Chapter 3 : Macroscopic properties of pure substances¶

Example 3.1 pg : 58¶

In [2]:

# Variables
V = 1.  			#ft**3
m = 30. 			#lbm

# Calculations
v = V/m
vf1 = 0.01665
vfg1 = 32.38 			#ft**3/lbm
x1 = 0.000515
uf1 = 169.92
ufg1 = 904.8
u1 = uf1+x1*ufg1
vfg = 0.0216
vfg2 = 0.4240
v2 = v
x2 = 0.0277
uf2 = 538.4
ufg2 = 571.
u2 = uf2+x2*ufg2
Q = m*(u2-u1)

# Results
print "Heat transfer  =  %d Btu"%(round(Q,-2))

Heat transfer  =  11500 Btu


Example 3.2 pg : 59¶

In [3]:

# Variables
V2 = 2.5 			#ft**3
V1 = 0.5 			#ft**3
P = 100. 			#psia
x1 = 0.5

# Calculations
W = -P*144*(V2-V1)
vf1 = 0.01774
vfg1 = 4.414
v1 = vf1+x1*vfg1
m = V1/v1
v2 = V2/m
uf1 = 298.08
ufg1 = 807.1
u1 = uf1+x1*ufg1
h2 = 1747.9
u2 = h2-P*144*v2/778
Q = m*(u2-u1)

# Results
print "Amount of heat  =  %d Btu"%(Q)
#The answer for u2 is given wrong in the textbook. Please use a calculator to find it

Amount of heat  =  188 Btu


Example 3.3 pg : 60¶

In [4]:

# Variables
V1 = 1.735*10**-4 			#ft**3
v1 = 0.016080    			#ft**3/lbm
h1 = 70.61 		    	    #B/lbm
P1 = 100. 		        	#psia
V2 = 1. 		    	    #ft**3

# Calculations
u1 = h1-P1*v1*144/778.
m = V1/v1
v2 = V2/m
vf2 = 0.01613
vfg2 = 350.3
x2 = (v2-vf2)/vfg2
hf2 = 67.97
hfg2 = 1037.2
h2 = hf2+x2*hfg2
P2 = 0.9492
u2 = h2- P2*144*v2/778.
Q = m*(u2-u1)

# Results
print "Enthalpy change  =  %.2f Btu"%(Q)

Enthalpy change  =  2.76 Btu


Example 3.4 pg : 64¶

In [6]:

# Variables
P = 20. 			#psia
V = 1.   			#ft**3
T = 560. 			#R
cv = 0.1715
Q = 10. 			#Btu

# Calculations
m = P*144*V/(53.35*T)
T2 = Q/(m*cv) +T
P2 = m*53.35*T2/V

# Results
print "Fina pressure  =  %d lbf/ft**2"%(P2)

# note : rounding off error.

Fina pressure  =  5990 lbf/ft**2


Example 3.5 pg : 66¶

In [7]:
from scipy.integrate import quad

# Variables
T1 = 560 			#R
T2 = 3460  			#R
m = 28.02 			#lb
cv = 0.248

# Calculations
def fun(T):
return 9.47 - 3.29*10**3 /T +1.07*10**6 /T**2

Q2 = m*cv*(T2-T1)
Error = (Q1-Q2)/Q1

# Results
print "Percentage error  =  %.1f percent"%(Error*100)

Percentage error  =  12.7 percent


Example 3.6 pg : 66¶

In [8]:

# Variables
import math
rate = 20. 			#gal/min
P1 = 20. 			#psia
P2 = 1000. 			#psia
T = 100.+460 			#R

# Calculations
vf = 0.01613
dv = -5.2*10**-5 			#ft**3/lbm
K = -dv/(vf*P2*144)
wt = K*vf*(P2**2 - P1**2)*144*144*10**4 /2
m = rate*8.33
Wt = wt*m
Wthp = Wt/33000

# Results
print "Pump power required  =  %d hp"%(Wthp)

Pump power required  =  188 hp