# Chapter 6 : Thermodynamic potentials¶

### Example 6.6 Page No : 218¶

In [1]:
# Variables
m = 0.1;			 #mass of superheated steam in the piston cylinder assembly in kg
P1 = 1.	    		 #initial pressure of superheated steam in MPa
T1 = 300.			 #initial temperature of superheated steam in degree celsius
P2 = 0.1			 #pressure of steam after expansion in MPa
T2 = 200.			 #temperature of steam after expansion in degree celsius

# Calculations
#For steam at P1 and T1:
h1 = 3052.1
v1 = 0.2580

#For steam at P2 and T2:
h2 = 2875.4
v2 = 2.1720

del_u = (((h1*10**3)-(P1*10**6*v1))-((h2*10**3)-(P2*10**6*v2)))*10**-3
W = m*(del_u)

# Results
print " The work done by steam = %0.2f kJ"%(W);

 The work done by steam = 13.59 kJ


### Example 6.8 Page No : 220¶

In [2]:
# Variables
P = 3.	    		 #pressure of superheated steam in MPa
Ti = 300.			 #temperature at which the steam enters the turbine in degree celsius
m = 1.		    	 #mass flow rate of steam in kg/s
Te = 60.			 #temperature of dry saturated steam when it leaves the turbine in degree celsius

# Calculations
#For steam at P and Ti:
h1 = 2995.1			 #specific entahlpy of steam in kJ/kg

#For saturated steam at Te:
h2 = 2609.7			 #specific enthalpy of saturated vapour in kJ/kg

Ws = m*(h1-h2)		 # Calculations of the power  Results of the turbine using Eq.(6.83) in kW

# Results
print " The power  Results of the turbine = %0.1f kW"%(Ws);

 The power  Results of the turbine = 385.4 kW


### Example 6.10 Page No : 223¶

In [3]:
# Variables
m = 0.1 			 #mass of superheated steam in the piston cylinder assembly in kg
P1 = 3.	    		 #initial pressure of superheated steam in MPa
T1 = 300.			 #initial temperature of superheated steam in degree celsius
T0 = 300.			 #temperature of the reservoir which is placed in thermal contact with the piston-cylinder assembly in degree celsius
P2 = 0.1			 #pressure of steam after expansion in MPa

#For steam at P1 and T1:
h1 = 2995.1
v1 = 0.08116
s1 = 6.5422

#For steam at P2 and T2:
h2 = 3074.5
v2 = 2.6390
s2 = 8.2166

# Calculations
T0 = T0+273.15

W = m*(h1-h2-(((P1*v1)-(P2*v2))*10**3)-(T0*(s1-s2)))

# Results
print " The maximum work obtained from steam = %0.2f kJ"%(W);

 The maximum work obtained from steam = 90.07 kJ


### Example 6.12 Page No : 226¶

In [4]:
import math

# Variables
P1 = 0.1			 #pressure at which air enters the compressor in MPa
T1 = 300.			 #temperature at which air enters the compressor in K
P2 = 1. 			 #pressure at which air leaves the compressor in MPa
T2 = 300.			 #temperature at which air leaves the compressor in K
T0 = 300.			 #ambient temperature in K
N = 1.  			 #molar flow rate of air in mol/s
gaamma = 1.4		 #ratio of specific heat capacities (no unit)
R = 8.314			 #universal gas constant in J/molK

# Calculations

Ws = (-N*T0*(-R*math.log (P2/P1)))*10**-3

# Results
print " The minimum power required to compress one mole per second of air = %0.3f kW"%(Ws);

 The minimum power required to compress one mole per second of air = 5.743 kW