Chapter 7 - Entropy¶
Example 1: pg 159¶
In [1]:
#pg 159
print('Example 7.1');
import math
# aim : To determine
# the specific enthalpy of water
# Given values
Tf = 273.+100;# Temperature,[K]
# solution
# from steam table
cpl = 4.187;# [kJ/kg K]
# using equation [8]
sf = cpl*math.log(Tf/273.16);# [kJ/kg*K]
print ' The specific entropy of water is (kJ/kg K) = ',round(sf,3)
# using steam table
sf = 1.307;# [kJ/kg K]
print ' From table The accurate value of sf in this case is (kJ/kg K) = ',sf
print "There is small error in book's final value of sf"
# End
Example 2: pg 160¶
In [2]:
#pg 160
print('Example 7.2');
# aim : To determine
# the specific entropy
import math
# Given values
P = 2.;# pressure,[MN/m^2]
x = .8;# dryness fraction
# solution
# from steam table at given pressure
Tf = 485.4;# [K]
cpl = 4.187;# [kJ/kg K]
hfg = 1888.6;# [kJ/kg]
# (a) finding entropy by calculation
s = cpl*math.log(Tf/273.16)+x*hfg/Tf;# formula for entropy calculation
print ' (a) The specific entropy of wet steam is (kJ/kg K) = ',round(s,2)
# (b) calculation of entropy using steam table
# from steam table at given pressure
sf = 2.447;# [kJ/kg K]
sfg = 3.89;# [kJ/kg K]
# hence
s = sf+x*sfg;# [kJ/kg K]
print ' (b) The specific entropy using steam table is (kJ/kg K) = ',s
# End
Example 3: pg 161¶
In [3]:
#pg 161
print('Example 7.3');
import math
# aim : To determine
# the specific entropy of steam
# Given values
P = 1.5;#pressure,[MN/m^2]
T = 273.+300;#temperature,[K]
# solution
# (a)
# from steam table
cpl = 4.187;# [kJ/kg K]
Tf = 471.3;# [K]
hfg = 1946.;# [kJ/kg]
cpv = 2.093;# [kJ/kg K]
# usung equation [2]
s = cpl*math.log(Tf/273.15)+hfg/Tf+cpv*math.log(T/Tf);# [kJ/kg K]
print ' (a) The specific entropy of steam is (kJ/kg K) = ',round(s,3)
# (b)
# from steam tables
s = 6.919;# [kJ/kg K]
print ' (b) The accurate value of specific entropy from steam table is (kJ/kg K) = ',s
# End
Example 4: pg 164¶
In [4]:
#pg 164
print('Example 7.4');
# aim : To determine
# the dryness fraction of steam
# Given values
P1 = 2.;# initial pressure, [MN/m^2]
t = 350.;# temperature, [C]
P2 = .28;# final pressure, [MN/m^2]
# solution
# at 2 MN/m^2 and 350 C,steam is superheated because the saturation temperature is 212.4 C
# From steam table
s1 = 6.957;# [kJ/kg K]
# for isentropic process
s2 = s1;
# also
sf2 = 1.647;# [kJ/kg K]
sfg2 = 5.368;# [kJ/kg K]
# using
# s2 = sf2+x2*sfg2, where x2 is dryness fraction of steam
# hence
x2 = (s2-sf2)/sfg2;
print ' The final dryness fraction of steam is x2 = ',round(x2,3)
# End
Example 5: pg 165¶
In [6]:
#pg 165
print('Example 7.5');
# aim : To determine
# the final condition of steam...
# the change in specific entropy during hyperbolic process
# Given values
P1 = 2;# pressure, [MN/m^2]
t = 250.;# temperature, [C]
P2 = .36;# pressure, [MN/m^2]
P3 = .06;# pressure, [MN/m^2]
# solution
# (a)
# from steam table
s1 = 6.545;# [kJ/kg K]
# at .36 MN/m^2
sg = 6.930;# [kJ/kg*K]
sf2 = 1.738;# [kJ/kg K]
sfg2 = 5.192;# [kJ/kg K]
vg2 = .510;# [m^3]
# so after isentropic expansion, steam is wet
# hence, s2=sf2+x2*sfg2, where x2 is dryness fraction
# also
s2 = s1;
# so
x2 = (s2-sf2)/sfg2;
# and
v2 = x2*vg2;# [m^3]
# for hyperbolic process
# P2*v2=P3*v3
# hence
v3 = P2*v2/P3;# [m^3]
print ' (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is (m^3/kg) = ',round(v3,2)
# (b)
# at this condition
s3 = 7.609;# [kJ/kg*K]
# hence
change_s23 = s3-sg;# change in specific entropy during the hyperblic process[kJ/kg*K]
print ' (b) The change in specific entropy during the hyperbolic process is (kJ/kg K) = ',change_s23
# In the book they have taken sg instead of s2 for part (b), so answer is not matching
# End
Example 6: pg 166¶
In [7]:
#pg 166
print('Example 7.6');
# aim : To determine the
# (a) heat transfer during the expansion and
# (b) work done durind the expansion
%matplotlib inline
import matplotlib
from matplotlib import pyplot
# given values
m = 4.5; # mass of steam,[kg]
P1 = 3.; # initial pressure,[MN/m^2]
T1 = 300.+273; # initial temperature,[K]
P2 = .1; # final pressure,[MN/m^2]
x2 = .96; # dryness fraction at final stage
# solution
# for state point 1,using steam table
s1 = 6.541;# [kJ/kg/K]
u1 = 2751;# [kJ/kg]
# for state point 2
sf2 = 1.303;# [kJ/kg/K]
sfg2 = 6.056;# [kJ/kg/k]
T2 = 273+99.6;# [K]
hf2 = 417;# [kJ/kg]
hfg2 = 2258;# [kJ/kg]
vg2 = 1.694;# [m^3/kg]
# hence
s2 = sf2+x2*sfg2;# [kJ/kg/k]
h2 = hf2+x2*hfg2;# [kJ/kg]
u2 = h2-P2*x2*vg2*10**3;# [kJ/kg]
# Diagram of example 7.6
x = ([s1, s2]);
y = ([T1, T2]);
pyplot.plot(x,y);
pyplot.title('Diagram for example 7.6(T vs s)');
pyplot.xlabel('Entropy (kJ/kg K)');
pyplot.ylabel('Temperature (K)');
x = ([s1,s1]);
y = ([0,T1]);
pyplot.plot(x,y);
x = ([s2,s2]);
y = ([0,T2]);
pyplot.plot(x,y);
pyplot.show()
# (a)
# Q_rev is area of T-s diagram
Q_rev = (T1+T2)/2*(s2-s1);# [kJ/kg]
# so total heat transfer is
Q_rev = m*Q_rev;# [kJ]
# (b)
del_u = u2-u1;# change in internal energy, [kJ/kg]
# using 1st law of thermodynamics
W = Q_rev-m*del_u;# [kJ]
print ' (a) The heat transfer during the expansion is (kJ) (received) = ',Q_rev
print ' (b) The work done during the expansion is (kJ) = ',W
# End
Example 7: pg 176¶
In [8]:
#pg 176
print('Example 7.7');
# aim : To determine the
# (a) change of entropy
# (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression
import math
# Given values
P1 = 140.;# initial pressure,[kN/m^2]
V1 = .14;# initial volume, [m^3]
T1 = 273.+25;# initial temperature,[K]
P2 = 1400.;# final pressure [kN/m^2]
n = 1.25; # polytropic index
cp = 1.041;# [kJ/kg K]
cv = .743;# [kJ/kg K]
# solution
# (a)
R = cp-cv;# [kJ/kg/K]
# using ideal gas equation
m = P1*V1/(R*T1);# mass of gas,[kg]
# since gas is following law P*V^n=constant ,so
V2 = V1*(P1/P2)**(1./n);# [m^3]
# using eqn [9]
del_s = m*(cp*math.log(V2/V1)+cv*math.log(P2/P1));# [kJ/K]
print ' (a) The change of entropy is (kJ/K) = ',round(del_s,3)
# (b)
W = (P1*V1-P2*V2)/(n-1);# polytropic work,[kJ]
Gamma = cp/cv;# heat capacity ratio
Q = (Gamma-n)/(Gamma-1)*W;# heat transferred,[kJ]
# Again using polytropic law
T2 = T1*(V1/V2)**(n-1);# final temperature, [K]
T_avg = (T1+T2)/2;# mean absolute temperature, [K]
# so approximate change in entropy is
del_s = Q/T_avg;# [kJ/K]
print ' (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression (kJ/K) = ',round(del_s,4)
# End
Example 8: pg 179¶
In [9]:
#pg 179
print('Example 7.8');
# aim : To determine
# the change of entropy
import math
# Given values
m = .3;# [kg]
P1 = 350.;# [kN/m^2]
T1 = 273.+35;# [K]
P2 = 700.;# [kN/m^2]
V3 = .2289;# [m^3]
cp = 1.006;# [kJ/kg K]
cv = .717;# [kJ/kg K]
# solution
# for constant volume process
R = cp-cv;# [kJ/kg K]
# using PV=mRT
V1 = m*R*T1/P1;# [m^3]
# for constant volume process P/T=constant,so
T2 = T1*P2/P1;# [K]
s21 = m*cv*math.log(P2/P1);# formula for entropy change for constant volume process
print ' The change of entropy in constant volume process is (kJ/kg K) = ',round(s21,3)
# 'For the above part result given in the book is wrong
V2 = V1;
# for constant pressure process
T3 = T2*V3/V2;# [K]
s32 = m*cp*math.log(V3/V2);# [kJ/kg K]
print ' The change of entropy in constant pressure process is (kJ/kg K) = ',round(s32,3)
print "there is misprint in the book's result"
# End
Example 9: pg 181¶
In [11]:
#pg 181
print('Example 7.9');
import math
# aim : To determine
# the change of entropy
# Given values
P1 = 700.;# initial pressure, [kN/m^2]
T1 = 273.+150;# Temperature ,[K]
V1 = .014;# initial volume, [m^3]
V2 = .084;# final volume, [m^3]
# solution
# since process is isothermal so
T2 = T1;
# and using fig.7.10
del_s = P1*V1*math.log(V2/V1)/T1 ;# [kJ/K]
#results
print ' The change of entropy is (kJ/kg K) = ',round(del_s,5)
# End