Chapter 7 : Thermodynamic Relations

Example 7.17 Page no : 370

In [1]:
# Variables
B = 5.*10**(-5); 			# /K
K = 8.6*10**(-12); 			# m**2/N
v = 0.114*10**(-3); 			#m**3/kg
p2 = 800.*10**5; 			#Pa
p1 = 20.*10**5; 			#Pa
T = 288.; 			#K

# Calculations and Results
W = -v*K/2*(p2**2-p1**2);
print ("(i) Work done on the copper  =  %.3f")%(W),("J/kg")

ds = -v*B*(p2-p1);
print ("(ii) Change in entropy  = %.3f")% (ds), ("J/kg K")

Q = T*ds;
print ("(iii) The heat transfer  = %.3f")%(Q), ("J/kg")

du = Q-W;
print ("(iv) Change in internal energy  = %.3f")%(du),("J/kg")

R = B**2*T*v/K;
print ("(v) cp – cv  = %.3f")%(R),("J/kg K")

(i) Work done on the copper  =  -3.135 J/kg
(ii) Change in entropy  = -0.445 J/kg K
(iii) The heat transfer  = -128.045 J/kg
(iv) Change in internal energy  = -124.909 J/kg
(v) cp – cv  = 9.544 J/kg K

Example 7.18 Page no : 371

In [1]:
# Variables
vg = 0.1274; 			#m**3/kg
vf = 0.001157; 			#m**3/kg
# dp/dT = 32; 			#kPa/K
T3 = 473; 			#K

# Calculations
h_fg = 32*10**3*T3*(vg-vf)/10**3;

# Results
print (" enthalpy of vapourisation = %.3f")%(h_fg),("kJ/kg")

 enthalpy of vapourisation = 1910.814 kJ/kg

Example 7.19 Page no : 372

In [2]:
import math

# Variables
h_fg = 334.; 			#kJ/kg
v_liq = 1.; 			#m**3/kg
v_ice = 1.01; 			#m**3/kg
T1 = 273.; 			#K
T2 = 263.; 			#K
p1 = 1.013*10**5; 			#Pa

# Calculations
p2 = (p1+h_fg*10**3/(v_ice-v_liq)*math.log(T1/T2))/10**5;

# Results
print ("pressure = %.3f")%(p2),("bar")

pressure = 13.477 bar

Example 7.20 Page no : 372

In [3]:
# Variables
h_fg = 294.54; 			#kJ/kg
p = 0.1; 			#bar
T = 523; 			#K

# Calculations
vg = h_fg*10**3/T/(2.302*3276.6*p*10**5/T**2 - 0.652*p*10**5/T);

# Results
print ("specific volume = %.3f")%(vg),("m**3/kg")

specific volume = 2.139 m**3/kg