Chapter 9 Law Of Thermodynamics

Example 1 Page No:165

In [1]:
#Input data
Qab=720   #Heat transfer of 1st processes in KJ 
Qbc=-80   #Heat transfer of 2nd processes in KJ
Qcd=40    #Heat transfer of 3rd processes in KJ
Qda=-640  #Heat transfer of 4th processes in KJ
Wab=-90   #Work transfer of 1st processes in KJ
Wbc=-50   #Work transfer of 2nd processes in KJ
Wcd=130   #Work transfer of 3rd processes in KJ


#Calculation
#From the 1st law of thermodynamic for close system undergoing a cycle.

#Work interaction during the 4th processes 
Wda=((Qab+Qbc+Qcd+Qda)-(Wab+Wbc+Wcd)) 

#Output
print("Work interaction during the 4th processes=",Wda,"KJ")

Work interaction during the 4th processes= 50 KJ

Example 2 Page No:166

In [2]:
#Input data
                      #During compression
W1=-9200              #Stroke work done by the piston in Nm
Nm1=-9.2              #Nm of work done
Q1=-50                #Heat rejected during copression in KJ
                      #During expansion
W2=8400               #Stroke work done by the piston in Nm
Nm2=8.4               #Nm of work done

#Calculation
                      #Quantity of heat transferred
Q2=-((Nm1+Nm2)+Q1)    #-sign for indicate heat is transferred


#Output
print("Quantity of heat transferred=",Q2,"KJ")

Quantity of heat transferred= 50.8 KJ

Example 3 Page No:166

In [3]:
#input data
W1=-20                  #Work interaction to the fluid in KJ
W2=42                   #Work interaction from the fluid in KJ
Q1=85                   #Heat interaction to the fluid in KJ
Q2=85                   #Heat interaction to the fluid in KJ
Q3=-50                  #Heat interaction from the fluid in KJ

#calculation
W3=((Q1+Q2+Q3)-(W1+W2)) #Magnitude and direction of the third heat interation


#output
print("Magnitude and direction of the third heat interation=",W3,"KJ")

Magnitude and direction of the third heat interation= 98 KJ

Example 4 Page No:168

In [4]:
#Input data
Q=-2100         #Non flow process losses heat in KJ
deltaU=420      #Gain heat

#Calculation
W=Q-deltaU      #Work done and compression process in KJ

#Output
print("Work done and compression process=",W,"KJ")

Work done and compression process= -2520 KJ

Example 5 Page No:168

In [5]:
#Input data
W=-2000      #Work input of panddle wheel in KJ
Q=-6000      #Heat transferred to the surrounding from tank

#Calculation
deltaU=Q-W   #Change in interval energy

#Output
print("change in interval energy drop=",deltaU,"KJ")

change in interval energy drop= -4000 KJ

Example 6 Page No:169

In [6]:
#Input data
U1=520       #internal energy in KJ/Kg
U2=350       #internal energy in KJ/Kg
W=-80        #work done by the air in the cylinder KJ/kg

#Calculation
deltaU=U2-U1
Q=deltaU+W   #Heat transferred during the process

#Output
print("Heat transferred during the process=",Q,"KJ")
    

Heat transferred during the process= -250 KJ

Example 7 Page No:169

In [7]:
#Input data
W1=800              #Power of turbine shaft Kw
W2=-5               #Work pump to feed in Kw  
Q1=2700             #Heat for steam generation KJ/Kg
Q2=-1800            #Condenser rejected heat KJ/Kg

#Calculation
m=((W1+W2)/(Q1+Q2)) #Steam flow rate in Kg/h


#Output
print("Steam flow rate=",round(m,4),"Kg/s")

Steam flow rate= 0.8833 Kg/s

Example 8 Page No:170

In [8]:
#input data
#Data consistent with first law pf thermodynamics
Qabcd=-22                    #In KJ
N=150                        #In Cycles/min
Qab=17580                    #In KJ/min
Qbc=0 
Qcd=-3660                    #In KJ/min
Wab=-8160                    #In KJ/min
Wbc=4170                    #In KJ/min 
DeltaUcd=-21630              #In KJ/min


#calculation
DeltaUab=Qab-Wab              #In KJ/min
DeltaUbc=Qbc-Wbc              #In KJ/min 
Wcd=Qcd-DeltaUcd              #In KJ/min
Qabcd1=-220*150               #In KJ/min
Qda=((Qabcd1)-(Qab+Qbc+Qcd))  #In KJ/min
Wda=((Qabcd1)-(Wab+Wbc+Wcd))     #In KJ/min
DeltaUabcd=0
DeltaUda=((DeltaUabcd)-(DeltaUab+DeltaUbc+DeltaUcd)) #In KJ/min
NWO=Qabcd1/60                   #In KW


#output
print("DeltaUab= ",DeltaUab,"KJ/min")
print("DeltaUbc= ",DeltaUbc,"KJ/min")
print("Wcd= ",Wcd,"KJ/min")
print("Qabcd1= ",Qabcd1,"KJ/min")
print("Qda= ",Qda,"KJ/min")
print("Wda= ",Wda,"KJ/min")
print("DeltaUabcd= ",DeltaUabcd,"KJ/min")
print("DeltaUda= ",DeltaUda,"KJ/min")
print("NWO=",NWO,"Kw")

DeltaUab=  25740 KJ/min
DeltaUbc=  -4170 KJ/min
Wcd=  17970 KJ/min
Qabcd1=  -33000 KJ/min
Qda=  -46920 KJ/min
Wda=  -46980 KJ/min
DeltaUabcd=  0 KJ/min
DeltaUda=  60 KJ/min
NWO= -550.0 Kw

Example 9 Page No:171

In [9]:
#Input data
Qab=-6500  #Heat transferred in 1st process KJ/min
Qbc=0      #Heat transferred in 2nd process 
Qcd=-10200 #Heat transferred in 3rd process KJ/min
Qda=32600  #Heat transferred in 4th process KJ/min
Wab=-1050  #Heat transferred in 1st process KJ
Wbc=-3450  #Heat transferred in 2nd process KJ
Wcd=20400  #Heat transferred in 3rd process KJ
Wda=0      #Heat transferred in 4th process

#Calculator
dQ=Qab+Qbc+Qcd+Qda #Net heat transfer in 1st cycle
dW=Wab+Wbc+Wcd+Wda #Net work done in 1st cycle
dW1=dW/60          #Net work done in 1st cycle
DeltaUab=Qab-Wab   #ab process
DeltaUbc=Qbc-Wbc   #bc processes
DeltaUcd=Qcd-Wcd   #cd processes
DeltaUda=Qda-Wda   #dc processes

#Output
print("Net heat transfer in 1st cycle= ",dQ,"KJ/min")
print("Net work done in 1st cycle= ",dW,"KJ/min")
print("Net work done in 1st cycle= ",dW1,"KW")
print("ab process= ",DeltaUab,"KJ/min")
print("bc processes= ",DeltaUbc,"KJ/min")
print("cd processes= ",DeltaUcd,"KJ/min")
print("dc processes= ",DeltaUda,"KJ/min")

Net heat transfer in 1st cycle=  15900 KJ/min
Net work done in 1st cycle=  15900 KJ/min
Net work done in 1st cycle=  265.0 KW
ab process=  -5450 KJ/min
bc processes=  3450 KJ/min
cd processes=  -30600 KJ/min
dc processes=  32600 KJ/min