Chapter 7: First Law of Thermodynamics¶

Exa 7.1¶

In [1]:
#An automobile with a mass of 3000 lb comes over the top of a hill 50 ft high
#with a velocity of 50 mph. Brakes are applied at the instant the automobile
#reaches the top, and it comes to rest at the bottom of the hill. How much
#energy is dissipated from the brakes?
import math
#initialisation of variables
m= 3000 											#lb
Z1= 50 												#ft
V1= 50 												#mph
gc= 32.2 											#ft/lbf s^2
V2= 0 												#mph
g= 32.2 											#ft/s^2
Z2= 0 												#ft
#CALCULATIONS
V1= V1*(73.3/50.) 									#Velocity
Q2= ((m*(V2*V2-V1*V1))/(2*gc))+((m*g)/gc)*(Z2-Z1)
#RESULTS
print '%s %.2e' % ('Energy dissipated from the brakes (ft lbf) = ',-Q2)
raw_input('press enter key to exit')

Energy dissipated from the brakes (ft lbf) =  4.00e+05
press enter key to exit

Out[1]:
''

Exa 7.2¶

In [2]:
#Steam at a pressure of 13 bar and a temperature of 300C flows adiabatically
#and with negligible velocity into an evacuated tank. Using a closed system
#analysis, determine the temperature of the steam in the tank reaches line
#pressure.
#initialisation of variables
P= 15 						#bar
T= 300 						#C
h1= 3043.1 					#J/gm
#CALCULATIONS
u2= h1
print '%s' %('From keenan and keyes steam tables')
T= 453.4  					#C temperature
#RESULTS
print '%s %.2f' % ('Temperature of the steam in the tank (C) = ',T)
raw_input('press enter key to exit')

From keenan and keyes steam tables
Temperature of the steam in the tank (C) =  453.40
press enter key to exit

Out[2]:
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Exa 7.3¶

In [3]:
#A 10 lb mass of air at 120F is contained in a rigid tank. How much heat
#is transferred to the tank to raise the air temperature of 275 F?
#initialisation of variables
m= 10 							#lbf
T= 120 							#F
T1= 275 						#F
u1= 98.9 						#Btu/lbm
u2= 125.6 						#Btu/lbm
#CALCULATIONS
Q= m*(u2-u1) 					#Heat transferred
#RESULTS
print '%s %.2f' % ('Heat transferred to the tank (Btu) = ',Q)
raw_input('press enter key to exit')

Heat transferred to the tank (Btu) =  267.00
press enter key to exit

Out[3]:
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Exa 7.4¶

In [4]:
#Fluid enters a turbine with a velocity of 1 m/s and an enthalpy of 2000 j/gm;
#it leaves with a velocity of 60 m/s and enthalpy of 1800 J/gm. Heat losses
#are 500 J/s, and the flow rate is 0.5 kg/s. If the inlet of the turbine is
#3 m higher than the outlet, what is the maximum theoretical power
# that can be developed?
#initialisation of variables
v0= 1 													#m/s
vi= 60 													#m/s
Q= -500 												#J/s
m= 500 													#gm/s
hi= 2000 												#J/gm
h0= 1800 												#J/gm
zi= 3 													#m
z0= 0 													#m
g= 9.8 													#m/s^2
gc= 1000. 												#gm m/Ns^2
#CALCULATIONS
W= Q+m*((hi-h0)+(vi*vi-v0*v0)/(2*gc)+(g/gc)*(zi-z0)) 	#Work
#RESULTS
print '%s %.2e' % ('Maximum theotrical power that can be devoloped (J/s) = ',W)
raw_input('press enter key to exit')

Maximum theotrical power that can be devoloped (J/s) =  1.00e+05
press enter key to exit

Out[4]:
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Exa 7.5¶

In [5]:
#To produce 0.3 lt/s hot water at 82 C, low pressure steam at 2.4 bar and
#80 percent quality is mixed with a stream of water at 16 C. What is the
#required steam flow rate?
#initialisation of variables
m= 0.3 								#lt/s
T= 82 								#C
P= 2.4 								#bar
p= 80.
Tw= 800 							#C
h1= 67.19 							#J/gm
h3= 343.3 							#J/gm
hf= 529.65 							#J/gm
hfg= 2185.4 						#J/gm
v3= 1.0305 							#cm^3/gm
V3= 300 							#cm^3/s
#CALCULATIONS
h2= hf+(p/100.)*hfg					#Enthalpy at 2
m3= V3/v3 							#Mass at 3
m2= (m3*(h3-h1))/(h2-h1) 			#Mass at 2
#RESULTS
print '%s %.2f' % ('Required steam flow rate (gm/s) = ',m2)
raw_input('press enter key to exit')

Required steam flow rate (gm/s) =  36.36
press enter key to exit

Out[5]:
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Exa 7.6¶

In [6]:
#Latent heat of transforation can be defined as the ratio of heat absorbed
#to the mass which undergoes a change of phase(L=Q/m). Show that the heat of
#transformation for any phase change equals the difference between the
#enthalphies of the system in the two phases.
#initialisation of variables
h2= 2 					#J/gm
h1= 1 					#J/gm
#CALCULATIONS
L= h2-h1 				#Difference between enthalpies
#RESULTS
print '%s %.2f' % ('Difference between the enthalpies of the system in the two phases ((h2-h1) J/gm) = ',L)
raw_input('press enter key to exit')

Difference between the enthalpies of the system in the two phases ((h2-h1) J/gm) =  1.00
press enter key to exit

Out[6]:
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