import math
T2 = -5.0 # Cold storage temperature in degree Celsius
T1 = 35.0 # Surrounding temperature in degree Celsius
COP = (T2+273)/((T1+273)-(T2+273))
ACOP = COP/3 # Actual COP
Q2 = 29.0 # Heat leakage in kW
W = Q2/ACOP
print "\n Example 14.1\n"
print "\n Power required to drive the plane is ",W ," kW"
#The answers vary due to round off error
import math
# At P = 0.14 MPa
h1 = 236.04 # Enthalpy at state 1 in kJ/kg
s1 = 0.9322 # Entropy at state 2 in kJ/kgK
s2 = s1 # Isenthalpic process
# At P = 0.8 MPa
h2 = 272.05 # Enthalpy at state 2 in kJ/kg
h3 = 93.42 # Enthalpy at state 3 in kJ/kg
h4 = h3 # Isenthalpic process
m = 0.06 # mass flow rate in kg/s
Q2 = m*(h1-h4) # Heat absorption
Wc = m*(h2-h1) # Compressor work
Q1 = m*(h2-h4) # Heat rejection in evaporator
COP = Q2/Wc # coefficient of performance
print "\n Example 14.2\n"
print "\n The rate of heat removal is ",Q2 ," kW"
print "\n Power input to the compressor is ",Wc ," kW"
print "\n The heat rejection rate in the condenser is ",Q1 ," kW"
print "\n COP is ",COP ," kW"
#The answers vary due to round off error
import math
h1 = 183.19 # Enthalpy at state 1 in kJ/kg
h2 = 209.41 # Enthalpy at state 2 in kJ/kg
h3 = 74.59 # Enthalpy at state 3 in kJ/kg
h4 = h3 # Isenthalpic process
T1 = 40.0 # Evaporator temperature in degree Celsius
T2 = -10.0 # Condenser temperature in degree Celsius
W = 5.0 # Plant capacity in tonnes of refrigeration
w = (W*14000/3600)/(h1-h4) # Refrigerant flow rate
v1 = 0.077 # Specific volume of vapor in m**3/kg
VFR = w*v1 # volume flow rate
T = 48.0 # Compressor discharge temperature in degree Celsius
P2 = 9.6066 # Pressure after compression
P1 = 2.1912 # Pressure before compression
rp = P2/P1 # Pressure ratio
Q1 = w*(h2-h3) # Heat rejected in condenser
hf = 26.87 # Enthalpy of fluid in kJ/kg
hfg = 156.31# Latent heat of vaporization in kJ/kg
x4 = (h4-hf)/hfg # quality of refrigerant
COP_v = (h1-h4)/(h2-h1) # Actual coefficient of performance of cycle
PI = w*(h2-h1) # Power input
COP = (T2+273)/((T1+273)-(T2+273)) # Ideal coefficient of performance
r = COP_v/COP
print "\n Example 14.3\n"
print "\n Refrigerant flow rate is ",w ," kg/s"
print "\n Volume flow rate is ",VFR ," m**3/s"
print "\n Compressor discharge temperature is ",T ," degree Celsius "
print "\n Pressure ratio is ",rp
print "\n Heat rejected to the condenser is ",Q1 ," kW"
print "\n Flash gas percentage is ",x4*100 ," percent"
print "\n COP is ",COP_v ," kW"
print "\n Power required to drive the compressor is ",PI ," kW"
print "\n Ratio of COP of cycle with Carnot refrigerator is ",r
#The answers vary due to round off error
import math
h3 = 882 # Enthalpy at state 3 in kJ/kg
h2 = 1034 # Enthalpy at state 2 in kJ/kg
h6 = 998 # Enthalpy at state 6 in kJ/kg
h1 = 1008 # Enthalpy at state 1 in kJ/kg
v1 = 0.084 # Specific volume at state 1 in m**3/kg
t4 = 25 # Temperature at state 4 in degree Celsius
m = 10 # mass flow rate in kg/s
h4 = h3-h1+h6
h5 = h4 # isenthalpic process
w = (m*14000)/((h6-h5)*3600) # in kg/s
VFR = w*3600*v1 # Volume flow rate in m**3/h
ve = 0.8 # volumetric efficiency
CD = VFR/(ve*60) # Compressor displacement in m**3/min
N = 900 # Number of strokes per minute
n = 2 # number of cylinder
D = ((CD*4)/(math.pi*1.1*N*n))**(1/3) # L = 1.1D L = length D = diameter
L = 1.1*D
COP = (h6-h5)/(h2-h1) # coefficient of performance
PI = w*(h2-h1) # Power input
print "\n Example 14.4\n"
print "\n Refrigeration effect is ",h6-h5 ," kJ/kg"
print "\n Refrigerant flow rate is ",w ," kg/s"
print "\n Diameter of cylinder is ",D*100 ," cm"
print "\n Length of cylinder is ",L*100 ," cm"
print "\n COP is ",COP
print "\n Power required to drive the compressor is ",PI ," kW"
#The answers vary due to round off error
import math
P2 = 1554.3 # Pressure at state 2 in kPa
P1 = 119.5# Pressure at state 1 in kPa
Pi = math.sqrt(P1*P2)
h1 = 1404.6 # Enthalpy at state1 in kJ/kg
h2 = 1574.3 # Enthalpy at state2 in kJ/kg
h3 = 1443.5 # Enthalpy at state3 in kJ/kg
h4 = 1628.1# Enthalpy at state4 in kJ/kg
h5 = 371.7 # Enthalpy at state5 in kJ/kg
h6 = h5 # Isenthalpic process
h7 = 181.5# Enthalpy at state7 in kJ/kg
w = 30 # capacity of plant in tonnes of refrigeration
m2_dot = (3.89*w)/(h1-h7) # mass flow rate in upper cycle
m1_dot = m2_dot*((h2-h7)/(h3-h6))# mass flow rate in lower cycle
Wc_dot = m2_dot*(h2-h1)+m1_dot*(h4-h3) # Compressor work
COP = w*3.89/Wc_dot # Coefficient of performance of cycle
# single stage
h1_ = 1404.6 #Enthalpy at state1 in kJ/kg
h2_ = 1805.1 # Enthalpy at state2 in kJ/kg
h3_ = 371.1 # Enthalpy at state3 in kJ/kg
h4_ = h3_ # Isenthalpic process
m_dot = (3.89*30)/(h1_-h4_) # mass flow rate in cycle
Wc = m_dot*(h2_-h1_) # Compressor work
COP_ = w*3.89/Wc # Coefficient of performance of cycle
IW = (Wc-Wc_dot)/Wc_dot # Increase in compressor work
ICOP = (COP-COP_)/COP_ # Increase in COP for 2 stage compression
print "\n Example 14.5\n"
print "\n Increase in work of compression for single stage is ",IW*100 ," percent"
print "\n Increase in COP for 2 stage compression is ",ICOP*100 ," percent"
#The answers vary due to round off error
import math
# Given that
te = -10 # Evaporator temperature in degree celsius
pc = 7.675 # Condenser pressure in bar
pf = 4.139 # Flash chamber pressure in bar
P = 100 # Power input to compressor in kW
print "\n Example 14.6\n"
# From the property table of R-134a,
h7 = 140.96 # In kJ/kg
hf = 113.29 # In kJ/kg
hfg = 300.5-113.29 # In kJ/kg
hg = 300.5 # In kJ/kg
h1 = 288.86 # In kJ/kg
s1 = 1.17189 # # In kJ/kgK
s2 =s1
#By interpolation
h2 = 303.468 # In kJ/kg
x8 = (h7-hf)/hfg
m1=x8
h5 = (1-m1)*h2 + m1*hg
# By interpolation
s5 = 1.7174 # In kJ/kgK
s6=s5
h6 = 315.79 # In kJ/kg
m = P/((h6-h5) + (1-m1)*(h2-h1))
m_e = (1-m1)*m
COP = m_e*(h1-hf)/P
print "\n The COP of the plant is ",COP ,", \n The mass flow rate of refrigerant in the evaporator is ",m_e ," kg/s"
import math
tsat = 120.2 # Saturation temperature in degree Celsius
hfg = 2201.9 # Latent heat of fusion in kJ/kg
T1 = 120.2 # Generator temperature in degree Celsius
T2 = 30 # Ambient temperature in degree Celsius
Tr = -10 # Operating temperature of refrigerator in degree Celsius
COP_max = (((T1+273)-(T2+273))*(Tr+273))/(((T2+273)-(Tr+273))*(T1+273)) # Ideal coefficient of performance
ACOP = 0.4*COP_max # Actual COP
L = 20 # Refrigeration load in tonnes
Qe = (L*14000)/3600 # Heat extraction in KW
Qg = Qe/ACOP # Heat transfer from generator
x = 0.9 # Quality of refrigerant
H = x*hfg # Heat extraction
SFR = Qg/H # Steam flow rate
print "\n Example 14.7\n"
print "\n Steam flow rate required is ",SFR ," kg/s"
#The answers vary due to round off error
import math
# Given that
tf = 5 # Temperature of flash chamber in degree celsius
x = 0.98 # Quality of water vapour living the evaporator
t2 = 14 # Returning temperature of chilled water in degree celsius
t0 = 30 # Make up water temperature in degree celsius
m = 12 # Mass flow rate of chilled water in kg/s
nc = 0.8 # Compressor efficiecy
pc = 0.1 # Condenser pressure in bar
print "\n Example 14.8\n"
#From the steam table
hf = 58.62 # In kJ/kg at 14 degree celsius
hf_ = 20.93 # In kJ/kg at 5 degree celsius
hf__ = 125.73 # In kJ/kg at 30 degree celsius
hv = x*2510.7
Rc = m*(hf-hf_)/3.5
m_v = Rc*3.5/(hv-hf__)
# At 0.10 bar
hg = 2800 # In kJ/kg
Win = m_v*(hg-hv)/nc
COP = Rc*3.5/Win
print "\nCOP of the system is ",COP
import math
T1 = 4.0 # Compressor inlet temperature in degree Celsius
T3 = 55.0 # Cooling limit in heat exchanger in degree Celsius
rp = 3.0 # Pressure ratio
g = 1.4 # Heat capacity ratio
cp = 1.005 # Constant volume heat capacity
L = 3.0 # Cooling load in tonnes of refrigeration
nc = 0.72 # compressor efficiency
T2s = (T1+273)*(rp**((g-1)/g)) # Ideal temperature after compression
T2 = (T1+273)+(T2s-T1-273)/nc # Actual temperature after compression
T4s = (T3+273)/(rp**((g-1)/g)) # Ideal temperature after expansion
T34 = 0.78*(T3+273-T4s) # Change in temperature during expansion process
T4 = T3+273-T34 # Actual temperature after expansion
COP = (T1+273-T4)/((T2-T1-273)-(T3+273-T4)) # Coefficient of performance of cycle
P = (L*14000)/(COP*3600) # Driving power required
m = (L*14000)/(cp*(T1+273-T4)) # Mass flow rate of air
print "\n Example 14.9\n"
print "\n COP of the refrigerator is ",COP
print "\n Driving power required is ",P ," kW"
print "\n Mass flow rate is ",m/3600 ," kg/s"
#The answers vary due to round off error
import math
P1 = 2.4 #Compressor inlet pressure in bar
T1 = 0 # Compressor inlet temperature in degree Celsius
h1 = 188.9 # Enthalpy of refrigerant at state 1 in kJ/kg
s1 = 0.7177 # Entropy of refrigerant at state 1 in kJ/kgK
v1 = 0.0703 # Specific volume at state 1 in m**3/kg
P2 = 9 # Compressor outlet pressure in bar
T2 = 60 # Compressor outlet pressure in degree Celsius
h2 = 219.37 # Actual compressor outlet enthalpy in kJ/kgK
h2s = 213.27 # Ideal compressor outlet enthalpy in kJ/kgK
h3 = 71.93 # Enthalpy of refrigerant at state 3 in kJ/kg
h4 = h3 # Isenthalpic process
A1V1 = 0.6/60 # volume flow rate in kg/s
m_dot = A1V1/v1 # mass flow rate
Wc_dot = m_dot*(h2-h1) # Compressor work
Q1_dot = m_dot*(h2-h3) # Heat extracted
COP = Q1_dot/Wc_dot # Coefficient of performance
nis = (h2s-h1)/(h2-h1) # Isentropic compressor efficiency
print "\n Example 14.10\n"
print "\n Power input is ",Wc_dot ," kW"
print "\n Heating capacity is ",Q1_dot ," kW"
print "\n COP is ",COP
print "\n The isentropic compressor efficiency is ",nis*100 ," percent"
#The answers vary due to round off error
import math
T1 = 275.0 # Temperature of air at entrance to compressor in K
T3 = 310.0 # Temperature of air at entrance to turbine in K
P1 = 1.0 # Inlet presure in bar
P2 = 4.0 # Outlet pressure in bar
nc = 0.8 # Compressor efficiency
T2s = T1*(P2/P1)**(.286) # Ideal temperature after compression
T2 = T1 + (T2s-T1)/nc # Actual temperature after compression
pr1 = 0.1 # Pressure loss in cooler in bar
pr2 = 0.08 #Pressure loss in condensor in bar
P3 = P2-0.1 # Actual pressure in condesor
P4 = P1+0.08 # Actual pressure in evaporator
PR = P3/P4 # Pressure ratio
T4s = T3*(1/PR)**(0.286) # Ideal temperature after expansion
nt = 0.85 # turbine efficiency
T4 = T3-(T3-T4s)*nt # Actual temperature after expansion
COP = (T1-T4)/((T2-T3)-(T1-T4)) # Coefficient of performance
print "\n Example 14.11\n"
print "\n Pressure ratio for the turbine is ",PR
print "\n COP is ",COP
#The answers vary due to round off error
import math
# Given that
L = 60.0 # Cooling load in kW
p = 1.0 # Pressure in bar
t = 20.0 # Temperature in degree celsius
v = 900.0 # Speed of aircraft in km/h
p1 = 0.35 # Pressure in bar
T1 = 255 # Temperature in K
nd = .85 # Diffuser efficiency
rp = 6.0 # Pressure ratio of compressor
nc = .85 # Copressor efficiency
E = 0.9 # Effectiveness of air cooler
nt = 0.88 # Turbine efficiency
p_ = 0.08 # Pressure drop in air cooler in bar
p5 = 1.08 # Pressure in bar
cp = 1.005 # Heat capacity of air at constant pressure in kJ/kgK
gama = 1.4 # Ratio of heat capacities of air
print "\n Example 14.12\n"
V = v*(5/18)
T2_ = T1 + (V**2)/(2*cp*1000)
T2 = T2_
p2_ = p1*((T2_/T1)**((gama/(gama-1))))
p2 = p1 + nd*(p2_-p1)
p3 = rp*p2
T3_ = T2*((p3/p2)**((gama-1)/gama))
T3 = T2 + (T3_-T2)/nc
P = cp*(T3-T2)
p4 = p3 - p_
T4 = T3 - E*(T3-T2)
T5_ = T4/((p4/p5)**(.286))
T5 = T4 - (T4-T5_)/nt
RE = cp*(t+273 - T5)
m = L/51.5
Pr = m*P
COP = L/Pr
print "\n Mass flow rate of air flowing through the cooling system is ",m
print "\n COP is ",COP
#The answers vary due to round off error