# Chapter 10 : Pumps compressors and turbines¶

### Example 10.1 page no : 362¶

In [3]:
#Calculate the efficiency of a pump

# variables
Q=50.                           #gal/min
P1=30.                          #psia 0r lbf/in**2
P2=100.                         #psia 0r lbf/in**2
dP=P2-P1                        #psia 0r lbf/in**2
power=2.8                       #hp

# calculation
#1 ft = 12 in
#1 hp.min = 33000 lbf.ft
#1 gal = 231 in**3
eta=(Q*dP/power)*(1/33000.0)*231*(1/12.0)           #dimentionless

# result
print "The efficiency of the pump is %f"%eta

The efficiency of the pump is 0.729167


### Example 10.2 page no : 366¶

In [2]:
#Calculate the maximum elevation above the lowest water level in sump at which pump inlet can be placed

# variables
P1=3.72                  #psia 0r lbf/in**2
P2=14.5                  #psia 0r lbf/in**2
dP=P2-P1                 #psia 0r lbf/in**2
rho=61.3                 #lbm/ft**3
g=32.2                   #ft/s**2

# calculation
#1 ft = 12 in
#1 lbf.s**2 = 32.2 lbm.ft
h_loss=4                 #ft
v=10                     #ft/s
h_max=(dP/rho/g)*144*32.2-(v**2/2.0/g)-h_loss                    #ft

# result
print "the maximum elevation above the lowest water level in sump at which pump inlet can be placed is %f ft"%h_max

the maximum elevation above the lowest water level in sump at which pump inlet can be placed is 19.770533 ft


### Example 10.3 page no : 370¶

In [2]:
#Calculate the work requird per pound mole for a 100% efficient isothermal and adiabatic compressor
import math

# variables
R=1.987                          #Btu/lbmol/R (universal gas constant)
T=528.                           #R (Rankine temperature scale)
ratio_P=10.                      #dimentionless

# calculations and results
#for isothermal compressor
W1=R*T*math.log(ratio_P)         #Btu/lbmol
print "The work required per pound mole for a 100 percent efficient isothermal compressor is %f "%W1,"Btu/lbmol\n"

gama=1.4#dimentionless
W2=(gama/(gama-1))*R*T*(ratio_P**((gama-1)/gama)-1)#Btu/lbmol
print "The work required per pound mole for a 100 percent efficient adiabatic compressor is %f "%W2,"Btu/lbmol\n"

The work required per pound mole for a 100 percent efficient isothermal compressor is 2415.724914  Btu/lbmol

The work required per pound mole for a 100 percent efficient adiabatic compressor is 3417.499724  Btu/lbmol



### Example 10.4 page no : 370¶

In [4]:
#Calculate the work requird per pound mole for a 100% efficient 2 stage adiabatic compressor

# variables
R=1.987                              #Btu/lbmol/R (universal gas constant)
T=528.0                              #R (Rankine temperature scale)
ratio_P1=3                           #dimentionless
ratio_P2=10/3.0                      #dimentionless
gama=1.4                             #dimentionless

# calculation
W=(gama/(gama-1))*R*T*((ratio_P1**((gama-1)/gama)-1)+(ratio_P2**((gama-1)/gama)-1))       #Btu/lbmol

# result
print "The work required per pound mole for a 100 percent efficient adiabatic compressor is %f "%W,
print "Btu/lbmol"

The work required per pound mole for a 100 percent efficient adiabatic compressor is 2861.592127  Btu/lbmol


### Example 10.5 page no : 373¶

In [5]:
#Calculate the pump head
import math

# variables
N=1750.                                       #rev/min
#1 min 60 sec
Q=100.                                        #gal/min
#1 gallon = 231 in**3
#1 ft =12 in
#1 min = 60 sec
d_inlet = 2.067                               #ft

# calculations
A_inlet=(math.pi)/4.*(d_inlet**2)             #ft**2
V1=(Q/A_inlet)*231/60.0/12.0                  #ft/s
d_outlet = 1.61                               #ft
A_outlet=(math.pi)/4*(d_outlet**2)            #ft**2
V2=(Q/A_outlet)*231/60.0/12.0                 #ft/s
g=32.2                                        #ft/s**2
d_inner=0.086                                 #ft
d_outer=0.336                                 #ft
h=(omega)**2/g*((d_outer**2)-(d_inner)**2)+(V2**2-V1**2)/2./g               #ft

#result
print "The pump head is %f ft"%h

The pump head is 112.471747 ft


### Example 10.6 page no : 378¶

In [2]:
#Calculate the pump head

# variables
rho=62.3                 #lbm/ft**3
g=32.2                   #ft/s**2
v=18.46                  #ft/s

# calculation
#1 lbf/s**2 = 32.2 lbm.ft
h=1/(rho*g) * (v**2/2)*32.2              #ft

# result
print "The pump head is %f ft"%h

The pump head is 2.734925 ft


### Example 10.7 page no : 381¶

In [7]:
#Calculate the estimated pressure rise in the first stage of mutisatge centrifugal compressor

# variables
rho=0.075                    #lbm/ft**3
d=2.                         #ft

# calculation
dP=(1/2.0)*(rho)*(omega*d/2.0)**2/32.2/144.0                 #psia
#1 lbf.s**2 = 32.2 lbm into feed
#1 ft = 144 in**2

# result
print "the estimated pressure rise in the first stage of mutisatge centrifugal compressor is %f psia"%dP

the estimated pressure rise in the first stage of mutisatge centrifugal compressor is 8.865562 psia


### Example 10.8 page no : 383¶

In [8]:
#Calculate the efficiency of a compressor and the change respective change in temperature

# variables
m=100.0                       #Kg/hr
M=29.                         #gm/mol
gama=1.4                      #dimentionless
R=8.314                       #J/mol/K
T=293.15                      #K
ratio_P=4.                    #dimentionless

# calculations
Po=(m/M)*R*T*(gama/(gama-1))*((ratio_P)**((gama-1)/gama)-1)/3600.0              #kW
P_real=5.3                    #kW
eta=Po/P_real                 #dimentionless

# result
print "The efficiency of the compressor is %f"%eta
Cp=29.1                       #J/mol/K
dT_real=P_real*(M/m)*3600/Cp                      #K
print "dT_real = %d K"%dT_real
dT_isentropic=Po*(M/m)*3600/Cp                    #K
print "dT_isentropic = %d K"%dT_isentropic

The efficiency of the compressor is 0.749243
dT_real = 190 K
dT_isentropic = 142 K