# Chapter 5 - Gases and Single Phase systems¶

## Example 1: pg 98¶

In [1]:
#pg 98
#calculate the new pressure and difference in two levels
print('Example 5.1');

#  aim : To determine
#  new pressure exerted on the air and the difference in two mercury column level

#  Given values
P1 = 765.;#  atmospheric pressure, [mmHg]
V1 = 20000.;#  [mm^3]
V2 = 17000.;#  [mm^3]

#  solution

#  using boyle's law P*V=constant
#  hence
P2 = P1*V1/V2;#  [mmHg]

del_h = P2-P1;#  difference in Height of mercury column level
#results
print ' The new pressure exerted on the air (mmHg) = ',P2
print ' The difference in the two mercury column level (mm) = ',del_h

#  End

Example 5.1
The new pressure exerted on the air (mmHg) =  900.0
The difference in the two mercury column level (mm) =  135.0


## Example 2: pg 99¶

In [2]:
#pg 99
#calculate the new volume
print('Example 5.2');

#  aim : To determine
#  the new volume

#  Given values
P1 = 300;#  original pressure,[kN/m^2]
V1 = .14;#  original volume,[m^3]

P2 = 60.;#  new pressure after expansion,[kn/m^2]

#  solution
#  since temperature is constant so using boyle's law P*V=constant
V2 = V1*P1/P2;#  [m^3]

#results
print ' The new volume after expansion (m^3) = ',V2

#  End

Example 5.2
The new volume after expansion (m^3) =  0.7


## Example 3: pg 101¶

In [3]:
#pg 101
#calculate the new volume
print('Example 5.3');

#  aim : To determine
#  the new volume of the gas

#  Given values
V1 = 10000;#  [mm^3]
T1 = 273.+18;#  [K]
T2 = 273.+85;#  [K]

#  solution
#  since pressure exerted on the apparatus is constant so using charle's law V/T=constant
#  hence
V2 = V1*T2/T1;#  [mm^3]

#results
print ' The new volume of the gas trapped in the apparatus (mm^3) = ',round(V2)

#  End

Example 5.3
The new volume of the gas trapped in the apparatus (mm^3) =  12302.0


## Example 4: pg 102¶

In [4]:
#pg 102
#calculate the final temperature
print('Example 5.4');

#  aim : To determine
#  the final temperature

#  Given values
V1 = .2;#  original volume,[m^3]
T1 = 273+303;# original temperature, [K]
V2 = .1;# final volume, [m^3]

#  solution
#  since pressure is constant, so using charle's law V/T=constant
#  hence
T2 = T1*V2/V1;#  [K]
t2 = T2-273;#  [C]
#results
print ' The final temperature of the gas (C) = ',t2

#  End

Example 5.4
The final temperature of the gas (C) =  15.0


## Example 5: pg 106¶

In [5]:
#pg 106
#calculate the new volume
print('Example 5.5');

#  aim : To determine
#  the new volume of the gas

#  Given values

#  initial codition
P1 = 140;#  [kN/m^2]
V1 = .1;#   [m^3]
T1 = 273+25;# [K]

#  final condition
P2 = 700.;#  [kN/m^2]
T2 = 273.+60;#  [K]

#  by charasteristic equation, P1*V1/T1=P2*V2/T2

V2=P1*V1*T2/(T1*P2);#  final volume, [m^3]

#results
print 'The new volume of the gas (m^3) = ',round(V2,4)

#  End

Example 5.5
The new volume of the gas (m^3) =  0.0223


## Example 6: pg 106¶

In [6]:
#pg 106
#calculate the new temperature and mass of gas
print('Example 5.6');

#  aim : To determine
#  the mas of the gas and new temperature

#  Given values
P1 = 350;#  [kN/m^2]
V1 = .03;#  [m^3]
T1 = 273+35;#  [K]
R = .29;#  Gas constant,[kJ/kg K]

#  solution
#  using charasteristic equation, P*V=m*R*T
m = P1*V1/(R*T1);#  [Kg]

#  Now the gas is compressed
P2 = 1050;#  [kN/m^2]
V2 = V1;
# since mass of the gas is constant so using, P*V/T=constant
#  hence
T2 = T1*P2/P1#  [K]
t2 = T2-273;#  [C]

#results
print ' The mass of the gas present (kg) = ',round(m,3)
print ' The new temperature of the gas (C) = ',t2

#  End

Example 5.6
The mass of the gas present (kg) =  0.118
The new temperature of the gas (C) =  651


## Example 7: pg 111¶

In [7]:
#pg 111
#calculate the final pressure and heat transferred
print('Example 5.7');

#  aim : To determine
#  the heat transferred to the gas and its final pressure

#  Given values
m = 2;#  masss of the gas, [kg]
V1 = .7;#  volume,[m^3]
T1 = 273+15;#  original temperature,[K]
T2 = 273+135;#  final temperature,[K]
cv = .72;#  specific heat capacity at constant volume,[kJ/kg K]
R = .29;#  gas law constant,[kJ/kg K]

#  solution
Q = m*cv*(T2-T1);#  Heat transferred at constant volume,[kJ]

#  Now,using P1*V1=m*R*T1
P1 = m*R*T1/V1;#  [kN/m^2]

#  since volume of the system is constant, so P1/T1=P2/T2
#  hence
P2 = P1*T2/T1;#  final pressure,[kN/m^2]
#results
print ' The heat transferred to the gas (kJ) = ',Q
print ' The final pressure of the gas (kN/m^2) = ',round(P2,2)

#  End

Example 5.7
The heat transferred to the gas (kJ) =  172.8
The final pressure of the gas (kN/m^2) =  338.06


## Example 8: pg 114¶

In [8]:
#pg 114
#calculate the heat transferred and work done
print('Example 5.8');

#  aim : To determine
#  the heat transferred from the gas and the work done on the gas

#  Given values
P1 = 275;#  pressure, [kN/m^2]
V1 = .09;#  volume,[m^3]
T1 = 273+185;#  initial temperature,[K]
T2 = 273+15;#  final temperature,[K]
cp = 1.005;#  specific heat capacity at constant pressure,[kJ/kg K]
R = .29;#  gas law constant,[kJ/kg K]

#  solution
#  using P1*V1=m*R*T1
m = P1*V1/(R*T1);#  mass of the gas

#  calculation of heat transfer
Q = m*cp*(T2-T1);#  Heat transferred at constant pressure,[kJ]

#  calculation of work done
#  Now,since pressure is constant so, V/T=constant
# hence
V2 = V1*T2/T1;#  [m^3]

W = P1*(V2-V1);#  formula for work done at constant pressure,[kJ]
#results
print ' The heat transferred to the gas (kJ) = ',round(Q,2)
print ' Work done on the gas during the process (kJ) = ',round(W,2)

#  End

Example 5.8
The heat transferred to the gas (kJ) =  -31.84
Work done on the gas during the process (kJ) =  -9.19


## Example 9: pg 117¶

In [9]:
#pg 117
#calculate the new pressure
print('Example 5.9');

#  aim : To determine
#  the new pressure of the gas

#  Given values
P1 = 300.;#  original pressure,[kN/m**2]
T1 = 273.+25;#  original temperature,[K]
T2 = 273.+180;#  final temperature,[K]

#  solution
#  since gas compressing according to the law,P*V**1.4=constant
#  so,for polytropic process,T1/T2=(P1/P2)**((n-1)/n),here n=1.4

#  hence
P2 = P1*(T2/T1)**((1.4)/(1.4-1));#  [kN/m**2]

#results
print ' The new pressure of the gas (kN/m^2) = ',round(P2)

#  End

Example 5.9
The new pressure of the gas (kN/m^2) =  1299.0


## Example 10: pg 118¶

In [10]:
#pg 118
#calculate the new temperature
print('Example 5.10');

#  aim : To determine
#  the new temperature of the gas

#  Given values
V1 = .015;#  original volume,[m**3]
T1 = 273.+285;#  original temperature,[K]
V2 = .09;#  final volume,[m**3]

#  solution
#  Given gas is following the law,P*V**1.35=constant
#  so process is polytropic with
n = 1.35; # polytropic index

# hence
T2 = T1*(V1/V2)**(n-1);#  final temperature, [K]

t2 = T2-273;#  [C]

#results
print ' The new temperature of the gas (C) = ',round(t2,1)

print ' there is minor error in book answer due to rounding off error'

#  End

Example 5.10
The new temperature of the gas (C) =  25.0
there is minor error in book answer due to rounding off error


## Example 11: pg 119¶

In [11]:
#pg 119
#calculate the final pressure, temperature and volume of gas
print('Example 5.11');

#  aim : To determine the
#  (a) original and final volume of the gas
#  (b) final pressure of the gas
#  (c) final temperature of the gas

#  Given values
m = .675;#  mass of the gas,[kg]
P1 = 1.4;#  original pressure,[MN/m**2]
T1 = 273+280;#  original temperature,[K]
R = .287;#gas constant,[kJ/kg K]

#  solution and results

#  (a)
#  using characteristic equation, P1*V1=m*R*T1
V1 = m*R*T1*10**-3/P1;#  [m**3]
#  also Given
V2 =  4*V1;# [m**3]
print ' (a) The original volume of the gas (m^3) = ',round(V1,4)
print '      and The final volume of the gas (m^3) = ',round(V2,3)

#  (b)
#  Given that gas is following the law P*V**1.3=constant
#  hence process is polytropic with
n = 1.3; #  polytropic index
P2 = P1*(V1/V2)**n;#  formula for polytropic process,[MN/m**2]
print ' (b) The final pressure of the gas (kN/m^2) = ',round(P2*10**3)

#  (c)
#  since mass is constant so,using P*V/T=constant
#  hence
T2 = P2*V2*T1/(P1*V1);#  [K]
t2 = T2-273;#  [C]
print ' (c) The final temperature of the gas (C) = ',round(t2)

#  End

Example 5.11
(a) The original volume of the gas (m^3) =  0.0765
and The final volume of the gas (m^3) =  0.306
(b) The final pressure of the gas (kN/m^2) =  231.0
(c) The final temperature of the gas (C) =  92.0


## Example 12: pg 120¶

In [12]:
#pg 120
#calculate the change in internal energy, work done and heat transfer
print('Example 5.12');

#  aim : T0 determine
#  (a) change in internal energy of the air
#  (b) work done
#  (c) heat transfer

#  Given values
m = .25;#  mass, [kg]
P1 = 140;#  initial pressure, [kN/m**2]
V1 = .15;#  initial volume, [m**3]
P2 = 1400;#  final volume, [m**3]
cp = 1.005;#  [kJ/kg K]
cv = .718;#  [kJ/kg K]

#  solution

#  (a)
#  assuming ideal gas
R = cp-cv;#  [kJ/kg K]
#  also, P1*V1=m*R*T1,hence
T1 = P1*V1/(m*R);#  [K]

#  given that process is polytropic with
n = 1.25; #  polytropic index
T2 = T1*(P2/P1)**((n-1)/n);#  [K]

#  Hence, change in internal energy is,
del_U = m*cv*(T2-T1);#  [kJ]
print ' (a) The change in internal energy of the air is del_U (kJ) = ',round(del_U,2)
if(del_U>0):
print('since del_U>0, so it is gain of internal energy to the air')
else:
print('since del_U<0, so it is gain of internal energy to the surrounding')
#  (b)
W = m*R*(T1-T2)/(n-1);#  formula of work done for polytropic process,[kJ]
print ' (b) The work done is (kJ) = ',round(W,1)
if(W>0):
print('since W>0, so the work is done by  the air')
else:
print('since W<0, so the work is done on the air')

#  (c)
Q = del_U+W;#  using 1st law of thermodynamics,[kJ]
print ' (c) The heat transfer is Q (kJ) = ',round(Q,2)
if(Q>0):
print('since Q>0, so the heat is received by  the air')
else:
print('since Q<0, so the heat is rejected by the air')

print 'The answer is a bit different from textbook due to rounding off error'
#  End

Example 5.12
(a) The change in internal energy of the air is del_U (kJ) =  30.73
since del_U>0, so it is gain of internal energy to the air
(b) The work done is (kJ) =  -49.1
since W<0, so the work is done on the air
(c) The heat transfer is Q (kJ) =  -18.4
since Q<0, so the heat is rejected by the air
The answer is a bit different from textbook due to rounding off error


## Example 13: pg 123¶

In [13]:
#pg 123
#calculate the final volume, work done and the change in internal energy
print('Example 5.13');

#  aim : To determine the
#  final volume, work done and the change in internal energy

#  Given values
P1 = 700.;#  initial pressure,[kN/m^2]
V1 = .015;# initial volume, [m^3]
P2 = 140.;#  final pressure, [kN/m^2]
cp = 1.046;#  [kJ/kg K]
cv = .752; #  [kJ/kg K]

#  solution

Gamma = cp/cv;
#  for adiabatic expansion, P*V^gamma=constant, so
V2 = V1*(P1/P2)**(1/Gamma);#  final volume, [m^3]
print '\n The final volume of the gas is V2 (m^3) = ',round(V2,3)

#  work done
W = (P1*V1-P2*V2)/(Gamma-1);#  [kJ]
print '\n The work done by the gas is (kJ) = ',round(W,2)

del_U = -W;#  [kJ]
print '\n The change of internal energy is (kJ) = ',round(del_U,2)
if(del_U>0):
print 'since del_U>0, so the the gain in internal energy of the gas '
else:
print 'since del_U<0, so this is a loss of internal energy from the gas'

print 'The answer is a bit different from textbook due to rounding off error'
#  End

Example 5.13

The final volume of the gas is V2 (m^3) =  0.048

The work done by the gas is (kJ) =  9.77

The change of internal energy is (kJ) =  -9.77
since del_U<0, so this is a loss of internal energy from the gas
The answer is a bit different from textbook due to rounding off error


## Example 14: pg 125¶

In [14]:
#pg 125
#calculate the heat transfer, change of internal energy and mass of gas
print('Example 5.14');
import math
#  aim : To determine the
#  (a)heat transfer
#  (b)change of internal energy
#  (c)mass of gas

#   Given values
V1 = .4;#  initial volume, [m^3]
P1 = 100.;#  initial pressure, [kN/m^2]
T1 = 273.+20;#  temperature, [K]
P2 = 450.;#  final pressure,[kN/m^2]
cp = 1.0;#  [kJ/kg K]
Gamma = 1.4; #  heat capacity ratio

#  solution

#  (a)
#  for the isothermal compression,P*V=constant,so
V2 = V1*P1/P2;#  [m^3]
W = P1*V1*math.log(P1/P2);#  formula of workdone for isothermal process,[kJ]

#  for isothermal process, del_U=0;so
Q = W;
print '\n (a) The heat transferred during compression is Q (kJ) = ',round(Q)

#  (b)
V3 = V1;
#  also

P3 = P2*(V2/V3)**Gamma;#  [kN/m^2]
W = -(P3*V3-P2*V2)/(Gamma-1);#  work done formula for adiabatic process,[kJ]
#  also, Q=0,so using Q=del_U+W
del_U = -W;#  [kJ]
print '\n (b) The change of the internal energy during the expansion is,del_U (kJ) = ',round(del_U,1)

#  (c)
#  for ideal gas
#  cp-cv=R, and cp/cv=gamma, hence
R = cp*(1-1/Gamma);#  [kj/kg K]

#  now using ideal gas equation
m = P1*V1/(R*T1);#  mass of the gas,[kg]
print '\n (c) The mass of the gas is,m (kg) = ',round(m,3)

print' There is calculation mistake in the book'

#  End

Example 5.14

(a) The heat transferred during compression is Q (kJ) =  -60.0

(b) The change of the internal energy during the expansion is,del_U (kJ) =  -45.2

(c) The mass of the gas is,m (kg) =  0.478
There is calculation mistake in the book


## Example 15: pg 128¶

In [15]:
#pg 128
#calculate the the heat transferred and polytropic specific heat capacity
print('Example 5.15');

#  aim : To determine
#  the heat transferred and polytropic specific heat capacity

#  Given values
P1 = 1;#  initial pressure, [MN/m^2]
V1 = .003;#  initial volume, [m^3]
P2 = .1;#  final pressure,[MN/m^2]
cv = .718;#  [kJ/kg*K]
Gamma=1.4;#  heat capacity ratio

#  solution
#  Given process is polytropic with
n = 1.3;#  polytropic index
#  hence
V2 = V1*(P1/P2)**(1/n);#  final volume,[m^3]
W = (P1*V1-P2*V2)*10**3/(n-1);#  work done,[kJ]
#  so
Q = (Gamma-n)*W/(Gamma-1);#  heat transferred,[kJ]

print ' The heat received or rejected by the gas during this process is Q (kJ) = ',round(Q,2)
if(Q>0):
print 'since Q>0, so heat is received by the gas'
else:
print 'since Q<0, so heat is rejected by the gas'

#  now
cn = cv*(Gamma-n)/(n-1);#  polytropic specific heat capacity,[kJ/kg K]
print '\n The polytropic specific heat capacity is cn (kJ/kg K)  = ',round(cn,3)

#  End

Example 5.15
The heat received or rejected by the gas during this process is Q (kJ) =  1.03
since Q>0, so heat is received by the gas

The polytropic specific heat capacity is cn (kJ/kg K)  =  0.239


## Example 16: pg 129¶

In [16]:
#pg 129

print('Example 5.16');

# To determine the
#  (a) initial partial pressure of the steam and air
#  (b) final partial pressure of the steam and air
#  (c) total pressure in the container after heating

#  Given values
T1 = 273.+39;#  initial temperature,[K]
P1 = 100.;#  pressure, [MN/m^2]
T2 = 273.+120.2;#  final temperature,[K]

#  solution

#  (a)
#  from the steam tables, the pressure of wet steam at 39 C is
Pw1 = 7;#  partial pressure of wet steam,[kN/m^2]
#  and by Dalton's law
Pa1 = P1-Pw1;#  initial pressure of air, [kN/m^2]

print '\n (a) The initial partial pressure of the steam is (kN/m^2) = ',Pw1
print '\n      The initial partial pressure of the air is (kN/m^2) = ',Pa1

#  (b)
#  again from steam table, at 120.2 C the pressure of wet steam is
Pw2 = 200.;#  [kN/m^2]

#  now since volume is constant so assuming air to be ideal gas so for air  P/T=contant, hence
Pa2 = Pa1*T2/T1 ;#  [kN/m^2]

print ' \n(b) The final partial pressure of the steam is (kN/m^2) = ',Pw2
print '\n     The final partial pressure of the air is (kN/m^2) = ',round(Pa2,2)

#  (c)
Pt = Pa2+Pw2;#  using dalton's law, total pressure,[kN/m^2]
print '\n (c) The total pressure after heating is (kN/m^2) = ',round(Pt,2)

#  End

Example 5.16

(a) The initial partial pressure of the steam is (kN/m^2) =  7

The initial partial pressure of the air is (kN/m^2) =  93.0

(b) The final partial pressure of the steam is (kN/m^2) =  200.0

The final partial pressure of the air is (kN/m^2) =  117.2

(c) The total pressure after heating is (kN/m^2) =  317.2


## Example 17: pg 130¶

In [17]:
#pg 130
print('Example 5.17');

#  aim : To determine
#  the partial pressure of the air and steam, and the mass of the air

#  Given values
P1 = 660.;#  vaccum gauge pressure on condenser [mmHg]
P = 765.;#  atmospheric pressure, [mmHg]
x = .8;#  dryness fraction
T = 273.+41.5;#  temperature,[K]
ms_dot = 1500.;#  condense rate of steam,[kg/h]
R = .29;#  [kJ/kg]

#  solution
Pa = (P-P1)*.1334;# absolute pressure,[kN/m^2]
# from steam table, at 41.5 C partial pressure of steam is
Ps = 8;#  [kN/m^2]
#  by dalton's law, partial pressure of air is
Pg = Pa-Ps;#  [kN/m^2]

print '\n The partial pressure of the air in the condenser is (kN/m^2) = ',round(Pg)
print '\n The partial pressure of the steam in the condenser is (kN/m^2) = ',Ps

#  also
vg = 18.1;#  [m^3/kg]
#  so
V = x*vg;#  [m^3/kg]
#  The air associated with 1 kg of the steam will occupiy this same volume
#  for air, Pg*V=m*R*T,so
m = Pg*V/(R*T);#  [kg/kg steam]
#  hence
ma = m*ms_dot;#  [kg/h]

print '\n The mass of air which will associated with this steam is (kg) = ',round(ma,1)

print' There is misprint in book'

#  End

Example 5.17

The partial pressure of the air in the condenser is (kN/m^2) =  6.0

The partial pressure of the steam in the condenser is (kN/m^2) =  8

The mass of air which will associated with this steam is (kg) =  1430.5
There is misprint in book


## Example 18: pg 130¶

In [18]:
#pg 130
print('Example 5.18');

#  aim : To determine the
#  (a) final pressure
#   (b) final dryness fraction of the steam

#  Given values
P1 = 130.;#  initial pressure, [kN/m^2]
T1 = 273.+75.9;#  initial temperature, [K]
x1 = .92;#  initial dryness fraction
T2 = 273.+120.2;#  final temperature, [K]

#  solution

#  (a)
#  from steam table, at 75.9 C
Pws = 40.;#  partial pressure of wet steam[kN/m^2]
Pa = P1-Pws;#  partial pressure of air, [kN/m^2]
vg = 3.99#  specific volume of the wet steam, [m^3/kg]
# hence
V1 = x1*vg;#  [m^3/kg]
V2 = V1/5;#  [m^3/kg]
#  for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so
P2 = Pa*V1*T2/(V2*T1);#  final pressure,[kN/m^2]

#  now for steam at 120.2 C
Ps = 200.;#  final partial pressure of steam,[kN/m^2]
#  so by dalton's law total pressure in cylindert is
Pt = P2+Ps;#  [kN/m^2]
print ' (a) The final pressure in the cylinder is (kN/m^2) = ',round(Pt,1)

#  (b)
#  from steam table at 200 kN/m^2
vg = .885;#  [m^3/kg]
#  hence
x2 = V2/vg;#  final dryness fraction of the steam
print ' (b) The final dryness fraction of the steam is  = ',round(x2,2)

#  End

Example 5.18
(a) The final pressure in the cylinder is (kN/m^2) =  707.1
(b) The final dryness fraction of the steam is  =  0.83


## Example 19: pg 131¶

In [19]:
#pg 131
print('Example 5.19')

#  aim : To determine the
#  (a) Gamma,
#  (b) del_U
import math
#  Given Values
P1 = 1400.;#  [kN/m^2]
P2 = 100.;#  [kN/m^2]
P3 = 220.;#  [kN/m^2]
T1 = 273.+360;#  [K]
m = .23;# [kg]
cp = 1.005;#  [kJ/kg*K]

#  Solution
T3 = T1;#  since process 1-3 is isothermal

#  (a)
#  for process 1-3, P1*V1=P3*V3,so
V3_by_V1 = P1/P3;
#  also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence
#  and process process 2-3 is iso-choric so,V3=V2 and
V2_by_V1 = V3_by_V1;
#  hence,
Gamma = math.log(P1/P2)/math.log(P1/P3); #  heat capacity ratio

print ' (a) The value of adiabatic index Gamma is  = ',round(Gamma,3)

#  (b)
cv = cp/Gamma;#  [kJ/kg K]
#  for process 2-3,P3/T3=P2/T2,so
T2 = P2*T3/P3;#  [K]

#   now
del_U = m*cv*(T2-T1);#  [kJ]
print ' (b) The change in internal energy during the adiabatic expansion is U2-U1  (This is loss of internal energy) (kJ) = ',round(del_U,2)
#   End

Example 5.19
(a) The value of adiabatic index Gamma is  =  1.426
(b) The change in internal energy during the adiabatic expansion is U2-U1  (This is loss of internal energy) (kJ) =  -55.97


## Example 20: pg 133¶

In [20]:
#pg 133
print('Example 5.20');

#  aim : To determine
#  the mass of oxygen and heat transferred

#  Given values
V1 = 300.;#  [L]
P1 = 3.1;#  [MN/m^2]
T1 = 273.+18;#  [K]
P2 = 1.7;#  [MN/m^2]
T2 = 273.+15;#  [K]
Gamma = 1.4; #  heat capacity ratio
#  density condition
P = .101325;# [MN/m^2]
T = 273.;# [K]
V = 1.;#  [m^3]
m = 1.429;# [kg]

#  hence
R = P*V*10**3/(m*T);#  [kJ/kg*K]
#  since volume is constant
V2 = V1;# [L]
#  for the initial conditions in the cylinder,P1*V1=m1*R*T1
m1 = P1*V1/(R*T1);#  [kg]

# after some of the gas is used
m2 = P2*V2/(R*T2);#  [kg]
#  The mass of oxygen remaining in cylinder is m2 kg,so
#  Mass of oxygen used is
m_used = m1-m2;#  [kg]
print ' The mass of oxygen used (kg) = ',round(m_used,1)

#  for non-flow process,Q=del_U+W
#  volume is constant so no external work is done so,Q=del_U
cv = R/(Gamma-1);#  [kJ/kg*K]

#  heat transfer is
Q = m2*cv*(T1-T2);#  (kJ)
print ' The amount of heat transferred through the cylinder wall is (kJ) = ',round(Q,2)

#  End

Example 5.20
The mass of oxygen used (kg) =  5.5
The amount of heat transferred through the cylinder wall is (kJ) =  13.28


## Example 21: pg 134¶

In [21]:
#pg 134
print('Example 5.21');

#  aim : To determine the
#  (a) work transferred during the compression
#  (b) change in internal energy
#  (c) heat transferred during the compression

#  Given values
V1 = .1;#  initial volume, [m^3]
P1 = 120.;#  initial pressure, [kN/m^2]
P2 = 1200.; # final pressure, [kN/m^2]
T1 = 273.+25;#  initial temperature, [K]
cv = .72;#  [kJ/kg*K]
R = .285;#  [kJ/kg*K]

#  solution

#  (a)
#  given process is polytropic with
n = 1.2;  # polytropic index
#  hence
V2 = V1*(P1/P2)**(1/n);#  [m^3]
W = (P1*V1-P2*V2)/(n-1);#  workdone formula, [kJ]
print ' (a) The work transferred during the compression is (kJ) = ',round(W,1)

#  (b)
#  now mass is constant so,
T2 = P2*V2*T1/(P1*V1);#  [K]
#  using, P*V=m*R*T
m = P1*V1/(R*T1);#  [kg]

#  change in internal energy is
del_U = m*cv*(T2-T1);#  [kJ]
print ' (b) The change in internal energy is (kJ) = ',round(del_U,1)

#  (c)
Q = del_U+W;#  [kJ]
print ' (c) The heat transferred during the compression is (kJ) = ',round(Q,0)

#  End

Example 5.21
(a) The work transferred during the compression is (kJ) =  -28.1
(b) The change in internal energy is (kJ) =  14.2
(c) The heat transferred during the compression is (kJ) =  -14.0


## Example 22: pg 135¶

In [22]:
#pg 135
print('Example 5.22');

#  aim : To determine the
#  (a) new pressure of the air in the receiver
#  (b) specific enthalpy of air at 15 C

#  Given values
V1 = .85;#  [m^3]
T1 = 15.+273;#  [K]
P1 = 275.;# pressure,[kN/m^2]
m = 1.7;#  [kg]
cp = 1.005;#  [kJ/kg*K]
cv = .715;#  [kJ/kg*K]

#  solution

#  (a)

R = cp-cv;#  [kJ/kg*K]
#  assuming m1 is original mass of the air, using P*V=m*R*T
m1 = P1*V1/(R*T1);#  [kg]
m2 = m1+m;#  [kg]
#  again using P*V=m*R*T
#  P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so
P2 = P1*m2/m1;#  [kN/m^2]
print ' (a) The new pressure of the air in the receiver is (kN/m^2) = ',round(P2)

#  (b)
#  for 1 kg of air, h2-h1=cp*(T1-T0)
#  and if 0 is chosen as the zero enthalpy, then
h = cp*(T1-273);#  [kJ/kg]
print ' (b) The specific  enthalpy of the air at 15 C is (kJ/kg) = ',h

#  End

Example 5.22
(a) The new pressure of the air in the receiver is (kN/m^2) =  442.0
(b) The specific  enthalpy of the air at 15 C is (kJ/kg) =  15.075


## Example 23: pg 136¶

In [23]:
#pg 136
print('Example 5.23');

#  aim : T determine the
#  (a) characteristic gas constant of the gas
#  (b) cp,
#  (c) cv,
#  (d) del_u
#  (e) work transfer

#  Given values
P = 1.;#  [bar]
T1 = 273.+15;#  [K]
m = .9;#  [kg]
T2 = 273.+250;#  [K]
Q = 175.;#  heat transfer,[kJ]

#  solution

#  (a)
#  using, P*V=m*R*T, given,
m_by_V = 1.875;
#  hence
R = P*100/(T1*m_by_V);#  [kJ/kg*K]
print ' (a) The characteristic gas constant of the gas is R (kJ/kg K) = ',round(R,3)

#  (b)
#  using, Q=m*cp*(T2-T1)
cp = Q/(m*(T2-T1));#  [kJ/kg K]
print ' (b) The specific heat capacity of the gas at constant pressure cp (kJ/kg K) = ',round(cp,3)

# (c)
#  we have, cp-cv=R,so
cv = cp-R;# [kJ/kg*K]
print ' (c) The specific heat capacity of the gas at constant volume cv  (kJ/kg K) = ',round(cv,3)

#  (d)
del_U = m*cv*(T2-T1);#  [kJ]
print ' (d) The change in internal energy is (kJ) = ',round(del_U)

#  (e)
# using, Q=del_U+W
W = Q-del_U;#  [kJ]
print ' (e) The work transfer is W (kJ) = ',round(W)

#  End

Example 5.23
(a) The characteristic gas constant of the gas is R (kJ/kg K) =  0.185
(b) The specific heat capacity of the gas at constant pressure cp (kJ/kg K) =  0.827
(c) The specific heat capacity of the gas at constant volume cv  (kJ/kg K) =  0.642
(d) The change in internal energy is (kJ) =  136.0
(e) The work transfer is W (kJ) =  39.0


## Example 24: pg 136¶

In [24]:
#pg 136
print('Example 5.24');

#  aim : To determine the
#  (a) work transfer,
#  (b)del_U  and,
#  (c)heat transfer

#  Given values
V1 = .15;#  [m^3]
P1 = 1200.;#  [kN/m^2]
T1 = 273.+120;# [K]
P2 = 200.;#  [kN/m^2]
cp = 1.006;#[kJ/kg K]
cv = .717;#  [kJ/kg K]

#  solution

#  (a)
#  Given, PV^1.32=constant, so it is polytropic process with
n = 1.32;#  polytropic index
# hence
V2 = V1*(P1/P2)**(1./n);#  [m^3]
#  now, W
W = (P1*V1-P2*V2)/(n-1);# [kJ]
print ' (a) The work transfer is W (kJ) = ',round(W,1)

#  (b)
R = cp-cv;#  [kJ/kg K]
m = P1*V1/(R*T1);#  gas law,[kg]
#  also for polytropic process
T2 = T1*(P2/P1)**((n-1)/n);#  [K]
#  now for gas,
del_U = m*cv*(T2-T1);#  [kJ]
print ' (b) The change of internal energy is del_U (kJ) = ',round(del_U,1)

#  (c)
Q = del_U+W;#  first law of thermodynamics,[kJ]
print ' (c) The heat transfer Q (kJ) = ',round(Q,1)

print 'The answer is a bit different due to rounding off error in textbook'
#  End

Example 5.24
(a) The work transfer is W (kJ) =  198.2
(b) The change of internal energy is del_U (kJ) =  -157.3
(c) The heat transfer Q (kJ) =  40.8
The answer is a bit different due to rounding off error in textbook


## Example 26: pg 141¶

In [25]:
#pg 141
print('Example 5.26');

#  aim : To determine
#  the volume of the pressure vessel and the volume of the gas before transfer

#  Given values

P1 = 1400.;#  initial pressure,[kN/m^2]
T1 = 273.+85;#  initial temperature,[K]

P2 = 700.;#  final pressure,[kN/m^2]
T2 = 273.+60;#  final temperature,[K]

m = 2.7;# mass of the gas passes,[kg]
cp = .88;#  [kJ/kg]
cv = .67;#  [kJ/kg]

#  solution

#  steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W  [1],
#  given, there is no kinetic energy change and neglecting potential energy term
W = 0;# no external work done
#  so final equation is,u1+P1*v1+Q=u2   [2]
# also u2-u1=cv*(T2-T1)
# hence Q=cv*(T2-T1)-P1*v1    [3]
#  and for unit mass P1*v1=R*T1=(cp-cv)*T1  [4]
#  so finally
Q = cv*(T2-T1)-(cp-cv)*T1;#  [kJ/kg]
#  so total heat transferred is
Q2 = m*Q;#  [kJ]

#  using eqn [4]
v1 = (cp-cv)*T1/P1;#  [m^3/kg]
#  Total volume is
V1 = m*v1;#  [m^3]

#  using ideal gas equation P1*V1/T1=P2*V2/T2
V2 = P1*T2*V1/(P2*T1);#  final volume,[m^3]

print 'heat transfer from the gas (kJ) = ',round(Q2,1)
print ' The volume of gas before transfer is (m^3) = ',round(V1,3)
print ' The volume of pressure vessel is (m^3) = ',round(V2,2)

#  End

Example 5.26
heat transfer from the gas (kJ) =  -248.2
The volume of gas before transfer is (m^3) =  0.145
The volume of pressure vessel is (m^3) =  0.27