# Chapter 12 : conventional power and refrigeration cycles¶

## Example 12.1 pg : 321¶

In [1]:

# Variables
h1 = 1357. 			#500 psia, 700 F
h2 = 935. 			#P2 = 2 psia
h3 = 93.99 			#sat liq at 2 psia
vf = 0.01613
P4 = 500. 			#psia
P3 = 2. 			#psia

# Calculations
dh4 = vf*(P4-P3)*144/778.
h4 = h3+dh4
eta =  ((h1-h2)-(h4-h3))/(h1-h4)

# Results
print "Thermal efficiency  =  %.1f percent "%(eta*100)
Thermal efficiency  =  33.3 percent

## Example 12.2 pg : 323¶

In [2]:

# Variables
h1 = 1357. 			#500 psia 700F
h2 = 1194. 			#P2 = 100 psia
h3 = 1379. 			#100 psia, 700 F
h4 = 1047. 			#p4 = 2 psia
h5 = 93.99 			#sat liq at 2 psia
h6 = 95.02 			#example 12.1

# Calculations
W = h1-h2+h3-h4-(h6-h5)
Q = (h1-h6)+(h3-h2)
eta = W/Q

# Results
print "Thermal efficiency  =  %.2f percent"%(eta*100)
Thermal efficiency  =  34.14 percent

## Example 12.3 pg : 327¶

In [3]:

# Variables
P = 100. 			#psia

# Calculations
h1 = 1357. 			#500 psia, 700F
h2 = 1194. 			#100 psia
h3 = 935.			#2 psia
h4 = 93.99 			#sat liq at 2 psia
vf = 0.01613
vf2 = 0.01774
P5 = 100. 			#psia
P4 = 2. 			#psia
dh4 = vf*(P5-P4)*144/778.
h5 = h4+dh4
h6 = 298.4
P7 = 500. 			#psia
P6 = 100. 			#psia
dh6 = vf2*(P7-P6)*144./778
h7 = dh6+h6
m = (h6-h5)/(h2-h5)
W = h1-h2 + (1-m)*(h2-h3) - (1-m)*(h5-h4) -(h7-h6)
Q = h1-h7
etath = W/Q

# Results
print "Thermal efficiency  =  %.1f percent"%(etath*100)
Thermal efficiency  =  35.2 percent

## Example 12.4 pg : 330¶

In [4]:

# Variables
x = 0.8

# Calculations
h1 = 1357. 			#500 psia 700F
h2 = 1194. 			#P2 = 100 psia
h3 = 1379. 			#100 psia, 700 F
h4 = 1047. 			#p4 = 2 psia
h5 = 93.99 			#sat liq at 2 psia
h6 = 95.02 			#example 12.1
h2d = h1- x*(h1-h2)
h4d = h3- x*(h3-h4)
W = (h1-h2d) +(h3-h4d) - (h6-h5)
Q = (h1-h6) + (h3-h2d)
eta = W/Q

# Results
print "Thermal efficiency  =  %d percent"%(eta*100+1)
Thermal efficiency  =  28 percent

## Example 12.5 pg : 335¶

In [5]:

# Variables
P4 = 50. 			#psia
P1 = 14.7 			#psia
P3 = 50. 			#psia
P2 = 14.7 			#psia
g = 1.4

# Calculations
V1r = (P4/P1)**(1/g)
V2r = (P3/P2)**(1/g)
#After solving,
V4 = 5.38    			#ft**3/min
V1 = 12.9 	    		#ft**3/min
V2 = 112.9  			#ft**3/min
PD = V2-V4
etavol = (V2-V1)/(V2-V4)
W32 = g*P2*144*V2*((P3/P2)**((g-1)/g) -1 ) /(1-g)
W41 = g*P4*144*V4*((P1/P4)**((g-1)/g) -1 ) /(1-g)
Wt = W32+W41

# Results
print "Total work  =  %.2e ft-lbf /min"%(Wt)
#The answer given in textbook is wrong . please verify it umath.sing a calculator
Total work  =  -3.10e+05 ft-lbf /min

## Example 12.6 pg : 337¶

In [6]:

# Variables
import math
P1 = 14.7 			#psia
P4 = 100. 			#psia
T1 = 530. 			#R
T3 = T1
g = 1.4
m = 10. 			#lbm
cp = 0.24

# Calculations
P2 = math.sqrt(P1*P4)
T2 = T1*(P2/P1)**((g-1)/g)
T4 = T2
W = 2*cp*(T2-T1)
Wt = W*m
hp = Wt*60./2545
Q = m*cp*(T2-T3)
T4 = T1*(P4/P1)**((g-1)/g)
W2 = m*cp*(T4-T1)

# Results
print "Work required in case 1   =  %d Btu/min"%(Wt+1)
print " Work required in case 2  =  %d Btu/min"%(W2+1)
Work required in case 1   =  802 Btu/min
Work required in case 2  =  928 Btu/min

## Example 12.7 pg : 342¶

In [8]:

# Variables
g = 1.4
r1 = 10.
r2 = 12.
r3 = 15.
Tl = 530. 			#R
Th = 1960. 			#R

# Calculations
eta1 = 1- (r1)**(1-g)
eta2 = 1- (r2)**(1-g)
eta3 = 1- (r3)**(1-g)
etac = 1-Tl/Th

# Results
print "Efficiency in case 1  =  %.1f percent"%(eta1*100)
print " Efficiency in case 2  =  %.1f percent"%(eta2*100)
print " Efficiency in case 3  =  %.1f percent"%(eta3*100)
print " Carnot efficiency  =  %.f percent"%(etac*100)
Efficiency in case 1  =  60.2 percent
Efficiency in case 2  =  63.0 percent
Efficiency in case 3  =  66.1 percent
Carnot efficiency  =  73 percent

## Example 12.8 pg : 344¶

In [10]:

# Variables
T1 = 70.+460 			#R
P1 = 14.7 			#psia
g = 1.4
r = 15.
rc = 2.
cp = 0.24
cp2 = 0.1715

# Calculations
T2 = T1*(r)**(g-1)
T3 = rc*T2
T4 = T3*(rc/r)**(g-1)
Qh = cp*(T3-T2)
Ql = cp2*(T4-T1)
W = Qh-Ql
eta = W/Qh

# Results
print "Work output  =  %.f B/lbm"%(W)
print " Efficiency  =  %.1f percent"%(eta*100)
Work output  =  227 B/lbm
Efficiency  =  60.4 percent

## Example 12.9 pg : 351¶

In [11]:

# Variables
P1 = 14.7 			#psia
P4 = 14.7 			#psia
T1 = 530. 			#R
T3 = 1960. 			#R
P2 = 60. 			#psia
P3 = P2
g = 1.4
eta1 = 0.85
eta2 = 0.9

# Calculations
T2 = T1*(P2/P1)**((g-1)/g)
T4 = T3*(P4/P3)**((g-1)/g)
T2d = (T2-T1)/eta1 + T1
T4d = -eta2*(T3-T4) +T3
Wact = 0.24*(T3-T4d - (T2d-T1))
Qh = 0.24*(T3-T2d)
etath = Wact/Qh

# Results
print "Thermal efficiency  =  %.1f percent"%(etath*100)
Thermal efficiency  =  24.5 percent

## Example 12.10 pg : 352¶

In [12]:

# Variables
e = 0.83

# Calculations
T1 = 530. 			#R
T2d = 838. 			#R
T6d = T2d
T3 = 1960. 			#R
T4d = 1375. 			#R
T5d = T4d
T5 = e*(T5d-T2d) +T2d
W = 0.24*((T3-T4d)- (T2d-T1))
Q = 0.24*(T3-T5)
eta = W/Q

# Results
print "Thermal efficiency  =  %d percent"%(eta*100+1)
Thermal efficiency  =  41 percent

## Example 12.11 pg : 354¶

In [14]:

# Variables
import math
T1 = 420. 			#R
T11 = 530. 			#R
T3 = 2460. 			#R
V1 = 300. 			#ft/sec
P1 = 5. 			#psia
P5 = P1
P2 = 50. 			#psia
P3 = 5. 			#psia
P4 = 50. 			#psia
g = 1.4
cp = 0.24
m = 1.

# Calculations
T2 = T1*(P2/P1)**((g-1)/g)
T4 = T3-T2+T11
T5 = T3*(P3/P4)**((g-1)/g)
V5 = math.sqrt(2*32.2*cp*(T4-T5)*778)
T = m*(V1-V5)/32.2
Qh = cp*(T3-T2)
P = -T*V1

# Results
print "Thrust  =  %.1f lbf"%(T)
print " Heat input  =  %d B/lbm"%(Qh)
print " Power  =  %d ft-lbf /sec"%(round(P,-3))
Thrust  =  -93.1 lbf
Heat input  =  395 B/lbm
Power  =  28000 ft-lbf /sec

## Example 12.12 pg : 360¶

In [15]:

# Variables
import math
h1 = 80.419 			#B/lbm
h3 = 36.013 			#B/lbm
h4 = h3
P3 = 172.35 			#psia
P2 = P3
m = 5.       			#tons
Q = 12000.

# Calculations
h2 = 91.5 			#B/lbm
COP = (h1-h4)/(h2-h1)
W = h2-h1
md = m*Q/(h1-h4)
Wt = md*(h2-h1)
Wt2 = Wt/2545.

# Results
print "Coefficient of performance  =  %.1f"%(COP)
print " Input work  =  %.1f hp"%(Wt2)
Coefficient of performance  =  4.0
Input work  =  5.9 hp