In [1]:

```
# Given:-
# At inlet 1:-
p1= 7.0 # pressure in bar
T2= 200.0 # temperature in degree celcius
m1dot= 40.0 # mass flow rate in kg/s
# At inlet 2:-
p2= 7.0 # pressure in bar
T2= 40.0 # temperature in degree celcius
A2= 25.0 # area in cm^2
# At exit:-
p3= 7.0 # pressure in bar
AV3= 0.06 # Volumetric flow rate through wxir in m^3/s
# From table A-3
v3 = (1.108)*(10**(-3)) # specific volume at the exit in m^3/kg
# from table A-2
v2= (1.0078)*(10**(-3)) # specific volume in state 2 in m^3/kg
# Calculation:-
m3dot= AV3/v3 # mass flow rate at exit
m2dot = m3dot-m1dot # mass flow rate at inlet 2
V2= (m2dot*v2)/(A2*(10**(-4)))
# Results:-
print '-> The mass flow rate at the inlet 2 is', round(m2dot,2),' kg/s.'
print '-> The mass flow rate at the exit is', round(m3dot,2),' kg/s.'
print '-> The velocity at the inlet is ', round(V2,2),'m/s.'
```

In [2]:

```
# Given:-
p1= 40.0 # pressure in bar
T1= 400.0 # temperature in degree celcius
V1= 10.0 # velocity m/s
# At exit:-
p2= 10.0 # pressure in bar
V2= 665.0 # velocity in m/s
mdot= 2.0 # mass flow rate in kg/s
# From table A-4
h1= 3213.6 # snpecific enthalpy in kJ/kg
v2 = 0.1627 # specific volume at the exit in m^3/kg
# Calculation:-
h2 = h1 + ((V1**2-V2**2)/2)/1000 # snpecific enthalpy in kJ/kg
A2=(mdot*v2)/V2 # Exit area
# Results:-
print '-> The exit Area of the nozzle is', round(A2,4),'m^2.'
```

In [3]:

```
# Given:-
m1dot = 4600.0 # mass flow rate in kg/h
Wcvdot= 1000.0 # turbine power output in kv
p1= 60.0 # pressure in bar
T1=400.0 # temperature in degree celc
V1= 10.0 # velocity in m/s
# At exit:-
p2= 0.10 # pressure in bar
q2= 0.90 # quality
V2= 50.0 # velocity in m/s
# From table A-2 and A-3:-
h1= 3177.2 # specific enthalpy at inlet in kJ/kg
hf2= 191.83
hg2= 2584.63
# Calculation:-
h2 = hf2+q2*(hg2-hf2) # specific enthalpy at exit in kJ/kg
Qcvdot = Wcvdot + m1dot*((h2-h1)+(V2**2- V1**2)/(2*1000))/3600
# Results:-
print '-> The rate of heat transfer between the turbine and surroundings is',round(Qcvdot,2),'kW.'
```

In [4]:

```
# Given:-
p1=1.00 # pressure in bar
t1= 290.00 # temperature in kelvin
A1= 0.1 # area in m^2
V1= 6.00 # velocity in m/s
# At exit:-
p2=7.00 # pressure in bar
t2= 450.00 # temperature in kelvin
V2= 2.00 # velocity in m/s
Qcvdot= -180.0 # heat transfer rate in kJ/min
R= 8.314 # universal gas constant in SI units
# from table A-22
h1= 290.16 # specific enthalpy in kJ/kg
h2= 451.8 # specific enthalpy in kJ/kg
# Calculations:-
v1 = (R*1000*t1)/(28.97*p1*10**5) # specific volume
mdot=(A1*V1)/v1 # mass flow rate
Wcvdot = Qcvdot/60 + mdot*((h1-h2)+(V1**2-V2**2)/(2*1000))
# Results:-
print '-> The power input to the compressor is ',round(Wcvdot,2),'kW.'
```

In [5]:

```
# Given:-
# At Entry:=
t1=20.0 # Temperatue in deg celcius
p1=1.0 # pressure in atm
AV1= 0.1 # volumetric flow rate in litre/s
D1=2.5 # Diameter of th hose in cm
# At Exit:=
t2=23.0 # temperatuer in deg celcius
p2=1.0 # pressure in atm
V2=50.0 # Velocity in m/s
Z2=5.0 # elevation in m
g= 9.8 # acceleration due to gravity in m/s^2
# from table A-2 and A-19:-
v= (1.0018)*((10.0)**(-3)) # specific volume in m^3/kg
c= 4.18
# Calculation:-
mdot = (AV1/1000)/v # mass flow rate in kg/s
V1= (AV1/1000)/(3.14*(D1/(2*100))**2) # Entry velocity in m/s
deltah = c*(t2-t1)+v*(p2-p1)
Wcvdot= ((mdot*10)/9)*(-deltah+(V1**2-V2**2)/(2*1000)+g*(0-Z2)/1000)
# Results:-
print '-> The power input to the motor is', round(Wcvdot,2),'KW.'
```

In [6]:

```
# Given:-
# Entering:-
p1=0.1 # pressure in bar
x1= 0.95 # Quality
p2= 0.1 # pressure in bar
t2= 45.0 # temperature in deg celcius
t3=20.0 # temperature of cooling entry in deg cel
t4=35.0 # temperature of cooling exit
# From table A-3
hf= 191.53 # Enthalpy in KJ/kg
hg= 2584.7 # Enthalpy in KJ/kg
h2=188.45 # Assumption at states 2,3 and 4, h is approx equal to hf(T), in kJ/kg
deltah4_3= 62.7 # Assumption 4, in kJ/kg
# Calculations:-
h1= hf + x1*(hg-hf)
ratio= (h1-h2)/(deltah4_3)
QRate= (h2-h1) # Part B
# Results:-
print '-> The rate of the mass flow rate of the cooling water to the mass flow rate of the condenstaing stream is (m3dot/m1dot)',round(ratio,2)
print '-> The rate of energy transfer from the condensing steam to the cooling water of the steam passing through the condenser is',round(QRate,2),'kJ/kg.'
```

In [7]:

```
# Given:-
T1 = 293.0 # In kelvin
P1= 1.01325 * (10**5) # In pascal
V1max= 1.3 # maximum velocity of entering air in m/s
T2max= 305.0 # maximum temperature at the exit in kelvin
pec= -80.0 # power received by electronic components in watt
Pf= -18.0 # Power received by fan in watt
R= 8.314 # Universal gas constant
M= 28.97*(10**(-3)) # Molar mass of air in kg
Qcvdot=0 # Heat transfer from the outer surface of the electronics enclosure to the surroundings is negligible.
Cp= 1.005*(10**3) # in j/kg*k
pi=3.14
# Calculations:-
Wcvdot = pec +Pf # total electric power provided to electronic components and fan in watt
mdotmin= (-Wcvdot)/(Cp*(T2max-T1)) # minimum mass flow rate
v1= ((R/M)*T1)/P1 # specific volume
A1min = (mdotmin*v1)/V1max
D1min = (4*A1min/(pi))**(0.5)
# Results:-
print '-> The smallest fan inlet diameter is',round(D1min*100,2),'cm.'
```

In [8]:

```
# Given:-
P1 = 20.0 # pressure in supply line in bars
P2 = 1.0 # exhaust pressure in bar
T2 = 120.0 # exhaust temperature in degree celcius
# from table A-3 at 20 bars
hf1 = 908.79 # Enthalpy in kj/kg
hg1 = 2799.5 # Enthalpy in kj/kg
# from table A-4, at 1 bar and 120 degree celcius
h2 = 2766.6 # in kj/kg
h1 = h2 # from throttling process assumption
# Calculations:-
x1 = (h1-hf1)/(hg1-hf1)
# Results:-
print '-> The quality of the steam in the supply line is',round(x1,2)
```

In [9]:

```
# Given:-
P1 = 1.0 # pressure of industrial discharge in bar
T1 = 478.0 # temperature of industrial discharge in kelvin
m1dot = 69.78 # mass flow rate of industrial discharge in kg/s
T2 = 400.0 # temperature of exit products from steam generator in kelvin
P2 = 1.0 # pressure of exit products from steam generator in bar
P3 = 0.275 # pressure of water stream entering the generator in Mpa
T3 = 38.9 # temperature of water stream entering the generator in degree celcius
m3dot = 2.079 # mass flow rate of water stream entering in kg/s
P5 = 0.07 # exit pressure of the turbine in bars
x5 = 0.93 # quality of turbine exit
# Part (a)
m2dot = m1dot # since gas and water streams do not mix
m5dot = m3dot # --DO
# from table A-22, A-2 and A-3:-
h1 = 480.3 # in kj/kg
h2 = 400.98 # in Kj/kg
h3 = 162.9 # assumption: h3 = hf(T3), units in Kj/kg
hf5 = 161.0 # in kj/kg
hg5 = 2571.72 # in kj/kg
# Part (b)
P4 = P3 # from the assumption that there is no pressure drop for water flowing through the steam generator
T4 = 180 # in degree celcius
# Calculations:-
h5 = hf5 + x5*(hg5-hf5)
Wcvdot = m1dot*h1 + m3dot*h3 - m2dot*h2 - m5dot*h5
h4 = h3 + (m1dot/m3dot)*(h1 -h2) # from steady state energy rate balance
# interpolating in table A-4, with these P4 and h4
# Results:-
print '-> The power developed by the turbine is ',round(Wcvdot,2),'kJ/s.'
print '-> Turbine inlet temperature is',round(T4,2),'degree celcius.'
```

In [10]:

```
# Given:-
V = 0.85 # volume of tank in m^3
T1 = 260.0 # initial temperature of the tank in degree celcius
X1 = 0.7 # initial quality
# from table A-2
uf1 = 1128.4 # in kg/kg
ug1 = 2599.0 # in kg/kg
vf1 = 1.2755e-3 # in m^3/kg
vg1 = 0.04221 # in m^3/kg
# for final state, from table A-2,
u2 = 2599.0 # units in KJ/kg
v2 = 42.21e-3 # units in m^3/Kg
he = 2796.6 # units in KJ/kg
# Calculations:-
u1 = uf1 + X1*(ug1-uf1) # in kj/kg
v1 = vf1 + X1*(vg1-vf1) # in m^3/kg
m1 = V/v1 # initial mass in kg
m2 = V/v2 # final mass in kg
U2 = m2*u2 # final internal energy in KJ
U1 = m1*u1 # initial internal energy in KJ
Qcv = (U2-U1) - he*(m2-m1)
# Results:-
print '-> The amount of heat transfer is',round(Qcv,2),'KJ.'
```

In [11]:

```
# Given:-
Pv = 15.0 # pressure in the vessel in bar
Tv = 320.0 # temperature in the vessel in degree celcius
Vt = 0.6 # volume of a tank in m^3
Tt = 400.0 # temperature in the tank in degree celcius when the tank is full
# Since the tank is initially empty:-
m1 = 0
u1 = 0
# From table A-4, at 15bar and 400 degree celcius:-
v2 = 0.203 # Volume in m^3/kg
m2 = Vt/v2 # mass within the tank at the end of the process in kg
hi = 3081.9 # in kj/kg
u2 = 2951.3 # in kj/kg
# Calculations:-
deltaUcv = m2*u2-m1*u1
Wcv = hi*(m2-m1)-deltaUcv
# Results:-
print '-> The amount of work developed by the turbine is ',round(Wcv,2),'kJ.'
```