AIR COMPRESSOR PDF

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Summary

Compressor is a machine used to compressed air or gas to a final pressure exceeding 241.25 KPa gage. TYPES OF COMPRESSORS  Centrifugal Compressors - For low pressure and high capacity application  Rotary Compressors - For medium pressure and low capacity application  Reciprocating Compressors - F...

Description

Compressor is a machine used to compressed air or gas to a final pressure exceeding 241.25 KPa gage. TYPES OF COMPRESSORS  Centrifugal Compressors - For low pressure and high capacity application  Rotary Compressors - For medium pressure and low capacity application  Reciprocating Compressors - For high pressure and low capacity application USES OF COMPRESSED AIR  Operation of small engines  Pneumatic tools  Air hoists  Industrial cleaning by air blast  Tire inflation  Paint Spraying  Air lifting of liquids  Manufacture of plastics and other industrial products  To supply air in mine tunnels  Other specialized industrial applications ANALYSIS OF CENTRIFUGAL AND ROTARY TYPE W 2 1 Assuming: KE = 0; and PE = 0 Q = h + KE + PE + W (Steady state - steady flow equation) Q = h + W For a compressor, work is done on the system; thus -W = h - Q; Let -W = W (compressor work) A) Isentropic Compression (PVk = C) P 2 W = -VdP PVk = c dP Q

V

1 V For Isentopic compression' Q = 0 W = h ; W = -VdP W = mCp(T2-T1) k 1   kmRT  P  k  1  2  1 KW W  k - 1  P    1   Q=0 PV1 = mRT1 P,V, and T relationship k 1 k

P  2  2 T P  1  1

T

k 1

V   1 V   2

where m - mass flow rate of the gas in kg/sec W - work in KW P - pressure in KPa T - temperature in K R - gas constant in KJ/kg-K V - volume flow rate in m3/sec B) Polytropic Compression (PVn = C) P 2 W = -VdP PVn =C

dP V

1 V For Polytropic compression, Q  0 W = h - Q ; W = -VdP h = mCp(T2 - T1) Q = mCn(T2 - T1) n1   nmRT  P  n  1  2 W  1 KW n - 1  P   1      k  n C  Cv  n  1 n  n1 n

P  2  2 T P  1  1

T

n1

V   1 V   2

C) Isothermal Compression (PV = C) P 2 W = -VdP PV = C

dP V

1 V W = -Q

P W  P V ln 2 1 1 P

1

P1V1 = mRT1

valves

cylinder

piston

piston-rod d

HE

CE

P P2

D

L 2

3

P1 4

1 V1’ V

CVD

VD

HE - Head end CE - Crank end L - length of stroke, m VD - Displacement volume, m3/sec D - diameter of bore, m d - diameter of piston rod, m A) For Isentropic Compression and RE-expansion process (PVK = C),no heat is removed from the gas. W = h ; W = -VdP W = mCp(T2-T1) k 1   kmRT1  P2  k    W  1 KW  k - 1  P1    Q=0 PV1' = mRT1 P,V, and T relationship k 1

 P  k  V1     2  V  T1  P1   2 where: V1' - volume flow rate measured at intake, m3/sec m - mass flow rate corresponding V1', kg/sec B) For Polytropic Compression and Re-expansion process (PVn = C), some amount of heat is removed from the gas. W = h - Q ; W = -VdP h = mCp(T2 - T1) Q = mCn(T2 - T1) T2

k 1

n 1   nmRT1  P2  n    W  1 KW  n - 1  P1    P1V1' = mRT1 k  n  Cn  Cv  KJ/kg-K 1n 

n 1

 P  n  V1     2  V  T1  P1   2 C) For Isothermal compression and re-expansion process (PV = C), an amount of heat equivalent to the compression work is removed from the gas. W = -Q P W  P1 V1'ln 2 P1 P1V1' = mRT1 T2

n 1

PERCENT CLEARANCE Clearance Volume C  Displaceme nt Volume V C  3 x 100% VD For compressor design, values of percent clearance C ranges from 3 to 10 %. V3 = CVD where: V3 - clearance volume VOLUMETRIC EFFICIENCY v = Volume flow rate at intake x 100% Displacement Volume V ηv  1' x 100% VD A) For Isentropic Compression (PVk= C) 1/k   P2   ηv  1  C  C   x 100% P     1   B) For Polytropic Compression (PVn = C) 1/n   P2   ηv  1  C  C   x 100% P     1   C) For Isothermal Compression (PV = C)   P  ηv  1  C  C 2  x 100%  P   1  

DISPLACEMENT VOLUME A) For single acting VD = LD2Nn' m3/sec 4(60) B) For Double acting without considering the volume of piston rod VD = 2LD2Nn' m3/sec 4(60) C) For Double acting considering volume of piston rod VD = LNn'[2D2 - d2] m3/sec 4(60) where: L - length of stroke, m D - diameter of bore, m d - diameter of piston rod, m n' - no. of cylinders ACTUAL VOLUMETRIC EFFICIENCY V ηva  a x 100 % VD Va - actual volume of air or gas drawn in MEAN EFFECTIVE PRESSURE W Pm  KPa VD W in KJ, KJ/kg, KW VD in m3, m3/kg, m3/sec PISTON SPEED PS = 2LN m/min PS = 2LN m/sec 60 EFFICIENCY A) COMPRESSION EFFICIENCY cn = Ideal Work x 100% Indicated Work B) MECHANICAL EFFICIENCY m = Indicated Work x 100% Brake Work C) COMPRESSOR EFFICIENCY c = cn = m = Ideal Work x 100% Brake work

MULTISTAGE COMPRESSION Multi staging is simply the compression of air or gas in two or more cylinders in place of a single cylinder compressor. It is used in reciprocating compressors when pressure of 300 KPa and above are desired, in order to:  Save power  Limit the gas discharge temperature  Limit the pressure differential per cylinder  Prevent vaporization of lubricating oil and to prevent its ignition if the temperature becomes too high. It is common practice for multi-staging to cool the air or gas between stages of compression in an intercooler, and it is this cooling that affects considerable saving in power.

A) 2 - Stage Compression without pressure drop in the intercooler Qx 1

suction

2

3

4

discharge

Intercooler

1stStage

2nd Stage

For an ideal multistage compression, with perfect inter-cooling and minimum work, the cylinder were properly designed so that:  the work at each stage are equal  the air in the intercooler is cooled back to the initial temperature  no pressure drop occurs in the intercooler  the pressure at each stage are equal W1 = W2 ; T1 = T3 ; P2 = P3 = Px where: W1 - work of the LP cylinder (1st stage) W2 - work of the HP cylinder (2nd stage) Px = ideal intercooler pressure, optimum pressure Assuming polytropic compression and expansion processes: P P4

5

4

PVn = C Px P1

6 7

3 2

8

W1 = W2 T1 = T3 P2 = P3 = P x W = W1 + W2 Px - the ideal intercooler pressure or optimum pressure Work 1st Stage: n 1   nmRT  P  n  1  2 W   1 KW 1 n - 1  P    1    Work 2nd Stage: n 1   nmRT  P  n  3  4 W   1 KW 2 n - 1  P   3    

1

Pressure Ratio: P2 P4  P1 P3 but P2 = P3 = Px Px P4  P1 Px then

Px  P1P4

Since W1 =W2, the total work W is; n 1   2nmRT1  P2  n    1 W     n - 1  P1   

KW

substituting Px to W, it follows that n 1   2nmRT1  P4  2n   W KW  1  n - 1  P1    By performing an energy balance on the inter-cooler Qx = mCP(T3 - T2) T-S Diagram: T P4

Px

P1 Qx

T2 = T4 T1 = T3

4

2

3

1

S B) 2 stage compressor with pressure drop in the intercooler For 2 stage compression with pressure drop in the intercooler, P2  P3.The air in the intercooler may or may not be cooled to the initial temperature, and the work at each stage may or may not be equal, thus the work W = W1 + W2 Work 1st Stage: n 1   nmRT1  P2  n    W1   1 KW  n - 1  P1    Work 2nd Stage: n 1   nmRT3  P4  n    W2   1 KW  n - 1  P3   

The total work W is; W = W1 + W2 The pressure, P2  P3, but the 1st stage may compress the air or gas to the optimum intercooler pressure P x, but a pressure drop will occur in the inter-cooler. P 5 4 P4 P2 P3

7

2

6

3

P1

8

1

V

Heat Rejected in the inter-cooler Qx = mCp(T3 - T2) C. Three-Stage compressor without pressure drop in the intercooler Qx Qy suction

1

2

3

4

LP Intercooler

1st Stage

5

6

discharge

HP Inercooler

2nd stage

3rd stage

Considering Polytropic compression and expansion processes and with perfect inter-cooling; Work of 1st stage cylinder: n 1   nmRT1  P2  n     1 KW W1   n - 1  P1    Work of the 2nd stage cylinder: n 1   nmRT3  P4  n    W2   1 KW  n - 1  P3   

Work of the 3rd stage cylinder: n 1   nmRT5  P6  n    W3   1 KW  n - 1  P5    For perfect inter-cooling: W1 = W2 = W3 T1 = T3 = T5

and

P2 P1



P4 P3



P6 P5

But P2 = P3 = Px (Ideal LP Intercooler pressure) P4 = P5 = Py (Ideal HP Intercooler pressure) Thus

Px P1



Py Px



P6 Py

By expressing Px and Py in terms of P1 & P6: Px  3 P1 P6 2

Py  3 P1P6

2

The total compressor work is equal to: W = W1 + W2 + W3 but: W1 = W2 = W3 ;therefore W = 3W1 n 1   3nmRT1  P2  n     1 KW W  n - 1  P1    then substituting Px andPy then simplify, the result is: n 1   3nmRT1  P6  3n    W  1 KW  n - 1  P1   

For multistage compression with minimum work and perfect inter-cooling and no pressure drop in the inter-coolers between stages, the following conditions apply: 1. the work at each stage are equal 2. the pressure ratio between stages are equal 3. the air temperature in the inter-coolers are cooled to the original temperature T1 4. the total work W is equal to

n 1   SnmRT1  P2S  Sn    W  1 KW  n - 1  P1   

where S - number of stages

Example An ideal 3-stage air compressor with intercoolers handles air at the rate of 2 kg/min. The suction pressure is 101 Kpa, suction temperature is 21C, delivery pressure is 5000 KPa. Assuming perfect inter-cooling and minimum work, calculate total power required if compressor efficiency is 60% and both compression and expansion processes are PVn=C, where n = 1.2. (21 KW) Given

m  2 kg / min P1  101 KPa

T1  21  273  294 K P6  5000 KPa ec  0.60

PV n  C

n  1 .2

3nmRT1 W n -1

 P  n 13n   6   1    P1   

KW

1.2  1    3(1.2)(2)( 0.287)(294 ) 5000  3(1.2)  1 W   60(1.2 - 1)   101    W  20.41 KW

Sample Problems 1. A 2-stage, double acting air compressor 41cm and 25cm bore x 18 cm running 600 RPM has a free air un-loader at each end for capacity control. It is driven by a 150 HP motor, 460V,3 phase, 60 hertz, 1175 RPM thru super-high capacity V-belts at sea level installation.P1 = 101 KPa and P4= 984 KPa, calculate the power assuming the LP cylinder discharges the air to the ideal pressure and intercooler cools the air to the initial temperature of 21C. 2. A single acting compressor has a volumetric efficiency of 87% operates at 500 RPM. It takes in air at 100 KPa and 30C and discharges it to 600 KPa. The air handled is 6 m3/min measured at discharge condition. If compression and expansion processes are isentropic, find the piston displacement per stroke. 3. A two-stage reciprocating air compressor is required to deliver 0.70 kg/sec of air from 98.6 Kpa and 305K to 1276

KPa. The compressor operates at 205 RPM, compression and re-expansion processes are PV1.25 = C, and both cylinders have 3.5% clearance. There is a 20 KPa pressure drop in the intercooler. The LP cylinder discharges at the optimum pressure into the intercooler. The air enters the HP cylinder at 310K. the intercooler is water cooled with the water entering at 295K and leaving at 305K. Determine the motor power required if m= 85%.( 260 KW) 4. A 36 cm x 36 cm , horizontal double acting air compressor w/ 5% clearance operates at 120 RPM and draws air at 100 KPa and 31C and discharges it to 397 KPa. Compression and expansion processes follows PV 1.3 = C. Determine compressor power required in KW.(22 KW) 5. An ideal 3-stage air compressor with intercoolers handles air at the rate of 2 kg/min. The suction pressure is 101 Kpa, suction temperature is 21C, delivery pressure is 5000 KPa. Assuming perfect inter-cooling and minimum work, calculate total power required if compressor efficiency is 60% and both compression and expansion processes are PVn=C, where n = 1.2. (21 KW) 6. A two stage compressor with 5% clearance delivers 40 kg/min of air to 965 Kpa. At intake the pressure is 98.6 KPa and the temperature is 16C. the compression is polytropic with n = 1.31 and the intercooler cools the air to 16C. Determine the mean effective pressure in KPa. 7. A double acting, 2 stage air compressor running at 150 RPM has a suction pressure and temperature of 100 KPa and 27C, respectively. The low pressure cylinder is 36 cm x 38 cm and connected in tandem to the high pressure cylinder. The LP cylinder discharges the air at 386 KPa. The intercooler cools back the air to its initial temperature and the air enters the HP cylinder at 370 KPa and leaves at 1 480 KPa. Both LP and HP cylinders have 4% clearance and compression follows PV1.3 = C. Neglect effect of piston rods. Standard air is at 101 KPa and 16C. Determine: a) the volume of free air based on apparent volumetric efficiency b) the heat removed by cooling water in the intercooler ( 24 KW) c) the diameter of the HP cylinder in cm (19 cm) d) the power required to drive the compressor if the aggregate losses is 20% of the power of the conventional diagram. (72 KW) 8. A turbine driven compressor handles 10 kg/sec of air from 100 KPa to 600 KPa with an inlet temperature of 300K and a discharge temperature of 530K. The inlet tubing has a 0.5 m inside diameter and the discharge piping a 0.20 m diameter. The compression is adiabatic. Determine: a) the air inlet and exit velocities in m/sec b) the isentropic compression efficiency c) the power required in KW 9. A large centrifugal compressor handles 9 kg/sec of air, compresses it from 100 KPa and 15C and with an initial velocity of 110 m/sec to discharge pressure. The velocity of the high pressure air stream is 90 m/sec. The pressure ratio is 4:1 the compressor has an isentropic compression efficiency of 80%. Determine a) the exit pressure in KPa b) the exit temperature in K c) the power required 10.A centrifugal compressor handling air draws 6 m3/sec of air at a pressure of 96.5 KPa and a temperature of 15.6C. The air delivered from the compressor at a pressure 482.5 KPa and a temperature of 73C. The area of the suction pipe is 0.2 m2, the area of the of the discharge pipe is 0.4 m2, the discharge pipe is located 6 m above the suction pipe. The weight of the jacket water that enters at 16C and leaves at 43C is 3.45 kg/sec. Find the power required to drive this compressor assuming no loss from radiation....