Chapter 5 : Air Compressors¶

Example 5.1 Page no : 250¶

In [1]:
# Variables
D = 0.2;			#Cylinder diameter in m
L = 0.3;			#Cylinder Stroke in m
P1 = 1.;			#Pressure at entry in bar
T1 = 300.;			#Temperature at entry in K
P2 = 8.;			#Pressure at exit in bar
n = 1.25;			#Adiabatic gas constant
N = 100.;			#Speed in rpm
R = 287.;			#Universal gas constant in J/kg-K

# Calculations
x = (n-1)/n;			#Ratio
V1 = (3.147*L*(D**2))/4;			#Volume of cylinder in m**3/cycle
W = (P1*(10**5)*V1*(((P2/P1)**x)-1))/x;			#Work done in J/cycle
Pc = (W*100)/(60*1000);			#Indicated power of compressor in kW
m = (P1*(10**5)*V1)/(R*T1);			#Mass of air delivered in kg/cycle
md = m*N;			#Mass delivered per minute in kg
T2 = T1*((P2/P1)**x);			#Temperature of air delivered in K

# Results
print 'Indicated power of compressor is %3.2f kW  \
\nMass of air delivered by compressor per minute is %3.2f kg  \
\nTemperature of air delivered is %3.1fK'%(Pc,md,T2)

Indicated power of compressor is 4.06 kW
Mass of air delivered by compressor per minute is 1.10 kg
Temperature of air delivered is 454.7K


Example 5.2 Page no : 251¶

In [2]:
import math

# Variables
IP = 37.;			#Indicated power in kW
P1 = 0.98;			#Pressure at entry in bar
T1 = 288.;			#Temperature at entry in K
P2 = 5.8;			#Pressure at exit in bar
n = 1.2;			#Adiabatic gas constant
N = 100.;			#Speed in rpm
Ps = 151.5;			#Piston speed in m/min
a = 2.;			#For double acting compressor

# Calculations
L = Ps/(2*N);			#Stroke length in m
x = (n-1)/n;			#Ratio
r = (3.147*L)/4;			#Ratio of volume to bore
D = math.sqrt((IP*1000*60*x)/(N*a*r*P1*(10**5)*(((P2/P1)**x)-1)));			#Cylinder diameter in m

# Results
print 'Stroke length of cylinder is %3.4f m  \
\nCylinder diameter is %3.4f m'%(L,D)

Stroke length of cylinder is 0.7575 m
Cylinder diameter is 0.3030 m


Example 5.3 Page no : 251¶

In [3]:
import math

# Variables
IP = 11.;			#Indicated power in kW
P1 = 1.;			#Pressure at entry in bar
P2 = 7.;			#Pressure at exit in bar
n = 1.2;			#Adiabatic gas consmath.tant
Ps = 150.;			#Piston speed in m/s
a = 2.; 			#For double acting compressor
r = 1.5;			#Storke to bore ratio

# Calculations
x = (n-1)/n;			#Ratio
y = 3.147/(4*(r**2));			#Ratio of volume to the cube of stroke
z = (P1*(10**2)*y*(((P2/P1)**x)-1))/x;			#Ratio of workdone to the cube of stroke
L = (math.sqrt(IP/(z*Ps)))*1000;			#Stroke in mm
D = (L/r);			#Bore in mm

# Results
print 'Stroke length of cylinder is %3.0f mm  \
\nBore diameter of cylinder is %3.0f mm'%(L,D)

Stroke length of cylinder is  30 mm
Bore diameter of cylinder is  20 mm


Example 5.4 Page no : 252¶

In [2]:
# Variables
x = 0.05            # ratio
P1 = 1.;			#Pressure at point 1 in bar
T1 = 310.;			#Temperature at point 1 in K
n = 1.2;			#Adiabatic gas constant
P2 = 7.;			#Pressure at point 2 in bar
Pa = 1.01325;			#Atmospheric pressure in bar
Ta = 288.;			#Atmospheric temperature in K

# Calculations
V1 = 1+x;			#Ratio of volume of air sucked to stroke volume
V4 = ((P2/P1)**(1/n))/20;			#Ratio of volume delivered to stroke volume
DV = V1-V4;			#Difference in volumes
nv1 = DV*100;			#Volumetric efficiency
V = (P1*DV*Ta)/(T1*Pa);			#Ratio of volumes referred to atmospheric conditions
nv2 = V*100;			#Volumetric efficiency referred to atmospheric conditions
W = (n*0.287*T1*((P2/P1)**((n-1)/n)-1))/(n-1);			#Work required in kJ/kg

# Results
print 'Volumetric efficiency is %3.1f percent  \
\nVolumetric efficiency referred to atmospheric conditions is %3.1f percent  \
\nWork required is %3.1f kJ/kg'%(nv1,nv2,W)

Volumetric efficiency is 79.7 percent
Volumetric efficiency referred to atmospheric conditions is 73.1 percent
Work required is 204.5 kJ/kg


Example 5.5 Page no : 253¶

In [4]:
# Variables
D = 0.2;			#Bore in m
L = 0.3;			#Stroke in m
P1 = 1.;			#Pressure at point 1 in bar
P2 = 7.;			#Pressure at point 2 in bar
n = 1.25;			#Adiabatic gas constant
lc = 0.015

# Calculations
V3 = (3.147*(D**2)*lc)/4.;			#Clearance volume in m**3
Vs = (3.147*(D**2)*L)/4.;			#Stoke volume in m**3
C = V3/Vs;			#Clearance ratio
nv = (1+C-(C*((P2/P1)**(1/n))))*100;			#Volumetric efficiency
DV = (nv*Vs)/100.;			#Volume of air taken in (m**3)/stroke

# Results
print 'Theoretical volume of air taken in per stroke is %.2e m**3/stroke'%(DV)

Theoretical volume of air taken in per stroke is 7.67e-03 m**3/stroke


Example 5.6 Page no : 254¶

In [5]:
import math

# Variables
D = 0.2;			#Bore in m
L = 0.3;			#Stroke in m
P1 = 1.;			#Pressure at point 1 in bar
r = 0.05            # ratio
T1 = 293.;			#Temperature at point 1 in K
P2 = 5.5;			#Pressure at point 2 in bar
n = 1.3;			#Adiabatic gas constant
N = 500.;			#Speed of compressor in rpm

# Calculations
x = (n-1)/n;			#Ratio
Vs = (3.147*L*(D**2))/4;			#Stroke volume in m**3
Vc = r*Vs;			#Clearance volume in m**3
V1 = Vc+Vs;			#Volume at point 1 in m**3
V4 = Vc*((P2/P1)**(1/n));			#Volume at point 4 in m**3
EVs = V1-V4;			#Effective swept volume in m**3
W = (P1*(10**5)*EVs*(((P2/P1)**x)-1))/x;			#Work done in J/cycle
MEP = (W/Vs)/(10**5);			#Mean effective pressure in bar
P = (W*N)/(60*1000);			#Power required in kW

# Results
print 'Mean effective pressure is %3.2f bar  \
\nPower required is %3.2f kW'%(MEP,P)

# rounding off error

Mean effective pressure is 1.81 bar
Power required is 14.21 kW


Example 5.7 Page no : 255¶

In [12]:
import math

# Variables
D = 0.2;			#Bore in m
L = 0.3;			#Stroke in m
P1 = 97.;			#Pressure at entry in kN/(m**2)
P4 = P1;			#Pressure at point 4 in kN/(m**2)
T1 = 293.;			#Temperature at point 1 in K
P2 = 550.;			#Compression Pressure in kN/(m**2)
P3 = P2;			#Pressure at point 3 in kN/(m**2)
n = 1.3;			#Adiabatic gas constant
N = 500.;			#Speed of compressor in rpm
Pa = 101.325;			#Air pressure in kN/(m**2)
Ta = 288.;			#Air temperature in K

# Calculations
x = (n-1)/n;			#Ratio
DV = (3.147*L*(D**2))/4;			#Difference in volumes in m**3
V3 = r*DV;			#Clearance volume in m**3
V1 = V3+DV;			#Volume at point 1 in m**3
V4 = V3*((P3/P4)**(1/n));			#Volume at point 4 in m**3
Vs = V1-V4;			#Effective swept volume in m**3
EVs = Vs*N;			#Effective swept volume per min
Va = (P1*EVs*Ta)/(Pa*T1);			#Free air delivered in (m**3)/min
nV = ((V1-V4)/(V1-V3))*100;			#Volumetric effciency
T2 = T1*((P2/P1)**x);			#Air delivery temperature in K
t2 = T2-273;			#Air delivery temperature in oC
W = (n*P1*(V1-V4)*(((P2/P1)**x)-1))*N/((n-1)*60);			#Cycle power in kW
Wiso = P1*V1*(math.log(P2/P1));			#Isothermal workdone
niso = (Wiso/(4.33*0.493))*100;			#Isothermal efficiency

# Results
print 'Free air delivered is %3.3f m**3/min  \
\nVolumetric efficiency is %3.0f percent  \
\nAir delivery temperature is %3.1f oC  \
\nCycle power is %3.0f kW  \
\nIsothermal efficiency is %3.1f percent'%(Va,nV,t2,W,round(niso,-1))

# rounding off error

Free air delivered is 3.820 m**3/min
Volumetric efficiency is  86 percent
Air delivery temperature is 164.3 oC
Cycle power is  14 kW
Isothermal efficiency is 80.0 percent


Example 5.8 Page no : 257¶

In [13]:
import math

# Variables
Ve = 30.;			#Volume of air entering compressor per hour in m**3
P1 = 1.;			#Presure of air entering compressor in bar
N = 450.;			#Speed in rpm
P2 = 6.5;			#Pressure at point 2 in bar
nm = 0.8;			#Mechanical efficiency
nv = 0.75;			#Volumetric efficiency
niso = 0.76;			#Isothermal efficiency

# Calculations
Vs = Ve/(nv*3600);			#Swept volume per sec in (m**3)/s
V = (Vs*60)/N;			#Swept volume per cycle in m**3
V1 = (Ve*60)/(3600*N);			#Volume at point 1 in m**3
Wiso = P1*100*V1*math.log(P2/P1);			#Isothermal workdone per cycle
Wact = Wiso/niso;			#Actual workdone per cycle on air
MEP = (Wact/V)/100;			#Mean effective pressure in bar
IP = (Wact*N)/60;			#Indicated power in kW
BP = IP/nm;			#Brake power in kW

# Results
print 'Mean effective pressure is %3.3f bar  \
\nBrake power is %3.2f kW'%(MEP,BP)

# rounding off error

Mean effective pressure is 1.847 bar
Brake power is 2.57 kW


Example 5.9 Page no : 258¶

In [22]:
import math

# Variables
Va = 15.;			#Volume of air in (m**3)/min
Pa = 1.01325;			#Pressure of air in bar
Ta = 302.;			#Air temperature in K
P1 = 0.985;			#Pressure at point 1 in bar
r = 0.04            # ratio
T1 = 313.;			#Temperature at point 1 in K
y = 1.3;			#Ratio of stroke to bore diameter
N = 300.;			#Speed in rpm
n = 1.3;			#Adiabatic gas constant
P2 = 7.5;			#Pressure at point 2 in bar

# Calculations
x=((P2/P1)**(1./n))-1;
a = x*r;			#Ratio of volume at point 4 to swept volume
nv = 1-a;			#Volumetric efficiency
V1 = (Pa*Va*T1)/(Ta*P1);			#Volume at point 1 in (m**3)/min
Vs = V1/(nv*N*2);			#Swept volume in m**3
D = ((Vs*4)/(math.pi*y))**(1./3);			#Bore in m
L = y*D;			#Stroke in m

# Results
print 'Cylinder bore in %3.3f m  \
\nCylinder stroke %3.3f m'%(D,L)

# rounding off error

Cylinder bore in 0.313 m
Cylinder stroke 0.407 m


Example 5.10 Page no : 259¶

In [23]:
import math

# Variables
P1 = 0.98;			#Pressure at point 1 in bar
P4 = P1;			#Pressure at point 4 in bar
P2 = 7.;			#Pressure at point 2 in bar
P3 = P2;			#Pressure at point 3 in bar
n = 1.3;			#Adiabatic gas consmath.tant
Ta = 300.;			#Air temperature in K
Pa = 1.013;			#Air pressure in bar
T1 = 313.;			#Temperature at point 1 in K
Va = 15.;			#Volume of air delivered in m**3
R = 0.287;			#Universal gas constant in kJ/kg-K
c = 0.04

# Calculations
x = (n-1)/n;			#Ratio
r = (P2/P1)**(1/n);			#Ratio of volumes
a = r*c;			#Ratio of volume at point 4 to swept volume
DV = 1+c-a;			#Difference in volumes
V = (P1*DV*Ta)/(T1*Pa);			#Volume of air delivered per cycle
nv = V*100;			#Volumetric efficiency
DV1 = (Pa*Va*T1)/(Ta*P1);			#Difference in volumes
T2 = T1*((P2/P1)**x);			#Temperature at point 2 in K
ma = (Pa*100*Va)/(R*Ta);			#Mass of air delivered in kg/min
IP = (ma*R*(T2-T1))/(x*60);			#Indicated power in kW
Piso = (ma*R*T1*math.log(P2/P1))/60;			#Isothermal indicated power in kW
niso = (Piso/IP)*100;			#Isothermal efficiency

# Results
print 'Volumetric efficiency is %3.1f percent  \
\nIndicated power is %3.2f kW  \
\nIsothermal efficiency is %3.0f percent'%(nv,IP,niso)

Volumetric efficiency is 79.6 percent
Indicated power is 65.74 kW
Isothermal efficiency is  79 percent


Example 5.11 Page no : 261¶

In [24]:
# Variables
V1 = 7.*(10**-3);			#Volume of air in (m**3)/s
P1 = 1.013;			#Pressure of air in bar
T1 = 288.;			#Air temperature in K
P2 = 14.;			#Pressure at point 2 in bar
n = 1.3;			#Adiabatic gas constant
nm = 0.82;			#Mechanical efficiency

# Calculations
x = (n-1)/n;			#Ratio
W = (P1*100*V1*(((P2/P1)**x)-1))/x;			#Work done by compressor in kW
P = W/nm;			#Power requred to drive compressor in kW

# Results
print 'Power requred to drive compressor is %3.2f kW'%(P)

Power requred to drive compressor is 3.12 kW


Example 5.12 Page no : 261¶

In [26]:
# Variables
L = 0.15;			#Stroke in mm
D = 0.15;			#Bore in mm
N = 8.;			#Speed in rps
P1 = 100.;			#Pressure at point 1 in kN/(m**2)
P2 = 550.;			#Pressure at point 2 in kN/(m**2)
n = 1.32;			#Adiabatic gas constant
C = 0.06            # RATIO

# Calculations
x = (n-1)/n;			#Ratio
nv = (1+C-(C*((P2/P1)**(1/n))))*100;			#Volumetric efficiency
DV = (3.147*(D**2)*L)/4;			#Difference in volumes at points 1 and 3
DV1 = (nv*DV)/100;			#Difference in volumes at points 1 and 4
V2 = DV1*((P1/P2)**(1/n))*N;			#Volume of air delivered per second
W = (P1*DV1*(((P2/P1)**x)-1))*N/x;			#Power of compressor in kW

# Results
print 'Theoretical volume efficiency is %3.1f percent  \
\nVolume of air delivered is %3.5f m**3/s  \
\nPower of compressor is %3.3f kW'%(nv,V2,W)

Theoretical volume efficiency is 84.2 percent
Volume of air delivered is 0.00491 m**3/s
Power of compressor is 3.774 kW


Example 5.13 Page no : 262¶

In [1]:
import math

# Variables
V = 16.;			#Volume of air compresssed in m**3
P1 = 1.;			#Pressure at point 1 in bar
P3 = 10.5;			#Pressure at point 3 in bar
T1 = 294.;			#Temperature at point 1 in K
Tc = 25.;			#Temperature of cooling water in oC
n = 1.35;			#Adiabatics gas constant
R = 0.287;			#Universal gas constant in kJ/kg-K
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K
Cw = 4.187;			#Specific heat of water in kJ/kg-K

# Calculations
x = (n-1)/n;			#Ratio
P2 = math.sqrt(P1*P3);			#Pressure at point 2 in bar
W1 = (2*P1*100*V*(((P2/P1)**x)-1))/(x*60);			#Indicated power of compressor from P1 to P2 in kW
W2 = (P1*100*V*(((P3/P1)**x)-1))/(x*60);			#Indicated power of compressor from P1 to P3 in kW
T4 = T1*((P2/P1)**x);			#Maximum temperature for two stage compression in K
T2 = T1*((P3/P1)**x);			#Maximum temperature for single stage compression in K
m = (P1*100*V)/(R*T1);			#Mass of air compressed in kg/min
Q = m*Cp*(T4-T1);			#Heat rejected by air in kJ/min
mc = Q/(Cw*Tc);			#Mass of cooling water in kg/min

# Results
print 'Minimum indicated power required for 2 stage compression is %3.1f kW  \
\nPower required for single stage compression is 18 percent more than that for \
two stage compression with perfect intercooling  \
\nMaximum temperature for two stage compression is %3.1f K  \
\nMaximum temperature for single stage compression is %3.1f K  \
\nHeat rejected by air is %3.1f kJ/min  \
\nMass of cooling water required is %3.1f kg/min'%(W1,T4,T2,Q,mc)

Minimum indicated power required for 2 stage compression is 73.3 kW
Power required for single stage compression is 18 percent more than that for two stage compression with perfect intercooling
Maximum temperature for two stage compression is 398.8 K
Maximum temperature for single stage compression is 540.9 K
Heat rejected by air is 1996.6 kJ/min
Mass of cooling water required is 19.1 kg/min


Example 5.14 Page no : 264¶

In [28]:
import math

# Variables
V = 0.2;			#Air flow rate in (m**3)/s
P1 = 0.1;			#Intake pressure in MN/(m**2)
P3 = 0.7;			#Final pressure in MN/(m**2)
T1 = 289.;			#Intake temperature in K
n = 1.25;			#Adiabatic gas constant
N = 10.;			#Compressor speed in rps

# Calculations
x = (n-1)/n;			#Ratio
P2 = math.sqrt(P1*P3);			#Intermediate pressure in MN/(m**2)
V1 = (V/N)*1000;			#Total volume of LP cylinder in litres
V2 = ((P1*V1)/P2);			#Total volume of HP cylinder in litres
W = ((2*P1*V*(((P2/P1)**x)-1))/x)*1000;			#Cycle power in kW

# Results
print 'Intermediate pressure is %3.3f MN/m**2  \
\nTotal volume of LP cylinder is %3.0f litres  \
\nTotal volume of HP cylinder is %3.1f litres  \
\nCycle power is %3.0f kW'%(P2,V1,V2,W)

Intermediate pressure is 0.265 MN/m**2
Total volume of LP cylinder is  20 litres
Total volume of HP cylinder is 7.6 litres
Cycle power is  43 kW


Example 5.15 Page no : 265¶

In [2]:
# Variables
P1 = 1.;			#Pressure at point 1 in bar
T1 = 290.;			#Temperature at point 1 in K
P3 = 60.;			#Pressure at point 3 in bar
P2 = 8.;			#Pressure at point 2 in bar
T2 = 310.;			#Temperature at point 2 in K
L = 0.2;			#Stroke in m
D = 0.15;			#Bore in m
n = 1.35;			#Adiabatic gas constant
N = 200.;			#Speed in rpm

# Calculations
x = (n-1)/n;	            		#Ratio
V1 = (3.147*(D**2)*L)/4;			#Volume at point 1 in m**3
V2 = (P1*V1*T2)/(T1*P2);			#Volume of air entering LP cylinder in m**3
W = ((P1*(10**5)*V1*(((P2/P1)**x)-1))/x)+((P2*(10**5)*V2*(((P3/P2)**x)-1))/x);			#Workdone by compressor per cycle in J
P = (W*N)/(60*1000);		    	#Power of compressor in kW

# Results
print 'Power of compressor is %3.2f kW'%(P)

# rounding off error

Power of compressor is 6.59 kW


Example 5.16 Page no : 265¶

In [8]:
import math

# Variables
N = 220.;			#Speed of compressor in rpm
P1 = 1.;			#Pressure entering LP cylinder in bar
T1 = 300.;			#Temperature at point 1 in K
Dlp = 0.36;			#Bore of LP cylinder in m
Llp = 0.4;			#Stroke of LP cylinder in m
Lhp = 0.4;			#Stoke of HP cylinder in m
P2 = 4.;			#Pressure leaving LP cylinder in bar
P5 = 3.8;			#Pressure entering HP cylinder in bar
T3 = 300.;			#Temperature entering HP cylinder in K
P6 = 15.2;			#Dicharge pressure in bar
n = 1.3;			#Adiabatic gas constant
Cp = 1.0035;			#Specific heat at constant pressure in kJ/kg-K
R = 0.287;			#Universal gas constant in kJ/kg-K
T5 = T1;			#Temperature at point 5 in K
C = 0.04
# Calculations
x = (n-1)/n;			#Ratio
Vslp = round((math.pi*(Dlp**2)*Llp*N*2)/4,2);			#Swept volume of LP cylinder in m**3/min
nv = round(1+C-(C*((P2/P1)**(1/n))),4);			#Volumetric efficiency
V1 = nv*Vslp;			#Volume of air drawn at point 1 in (m**3)/min
m = round((P1*100*V1)/(R*T1),2);			#Mass of air in kg/min
T2 = round(T1*((P2/P1)**x));			#Temperature at point 2 in K
QR = m*Cp*(T2-T5);			#Heat rejected in kJ/min
V5 = (m*R*T5)/(P5*100);			#Volume of air drawn in HP cylinder M**3/min
Plp = P2/P1;			#Pressure ratio of LP cylinder
Php = P6/P5;			#Pressure ratio of HP cylinder
Vshp = V5/nv;			#Swept volume of HP cylinder in m**3/min
Dhp = math.sqrt((Vshp*4)/(3.147*Lhp*N*2));			#Bore of HP cylinder in m
P = (m*R*(T2-T1))/(x*60);			#Power required for HP cylinder in kW

print V5,Plp,Php,Vshp,Dhp,P
# Results
print 'Heat rejected in intercooler is %3.1f kJ/min  \
\nDiameter of HP cylinder is %3.4f m  \
\nPower required for HP cylinder is %3.0f kW'%(QR,Dhp,P)

# rounding off error. please check

4.35484736842 4.0 4.0 4.71405863652 0.184511219993 45.0178314444
Heat rejected in intercooler is 2179.5 kJ/min
Diameter of HP cylinder is 0.1845 m
Power required for HP cylinder is  45 kW


Example 5.17 Page no : 267¶

In [41]:
import math

# Variables
P1 = 1.;			#Pressure at point 1 in bar
P3 = 30.;			#Pressure at point 3 in bar
T1 = 300.;			#Temperature at point 1 in K
n = 1.3;			#Adiabatics gas constant

# Calculations
P2 = math.sqrt(P1*P3);			#Intermediate pressure in bar
rD = math.sqrt(P2/P1);			#Ratio of cylinder diameters

# Results
print 'Ratio of cylinder diameters is %3.2f'%(rD)

Ratio of cylinder diameters is 2.34


Example 5.18 Page no : 268¶

In [42]:
# Variables
P1 = 1.013;			#Pressure at point 1 in bar
T1 = 288.;			#Temperaturea at point 1 in K
v1 = 8.4;			#free air delivered by compressor in m**3
P4 = 70.;			#Pressure at point 4 in bar
n = 1.2;			#Adiabatic gas constant
Cp = 1.0035;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
x = (n-1)/n;			#Ratio
P2 = P1*((P4/P1)**(1./3));			#LP cylinder delivery pressure in bar
P3 = P2*((P4/P1)**(1./3));			#IP cylinder delivery pressure in bar
r = P2/P1;			#Ratio of cylinder volumes
r1 = P3/P2;			#Ratio of cylinder volumes
r2 = r*r1;			#Ratio of cylinder volumes
V3 = 1;			#Volume at point 3 in m**3
T4 = T1*((P2/P1)**x);			#Three stage outlet temperature in K
QR = Cp*(T4-T1);			#Heat rejected in intercooler in kJ/kg of air
W = ((3*P1*100*v1*(((P4/P1)**(x/3))-1))/(x*60));			#Total indiacted power in kW

# Results
print 'LP cylinder delivery pressure is %3.3f bar  \
\nIP cylinder delivery pressure is %3.2f bar  \
\nRatio of cylinder volumes is %3.2f:%3.1f:%3.0f  \
\nTemperature at end of each stage is %3.2f K  \
\nHeat rejected in each intercooler is %3.1f kJ/kg of air  \
\nTotal indicated power is %3.2f kW'%(P2,P3,r2,r1,V3,T4,QR,W)

LP cylinder delivery pressure is 4.157 bar
IP cylinder delivery pressure is 17.06 bar
Ratio of cylinder volumes is 16.84:4.1:  1
Temperature at end of each stage is 364.41 K
Heat rejected in each intercooler is 76.7 kJ/kg of air
Total indicated power is 67.72 kW


Example 5.19 Page no : 269¶

In [43]:
# Variables
D = 0.45;			#Bore in m
L = 0.3;			#Stroke in m
P1 = 1.;			#Pressure at point 1 inn bar
T1 = 291.;			#Temperature at point 1 in K
P4 = 15.;			#Pressure at point 4 in bar
n = 1.3;			#Adiabatic gas constant
R = 0.29;			#Universal gas constant in kJ/kg-K

# Calculations
x = (n-1)/n;			#Ratio
k = (P4/P1)**(1./3);			#Pressure ratio
P2 = k*P1;			#Pressure at point 2 in bar
P3 = k*P2;			#Pressure at point 1 in bar
Vslp = (3.147*(D**2)*L)/4;			#Swept volume of LP cylinder
V7 = C*Vslp;			#Volume at point 7 in m**3
V1 = Vslp+V7;			#Volume at point 1 in m**3
V8 = V7*(k**(1/n));			#Volume at point 8 in m**3
EVs = (V1-V8)*1000;			#Effective swept volume in litres
T4 = T1*(k**x);			#Temperature at point 4 in K
t4 = T4-273;			#Delivery temperature in oC
DV = ((P1*T4*(V1-V8))/(P4*T1))*1000;			#Delivery volume per stroke in litres
W = (3*R*T1*((k**x)-1))/x;			#Workdone per kg of air in kJ

# Results
print 'Intermediate pressures are %3.3f bar and %3.3f bar  \
\nEffective swept volume of LP cylinder is %3.2f litres  \
\nTemperature of air delivered per stroke is %3.1f oC  \
\nVolume of air delivered per stroke is %3.2f litres  \
\nWork done per kg of air is %3.1f kJ'%(P2,P3,EVs,t4,DV,W)

Intermediate pressures are 2.466 bar and 6.082 bar
Effective swept volume of LP cylinder is 44.92 litres
Temperature of air delivered per stroke is 85.4 oC
Volume of air delivered per stroke is 3.69 litres
Work done per kg of air is 254.1 kJ


Example 5.20 Page no : 271¶

In [51]:
import math

# Variables
P1 = 1.;			#Pressure at point 1 in bar
Pns = 100.;			#Maximum pressure in bar
p = 4.; 			#Pressure ratio

# Calculations
Ns = math.log(Pns)/math.log(p);			#Number of stages
y = math.ceil(Ns);	                		#Rounding off to next higher integer
ps = (Pns/P1)**(1/y);			        #Exact stage pressure ratio
P2 = ps*P1;			#Pressure at point 2 in bar
P3 = ps*P2;			#Pressure at point 3 in bar
P4 = ps*P3;			#Pressure at point 4 in bar

# Results
print 'Number of stages are %3.2f  \
\nExact stage pressure ratio is %3.3f  \
\nIntermediate pressures are %3.3f bar, %3.2f bar, %3.2f bar'%(y,ps,P2,P3,P4)

# rounding off error

Number of stages are 4.00
Exact stage pressure ratio is 3.162
Intermediate pressures are 3.162 bar, 10.00 bar, 31.62 bar