Chapter 5: Airfoils, Wings and Other Aerodynamic shapes

Example 5.1

In [1]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.1
''' A car of mass 1100 kg drives with a velocity such that it has a kinetic energy of 400 kJ (see
Fig. 5.4). Find the velocity. If the car is raised with a crane, how high should it be lifted in
the standard gravitational field to have a potential energy that equals the kinetic energy? '''

#Variable Declaration: 
m = 1100 		#Mass of car in kg
ke = 400 		#Kinetic energy of car in kJ
g = 9.807 		#Acc. due to gravity in m/s**2

#Calculations:
V = (2*ke*1000/m)**0.5 #Velocity of car in m/s
H = ke*1000/(m*g) 	#Height to which the car should be lifted so that its potential energy equals its kinetic energy

#Results:
print "Required height of car: ", round(H,1),"m"
Required height of car:  37.1 m

Example 5.2

In [2]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.2
'''A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel
is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the fluid
inside a control surface and determine the change in internal energy of this control mass.  '''

#Variable Declaration: 
W = -5090 		#Work input to paddle wheel in kJ
Q = -1500 		#Heat transfer from tank in kJ

#Calculations:
dU = Q-W 		#Change in internal energy in kJ

#Results:
print "Change in internal energy is: ",round(dU),"KJ"
Change in internal energy is:  3590.0 KJ

Example 5.3

In [3]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.3
''' Consider a stone having a mass of 10 kg and a bucket containing 100 kg of liquid water.
Initially the stone is 10.2 m above the water, and the stone and the water are at the same
temperature, state 1. The stone then falls into the water.
Determine U, KE, PE, Q, and W for the following changes of state, assuming
standard gravitational acceleration of 9.806 65 m/s2.
a. The stone is about to enter the water, state 2.
b. The stone has just come to rest in the bucket, state 3.
c. Heat has been transferred to the surroundings in such an amount that the stone and
water are at the same temperature, T1, state 4. '''

#Variable Declaration: 
g = 9.806 		#Acceleration due to gravity in m/s**2
m = 10 		#Mass of stone in kg
H1 = 10.2 		#Initial height of stone above water in metres
H2 = 0 		#Final height in metres

#Calculations:
dKE1 = -m*g*(H2-H1) #Change in kinetic energy when stone enters state 2 in J
dPE1 = -1 		#Change in potential energy when stone enters state 2 in J
dPE2 = 0 		#Change in potential energy when stone enters state 3 in JQ2 = 0 		#no heat transfer when stone enters state 3 in J
W2 = 0 		#no work done when stone enters state 3 in J
dKE2 = -1 		#Change in kinetic energy when stone enters state 3
dU2 = -dKE2 	#Change in internal energy when stone enters state 3 in J
dKE3 = 0 		#Change in kinetic energy when stone enters state 4
dPE = 0 		#Change in potential energy when stone enters state 4 in J
W3 = 0 		#No work done when stone enters state 4 in J
dU3 = -1 		#Change in internal energy when stone enters state 4 in J
Q3 = dU3 		#Heat transfer when stone enters state 4 in J

#Results:
print "When stone is about to enter state 2, dKE: ",round(dKE1/1000,2),"KJ and dPE :",round(dPE1,2),"KJ"
print "When stone has just come to rest in the bucket i.e. state 3, W = 0, dPE = 0, dKE1:",round(dKE2,2),"J and dU :",round(dU2,2),"J"
print "When stone has entered state 4, dPE = 0, W3 = 0,dKE = 0, dU: ",round(dU3,2),"J Q3: ",round(Q3,2),"J"
When stone is about to enter state 2, dKE:  1.0 KJ and dPE : -1.0 KJ
When stone has just come to rest in the bucket i.e. state 3, W = 0, dPE = 0, dKE1: -1.0 J and dU : 1.0 J
When stone has entered state 4, dPE = 0, W3 = 0,dKE = 0, dU:  -1.0 J Q3:  -1.0 J

Example 5.4

In [4]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.4
'''Determine the missing property (P, T, or x) and v forwater at each of the following states:
a. T = 300◦C, u = 2780 kJ/kg
b. P = 2000 kPa, u = 2000 kJ/kg'''

#Variable Declaration: 
T1 = 300 		#Given temp. in Celsius
u1 = 2780 		#Given specific internal enrgy in kJ/kg
P2 = 2000 		#Given pressure in kPa
u2 = 2000 		#Given specific intrernal energy in kJ/kg
uf = 906.4 		#in kJ/kg
ug = 2600.3 	#in kJ/kg 
vf = 0.001177 	#in m**3/kg 
vg = 0.099627 	#in m**3/kg

#Calculation:
x2 = (u2-906.4)/(ug-uf) 
v2 = vf+x2*(vg-vf)#Specific volume for water in m**3/kg

#Result:
print "Specific volume for water is: ",round(v2,5),"m**3/Kg"
print "Therefore ,this state is in the two phase region with quality: ",round(x2,4)
Specific volume for water is:  0.06474 m**3/Kg
Therefore ,this state is in the two phase region with quality:  0.6456

Example 5.5

In [5]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.5
'''A vessel having a volume of 5 m3 contains 0.05 m3 of saturated liquid water and 4.95 m3
of saturated water vapor at 0.1 MPa. Heat is transferred until the vessel is filled with
saturated vapor. Determine the heat transfer for this process.'''

#Variable Declaration: 
Vliq = 0.05 		#Volume of saturated liquid in m**3
vf = 0.001043 		#in m**3/kg
Vvap = 4.95 		#Volume of saturated water vapour in m**3
vg = 1.6940 		#in m**3/kg
u1liq = 417.36 		#Specific internal energy of liquid in kJ/kg
u1vap = 2506.1 		#Specific internal energy of vapors in kJ/kg
V = 5 		      #Total volume in m**3
u2 = 2600.5 		#Specific internal energy at final state in kJ/kg

#Calculation:
m1liq = Vliq/vf 		#Mass of liquid in kg
m1vap = Vvap/vg 		#Mass of vapors in kg
U1 = m1liq*u1liq+m1vap*u1vap #Total internal energy in kJ
m = m1liq+m1vap 		#Total mass in kg
v2 = V/m 		      #Final specific volume in m**3/kg
U2 = m*u2 		      #Internal energy at final state in kJ
Q = U2-U1 		      #Heat transfer for the process in kJ

print 'By interplotation we find that for steam, if vg = 0.09831 m**3/kg then pressure is 2.03 MPa'
print "Heat transfer for the process is: ",round(Q,2),"KJ"
By interplotation we find that for steam, if vg = 0.09831 m**3/kg then pressure is 2.03 MPa
Heat transfer for the process is:  104932.6 KJ

Example 5.6

In [6]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.6
'''A cylinder fitted with a piston has a volume of 0.1 m3 and contains 0.5 kg of steam at 0.4
MPa. Heat is transferred to the steam until the temperature is 300◦C, while the pressure
remains constant.'''

#Variable Declaration: 
V1 = 0.1 		      #Volume of cylinder in m**3
m = 0.5 		      #Mass of steam in kg
vf = 0.001084 		#m**3/kg
vfg = 0.4614 		#m**3/kg
hf = 604.74 		#kJ/kg
hfg = 2133.8		#kJ/kg
h2 = 3066.8 		#Final specific heat enthalpy in kJ/kg
P = 400 		      #Pressure inside cylinder in kPa
v2 = 0.6548 		#Specific enthalpy in m**3/kg

#Calculation:
v1 = V1/m 		      #Specific volume of steam in m**3/kg
x1 = (v1-vf)/vfg 		#Quality
h1 = hf+x1*hfg 		#Initial specific enthalpy in kJ/kg
Q = m*(h2-h1) 		#Heat transfer for this process in kJ
W = m*P*(v2-v1) 		#Work done for the process in kJ

#Result:
print "Work done for the process: ",round(W), "KJ"
print "Heat transfer: ",round(Q,1), "KJ"
Work done for the process:  91.0 KJ
Heat transfer:  771.1 KJ

Example 5.8

In [7]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.8
'''Calculate the change of enthalpy as 1 kg of oxygen is heated from 300 to 1500 K. Assume
ideal-gas behavior.'''
from sympy import symbols,integrate

#Variable Declaration: 
h1 = 273.2 		#Specific heat enthalpy for oxygen at 300 K
h2 = 1540.2 	#Specific heat enthalpy for oxygen at 1500 K
T1 = 300 		#Initial temperature in K
T2 = 1500 		#Final temparature in K
x = symbols('x')

#Calculation:
Cp = 0.88-0.00001*x+0.54*x**2-0.33*x**3 #Expression for constant pressure specific heat enthalpy for oxygen
dh1 = h2-h1 	#This change in specific heat enthalpy is calculated using ideal gas tables 
dh2 = 1000*integrate(Cp,(x,T1/1000,T2/1000)) #Using empirical equation
dh3 = 0.922*(T2-T1) #It is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K
dh4 = 1.0767*(T2-T1)#It is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature

#Result:
print "Change in specific heat enthalpy if ideal gas tables are used is: ",round( dh1,2),"KJ/Kg"
print "If empirical equations are used: ",round( dh2,2),"KJ/Kg"
print "If specific heat is assumed to be constant and using its value at T1: ",round(dh3,2),"KJ/Kg"
print "If specific heat is assumed to be constant at its value at (T1+T2)/2: ",round(dh4),"KJ/Kg"
Change in specific heat enthalpy if ideal gas tables are used is:  1267.0 KJ/Kg
If empirical equations are used:  1241.64 KJ/Kg
If specific heat is assumed to be constant and using its value at T1:  1106.4 KJ/Kg
If specific heat is assumed to be constant at its value at (T1+T2)/2:  1292.0 KJ/Kg

Example 5.9

In [8]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.9
'''A cylinder fitted with a piston has an initial volume of 0.1 m3 and contains nitrogen at 150
kPa, 25◦C. The piston is moved, compressing the nitrogen until the pressure is 1 MPa and
the temperature is 150◦C. During this compression process heat is transferred from the
nitrogen, and the work done on the nitrogen is 20 kJ. Determine the amount of this heat
transfer.'''

#Variable Declaration: 
P = 150 		#Pressure of nitrogen in cylinder in kPa
V = 0.1 		#Initial volume of cylinder in m**3
T1 = 25 		#Initial temperature of nitrogen in celsius
T2 = 150 		#Final tempareture of nitrogen in celsius
R = 0.2968 		#In kJ/kg-K
Cv = 0.745 		#Constant volume specific heat for nitrogen in kJ/kg-K
W = -20 		#Work done on nitrogen gas in kJ

#Calculation:
m = P*V/(R*(T1+273))#Mass of nitrogen in kg
Q = m*Cv*(T2-T1)+W #Heat transfer during the process in kJ

#Result:
print "Heat transfer for the above process is: ",round( Q,2),"KJ"
Heat transfer for the above process is:  -4.21 KJ

Example 5.10

In [9]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.10
'''During the charging of a storage battery, the current i is 20 A and the voltage e is 12.8
V. The rate of heat transfer from the battery is 10 W. At what rate is the internal energy
increasing?'''

#Variable Declaration: 
W = -12.8*20 	#Power consumed in  J/s
Q = -10 		#Heat transfer rate from battery in J/s

#Calculation:
r = Q-W 		#Rate of increase of internal energy

#Result:
print "Rate of increase of internal energy is: ",round( r,2),"J/s"
Rate of increase of internal energy is:  246.0 J/s

Example 5.11

In [10]:
# -*- coding: utf8 -*-
from __future__ import division
#Example: 5.11
'''A 25-kg cast-iron wood-burning stove, shown in Fig. 5.12, contains 5 kg of soft pine wood
and 1 kg of air. All the masses are at room temperature, 20◦C, and pressure, 101 kPa. The
wood now burns and heats all the mass uniformly, releasing 1500W. Neglect any air flow
and changes in mass of wood and heat losses. Find the rate of change of the temperature
(dT/dt) and estimate the time it will take to reach a temperature of 75◦C.'''

#Variable Declaration: 
Q = 1500 		#Power produced by burning wood in J/s
mair = 1 		#Mass of air in kg
mwood = 5 		#Mass of soft pine wood in kg 
miron = 25 		#Mass of cast iron in kg
Cvair = 0.717 	#Constant volume specific heat for air in kJ/kg
Cwood = 1.38 	#Constant volume specific heat for wood in kJ/kg
Ciron = 0.42 	#Constant volume specific heat for iron in kJ/kg

#Calculations:
dT = 75-20 		#Increase in temperature in Celsius
T = (Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) 	#Rate of change of temperature in K/s
dt = (dT/T)/60 	#in minutes

#Results:
print "Rate of change of temperature is: ",round(T,4),"K/s"
print "Time taken to reach a temperature of T: ",round(dt),"min"
Rate of change of temperature is:  0.0828 K/s
Time taken to reach a temperature of T:  11.0 min