For any powerplant system, exergy and energy analysis play a vital to determine the efficiency of the cycle and various important parameters associated with the cycle. In this article, overall and component-wise exergy and energy analysis of a combined cycle powerplant will be discussed. The combined cycle is comprised of two different cycles: the upper cycle is a Brayton cycle and the bottoming cycle is a steam Rankine cycle. The analysis will be performed in the python programming language and the thermodynamic properties will be calculated via the CoolProp package.
Cycle Layout:
Ref: Fundamentals of Engineering Thermodynamics by Michael J. Moran , Howard N. Shapiro , Daisie D. Boettner , Margaret B. Bailey
This article is intended to implement this problem in python. Detailed explanation of thermodynamic principles of the cycle can be found in any standard thermodynamics book.
First,lets review some basic thermodynamic formula which will be used in this simulation.
General Formula:
Exergy at any point
E= m (( h-h0 )-T0 ( s-s0 ))
1st law of thermodynamics for a steady-state flow
0 = Q-W+mhin-mhout
2nd law of thermodynamics for a steady-state flow
0=Φ-W+mΨin-mΨout-I
Component Wise Exergy and Energy Analysis - Python Implementation
Brayton Cycle
Air Compressor:
First import the necessary package:
from CoolProp.CoolProp import PropsSI import numpy as np import matplotlib.pyplot as plt T0=300 #environment temperature m_a=1 #mass flow rate of air
The exergy and energy analysis of the compressor
# Air Compressor n_c=0.84 #isentropic efficiency Pin_c=100000 # Inlet pressure Tin_c=300 #inlet temperature Pout_c=1200000 #outlet pressure Q_c=0 #heat gain in compressor Fi_c=0 # Exergy input by heat in compressor hin_c=PropsSI('H','P',Pin_c,'T',Tin_c,'Air') sin_c=PropsSI('S','P',Pin_c,'T',Tin_c,'Air') sout_cs=sin_c Pout_cs=Pout_c hout_cs=PropsSI('H','P',Pout_cs,'S',sout_cs,'Air') hout_c=hin_c+(hout_cs-hin_c)/n_c sout_c=PropsSI('S','P',Pout_c,'H',hout_c,'Air') ########################################################### #1st law W_c=Q_c+m_a*(hin_c-hout_c) #2nd law I_c=Fi_c-W_c+m_a*(hin_c-hout_c-T0*(sin_c-sout_c)) n_2_c= (W_c+I_c)/W_c #printig value: print('compressor input',W_c,'\n') print('irreversibility',I_c,'\n') print('Exergy efficiency',n_2_c)
compressor input -370361.9987783382 irreversibility 28153.0650389996 Exergy efficiency 0.9239850062051068
Notice that, here heat generation or exergy associated with it is considered zero. But in practical application, there is a little heat generation or exergy produced.
Combustor
#Combustor hin_com=hout_c Pin_com=Pout_c Tout_com=1400 Pout_com=Pin_com W_com=0 I_com=0 hin_com=PropsSI('H','P',Pin_com,'H',hin_com,'Air') sin_com=PropsSI('S','P',Pin_com,'H',hin_com,'Air') hout_com=PropsSI('H','P',Pout_c,'T',Tout_com,'Air') sout_com=PropsSI('S','P',Pout_c,'T',Tout_com,'Air') ######################################## #1st law Q_com=W_com-m_a*(hin_com-hout_com) #2nd law Fi_com=W_com-m_a*(hin_com-hout_com-T0*(sin_com-sout_com)) n_2_com=(Fi_com-I_com)/Fi_com #printing value print('heat input',Q_com) print('Exergy input',Fi_com) print("Exergy efficiency",n_2_com)
heat input 846037.5661962205 Exergy input 589974.4347588865 Exergy efficiency 1.0
Notice that, heat input is greater than the exergy input associated with heat. As we have assumed no exergy destruction in the combustor, the exergy efficiency is found 100% which in real case is less than that because of some internal exergy destruction associated with it.
Air turbine
# Air Turbine n_t_a=0.88 #efficiency Pin_t_a=Pout_com # Inlet pressure Tin_t_a=Tout_com #inlet temperature Pout_t_a=100000 #outlet pressure Q_t_a=0 #heat gain in turbine Fi_t_a=0 # Exergy input by heat in turbine hin_t_a=PropsSI('H','P',Pin_t_a,'T',Tin_t_a,'Air') sin_t_a=PropsSI('S','P',Pin_t_a,'T',Tin_t_a,'Air') sout_ts_a=sin_t_a Pout_ts_a=Pout_t_a hout_ts_a=PropsSI('H','P',Pout_ts_a,'S',sout_ts_a,'Air') hout_t_a=hin_t_a+(hout_ts_a-hin_t_a)*n_t_a sout_t_a=PropsSI('S','P',Pout_t_a,'H',hout_t_a,'Air') ###################################################################### #1st law W_t_a=Q_t_a+m_a*(hin_t_a-hout_t_a) #2nd law I_t_a=Fi_t_a-W_t_a+m_a*(hin_t_a-hout_t_a-T0*(sin_t_a-sout_t_a)) n_2_t_a= W_t_a/(W_t_a+I_t_a) #printig value: print('turbine outputput',W_t_a) print('irreversibility',I_t_a) print('Exergy efficiency',n_2_t_a) hin_com=PropsSI('H','P',Pin_com,'H',hin_com,'Air') sin_com=PropsSI('S','P',Pin_com,'H',hin_com,'Air') hout_com=PropsSI('H','P',Pout_c,'T',Tout_com,'Air') sout_com=PropsSI('S','P',Pout_c,'T',Tout_com,'Air') ######################################## #1st law Q_com=W_com-m_a*(hin_com-hout_com) #2nd law Fi_com=W_com-m_a*(hin_com-hout_com-T0*(sin_com-sout_com)) n_2_com=(Fi_com-I_com)/Fi_com #printing value print('heat input',Q_com) print('Exergy input',Fi_com) print("Exergy efficiency",n_2_com)
turbine outputput 658286.9448573011 irreversibility 34038.984400560264 Exergy efficiency 0.9508338732350406
Notice that, like compressor, no heat gain or rejection is considered here. Also, the exergy efficiency is greater than the isentropic efficiency.
Rankine Cycle
Pump and Heat Recovery Steam generator
To calculate the work and heat in the bottom steam Rankine cycle, we need to know the mass flow rate of water which depends on the heat gain in heat recovery steam generator (HRSG). Again, to calculate the state point of HRSG, we need to calculate the state point of the outlet of the pump first. So,we will calculate them in a specific order.
It is to be noted that pump inlet state is considered to be saturated water, which is a common practice in thermodynamic analysis of power plant.
#Pump and HRSG #Pump n_p=0.80 #efficiency Pin_p=8000 # Inlet pressure Qin_p=0 #saturated water Pout_p=8000000 Q_p=0 #heat gain in compressor Fi_p=0 # Exergy input by heat in compressor hin_p=PropsSI('H','P',Pin_p,'Q',Qin_p,'Water') sin_p=PropsSI('S','P',Pin_p,'Q',Qin_p,'Water') sout_ps=sin_p Pout_ps=Pout_p hout_ps=PropsSI('H','P',Pout_ps,'S',sout_ps,'Water') hout_p=hin_p+(hout_ps-hin_p)/n_p sout_p=PropsSI('S','P',Pout_p,'H',hout_p,'Water') #HRSG Q_hrsg=0 W_hrsg=0 Fi_hrsg=0 Tout_hrsg_air=400 Pout_hrsg_air=Pout_t_a Tout_hrsg_water=400+273 Pout_hrsg_water=Pout_p hin_hrsg_air=hout_t_a sin_hrsg_air=sout_t_a hout_hrsg_air=PropsSI('H','P',Pout_hrsg_air,'T',Tout_hrsg_air,'Air') sout_hrsg_air=PropsSI('S','P',Pout_hrsg_air,'T',Tout_hrsg_air,'Air') hin_hrsg_water=hout_p sin_hrsg_water=sout_p hout_hrsg_water=PropsSI('H','P',Pout_hrsg_water,'T',Tout_hrsg_water,'Water') sout_hrsg_water=PropsSI('S','P',Pout_hrsg_water,'T',Tout_hrsg_water,'Water') ##################################################################################### #1st law m_w=m_a*(hin_hrsg_air-hout_hrsg_air)/(hout_hrsg_water-hin_hrsg_water) print("mass flow rate of water",m_w) #2nd law I_hrsg=Fi_hrsg-W_hrsg+m_a*(hin_hrsg_air-hout_hrsg_air-T0*(sin_hrsg_air-sout_hrsg_air))+m_w*(hin_hrsg_water-hout_hrsg_water-T0*(sin_hrsg_water-sout_hrsg_water)) flow_exergy_hrsg=m_a*(hin_hrsg_air-hout_hrsg_air-T0*(sin_hrsg_air-sout_hrsg_air)) n_2_hrsg=(flow_exergy_hrsg-I_hrsg)/flow_exergy_hrsg print('Irreversibility in HRSG',I_hrsg) print('Exergy efficiency of hrsg',n_2_hrsg) #pump #1st law W_p=Q_p+m_w*(hin_p-hout_p) print('pump work input', W_p) #2nd law I_p=Fi_p-W_p+m_w*(hin_p-hout_p-T0*(sin_p-sout_p)) n_2_p=(W_p+I_p)/W_p print('irreversibility of pump',I_p) print('Exergy efficiency of pump',n_2_p)
mass flow rate of water 0.15470405199984053 Irreversibility in HRSG 36478.14657086009 Exergy efficiency of hrsg 0.8386032865304848 pump work input -1555.9450282783123 irreversibility of pump 296.2364333429248 Exergy efficiency of pump 0.8096099618180489
Turbine
# Water Turbine n_t_w=0.90 #efficiency Pin_t_w=Pout_hrsg_water # Inlet pressure Tin_t_w=Tout_hrsg_water #inlet temperature Pout_t_w=Pin_p #outlet pressure Q_t_w=0 #heat gain in turbine Fi_t_w=0 # Exergy input by heat in turbine hin_t_w=PropsSI('H','P',Pin_t_w,'T',Tin_t_w,'Water') sin_t_w=PropsSI('S','P',Pin_t_w,'T',Tin_t_w,'Water') sout_ts_w=sin_t_w Pout_ts_w=Pout_t_w hout_ts_w=PropsSI('H','P',Pout_ts_w,'S',sout_ts_w,'Water') hout_t_w=hin_t_w+(hout_ts_w-hin_t_w)*n_t_w sout_t_w=PropsSI('S','P',Pout_t_w,'H',hout_t_w,'Water') ###################################################################### #1st law W_t_w=Q_t_w+m_w*(hin_t_w-hout_t_w) #2nd law I_t_w=Fi_t_w-W_t_w+m_w*(hin_t_w-hout_t_w-T0*(sin_t_w-sout_t_w)) n_2_t_w=W_t_w/(W_t_w+I_t_w) #printig value: print('turbine outputput',W_t_w) print('irreversibility',I_t_w) print('Exergy efficiency',n_2_t_w)
turbine outputput 159935.29122534447 irreversibility 16942.72322606985 Exergy efficiency 0.9042123845712676
Condenser:
#Condenser hin_con=hout_t_w Pin_con=Pout_t_w Qout_con=0 Pout_con=Pin_p W_con=0 I_con=0 hin_con=PropsSI('H','P',Pin_con,'H',hin_con,'Water') sin_con=PropsSI('S','P',Pin_con,'H',hin_con,'Water') hout_con=PropsSI('H','P',Pout_con,'Q',Qout_con,'Water') sout_con=PropsSI('S','P',Pout_con,'Q',Qout_con,'Water') ######################################## #1st law Q_con=W_con-m_w*(hin_con-hout_con) #2nd law Fi_con=W_con-m_w*(hin_con-hout_con-T0*(sin_con-sout_con)) flow_exergy_con=m_w*(hin_con-hout_con-T0*(sin_con-sout_con)) n_2_con=(flow_exergy_con-I_con)/flow_exergy_con #printing value print('heat output',Q_con) print('Exergy output',Fi_con) print('Exergy efficiency',n_2_con)
heat output -298778.6596625886 Exergy output -13918.97122470541 Exergy efficiency 1.0
Note that, here unlike the HRSG, exergy efficiency is assumed to be zero, as there is no coolant source given. In practice application, condenser produces the biggest exergy destruction and should be considered.
Energy and Exergy Balance
Energy Balance
If we consider the total combined cycle as a closed volume system and apply 1st law of thermodynamics,
#energy balance of the cycle Q_net=Q_com+Q_con W_net=W_t_a+W_t_w+W_c+W_p Q_net-W_net+m_a*(hin_c-hout_hrsg_air)
-5.820766091346741e-11
Which is almost 0. So, the energy of the cycle is balanced.
Exergy balance:
#Exergy balance of the cycle Fi_net=Fi_com+Fi_con I_total=I_c+I_com+I_t_a+I_p+I_hrsg+I_t_w+I_con Fi_net-W_net+m_a*((hin_c-hout_hrsg_air)-T0*(sin_c-sout_hrsg_air))-I_total
-8.731149137020111e-11
So, the exergy of the cycle is also balanced.
Overall efficiency
#overall efficiency n_1=W_net/Q_com n_2=W_net/Fi_com print('Thermal/1st law efficiency',n_1) print('Exergy/2nd law efficiency',n_2)
Thermal/1st law efficiency 0.527523020381483 Exergy/2nd law efficiency 0.7564807320141369
Componentwise exergy destruction
#Componentwise exergy destruction objects = ('Compressor', 'Combustor', 'Air turbine', 'Pump', 'HRSG', 'Steam turbine','Condenser') y_pos = np.arange(len(objects)) performance = [I_c,I_com,I_t_a,I_p,I_hrsg,I_t_w,I_con] plt.bar(y_pos, performance, align='center') plt.xticks(y_pos, objects) plt.ylabel('Exergy Destruction(W)') plt.title('Exergy Destruction') fig=plt.gcf() fig.set_size_inches(8,6) plt.show()
Componentwise exergy efficiency
# Componentwise exergy efficiency objects = ('Compressor', 'Combustor', 'Air turbine', 'Pump', 'HRSG', 'Steam turbine','Condenser','Overall') y_pos = np.arange(len(objects)) performance = [n_2_c,n_2_com,n_2_t_a,n_2_p,n_2_hrsg,n_2_t_w,n_2_con] plt.bar(y_pos, performance, align='center') plt.xticks(y_pos, objects) plt.ylabel('Exergy Efficiency') plt.title('Componentwise Exergy Efficiency') fig=plt.gcf() fig.set_size_inches(8,6) plt.show()
Heat and work input-output
# Heat and work input output objects = ('Compressor', 'Combustor', 'Air turbine', 'Pump', 'Steam turbine','Condenser') y_pos = np.arange(len(objects)) performance = [-W_c,Q_com,W_t_a,-W_p,W_t_w,-Q_con] plt.bar(y_pos, performance, align='center') plt.xticks(y_pos, objects) plt.ylabel('Energy(W)') plt.title('Heat and work input?output') fig=plt.gcf() fig.set_size_inches(8,6) plt.show() plt.bar(y_pos, performance, align='center') plt.xticks(y_pos, objects) plt.ylabel('Exergy Efficiency') plt.title('Componentwise Exergy Efficiency') fig=plt.gcf() fig.set_size_inches(8,6) plt.show()
Overview of the cycle parameter
The full code is available here.
If you want to know the basics of Coolprop, go to this link
thank you so much
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