# Chapter 4 : Steam nozzles and Steam turbines¶

## Example 4.1 Page no : 161¶

In [1]:
# Variables
P1 = 3.5;			#Pressure at entry in MN/(m**2)
T1 = 773.;			#Temperature at entry in K
P2 = 0.7;			#Pressure at exit in MN/(m**2)
ma = 1.3;			#mass flow rate of air in kg/s
y = 1.4;			#Ratio of specific heats
R = 0.287;			#Universal gas constant in KJ/Kg-K

# Calculations
c = y/(y-1);        			#Ratio
Pt = ((2/(y+1))**c)*P1;			#Throat pressure in MN/(m**2)
v1 = (R*T1)/(P1*1000);			#Specific volume at entry in (m**3)/kg
Ct = ((2*c*P1*v1*(1-((Pt/P1)**(1/c))))**0.5)*1000;			#Velocity at throat in m/s
vt = v1*((P1/Pt)**(1/y));			#Specific volume at throat in (m**3)/kg
At = ((ma*vt)/Ct)*(10**6);			#Area of throat in (mm**2)
C2 = ((2*c*P1*v1*(1-((P2/P1)**(1/c))))**0.5)*1000;			#Velocity at exit in m/s
v2 = v1*((P1/P2)**(1/y));			#Specific volume at exit in (m**3)/kg
A2 = ((ma*v2)/C2)*(10**6);			#Area of exit in (mm**2)
M = C2/Ct;			            #Mach number at exit

# Results
print 'Throat area is %3.0f mm**2  \
\nExit area is %3.0f mm**2  \
\nMach number at exit is %3.2f'%(At,A2,M)

# rounding off error

Throat area is 255 mm**2
Exit area is 344 mm**2
Mach number at exit is 1.49


## Example 4.2 Page no : 163¶

In [2]:
# Variables
T1 = 273.;			#Temperature at section 1 in K
P1 = 140.;			#Pressure at section 1 in KN/(m**2)
v1 = 900.;			#Velocity at section 1 in m/s
v2 = 300.;			#Velocity at section 2 in m/s
Cp = 1.006;			#Specific heat at constant pressure in kJ/kg-K
Cv = 0.717;			#Specific heat at constant volume in kJ/kg-K
y = 1.4;			#Ratio of specific heats

# Calculations
c = y/(y-1);			#Ratio
R = Cp-Cv;			#Universal gas constant in KJ/Kg-K
T2 = T1-(((v2)**2-(v1)**2)/(2000*c*R));			#Temperature at section 2 in K
DT = T2-T1;			#Increase in temperature in K
P2 = P1*((T2/T1)**c);			#Pressure at section 2 in KN/(m**2)
DP = (P2-P1)/1000;			#Increase in pressure in MN/(m**2)
IE = Cv*(T2-T1);			#Increase in internal energy in kJ/kg

# Results
print 'Increase in temperature is %3.0f K  \
\nIncrease in pressure is %3.2f MN/m**2  \
\nIncrease in internal energy is %3.0f kJ/kg'%(DT,DP,IE)

Increase in temperature is 356 K
Increase in pressure is 2.46 MN/m**2
Increase in internal energy is 255 kJ/kg


## Example 4.3 Page no : 163¶

In [2]:
# Variables
P1 = 2;			#Pressure at entry in MN/(m**2)
T1 = 598;			#Temperature at entry in K
P2 = 0.36;			#Pressure at exit in MN/(m**2)
m = 7.5;			#mass flow rate of steam in kg/s
n = 1.3;			#Adiabatic gas constant
v1 = 0.132;			#Volume at entry in (m**3)/kg from steam table
Ts = 412.9;			#Saturation temperature in K

# Calculations
c = n/(n-1);			#Ratio
Pt = ((2/(n+1))**c)*P1;			#Throat pressure in MN/(m**2)
Ct = ((2*c*P1*v1*(1-((Pt/P1)**(1/c))))**0.5)*1000;			#Velocity at throat in m/s
vt = v1*((P1/Pt)**(1/n));			#Specific volume at throat in (m**3)/kg
At = ((m*vt)/Ct)*(10**6);			#Area of throat in (mm**2)
C2 = ((2*c*P1*v1*(1-((P2/P1)**(1/c))))**0.5)*1000;			#Velocity at exit in m/s
v2 = v1*((P1/P2)**(1/n));			#Specific volume at exit in (m**3)/kg
A2 = ((m*v2)/C2)*(10**6);			#Area of exit in (mm**2)
T2 = T1*((P2/P1)**(1/c));			#Temperature at exit in K
D = Ts-T2;			#Degree of undercooling at exit in K

# Results
print 'Throat area is %3.0f mm**2  \
\nExit area is %3.0f mm**2  \
\nDegree of undercooling at exit is %3.1f K'%(At,round(A2,-1),D)

Throat area is 2888 mm**2
Exit area is 4280 mm**2
Degree of undercooling at exit is 10.3 K


## Example 4.4 Page no : 165¶

In [3]:
# Variables
P1 = 2.2;			#Pressure at entry in MN/(m**2)
T1 = 533.;			#Temperature at entry in K
P2 = 0.4;			#Pressure at exit in MN/(m**2)
m = 11.;			#mass flow rate of steam in kg/s
n = 0.85;			#Efficiency of expansion
h1 = 2940.;			#Enthalpy at entrance in kJ/kg from Moiller chart
ht = 2790.;			#Enthalpy at throat in kJ/kg from Moiller chart
h2s = 2590.;			#Enthalpy below exit level in kJ/kg from Moiller chart
vt = 0.16;			#Throat volume in (m**3)/kg
v2 = 0.44;			#Volume at exit in (m**3)/kg

# Calculations
Ct = (2000*(h1-ht))**0.5;			#Throat velocity in m/s
h2 = ht-(0.85*(ht-h2s));			#Enthalpy at exit in kJ/kg
C2 = (2000*(h1-h2))**0.5;			#Exit velocity in m/s
At = ((m*vt)/Ct)*(10**6);			#Area of throat in (mm**2)
A2 = ((m*v2)/C2)*(10**6);			#Area of exit in (mm**2)

# Results
print 'Throat velocity is %3.0f m/s  \
\nExit velocity is %3.0f m/s  \
\nThroat area is %3.0f mm**2  \
\nExit area is %3.0f mm**2 '%(Ct,C2,round(At,-1),A2)

Throat velocity is 548 m/s
Exit velocity is 800 m/s
Throat area is 3210 mm**2
Exit area is 6050 mm**2


## Example 4.5 Page no : 166¶

In [5]:
# Variables
P1 = 35.;			#Pressure at entry in bar
T1 = 573.;			#Temperature at entry in K
P2 = 8.;			#Pressure at exit in bar
Ts = 443.4;			#Saturation temperature in K
Ps = 3.1;			#Saturation pressure in bar
m = 5.2;			#mass flow rate of steam in kg/s
n = 1.3;			#Adiabatic gas consmath.tant
v1 = 0.06842;			#Specific volume at entry in (m**3)/kg from steam table
v3 = 0.2292;			#Specific volume at exit in (m**3)/kg from steam table
h1 = 2979.;			#Enthalpy in kJ/kg from Moiller chart
h3 = 2673.3;			#Enthalpy in kJ/kg from Moiller chart

# Calculations
c = n/(n-1);			#Ratio
C2 = ((2*c*P1*(10**5)*v1*(1-((P2/P1)**(1/c))))**0.5);			#Velocity at exit in m/s
v2 = v1*((P1/P2)**(1/n));			#Specific volume at exit in (m**3)/kg
A2 = ((m*v2)/C2)*(10**4);			#Area of exit in (cm**2)
a = ((A2/18)**0.5)*10;			#Length in mm
b = 3*a;			#Breadth in mm
T2 = T1*((P2/P1)**(1/c));			#Temperature at exit in K
D = Ts-T2;			#Degree of undercooling in K
Ds = P2/Ps;			#Degree of supersaturation
hI = h1-h3;			#Isentropic enthalpy drop in kJ/kg
ha = (C2**2)/2000;			#Actual enthalpy drop in kJ/kg
QL = hI-ha;			#Loss in available heat in kJ/kg
DS = QL/Ts;			#Increase in entropy in kJ/kg-K
C3 = (2000*(h1-h3))**0.5;			#Exit velocity from nozzle
mf = ((A2*C3*(10**-4))/v3);			#Mass flow rate in kg/s
Rm = m/mf;			#Ratio of mass rate

# Results
print 'Cross section of nozzle is %3.1f mm * %3.1f mm  \
\nDegree of undercooling is %3.1f K and Degree of supersaturation is %3.2f  \
\nLoss in available heat drop due to irreversibility is %3.2f kJ/kg  \
\nIncrease in entropy is %3.5f kJ/kg-K  \
\nRatio of mass flow rate with metastable expansion to the thermal expansion is %3.3f'%(b,a,D,Ds,QL,DS,Rm)

Cross section of nozzle is 26.7 mm * 8.9 mm
Degree of undercooling is 35.8 K and Degree of supersaturation is 2.58
Loss in available heat drop due to irreversibility is 6.16 kJ/kg
Increase in entropy is 0.01390 kJ/kg-K
Ratio of mass flow rate with metastable expansion to the thermal expansion is 1.065


## Example 4.6 Page no : 169¶

In [1]:
import math

# Variables
m = 14.;			#Mass flow rate of steam in kg/s
P1 = 3.;			#Pressure of Steam in MN/(m**2)
T1 = 300.;			#Steam temperature in oC
h1 = 2990.;			#Enthalpy at point 1 in kJ/kg
h2s = 2630.;			#Enthalpy at point 2s in kJ/kg
ht = 2850.;			#Enthalpy at point t in kJ/kg
n = 1.3;			#Adiabatic gas consmath.tant
C2 = 800.;			#Exit velocity in m/s
v2 = 0.4;			#Specific volume at exit in (m**3)/kg

# Calculations
x = n/(n-1);			#Ratio
Pt = ((2/(n+1))**x)*P1;			#Temperature at point t in MN/(m**2)
h2 = h1-((C2**2)/2000);			#Exit enthalpy in kJ/kg
nN = ((h1-h2)/(h1-h2s))*100;			#Nozzle efficiency
A2 = ((m*v2)/C2)*(10**6);			#Exit area in (mm**2)
Ct = math.sqrt(2*(h1-ht)*10**3);			#Throat velocity in m/s

# Results
print 'Nozzle efficiency is %3.1f percent  \
\nExit area is %3.0f mm**2  \
\nThroat velocity is %3.0f m/s'%(nN,A2,Ct)

Nozzle efficiency is 88.9 percent
Exit area is 7000 mm**2
Throat velocity is 529 m/s


## Example 4.7 Page no : 170¶

In [6]:
import math

# Variables
P1 = 10.;			#Pressure at point 1 in bar
P2 = 0.5;			#Pressure at point 2 in bar
h1 = 3050.;			#Enthalpy at point 1 in kJ/kg
h2s = 2480.;			#Enthalpy at point 2s in kJ/kg
ht = 2910.;			#Enthalpy at throat in kJ/kg
n = 1.3;			#Adiabatic gas constant
r = 0.1;			#Total available heat drop
v1 = 0.258;			#Specific volume at point 1 in (m**3)/kg
h2f = 340.6;			#Enthalpy for exit pressure from steam tables in kJ/kg
hfg = 2305.4;			#Enthalpy for exit pressure from steam tables in kJ/kg
m = 0.5;			#Mass flow rate in kg/s

# Calculations
x = n/(n-1);			#Ratio
Pt = ((2/(n+1))**x)*P1;			#Temperature at throat in bar
h2 = h2s+(r*(h1-h2s));			#Enthalpy at point 2 in kJ/kg
vt = ((P1/Pt)**(1/n))*v1;			#Specific volume at throat in (m**3)/kg
v2 = ((P1/P2)**(1/n))*v1;			#Specific volume at point 2 in (m**3)/kg
Ct = math.sqrt(2000*(h1-ht));			#Throat velocity in m/s
At = ((m*vt)/Ct)*(10**6);			#Throat area in (mm**2)
C2 = math.sqrt(2000*(h1-h2));			#Exit velocity in m/s
A2 = ((m*v2)/C2)*(10**6);			#Exit area in (mm**2)
x2 = ((h2-h2f)/hfg)*100;			#Steam quality at exit

# Results
print 'Throat area is %d mm**2  \
\nExit area is %d mm**2  \
\nSteam quality at exit is %3.0f percent'%(At,A2,x2)

# rounding off error

Throat area is 388 mm**2
Exit area is 1275 mm**2
Steam quality at exit is  95 percent


## Example 4.8 Page no : 171¶

In [7]:
import math

# Variables
P1 = 3.5;			#Dry saturated steam in bar
P2 = 1.1;			#Exit pressure in bar
At = 4.4;			#Throat area in cm**2
h1 = 2731.6;			#Enthalpy at P1 in kJ/kg
v1 = 0.52397;			#Specific volume at P1 in m**3/kg
n = 1.135;			#Adiabatic gas constant
ht = 2640.;			#Enthalpy at Pt in kJ/kg
vt = 0.85;			#Specific volume at throat in m**3/kg
h2 = 2520.;			#Enthalpy at P2 in kJ/kg
v2 = 1.45;			#Specific volume at P2 in m**3/kg

# Calculations
x = n/(n-1);			#Ratio
Pt = ((2/(n+1))**x)*P1;			#Throat pressure in bar
Ct = math.sqrt(2000*(h1-ht));			#Throat velocity in m/s
mmax = ((At*Ct*(10**-4))/vt)*60;			#Maximum discharge in kg/min
C2 = math.sqrt(2000*(h1-h2));			#Exit velocity in m/s
A2 = ((mmax*v2)/(C2*60))*(10**6);			#Exit area in mm**2

# Results
print 'Maximum discharge is %3.3f kg/min  \
\nExit area is %3.1f mm**2'%(mmax,A2)

Maximum discharge is 13.294 kg/min
Exit area is 493.8 mm**2


## Example 4.9 Page no : 172¶

In [5]:
import math

# Variables
P1 = 10.;			#Pressure at point 1 in bar
T1 = 200.;			#Temperature at point 1 in oC
P2 = 5.;			#Pressure at point 2 in bar
n = 1.3;			#Adiabatic gas consmath.tant
h1 = 2830.;			#Enthalpy at P1 in kJ/kg
ht = 2710.;			#Enthalpy at point Pt in kJ/kg
vt = 0.35;			#Specific volume at Pt in m**3/kg
m = 3.  			#Nozzle flow in kg/s

# Calculations
x = n/(n-1);			#Ratio
Pt = ((2/(n+1))**x)*P1;			#Throat pressure in bar
Ct = math.sqrt(2000*(h1-ht));			#Throat velocity in m/s
At = (m*vt)/Ct;			#Throat area in m**2

# Results
print 'Since throat pressure is greater than exit pressure,nozzle used is\
convergent-divergent nozzle  \
\nMinimum area of nozzle required is %.2e m**2'%(At)

Since throat pressure is greater than exit pressure,nozzle used is convergent-divergent nozzle
Minimum area of nozzle required is 2.14e-03 m**2


## Example 4.10 Page no : 173¶

In [1]:
import math

# Variables
P1 = 10.5;			#Pressure at point 1 in bar
x1 = 0.95;			#Dryness fraction
n = 1.135;			#Adiabatic gas constant
P2 = 0.85;			#Pressure at point 2 in bar
vg = 0.185;			#Specific volume in m**3/kg

# Calculations
c = n/(n-1);			#Ratio
Pt = round(((2/(n+1))**c)*P1,2);			#Throat pressure in MN/(m**2)
v1 = round(x1*vg,3);			#Specific volume at point 1 in m**3/kg
Ct = round(math.sqrt((2*n*P1*v1*(10**5)/(n+1))),2);			#Velocity at throat in m/s
vt = round(((P1/Pt)*(v1**n))**(1/1.135),3);			#Specific volume at throat in m**3/kg
m = Ct/vt;			#Mass flow rate per unit throat area in kg/(m**2)

# Results
print 'Throat velocity is %3.2f m/s  \
\nMass flow rate of steam is %3.2f kg/m**2'%(Ct,m)

Throat velocity is 443.27 m/s
Mass flow rate of steam is 1549.90 kg/m**2


## Example 4.11 Page no : 174¶

In [2]:
# Variables
P1 = 10.;			#Pressure at point 1 in bar
T1 = 452.9;			#Temperature at point 1 in K
P2 = 4.;			#Pressure at point 2 in bar
n = 1.3;			#Adiabatic gas constant
Ps = 0.803;			#Saturation pressure at T2 in bar
Ts = 143.6;			#Saturation temperature at P2 in oC
# Calculations
x = (n-1)/n;			#Ratio
T2 = ((P2/P1)**x)*T1;			#Temperature at point 2 in K
Ds = P2/Ps;			#Degree of supersaturation
Du = Ts-(T2-273);			#Degree of undercooling

# Results
print 'Degree of supersaturation is %3.2f  \
\nDegree of undercooling %3.0f C'%(Ds,Du)

Degree of supersaturation is 4.98
Degree of undercooling  50 C


## Example 4.12 Page no : 174¶

In [14]:
import math

# Variables
P1 = 9.;			#Pressure at point 1 in bar
P2 = 1.;			#Pressure at point 2 in bar
Dt = 0.0025;			#Throat diameter in m
nN = 0.9;			#Nozzle efficiency
n = 1.135;			#Adiabatic gas constant
h1 = 2770.;			#Enthalpy at point 1 in kJ/kg
ht = 2670.;			#Throat enthlapy in kJ/kg
h3 = 2400.;			#Enthlapy at point 2 in kJ/kg
x2 = 0.96;			#Dryness fraction 2
vg2 = 0.361;			#Specific volume in m**3/kg

# Calculations
x = n/(n-1);			#Ratio
Pt = ((2/(n+1))**x)*P1;			#Throat pressure in bar
Ct = math.sqrt(2000*(h1-ht)*nN);			#Throat velocity in m/s
At = (3.147*2*(Dt**2))/4;			#Throat area in m**2
vt = x2*vg2;			#Specific volume at throat in m**3/kg
m = (At*Ct)/vt;			#Mass flow rate of steam in kg/s
hact = nN*(h1-h3);			#Actual enthalpy drop in kJ/kg
C2 = math.sqrt(2000*hact);			#Exit velocity of steam in m/s

# Results
print 'Quantity of steam used per second is %3.3f kg/s  \
\nExit velocity of steam is %3.2f m/s'%(m,C2)

Quantity of steam used per second is 0.012 kg/s
Exit velocity of steam is 816.09 m/s


## Example 4.13 Page no : 202¶

In [15]:
# Variables
C1 = 1000.;			#Steam velocity in m/s
a1 = 20.;			#Nozzle angle in degrees
U = 400.;			#Mean blade speed in m/s
m = 0.75;			#Mass flow rate of steam in kg/s
b1 = 33.;			#Blade angle at inlet from the velocity triangle in degrees
b2 = b1;			#Blade angle at exit from the velocity triangle in degrees
Cx = 1120.;			#Change in whirl velocity from the velocity triangle in m/s
Ca = 0;		    	#Change in axial velocity from the velocity triangle in m/s

# Calculations
Fx = m*Cx;		    	    #Tangential force on blades in N
Fy = m*Ca;			        #Axial thrust in N
W = (m*Cx*U)/1000;			#Diagram power in kW
ndia = ((2*U*Cx)/(C1**2))*100;			#Diagram efficiency

# Results
print 'Blade angles are %3.0f degrees, %3.0f degrees  \
\nTangential force on blades is %3.0f N  \
\nAxial thrust is %3.0f  \
\nDiagram power is %3.0f kW  \
\nDiagram efficiency %3.1f percent'%(b1,b2,Fx,Fy,W,ndia)

Blade angles are  33 degrees,  33 degrees
Tangential force on blades is 840 N
Axial thrust is   0
Diagram power is 336 kW
Diagram efficiency 89.6 percent


## Example 4.14 Page no : 203¶

In [14]:
# Variables
D = 2.5;			#Mean diameter of blade ring in m
N = 3000.;			#Speed in rpm
a1 = 20.;			#Nozzle angle in degrees
r = 0.4;			#Ratio blade velocity to steam velocity
Wr = 0.8;			#Blade friction factor
m = 10.;			#Steam flow in kg/s
x = 3.;	    		#Sum in blade angles in degrees
b1 = 32.5;			#Blade angle at inlet from the velocity triangle in degrees
W1 = 626.7;			#Relative velocity at inlet from the velocity triangle in m/s
Cx = 967.;			#Change in whirl velocity from the velocity triangle in m/s

# Calculations
U = (3.147*D*N)/60;			#Blade velocity in m/s
C1 = U/r;			#Steam velocity in m/s
b2 = b1-x;			#Blade angle at exit in degrees
W2 = Wr*W1;			#Relative velocity at outlet from the velocity triangle in m/s
W = (m*Cx*U)/1000;			#Power developed in kW
sc = (m*3600)/W;			#Steam consumption in kg/kWh

# Results
print 'Power developed is %3.0f kW  \
\nBlade efficiency is %3.1f percent  \
\nSteam consumed is %3.2f kg/kWh'%(round(W,-1),ndia,sc)

Power developed is 3800 kW
Steam consumed is 9.46 kg/kWh


## Example 4.15 Page no : 204¶

In [17]:
# Variables
m = 3.;	    		#Mass flow rate of steam in kg/s
C1 = 425.;			#Steam velocity in m/s
r = 0.4;			#Ratio of blade speed to jet speed
W = 170.;			#Stage output in kW
IL = 15.;			#Internal losses in kW
a1 = 16.;			#Nozzle angle in degrees
b2 = 17.;			#Blade angle at exit in degrees
W1 = 265.;			#Relative velocity at inlet from the velocity triangle in m/s
W2 = 130.;			#Relative velocity at outlet from the velocity triangle in m/s

# Calculations
U = C1*r;			#Blade speed in m/s
P = (W+IL)*1000;			#Total power developed in W
Cx = P/(m*W);			#Change in whirl velocity in m/s
Wr = W2/W1;			#Blade velocity co-efficient

# Results
print 'Blading efficiency is %3.1f percent  \

Blading efficiency is 68.3 percent


## Example 4.16 Page no : 205¶

In [18]:
# Variables
C1 = 375.;			#Steam velocity in m/s
a1 = 20.;			#Nozzle angle
U = 165.;			#Blade speed in m/s
m = 1.;			#Mass flow rate of steam in kg/s
Wr = 0.85;			#Blade friction factor
Ca1 = 130.;			#Axial velocity at inlet from the velocity triangle in m/s
Ca2 = Ca1;			#Axial velocity at outlet in m/s
W1 = 230.;			#Relative velocity at inlet from the velocity triangle in m/s
Cx = 320.;			#Change in whirl velocity from the velocity triangle in m/s

# Calculations
b2 = 41;			#Blade angle at exit from the velocity triangle in degrees
b1 = 34;			#Blade angle at exit from the velocity triangle in degrees
W = (m*Cx*U)/1000;			#Power developed by turbine in kW

# Results
print 'Blade angles assumed are %3.0f degrees, %3.0f degrees  \
\nPower developed by turbine is %3.1f kW'%(b1,b2,W)

Blade angles assumed are  34 degrees,  41 degrees
Power developed by turbine is 52.8 kW


## Example 4.17 Page no : 206¶

In [19]:
# Variables
m = 2.;			#Mass flow rate of steam in kg/s
W = 130.;			#Turbine power in kW
U = 175.;			#Blade velocity in m/s
C1 = 400.;			#Steam velocity in m/s
Wr = 0.9;			#Blade friction factor
W1 = 240.;			#Realtive velocity at inlet from the velocity triangle in m/s

# Calculations
Cx1 = (W*1000)/(m*U);			#Whirl velocity at inlet in m/s
W2 = Wr*W1;			#Realtive velocity at outlet from the velocity triangle in m/s
a1 = 19;			#Nozzle angle from the velocity triangle in degrees
b1 = 33;			#Blade angle at inlet from the velocity triangle in degrees
b2 = 36;			#Blade angle at outlet from the velocity triangle in degrees

# Results
print 'Nozzle angle is %3.0f degrees  \
\nBlade angles are %3.0f degrees, %3.0f degrees'%(a1,b1,b2)

Nozzle angle is  19 degrees
Blade angles are  33 degrees,  36 degrees


## Example 4.18 Page no : 207¶

In [20]:
# find Diagram efficiency

# Variables
U = 150.;			#Blade speed in m/s
m = 3.;			#Mass flow rate of steam in kg/s
P = 10.5;			#Pressure in bar
r = 0.21;			#Ratio blade velocity to steam velocity
a1 = 16.;			#Nozzle angle in first stage in degrees
b2 = 20.;			#Blade angle at exit in first stage in degrees
a3 = 24.;			#Nozzle angle in second stage in degrees
b4 = 32.;			#Blade angle at exit in second stage in degrees
Wr = 0.79;			#Blade friction factor for first stage
Wr2 = 0.88;			#Blade friction factor for second stage
Cr = 0.83;			#Blade velocity coefficient
W1 = 570.;			#Relative velocity at inlet from the velocity triangle for first stage in m/s
C2 = 375.;			#Velocity in m/s
W3 = 185.;			#Relative velocity at inlet from the velocity triangle for second stage in m/s

# Calculations
C1 = U/r;			#Steam speed at exit in m/s
W2 = Wr*W1;			#Relative velocity at outlet for first stage in m/s
C3 = Cr*C2;			#Steam velocity at inlet for second stage in m/s
W4 = Wr2*W3;			#Relative velocity at exit for second stage in m/s
DW1 = W1+W2;			#Change in relative velocity for first stage in m/s
DW2 = 275;			#Change in relative velocity from the velocity triangle for second stage in m/s
ndia = ((2*U*(DW1+DW2))/(C1**2))*100;			#Diagram efficiency

# Results
print 'Diagram efficiency is %3.1f percent'%(ndia)

Diagram efficiency is 76.2 percent


## Example 4.19 Page no : 208¶

In [21]:
import math
# Variables
b1 = 30.;			#Blade angle at inlet in first stage in degrees
b2 = 30.;			#Blade angle at exit in first stage in degrees
b3 = 30.;			#Blade angle at inlet in second stage in degrees
b4 = 30.;			#Blade angle at exit in second stage in degrees
t1 = 240.;			#Temperature at entry in oC
P1 = 11.5;			#Pressure at entry in bar
P2 = 5.;			#Pressure in wheel chamber in bar
vl = 10.;			#Loss in velocity in percent
h = 155.;			#Enthalpy at P2 in kJ/kg
W4 = 17.3;			#Relative velocity at exit from the velocity triangle for second stage in m/s
a4 = 90.;			#Nozzle angle in second stage in degrees
C3 = 33.;			#Steam velocity at inlet from the velocity triangle for second stage in m/s
W2 = 49.;			#Relative velocity at outlet from the velocity triangle for first stage in m/s
x = 15.;			#Length of AB assumed for drawing velocity triangle in mm
y = 67.;			#Length of BC from the velocity triangle in mm

# Calculations
C1 = math.sqrt(2000*h);			#Velocity of steam in m/s
W3 = W4/0.9;			#Relative velocity at inlet for second stage in m/s
C2 = C3/0.9;			#Velocity in m/s
W1 = W2/0.9;			#Relative velocity at inlet for first stage in m/s
C1n = C1/y;			#Velocity of steam in m/s
U = x*C1n;			#Blade speed in m/s
a3 = 17.;			#Nozzle angle in second stage from the velocity triangle in degrees
a2 = 43.;			#Nozzle angle from the velocity triangle in degrees
DW1 = 731.5;			#Change in relative velocity from the velocity triangle for first stage in m/s
DW2 = 257.5;			#Change in relative velocity from the velocity triangle for second stage in m/s
ndia = ((2*U*(DW1+DW2))/(C1**2))*100;			#Diagram efficiency

# Results
print 'Blade speed is %3.1f m/s  \
\nBlade tip angles of the fixed blade are %3.0f degrees and %3.0f degrees  \
\nDiagram efficiency is %3.1f percent'%(U,a3,a2,ndia)

Blade speed is 124.7 m/s
Blade tip angles of the fixed blade are  17 degrees and  43 degrees
Diagram efficiency is 79.5 percent


## Example 4.20 Page no : 210¶

In [17]:
# Variables
C1 = 600.;			#Steam velocity in m/s
b1 = 30.;			#Blade angle at inlet in first stage in degrees
b2 = 30.;			#Blade angle at exit in first stage in degrees
b3 = 30.;			#Blade angle at inlet in second stage in degrees
b4 = 30.;			#Blade angle at exit in second stage in degrees
a4 = 90.;			#Nozzle angle in second stage in degrees
m = 3.;			#Mass of steam in kg/s
x = 15.;			#Length for drawing velocity triangle in mm
y = 56.;			#Length of BC from the velocity triangle in mm

# Calculations
C1n = round(C1/y,1);			#Velocity of steam in m/s
U = round(x*C1n,1);			#Blade speed in m/s
l = 103.;			#Length from velocity triangle in mm
P = (m*l*C1n*U)/1000;			#Power developed in kW

# Results
print 'Blade speed is %3.1f m/s  \
\nPower developed by the turbine is %3.2f kW'%(U,P)

Blade speed is 160.5 m/s
Power developed by the turbine is 530.66 kW


## Example 4.21 Page no : 211¶

In [18]:
import math
# Variables
N = 400.;			#Speed in rpm
m = 8.33;			#Mass of steam in kg/s
P = 1.6;			#Pressure of steam in bar
x = 0.9;			#Dryness fraction
W = 10.;			#Stage power in kW
r = 0.75;			#Ratio of axial flow velocity to blade velocity
a1 = 20.;			#Nozzle angle at inlet in degrees
a2 = 35.;			#Nozzle angle at exit in degrees
b1 = a2;			#Blade tip angle at exit in degrees
b2 = a1;			#Blade tip angle at inlet in degrees
a = 25.;			#Length of AB from velocity triangle in mm
vg = 1.091;			#Specific volume of steam from steam tables in (m**3)/kg

# Calculations
Cx = 73.5;			#Change in whirl velocity from the velocity triangle by measurement in mm
y = Cx/a;			#Ratio of change in whirl velocity to blade speed
U = math.sqrt((W*1000)/(m*y));			#Blade speed in m/s
D = ((U*60)/(3.147*N))*1000;			#Mean diameter of drum in mm
v = m*x*vg;			#Volume flow rate of steam in (m**3)/s

# Results
print 'Mean diameter of drum is %3.0f mm  \
\nVolume of steam flowing per second is %3.2f m**3/s'%(D,v)

# rounding off error

Mean diameter of drum is 963 mm
Volume of steam flowing per second is 8.18 m**3/s


## Example 4.22 Page no : 212¶

In [24]:
import math

# Variables
N = 300.;			#Speed in rpm
m = 4.28;			#Mass of steam in kg/s
P = 1.9;			#Pressure of steam in bar
x = 0.93;			#Dryness fraction
W = 3.5;			#Stage power in kW
r = 0.72;			#Ratio of axial flow velocity to blade velocity
a1 = 20.;			#Nozzle angle at inlet in degrees
b2 = a1;			#Blade tip angle at inlet in degrees
l = 0.08;			#Tip leakage steam
vg = 0.929;			#Specific volume of steam from steam tables in (m**3)/kg

# Calculations
mact = m-(m*l);			#Actual mass of steam in kg/s
a = (3.147*N)/60;			#Ratio of blade velocity to mean dia
b = r*a;			#Ratio of axial velocity to mean dia
c = 46;			#Ratio of change in whirl velocity to mean dia
D = math.sqrt((W*1000)/(mact*c*a));			#Mean dia in m
Ca = b*D;			#Axial velocity in m/s
h = ((mact*x*vg)/(3.147*D*Ca))*1000;			#Blade height in mm
D1 = D-(h/1000);			#Drum dia in m

# Results
print 'Drum diameter is %3.3f m  \

Drum diameter is 1.030 m


## Example 4.23 Page no : 214¶

In [27]:
import math
# Variables
P0 = 800.;			#Steam pressure in kPa
P2 = 100.;			#Pressure at point 2 in kPa
T0 = 973.;			#Steam temperature in K
a1 = 73.;			#Nozzle angle in degrees
ns = 0.9;			#Steam efficiency
m = 35.;			#Mass flow rate in kg/s
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K
y = 1.4;			#Ratio of specific heats

# Calculations
b1 = math.degrees(math.atan(tanb1))
b2 = b1;			#Blade angle at exit in degrees
Dh = ns*Cp*T0*(1-((P2/P0)**((y-1)/y)));			#Difference in enthalpies in kJ/kg
W = (m*Dh)/1000;			#Power developed in MW

# Results
print 'Rotor blade angles are %3.2f degrees and %3.2f degrees  \
\nFlow coefficient is %3.3f  \
\nPower developed is %3.1f MW'%(b1,b2,p,s,W)


Rotor blade angles are 58.56 degrees and 58.56 degrees
Flow coefficient is 0.611
Power developed is 13.8 MW


## Example 4.24 Page no : 215¶

In [10]:
import math

# Variables
P0 = 100.;			#Steam pressure in bar
T0 = 773.;			#Steam temperature in K
a1 = 70.;			#Nozzle angle in degrees
ns = 0.78;			#Steam efficiency
m = 100.;			#Mass flow rate of steam in kg/s
D = 1.;			#Turbine diameter in m
N = 3000.;			#Turbine speed in rpm
h0 = 3370.;			#Steam enthalpy from Moiller chart in kJ/kg
v2 = 0.041;			#Specific volume at P2 from steam tables in (m**3)/kg
v4 = 0.05;			#Specific volume at P4 from steam tables in (m**3)/kg

# Calculations
U = (3.147*D*N)/60;			#Blade speed in m/s
C1 = (2*U)/math.sin(math.radians(a1));			#Steam speed in m/s
b1 = math.tan(math.radians(a1))/2;			#Blade angle at inlet for first stage in degrees
b1 = math.degrees(math.atan(b1))
b2 = b1;			#Blade angle at exit for first stage in degrees
b3 = b1;			#Blade angle at inlet for second stage in degrees
b4 = b2;			#Blade angle at exit for second stage in degrees
Wt = (4*m*(U**2))/(10**6);			#Total workdone in MW
Dh = (2*(U**2))/1000;			#Difference in enthalpies in kJ/kg
Dhs = Dh/ns;			#Difference in enthalpies in kJ/kg
h2 = h0-Dh;			#Enthalpy at point 2 in kJ/kg
h2s = h0-Dhs;			#Enthalpy at point 2s in kJ/kg
Dh2 = (2*(U**2))/1000;			#Difference in enthalpies in kJ/kg
Dh2s = Dh2/ns;			#Difference in enthalpies in kJ/kg
h4 = h2-Dh2;			#Enthalpy at point 4 in kJ/kg
h4s = h2-Dh2s;			#Enthalpy at point 4s in kJ/kg
Ca = C1*math.cos(math.radians(a1));			#Axial velocity in m/s
hI = (m*v2)/(math.pi*D*Ca);			#Blade height at first stage in m/s
hII = (m*v4)/(math.pi*D*Ca);			#Blade height at second stage in m/s

# Results
print 'Rotor blade angles for first stage are %3.2f degrees and %3.2f degrees  \
\nRotor blade angles for second stage are %3.2f degrees and %3.2f degrees  \
\nPower developed is %3.2f MW  \
\nFinal state of steam at first stage is %3.2f kJ/kg  \
\nFinal state of steam at second stage is %3.2f kJ/kg  \
\nBlade height at first stage is %3.4f m  \
\nBlade height at second stage is %3.4f m'%(b1,b2,b3,b4,Wt,h2s,h4s,hI,hII)

# rounding off error

Rotor blade angles for first stage are 53.95 degrees and 53.95 degrees
Rotor blade angles for second stage are 53.95 degrees and 53.95 degrees
Power developed is 9.90 MW
Final state of steam at first stage is 3306.52 kJ/kg
Final state of steam at second stage is 3257.00 kJ/kg
Blade height at first stage is 0.0114 m
Blade height at second stage is 0.0139 m


## Example 4.25 Page no : 218¶

In [4]:
import math

# Variables
P0 = 100.;			#Steam pressure in bar
T0 = 773.;			#Steam temperature in K
a1 = 70.;			#Nozzle angle in degrees
ns = 0.78;			#Steam efficiency
m = 100.;			#Mass flow rate of steam in kg/s
D = 1.;			#Turbine diameter in m
N = 3000.;			#Turbine speed in rpm
h0 = 3370.;			#Steam enthalpy from Moiller chart in kJ/kg
P4 = 27.;			#Pressure at point 4 in bar
T4 = 638.;			#Temperature at point 4 in K
v4 = 0.105;			#Specific volume at P4 from mollier chart in (m**3)/kg
ns = 0.65;			#Stages efficiency

# Calculations
U = (3.147*D*N)/60;			#Blade speed in m/s
C1 = (4*U)/math.sin(math.radians(a1));			#Steam speed in m/s
Ca = C1*math.cos(math.radians(a1));			#Axial velocity in m/s
tanb1 = (3*U)/Ca;			#Blade angle at inlet for first stage in degrees
b1 = math.degrees(math.atan(tanb1))
b2 = b1;			#Blade angle at exit for first stage in degrees
b4 = math.degrees(math.atan(U/Ca));			#Blade angle at exit for second stage in degrees
b3 = b4;			#Blade angle at inlet for second stage in degrees
WI = m*6*(U**2);			#Power developed in first stage in MW
WII = m*2*(U**2);			#Power developed in second stage in MW
W = (WI+WII)/(10**6);			#Total power developed in MW
Dh = (W*1000)/100;			#Difference in enthalpies in kJ/kg
Dhs = (W*1000)/(ns*100);			#Difference in enthalpies in kJ/kg
h4 = h0-Dh;			#Enthalpy at point 4 in kJ/kg
h4s = h0-Dhs;			#Enthalpy at point 4s in kJ/kg
h = (m*v4)/(3.147*D*Ca);			#Rotor blade height in m

# Results
print 'Rotor blade angles for first stage are %3.2f degrees and %3.2f degrees  \
\nRotor blade angles for second stage are %3.2f degrees and %3.2f degrees  \
\nPower developed is %3.2f MW  \
\nFinal state of steam at first stage is %3.1f kJ/kg  \
\nFinal state of steam at second stage is %3.2f kJ/kg  \
\nRotor blade height is %3.4f m'%(b1,b2,b3,b4,W,h4,h4s,h)

# rounding off error

Rotor blade angles for first stage are 64.11 degrees and 64.11 degrees
Rotor blade angles for second stage are 34.48 degrees and 34.48 degrees
Power developed is 19.81 MW
Final state of steam at first stage is 3171.9 kJ/kg
Final state of steam at second stage is 3065.27 kJ/kg
Rotor blade height is 0.0146 m


## Example 4.26 Page no : 221¶

In [40]:
import math

# Variables
a1 = 30.;			#Nozzle angle in degrees
Ca = 180.;			#Axial velocity in m/s
U = 280.;			#Rotor blade speed in m/s
R = 0.5;			#Degree of reaction

# Calculations
a1n = 90-a1;			#Nozzle angle measured from axial direction in degrees
Cx1 = Ca*math.tan(math.radians(a1n));			#Whirl velocity in m/s
b1 = math.degrees(math.atan((Cx1-U)/Ca));			#Blade angle at inlet in degrees
b2 = a1n;			#Blade angle at exit in degrees

# Results
print 'Blade angle at inlet is %3.0f degrees  \
\nBlade angle at exit is %3.0f degrees'%(b1,b2)

Blade angle at inlet is  10 degrees
Blade angle at exit is  60 degrees


## Example 4.27 Page no : 222¶

In [30]:
import math

# Variables
P0 = 800.;			#Steam pressure in kPa
T0 = 900.;			#Steam temperature in K
a1 = 70.;			#Nozzle angle in degrees
ns = 0.85;			#Steam efficiency
m = 75.;			#Mass flow rate of steam in kg/s
R = 0.5;			#Degree of reaction
U = 160.;			#Blade speed in m/s

# Calculations
C1 = U/math.sin(a1);			#Steam speed in m/s
Ca = C1*math.cos(a1);			#Axial velocity in m/s
b1 = 0;			                #Blade angle at inlet from velocity triangle in degrees
b2 = a1;            			#Blade angle at exit in degrees
a2 = b1;			            #Nozzle angle in degrees
W = (m*(U**2))/(10**6);			#Power developed in MW
Dhs = (W*1000)/(ns*m);			#Isentropic enthalpy drop in kJ/kg

# Results
print 'Rotor blade angles are %3.0f degrees and %3.0f degrees  \
\nPower developed is %3.2f MW  \
\nIsentropic enthalpy drop is %3.2f kJ/kg'%(b1,b2,W,Dhs)

Rotor blade angles are   0 degrees and  70 degrees
Power developed is 1.92 MW
Isentropic enthalpy drop is 30.12 kJ/kg