Chapter 7 : equations of state and general thermodynamic relations¶

Example 7.1 pg : 158¶

In [1]:

# Variables
T1 = 1160.  			#R
h1 = 281.14 			#B/lbm
Pr1 = 21.18
P2 = 30.     			#psia
P1 = 100.    			#psia

# Calculations
Pr2 = Pr1*P2/P1
T2 = 833.    			#R
h2 = 199.45 			#B/lbm
dh = h2-h1

# Results
print "Change in enthalpy  =  %.2f B/lbm"%(dh)

Change in enthalpy  =  -81.69 B/lbm


Example 7.2 pg : 159¶

In [2]:
import math

# Variables
T2 = 860. 			#R
phi1 = 0.78767
phi2 = 0.71323
P2 = 30. 			#psia
P1 = 100. 			#psia

# Calculations
dS = phi2-phi1- 53.35/778 *math.log(P2/P1)

# Results
print "Net change of entropy  =  %.5f B/lbm R"%(dS)

Net change of entropy  =  0.00812 B/lbm R


Example 7.3 pg : 159¶

In [3]:

# Variables
T1 = 540. 			#R
T2 = 960. 			#R
h2 = 231.06 			#B/lbm
h1 = 129.06 			#B/lbm
cp = 0.24

# Calculations
W = h2-h1
dh = cp*(T2-T1)

# Results
print "Change in enthalpy  =  %.1f B/lbm"%(dh)

Change in enthalpy  =  100.8 B/lbm


Example 7.4 pg : 161¶

In [4]:
import math

# Variables
T1 = 420. 			#R
T2 = 380. 			#R
hig = 1221.2
P1 = 0.0019

# Calculations
lnp = hig*778*(1/T1 - 1/T2)/85.6
pra = math.exp(lnp)
P2 = pra*P1

# Results
print "Final pressure  =  %.3e psia"%(P2)

Final pressure  =  1.177e-04 psia


Example 7.5 pg : 170¶

In [5]:

# Variables
pc = 482.			#psia
Tc = 227. 			#R
vc = 1.44 			#ft**3/lbm mol
P = 600. 			#psia
T = 310. 			#R

# Calculations
Pr = P/pc
Tr = T/Tc
Z = 0.83
v = Z*55.12*T/(P*144)
rho = 1/v

# Results
print "Density  =  %.1f lbm/ft**3"%(rho)

Density  =  6.1 lbm/ft**3


Example 7.6 pg : 174¶

In [6]:

# Variables
T = -150.+460 			#R
v = 0.6 		    	#ft**3/lbm
vc = 1.44
Tc = 227. 	    		#R
Pc = 482.    			#psia

# Calculations
vr = v/vc
Tr = T/Tc
Pr = 1.75
P = Pr*Pc

# Results
print "Final pressure  =  %d psia"%(P)

Final pressure  =  843 psia


Example 7.7 pg : 177¶

In [8]:

# Variables
Tc = 344. 			#R
Pc = 673. 			#psia
P1 = 20. 			#psia
P2 = 500. 			#psia
M = 16.
T = 560. 			#R

# Calculations
pr1 = P1/Pc
pr2 = P2/Pc
Tr = T/Tc
dh2 = 0.65*Tc
dsp = 0.35 			#B/lbm mol R
dsp2 = 0.018-dsp- 1545/778 *math.log(P2/P1)
W = dh2-dsp2*T
W2 = W/M

# Results
print "Work per pound mass  =  %d B/lbm"%(W2)

#The answer is a bit different due to rounding off error. check using calculator.

Work per pound mass  =  138 B/lbm


Example 7.8 pg : 179¶

In [9]:

# Variables
P = 1000. 	        		#psia
T1 = 100. + 460 			#R
T2 = 800. + 460 			#R

# Calculations
pc = 1070. 			#psia
Tc = 548. 			#R
pr1 = P/pc
Tr1 = T1/Tc
Tr2 = T2/Tc
M = 44.
h1 = 4235.8 			#B/lbm mol
h2 = 11661 			#B/lbm mol
h2bar = 3.5 			#B/lbm mol
h1bar = 0.48 			#B/lbm mol
dhbar = Tc*(h2bar-h1bar) + h2-h1
Q = dhbar/M
cp = 0.202 			#B/lbm F
Q2 = cp*(T2-T1)
Error = (Q-Q2)/Q

# Results
print "Error in calculation  =  %d percent"%(Error*100)

Error in calculation  =  31 percent