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 Q

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 dP PVk = c V

1 V For Isentopic compression' Q = 0

W = h ; W = -VdP W = mCp(T2-T1)

KW 1k

1k

P

P

1-k1

kmRTW

1

2

Q = 0 PV1 = mRT1

P,V, and T relationship

1k

2

1k

1k

1

2

1

2

V

V

P

P

T

T

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 dP PVn =C V

1 V

For Polytropic compression, Q 0

W = h - Q ; W = -VdP

h = mCp(T2 - T1) Q = mCn(T2 - T1)

KW 1P

P

1-n

nmRTW

n1n

1

21

n1

nkCC v

n

1n

2

1n

1n

1

2

1

2

V

V

P

P

T

T

C) Isothermal Compression (PV = C) P 2

W = -VdP dP PV = C V

1 V W = -Q

1

211 P

PlnVPW

P1V1 = mRT1

valves cylinder piston piston-rod d D HE CE P L P2 3 2 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)

KW1-k

kmRTW

1

1P

P k1k

1

2

Q = 0 PV1' = mRT1

P,V, and T relationship

1k

2

1k

1k

1

2

1

2

V

V

P

P

T

T

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)

KW1-n

nmRTW

1

1P

P n1n

1

2

P1V1' = mRT1

n1

nkCC v

n KJ/kg-K

1n

2

1n

1n

1

2

1

2

V

V

P

P

T

T

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

1

2

1'1P

PlnVPW

P1V1' = mRT1 PERCENT CLEARANCE

Volume ntDisplaceme

Volume Clearance C

100% x V

V C

D

3

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

100%xV

Vη

D

1'

v

A) For Isentropic Compression (PVk= C)

100%xP

PCC1η

1/k

1

2

v

B) For Polytropic Compression (PVn = C)

100%xP

PCC1η

1/n

1

2

v

C) For Isothermal Compression (PV = C)

100%xP

PCC1η

1

2

v

DISPLACEMENT VOLUME A) For single acting

VD = LD2Nn' m3/sec 4(60) B) For Double acting without considering the volume of piston rod

VD = 2LD2Nn' 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

% 100 x V

Vη

D

a

va

Va - actual volume of air or gas drawn in MEAN EFFECTIVE PRESSURE

D

mV

WP KPa

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 suction 1 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 6 7 3 2

P1 8

1

W1 = W2 T1 = T3 P2 = P3 = Px W = W1 + W2

Px - the ideal intercooler pressure or optimum pressure Work 1st Stage:

KW 1P

P

1-n

nmRTW

n1n

1

211

Work 2nd Stage:

KW 1P

P

1-n

nmRTW

n1n

3

432

Pressure Ratio:

3

4

1

2

P

P

P

P

but P2 = P3 = Px

41x

x

4

1

x

PPP

then

P

P

P

P

Since W1 =W2, the total work W is;

KW 1P

P

1-n

2nmRTW

n1n

1

21

substituting Px to W, it follows that

KW 1P

P

1-n

2nmRTW

2n1n

1

41

By performing an energy balance on the inter-cooler Qx = mCP(T3 - T2) T-S Diagram: T P4 Px P1 Qx T2 = T4 4 2 T1 = T3 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:

KW1-n

nmRTW

1

1

1P

P n1n

1

2

Work 2nd Stage:

KW1-n

nmRTW

3

2

1P

P n1n

3

4

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 Px, but a pressure drop will occur in the inter-cooler. P

P4 5

4

P2 7 2

P3 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 5 6 discharge

LP Intercooler HP Inercooler

1st Stage 2nd stage 3rd stage

Considering Polytropic compression and expansion processes and with perfect inter-cooling; Work of 1st stage cylinder:

KW1-n

nmRTW

1

1

1P

P n1n

1

2

Work of the 2nd stage cylinder:

KW1-n

nmRTW

3

2

1P

P n1n

3

4

Work of the 3rd stage cylinder:

KW1-n

nmRTW

5

3

1P

P n1n

5

6

For perfect inter-cooling: W1 = W2 = W3 T1 = T3 = T5

and

5

6

3

4

1

2

P

P

P

P

P

P

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

y

6

x

y

1

x

P

P

P

P

P

P

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

3 2

3

61y

6

2

1x

PPP

PPP

The total compressor work is equal to: W = W1 + W2 + W3 but: W1 = W2 = W3 ;therefore W = 3W1

KW1-n

3nmRTW

1

1P

P n1n

1

2

then substituting Px andPy then simplify, the result is:

KW1-n

3nmRTW

1

1P

P n31n

1

6

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

KW1-n

SnmRTW

1

1P

P Sn1n

1

S2

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 21C, 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)

2.1n

CPV

0.60 e

KPa 5000P

K 29427321T

KPa 101P

min/kg 2m

Given

n

c

6

1

1

KW 41.20W

1)2.1(3

12.1

101

5000

1P

P n31n

1

6

1)-60(1.2

)0.287)(2943(1.2)(2)(W

KW1-n

3nmRTW

1

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 21C. 2. A single acting compressor has a volumetric efficiency of 87% operates at 500 RPM. It takes in air at 100 KPa and

30C 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 305K 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 310K. the intercooler is water cooled with the

water entering at 295K and leaving at 305K. 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 31C and discharges it to 397 KPa. Compression and expansion processes follows PV1.3 = 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 21C, 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 16C. the compression is polytropic with n = 1.31 and the intercooler cools the air to 16C. 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

27C, 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 16C. 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 300K and

a discharge temperature of 530K. 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 15C 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 15.6C.

The air delivered from the compressor at a pressure 482.5 KPa and a temperature of 73C. 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 16C and leaves at 43C is 3.45 kg/sec. Find the power required to drive this compressor assuming no loss from radiation.