Chapter 21 : Energy Terminology Concepts and Units

Example 21.1 Page no : 616

In [1]:
from scipy.integrate import quad

# Variables
V1 = 0.1 ;			# Volume of gas initially -[cubic metres]
V2 = 0.2 ;			# Volume of gas finally -[cubic metres]
T1 = 300 ;			# Temperature of gas initially -[K]
P1 = 200 ;			# Pressure of gas finally -[kPa]
R = 8.314 ;			# Universal gas constant 
n = (P1*V1)/(T1*R) ;			# Moles of gas taken-[kg mol]
#You are asked to calculate work by eqn. 21.1 , but you do not know the F(force) exerted by gas , so write F = P.A, multiply divide A and eqn 21.1 reduces to W= integate(P.dv)

# Calculations and Results
# Isobaric process see fig E21.1b to see the path followed
def f(V):
    return -(P1)
W= quad(f,V1,V2)[0] ;			# Work done by gas on piston -[kJ]
print ' (a)Work done by gas on piston for isobaric process is %.0f kJ . ',W

def f1(V):
    return -(T1*R*n/V)
W= quad(f1,V1,V2)[0] ;			# Work done by gas on piston -[kJ]
print '(b)Work done by gas on piston for isothermal process is %.2f kJ . ',W

 (a)Work done by gas on piston for isobaric process is %.0f kJ .  -20.0
(b)Work done by gas on piston for isothermal process is %.2f kJ .  -13.8629436112

Example 21.2 page no. 624

In [2]:
# Variables
id_ = 3. ;			# Internal diameter of tube-[cm]
Vf = 0.001 ;			# Volume flow rate of water in tube-[cubic meter/s]
rho = 1000. ;			# Assumed density of water-[kg/cubic meter] 

# Calculations
rad = id_/2. ;			# Radius of tube -[ cm]
a = 3.14*rad**2 ;			# Area of flow of tube -[squqre centimeter]
v = Vf*(100)**2/a ;			# Velocity of water in tube - [m/s]
KE = v**2/2. ;			# Specific(mass=1kg) kinetic energy of water in tube -[J/kg]

# Results
print 'Specific kinetic energy of water in tube is %.2f J/kg . '%KE

Specific kinetic energy of water in tube is 1.00 J/kg .

Example 21.3 page no. 626

In [3]:
# Variables
# Let water level in first reservoir be the reference plane
h = 40. ;			# Difference of water-[ft]
g = 32.2 ;			# acceleration due to gravity-[ft/square second]

# Calculations
PE=g*h/(32.2*778.2) ;			#			# Specific(mass=1kg) potential energy of water -[Btu/lbm]

# Results
print 'Specific potential energy of water is %.4f Btu/lbm . '%PE

Specific potential energy of water is 0.0514 Btu/lbm .

Example 21.4 page no : 629

In [4]:
# Variables
#Constant volume process 
mol_air = 10. ;			# Moles of air-[kg mol]
T1 = 60.+273 ;			# Initial temperature of air-[K]
T2 = 30.+273 ;			# final temperature of air-[K]
# Additional data needed
Cv = 2.1*10.**4 ; 			# Specific heat capacity of air at constant volume-[J/(kg mol*C)]

# Calculations
def f(T):
    return mol_air*Cv
del_U = quad(f,T1,T2)[0] ;			#Change in internal energy-[J]

# Results
print 'Change in internal energy is %.1e J . '%del_U

Change in internal energy is -6.3e+06 J .

Example 21.7 page no : 633

In [5]:
# Variables
#Constant pressure process 
mol_air = 10. ;			# Moles of air-[kg mol]
T1 = 60+273 ;			# Initial temperature of air-[K]
T2 = 30+273 ;			# final temperature of air-[K]

# Calculations
# Additional data needed
Cp = 2.9*10**4  ;			# Specific heat capacity of air at constant pressure-[J/(kg mol*C)]
# Use eqn. 21.11 for del_H
def f(T):
    return mol_air*Cp

del_H = quad(f,T1,T2)[0] ;			#Change in enthalpy-[J]
print 'Change in enthalpy is %.1e J . '%del_H

Change in enthalpy is -8.7e+06 J .
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