Chapter 4 : The First Law of Thermodynamics for Systems-Pure Substances

Example 4.1 Page No : 93

In [1]:
			
# Variables
#four heat transfer
Q_1 = 900 			#J
Q_2 = 80 			#J
Q_3 = -800 			#J
Q_4 = 150 			#J
#four work interactions
W_1 = 200 			#J
W_2 = 150 			#J
W_3 = 300 			#J
			#W_4
			
# Calculations and Results
W_4 = Q_1 +Q_2 +Q_3 +Q_4 -W_1 -W_2 -W_3  
print "Magnitude and Direction of the fourth work interaction, W4 = %.0f J"%(W_4)

Magnitude and Direction of the fourth work interaction, W4 = -320 J

Example 4.2 Page No : 94

In [2]:
			
# Variables
Q_a = -50 			#KJ 			#heat transferred from the system along path A
W_a = -65 			#KJ 			#work done along path A
Q_b = 0 			#KJ 			#heat transferred from the system along path B
			#W_b work done along path B

			
# Calculations and Results

#Part(a)
print "Part a";
delE_a = Q_a - W_a 			#KJ 			#Change in energy along path A
print "Change in energy of the system = %.0f KJ"%(delE_a);

#Part(b)
print "Part b";
delE_b = -1*delE_a 			#KJ 			#Change in energy along path B
W_b = delE_b - Q_b 			#KJ 			#work done along path B
print "Magnitude and direction of work done during B, W_b = %.0f KJ"%(W_b)

Part a
Change in energy of the system = 15 KJ
Part b
Magnitude and direction of work done during B, W_b = -15 KJ

Example 4.3 Page No : 99

In [4]:
# Variables
m = 2.3 			#kg 			#mass of substance
u = 21 * 10**3 			#J/kg 			#internal energy
V = 110. 			#m/s 			#velocity 
z = 1500. 			#m 			#elevation above sea level
g = 9.81 			#m/s**2 			#acceleration due to gravity


			
# Calculations and Results
E = m*(g*z + V**2/2 + u) 			#J/kg 			#Total energy of the system
print "The total energy of the system with respect to an observer at rest at sea level, E = %.4f KJ"%(0.001*E);

The total energy of the system with respect to an observer at rest at sea level, E = 96.0595 KJ

Example 4.4 Page No : 103

In [6]:
from numpy import *
from sympy import Derivative
			
# Variables
t = poly1d(0); 			#C 			#Temperature in C
u = 196. + .718*t; 			#KJ/kg 			#specific internal energy
pv = 287*(t+273.); 			#Nm/kg 			#p is pressure and v = specific volume

			
# Calculations and Results
Cv = poly(u);
print "Specific heat at constant volume,Cv = %.3f kJ/kgK"%(Cv[0])

h = u + pv*.001 			#KJ/kg 			#enthalpy
Cp = poly(h);
print "Specific heat at constant pressure,Cp = %.3f kJ/kgK"%(Cp[0])

# Note: Poly function gives different result then book has.

Specific heat at constant volume,Cv = 1.000 kJ/kgK
Specific heat at constant pressure,Cp = 1.000 kJ/kgK