method gives ~1008 kWh/day).
Confirmed vs Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers — Printed worked solution: average compressor power 70 kW, SEC 0.162 kW/cfm, ~1007-1008 kWh/day for cleaning air.
13. Calculate the free air delivery (FAD) capacity of a compressor in m3/min for the following data: Receiver capacity 0.5 m3, Initial pressure 0 kg/cm2(g), Final pressure 7 kg/cm2(g), Initial air temperature 32°C, Final air temperature 51°C, Additional holdup volume 0.03 m3, Pump up time 4.5 minutes, Atmospheric pressure 1.026 kg/cm2 absolute.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Standard pump-up FAD test formula correcting for pressure rise, receiver+holdup volume, pump-up time and temperature ratio.
14. A compressed air leakage test was conducted in an industry running 3 nos. of 500 cfm reciprocating compressors, maintained at loading-unloading settings of 6.6 and 7.0 kg/cm2g. Trial 1 (before): On load 30 secs, Unload 110 secs. Trial 2 (after attending leaks): On load 18 secs, Unload 145 secs. Average power was 71 kW during load and 16 kW during unload. Calculate the annual cost savings for 4000 hr/year operation at energy charge Rs. 6.00 per kWh.
Model answer: Leakage (trial 1) = (30×500)/(30+110) = 107 cfm. Leakage (trial 2) = (18×500)/(18+145) = 55 cfm. Specific power consumption = 71/(500×60) = 0.0023666 kW/ft³. Reduction in leakage = 107−55 = 52 cfm = 3120 cfh. Energy saving per hour = 3120 × 0.0023666 = 7.3838 kWh. Annual cost saving = 7.3838 × 4000 × 6 = Rs. 1,77,211 per annum.
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — leak quantification — Leakage quantity from load fraction × FAD; specific power per ft³; multiply leakage reduction by specific power, operating hours and tariff.
15. Calculate the free air delivery (FAD) of a compressor in m3/hr from a pump-up test: Receiver 0.5 m3; initial 0.0 kg/cm2(g); final 7.0 kg/cm2(g); atmospheric 1.026 kg/cm2(a); ambient air 40°C; final compressed air 60°C; additional holdup volume 0.005 m3; pump-up time 5 min 30 s.
Model answer: Q = [(P2 - P1)/P0] x [V/t] x [(273+t1)/(273+t2)]
V = 0.5 + 0.005 = 0.505 m3; t = 5 min 30 s = 5.5 min; P2 - P1 = 7.0 - 0.0 = 7.0 kg/cm2; P0 = 1.026 kg/cm2(a); t1 = 40 degC (ambient), t2 = 60 degC (compressed air).
Q = (7.0/1.026) x (0.505/5.5) x (313/333) = 6.8226 x 0.09182 x 0.9399 = 0.5888 m3/min
FAD = 0.5888 x 60 = 35.33 m3/hr. (2 marks for the formula, 3 marks for the calculation.)
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Q = [(P2-P1)/P0] x [V/t] x [(273+t1)/(273+t2)]; the answer now carries the full working rather than the bare result.
16. A small foundry has installed a reciprocating air compressor of 14.25 m3/min. The plant could not meet the compressed air requirement and hence conducted a capacity test to determine the derating in the compressor capacity. Calculate the actual FAD delivered after considering necessary temperature correction in m3/min, and the percentage derating. Operating parameters: Volume of air receiver including pipe and cooler 9 m3; Atmospheric temperature (T1) 35degC; Receiver temperature (T2) 44degC; Initial Pressure 0.5 kg/cm2(g); Final Pressure 7.0 kg/cm2(g); Atmospheric pressure 1.026 kg/cm2(a); Time taken to build up the pressure 5 minutes.
Model answer: FAD = [(P2 - P1)/Pa] x [(receiver + hold-up volume)/time] x temperature correction factor.
P1 (initial) = 0.5 kg/cm2(g); P2 (final) = 7.0 kg/cm2(g); Pa = 1.026 kg/cm2(a); receiver and holding volume = 9 m3; pump-up time = 5 min.
Uncorrected output = [(7.0 - 0.5) x 9] / (1.026 x 5) = 11.40 m3/min.
Temperature correction factor = (273 + T1)/(273 + T2) = (273 + 35)/(273 + 44) = 308/317 = 0.972.
FAD after correction = 11.40 x 0.972 = 11.08 m3/min.
Capacity shortfall = 14.25 - 11.08 = 3.17 m3/min; % de-rating = (3.17/14.25) x 100 = 22.24%. As this is far above the 10% deviation the book allows, corrective action on the compressor is called for.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD pump-up formula with the temperature correction (273+T1)/(273+T2); the P1/P2 labels in the model answer had been swapped and are now correct (initial 0.5, final 7.0 kg/cm2 g).
Source: Mar 2021 (Set B)
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19); Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
17. A plant has installed a refrigerant dryer for supplying dry air for their process applications and dryer coil is maintained at 5 °C & 100% RH. The average air flow through the dryer is 100 kg/min. The air properties are given below — Inlet at 35 °C & 50% RH: Enthalpy 81 kJ/kg of dry air, Absolute Humidity 18 grams/kg of dry air; Dryer coil 5 °C & 100% RH: Enthalpy 19 kJ/kg of dry air, Absolute Humidity 5.5 grams/kg of dry air. (i) Calculate the moisture removed per hour. (ii) Cooling capacity of coil in TR. (iii) List down any three energy saving measures in compressed air systems.
Model answer: (i) Moisture removed = mass flow x change in absolute humidity = 100 x (18 - 5.5) = 1250 grams/min = 1.25 kg/min = 75 kg/hr.
(ii) Cooling load = 100 x (81 - 19) = 6200 kJ/min = 6200/4.186 = 1481.1 kcal/min = 88,868 kcal/hr; TR = 88,868/3024 = 29.39 TR.
(iii) Three energy saving measures in compressed air systems (Book-3 §3.7 checklist): (1) carry out periodic leak tests and arrest leaks, since 40-50% leakage is not uncommon; (2) reduce the compressor delivery pressure to the minimum the plant needs, saving about 6-10% of power per bar; (3) draw cool, clean intake air from outside the compressor room and clean the inlet filters regularly — every 4°C rise in inlet temperature costs about 1% more power and every 250 mmWC filter pressure drop about 2%. (Also acceptable: eliminate misuse of compressed air, retrofit VSDs to eliminate unloaded running, or size piping generously as a ring main.)
Confirmed vs Book-3 §3.5 Air Dryers — Moisture removed = mass flow x change in absolute humidity; cooling load from the enthalpy difference (1 TR = 3024 kcal/h). Part (iii) now lists three actual measures instead of a page reference.
Source: Jul 2022
📖 Book-3 §3.5 Sizing of Compressed Air Piping
18. In an engineering industry, compressed air delivered is 500 CFM (FAD) and the compressor discharge pressure is 6 kg/cm² (gauge). Calculate the size of the header by considering velocity of compressed air 6 m/s. Assume Temperature remains constant.
Model answer: Quantity of air = 500 CFM = 500/35.31 = 14.16 m3/min of free air.
Working pressure = 6 kg/cm2(g) = 7.013 kg/cm2(a); atmospheric = 1.013 kg/cm2(a).
Applying Boyle's law at constant temperature, P1V1 = P2V2: V2 = 14.16 x 1.013 / 7.013 = 2.05 m3/min = 0.0341 m3/s (the compressed volume actually flowing in the header).
Quantity of air flow = area x velocity, so (pi/4) x D2 x 6 = 0.0341, giving D2 = 0.00724 m2 and D = 0.085 m = 85 mm (about 3.35 inch). A standard 3" NB header would be selected, checking that the velocity stays in the usual 6-10 m/s band.
Confirmed vs Book-3 §3.5 Sizing of Compressed Air Piping — Convert FAD to the compressed volume by Boyle's law, then size from Q = area x velocity at 6 m/s; the garbled area step has been written out, giving D = 85 mm.
Source: Mar 2023
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19)
19. Discuss in brief about working principal Membrane Dryer for Compressed Air.
Model answer: A membrane dryer removes water vapour from compressed air by SELECTIVE PERMEATION of the gas components (Book-3 Figure 3.13).
Construction: the dryer is a cylinder housing thousands of tiny hollow polymer fibres with an inner coating; the coating is selectively permeable to water vapour.
Working: filtered wet compressed air enters the cylinder and passes through the bores of the fibres. The membrane coating allows water vapour to permeate through the fibre wall and collect between the fibres, while the dry air continues through the fibres and leaves at almost the same pressure as the incoming wet air. The permeated water is vented to the atmosphere outside the cylinder. The driving force for the separation is the difference in the partial pressure of the water vapour between the inside and the outside of the hollow fibre.
Features: simple to operate, silent, no moving parts, low power consumption and minimal servicing — mainly the filters upstream of the dryer.
Confirmed vs Book-3 §3.5 Air Dryers — Descriptive answer supplied from the book: selective permeation through coated hollow polymer fibres, driven by the water-vapour partial-pressure difference across the fibre wall.
Source: Mar 2023
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System (end-of-chapter Short question S-4)
20. S-4 List down any 5 energy conservation opportunities in compressed air system.
Model answer: Any five of the following, from the Book-3 §3.7 checklist for energy efficiency in compressed air systems (pp.101-102):
1. Cool, clean, dry intake air — locate the compressor in a well-ventilated area or duct cold air in from outside, since every 4 degC rise in inlet air temperature increases power consumption by about 1% (Table 3.3).
2. Clean the inlet air filters regularly and fit manometers across them — compressor efficiency falls about 2% for every 250 mmWC of pressure drop across a choked filter (Table 3.4).
3. Reduce the delivery pressure to the minimum the plant actually needs and keep the smallest possible range between load and unload settings — a 1 bar reduction saves about 6-10% of input power (Table 3.9).
4. Carry out periodic leak tests and arrest leaks — leakage of 40-50% is not uncommon; % leakage = load time/(load time + unload time) x 100, and a single 3.1 mm orifice at 7 bar wastes about 3 kW (Table 3.17).
5. Minimise low-load and unloaded running — if demand is below 50% of capacity change over to a smaller compressor or reduce speed by trimming the motor pulley, and retrofit variable speed drives on machines above about 100 kW, since unloading consumes up to 30% of full-load power.
Other acceptable answers: avoid misuse of compressed air (body/floor cleaning, agitation, drying) and replace pneumatic conveying (about 8 times more energy) and pneumatic tools (about 20 times more energy) with blowers or electric equipment where safe; inspect compressor valves every six months, as worn valves can cost up to 50% of efficiency; clean inter-coolers and after-coolers periodically; carry out periodic FAD tests and correct deviations beyond 10%; recover heat from hot compressed air; size pipework generously and lay it out as a ring main (0.3 bar drop in the header, 0.5 bar in distribution); sequence multiple compressors so that only one small machine modulates; and fit interlocked solenoid cut-off valves so idle machines draw no air.
Confirmed vs Book-3 §3.7 — this end-of-chapter Short question S-4 was missing from the bank and has been added from the printed book text.
The expected answer is drawn straight from the §3.7 checklist; a full-mark answer states five distinct measures WITH the book's quantified figures (4 degC/1%, 250 mmWC/2%, 1 bar/6-10%, 40-50% leakage, unloading up to 30% of full-load power).
The book does not print a model answer for S-4, so this record is marked unverified.
Source: Book EOC
Long questions (10 marks) — 12
📖 §3.7 Checklist for Energy Efficiency in Compressed Air Systems
1. List five energy efficiency measures for a compressed air system.
Model answer: Five measures from the Book-3 §3.7 checklist (each with the book's own figure):
1) Find and fix air leaks — the book notes leakage of 40–50% is not uncommon; carry out periodic leak tests using the load/unload method (% leakage = T/(T+t) × 100) and ultrasonic leak detectors.
2) Reduce the compressor delivery pressure to the minimum the process actually needs — a 1 bar reduction cuts power by 6–10%; keep the load/unload pressure band as narrow as possible.
3) Ensure a cool, clean, dry air intake — every 4 °C drop in intake temperature saves about 1% power, and clean inlet filters, since every 250 mm WC of filter pressure drop costs about 2% more power.
4) Match capacity to demand — retrofit a variable speed drive on large compressors (over 100 kW) to eliminate unloaded running (unloading can consume up to 30% of full-load power); if several compressors feed a common header, let only one small compressor modulate while the rest run fully loaded.
5) Improve the system — use two-stage/multi-stage machines with effective inter-cooling, liberal pipe sizing in a ring-main layout (design for a maximum 5% pressure drop from compressor to the most distant point of use), and recover the heat of compression.
Other valid points: use blowers instead of compressed air for low-pressure duties; stop misuse for body/floor cleaning; replace pneumatic tools where practical, since the book notes they use about 20 times more energy than motor-driven tools.
Confirmed vs Book-3 §3.7 — all figures quoted (40–50% leakage, 1 bar = 6–10%, 4 °C = 1%, 250 mm WC = 2%, VSD above 100 kW, 5% pressure drop, 20× for pneumatic tools) are the book's own checklist values. Quote the numbers in the exam — the marks are in the figures, not the generalities.
2. A 2-stage reciprocating compressor is supplying nitrogen from low pressure header to high pressure vessel. This high-pressure nitrogen is only used during any process upset. Compressor is cut-off once vessel pressure reaches 45 barg, and started when vessel pressure comes down to 35 barg. During energy audit, it was observed that compressor is started at gap of every 36 hrs when there is no intended consumption. Other data:
Vol. of high pressure N2 vessel: 11.5 m3
Vessel temperature: 35.0 Deg.C
Initial gas density: 50.3 kg/m3
End gas density: 39.4 kg/m3
Compressor load kW drawn: 30.0 kW
Compressor capacity at constant suction pressure: 250.0 kg/hr
i. Estimate the leak rate (kg/hr).
ii. Estimate the energy saving potential (kWh/Annum), if all leaks are attended. Consider operating time of 8760 hrs/annum.
Model answer: Initial Vessel Pressure = 45.0 barg; End vessel pressure = 35.0 barg
Initial gas density = 50.3 kg/m3; End gas density = 39.4 kg/m3
Change in gas quantity in 36 hrs = (50.3 − 39.4) × 11.5 = 125.7 kg
N2 leakage rate = 125.7 / 36 = 3.5 kg/hr
Time needed for compressor run = 125.7 / 250 = 0.51 hrs or 30.6 min
% time of compressor running = 0.51 / 36 = 1.40 %
Running time of compressor per annum = 1.40% × 8760 = 122.3 hrs
Power consumption per annum due to air leakage = 122.3 × 30 = 3670.0 kWh/Annum
Energy Saving Potential = 3670.00 kWh/Annum
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — leak quantification — Leak = (Δdensity × volume)/time; compressor run time to replace leak = leaked mass/capacity; annual energy = % run time × hours × kW.
Source: Sep 2024
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
3. List down any 5 energy conservation opportunities in a compressed air system.
Model answer: 1. Locate compressor intake in a cool, well-ventilated area (every 4°C rise raises power 1%). 2. Clean air-inlet filters regularly (2% efficiency loss per 250 mmWC drop). 3. Inspect/maintain compressor valves every six months. 4. Install manometers across filters to monitor pressure drop. 5. Minimize low-load operation; switch to a smaller compressor or reduce speed if demand <50%. (Also: use regenerative air dryers, clean inter-coolers, periodic FAD tests, optimal multi-compressor sequencing.)
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — Standard compressed air ENCON checklist from BEE guidebook.
4. A pump-up test on a reciprocating compressor gave: receiver + holdup volume 4100 litres, initial pressure 1 kg/cm2(g), final pressure 8.5 kg/cm2(g), atmospheric pressure 1.026 kg/cm2(a), ambient 32 degC, final compressed air temp 52 degC, pump-up time 65 sec. (a) Calculate the FAD in cfm. (b) A leakage test on the same system: on load 3 min, unloaded 13 min, drawing 145 kW on load. Calculate (i) % leakage, (ii) leakage quantity, (iii) specific power consumption, (iv) power lost due to leakage.
Model answer: (a) V = 4100 L = 4.1 m3. FAD = [(P2 - P1)/P_atm] x V / t = [(8.5 - 1.0)/1.026] x 4.1 / (65/60) min = (7.5/1.026) x 4.1 / 1.0833 = 7.310 x 4.1 / 1.0833 = 27.67 m3/min, with temperature correction (273+32)/(273+52)=305/325=0.938 gives ~25.96 m3/min = about 916 cfm. (b) % leakage = load time/(load + unload) x 100 = 3/(3+13) x 100 = 18.75%. (ii) Leakage quantity = 18.75% of FAD ≈ 0.1875 x 25.96 = 4.87 m3/min (≈172 cfm). (iii) Specific power consumption = 145 kW / FAD; using ~25.96 m3/min = 1557 m3/hr, SPC = 145/1557 = 0.093 kW per m3/hr. (iv) Power lost due to leakage = 18.75% of 145 kW = 27.2 kW.
Confirmed vs Book-3 §3.7 Solved Example — FAD from pump-up formula (temperature corrected); leakage % = load/(load+unload); leakage qty and power loss scale by that %; SPC = power/FAD.
5. A free air delivery test was carried out before conducting a leakage test on a reciprocating air compressor in an engineering industry, with the following observations: Receiver capacity 8.0 m3; Initial pressure 0.1 kg/cm2(g); Final pressure 7.0 kg/cm2(g); Additional hold-up volume 0.3 m3; Atmospheric pressure 1.026 kg/cm2 abs; Compressor pump-up time 3.5 minutes. During the subsequent leakage test at lunch time, with no pneumatic equipment or control valves in operation: (a) compressor on-load time 24 seconds at an unloading pressure of 7 kg/cm2(g); (b) average power drawn during loading 92 kW; (c) compressor unload time 79 seconds and loading pressure 6.6 kg/cm2(g). Find: (i) compressor output in m3/hr (neglect temperature correction); (ii) specific power consumption in kW/(m3/hr); (iii) % air leakage in the system; (iv) leakage quantity in m3/hr; (v) power lost due to leakage.
Model answer: Total system volume V = 8.0 + 0.3 = 8.3 m3.
(i) Compressor output Q = [(P2 - P1)/P0] x V/t = [(7.0 - 0.1)/1.026] x (8.3/3.5) = 6.725 x 2.371 = 15.94 m3/min = 956.6 m3/hr (temperature correction neglected as stated).
(ii) Specific power consumption = 92 kW / 956.6 m3/hr = 0.0962 kW per m3/hr.
(iii) % air leakage = T/(T+t) x 100 = 24/(24+79) x 100 = 23.30%.
(iv) Leakage quantity = 0.2330 x 956.6 = 222.9 m3/hr (= 3.71 m3/min).
(v) Power lost due to leakage = leakage quantity x specific power consumption = 222.9 x 0.0962 = 21.4 kW.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — L-2 as printed: FAD = [(P2-P1)/Patm] x (receiver + hold-up)/pump-up time, then % leakage = load/(load+unload), leakage quantity = % x FAD and power lost = leakage quantity x specific power. The stem was restored from the book (pump-up time 3.5 min and the full leakage data were missing) and the item re-typed as Long, per its own source line.
6. a) Pump-up test on a reciprocating compressor: receiver+holdup 4100 litres, initial 1 kg/cm²(g), final 8.5 kg/cm²(g), atmospheric 1.026 kg/cm²(a), ambient 32°C, final compressed air temp 52°C, pump-up time 65 s. Calculate FAD in cfm. b) Leakage test: load 3 min, unload 13 min, drawing 145 kW on load. Find i) % leakage, ii) leakage quantity, iii) specific power consumption, iv) power lost due to leakage.
Model answer: a) FAD Q = [(P2-P1)/Pa] x [V/t] x [(273+t1)/(273+t2)] = [(8.5-1)/1.026] x [4.1/1.0833] x [305/325] = 25.96 m³/min = 25.96 x 3.28³ = 916 cfm. b) i) % leakage = T/(T+t) x 100 = 3/(3+13) x 100 = 18.75%. ii) Leakage quantity = 0.1875 x 916 = 171.75 cfm. iii) Specific power consumption = 145/916 = 0.1583 kW/cfm. iv) Power lost to leakage = 171.75 x 0.1583 = 27.19 kW.
Confirmed vs Book-3 §3.7 Solved Example — FAD from receiver pump-up with temperature correction; leakage from load/unload duty cycle.
Source: 14th Exam
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method) (part a); Book-3 Ch-4 HVAC & Refrigeration — VCR vs VAR (part b)
7. a) Calculate FAD in m³/min: receiver capacity 0.25 m³, initial 1 kg/cm²(g), final 7 kg/cm²(g), initial temp 32°C, final temp 52°C, additional holdup 0.05 m³, pump-up time 2.1 min. b) Identify as VCR or VAR: I) system operates under vacuum; II) uses water as refrigerant; III) uses large amount of high-grade energy; IV) COP decreases considerably with decrease in evaporator pressure; V) can work on lower evaporator pressures without affecting COP.
Model answer: a) Q = [(P2-P1)/Pa] x [V/t] x [(273+t1)/(273+t2)] = [(7-1)/1.026] x [(0.25+0.05)/2.1] x [(273+32)/(273+52)] = 0.784 m³/min. b) I) VAR; II) VAR; III) VCR; IV) VCR; V) VAR.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD from receiver pump-up with temperature correction; VCR/VAR classification.
8. The rated compressor capacity is 15 m3/min. Evaluate if there is any capacity de-rating using the air-receiver tank filling method. Data: Receiver volume (incl. pipe & cooler) = 9 m3; Initial Pressure = 0.5 kg/cm2; Final Pressure = 7.0 kg/cm2; Atmospheric pressure = 1.026 kg/cm2; Time to build pressure = 5 minutes. (b) What is the deficiency in this calculation and how can it be corrected?
Model answer: (a) Compressor output = [(P2 - P1) x V] / (Pa x t) = [(7.0 - 0.5) x 9] / (1.026 x 5) = 58.5/5.13 = 11.40 m3/min.
Capacity shortfall against the rated 15 m3/min = 15 - 11.40 = 3.60 m3/min, i.e. (3.60/15) x 100 = 24% de-rating. Since this far exceeds the 10% the book allows before corrective action, the compressor must be investigated (worn valves alone can cost up to 20% of capacity).
(b) Deficiency: the formula as used assumes the compressed air temperature equals the ambient temperature, i.e. perfect isothermal compression. In practice the discharge/receiver temperature t2 is higher than the ambient t1, so the measured volume is overstated. Correction: multiply the result by the factor (273 + t1)/(273 + t2), where t1 is the ambient/suction temperature and t2 the compressed air (receiver) temperature; this factor is always less than 1.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Tank-filling FAD method Q = (P2-P1) x V/(Pa x t), compared against rated capacity. The model answer's temperature correction had been written inverted; the book's factor is (273+t1)/(273+t2), always less than 1 when the discharge is hotter than ambient.
Source: 17th Sep-2016
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19); Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)
9. L-2: Write short notes on the following with respect to the compressed air system: (a) Refrigeration drier, (b) Heat of compression drier, (c) Role of air receiver, (d) Dew point.
Model answer: (a) Refrigeration dryer: straight mechanical refrigeration in which the dew point of the air is reduced by chilling; a second heat exchanger lets the outgoing cold air pre-cool the incoming compressed air. The achievable atmospheric dew point is about -20 degC. It is the most economical process for roughly 90% of all applications, separates almost 100% of solid particles and water droplets larger than 3 micron, and costs about 0.2 bar in pressure loss (2.9 kW per 1000 m3/hr, Table 3.19).
(b) Heat of compression (HOC) dryer: a twin-tower adsorption dryer in which compressed air taken directly from the compressor discharge before the after-cooler, at about 135 degC for a reciprocating machine, regenerates the desiccant. There are no electrical heaters and no purge loss, so operating cost is zero to very minimal (0.8 kW per 1000 m3/hr) and the atmospheric dew point achieved is -40 degC. Vessel A is in service for 4 hours while vessel B is heated for 2.5 hours and cooled for 1.5 hours, then they change over. Capacities range from 400 to 5000 cfm.
(c) Role of the air receiver: it dampens the pulsations leaving the compressor discharge, acts as a reservoir for sudden or unusually heavy demands in excess of compressor capacity, prevents too frequent loading and unloading (short cycling), and separates moisture and oil vapour by allowing carry-over from the after-cooler to precipitate. Per IS 7938-1976 its volume in m3 should be 1/10th to 1/6th of the output in m3/min, and a local receiver near a point of high cyclic demand avoids having to add compressor capacity.
(d) Dew point: the temperature at which the moisture present in the air starts condensing. The extent of drying is expressed as the ATMOSPHERIC dew point (at atmospheric pressure); the lower the dew point, the drier the air — air at -40 degC atmospheric dew point holds only 80 ppm of moisture against 3800 ppm at 0 degC (Table 3.18). Dryer performance is quoted as PRESSURE dew point, and raising the pressure of a gas raises its dew point temperature, since the partial pressure of the water vapour rises in proportion (Dalton's law).
Confirmed vs Book-3 §3.5 Air Dryers — Short notes from §3.5: the refrigerant dryer's achievable atmospheric dew point is -20 degC (the earlier +3 degC figure contradicted Table 3.19 and has been corrected), the HOC dryer reaches -40 degC at 0.8 kW/1000 m3/hr, plus the receiver's four duties and the atmospheric-vs-pressure dew point distinction.
Source: 19th Exam
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14); Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
10. L-5: A 220 cfm screw compressor: Shift I load 60 s/unload 10 s; Shift II load 45/unload 25; Shift III load 25/unload 45 (8 hrs/shift). Load power 37 kW, unload power 11 kW. Calculate: (1) energy loss per day; (2) shift-wise average air requirement in cfm; (3) energy savings after installing VFD with VFD loss 3% of load power.
Model answer: Shift I = ((60/70)x37 + (10/70)x11) x 8 = (31.71+1.57)x8 = 266.24 kWh. Shift II = (0.64x37 + 0.36x11)x8 = 221.12 kWh. Shift III = (0.36x37 + 0.64x11)x8 = 162.88 kWh. Daily total = 650.24 kWh. Daily unloading (loss) energy = (1.57+3.96+7.04)x8 = 100.56 kWh. Load-cycle energy = 549.68 kWh. With VFD = 549.68/0.97 = 566.68 kWh; VFD loss = 17 kWh; net VFD savings = 100.56 - 17 = 83.56 kWh/day. Air requirement: Shift I = 0.86x220 = 189.2 cfm; Shift II = 0.64x220 = 140.8 cfm; Shift III = 0.36x220 = 79.2 cfm.
Confirmed vs Book-3 §3.5 Capacity Control of Compressors — Printed solution: daily energy loss (unloading) 100.56 kWh; net VFD savings 83.56 kWh/day; shift air 189.2/140.8/79.2 cfm.
11. L-3: A free air delivery test was carried out before a leakage test on a reciprocating air compressor. Receiver capacity = 12 m3; Initial pressure = 0.2 kg/cm2(g); Final pressure = 7.0 kg/cm2(g); Additional hold-up volume = 0.3 m3; Atmospheric pressure = 1.026 kg/cm2(a); Compressor pump-up time = 4.8 minutes. Leakage test (lunch time): (a) on load time 40 s, unloading pressure 7 kg/cm2(g); (b) average power during loading 95 kW; (c) unload time and loading pressure are 90 s and 6.6 kg/cm2(g). Find (i) compressor output m3/hr, (ii) specific power consumption kW/(m3/hr), (iii) % air leakage, (iv) leakage quantity m3/hr, (v) power lost due to leakage.
Model answer: (i) Compressor output = [Total Volume x (P2-P1)/Atm.Pressure] / Pump-up time = [(12+0.3) x (7.0-0.2)/1.026] / 4.8 = [12.3 x 6.8/1.026]/4.8 = 16.9834 m3/minute = 1019 m3/hr. (ii) Specific power consumption = 95/1019 = 0.093228 kW/m3/hr. (iii) % leakage = T/(T+t) x 100 = 40/(40+90) x 100 = 30.77%. (iv) Leakage quantity = 0.3077 x 1019 = 313.54 m3/hr. (v) Power lost due to leakage = leakage quantity x specific power consumption = 313.54 x 0.093228 = 29.23 kW.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD via receiver pump-up formula; leakage % from load/(load+unload) time; power loss = leakage volume x specific power.
Source: 11th Feb-2011
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power); Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
12. L-6: An engineering industry operating three shifts per day has replaced its old reciprocating compressors with 1000 CFM screw compressors. During the energy audit, run hours counter readings: Load Hours start 7956, end 8401 (166 Loading kW); Un-Load Hours start 4918, end 5121 (58.1 un-loading kW). Calculate: 1. Capacity Utilization (%) of the compressor; 2. Monthly energy consumption for the present loading of the compressor; 3. Plant management is considering to install a 750 CFM compressor for energy savings. Estimate the energy savings for the same operating load, if loading power is 125 kW and unloading power is 43.75 kW; 4. To meet the present air requirement, if VFD is to be installed in the 1000 CFM compressor, what should be the percentage reduction in speed.