# Chapter 8 : Thermodynamic Cycles¶

### Example 8.1 Page Number : 287¶

In [1]:

import math

# Variables
P_1 = 30;			#[bar]
P_2 = 0.04;			#[bar]

#(1).Carnot cycle
#It has been reported in the book that at 30 bar pressure (saturated) :
H_liq_1 = 1008.42;			#[kJ/kg]
H_vap_1 = 2804.2;			#[kJ/kg]
S_liq_1 = 2.6457;			#[kJ/kg-K]
S_vap_1 = 6.1869;			#[kJ/kh-K]
#Therefore, H_1 = H_liq_1, H_2 = H_vap_1, S_1 = S_liq_1 and S_2 = S_vap_1
H_1 = H_liq_1;
H_2 = H_vap_1;
S_1 = S_liq_1;
S_2 = S_vap_1;

#At 0.04 bar pressure (saturated) :
H_liq_2 = 121.46;			#[kJ/kg]
H_vap_2 = 2554.4;			#[kJ/kg]
S_liq_2 = 0.4226;			#[kJ/kg-K]
S_vap_2 = 8.4746;			#[kJ/kh-K]

# Calculations and Results
#Dryness fraction at state 3 can be found the fact that S_3 = S_2
x_3 = (S_2 - S_liq_2)/(S_vap_2 - S_liq_2);
H_3 = H_liq_2*(1 - x_3) + x_3*H_vap_2;			#[kJ/kg]

#Dryness fraction at state 4 can be found the fact that S_4 = S_1
x_4 = (S_1 - S_liq_2)/(S_vap_2 - S_liq_2);
H_4 = H_liq_2*(1 - x_4) + x_4*H_vap_2;			#[kJ/kg]

#Work done by turbine W_tur = -delta_H = -(H_3 - H_2)
W_tur = H_2 - H_3;			#[kJ/kg]

#Work supplied by boiler,
q_H = H_2 - H_1;			#[kJ/kg]

#Work transfer in compressor is given by
W_com = -(H_1 - H_4);			#[kJ/kg]

#Efficiency can now be calculated as
#n = (Net work done/Work supplied by boiler)
n_carnot = (W_tur + W_com)/q_H;

#Efficiency of the Carnot cycle can also be determined from the formula
# n = 1 - (T_L/T_H), Where T_L is saturated temperature at 0.04 bar and T_H is saturated temperature at 30 bar

print "1.Carnot cycle";
print "The work done by the turbine is %f kJ/kg"%(W_tur);
print "The heat transfer in the boiler is %f kJ/kg"%(q_H);
print "The cycle efficiency is %f"%(n_carnot);

#(2).Rankine cycle
#The enthalpies at state 2 and 3 remain as in the Carnot cycle
#Saturated liquid enthalpy at 0.04 bar is
H_4_prime = H_liq_2;

#Saturated liquid volume at 0.04 bar as reported in the book is
V_liq = 0.001004;			#[m**(3)/kg]
#Work transfer in pump can be calculated as
W_pump = -V_liq*(P_1 - P_2)*100;			#[kJ/kg]

#Work transfer around pump gives, W_pump = -delta_H =  -(H_1_prime - H_4_prime);
H_1_prime = H_4_prime - W_pump;			#[kJ/kg]

#Heat supplied to boiler is
q_H_prime = H_2 - H_1_prime;			#[kJ/kg]

#Work done by turbine is
W_tur_prime = H_2 - H_3;			#[kJ/kg]

#Efficiency can now be calculated as
#n = (Net work done/Heat input)
n_rankine = (W_tur_prime + W_pump)/q_H_prime;			#

print "2.Rankine cycle";
print "The work done by the turbine is %f kJ/kg"%(W_tur_prime);
print "The heat transfer in the boiler is %f kJ/kg"%(q_H_prime);
print "The cycle efficiency is %f"%(n_rankine);

1.Carnot cycle
The work done by the turbine is 941.036567 kJ/kg
The heat transfer in the boiler is 1795.780000 kJ/kg
The cycle efficiency is 0.404166
2.Rankine cycle
The work done by the turbine is 941.036567 kJ/kg
The heat transfer in the boiler is 2679.732016 kJ/kg
The cycle efficiency is 0.350046


### Example 8.2 Page Number : 288¶

In [2]:

T_max = 700+273.15;			#[K] - Maximum temperature.
P_boiler = 10*10**(6);			#[Pa] - Constant pressure in the boiler
P_condenser = 10*10**(3);			#[Pa] - Constant pressure in the condenser

#At state 2 i.e, at 700 C and 10 MPa,it has been reported in the book that from steam table
S_2 = 7.1687;			#[kJ/kg-K] - Entropy
H_2 = 3870.5;			#[kJ/kg] - Enthalpy

#At state 3 i.e, at 700 C and 10 KPa,
S_3 = S_2;			#[kJ/kg-K]- Entropy

#For sturated steam at 10 kPa, it has been reported in the book that from steam table
S_liq = 0.6493;			#[kJ/kg-K]- Entropy of saturated liquid
S_vap = 8.1502;			#[kJ/kg-K] - Enthalpy of saturated liquid
#Therefore steam is saturated and its dryness factor can be calculated as
x = (S_2 - S_liq)/(S_vap - S_liq);

# Calculations and Results
#The enthalpy at state 3 is now calculated. For steam at 10 kPa,it has been reported in the book that from steam table
H_liq = 191.83;			#[kJ/kg]
H_vap = 2584.7;			#[kJ/kg]
#Therefore enthalpy at state 3 is
H_3 = H_liq*(1-x) + H_vap*x;			#[kJ/kg]

#Work done by the turbine
W_tur = -(H_3 - H_2);			#[kJ/kg]

#Now we have to calculate work input to the pump
#State 4:Saturated liquid at 10 kPa
#State 4:Compressed liquid at 10 MPa
#Since volume of liquid does not get affected by pressure we take volume of saturated liquid at 10 kPa,
V_liq = 0.001010;			#[m**(3)/kg]

#Work transfer in the pump is
W_pump = -V_liq*(P_boiler - P_condenser)*10**(-3);			#[kJ/kg]

#Energy balance around pump gives, W_pump = -delta_H =  -(H_1 - H_4)
H_4 = H_liq;			# Enthalpy at state 4 (saturated liquid at 10 kPa)
H_1 = H_4 - W_pump;			#[kJ/kg]

#Heat supplied to boiler is
q_H = H_2 - H_1;			#[kJ/kg]

#Efficiency can now be calculated as
#n = (Net work done/Heat input)
n_rankine = (W_tur + W_pump)/q_H;

print "The efficiency of the Rankine cycle is found to be %f"%(n_rankine);

#Now let us determine the efficiency of Carnot cycle. The maximun temperature is 700 C and minimum temperature is that of saturated steam at 10 kPa,
T_min = 45.81 + 273.15;			#[K] - From steam table as reported in the book
n_carnot = 1-(T_min/T_max);
#Note that the efficiency of Rankine cycle is less than that of carnot cycle

The efficiency of the Rankine cycle is found to be 0.433088


### Example 8.3 Page Number : 291¶

In [3]:

W = 1.1;			#[kW] - Work done per ton of refrigeration
#1 ton refrigeration = 3.517 kW, therefore
H = 3.517;			#[kW] - Heat absorbed
T_low = -30 + 273.15;			#[K] - Low temperature maintained

# Calculations
#COP can be calculated as
#COP = (Heat absorbed/Work done)
COP = H/W;

#For reversed carnot cycle, COP = T_low/(T_high - T_low). Solving this we get
T_high = (T_low/COP) + T_low;			#[K] - Higher temperature

#Heat rejected is
H_rej = W + H;			#[kW];

# Results
print "The COP is %f"%(COP);
print "The higher temperature of the cycle is %f K"%(T_high);
print "The heat rejected per ton of refrigeration is %f kW"%(H_rej);

The COP is 3.197273
The higher temperature of the cycle is 319.199190 K
The heat rejected per ton of refrigeration is 4.617000 kW


### Example 8.4 Page Number : 292¶

In [4]:

T_high = 20 + 273.15;			#[K] - High temperature
T_low = 0 + 273.15;			#[K] - Low temperature
Q_H = 10;			#[kW] - Heat supplied

# Calculations
#If 'Q_H' is the rate at which heat is taken from surrounding and 'W' is the rate at which work is done,then
# Q_H = W + Q_L
#(Q_H/Q_L) = (T_high/T_low)
#Also for a reversible cycle, (Q_H/Q_L) = 1 + (W/Q_L). Solving we get,
Q_L = (T_low/T_high)*Q_H;			#[kW]
W = (Q_H - Q_L) ;			#[kW]

# Results
print "The minimum power required is %f kW"%(W);

The minimum power required is 0.682245 kW


### Example 8.5 Page Number : 292¶

In [5]:

T_high = 40 + 273.15;			#[K] - High temperature
T_low = -20 + 273.15;			#[K] - Low temperature
C = 10;			#[tons of refrigeration] - Capacity
#1 ton refrigeration = 3.517 kW, therefore
H = C*3.517;			#[kW] - Heat absorbed

# Calculations
#For reversed carnot cycle, COP = T_low/(T_high - T_low)
COP = T_low/(T_high - T_low);

# COP = (Refrigerating effect)/(Work input), therefore power required is given by
P = (H/COP);			#[kW]

# Results
print "The COP is %f"%(COP);
print "The power required is %f kW"%(P);

The COP is 4.219167
The power required is 8.335769 kW


### Example 8.6 Page Number : 292¶

In [6]:

COP = 4;			#Coefficient of performance
P = 10;			#[kW] - Work done on the cycle

# Calculations
#For reversed carnot cycle, COP = T_low/(T_high - T_low)
#ratio = (T_high/T_low),therefore
ratio = -1/(COP + 1);

# Refrigerating effect = (COP)*Work input, therefore refrigeration is given by
H = COP*P;			#[kW]

#Maximum refrigearation in tons is given by
H_max = (H/3.517);

# Results
print "The maximum refrigeration value is %f ton"%(H_max);

The maximum refrigeration value is 11.373330 ton


### Example 8.7 Page Number : 292¶

In [7]:

m = 0.6;			#[kg/s] - mass flow rate
T_low = -20+273.15;			#[K] - Temperature at which vapour enters the compressor
T_high = 30+273.15;			#[K] - Temperature at which vapour leaves the condenser

#From saturated refrigeration-12 tables we get,at -20 C
H_1 = 178.74;			#[kJ/kg] - (H_1 = H_vap)
P_1 = 0.15093;			#[MPa] - (P_1 = P_sat)
P_4 = P_1;
S_1 = 0.7087;			#[kJ/kg-K] - (S_1 = S_vap)
S_2 = S_1;

#At 30 C
P_2 = 0.7449;			#[MPa] - (P_2 = P_sat)
P_3 = P_2;
H_3 = 64.59;			#[kJ/kg] - (H_3 = H_liq)
H_4 = H_3;
S_3 = 0.24;			#[kJ/kg-K] - (S_3 = S_liq)

# Calculations and Results
#It is assumed that presssure drop in the evaporator and condenser are negligible. The heat transfer rate in the evaporator is
Q_L = m*(H_1 - H_4);

print "The heat transfer rate in the evaporator is %f kW"%(Q_L);

#At state 2 (P = 0.7449 MPa and S = 0.7087 kJ/kg-K) and looking in the superheated tables we have to calculate the enthalpy at state 2

#At P = 0.7 MPa and S = 0.6917 kJ/kg-K,
H_11 = 200.46;			#[kJ/kg]

#At P = 0.7 MPa and S = 0.7153 kJ/kg-K,
H_12 = 207.73;			#[kJ/kg]

#Thus at P = 0.7 MPa and S = 0.7087 kJ/kg-K, enthalpy is given by
H_13 = ((S_2 -0.6917)/(0.7153 - 0.6917))*(H_12 - H_11) + H_11;			#[kJ/kg]

#At P = 0.8 MPa and S = 0.7021 kJ/kg-K,
H_21 = 206.07;			#[kJ/kg]

#At P = 0.8 MPa and S = 0.7253 kJ/kg-K,
H_22 = 213.45;			#[kJ/kg]

#Thus at P = 0.8 MPa and S = 0.7087 kJ/kg-K, enthalpy is given by
H_23 = ((S_2 -0.7021)/(0.7253 - 0.7021))*(H_22 - H_21) + H_21;			#[kJ/kg]

#At P = 0.7449 MPa, S = 0.7087 kJ/kg-K, the enthalpy is
H_2 = ((0.7449 - 0.7)/(0.8 - 0.7))*(H_23 - H_13) + H_13;			#[kJ/kg]

#Power consumed by the compressor is
W_comp = m*(H_2 - H_1);			#[kW]

print "The power consumed by the compressor is %f kW"%(W_comp);

#Heat removed in evaporator/work done on compressor
COP_R = Q_L/W_comp;

print "The COP the refrigerator is %f kW"%(COP_R);

#At -20 C,saturated conditions
H_liq = 17.82;			#[kJ/kg]
H_vap = 178.74;			#[kJ/kg]
x_4 = (H_4 - H_liq)/(H_vap - H_liq);

print "The dryness factor of refrigerant after the expansion valve is %f"%(x_4);

#The heat transfer rate in the condenser is
Q_H = m*(H_3 - H_2);			#[kW]

print "The heat transfer rate in the condenser is %f kW"%(Q_H);

#If the cycle would have worked as a pump then,
#COP_HP = (Heat supplied from condenser/Work done on compressor)
COP_HP = (-Q_H)/W_comp;

print "The COP if cycle would work as a heat pump is %f kW"%(COP_HP);

#If the cycle would have been a reversed Carnot cycle then
COP_C = T_low/(T_high - T_low);

print "The COP if cycle would run as reversed Carnot cycle is %f kW"%(COP_C);

The heat transfer rate in the evaporator is 68.490000 kW
The power consumed by the compressor is 16.840242 kW
The COP the refrigerator is 4.067044 kW
The dryness factor of refrigerant after the expansion valve is 0.290641
The heat transfer rate in the condenser is -85.330242 kW
The COP if cycle would work as a heat pump is 5.067044 kW
The COP if cycle would run as reversed Carnot cycle is 5.063000 kW


### Example 8.8 Page Number : 300¶

In [1]:

from scipy.optimize import fsolve
import math

# Variables
H_1 = 310.38;			#[kJ/kg]
H_2 = 277.7;			#[kJ/kg]
H_5 = -122.6;			#[kJ/kg]
H_6 = 77.8;			#[kJ/kg]

#The enthalpy at point 3 is same at point 4 as the expansion is isenthalpic

# Calculations and Results
#The mass condensed is 1 kg and therefore m_1 = m+6 + 1

#Enthalpy balance around heat exchanger
#m_2*H_2 + m_2*H_6 = m_3*H_3 + m_7*H_7

#Enthalpy balance around separator
#m_4*H_4 = m_5*H_5 + m_6*H_6
#It can be seen that m_1 = m_2 = m_3 = m_4
#and m_6 = m_7 = m_1 - 1

#Substituting the values for enthalpy balance around heat exchanger we get,
#m_1*H_2 + (m_1 - 1)*(H_6) = m_1*H_3 + (m_1 - 1)*H_1
#and substituting the values for enthalpy balance around seperator we get
#m_1*H_3 = (1)*(-122.6) + (m_1 - 1)*77.8
#H_3 = ((1)*(-122.6) + (m_1 - 1)*77.8)/m_1
#Substituting the expression for 'H_3' in the above equation and then solving for m_1, we get
def f(m_1):
return m_1*H_2+(m_1-1)*(H_6)-m_1*(((1)*(-122.6) + (m_1 - 1)*77.8)/m_1)-(m_1-1)*H_1
m_1 = fsolve(f,4)
#Thus to liquify 1 kg of air compression of m_1 kg of air is carried out.

#Now substituting this value of m_1 to get the value of H_3,
H_3 = ((1)*(-122.6) + (m_1 - 1)*77.8)/m_1;			#[kJ/kg]

#From given compressed air table we see at 200 bar and 160 K,
H_3_1 = 40.2;			#[kJ/kg]

#At 200 bar and 180 K,
H_3_2 = 79.8;			#[kJ/kg]
#By interpolation we get,
T_3 = ((H_3 - H_3_1)*(180 - 160))/(H_3_2 - H_3_1) + 160;			#[K]

print "Temperature before throttling is %f K"%(T_3);

Temperature before throttling is 171.350719 K


### Example 8.9 Page Number : 304¶

In [9]:

# Variables
#At 1 bar, 310 K
H_1 = 310.38;			#[kJ/kg]
#At 200 bar, 310 K
H_2 = 277.7;			#[kJ/kg]
#At 1 bar, Saturated liquid
H_7 = -122.6;			#[kJ/kg]
#At 1 bar, Saturated vapour
H_8 = 77.8;			#[kJ/kg]
#At 200 bar, 200 K
H_3 = 117.6;			#[kJ/kg]
#At 1 bar, 100 K
H_11 = 98.3;			#[kJ/kg]

# Calculations and Results
#For 1 kg of liquid air obtained,the overall enthalpy balance is
#m_2*H_2 = W - 122.6 + (m_2 - 1)*H_1
#W = - 0.8*m_2*(H_11 - H_3)
#Overall enthalpy balance equation becomes
#H_2*m_2 = 15.44*m_2 - H_7 + (m_2 - 1)*H_1, solving
m_2_prime = (H_7 - H_1)/(H_2 - 15.44 - H_1);

print "The number of kimath.lograms of air compressed per kg of liquid air produced is %f kg"%(m_2_prime);

#(2)
#Enthalpy balance around separator is
#0.2*m_2*H_5 = -H_7 + (0.2*m_2 - 1)*H_8, solving
m_2 = m_2_prime;
H_5_prime = ((0.2*m_2-1)*H_8 - H_7)/(0.2*m_2);

#At point 5, P = 200 bar and enthalpy is
H_5_1 = -33.53;			#[kJ/kg]
#From compressed air tables at 200 bar and 140 K,
H_5_2 = 0.2;			#[kJ/kg]
#At 200 bar and 120 K,
H_5_3 = -38.0;			#[kJ/kg]
#Solving by interpolation we get
T_5 = ((H_5_1 - H_5_3)*(140 - 120))/(H_5_2 - H_5_3) + 120;			#[K]

print "The temperature of air before throttling is %f K"%(T_5);

#(3)
#During mixing of streams 8 and 11 to produce stream 9, the enthalpy balance is
# (0.2*m_2 - 1)*H_8 + 0.8*m_2*H_11 = (m_2 - 1)*H_9,Solving for H_9

H_9_prime = ((0.2*m_2-1)*H_8+0.8*m_2*H_11)/(m_2 - 1);

#From given compressed air tables at 1 bar and 100 K,
H_9_1 = H_11;
#At 1 bar and 90 K
H_9_2 = 87.9;			#[kJ/kg]
#Solving by interpolation we get
T_9 = ((H_9_prime - H_9_2)*(100 - 90))/(H_9_1 - H_9_2) + 90;			#[K]

print "The temperature of stream entering second heat exchanger is %f K"%(T_9);

#(4)
#Enthalpy balance around first heat exchanger is
#H_2*m_2 + (m_2 - 1)*H_10 = H_3*m-2 + (m-2 - 1)*H_1, solving for H_10

H_10_prime = ((m_2 - 1)*H_1 + H_3*m_2 - H_2*m_2)/(m_2 - 1);

#From given compressed air tables at 1 bar and 140 K,
H_10_1 = 139.1;			#[kJ/kg]
#At 1 bar and 120 K
H_10_2 = 118.8;			#[kJ/kg]
#Solving by interpolation we get
T_10 = ((H_10_prime - H_10_2)*(140 - 120))/(H_10_1 - H_10_2) + 120;			#[K]

print "The temperature of stream exiting second heat exchanger is %f K"%(T_10);

The number of kimath.lograms of air compressed per kg of liquid air produced is 8.997922 kg
The temperature of air before throttling is 122.340314 K
The temperature of stream entering second heat exchanger is 98.029358 K
The temperature of stream exiting second heat exchanger is 131.292906 K


### Example 8.10 Page Number : 307¶

In [10]:

P_high = 40;			#[bar]
P_low = 5;			#[bar]
m_1 = 0.5;			#[kg/s] - Rate of mass moving through the expander
m_2 = 0.1;			#[kg/s] - Rate of mass of vapour mixing with air
e = 0.7;			#Efficiency

#At state 3,(40 bar and 200 K),enthalpy and entropy is given by
H_3 = 179.7;			#[kJ/kg]
S_3 = 5.330;			#[kJ/kg-K]

#If isentropic conditions exits in the turbine then state 11 is at 5 bar
S_11 = 5.330;			#[kJ/kg-K]
#From given compressed air tables at 5 bar and 120 K,
H_11_1 = 113.6;			#[kJ/kg]
S_11_1 = 5.455;			#[kJ/kg-K]
#At 5 bar and 100 K
H_11_2 = 90.6;			#[kJ/kg]
S_11_2 = 5.246;			#[kJ/kg-K]
#The enthalpy has to be determined when S = S_3

# Calculations and Results
#Solving by interpolation we get
H_11_s = ((H_11_1 - H_11_2)*(S_3 - S_11_2))/(S_11_1 - S_11_2) + H_11_2;			#[kJ/kg]

#The adiabatic efficiency of tyrbine is given by
#(H_3 - H_11_a)/(H_3 - H_11_s) = e
H_11_a = H_3 - e*(H_3 - H_11_s);			#[kJ/kg] - Actual enthalpy

#At 5 bar,the saturated enthalpy is given to be
H_8 = 88.7;			#[kJ/kg]
#From enthalpy balance during mixing we get,
#0.1*H_8 + 0.5*H_11_a = 0.6*H_9
H_9 = (m_2*H_8 + m_1*H_11_a)/(m_1 + m_2);			#[kJ/kg]

#From given compressed air tables at 5 bar and 140 K,
H_9_1 = 135.3;			#[kJ/kg]
#At 5 bar and 120 K
H_9_2 = 113.6;			#[kJ/kg]
#By interpolation we get
T_9 = ((H_9 - H_11_1)*(140 - 120))/(H_9_1 - H_11_1) + 120;			#[K]

print " The temperature of air entering the second heat exchanger is %f K"%(T_9);

 The temperature of air entering the second heat exchanger is 124.009841 K