# Chapter 13 - Steam turbines¶

## Example 1: pg 380¶

In [1]:
#pg 380
print('Example 13.1');

#  aim : To determine
#  the power developed for a steam flow of 1 kg/s at the blades and the kinetic energy of the steam finally leaving the wheel

#  Given values
alfa = 20;#  blade angle, [degree]
Cai = 375;# steam exit velocity in the nozzle,[m/s]
U = 165;# blade speed, [m/s]
loss = .15;#  loss of velocity due to friction

#  solution
#  using Fig13.12,
Cvw = 320;# change in velocity of whirl, [m/s]
cae = 132.5;# absolute velocity at exit, [m/s]
Pds = U*Cvw*10**-3;# Power developed for steam flow of 1 kg/s, [kW]
Kes = cae**2/2*10**-3;# Kinetic energy change of steam, [kW/kg]

#results
print ' The power developed for a steam flow of 1 kg/s is (kW) = ',Pds
print ' The energy of steam finally leaving the wheel is (kW/kg) = ',round(Kes,3)

#  End
Example 13.1
The power developed for a steam flow of 1 kg/s is (kW) =  52.8
The energy of steam finally leaving the wheel is (kW/kg) =  8.778

## Example 2: pg 382¶

In [2]:
#pg 382
print('Example 13.2');

# aim : To determine
# (a) the entry angle of  the blades
# (b) the work done per kilogram of steam per second
# (c) the diagram efficiency
# (d) the end-thrust per kilogram of steam per second

# given values
Cai = 600.;# steam velocity, [m/s]
sia = 25.;# steam inlet angle with blade, [degree]
U = 255.;# mean blade speed, [m/s]
sea = 30.;# steam exit angle with blade,[degree]

# solution
# (a)
# using Fig.13.13(diagram for example 13.2)
eab = 41.5;# entry angle of blades, [degree]
print ' (a) The angle of blades is (degree) = ',eab

# (b)
Cwi_plus_Cwe = 590;# velocity of whirl, [m/s]
W = U*(Cwi_plus_Cwe);# work done on the blade,[W/kg]
print ' (b) The work done on the blade is (kW/kg) = ',W*10**-3

# (c)
De = 2*U*(Cwi_plus_Cwe)/Cai**2;# diagram efficiency
print ' (c) The diagram efficiency is (percent) = ',round(De*100,1)

# (d)
# again from the diagram
Cfe_minus_Cfi = -90;# change invelocity of flow, [m/s]
Eth = Cfe_minus_Cfi;# end-thrust, [N/kg s]
print ' (d) The End-thrust is (N/kg) = ',Eth

#  End
Example 13.2
(a) The angle of blades is (degree) =  41.5
(b) The work done on the blade is (kW/kg) =  150.45
(c) The diagram efficiency is (percent) =  83.6
(d) The End-thrust is (N/kg) =  -90

## Example 3: pg 384¶

In [3]:
#pg 384
print('Example 13.3');

# aim : To determine
# (a) the power output of the turbine
# (b) the diagram efficiency

# given values
U = 150.;# mean blade speed, [m/s]
Cai1 = 675.;# nozzle speed, [m/s]
na = 20.;# nozzle angle, [degree]
m_dot = 4.5;# steam flow rate, [kg/s]

# solution
# from Fig. 13.15(diagram 13.3)
Cw1 = 915.;# [m/s]
Cw2 = 280.;# [m/s]

# (a)
P = m_dot*U*(Cw1+Cw2);# power of turbine,[W]
print ' (a) The power of turbine is (kW) = ',P*10**-3

# (b)
De = 2*U*(Cw1+Cw2)/Cai1**2;# diagram efficiency
print ' (b) The diagram efficiency is (percent) = ',round(De*100,1)

#  End
Example 13.3
(a) The power of turbine is (kW) =  806.625
(b) The diagram efficiency is (percent) =  78.7

## Example 4: pg 386¶

In [4]:
#pg 386
print('Example 13.4');
import math
# aim : To determine
# (a) the power output of the stage
# (b) the specific enthalpy drop in the stage
# (c) the percentage increase in relative velocity in the moving blades due to expansion in the bladse

# given values
N = 50.;# speed, [m/s]
d = 1.;# blade ring diameter, [m]
nai = 50.;# nozzle inlet angle, [degree]
nae = 30.;# nozzle exit angle, [degree]
m_dot = 600000.;# steam flow rate, [kg/h]
se = .85;# stage efficiency

# solution
# (a)
U = math.pi*d*N;# mean blade speed, [m/s]
# from Fig. 13.17(diagram 13.4)
Cwi_plus_Cwe = 444;# change in whirl speed, [m/s]
P = m_dot*U*Cwi_plus_Cwe/3600;# power output of the stage, [W]
print ' (a) The power output of the stage is (MW) = ',round(P*10**-6,1)

# (b)
h = U*Cwi_plus_Cwe/se;# specific enthalpy,[J/kg]
print ' (b) The specific enthalpy drop in the stage is (kJ/kg) = ',round(h*10**-3)

# (c)
# again from diagram
Cri = 224.;# [m/s]
Cre = 341;# [m/s]
Iir = (Cre-Cri)/Cri;# increase in relative velocity
print ' (c) The increase in relative velocity is (percent) = ',round(Iir*100,1)

#  End
Example 13.4
(a) The power output of the stage is (MW) =  11.6
(b) The specific enthalpy drop in the stage is (kJ/kg) =  82.0
(c) The increase in relative velocity is (percent) =  52.2

## Example 5: pg 389¶

In [5]:
#pg 389
print('Example 13.5');

# aim : To determine
# (a) the blade height of the stage
# (b) the power developed in the stage
# (c) the specific enthalpy drop at the stage
from math import sqrt,pi
# given values
U = 60.;# mean blade speed, [m/s]
P = 350.;# steam pressure, [kN/m**2]
T = 175.;# steam temperature, [C]
nai = 30.;# stage inlet angle, [degree]
nae = 20.;# stage exit angle, [degree]

# solution
# (a)
m_dot = 13.5;# steam flow rate, [kg/s]
# at given T and P
v = .589;# specific volume, [m**3/kg]
# given H=d/10, so
H = sqrt(m_dot*v/(pi*10*60));# blade height, [m]
print ' (a) The blade height at this stage is (mm) = ',round(H*10**3)

# (b)
Cwi_plus_Cwe = 270;# change in whirl speed, [m/s]
P = m_dot*U*(Cwi_plus_Cwe);# power developed, [W]
print ' (b) The power developed is (kW) = ',P*10**-3

# (c)
s = .85;# stage efficiency
h = U*Cwi_plus_Cwe/s;# specific enthalpy,[J/kg]
print ' (a) The specific enthalpy drop in the stage is (kJ/kg) = ',round(h*10**-3,3)

#  End
Example 13.5
(a) The blade height at this stage is (mm) =  65.0
(b) The power developed is (kW) =  218.7
(a) The specific enthalpy drop in the stage is (kJ/kg) =  19.059