Chapter 19: Gas Compressors

Ex19.1:pg-818

In [2]:
import math
T2 = 488.0
T1 = 298.0 
n = 1.3 
R =8314.0/44.0
rp = (T2/T1)**(n/(n-1))

b = 0.12 # Bore of compressor
L = 0.15 # Stroke of compressor
V1 = (math.pi/4)*(b)**2*L  
P1 = 120e03 # in kPa
W = ((n*P1*V1)/(n-1))*(((rp)**((n-1)/n))-1)
P = (W*1200*0.001)/60  

V1_dot = V1*(1200.0/60.0)
m_dot = (P1*V1_dot)/(R*T1)

rp_1 = rp**2
V2 = (1/rp)**(1/n)*V1
d = math.sqrt((V2*4)/(L*math.pi))
print "\n Example 19.1\n"
print "\n Pressure ratio is ",rp
print "\n Indicated power is ",P ," kW"
print "\n Shaft power is ",P/0.8 ," kW"
print "\n Mass flow rate is ",m_dot ," kg/s"
print "\n Pressure ratio when second stage is added is ",rp_1
print "\n Volume derived per cycle is V2 ",V2 ," m**3"
print "\n Second stage bore would be ",d*1000 ," mm"
#The answers vary due to round off error
 Example 19.1


 Pressure ratio is  8.4764775804

 Indicated power is  11.2490101513  kW

 Shaft power is  14.0612626891  kW

 Mass flow rate is  0.0723071537289  kg/s

 Pressure ratio when second stage is added is  71.8506721711

 Volume derived per cycle is V2  0.000327741753347  m**3

 Second stage bore would be  52.7442736748  mm

Ex19.2:pg-819

In [3]:
import math
c = 0.05 # Clearance volume
p1 = 96.0 # Inlet ressure in bar
p2 = 725.0 # Outlet pressure in bar
pa = 101.3 # Atmospheric pressure
Ta = 292.0 # Atmospheric temperature in kelvin
T1 = 305.0 # Inlet temperature in Kelvin
n = 1.3 # polytropic index
print "\n Example 19.2 \n "
n_v = (1+c-c*((p2/p1)**(1/n)))*(p1/pa)*(Ta/T1)
print "\n Volumetric efficiency of system is ",n_v*100 ," percent"
# Answer is not mentioned in book
 Example 19.2 
 

 Volumetric efficiency of system is  73.7793963433  percent

Ex19.3:pg-819

In [5]:
import math
P1 = 101.3e03 
P4 = P1 # in Pa
P2 = 8*P1  
P3 = P2
T1 = 288 
Vs = 2000
V3 = 100 
Vc = V3
V1 = Vs + Vc 
n = 1.25 
R = 287
V4 = ((P3/P4)**(1/n))*V3
W = ((n*P1*(V1-V4)*1e-06)/(n-1))*(((P2/P1)**((n-1)/n))-1)
P = (W*800*0.001)/60  

m = (P1*(V1-V4)*1e-06)/(R*T1)
m_dot = m*800

FAD = (V1-V4)*1e-06*800

Wt = P1*(V1-V4)*1e-06*math.log(P2/P1)
n_isothermal = (Wt*800*0.001)/(P*60)

Pi = P/0.85
n_v =100*(V1-V4)/Vs
print "\n Example 19.3\n"
print "\n Indicated poer is ",P ," kW"
print "\n Volumetric efficiency is ",n_v ," percent"
print "\n Mass flow rate is ",m_dot ," kg/min"
print "\n Free air delivery is ",FAD ," m**3/min"
print "\n Isothermal efficiency is ",100*n_isothermal ," percent"
print "\n Input power is ",Pi ," kW"

#The answers vary due to round off error
 Example 19.3


 Indicated poer is  5.47565638255  kW

 Volumetric efficiency is  78.6098417845  percent

 Mass flow rate is  1.54145895718  kg/min

 Free air delivery is  1.25775746855  m**3/min

 Isothermal efficiency is  80.6428056306  percent

 Input power is  6.44194868535  kW

Ex19.4:pg-819

In [6]:
import math
# Given that
m = 3.0 # Mass flow rate in kg/min
p1 = 1.0 # Initial pressure in bar
T1 = 300.0 # Initial temperature in K
p3 = 6.0 # Pressure after compression in bar
p5 = 15.0 # Maximum pressure in bar
N = 300.0 # Rpm of compressure
n = 1.3 # Index of compression and expansion 
r = 1.5 # Stroke to bore ratio
R = 287.0 # Gas constant of air
t = 15.0 # Temperature in degree centigrade
print "\n Example 19.4\n"
T = t+273
Wc = (n/(n-1))*(m/60)*(R*(1e-3)*T1)*(((p3/p1)**((n-1)/n))-1)
r1 = (p5/p1)**(1.0/n)# Where r1 = V1/Vc
r2 = r1-1 # Where r2 = Vs/Vc
r3 = (p3/p1)**(1.0/n)
n_vol = (r1-r3)*(T/T1)/r2
V = m*R*T/(2*(1e5)*N)
Vs = V/n_vol
d = (Vs*4/(math.pi*r))**(1.0/3.0)
l = r*d
print "\n Power input is ",Wc ," kW, \n Volumetric efficiency is ",n_vol*100 ," percent, \n Bore of the cylinder is ",d ," m, \n Stroke of the cylinder is ",l ," m"
#The answers vary due to round off error
 Example 19.4


 Power input is  9.55276123312  kW, 
 Volumetric efficiency is  55.4657309635  percent, 
 Bore of the cylinder is  0.184932327621  m, 
 Stroke of the cylinder is  0.277398491431  m

Ex19.5:pg-820

In [9]:
import math
# Given that
d = 15.0 # Diameter in cm
l = 18.0 # Stroke in cm
C = 0.04 # Ratio of clearance volume and sweft volume
p1 = 1.0 # Pressure in bar
t1 = 25.0 # Temperature in degree centigrade
p2 = 8.0# Pressure in bar
N = 1200.0 # Rpm of compressure 
W = 18.0 # Actual power input in kW
m = 4.0 # Mass flow rate in kg/min
R = 0.287
print "\n Example 19.5\n"
T1 = t1+273
v = R*T1/(p1*100)
V = m*v
Vs = (math.pi/4)*((d*(1e-2))**2)*(l*1e-2)*N
n_vol = V/Vs
n = (math.log(p2/p1))/(math.log((1+C-n_vol)/C))
# The value of n given in the example is wrong
n = 1.573
T2 = T1*(p2/p1)**((n-1)/n)
Wc = (n/(n-1))*(m*R/60)*(T2-T1)
n_mech = Wc/W
W_isothermal = m*R*T1*math.log(p2/p1)/60
n_iso = W_isothermal/W
print "\n Power required to drive the unit is ",Wc ," kW,\n Isothermal efficiency is ",n_iso*100 ," percent,\n Mechanical efficiency is ",n_mech*100 ," percent"
#The answers vary due to round off error
 Example 19.5


 Power required to drive the unit is  17.7326053799  kW,
 Isothermal efficiency is  65.8690064051  percent,
 Mechanical efficiency is  98.5144743328  percent

Ex19.6:pg-820

In [12]:
import math
# Given that
d = 40.0 # Diameter in cm
l = 50.0 # Stroke in cm
D = 5.0 # Piston rod diameter in cm
C = 0.04 # Ratio of clearance volume and sweft volume
p1 = 1.0 # Pressure in bar
t1 = 15.0 # Temperature in degree centigrade
p2 = 7.5# Pressure in bar
N = 300.0 # Rpm of compressure 
n_vol = 0.8 # Volumetric efficiency
n_mech = 0.95 # Mechanical efficiency
n_iso = .7 # Isothermal efficiency
R = 0.287
print "\n Example 19.6\n"
Vs = (math.pi/4)*((d*(1e-2))**2)*(l*(1e-2))
Vs_ = (math.pi/4)*(((d*(1e-2))**2)-(D*(1e-2))**2)*(l*1e-2)
Vs_min = (Vs+Vs_)*2*N
V1 = Vs_min*n_vol
W_iso = p1*V1*(math.log(p2/p1))
Win = W_iso/n_iso
Wc = Win/n_mech
print "\n Power required to drive the compressure is ",Wc ," kW"
#The answers vary due to round off error
 Example 19.6


 Power required to drive the compressure is  181.333212391  kW

Ex19.7:pg-820

In [14]:
import math
# Given that
p1 = 1.0 # Pressure in bar
t1 = 27.0 # Temperature in degree centigrade
n = 1.3 # Index of the compression process
p3 = 9.0# Pressure in bar
R = 0.287
print "\n Example 19.7\n"
T1 = t1+273
p2 = math.sqrt(p1*p3)
Wc = ((2*n*R*T1)/(n-1))*(((p2/p1)**((n-1)/n))-1)
T2 = T1*((p2/p1)**((n-1)/n))
H = 1.005*(T2-T1)
print "\n Minimum work done is ",Wc ," kJ/kg,\n Heat rejected to intercooler is ",H ," kJ/kg"
#The answers vary due to round off error
 Example 19.7


 Minimum work done is  215.324046  kJ/kg,
 Heat rejected to intercooler is  87.0010719231  kJ/kg

Ex19.8:pg-820

In [18]:
import math
# Given that
V = 4.0 # Volume flow rate in m**3/min
p1 = 1.013 # Pressure in bar
t1 = 15.0 # Temperature in degree centigrade
N = 250.0 # Speed in RPM
p4 = 80.0# Delivery pressure in bar
v = 3.0 #Speed of piston in m/sec
n_mech = .75 # Mechanical efficiency 
n_vol = .8 # Volumetric efficiency
n = 1.25 # Polytropic index
print "\n Example 19.8\n"
T1 = t1+273
p2 = math.sqrt(p1*p4)
W = (2*n/(n-1))*(p1*100/n_mech)*(V/60)*((p2/p1)**((n-1)/n) - 1)
L = v*60/(N*2)
Vs = V/N
D_LP = math.sqrt(Vs*V/(math.pi*L*n_vol))
D_HP = D_LP*math.sqrt(p1/p2)
print "\n Minimum power required by the compressure is ",W ," kW,\n Bore of the compressure in low pressure side is ",D_LP*100 ," cm,\n Bore of the compressure in high pressure side is ",D_HP*100 ," cm,\n Stroke of the compressure is ",L*100 ," cm"
#The answers vary due to round off error
 Example 19.8


 Minimum power required by the compressure is  49.3370051888  kW,
 Bore of the compressure in low pressure side is  26.5961520268  cm,
 Bore of the compressure in high pressure side is  8.92172168806  cm,
 Stroke of the compressure is  36.0  cm

Ex19.9:pg-820

In [20]:
import math
# Given that
p1 = 1.0 # Pressure in bar
T1 = 300.0 # Temperature in K
p4 = 9.0# Compressed pressure in bar
n = 1.3 # Polytropic index
R = 0.287 # Gas constant in kJ/kgK
cp = 1.042 # Heat capapcity in kJ/kgK
print "\n Example 19.9\n"
p2 = math.sqrt(p1*p4)
T2 =T1*((p2/p1)**((n-1)/n))
Wc = (2*n/(n-1))*R*1*(T2-T1)
Wc_ = Wc/2
Q = 1*cp*(T2-T1)
Q_ = cp*(T1-T2)+Wc_
H = Q+2*Q_
print "\n Compressor work = ",Wc_ ," kJ/kg,\n Total heat transfer to the surrounding = ",H ," kJ/kg"
#The answers given in the book contain calculation error
 Example 19.9


 Compressor work =  107.662023  kJ/kg,
 Total heat transfer to the surrounding =  125.119949539  kJ/kg

Ex19.10:pg-820

In [23]:
import math
# Given that
N = 300.0 # Speed in RPM
# Intake condition of compressor
p1 = 0.98 # Pressure in bar
T1 = 305.0 # Temperature in K

p6 = 20.0# Delivery pressure in bar
p3 = 5.0 # Intermediate pressure in bar
C = .04 # Ratio of clearance volume to the stroke volume
v = 3.0 # Volume flow rate of compressure in m**3/min
p = 1.0 # pressure in bar
t = 25.0 # Temperautre in degree centigrade
n = 1.3 # Polytropic index
R = 0.287 # Gas constant in kJ/kgK
print "\n Example 19.10\n"
T = t+273
r0 = 1+C # Where r0 = v1/vs
r1 = C*(p3/p1)**(1/n)# Where r1 = v4/vs
r2=r0-r1#Where r2 is the ratio of volume of air taken at 0.98 bar,305 k and vs
r3 = r2*(T/T1)*p1/p # Where r3 is the ratio of volume of air taken at free air conditions and vs
n_vol = r3
m = p*(1e5)*(v/60)/(R*1000*T)
T2 = T1*((p3/p1)**((n-1)/n))
# For perfect intercooling
T5 = T1
p5 = p3
T6 = T5*((p6/p5)**((n-1)/n))
Wc = (n/(n-1))*m*R*((T2-T1)+(T6-T5))
m_a_s = m*60/N
v_fa_s = m_a_s *(R*1000)*T/(p*1e5)
d = ((v_fa_s/n_vol)*(4/math.pi))**(1.0/3.0)
l = d # As given in the question
P_iso = m*R*T1*(math.log(p6/p1))
n_iso = P_iso/Wc
print "\n Diameter of cylinder = ",Wc,d*100 ," cm, \n Storke of the cylinder = ",l*100 ," cm,\n Isothermal efficiency = ",n_iso*100 ," percent"
#The answers given in the book contain calculation error
 Example 19.10


 Diameter of cylinder =  18.484702902 24.5391705107  cm, 
 Storke of the cylinder =  24.5391705107  cm,
 Isothermal efficiency =  83.4955018622  percent

Ex19.11:pg-820

In [24]:
import math
# Given that
p1 = 1 # Intake pressure of compressor in bar
T1 = 298 # Intake temperature in K
p_d = 36 # Delivery pressure in bar
T2 = 390 # Maximum temperature in any stage in K
n = 1.3 # Polytropic index
R = 0.287
print "\n Example 19.11\n"
r = (T2/T1)**(n/(n-1))
N = math. ceil(r)
p2 = (p_d/p1)**(1/N)
p3 = (p_d/p1)**(2/N)
p4 = (p_d/p1)**(3/N)
Wc = (N*n*R*T1/(n-1))*((p_d/p1)**((n-1)/(N*n))-1)
Wc_ = (n/(n-1))*(1*R*T1)*((p_d/p1)**((n-1)/n)- 1)
T = T1*((p2/p1)**((n-1)/n))
print "\n No of stages for min power input = ",N ,",\n Power required = ",Wc ," kW/kg air,\n The power required for a single stage compressor = ",Wc_ ," kW,\n Maximum temperature in any stage = ",T ," K"
#The answers given in the book contain round off error
 Example 19.11


 No of stages for min power input =  1.0 ,
 Power required =  476.74544125  kW/kg air,
 The power required for a single stage compressor =  476.74544125  kW,
 Maximum temperature in any stage =  681.338601917  K

Ex19.12:pg-820

In [25]:
import math
# Given that
p1 = 700.0 # Intake pressure of compressor in kPa
t1 = 38.0 # Intake temperature in degree centigrade
c = 0.4 # Ratio of cutoff volume to stroke volume
p3 = 112.0 # Back pressure in kPa
r = 0.85 # Ratio of area of actual indicator diagram to the outlined in the question
n = 1.3 # Polytropic index
R = 0.287
m = 1.25 # Air mass in kg
print "\n Example 19.12\n"
T1 = t1+273
T2 = T1/((1/c)**(n-1))
p2 = p1*(c**n)
V2 = m*R*T2/p2
v2 = V2/m
A = R*T1 + R*(T1-T2)/(n-1) - p3*v2
Io = A*r*m
print "\n Indicated output = ",Io ," kJ"
# The answer given in the book vary due to round off error
 
 Example 19.12


 Indicated output =  132.877965499  kJ

Ex19.13:pg-820

In [26]:
import math
# Given that
d = 450.0 # Bore of low pressure cylinder in mm
l = 300.0 # Stroke in mm
c = 0.05 # Ratio of clearance volume to sweft volume
p1 = 1.0 # Intake pressure in bar
t1 = 18.0 # Intake temperature in degree centigrade
p4 = 15.0 # Delivery pressure in bar
n = 1.3 # Compression and expansion index
R = 0.29 # Gas constant in kJ/kgK
print "\n Example 19.13\n"
T1 = t1+273
r = (p4/p1)**(1.0/3.0)
p2 = p1*r
p3 = p2*r
Vs = (math.pi/4)*((d*1e-3)**2)*(l*1e-3)
V11 = c*Vs
V1 = Vs +V11
V12 = V11*((r)**(1.0/n))
Vs_e = V1 - V12
T3 = T1
T5 = T3
T6 = T1*(r**((n-1)/n))
t6 = T6-273
V6_7 = (p1/p4)*(T6/T1)*(V1 - V12)
W = (3*n*R*T1/(n-1))*((p2/p1)**((n-1)/n)-1)
print "\n The intermediate pressure are - \n p2 = ",p2 ," bar,\n p3 = ",p3 ," bar,\n The effective sweft volume = ",Vs ," m**3,\n Temperature of air delivered per stroke at 15 bar = ",t6 ," degree centigrade,\n The work done per kg of air = ",W ," kJ"
# The answers given in the book vary due to round off error
 Example 19.13


 The intermediate pressure are - 
 p2 =  2.46621207433  bar,
 p3 =  6.08220199557  bar,
 The effective sweft volume =  0.0477129384264  m**3,
 Temperature of air delivered per stroke at 15 bar =  85.3946742162  degree centigrade,
 The work done per kg of air =  254.077921795  kJ

Ex19.14:pg-820

In [27]:
import math
# Given that
p1 = 1.013 # Inlet pressure in bar
r = 1.5 # Pressure ratio
Vs = 0.03 # Induce volume of air in m**3/rev
gama = 1.4 
print "\n Example 19.14\n"
p2 = p1*r
W = (p2-p1)*Vs*100
pi = (p1+p2)/2
A_A = (gama/(gama-1))*(p1*Vs)*((pi/p1)**((gama-1)/gama)-1)*100
Vb = Vs *((p1/pi)**(1/gama))
A_B = (p2-pi)*Vb*100
Wr = A_A + A_B
print "\n Work input = ",W ," kJ/rev,\n Work input for a vane-type compressor = ",Wr ," kJ/rev"
# The answers given in the book vary due to round off error
 
 Example 19.14


 Work input =  1.5195  kJ/rev,
 Work input for a vane-type compressor =  1.34802979062  kJ/rev

Ex19.15:pg-820

In [28]:
import math
# Given that
m = 1.0 # Mass flow rate in kg/s
r = 2.0 # Prssure ratio of blower 
t1 = 70.0 # Inlet temperature in degree centigrade
p1 = 1.0 # Inlet pressure in bar
R = 0.29 # Gas constant in kJ/kgK
x = 0.7 # Reduction in pressure ratio and intake volume 
gama = 1.4
print "\n Example 19.15\n"
T1 = t1+273
V = m*R*T1/(p1*100)
P = V*(p1*r-p1)*100
p2 = p1*((1/x)**(gama))
V2 = x*V
P_ = (gama/(gama-1))*(p1*100*V)*((p2/p1)**((gama-1)/gama)-1) + V2*(p1*r-p2)*100

print "\n Power required to drive the blower = ",P ," kW,\n Power required = ",P_ ," kW"
# The answers given in the book vary due to round off error
 
 Example 19.15


 Power required to drive the blower =  99.47  kW,
 Power required =  77.9220893777  kW

Ex19.16:pg-820

In [29]:
import math
# Given that
r1 = 2.5 # Pressure ratio of compressor for first stage
r2 = 2.1 # Pressure ratio of compressor for second stage
m = 5.0 # Mass flow rate of air in kg/s 
t1 = 10.0 # Inlet temperature in degree centigrade
p1 = 1.013 # Inlet pressure in bar
td = 50.0 # Temperature drop in intercooler in degree centigreade
n_iso = .85 # Isentropic efficiency
cp = 1.005 # Heat capacity of air in kJ/kgK
x = 0.7 # Reduction in pressure ratio and intake volume 
gama = 1.4 # Ratio of heat capacities for air
print "\n Example 19.16\n"
T1 = t1+273
T2s = T1*((r1)**((gama-1)/gama))
T2 = T1 + (T2s-T1)/n_iso
T3 = T2 - td
T4s = T3*((r2)**((gama-1)/gama))
T4 = T3 + (T4s-T3)/n_iso
P = m*cp*((T2-T1)+(T4-T3))
print "\n Actual temperature at the end of first stage = ",T2 ," K,\n Actual temperature at the end of second stage = ",T4 ," K,\n The total compressor power = ",P ," kW"
# The answers given in the book vary due to round off error
 Example 19.16


 Actual temperature at the end of first stage =  382.63704941  K,
 Actual temperature at the end of second stage =  425.041961043  K,
 The total compressor power =  965.01085424  kW

Ex19.17:pg-821

In [30]:
import math
# Given that
r = 2.5 # Static pressure ratio of supercharger 
p1 = 0.6 # Static inlet pressure in bar
t1 = 5 # Static inlet temperature in degree centigrade
A_r = 13.0 # Air-fuel ratio
m = 0.04 # The rate of fuel consumed by the engine in kg/s
gama= 1.39 # For air-fuel mixture 
cp = 1.005 # Heat capacity for air-fuel mixture in kJ/kgk
n_iso = .84 # Isentropic efficiency of compressor 
v = 120.0 # Exit velocity from the compressor in m/s
print "\n Example 19.17\n"
T1 = t1+273
T2s = T1*((r)**((gama-1)/gama))
T2 = T1 +(T2s-T1)/n_iso
m_g = m*(A_r+1)
P = m_g*cp*(T2-T1)
T02 = T2 + (v**2)/(2*cp*1000)
t02 = T02-273
p02 = p1*r*((T02/T2)**(gama/(gama-1)))*100
print "\n Power required to drive the compressor = ",P ," kW,\n Stagnatio temperature = ",t02 ," degree centigrade,\n Stagnation pressure = ",p02 ," kPa"
# The answers given in the book vary due to round off error
 
 Example 19.17


 Power required to drive the compressor =  54.6039650117  kW,
 Stagnatio temperature =  109.18614963  degree centigrade,
 Stagnation pressure =  160.465577551  kPa

Ex19.18:pg-821

In [31]:
import math
# Given that
N = 10000 # Speed in RPM
V = 1.2 # Volume flow rate of free air in m**3/s
p1 = 1.0 # Inlet pressure in bar
t1 = 27.0 # Inlet temperature in degree centigrade
r = 5.0 # Pressure ratio
vf = 60.0 # Velocity flow rate in m/s
sigma = 0.9 # Slip factor
n_iso = 0.85 # Isentropic efficiency
gama = 1.4
R = 0.287
cp = 1.005
print "\n Example 19.18\n"
T1 = t1+273
T2s = T1*((r)**((gama-1)/gama))
T2 = T1 +(T2s-T1)/n_iso
m = p1*100*V/(R*288)
Wc = m*cp*(T2-T1)
Vb2 = (Wc*1000/(m*sigma))**(1.0/2.0)
D = Vb2*60/(math.pi*N)
Vb1 = Vb2/2
beta1 = math.atan(vf/Vb1)
alpha = math.atan(vf/(sigma*Vb2))
print "\n The temperature of air at outlet = ",T2-273 ," degree centigrade,\n Power input = ",Wc ," kW,\n Diameter of impeller = ",D ," m, \n Blade inlet angle = ",beta1 ," degree,\n Diffuser inlet angle = ",alpha ," degree "
# The answers given in the book vary due to round off error
 
 Example 19.18


 The temperature of air at outlet =  233.053979565  degree centigrade,
 Power input =  300.644961473  kW,
 Diameter of impeller =  0.916122726914  m, 
 Blade inlet angle =  0.245135262084  degree,
 Diffuser inlet angle =  0.138096713577  degree 

Ex19.19:pg-821

In [36]:
import math
# Given that
N = 264 # Speed in RPS
sigma = 0.91 # Slip factor
d = 0.482 # Impeller diameter in m
D = 0.306 # Impeller eye diameter
D_ = 0.153 # Impeller root eye diameter in m
vf = 138 # Uniform axial inlet velocity in m/s
V = 1.2 # Volume flow rate of free air in m**3/s
m = 9.1 # Air mass flow rate in kg/s
T1 = 294 # Inlet air stagnation temperature in K
n_iso = 0.8 # Total head isentropic efficiency
n_mech = 0.98 # Mechanical efficiency
gama = 1.4 # Ratio of heat capacities
cp = 1.006 # Heat capacity in kJ/kgK
print "\n Example 19.19\n"
Wc = m*sigma*(2*math.pi*d*N/2)/1000
P_e = Wc/n_mech
delta_T = Wc/(m*cp)
delta_T_ideal = delta_T*n_iso
T2_i = delta_T_ideal + T1
r = (T2_i/T1)**(gama/(gama-1)) # Where r = p02/p01
Vb = 2*math.pi*N*D/2
V_er = (2*math.pi*N*D_/2)
beta1 = math.atan(vf/Vb)
beta2 = math.atan(vf/V_er)
beta1_ = (beta1 - math.floor(beta1))*60
beta2_ = (beta2 - math.floor(beta2))*60
print "\n Total head pressure ratio = ",r ,", \n The required power at input shaft = ",P_e ," kW,\n Inlet angle at the root = ",math.floor(beta1) ," degree and ",beta1_ ," minute,\n Inlet angle at the tip = ",math.floor(beta2) ," degree and ",beta2_ ," minute"
# The answers given in the book for total head pressure ratio and required power at input shaft contain calculation error
 
 
 Example 19.19


 Total head pressure ratio =  1.00344817308 , 
 The required power at input shaft =  3.37798367776  kW,
 Inlet angle at the root =  0.0  degree and  29.8821913183  minute,
 Inlet angle at the tip =  0.0  degree and  49.6377044903  minute

Ex19.20:pg-821

In [38]:
import math
# Given that
N = 16000.0 # Speed in RPM
t1 = 17.0 # Intake temperture of gas in degree centigrade
rp = 4.0 # Pressure ratio
sigma = 0.85# Slip factor
n_iso = 0.82 # Isentropic efficiency
alpha_wirl = 20.0 # Pre-wirl angle in degree
d1 = 200.0 # Mean diameter of impeller eye in mm
V1 = 120.0 #Absolute air velocity in m/s
gama = 1.4 # Ratio of heat capacities
cp = 1.005 # Heat capacity in kJ/kgK
print "\n Example 19.20\n"
T1 = t1 + 273
T2s = T1*((rp)**((gama-1)/gama))
delta_Ts = T2s-1
delta_T = delta_Ts/n_iso
Wc = 1 *cp*delta_T
Vb1 = (math.pi*d1*(1e-3)*N)/60
Vw1 = V1*math.sin(alpha_wirl)
Vb2 = 459.78 # By solving quadratic equation 172.81e3=0.85*Vb2**2-167.55*41.05
d2 = Vb2*60/(math.pi*N)

print "\n Impeller tip diameter = ",d2*1000 ," mm"
# The answer given in the book varies due to round off error
 
 Example 19.20


 Impeller tip diameter =  548.821948011  mm

Ex19.21:pg-821

In [41]:
import math
# Given that
m = 2.5 # Mass flow rate in kg/s
p1 = 1.0 # Inlet pressure in bar
T1 = 300.0 # Inlet temperature in bar
n_s = 0.88 # Stage efficiency
Wc = 600.0 # Power input in kW
delta_t = 21.0 # Temperature rise in first stage in degree centigrade
gama = 1.4 # Ratio of heat capacities 
cp = 1.005 # Heat capacity in kJ/kgK
print "\n Example 19.21\n"
x = n_s*gama/(gama-1)# Where x = (n/(n-1))
T = Wc/(m*cp)+T1
p = p1*((T/T1)**(x))
T2 = T1 + n_s*delta_t
r = ((T2/T1)**(gama/(gama-1)))# Where r = p2/p1
N = math.log(p/p1)/math.log(r)
N_ = math. ceil(N)
Ts = T1*(p/p1)**((gama-1)/gama)
n_inter = (Ts-T1)/(T-T1)
print "\n The delivery pressure = ",p ," bar,\n The no of stages = ",N_ ,",\n The internal efficiency = ",n_inter ," "
 Example 19.21


 The delivery pressure =  6.07125291521  bar,
 The no of stages =  9.0 ,
 The internal efficiency =  0.84689822539  

Ex19.22:pg-821

In [42]:
import math
# Given that
D = 0.5 # Mean diameter of impeller in m
N = 15000.0 # Speed in RPM
Vf = 230.0 # Velocity of flow in m/s
p1 = 1.0  # Inlet pressure in bar
T1 = 300.0 # Inlet temperature in K
Vw1 = 80.0 # Velocity of whirl at inlet in m/s
n_s = 0.88 # Stage efficiency
rp = 1.5 # Pressure ratio
gama = 1.4 
cp = 1.0005
print "\n Example 19.22\n"
Vb = (math.pi*D*N/60)
Ts = T1*((rp)**((gama-1)/gama))
T = T1 + (Ts-T1)/n_s
Wc = cp*(T-T1)
Vw2 = Vw1 + (Wc*1000)/(Vb)
beta1 = math.atan(Vf/(Vb-Vw1))
beta2 =  math.atan(Vf/(Vb-Vw2))
theta = beta2-beta1
R = 1-((Vw1+Vw2)/(2*Vb))

print "\n Fluid deflection angle = ",theta ," degree,\n Power input = ",Wc ," kJ/kg,\n The degree of reaction = ",R*100 ," percent"
# The answers given in the book vary because of round off error
 
 Example 19.22


 Fluid deflection angle =  0.206163966177  degree,
 Power input =  41.8928434516  kJ/kg,
 The degree of reaction =  66.0453433333  percent

Ex19.23:pg-821

In [43]:
import math
# Given that
v = 5.0 #olume flow rate in m**3/s
d = 1.0 #ean impeller diameter in m
D = 0.6 # Hub diameter in m
N = 600.0 #otational speed in RPM
h = 35.0 #heoratical head in mm
rho = 1.2 # Density of air in kg/m**3
rho_w = 1000.0 #ensity of water in kg/m**3
print "\n Example 19.23\n"
Vf = v*4/(math.pi*(d**2 - D**2))
Vb = (math.pi*d*N/60)
Vb_ = (math.pi*D*N/60)
H = h/rho
Vw2 = H*9.81/(Vb)
Vw2_ = H*9.81/(Vb_)
beta_tip = (Vf/(Vb_-Vw2))
beta_hub = (Vf/(Vb_-Vw2_))
print "\n Blade angle at the tip = ",beta_tip ," degree,\n Blade angle at the hub = ",beta_hub ," degree"
# The answers given in the book vary because of round off error
 
 Example 19.23


 Blade angle at the tip =  1.02107077046  degree,
 Blade angle at the hub =  2.71029118833  degree

Ex19.24:pg-821

In [45]:
import math
# Given that
N0 = 9000.0 # Rotational speed in RPM
Q = 6.0 # Volume flow rate in m**3/s
p1 = 1.0 # Initial pressure in bar
t1 = 25.0 # Initial temperature in degree centigrade
p2 = 2.2 # Compressed pressure in bar
n = 1.33 # Compression index
Vf = 75.0 # Velocity of flow in m/s
beta1 = 30.0 # Blade angle at inlet in degree
beta2 = 55.0 # Blade angle at outlet in degree
d = 0.75 # Diameter of impeller in m
cp = 1.005 
print "\n Example 19.24\n"
T1 = t1+273
T2 = T1*(p2/p1)**((n-1)/n)
Wc = cp*(T2-T1)
x = Wc # Where x = Vw2*Vb2
y = Vf/math.tan(beta2)# Where y = Vb2-Vw2(Equation 1)
z = (y**2 +4*x*1000)**(0.5) # Where z = Vw2+Vb2(Equation 2)
# By solving Equation 1 and Equation 2
Vb2 = (y+z)/2
Vw2 = ((z-y)/2)
N = Vb2*60/(math.pi*d)
Vb1 = Vf/math.tan(beta1)
D1 = Vb1*60/(math.pi*N)
b1 = Q/(math.pi*D1*Vf)
Q_ = Q* (1/p2)*(T2/T1)
b2 = Q_/(math.pi*d*Vf)
print "\n Speed of impeller = ",N ," RPM,\n Impeller width at inlet = ",b1*100 ," cm,\n Impeller width at outlet = ",b2*100 ," cm,"
# The answers given in the book vary because of round off error
 Example 19.24


 Speed of impeller =  6456.85894335  RPM,
 Impeller width at inlet =  -73.5259022616  cm,
 Impeller width at outlet =  1.87680083777  cm,