# Chapter 9 : Properties of Pure Substances¶

## Example 9.1 Page No : 295¶

In :
# Variables
tsat = 179.91; # celsius
vf = 0.001127; # m**3/kg
vg = 0.19444;  #m**3/kg

# Calculation
vfg = vg-vf;
sf = 2.1387;
sg = 6.5865;
sfg = sg-sf;

# Results
print "At 1 Mpa saturation temperature is",tsat,"degree"
print "Changes in specific volume is",vfg,"m3/kg"
print "Change in entropy during evaporation is",sfg,"kJ/kg K"

At 1 Mpa saturation temperature is 179.91 degree
Changes in specific volume is 0.193313 m3/kg
Change in entropy during evaporation is 4.4478 kJ/kg K


## Example 9.3 Page No : 295¶

In :
# Variables
v = 0.09
vf = 0.001177
vg = 0.09963;

# Calculation
x = (v-vf)/(vg-vf);
hf = 908.79; hfg = 1890.7;
sf = 2.4474; sfg = 3.8935;
h = hf+(x*hfg);
s = sf+(x*sfg);

# Results
print "The enthalpy and entropy of the system are %.4f and %.2f kJ/kg and kJ/kg K respectively"%(s,h)

# rounding off error. please check.

The enthalpy and entropy of the system are 5.9601 and 2614.55 kJ/kg and kJ/kg K respectively


## Example 9.4 Page No : 296¶

In :
from  numpy import interp

# Variables
# for T = 350 degree
T1 = 350;
v1 = 0.2003;
h1 = 3149.5;
s1 = 7.1369;
# for T = 400 degree
T2 = 400;
v2 = 0.2178;
h2 = 3257.5;
s2 = 7.3026;

# Calculation
# Interpolation for T = 380;
T = [T1 ,T2];
v = [v1 ,v2];
h = [h1 ,h2];
s = [s1 ,s2];
v3 = interp(380,T,v);
h3 = interp(380,T,h);
s3 = interp(380,T,s);

# Results
print "The entropy, enthalpy and volume of stem at 1.4MPa and 380 degree is %.4f kJ/Kg, %.1f kJ/Kg, %.4f kJ/Kg  respectively"%(s3,h3,v3)

The entropy, enthalpy and volume of stem at 1.4MPa and 380 degree is 7.2363 kJ/Kg, 3214.3 kJ/Kg, 0.2108 kJ/Kg  respectively


## Example 9.5 Page No : 296¶

In :
# Variables
Psat = 3.973e06;
vf = 0.0012512
vg = 0.05013;
hf = 1085.36
hfg = 1716.2;
sf = 2.7927
sfg = 3.2802;
mf = 9           # liquid in kg
V = 0.04;        # volume

# Calculation
Vf = mf*vf;
Vg = V-Vf;
mg = Vg/vg;
m = mf+mg;
x = mg/m;
v = vf+x*(vg-vf);
h = hf+x*hfg;
s = sf+(x*sfg);
u = h-Psat*v*1e-03;

# at T = 250
uf = 1080.39
ufg = 1522;
u_ = uf+x*ufg;

# Results
print "The pressure is %.3f Mpa"%(Psat/1000000)
print "The mass is %.3f Kg"%m
print "Specific volume is %.5f m3/Kg"%v
print "Enthalpy is %.2f kJ/Kg"%h
print "The entropy is %.4f kJ/Kg K"%s
print "The interal energy is %.2f kJ/Kg"%u
print "u = %.2f kJ/kg"%u_   #incorrect answer in the textbook


The pressure is 3.973 Mpa
The mass is 9.573 Kg
Specific volume is 0.00418 m3/Kg
Enthalpy is 1188.13 kJ/Kg
The entropy is 2.9891 kJ/Kg K
The interal energy is 1171.53 kJ/Kg
u = 1171.53 kJ/kg


## Example 9.6 Page No : 297¶

In :
import math
from numpy import interp

# Part (a)
vg1_ = 0.8919
T1_ = 120;
vg2_ = 0.77076
T2_ = 125;
vg_ = [vg1_, vg2_]
T_ = [T1_, T2_];
v1 = 0.7964;
h1 = 2967.6;
P1 = 0.3e03; 			# in Kpa

# Calculation and Results
T1 = interp(v1,vg_,T_);
print "The steam become saturated at ",T1,"degree"

# Part (b)
vf = 0.001029
vg = 3.407;
hf = 334.91
hfg = 2308.8;
Psat = 47.39; 			# In kPa
v2 = v1;
x2 = (v1-vf)/(vg-vf);
h2 = hf+x2*hfg;
P2 = Psat;
Q12 = (h2-h1)+v1*(P1-P2);

print "The quality factor at t=80 degree is %.4f"%x2
print "The heat transfered per kg of steam in cooling from 250 degree to 80 degree %.2f kJ/Kg"%Q12

# rounding off error. interp function gives slightly different answer.

The steam become saturated at  125.0 degree
The quality factor at t=80 degree is 0.2335
The heat transfered per kg of steam in cooling from 250 degree to 80 degree -1892.35 kJ/Kg


## Example 9.7 Page No : 298¶

In :
# Variables
# At T = 40 degree
Psat = 7.384e06;
sf = 0.5725
sfg = 7.6845;
hf = 167.57
hfg = 2406.7;
s1 = 6.9189
h1 = 3037.6;

# Calculation
x2 = round((s1-sf)/sfg,3) ;
h2 = hf+(x2*hfg);
W = h1-h2;

# Results
print "The ideal work output of the turbine is %.2f kJ/Kg"%W

# rounding off error. please check using calculator.

The ideal work output of the turbine is 882.10 kJ/Kg


## Example 9.8 Page No : 299¶

In :
import math
from numpy import interp

# Variables
w1 = 1.0;
w2 = w3-w1;
h1 = 2950.0;

# At 0.8MPa, 0.95 dry
x = 0.95;
hf = 721.11
hfg = 2048;

# Calculation
h2 = hf + (x*hfg);
h3 = ((w1*h1)+(w2*h2))/w3;
# Interpolation
H = [2769.1, 2839.3];
T = [170.43, 200];
t3 = interp(2790,H,T);
s3 = 6.7087;
s4 = s3;
x4 = (s3-1.7766)/5.1193;
h4 = 604.74+(x4*2133.8);
V4 = math.sqrt(2000*(h3-h4));

# Results
print "The condition of superheat after mixing %.2f degree"%(t3-T)
print "The velocity of steam leaving the nozzle is %.1f m/sec"%V4

# rounding off error. please check.

The condition of superheat after mixing 8.80 degree
The velocity of steam leaving the nozzle is 508.7 m/sec


## Example 9.9 Page No : 300¶

In :
# Variables
h2 = 2716.2
hf = 844.89
hfg = 1947.3;

# Calculation
x1 = (h2-hf)/hfg;
h3 = 2685.5;
x4 = (h3-hf)/hfg;

# Results
print "The quality of steam in pipe line is %.3f"%x1
print "Maximum moisture is %.2f %%"%(100-(x4*100))

# rounding off error.

The quality of steam in pipe line is 0.961
Maximum moisture is 5.48 %


## Example 9.10 Page No : 301¶

In :
# Variables
# At 0.1Mpa, 110 degree
h2 = 2696.2
hf = 844.89
hfg = 1947.3;

# Calculation
x2 = (h2-hf)/hfg;
vf = 0.001023; 			# at T = 70 degree
V = 0.000150; 			# In m3
m1 = V/vf;
m2 = 3.24;
x1 = (x2*m2)/(m1+m2);

# Results
print "The quality of the steam in the pipe line is %.3f"%x1

The quality of the steam in the pipe line is 0.910


## Example 9.11 Page No : 302¶

In :
# Variables
# P = 1MPa
vf = 0.001127
vg = 0.1944;
hg = 2778.1
uf = 761.68;
ug = 2583.6
ufg = 1822;
# Initial anf final mass
Vis = 5.
Viw = 5.;
Vfs = 6.
Vfw = 4.

# Calculation
ms = ((Viw/vf)+(Vis/vg)) - ((Vfw/vf)+(Vfs/vg)) ;
U1 =  ((Viw*uf/vf)+(Vis*ug/vg));
Uf =  ((Vfw*uf/vf)+(Vfs*ug/vg));
Q = Uf-U1+(ms*hg)

# Results
print "The heat transfer during the process is",round(Q/1000,3),"kJ"


The heat transfer during the process is 1788.192 kJ


## Example 9.12 Page No : 303¶

In :
import math

# Variables
m = 0.02
d = 0.28
l = 0.305;
P1 = 0.6e06
P2 = 0.12e06;

# Calculation and Results
# At 0.6MPa, t = 200 degree
v1 = 0.352
h1 = 2850.1;
V1 = m*v1;
Vd = (math.pi/4)*d**2*l;
V2 = V1+Vd ;
n = math.log(P1/P2)/math.log(V2/V1);
W12 = ((P1*V1)-(P2*V2))/(n-1);

print "The value of n is %.2f"%n
print "The work done by the steam is %.1f kJ"%(W12/1000)

v2 = V2/m;
vf = 0.0010476
vfg = 1.4271;
x2 = (v2-vf)/vfg ;
# At 0.12MPa
uf = 439.3
ug = 2512.0;
u2 = uf + (x2*(ug-uf));
u1 = h1-(P1*v1*1e-03);
Q12 = m*(u2-u1)+ (W12/1000);

print "The heat transfer is %.3f kJ"%Q12

# rounding off error. please check.

The value of n is 1.24
The work done by the steam is 4.7 kJ
The heat transfer is -1.801 kJ


## Example 9.13 Page No : 305¶

In :
# Variables
x1 = 1.
x2 = 0.8;
# at 0.2MPa
vg = 0.8857
v1 = vg
hg = 2706.7
h1 = hg;
m1 = 5.
V1 = m1*v1;

# Calculation
# at 0.5MPa
m2 = 10;
hf = 640.23
hfg = 2108.5
vf = 0.001093
vfg = 0.3749;
v2 = vf+(x2*vfg);
V2 = m2*v2;

Vm = V1+V2;
m = m1+m2;
vm = Vm/m;
u1 = h1;
h2 = hf+(x2*hfg);
u2 = h2;
m3 = m;
h3 = ((m1*u1)+(m2*u2))/m3;
u3 = h3;
v3 = vm;
# From mollier diagram
x3 = 0.870
p3 = 3.5
s3 = 6.29;
s1 = 7.1271;
sf = 1.8607
sfg = 4.9606;
s2 = sf+(x2*sfg);
E = m3*s3-((m1*s1)+(m2*s2));

# Results
print "Final pressure is ",p3,"bar"
print "Steam quality is",x3
print "Entropy change during the process is %.2f kJ/K"%E

Final pressure is  3.5 bar
Steam quality is 0.87
Entropy change during the process is 0.42 kJ/K


## Example 9.14 Page No : 306¶

In :
# Variables
# At 6 MPa, 400 degree
h1 = 3177.2
s1 = 6.5408;

# At 20 degree
h0= 83.96
s0 = 0.2966;
T0 = 293.

# Calculation
f1 = (h1-h0)-T0*(s1-s0);
# By interpolation
t2 = 273. + 393;
s2 = 6.63;
h2 = h1;
f2 = (h2-h0)-T0*(s2-s0);
df = f1-f2;
x3s = (s2-1.5301)/(7.1271-1.5301);
h3s = 504.7+(x3s*2201.9);
eis = 0.82;
h3 = h2-eis*(h1-h3s);
x3 = (h3-504.7)/2201.7;
s3 = 1.5301+(x3*5.597);
f3 = (h3-h0)-T0*(s3-s0);

# Results
print "The availability of the steam before the throttle valve %.1f kJ/Kg"%f1
print "The availability of the steam after the throttle valve %.2f Kj/Kg"%f2
print "The availability of the steam at the turbine exhaust %.2f kJ/Kg"%f3
print "The specific work output from the turbine is %.1f kJ/Kg"%(h2-h3)

The availability of the steam before the throttle valve 1263.7 kJ/Kg
The availability of the steam after the throttle valve 1237.55 Kj/Kg
The availability of the steam at the turbine exhaust 601.85 kJ/Kg
The specific work output from the turbine is 546.3 kJ/Kg


## Example 9.15 Page No : 308¶

In :
# Variables
# At 25 bar, 350 degree
h1 = 3125.87
s1 = 6.8481;
# 30 degree
h0 = 125.79
s0 = 0.4369;
h2 = 2865.5
s2 = 7.3115;
# At 0.2 bar 0.95 dry
hf = 251.4
hfg = 2358.3;
sf = 0.8320
sg = 7.0765;

# Calculation
h3 = hf+0.92*hfg;
s3 = sf+(0.92*sg);
# Part (a)
T0 = 303;
f1 = (h1-h0)-(T0*(s1-s0));
f2 = (h2-h0)-(T0*(s2-s0));
f3 = (h3-h0)-(T0*(s3-s0));

# Results
print "Availability of steam entering at state 1 is %.2f kJ/Kg"%f1
print "Availability of steam leaving  at state 2 is %.2f kJ/Kg"%f2
print "Availability of steam leaving  at state 3 is %.2f kJ/Kg"%f3

# Part (b)
m2m1 = 0.25
m3m1 = 0.75;
Wrev = f1-(m2m1*f2)-(m3m1*f3);
print "Maximum work is %.2f kJ/Kg"%Wrev

# Part (c)
w1 = 600.
w2 = 150.
w3 = 450.;
Q = -10.*3600;  			# For 1 hour
I = T0*(w2*s2+w3*s3-w1*s1)-Q;
print "Irreversibility is %.2f MJ/h"%(I/1000)

# rounding off error. please check using calcultor.

Availability of steam entering at state 1 is 1057.49 kJ/Kg
Availability of steam leaving  at state 2 is 656.71 kJ/Kg
Availability of steam leaving  at state 3 is 202.89 kJ/Kg
Maximum work is 741.15 kJ/Kg
Irreversibility is 124.46 MJ/h


## Example 9.16 Page No : 309¶

In :
# Variables
# At dead state of 1 bar, 300K
u0 = 113.1
h0 = 113.2;
v0 = 0.001005
s0 = 0.0395;
T0 = 300.
P0 = 100.;

# Calculation and Results
K = h0-(T0*s0);
# Part (a)
u = 376.9
h = 377;
v = 0.001035
s = 1.193;
m = 3
a = h - 300 * s
fi = m*(a - (-5.3)) 			# As P = P0 = 1 bar
print "Energy of system in Part (a) is",fi,"kJ"

# Part (b)
u = 3099.8
h = 3446.3
v = 0.08637
s = 7.090; 			# At P = 4 Mpa, t = 500 degree
m = 0.2;
b = u +100* v - 300 * s
fib = m*(b-(-5.3));
print "Energy of system in Part (b) is",fib,"kJ"

# Part (c)
m = 0.4;
x = 0.85; 			# Quality
u = 192+x*2245;
h = 192+x*2392;
s = 0.649+x*7.499;
v = 0.001010+x*14.67;
c = round(u + 100*v - 300*s,1)
fic = m*(c - (-5.3));
print "Energy of system in Part (c) is",round(fic,1),"kJ"

# Part (d)
m = 3;
h = -354.1; s = -1.298; 			# at 1000kPa, -10 degree
fid = m*(h-h0-T0*(s-s0))#      ((h-h0)-T0*(s-s0));
print "Energy of system in Part (d) is",fid,"kJ"


Energy of system in Part (a) is 73.2 kJ
Energy of system in Part (b) is 197.3474 kJ
Energy of system in Part (c) is 498.3 kJ
Energy of system in Part (d) is -198.15 kJ


## Example 9.17 Page No : 310¶

In :
# Variables
# Given
th1 = 90.+273;
tc1 = 25.+273;
tc2 = 50.+273;
mc = 1.
T0 = 300.;
th2p = 60.+273; 			# Parallel
th2c = 35.+273; 			# Counter

# Calculation
mhp = (tc2-tc1)/(th1-th2p); 			# Parallel
mhc = (tc2-tc1)/(th1-th2c); 			# Counter
h0 = 113.2
s0 = 0.395
T0 = 300; 			# At 300 K
h1 = 376.92
s1 = 1.1925; 			# At 90 degree
af1 = mhp*((h1-h0)-T0*(s1-s0));

# Parallel Flow
h2 = 251.13; s2 =0.8312; 			# At 60 degree
h3 = 104.89; s3 = 0.3674; 			# At 25 degree
h4 = 209.33; s4 = 0.7038; 			# At 50 degree
REG = mc*((h4-h3)-T0*(s4-s3)); 			# Rate of energy gain
REL = mhp*((h1-h2)-T0*(s1-s2)); 			# Rate of energy loss
Ia = REL-REG; 			# Energy destruction
n2a = REG/REL; 			# Second law efficiency

# Results
print ("In parallel flow")
print "The rate of irreversibility is",Ia,"kW"
print "The Second law efficiency is %.2f %%"%(n2a*100)

# Counter flow
h2 = 146.68
s2 = 0.5053; 			# At 35 degree
REG_b = REG; 			# Rate of energy gain by hot water is same in both flows
REL_b = mhc*((h1-h2)-T0*(s1-s2));
Ib = REL_b-REG_b; 			# Energy destruction
n2b = REG_b/REL_b; 			# Second law efficiency

print ("In Counter flow")
print "The rate of irreversibility is %.2f kW"%Ib
print "The Second law efficiency is %.2f %%"%(n2b*100)

In parallel flow
The rate of irreversibility is 10.98 kW
The Second law efficiency is 24.28 %
In Counter flow
The rate of irreversibility is 7.43 kW
The Second law efficiency is 32.16 %


## Example 9.18 Page No : 312¶

In :
# Variables
m = 50 ; 			# in kg/h
Th = 23.+273; 			# Home temperature

# State 1
T1 = 150.+273;
h1 = 2746.4;
s1 = 6.8387;

# State 2
h2 = 419.0;
s2 = 1.3071;
T0 = 318;

# Calculation
b1 = h1-(T0*s1);
b2 = h2-(T0*s2);
Q_max = m*(b1-b2)/(T0/Th-1);

# Results
print "The maximum cooling rate is %.0f kW"%(Q_max/3600)

The maximum cooling rate is 106 kW