Chapter 3 :- Evaluating Properties¶
Example 3.1 Page no-80¶
In [49]:
# Given:-
# Those with 1 are of state 1 and 2 are with state 2
# State 1
p1 = 10**5 # initial pressure in pascal
x1 = 0.5 # initial quality
T1 = 99.63 # temperature in degree celcius, from table A-3
v = 0.5 # volume of container in m3
vf1 = 1.0432*(10**(-3)) # specific volume of fluid in state 1 in m3/Kg(from table A-3)
vg1 = 1.694 # specific volume of gas in state 1 in m3/kg(from table A-3)
# State 2
p2 = 1.5*(10**5) # pressure after heating in pascal
T2 = 111.4 # temperature in degree celcius in state 2, from A-3
vf2 = 1.0582*(10**(-3)) # specific volume of fluid in state 2 in m3/Kg, from A-3
vg2 = 1.159 # specific volume of gas in state 2 in m3/Kg,from A-3
# Calculations
v1 = vf1 + x1*(vg1-vf1) # specific volume in state 1 in m3/Kg
v2 = v1 # specific volume in state 2 in m3/Kg
m = v/v1 # total mass in Kg
mg1 = x1*m # mass of vapour in state 1 in Kg
x2 = (v1-vf2)/(vg2-vf2) # quality in state 2
mg2 = x2*m # mass of vapor in state 2 in Kg
# State 3
p3 = 2.11 # pressure in state 3 from table A-3
# Results
print '-> The temperature in state 1 is ',round(T1,2),'degree celcius.'
print '-> The temperature in state 2 is ',round(T2,2),'degree celcius.'
print '-> The mass of vapour in state 1 is',round(mg1,2),'kg.'
print '-> The mass of vapour in state 2 is',round(mg2,2),'kg.'
print '-> The pressure corresponding to state 3 is',round(p3,2),'bar.'
Example 3.2 Page no-81
In [50]:
# Given:-
m = 0.05 # mass of ammonia in kg
p1 = 1.5*(10**5) # initial pressure of ammonia in pascal
v1 = 0.7787 # specific volume in state 1 in m3/kg from table A-14
v2 = 0.9553 # specific volume in state 2 in m3/kg from table A-15
T2 = 25.0 # final temperature in degree celcius
# Calculations
V1 = m*v1 # volume occupied by ammonia in state 1 in m3
V2 = m*v2 # volume occupied by ammonia in state 2 in m3
w = (p1*(V2-V1))/1000 # work in KJ
# Results
print '-> The volume occupied by ammonia in state 1 is ',round(V1,2),'m^3.'
print '-> The volume occupied by ammonia in state 2 is ',round(V2,2),'m^3.'
print '-> The work done for the process is',round(w,2),'KJ.'
Example 3.3 Page no-87¶
In [51]:
# Given:-
V = 0.25 # volume of tank in m3
v = 1.673 # specific volume in m3/kg obtained using table A-2
# State 1
T1 = 100.0 # initial temperature in degree celcius
u1 = 2506.5 # specific internal energy in state 1 in KJ/Kg obtained from table A-2
# State 2
p2 = 1.5 # final pressure in bars
T2 = 273.0 # temperature in state 2 in degree celcius obtained from table A-4
u2 = 2767.8 # specific internal energy in state 2 in KJ/Kg obtained from table A-4
# Calculations
m = V/v # mass of the system in kg
DeltaU = m*(u2-u1) # change in internal energy in KJ
W = - DeltaU # from energy balance
# Results
print '-> The temperature at the final state in is',round(T2,2),' degree celcius.'
print '-> The work during the process is',W,'KJ.'
Example 3.4 Page no-88
In [52]:
# Given:-
# State
P1 = 10*(10**5) # initial pressure in pascal
T1 = 400.0 # initial temperature in degree celcius
v1 = 0.3066 # specific volume in state 1 in m3/kg obtained from table A-4
u1 = 2957.3 # specific internal energy in state 1 in KJ/Kg obtained from table A-4
# State 2
v2 = 0.1944 # specific volume in state 2 in m3/kg obtained from table A-3
w2to3 = 0 # work in process 2-3
# State 3
v3 = v2
vf3 = 1.0905*(10**(-3)) # specific volume of fluid in state 3 from table A-2
vg3 = 0.3928 # specific volume of gas in state 3 from table A-2
uf3 = 631.68 # specific internal energy for fluid in state 3 from table A-2
ug3 = 2559.5 # specific internal energy for gas in state 3 from table A-2
# Calculations
w1to2 = (P1*(v2-v1))/1000 # work in KJ/Kg in process 1-2
W = w1to2 + w2to3 # net work in KJ/kg
x3 = (v3-vf3)/(vg3-vf3)
u3 = uf3+x3*(ug3-uf3) # specific internal energy in state 3 in Kj/Kg
q = (u3-u1) + W # heat transfer in Kj/Kg
# Results
print '-> The work done in the overall process is',W,'KJ/kg.'
print '-> The heat transfer in the overall process is',q,'KJ/kg.'
Example 3.6 Page no-98
In [53]:
# Given:-
# State 1
p1 = 20.0 # initial pressure in MPa
T1 = 520.0 # initial temperature in degree celcius
Z1 = 0.83 # compressibility factor
R = 8.314 # universal gas constant in SI unit
n = 1000.0/18.02 # number of moles in a kg of water
# State 2
T2 = 400.0 # final temperature in degree celcius
# From table A-1
Tc = 647.3 # critical temperature in kelvin
pc = 22.09 # critical pressure in MPa
# Calculations
Tr = (T1+273)/Tc # reduced temperature
Pr = p1/pc # reduced pressure
v1 = (Z1*n*R*(T1+273))/(p1*(10**6))
vr = v1*(pc*(10**6))/(n*R*Tc)
Tr2 = (T2+273)/Tc
PR = 0.69 # at above vr and Tr2
P2 = pc*PR
# Results
print '-> The specific volume in state1 is',v1,'m3/kg and the corresponding value obtained from table A-4 is .01551 m^3/Kg'
print '-> The pressure in MPa in the final state is',P2,' MPa and the corresponding value from the table is 15.16Mpa'
Example 3.7 Page no:-101
In [54]:
%matplotlib inline
In [55]:
# Given:-
T1 = 300.00 #temperature in state 1 in kelvin
P1 = 1.00 #pressure in state 1 in bar
P2 = 2.00 #pressure in state 2 in bar
R = 287.00 #gas constant of air in SI units
# Calculations
v1 = (R*T1)/(P1*10**5) #specific volume in state 1
v = []
P = []
vv = []
Pa = []
pp = []
pcommon = []
vcommon = []
from numpy import linspace
from pylab import *
P = linspace(1,2,50)
for i in range(0,50):
v.append(i)
v[i] = v1
T2 = (P2*10**5*v1)/R
v3 = (R*T2)/(P1*10**5)
vv = linspace(v1,v3,50)
for i in range(0,50):
Pa.append(i)
Pa[i] = P1
#function[out]= f(inp)
#out = (R*T2)/(inp
VV = linspace(v1,v3,50)
for j in range(0,50):
pp.append(j)
pp[j] = (R*T2)/((VV[j])/(10**5))
vcommon = numpy.concatenate((v, VV))
pcommon = numpy.concatenate((P, pp))
plot(vcommon,pcommon)
xlabel('v')
ylabel('p(bar)')
show()
plot(vv,Pa)
xlabel('v')
ylabel('p(bar)')
show()
#The two steps are shown in one graph and the other on is shown in the other graph"""
print 'The temperature in kelvin in state 2 is T2 = ',T2
print 'The specific volume in state 3 in m^3/kg is v = ',v3
Example 3.8 Page no-108
In [56]:
# Given:-
# State 1
m = 0.9 # mass of air in kg
T1 = 300.0 # initial temperature in kelvin
P1 = 1.0 # initial pressure in bar
# State 2
T2 = 470.0 # final temperature in kelvin
P2 = 6.0 # final pressure in bar
Q = -20.0 # heat transfer in kj
# From table A-22
u1 = 214.07 # in KJ/kg
u2 = 337.32 # in KJ/Kg
# Calculations
deltaU = m*(u2-u1) # change in internal energy in kj
W = Q - deltaU # in KJ/kg
# Results
print '-> The work during the process is ',W,' KJ.'
Example 3.9 Page no-109
In [57]:
# Given:-
# State 1
m1 = 2.0 # initial mass of gas in tank 1 in kg
T1 = 350.0 # initial temperature in kelvin in tank1
p1 = 0.7 # initial pressure in bar in tank 1
# State 2
m2 = 8.0 # initial mass of gas in tank 2 in kg
T2 = 300.0 # initial temperature in kelvin in tank 2
p2 = 1.2 # initial pressure in bar in tank 2
Tf = 315.0 # final equilibrium temperature in kelvin
# From table A-20
Cv = 0.745 # in KJ/Kg.k
# Calculations
pf = ((m1+m2)*Tf)/((m1*T1/p1)+(m2*T2/p2))
Ui = (m1*Cv*T1)+(m2*Cv*T2)
Uf = (m1+m2)*Cv*Tf
deltaU = Uf-Ui
Q = deltaU
# Results
print '-> The final equilibrium pressure is',pf,'bar.'
print '-> The heat transfer for the process is',Q,'KJ.'
Example 3.11 Page no-113
In [58]:
# Given:-
p1 = 1.0 # initial pressure in bar
T1 = 295.0 # initial temperature in kelvin
p2 = 5.0 # final pressure in bar
n = 1.3 # polytropic constant
R = 8314/28.97 # gas constant for air in SI units
# From table A-22
u2 = 306.53
u1 = 210.49
# Calculations
T2 = T1*(p2/p1)**((n-1)/n)
w = R*(T2-T1)/(1-n)
Q = u2-u1+w/1000
# Results
print '-> The work done per unit mass is ',w/1000,'KJ/kg.'
print '-> The heat transfer per unit mass is',Q,'KJ/kg.'