# -*- coding: utf8 -*-
from __future__ import division
#Example: 11.1
'''Determine the efficiency of a Rankine cycle using steam as the working fluid in which the
condenser pressure is 10 kPa. The boiler pressure is 2 MPa. The steam leaves the boiler
as saturated vapor.
In solving Rankine-cycle problems, we let w p denote the work into the pump per
kilogram of fluid flowing and q L denote the heat rejected from the working fluid per
kilogram of fluid flowing.
To solve this problem we consider, in succession, a control surface around the pump,
the boiler, the turbine, and the condenser. For each, the thermodynamic model is the steam
tables, and the process is steady state with negligible changes in kinetic and potential
energies. First, consider the pump:
Control volume:
Inlet state:
Exit state:
Pump.
P 1 known, saturated liquid; state fixed.
P 2 known.'''
#Variable Declaration:
P2 = 2000 #Exit pressure in kPa
P1 = 10 #Inlet pressure in kPa
v = 0.00101 #specific weight of water in m**3/kg
h1 = 191.8 #Enthalpy in kJ/kg from table
h3 = 2799.5 #Enthalpy in kJ/kg
s4 = 6.3409 #kJ/kg
sf = 0.6493 #Entropy at liquid state in kJ/kg
sfg = 7.5009 #Entropy difference for vapor and liquid state in kJ/kg
hfg = 2392.8 #Enthalpy difference in kJ/kg for turbine
#Calculations:
wp = v*(P2-P1) #work done in pipe in kJ/kg
h2 = h1+wp #enthalpy in kJ/kg
x4 = (s4-sf)/sfg #x-factor
h4 = h1+x4*hfg #Enthalpy in kJ/kg
nth = ((h3-h2)-(h4-h1))/(h3-h2)
#Results:
print 'Percentage efficiency: ',round(nth*100,1),"%"
# -*- coding: utf8 -*-
from __future__ import division
#Example: 11.2
''' In a Rankine cycle, steam leaves the boiler and enters the turbine at 4 MPa and 400
◦
C.
The condenser pressure is 10 kPa. Determine the cycle efficiency.
To determine the cycle efficiency, we must calculate the turbine work, the pump
work, and the heat transfer to the steam in the boiler. We do this by considering a control
surface around each of these components in turn. In each case the thermodynamic model
is the steam tables, and the process is steady state with negligible changes in kinetic and
potential energies.
Control volume:
Inlet state:
Exit state:
Pump.
P 1 known, saturated liquid; state fixed.
P 2 known.'''
#Variable Declaration:
P2 = 4000 #Exit pressure in kPa
P1 = 10 #Inlet pressure in kPa
v = 0.00101 #specific weight of water in m**3/kg
h1 = 191.8 #Enthalpy in kJ/kg from table
h3 = 3213.6 #Enthalpy in kJ/kg from table
s4 = 6.7690 #Entropy in kJ/kg from table
sf = 0.6493 #Entropy at liquid state in kJ/kg from table
sfg = 7.5009 #Entropy difference for vapor and liquid state in kJ/kg from table
hfg = 2392.8 #Enthalpy difference in kJ/kg for turbine
#Calculations:
wp = v*(P2-P1) #work done in pipe in kJ/kg
h2 = h1+wp #Enthalpy in kJ/kg
x4 = (s4-sf)/sfg #x-factor
h4 = h1+x4*hfg #Enthalpy in kJ/kg
nth = ((h3-h2)-(h4-h1))/(h3-h2)
#Results:
print 'Percentage efficiency: ',round(nth*100,1),'%'
# -*- coding: utf8 -*-
from __future__ import division
#Example: 11.3
''' Consider a reheat cycle utilizing steam. Steam leaves the boiler and enters the turbine at
4 MPa, 400
◦
C. After expansion in the turbine to 400 kPa, the steam is reheated to 400
◦
C
and then expanded in the low-pressure turbine to 10 kPa. Determine the cycle efficiency.
For each control volume analyzed, the thermodynamic model is the steam tables,
the process is steady state, and changes in kinetic and potential energies are negligible.
For the high-pressure turbine,
Control volume:
Inlet state:
Exit state:
High-pressure turbine.
P 3 , T 3 known; state fixed.
P 4 known.'''
#Variable Declaration:
P2 = 4000 #Exit pressure in kPa
P1 = 10 #Inlet pressure in kPa
v = 0.00101 #specific weight of water in m**3/kg
h1 = 191.8 #Enthalpy in kJ/kg from table
h3 = 3213.6 #Enthalpy in kJ/kg from table
s4 = 6.7690 #Entropy in kJ/kg from table
sf1 = 1.7766 #Entropy at liquid state in kJ/kg from table
sfg1 = 5.1193 #Entropy difference for vapor and liquid state in kJ/kg from table
hf = 604.7 #Enthalpy of liquid state in kJ/kg
hfg1 = 2133.8 #Enthalpy difference in kJ/kg for turbine
sf = 0.6493 #Entropy in liquid state in kJ/kg for turbine
h5 = 3273.4 #Enthalpy in kJ/kg
s5 = 7.8985 #Entropy in kJ/kg
sfg = 7.5009 #Entropy diff in kJ/kg
hfg = 2392.8 #Enthalpy difference for low pressure turbine in kj/kg
#Calculations:
wp = round(v*(P2-P1))#Work done in pipe in kJ/kg
h2 = h1+wp #Enthalpy in kJ/kg
x4 = (s4-sf1)/sfg1 #x-factor
h4 = hf+x4*hfg1 #Enthalpy in kJ/kg
x6 = (s5-sf)/sfg #x-factor
h6 = h1+x6*hfg #Entropy in kg/kg
wt = (h3-h4)+(h5-h6)#Work output in kJ/kg
qh = (h3-h2)+(h5-h4)
nth = (wt-wp)/qh
#Results:
print 'Percentage efficiency: ',round(nth*100,1),'%'
# -*- coding: utf8 -*-
from __future__ import division
#Example: 11.4
''' Consider a regenerative cycle using steam as the working fluid. Steam leaves the boiler
and enters the turbine at 4 MPa, 400
◦
C. After expansion to 400 kPa, some of the steam is
extracted from the turbine to heat the feedwater in an open feedwater heater. The pressure
in the feedwater heater is 400 kPa, and the water leaving it is saturated liquid at 400 kPa.
The steam not extracted expands to 10 kPa. Determine the cycle efficiency.
The line diagram and T–s diagram for this cycle are shown in Fig. 11.12.
As in previous examples, the model for each control volume is the steam tables, the
process is steady state, and kinetic and potential energy changes are negligible.
From Examples 11.2 and 11.3 we have the following properties:
h 5 = 3213.6 h 6 = 2685.6
h 7 = 2144.1 h 1 = 191.8
For the low-pressure pump,
Control volume:
Inlet state:
Exit state:
Low-pressure pump.
P 1 known, saturated liquid; state fixed.
P 2 known.'''
#Variable Declaration:
h1 = 191.8 #kJ/kg
h5 = 3213.6 #kg/kg
h6 = 2685.7 #kJ/kg
h7 = 2144.1 #kJ/kg
h3 = 604.7 #kJ/kg
P2 = 400 #Exit pressure in kPa
P1 = 10 #Inlet pressure in kPa
v = 0.00101 #Specific weight of water in m**3/kg
h1 = 191.8 #Enthalpy in kJ/kg from table
P3 = 400 #KPa
P4 = 4000 #KPa
v = 0.001084 #Specific heat for 3-4 process in m**3/kg
#Calculations:
wp1 = v*(P2-P1) #work done for low pressure pump in kJ/kg
h2 = h1+wp1 #Enthalpy in kJ/kg
y = (h3-h2)/(h6-h2) #Extraction fraction
wt = (h5-h6)+(1-y)*(h6-h7)#Turbine work in kJ/kg
wp2 = v*(P4-P3) #Work done for high pressure pump
h4 = h3+wp2 #Enthalpy in kJ/kg
wnet = wt-(1-y)*wp1-wp2
qh = h5-h4 #Heat output in kJ/kg
nth = wnet/qh
#Results:
print 'Refrigerator Efficiency: ',round(nth*100,1),'%'
# -*- coding: utf8 -*-
from __future__ import division
#Example: 11.5
''' A steam power plant operates on a cycle with pressures and temperatures as designated in
Fig. 11.17. The efficiency of the turbine is 86%, and the efficiency of the pump is 80%.
Determine the thermal efficiency of this cycle.'''
#Variable Declaration:
h5 = 3169.1 #kJ/kg
s5 = 6.7235 #kJ/kg
sf = 0.6493 #Entropy for liquid state in kJ/kg
sfg = 7.5009 #Entropy difference in kJ/kg
hf = 191.8 #kJ/kg
hfg = 2392.8 #Enthalpy difference in kJ/kg
nt = 0.86 #turbine efficiency given
np = 0.80 #pump efficiency given
v = 0.001009 #specific heat in m**3/kg
P2 = 5000 #kPa
P1 = 10 #kPa
h3 = 171.8 #in kJ/kg from table
h4 = 3213.6 #kJ/kg from table
#Calculations:
s6s = s5
x6s = (s6s-sf)/sfg #x-factor
h6s = hf+x6s*hfg #kJ/Kg at state 6s
wt = nt*(h5-h6s)
wp = v*(P2-P1)/np #Work done in pump in kJ/kg
wnet = wt-wp #net work in kJ/kg
qh = h4-h3
nth = wnet/qh
#Results:
print 'Cycle Efficiency: ',round(nth*100,1),'%'
# -*- coding: utf8 -*-
from __future__ import division
#Example: 11.6
''' Consider an ideal refrigeration cycle that uses R-134a as the working fluid. The temperature
of the refrigerant in the evaporator is −20
◦
C, and in the condenser it is 40
◦
C. The refrigerant
is circulated at the rate of 0.03 kg/s. Determine the COP and the capacity of the plant in
rate of refrigeration.
The diagram for this example is shown in Fig. 11.21. For each control volume
analyzed, the thermodynamic model is as exhibited in the R-134a tables. Each process is
steady state, with no changes in kinetic or potential energy.
Control volume:
Inlet state:
Exit state:
Compressor.
T 1 known, saturated vapor; state fixed.
P 2 known (saturation pressure at T 3 ).
'''
#Variable Declaration:
P2 = 1017 #KJ/Kg
h1 = 386.1 #KJ/Kg
s1 = 1.7395 #KJ/Kg
T2 = 47.7 #°C
h2 = 428.4 #KJ/Kg
h3 = 256.5 #KJ/Kg
#Calculations:
s2 = s1
wc = h2-h1 #Work done by compressor in KJ/Kg
h4 = h3
qL = h1-h4 #From Energy equation in KJ/Kg
B = round(qL/wc)
RC = qL*B/100 #Refrigeration capacity in KW
#Results:
print 'Refrigeration Capacity: ', round(RC,2), 'KW'
# -*- coding: utf8 -*-
from __future__ import division
#Example: 11.7
'''A refrigeration cycle utilizes R-134a as the working fluid. The following are the properties
at various points of the cycle designated in Fig. 11.22:
P 1 = 125 kPa,
P 2 = 1.2 MPa,
P 3 = 1.19 MPa,
P 4 = 1.16 MPa,
P 5 = 1.15 MPa,
P 6 = P 7 = 140 kPa,
P 8 = 130 kPa,
T 1 = −10 ◦C
T 2 = 100◦C
T 3 = 80◦C
T 4 = 45◦C
T 5 = 40◦C
x 6 = x 7
T 8 = −20◦C
The heat transfer from R-134a during the compression process is 4 kJ/kg. Determine
the COP of this cycle.
For each control volume, the R-134a tables are the model. Each process is steady
state, with no changes in kinetic or potential energy.
As before, we break the process down into stages, treating the compressor, the
throttling value and line, and the evaporator in turn.
Control volume:
Inlet state:
Exit state:
Compressor.
P 1 , T 1 known; state fixed.
P 2 , T 2 known; state fixed.
'''
#Variable Declaration:
h1 = 394.9 #KJ/Kg
h2 = 480.9 #KJ/Kg
q = -4 #KJ/Kg
h5 = 256.4 #KJ/Kg
h8 = 386.6 #KJ/Kg
#Calculations:
wc = h2-h1-q #Work done by compressor in KJ/Kg
h6 = h5
h7 = h6
qL = h8-h7
B = qL/wc #Coefficient of performance
#Results:
print 'COP of refrigerator : ', round(B,2)