Reciprocating Compressor

742 ________________________________________________________ Applied Thermodynamics EXAMPLES 1. A reciprocating air compressor has cylinder with 24 cm bore and 36 cm stroke. Compressor admits air at 1 bar, 17°C and compresses it up to 6 bar. Compressor runs at 120 rpm. Considering compressor to be single acting and single stage determine mean effective pressure and the horse power required to run compressor when it compresses following the isothermal process and polytropic process with index of 1.3. Also find isothermal efficiency when compression is of polytropic and adiabatic type. Solution: Compression ratio =

P2 =6=r P1

From cylinder dimensions the stroke volume =

p ´ (0.24)2 ´ (0.36) 4

= 0.01628 m3 Volume of air compressed per minute = 0.01628 ´ 120 = 1.954 m3/min Let us neglect clearance volume. Work done in isothermal process Wiso = P1 V1 ln r Mean effective pressure in isothermal process

p1V1 ln r = P1 ln r V1 = 1 ´ 102 ´ ln 6 = 179.18 kPa

mepiso =

Work done in polytropic process with index n = 1.3, i.e. PV 1.3 = C Wpoly

n = P V n-1 1 1

Mean effective pressure in polytropic process,

LMF P I MNGH P JK 2

n-1 n

1

F n I PV L H n - 1 K MF P I - 1OP = V MNH P K PQ F n I P LMF P I - 1OP = H n - 1 K MNH P K PQ F 1.3 IJ ´ 1 ´ 10 L(6) - 1O = G H 1.3 - 1 K NM QP 1 1

meppoly

meppoly

OP PQ

-1

2

1

n -1 n

1

2

n -1 n

1

1

2

meppoly = 221.89 kPa

1. 3 - 1 1.3

Reciprocating and Rotary Compressor ______________________________________________ 743

Fig. 16.29

p-V representation

LMF P I MNH P K

g Work done in adiabatic process, Wadibatic = P V g -1 1 1

2 1

Mean effective pressure in adiabatic process, mepadiabatic =

g mepadiabatic = P g -1 1 =

LMF P I MNH P K 2

g -1 g

1

OP PQ

Wadiabatic V1

-1

FG 1.4 IJ ´ 1 ´ 10 L(6) NM H 1.4 - 1 K 2

1. 4 - 1 1. 4

mepadiabatic = 233.98 kPa Horse power required for isothermal process, HPiso =

mepiso ´ Volume per minute 0.7457 ´ 60

(As 1 hp = 0.7457 kW)

179.18 ´ 1.954 0.7457 ´ 60 HPiso = 7.825 hp =

Horse power required for polytropic process, HPpoly =

221.89 ´ 1.954 0.7457 ´ 60 = 9.69 hp

= HPpoly

meppoly ´ Volume per minute 0.7457 ´ 60

g -1 g

OP Q

-1

OP PQ

-1

744 ________________________________________________________ Applied Thermodynamics Horse power required for adiabatic process,

mepadiabatic ´ Volume per minute 0.7457 ´ 60

HPadiabatic =

233.98 ´ 1.954 0.7457 ´ 60 = 10.22 hp =

Isothermal process power required Actual power required

Isothermal efficiency =

HPiso 7.825 = = 0.8075 or 80.75% 9.69 HPpoly

Isothermal efficiencypoly =

HPadiabatic 7.825 = = 0.7657 or 76.57% HPpoly 10.22

Isothermal efficiencyadiabatic =

mep: 179.18 kPa for isothermal, 221.89 kPa for polytropic process HP required: 7.825 HP for isothermal, 9.69 HP for polytropic Isothermal efficiency: 80.75% for polytropic process, 76.57% for adiabatic process

Ans.

2. A single stage single acting reciprocating air compressor has air entering at 1 bar, 20°C and compression occurs following polytropic process with index 1.2 upto the delivery pressure of 12 bar. The compressor runs at the speed of 240 rpm and has L/D ratio of 1.8. The compressor has mechanical efficiency of 0.88. Determine the isothermal efficiency and cylinder dimensions. Also find out the rating of drive required to run the compressor which admits 1 m3 of air perminute. Solution: Using perfect gas equation the mass of air delivered per minute can be obtained as, m=

P1V1 RT1

b1 ´ 10 ´ 1g 2

=

( 0.287 ´ 293)

= 1.189 kg/min Compression process follows PV1.2 = constt. Temperature at the end of compression; T2 T2

FPI = T HPK 12 = 293 FH IK 2

1

n -1 n

1

1 T2 = 443.33 K

1. 2 - 1 1. 2

Reciprocating and Rotary Compressor ______________________________________________ 745

FG n IJ ´ mR (T – T ) H n -1K F 1.2 I ´ 1.189 ´ 0.287 (443.33 – 293) H 1.2 - 1 K

Work required during compression process, W = =

2

W = 307.79 kJ/min =

1

307.79 hp 60 ´ 0.7457

W = 6.88 hp

6.88 = 7.82 hp 0.88 Isothermal work required for same compression,

Capacity of drive required to run compressor =

Wiso = m RT1 ln

FP I HPK 2 1

= (1.189 ´ 0.287 ´ 293) × ln

FH 12 IK 1

Wiso = 248.45 kJ/min Isothermal efficiency = Volume of air entering per cycle = Volume of cylinder = = Bore,

D = Stroke length L =

Isothermal work 248.45 = = 0.8072 307.79 Actual work 1 = 4.167 ´ 10–3 m3/cycle 240 p 2 4.167 ´ 10–3 = D L 4 p 4.167 ´ 10–3 = ´ D2 ´ 1.8D 4 0.1434 m or 14.34 cm 1.8 ´ D = 1.8 ´ 14.34 = 25.812 cm

Isothermal efficiency = 80.72 % Cylinder dimension, D = 14.34 cm L = 25.812 cm Rating of drive = 7.82 hp

Ans.

3. A reciprocating compressor of single stage, double acting type delivers 20 m3/min when measured at free air condition of 1 bar, 27°C. The compressor has compression ratio of 7 and the conditions at the end of suction are 0.97 bar, 35°C. Compressor runs at 240 rpm with clearance volume of 5% of swept volume. The L/D ratio is 1.2. Determine the volumetric efficiency and dimensions of cylinder and isothermal efficiency taking the index of compression and expansion as 1.25. Also show the cycle on P-V diagram. Solution: P1 = 0.97 bar, T1 = 273 + 35 = 308 K,

P2 = 7, N = 240 rpm P1

746 ________________________________________________________ Applied Thermodynamics V3 = 0.05 Vs, Vs = V1 – V3, V1 = 1.05 Vs For free air condition of 1 bar, 27°C the mass of air delivered,

1 ´ 10 2 ´ 20 PV = 0.287 ´ 300 RT m = 23.23 kg/min m=

Fig. 16.30 P-V representation

For compression process 1–2, PV 1.25 = Constant T2 = T1

FP I ´ HPK 2

1. 25 - 1 1. 25

1

= 308 ´ (7)0.2 T2 = 454.54 K For expansion process, 3–4, PV 1.25 = Constant V4 = V3

FPI ´ HP K 3

1 1. 25

4

V4 = 0.05 Vs (7)1/1.25 V4 = 0.273 Vs V1 – V4 = 1.05 Vs – 0.237 Vs = 0.813 Vs Volume of air corresponding to the suction conditions can be obtained using the volume of free air delivered

0.97 ´ 300 0.97 ´ 300 ´ (V1 – V4) = ´ 0.813 Vs 1 ´ 308 1 ´ 308 = 0.768 Vs

=

Volumetric efficiency =

0.768 Vs = 0.768 or 76.8% Vs

Volumetric efficiency = 76.8%

Ans.

Reciprocating and Rotary Compressor ______________________________________________ 747 In order to find out the dimension of cylinder the volume of air sucked in one cycle. Volume of air 20 sucked in a cycle = = 0.0208 m3/cycle. 2 ´ 240 Volume per cycle = Swept volume

p 2 p D ×L= D 2 × (1.2 D) 4 4 D = 0.2805 m or 28.05 cm L = 33.66 cm

0.0208 = Þ

Bore = 28.05 cm Stroke = 33.66 cm

Ans.

Work required in reciprocating compressor W= W=

n mR (T2 – T1) n -1

FG 1.25 IJ ´ 23.23 ´ 0.287 (454.54 – 308) H 1.25 - 1 K

= 4884.92 kJ/min

4884.92 , {as 1 hp = 0.7457 kW] 60 ´ 0.7457 W = 109.18 hp =

Work required when compression is isothermal, Wiso = m RT1 ln

FP I HPK 2 1

= 23.23 ´ 0.287 ´ 308 ln (7) = 3995.81 kJ/min

3995.81 60 ´ 0.7457 = 89.31 hp

= Wiso Isothermal efficiency =

Wiso 89.31 = = 0.8133 or 81.33% W 109.81

Isothermal efficiency = 81.33%

Ans.

4. A reciprocating compressor of single stage and double acting type is running at 200 rpm with mechanical efficiency of 85%. Air flows into compressor at the rate of 5 m3/min measured at atmospheric condition of 1.02 bar, 27°C. Compressor has compressed air leaving at 8 bar with compression following polytropic process with index of 1.3. Compressor has clearance volume of 5% of stroke volume. During suction of air from atmosphere into compressor its temperature rises by 10°C. There occurs pressure loss of 0.03 bar during suction and pressure loss of 0.05 bar during discharge passage through valves. Determine the dimensions of cylinder, volumetric efficiency and power input required to drive the compressor if stroke to bore ratio is 1.5.

748 ________________________________________________________ Applied Thermodynamics Solution: Considering the losses at suction and discharge, the actual pressure at suction and delivery shall be as under. Atmospheric pressure, Pa = 1.02 bar, Ta = 273 + 27 = 300 K, Va = 5 m3/min Pressure at suction, P1 = 1.02 – 0.03 = 0.99 bar T1 = 300 + 10 = 310 K Pressure at delivery, P2 = 8 + 0.05 = 8.05 bar Volume corresponding to suction condition of P1, T1, V1 =

Pa × T1 × Va 1.02 ´ 310 ´ 5 = = 5.32 m3/min P1 × Ta 0.99 ´ 300

Therefore, work required for compression,

n W= P V n -1 1 1

LMF P I MNH P K 2

n-1 n

OP PQ

-1

1

F 1.3 IJ ´ 0.99 ´ 10 ´ 5.32 LMF 8.05 I = G H 1.3 - 1 K 60 MNH 0.99 K 2

= 23.66 kW or 31.73 hp Power input required =

Here,

OP PQ

-1

31.73 = 37.33 hp 0.85 Power input = 37.33 hp

Volumetric efficiency,

0 .3 1.3

hvol =

LM MN

F I OP H K PQ

P1Ta P 1+ C - C 2 Pa T1 P1

C = 0.05, so hvol =

Ans. 1 n

0.99 ´ 300 1.02 ´ 310

= 0.7508 or 75.08% Stroke volume per cycle =

5 = 0.0125 m3/cycle 2 ´ 200

Actual stroke volume taking care of volumetric efficiency =

0.0125 0.7508

= 0.0167 m3/cycle Stroke volume =

p 2 D L = 0.0167 4

p 2 D ´ 1.5D = 0.0167 4 D = 0.2420 m or 24.20 cm Stroke L = 1.5 D = 36.3 cm

LM1 + 0.05 - 0.05F 8.05 I H 0.99 K MN

1 1.3

OP PQ

Reciprocating and Rotary Compressor ______________________________________________ 749 Cylinder dimensions = bore of 24.20 cm, stroke of 36.3 cm. Volumetric efficiency = 75.08%

Ans.

5. In a reciprocating air compressor the air is compressed at the rate of 4 m3/min at 1 bar, 27°C up to the pressure of 8 bar following index of compression as 1.2. The compression occurs in two stages with intercooling at optimum intercooler pressure and perfect intercooling. Compare the work input required if the same compression occurs in single stage. Also compare the work input if same compression occurs in two stages with imperfect intercooling up to 30°C at the optimum intercooling pressure. Consider Cp = 1.0032 kJ/kg × K and R = 0.287 kJ/kg × K. Solution: Mass of air compressed per minute; m=

1 ´ 10 2 ´ 4 = 4.65 kg/min 0.287 ´ 300

Fig. 16.31 P-V diagram

Different types of compression are shown on P–V diagram For single stage compression from 1 bar to 8 bar, process 1–2–3, PV 1.2 = Constt.

n Work input, WI = PV n-1 1 1

LMF P I MNH P K 3

OP PQ L8 ´ 4 M FH IK MN 1

n -1 n

1

-1

O F 1.2 IJ ´ 1 ´ 10 = G - 1P H 1.2 - 1 K PQ F 994.113 IJ = 22.22 hp W = 994.113 kJ/min or G H 60 ´ 0.7457 K 2

1. 2 - 1 1.2

I

Optimum intercooling pressure = 8 ´ 1 = 2.83 bar For two stage compression with perfect intercooling; Work input, WII, I/C

n = 2´ PV n -1 1 1

LMF P I MNH P K 3 1

n -1 2n

OP PQ

-1

750 ________________________________________________________ Applied Thermodynamics

F 1.2 IJ ´ 1 ´ 10 = 2´G H 1.2 - 1 K

2

WII,

I/C

L8 ´ 4 ´ M FH IK MN 1

1. 2 - 1 2.4

OP PQ

-1

= 908.19 kJ/min or 20.29 hp

When there is two stage compression with imperfect intercooling: Intercooler pressure, P2 = 2.83 bar, T2¢ = 273 + 30 = 303 K Volume of air at inlet of HP cylinder, V2¢ = V2¢ =

P1V1 T2 ¢ ´ T1 P2

1 ´ 4 ´ 303 = 1.43 m3/min 300 ´ 2.83

Hence, work required for two stage compression, W¢II,I/C = W¢HP + W¢LP

L n R|F P I = M MN n - 1 P V S|TH P K 1.2 I = LM F HN 1.2 - 1 K ´ 1 ´ 10 2

1 1

1

W¢II,I/C

R|F P I U|O L n U|O - 1V P + M - 1VP P V¢ S |WPQ MN n - 1 |TH P K |WPQ ´ 4 {( 2.83) - 1}OP Q LF 1.2 I ´ 2.83 ´ 10 ´ 1.43 ´ R|F 8 I + MG S|H 2.83 K MNH 1.2 - 1 JK T

n -1 n

3

n -1 n

2 2

2

(1. 2 - 1) 1. 2

2

2

1. 2 - 1 1.2

U|OP V|P WQ

-1

= [454.3] + [459.2] = 913.5 kJ/min or 20.42 hp % saving in work when compression occur with perfect intercooling as compared to single stage compression = 100 ´

FG W - W H W I

II , I / C

I

IJ K

= 8.69% Ans. % excess work to be done when two stage compression occurs with imperfect intercooling as compared to two stage compression with perfect intercooling: = 100 ´

F W¢ GH

II , I / C

- WII , I / C

W ¢ II , I / C

I JK

= 0.636% Ans. 6. A reciprocating air compressor has four stage compression with 2 m3/min of air being delivered at 150 bar when initial pressure and temperature are 1 bar, 27°C. Compression occur polytropically following polytropic index of 1.25 in four stages with perfect intercooling between stages. For the optimum intercooling conditions determine the intermediate pressures and the work required for driving compressor. Solution: Here there is four stage compression with perfect intercooling at optimum intercooling conditions. So optimum stage pressure ratio = (150 )1/ 4 = 3.499 » 3.5

Reciprocating and Rotary Compressor ______________________________________________ 751 Intermediate pressure shall be as follows: Between Ist and IInd stage = 3.5 bar Between IInd and IIIrd stage = 12.25 bar Between IIIrd and IVth stage = 42.87 bar Intermediate pressure: 3.5 bar, 12.25 bar, 42.87 bar

Ans.

Since it is perfect intercooling so temperature at inlet of each stage will be 300 k. So temperature at the end of fourth stage, T = T1

FPI ´ HPK 2 1

n -1 n

F 1. 25 - 1 IK

= 300 ´ ( 3.5) H T = 385.42 K Mass of air, kg/min, m =

1. 25

150 ´ 10 2 ´ 2 PV = = 271.21 kg/min 0.287 ´ 385.42 RT

Work required for driving compressor,

F n IJ m RT LMF P I - 1OP ´ 4 W= G H n -1K MNH P K PQ F 1.25 IJ ´ 271.21 ´ 0.287 ´ 300 L(3.5) =G - 1O ´ 4 H 1.25 - 1 K QP NM 132978.04 I = 132978.04 kJ/min or 2972.11 hp F = H 60 ´ 0.7457 K 2

1

n-1 n

1

(1.25 - 1) 1. 25

.

Work input = 2972.11 hp

Ans.

7. A two stage reciprocating compressor has air entering at 1 bar, 17°C into LP compressor and leaving HP stage at 16 bar. An intercooler working at 4 bar pressure is provided between the HP and LP stages. The compression process follows the process given by PV1.3 = constant. The bore diameters of HP and LP cylinder are 6 cm and 12 cm respectively while stroke lengths are equal. For perfect intercooling determine the work done in compressing per unit mass of air. Also state whether the intercooler pressure will rise, fall or no change if the volumetric efficiency is taken as 0.90 for LP cylinder. Neglect clearance volumes. Solution: For perfect intercooling the amount of work required shall be;

n´m RT1 W= n -1

LMF P I MNGH P JK 2 1

n -1 n

FPI +G J HP K 3

n -1 n

2

LM MN

-2

OP PQ

e j

(1.3 - 1) 1.3 ´ 1 16 = ´ 0.287 ´ (273 + 17) ( 4 ) 1. 3 + 4 1.3 - 1

1. 3 - 1 1.3

-2

OP PQ

752 ________________________________________________________ Applied Thermodynamics = 271.94 kJ/kg .

Work = 271.94 kJ/kg

Ans.

The ratio of volumes handled by HP and LP stages. Using the perfect intercooling.

Fig. 16.32 Two stage compression on P-V diagram

State at inlet of LP cylinder = P1 V1 State at inlet of HP cylinder = P2 V2¢ Since state 2¢ lies on isothermal compression process line so, P1 V1 = P2 V2¢ or,

V1 = V2¢

P2 P1

=

FH 4 IK = 4 1

Volume at inlet to LP = 4 Volume at inlet to HP From given cylinder dimensions, for given bore diameters and common stroke length, Ratio of effective cylinder volumes =

Effective volume of LP cylinder Effective volume of HP cylinder p 2 D ×L 4 LP p 2 D ×L 4 HP

0.90 ´ =

0 .9 ´ ( 0.12 ) 2 = 3.6 ( 0.06 ) 2 Theoretically, the volume ratio is 4 while considering volumetric efficiency the ratio of effective cylinder volumes comes out to be 3.6 which is less than the theoretical volume ratio. Therefore, it can be concluded that less amount of air is given into HP cylinder than its’ capacity. Thus, the HP cylinder would inhale volume equal to its’ capacity and since it has larger capacity than volume available so the pressure of intercooler shall drop. 8. In a two stage reciprocating air compressor running at 200 rpm the air is admitted at 1 bar, 17°C and discharged at 25 bar. At low pressure stage suction conditions the rate of air flow is 4 kg/minute. The low pressure cylinder and high pressure cylinders have clearance volumes of 4% and 5% of respective cylinder stroke volumes. The index for compression and expansion processes in two stages are same

=

Reciprocating and Rotary Compressor ______________________________________________ 753 as 1.25. Considering an optimum and perfect intercooling in between two stages determine the power required, isothermal efficiency, free air delivered, heat transferred in each cylinder and the cylinder volumes. Solution: For the optimum intercooling the pressure ratio in each stage =

25 =5 1

Fig. 16.33 P-V representation

P2 P = 6 =5 P1 P5 Perfect intercooling indicates, T1 = T5 = 273 + 17 = 290 K T2 = T1

FG P IJ HPK FG P IJ HPK 2

n -1 n

= 400.12 K

1

T6 = T5

6

n -1 n

= 400.12 K

5

Actual compression work requirement, W = WHP + WLP

F n IJ m RT LMFG P IJ - 1OP = 2G H n -1K MNH P K PQ F 1.25 IJ ´ 4 ´ 0.287 ´ 290 L(5) = 2´G MN H 1.25 - 1 K 2

n-1 n

1

1

(1.25 - 1) 1. 25

W = 1264.19 kJ/min or 28.25 hp Work requirement if the process is isothermal compression, Wiso = m RT1 ln

FG P IJ = 4 ´ 0.287 ´ 290 ln (25) HPK 6 1

Wiso = 1071.63 kJ/min

OP Q

-1

754 ________________________________________________________ Applied Thermodynamics Wiso = 0.8477 or 84.77% W mRT1 4 ´ 0.287 ´ 290 = = 3.33 m3/min Free air delivered = 1 ´ 10 2 P1

Isothermal efficiency =

Heat transferred in HP cylinder = Heat transferred in LP cylinder = Q (Due to optimum and perfect intercooling)

FH W IK – m C (T – T ) 2 1264.19 I = FH K – 4 ´ 1.0032 ´ (400.12 – 290)

Q=

p

2

1

2 Q = 190.21 kJ/min

LM MN

F I OP H K PQ 1 n

PT P Volumetric efficiency, hvol = 1 a 1 + C - C 2 Pa × T1 P1

Here the ambient conditions and suction conditions are same so expression gets modified as, hvol

FP I = 1+C–CG J HPK 2

1 n

1

Volumetric efficiency of HP, hvol,

HP

= 1 + CHP – CHP

FG P IJ HP K 6

1 n

5

CHP = 0.04 = 1 + 0.04 – 0.04 (5)1/1.25 hvol, HP = 0.895 or 89.5% Volumetric efficiency of LP, hvol,

LP

= 1 + CLP – CLP

FG P IJ HPK 2

1 n

1

CLP = 0.05 = 1 + 0.05 – 0.05 (5)1/1.25 = 0.8688 or 86.88% Stroke volume of HP cylinder = Vs,

HP

=

Free air delivery Pressure ratio ´ speed ´ h vol, HP

3.33 = 3.721 ´ 10–3 m3 5 ´ 200 ´ 0.895

Clearance volume, Vc, HP = 0.05 ´ 3.721 ´ 10–3 = 1.861 ´ 10–4 m3 Total HP cylinder volume, VHP = Vs, HP + Vc, HP = 3.907 ´ 10–3 m3 ; Vc, HP = Clearance volume of HP Stroke volume of LP cylinder =

Free air delivery 3.33 = Speed ´ h vol, LP 200 ´ 0.8688

Reciprocating and Rotary Compressor ______________________________________________ 755 Vs, LP = 0.01916 m3 Clearance volume, Vc, LP = 0.04 ´ Vs, LP = 7.664 ´ 10–4 m3 Total LP cylinder volume, VLP = Vs, LP + Vc, LP = 0.019926 m3 Power required = 28.25 hp, Isothermal efficiency = 84.77% Free air delivered = 3.33 m3/min, Heat transfer in HP cylinder = 190.21 kJ/min Heat transferred in LP cylinder = 190.21 kJ/min, HP cylinder volume = 3.907 ´ 10–3 m3 LP cylinder volume = 0.019926 m3

Ans.

9. A two stage double acting reciprocating air compressor running at 200 rpm has air entering at 1 bar, 25°C. The low pressure stage discharges air at optimum intercooling pressure into intercooler after which it enters at 2.9 bar, 25°C into high pressure stage. Compressed air leaves HP stage at 9 bar. The LP cylinder and HP cylinder have same stroke lengths and equal clearance volumes of 5% of respective cylinder swept volumes. Bore of LP cylinder is 30 cm and stroke is 40 cm. Index of compression for both stages may be taken as 1.2. Determine, (i) the heat rejected in intercooler, (ii) the bore of HP cylinder, (iii) the hp required to drive the HP cylinder. Solution: Optimum intercooling pressure = 9 = 3 bar LP stage pressure ratio = HP stage pressure ratio = 3 From the given dimensions of LP cylinder, the volume of LP cylinder, in m3/min

p ´ (0.30)2 ´ (0.40) ´ 200 ´ 2 4 = 11.31 m3/min

VLP = VLP

Volumetric efficiency of LP compressor, here ambient and suction conditions are same, hvol, LP

FPI = 1+C–CG J HPK 2

1 n

1

hvol, LP = 0.9251 or 92.51%

F 3 I = 1 + 0.05 – G 0.05 FH IK J H 1 K

Fig. 16.34 P-V diagram

1 1. 2

756 ________________________________________________________ Applied Thermodynamics Volume of air inhaled in LP stage = VLP ´ hvol, LP = 11.31 ´ 0.9251 = 10.46 m3/min Mass of air per minute,

m= =

P1V1 RT1 1 ´ 10 2 ´ 10.46 = 12.23 kg/min 0.287 ´ 298

Temperature after compression in LP stage, T2

FPI = T ´G J HPK 3 = 298 ´ FH IK 2

n -1 n

1

1

1. 2 - 1 1. 2

1 T2 = 357.88 K

Volume of air going into HP cylinder

mRT5 P5 After intercooling, T5 = 298 K, P5 = 2.9 bar, V5 =

12.23 ´ 0.287 ´ 298 2.9 ´ 10 2 V5 = 3.61 m3/min

V5 =

Since the clearance volume fraction and pressure ratio for both HP and LP stages are same so the volumetric efficiency of HP stage referred to LP stage suction condition shall be same hvol, HP = hvol, LP = 0.9251 Hence, the volume of HP cylinder/min =

V5 h vol,HP

=

3.61 = 3.902 m3/min 0.9251

Let bore of HP cylinder be DHP,

p ´ (DHP)2 ´ 0.40 ´ 2 ´ 200 4 DHP = 0.1762 m or 17.62 cm Heat rejected in intercooler, Q = mCp (T2 – T5) = 12.23 ´ 1.0032 ´ (357.88 – 298) = 734.68 kJ/min 3.902 =

In HP stage,

FG IJ H K

n -1

T6 P6 n = Þ T6 = 298 ´ P5 T5 T6 = 359.91 K

FH 9 IK 2.9

1. 2 - 1 1.2

Reciprocating and Rotary Compressor ______________________________________________ 757 Work input required for HP stage, WHP = =

or

FG n IJ mR (T H n -1K

6

– T5)

FG 1.2 IJ ´ 12.23 ´ 0.287 × (359.9 – 298) H 1.2 - 1 K

WHP = 1303.62 kJ/min WHP = 29.14 hp Heat rejected in intercooler = 734.68 kJ/min Bore of HP cylinder = 17.62 cm Horse power required to drive HP stage = 29.14 hp

Ans.

10. During an experiment on reciprocating air compressor the following observations are being taken; Barometer reading = 75.6 cm Hg, Manometer reading across orifice = 13 cm Hg. Atmospheric temperature = 25°C. Diameter of orifice = 15 mm. Coefficient of discharge across the orifice = 0.65 Take density of Hg = 0.0135951 kg/cm3 Determine the volume of free air handled by compressor in m3/min. Solution:

p ´ (15 ´ 10–3)2 = 1.77 ´ 10–4 m2 4 Atmospheric pressure = 75.6 ´ 0.0135951 ´ 9.81 ´ 104 ´ 10–3 = 100.83 kPa

Cross-sectional area of orifice,

A=

Specific volume of air per kg at atmospheric conditions, v=

1 ´ 0.287 ´ 298 RT = = 0.848 m3/kg 100.83 P

Density of air = 1 = 1.18 kg/m3 v

Pressure difference across orifice = 13 ´ 0.0135951 ´ 9.81 ´ 104 ´ 10–3 = 17.34 kPa Height of air column for pressure difference across orifice. ra ´ ha ´ g = 17.34 ´ 103 Put, ra = 1.18 kg/m3 Þ ha = 1497.95 m Free air delivery = Cd × A ×

2gha

= 0.65 ´ 1.77 ´ 10–4 2 ´ 9.81 ´ 1497.95 = 0.01972 m3/s or 1.183 m3/min .

Free air delivery = 1.183 m3/min

Ans.

11. During a trial on single acting single stage compression the following observations are made; Dimensions of cylinder: 10 cm bore and 8 cm stroke. Speed of rotation: 500 rpm.

758 ________________________________________________________ Applied Thermodynamics Barometer reading: 76 cm Hg Atmospheric temperature: 27°C Delivery air temperature = 130°C Free air delivery = 15 m3/hr Spring balance of dynamometer type (electric motor) reading: 10 kg Radius of arm of spring balance: 30 cm Take mechanical efficiency = 0.90. Determine the volumetric efficiency, shaft output per m3 of free air per minute. Solution: Free air delivery = 15 m3/hr = 0.25 m3/min

p ´ (0.10)2 ´ (0.08) = 6.28 ´ 10–4 m3 4 (15 60) Volumetric efficiency = = 0.7962 or 79.62% Ans. 6.28 ´ 10 - 4 ´ 500 Volume of cylinder =

b

Shaft output =

2p NT 60

g

2 ´ p ´ 500 ´ 10 ´ 9.81 ´ 0.30 ´ 10 - 3 60 = 15.41 kJ/s or 20.66 hp

Shaft output =

20.66 0.25 = 82.64 hp per m3 of free air per minute. Ans. 12. Determine the minimum number of stages required in an air compressor which admits air at 1 bar, 27°C and delivers at 180 bar. The maximum discharge temperature at any stage is limited to 150°C. Consider the index for polytropic compression as 1.25 and perfect and optimum intercooling in between the stages. Neglect the effect of clearance. Solution: Let there be ‘i’ number of stages. So the overall pressure ratio considering inlet state as Pa and Ta and delivery state pressure as Pi.

Shaft output per m3 of free air per minute =

Pi P P P P = 1 × 2 × 3 × ⋅⋅⋅× i Pa P1 P2 Pi − 1 Pa When perfect and optimum intercooling is considered then pressure ratio in each stage will be same.

P P1 P P = 2 = 3 =¼= i =r Pa P1 P2 Pi - 1 Pi = (r)i, for any stage, say second stage, Pa T1 = 273 + 27 = 300 K

Reciprocating and Rotary Compressor ______________________________________________ 759 and

T2 = 273 + 150 = 423 K

FG T IJ HT K FH 180 IK = FH 423 IK P2 = P1

and

1

2

n n -1

1

i ´ 1. 25 1. 25 - 1

300

ln 180 =

P , i = (r)i = Pa

FG T IJ HT K 2

in n -1

1

, Taking log for solving,

FH 1.25i IK ln FH 423 IK 0.25 300

Upon solving, i = 3.022 » say 3 stages 3 stages

Ans.

13. In a triple stage reciprocating compressor of single acting type the air enters at 1 bar, 27°C. The compressor has low pressure cylinder with bore of 30 cm and stroke of 20 cm. Clearance volume of LP cylinder is 4% of the swept volume. The final discharge from compressor takes place at 20 bar. The expansion and compression index may be taken uniformly as 1.25 for all the stages. The intercooling between the stages may be considered to be at optimum intercooling pressure and perfect intercooling. Determine, the interstage pressures, effective swept volume of low pressure cylinder, temperature and volume of air delivered in each stroke and the work done per kg of air. Solution: Here P1 = 1 bar, T1 = 300 K, C = 0.04, P10 = 20 bar, n = 1.25, See Fig. 16.35 For optimum and perfect intercooling,

P2 P P = 6 = 10 = P1 P2 P6

FH 20 IK 1

1 3

= 2.714

P2 = 2.714 bar, T5 = T1 = 300 K P6 = 7.366 bar T9 = T1 = 300 K Volumetric efficiency of LP stage, hvol, LP

FP I = 1+C–CG J HPK 2

1 1. 25

1

= 1 + 0.04 – 0.04 (2.714)1/1.25 = 0.9511 or 95.11%

p 2 p D L= ´ (0.30)2 ´ 0.20 = 0.01414 m3 4 4 Effective swept volume of LP cylinder, V1 – V4 = hvol, LP ´ (V1 – V3) LP swept volume, V1 – V3 =

V1 – V4 = 0.9511 ´ 0.01414 = 0.01345 m3 Temperature of air delivered, T10 = T9

F P IJ ´G HPK 10

1. 25 - 1 1. 25

= 300 ´ ( 2.714 )

6

= 366.31 K

(1.25 - 1) 1. 25

760 ________________________________________________________ Applied Thermodynamics

Fig. 16.35 P-V representation

For the compression process of air as perfect gas;

a

P1 ´ V1 - V4 T1

f=

V10 – V11 = =

a

P10 ´ V10 - V11 T10

aV - V f ´ T 1

4

10

f

´ P1

T1 ´ P10

0.01345 ´ 366.31 ´ 1 ´ 10 2 300 ´ 20 ´ 10 2

Volume of air delivered = V10 – V11 = 8.2115 ´ 10–4 m3 Total Work done per kg air,

R| n F F P I I U| W= 3 ´ S - 1J V RT G G J JK |W |T n - 1 GH H P K 1.25 I = 3 ´ RSF HT 1.25 - 1 K ´ 0.287 ´ 300 ´ FH (2.714) 2

1. 25 - 1 1.25

1

1

(1. 25 - 1) 1.25

= 285.44 kJ/kg of air

Intermediate pressure = 2.714 bar, 7.366 bar Effective swept volume of LP cylinder = 0.01345 m3 Temperature of air delivered = 366.31 K Volume of air delivered = 8.2115 ´ 10–4 m3 Work done = 285.44 kJ/kg of air

IK UV W

-1

Ans.

14. A two stage reciprocating air compressor has air being admitted at 1 bar, 27°C and delivered at 30 bar, 150°C with interstage pressure of 6 bar and intercooling up to 35°C. Compressor delivers at the rate of 2 kg/s. Clearance volumes of LP and HP cylinders are 5% and 7% of stroke volume respectively. The index of compression and expansion are same throughout. Determine the swept volume of both cylinders in m3/min, amount of cooling required in intercooler and total power required. Also estimate the amount of cooling required in each cylinder.

Reciprocating and Rotary Compressor ______________________________________________ 761 Solution: Given: P1 = 1 bar, T1 = 300 K, P2 = 6 bar, P6 = 30 bar, T6 = 273 + 150 = 423 K, T5 = 273 + 35 = 308 K, CLP = 0.05, CHP = 0.07, m = 2 kg/s

Fig. 16.36 P-V representation

For process 5–6, P2 = P5

FG T IJ HT K FH 30 IK = FH 423 IK P6 = P5

Þ

6 Taking log of both sides,

Þ

6

n n -1

5

n n -1

308

ln (5) =

n ln (1.3734) n -1

Upon solving we get, n = 1.245 Volumetric efficiency of LP cylinder, hvol, LP

FP I = 1+C –C G J HPK 6 = 1 + 0.05 – 0.05 FH IK

1 1. 245

2

LP

LP

1

1 1. 245

1 = 0.8391 or 83.91%

Volumetric efficiency of HP cylinder, hvol, HP

FPI = 1+C –C G J HPK 30 = 1 + 0.07 – 0.07 F H IK 6

HP

1 1. 245

HP

5

6

hvol, HP = 0.815 or 81.50%

1 1. 245

762 ________________________________________________________ Applied Thermodynamics For suction of LP cylinder P1 × (V1 – V4) = mRT1 (V1 – V4) = hvol, LP = =

2 ´ 0.287 ´ 300 = 1.722 m3/s or 103.32 m3/min 1 ´ 10 2 V1 - V4 = 0.8391 Þ (V1 – V3) V1 - V3 103.32 = 123.13 m3/min 0.8391

Swept volume of LP cylinder = 123.13 m3/min

Ans.

For HP cylinder, P2 ´ (V5 – V8) = m RT5 Þ

(V5 – V8) = = hvol, HP =

mRT5 P2 2 ´ 0.287 ´ 308 = 0.2946 m3/s or 17.676 m3/min 6 ´ 10 2 V5 - V8 17.676 Þ (V6 – V7) = = 21.69 m3/min V6 - V7 0.815

Swept volume of HP cylinder = 21.69 m3/min

Ans.

For compression in LP stage,

P  T2 = T1  2   P1 

n −1 n

6 = 300 ´ FH IK 1

1. 245 - 1 1. 245

T2 = 426.83 K Cooling required in intercooler, QI/C = m ´ Cp ´ (T2 – T5) = 2 ´ 1.0032 ´ (426.83 – 308) QI/C = 238.42 kJ/s Heat picked in intercooler = 238.42 kW

Ans.

Work input required = WLP + WHP

LMF P I MNGH P JK L R|F P I n mR M T S MN T|H P K n -1

n = m RT1 n -1

=

2

n-1 n

5

1

2

1

1

O n - 1P + PQ n - 1 m RT U| R|F P I - 1V + T SG J W| |TH P K

n-1 n

6

5

5

n -1 n

LMF P I MNGH P JK U|O - 1V P |WPQ 6 5

n -1 n

OP PQ

-1

Reciprocating and Rotary Compressor ______________________________________________ 763 =

FG 1.245 IJ ´ 2 ´ 0.287 LM300R|SF 6 I H 1.245 - 1 K MN |TH 1 K

1. 245 - 1 1.245

= 704.71 kJ/s

Total work required = 704.71 kW

U| R| 30 I - 1V + 308 SFH |W |T 6 K

1.245 - 1 1. 245

U|OP V|P WQ

-1

Ans.

Heat transferred in LP cylinder = Amount of cooling required in LP cylinder

F g - n I ´ C ´ (T – T ) H n -1 K F 1.4 - 1.245 IJ ´ 0.72 ´ (426.83 – 300) = 2´G H 1.245 - 1 K

QLP = m

v

2

1

= 115.55 kJ/s .

Amount of cooling required in LP cylinder = 115.55 kW

Ans.

Heat transferred in HP cylinder = Amount of cooling required in HP cylinder

F g - n I ´ C ´ (T – T ) H n -1 K F 1.4 - 1.245 IJ ´ 0.72 ´ (423 – 308) = 2´G H 1.245 - 1 K

QHP = m

v

6

5

= 104.77 kJ/s Amount of cooling required in HP cylinder = 104.77 kW

Ans.

15. A roots blower handles free air of 0.5 m3/s at 1 bar and 27°C and delivers air at pressure of 2 bar. Determine indicated power required to drive compressor and isentropic efficiency. Solution: Indicated power required = (P2 – P1) ´ V1 Wroots = (2 – 1) ´ 102 ´ 0.5 = 50 kJ/s Wroots = 50 kW or 67.05 hp Indicated power when isentropic compression occurs, Wisentropic

g = ´ P1 V1 g -1 =

R|F P I S|H P K T 2 1

FG 1.4 IJ ´ 1 ´ 10 H 1.4 - 1 K

2

U| V| W R| 2 ´ 0.5 SFH IK |T 1

= 38.33 kW or 51.4 hp Fig. 16.37

( g - 1) g

-1

1. 4 - 1 1. 4

U| V| W

-1

764 ________________________________________________________ Applied Thermodynamics Isentropic efficiency of roots blower =

Wisentropic Wroots

38.33 = 0.7666 or 76.66% 50

=

Indicated power of roots blower = 67.05 hp, Isentropic efficiency = 76.66%

Ans.

16. A vaned compressor handles free air of 0.6 m3/s at 1 bar and compresses up to 2.3 bar. There occurs 30% reduction in volume before the back flow occurs. Determine the indicated power required and isentropic efficiency. Solution: Here on P–V diagram the state 2 indicates the point at which delivery occurs. While 2¢ is the point up to which air is compressed inside. V2 = 0.7 ´ V1

F V I = FG V IJ H V K H 0.7 V K 1 I = 1 ´ FH = 1.65 bar 0. 7 K g

P2 = P1

1. 4

1

1 2

1

1. 4

Þ

P2

Fig. 16.38 P-V diagram

Indicated power required for vaned compressor Wvane

F g I P V R|SF P I - 1U|V + (P = H g - 1K |TH P K |W F 1.4 IJ ´ 1 ´ 10 ´ 0.6 {(1.65) =G H 1.4 - 1 K 2

1

1

g -1 g

2¢

1

(1. 4 - 1) 1. 4

2

– P2) ´ V2

}

- 1 + (2.3 – 1.65) ´ 102 ´ (0.7 ´ 0.6)

= 59.60 kJ/s or 79.925 hp Power requirement when compression occurs isentropically,

g Wisentropic = P V g -1 1 1

R|F P I S|H P K T 2¢ 1

g -1 g

U| V| W

-1

Reciprocating and Rotary Compressor ______________________________________________ 765

F 1.4 IJ ´ 1 ´ 10 =G H 1.4 - 1 K

2

R| 2.3 I ´ 0.6 SFH |T 1 K

1. 4 - 1 1. 4

= 56.42 kJ/s or 75.66 hp

U| V| W

-1

Wisentropic 75.66 = Wvane 79.925

Isentropic efficiency of Vane compressor = = 0.9466 or 94.66%

Indicated power required = 79.925 hp, isentropic efficiency = 94.66%

Ans.

17. A centrifugal compressor delivers free air of 18 kg/min. Air is sucked at static states of 1 bar, 27°C with inlet velocity of 50 m/s. The total head pressure ratio is 4 and isentropic efficiency of compressor is 0.75. The mechanical efficiency of motor attached to it is 0.90. Determine total head temperature of air at exit of compressor and brake power required to drive compressor. Solution: Stagnation temperature at inlet, T01= T1 +

V12 2Cp

T01 = 300 +

( 50 ) 2

2 ´ 1.0032 ´ 10 3

= 301.25 K

Isentropic efficiency of compressor, hisen =

T02 ¢ - T01 T02 - T01

T02 ¢ = T01

FG P IJ HP K

For process 1–2,

Þ

02

g -1 g

01

T02¢ = T01 ´ ( 4 )

(1. 4 - 1 ) 1. 4

T02¢ = 301.25 ´ ( 4 ) 0 . 4/1. 4 = 447.66 K

Fig. 16.39

Substituting temperature values in expression of isentropic efficiency,

766 ________________________________________________________ Applied Thermodynamics T02 =

aT

02 ¢

- T01

h isen

f +T

01

( 447.66 - 301.25) + 301.25 0.75 = 496.46 K

= T02

Total head temperature at exit = 496.46 K Brake power of drive required

=

Ans.

a

m ´ C p ´ T02 - T01 h mech

f

18 ´ 1.0032 ´ ( 496.46 - 301.25) 60 ´ 0.9 = 65.28 kW or 87.54 hp =

Brake power required = 87.54 hp

Ans.

18. A double-acting single cylinder reciprocating air compressor has a piston displacement of 0.015 m3 per revolution, operates at 500 r.p.m. and has a 5% clearance. The air is received at 1 bar and delivered at 6 bar. The compression and expansion are polytropic with n = 1.3. Determine, (i) the volumetric efficiency (ii) the power required (iii) the heat transferred and its direction, during compression if inlet temperature of air is 20°C. [U.P.S.C., 1998] Solution: V = 0.015 m3 per revolution, N = 500 r.p.m., C = 5% or 0.05, n = 1.3, P1 = 1 bar, P2 = 6 bar, T1 = 20°C Volumetric efficiency, hvol

FP I =1+C–CG J HPK 2

1 n

1

6 = 1 + 0.05 – 0.05 FH IK 1

1 1.3

= 0.8516 or 85.16% Ans.

Volumetric efficiency = 85.16%

n Power required = P1V1 n-1 Here swept volume in cylinder = = Vs = Actual air inhaled = V1 = V1 =

LMF P I MNGH P JK 2 1

Vs = V ´ 2N 0.015 ´ 2 ´ 500 15 m3/min 15 ´ 0.85 12.75 m3/min

n-1 n

OP PQ

-1

Reciprocating and Rotary Compressor ______________________________________________ 767 Mass of air entering, m =

P1V1 1 ´ 10 2 ´ 12.75 = = 15.16 kg/min. 0.287 ´ 293 RT1

Power required =

(1.3)

(1.3 - 1)

´1´

102

= 2829.21 kJ/min

L6 ´ 12.75 M F I MNH 1 K

1.3 - 1 1.3

OP PQ

-1

Ans. Power required = 2829.21 kJ/min Heat transferred during compression, (this is heat rejected), for a polytropic process Q = m Cv × During compression process

T2 = T1

FG P IJ HPK 2 1

T2 = 298 ´

F g - n I (T H n -1 K

2

– T1)

n -1 n

FH 6 IK 1

1.3 - 1 1. 3

T2 = 450.59 K Substituting in heat transferred Q, Q = 15.16 ´ 0.718 ´ Q = 571.78 kJ/min

FG 1.4 - 1.3 IJ (450.59 – 293) H 1.3 - 1 K

Heat rejected during compression = 571.78 kJ/min

Ans.

-:-4+1516.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9

Classify the compressors. Discuss the applications of compressed air to highlight the significance of compressors. Describe the working of single stage reciprocating compressor. Discuss the indicator diagram for reciprocating compressor. Also describe the factors responsible for deviation of hypothetical indicator diagram to actual indicator diagram. Obtain the volumetric efficiency of single stage reciprocating compressor with clearance volume and without clearance volume. Discuss the effects of clearance upon the performance of reciprocating compressor. Define isothermal efficiency. Also discuss its significance. What do you understand by multistage compression? What are its’ merits over single stage compression? Show that the volumetric efficiency with respect to free air delivery is given by, hvol. =

1   P1T a   P n 1+ C − C  2    Pa ⋅ T1   P1    

where all the terms of expression have their usual meanings.

768 ________________________________________________________ Applied Thermodynamics 16.10 Discuss the significance of intercooling upon the performance of multi-stage compression. 16.11 What is the optimum pressure ratio for perfect intercooling in between two stages of compression? The inlet and outlet pressures may be taken as P1 and P3. 16.12 Discuss the control of reciprocating air compressor. 16.13 Discuss the working of positive displacement rotary compressors. 16.14 Describe the working of centrifugal compressors. 16.15 What do you understand by surging and choking phenomenon? 16.16 Explain the stalling and its effect on the compressor performance. 16.17 Describe the characteristics of centrifugal compressor. 16.18 Compare the axial flow compressor with centrifugal compressors. 16.19 Show that the heat rejected in each stage of a reciprocating compressor with perfect intercooling is given by,

16.20

16.21

16.22

16.23

16.24

16.25

16.26

  γ − n  Q =  C p + Cv    (T2 – T1)  n − 1   Write short notes on the following: (i) Free air delivery (ii) Volumetric efficiency (iii) Axial flow compressors (iv) Air flow rate measurement in reciprocating compressors. A single stage single cylinder reciprocating compressor has 60 m3/hr air entering at 1.013 bar, 15°C and air leaves at 7 bar. Compression follows polytropic process with index of 1.35. Considering negligible clearance determine mass of air delivered per minute, delivery temperature, indicated power and isothermal efficiency. [1.225 kg/min, 202.37°C, 4.23 kW, 77.1%] A reciprocating compressor of single stage and double acting type has free air delivered at 14 m3/min measured at 1.013 bar, 288 K. Pressure and temperature at suction are 0.95 bar and 305 K. The cylinder has clearance volume of 5% of swept volume. The air is delivered at pressure of 7 bar and expansion and compression follow the common index of 1.3. Determine the indicated power required and volumetric efficiency with respect to free air delivery. [63.55 kW, 72.4%] 3 A single stage double acting reciprocating compressor delivers 14 m /min measured at suction states of 1 bar and 20°C. Compressor runs at 300 rpm and air is delivered after compression with compression ratio of 7. Compressor has clearance volume of 5% of swept volume and compression follows polytropic process with index 1.3. Determine the swept volume of cylinder and indicated power in hp. [0.028 m3, 76.86 hp] 3 A single stage single acting reciprocating air compressor handles 0.5 m /min of free air measured at 1 bar. Compressor delivers air at 6.5 bar while running at 450 rpm. The volumetric efficiency is 0.75, isothermal efficiency is 0.76 and mechanical efficiency is 0.80. Determine indicated mean effective pressure and power required to drive the compressor. [0.185 MPa, 3.44 hp] A single stage single acting reciprocating air compressor compresses air by a ratio of 7. The polytropic index of both compression and expansion is 1.35. The clearance volume is 6.2% of cylinder volume. For volumetric efficiency of 0.8 and stroke to bore ratio of 1.3 determine the dimensions of cylinder. [14.67 cm and 19.08 cm] A single stage single acting reciprocating air compressor runs with air entering at 1 bar and leaving at 7 bar following PV1.3 = constant. Free air delivery is 5.6 m3/minute and mean piston 1 th of swept speed is 150 m/min. Take stroke to bore ratio of 1.3 and clearance volume to be 15

Reciprocating and Rotary Compressor ______________________________________________ 769

16.27

16.28

16.29

16.30

volume per stroke. The suction pressure and temperature are equal to atmospheric air pressure and temperature. Determine volumetric efficiency, speed of rotation, stroke and bore. Take mean piston speed = 2 ´ stroke ´ rpm. [76.88%, 164 rpm, 45.7 cm, 35.1 cm] A reciprocating compressor of single acting type has air entering at 1.013 bar, 15°C and leaving at 8 bar. Compressor is driven by electric motor of 30.84 hp and the mechanical efficiency is 0.87. The clearance volume is 7% of swept volume and the bore is equal to stroke. The compression and expansion follow PV1.3 = constant. Determine (i) free air delivered in m3/min, (ii) volumetric efficiency, and (iii) cylinder dimensions. [4.47 m3/min, 72.68%, L = D = 29.7 cm] A reciprocating compressor has two stages with inlet air going into LP stage at 1 bar, 16°C and at the rate of 12 m3/min. Air is finally delivered at 7 bar and there is perfect intercooling at optimum pressure between the stages. The index for compression is 1.25 and compressor runs at 600 rpm. Neglecting clearance volume determine intermediate pressure, total volume of each cylinder and total work required. [2.645 bar, 0.02 m3, 0.0075 m3, 57.6 hp] A two stage reciprocating air compressor delivers 4.2 kg of free air per min at 1.01325 bar and 15°C. The suction conditions are 0.95 bar, 22°C. Compressor delivers air at 13 bar. Compression throughout occurs following PV1.25 = C. There is optimum and perfect intercooling between the two stages. Mechanical efficiency is 0.75. Neglecting clearance volume determine (i) the heat transfer in intercooler per second. (ii) the capacity of electric motor. (iii) the % saving in work if two stage intercooling is compared with single stage compressor between same limits. [7.6 kJ/s, 44.65 hp, 13%] A single acting reciprocating air compressor has two stages with the optimum and perfect intercooling in between. Compressor has air sucked at 1 bar and at the rate of 2.4 m3/min when measured at 1.013 bar, 288 K. Compressor delivers air at 70 bar. Temperature at the end of suction stroke is 32°C. The compression and expansion follows polytropic process PV1.25 = C uniformly. The clearance volume is 3% of swept volume in each HP and LP cylinder. Compressor runs at 750 rpm. If the mechanical efficiency is 0.85 then determine the power of drive required, swept volumes of each cylinder, % saving in power as compared to single stage compression within limits. [35.8 hp, 3963 cm3, 473 cm3, 20.89%]