# Chapter4-Axial-flow Turbines:Two-dimensional Theory¶

## Ex1-pg101¶

In [1]:
import math
#calculate the

##given data
phi = 0.4;
epsilon = 28.6;##in deg

##calculations
alpha2 = (180./math.pi)*math.atan(1./phi);##in deg
zeta = 0.04*(1+ 1.5*(alpha2/100.)**2);
eta = 1 + (phi**2)*(zeta*((1./math.cos(math.pi*alpha2/180.))**2) +0.5);

##results
print'%s %.2f %s'%('The efficiency = ',1/eta,'');
print('This value appears to be the same as the peak value of efficiency curve.\n');

The efficiency =  0.86
This value appears to be the same as the peak value of efficiency curve.



## Ex2-pg105¶

In [2]:
import math
#calculate the

##given data
alpha2 = 70.;##in deg
p01 = 311.;##in kPa
T01 = 850.;##in degC
p3 = 100.;##in kPa
eff_tot_stat = 0.87;
U = 500.;##in m/s
Cp = 1.148;##in kJ/(kgC)
gamma = 1.33;

##Calculations
delW = eff_tot_stat*Cp*(T01+273.15)*(1.-(p3/p01)**((gamma-1.)/gamma));##specific work
cy2 = delW*1000./U;##in m/s
c2 = cy2/math.sin(math.pi*alpha2/180.);##in m/s
T2 = (T01+273.15) - 0.5*(c2**2)/(Cp*1000.);##Nozzle exit temperature in K
M2 = c2/math.sqrt(gamma*287.*T2);##Nozzle exit mach number
cx = c2*math.cos(math.pi*alpha2/180.);##axial velocity in m/s
eff_tot_tot = 1./((1./eff_tot_stat)-((cx**2)/(2.*1000.*delW)));##Total to total efficiency
R = 1. - 0.5*(cx/U)*math.tan(math.pi*alpha2/180.);##stage reaction

##results
print'%s %.2f %s'%('(i) The specific work done =',delW,' kJ/kg.\n');
print'%s %.2f %s'%('(ii) The Mach number leaving the nozzle = ',M2,'');
print'%s %.2f %s'%('(iii) The axial velocity = .\n',cx,'m/s');
print'%s %.2f %s'%('(iv) The total-to-total efficiency = .\n',eff_tot_tot,'');
print'%s %.2f %s'%('(v) The stage reaction = .\n',R,'');

##there are small errors in the answers given in the book

(i) The specific work done = 275.24  kJ/kg.

(ii) The Mach number leaving the nozzle =  0.96
(iii) The axial velocity = .
200.36 m/s
(iv) The total-to-total efficiency = .
0.93
(v) The stage reaction = .
0.45


## Ex3-pg106¶

In [3]:
import math
#calculate the

##given data
H_b = 5.0;##average bladeaspect ratio for the stage
t_c = 0.2;##max. blade thickness to chord ratio
Re = 1*10**5;##average Reynolds number
cx = 200.;##in m/s
cy2 = 552.;##in m/s
U = 500.;##in m/s
c2 = 588.;##in m/s
delW = 276.;##in kJ
c3 = 200.;##in m/s
Cp = 1.148;##in kJ/(kgC)
T2 = 973.;##in K
T01 = 1123.;##in K
alpha1 = 0.;##in deg
alpha2 = 70.;##in deg

##calculations
eps = alpha1 + alpha2;##in deg
zetaN = 0.04*(1. + 1.5*(eps/100.)**2);
zetaN1 = (1.+zetaN)*(0.993 + 0.021/H_b) - 1;
beta2 = (180./math.pi)*math.atan((cy2-U)/cx);
beta3 = (180./math.pi)*math.atan(U/cx);
epsR = beta2 + beta3;
zetaR = 0.04*(1. + 1.5*(epsR/100.)**2);
zetaR1 = (1.+zetaR)*(0.975 + 0.075/H_b) - 1;
w3_U = math.sqrt(1.+(cx/U)**2);
eff_ts = 1./(1. + (zetaR1*w3_U + zetaN1*((c2/U)**2) + (cx/U)**2)/(2.*cy2/U));
T3 = T01 - (delW*1000. + 0.5*c3**2.)/(Cp*1000.);
eff_ts1 = 1/(1. + (zetaR1*(w3_U)**2 + (T3/T2)*zetaN1*((c2/U)**2.) + (cx/U)**2.)/(2.*cy2/U));

##Results
print'%s %.2f %s'%('The total-to static efficiency = ',eff_ts,'');
print('\n The result is very close to the value assumed in first example.')
print'%s %.2f %s'%('\n The total-to-static efficiency after including the temperature ratio in the equation = ',eff_ts1,'');

##there are small errors in the answers given in the book

The total-to static efficiency =  0.87

The result is very close to the value assumed in first example.

The total-to-static efficiency after including the temperature ratio in the equation =  0.87


## Ex4-pg119¶

In [4]:
import math
#calculate the

##given data
T02 = 1200.;##in K
p01 = 4.0;##in bar
dt = 0.75;##tip diameter in m
hb = 0.12;##blade height in m
v = 10500.;##shaft speed in rev/min
R = 0.5;##degree of reaction at mean radius
phi = 0.7;##flow coefficient
eff_noz = 0.96;##Nozzle efficiency
Cp = 1160.;##in kJ/(kgC)
gamma = 1.33;
Rg = 287.8;##specific gas constant
A2 = 0.2375;##in m^2
K = 2/3.;##stress taper factor
rho = 8000.;##in kg/m^3

##calculations
beta3 = (180./math.pi)*math.atan((0.5*psi + R)/phi);
beta2 = (180./math.pi)*math.atan((0.5*psi - R)/phi);
alpha2 = beta3;
alpha3 = beta2;
rm = (dt-hb)/2.;
Um = (v/30.)*math.pi*rm;
cx = phi*Um;
c2 = cx/(math.cos(alpha2*math.pi/180.));
T2 = T02 - 0.5*(c2**2)/Cp;
p2 = p01*((1-((1.-(T2/T02))/eff_noz))**(gamma/(gamma-1.)));
mdot = ((p2*10**5)/(Rg*T2))*A2*cx;
Ut = (v/30.)*math.pi*0.5*dt;
sig_rho = K*0.5*(Ut**2)*(1-((dt-2.*hb)/dt)**2);
sig1 = rho*sig_rho;
Tb = T2 + 0.85*((cx/math.cos(beta2*math.pi/180.))**2.)/(2.*Cp);

##Results
print'%s %.2f %s %.2f %s'%('(i)The relative and absolute angles for the flow: \n beta3 = ',beta3,' deg' and 'beta2 = ',beta2,' deg.');
print'%s %.2f %s %.2f %s'%(' alpha2 = ',alpha2,' deg' and 'alpha3 = ',alpha3,'deg.');
print'%s %.2f %s'%('\n (ii) The velocity at nozzle exit = ',c2,' m/s');
print'%s %.2f %s %.2f %s %.2f %s '%('\n (iii)The static temperature and pressure at nozzle exit assuming a nozzle efficiency of ',eff_noz,''and ': \n T2 = ',T2,'K'and '\n p2 =',p2,' bar');
print'%s %.2f %s' %('\n and mass flow = ',mdot,'kg/s');
print'%s %.2f %s %.2f %s '%('\n (iv)The rotor blade root stress assuming the blade is tapered with a stress taper factor K of 2/3 and \n the blade material density is ',rho,' kg/m2'and ' =',sig1/(10**6),' MPa');
print'%s %.2f %s'%('\n (v) The approximate average mean blade temperature is Tb = ',Tb,' K');

#

##there are very small errors in the answers given in textbook

(i)The relative and absolute angles for the flow:
beta3 =  68.20 beta2 =  46.97  deg.
alpha2 =  68.20 alpha3 =  46.97 deg.

(ii) The velocity at nozzle exit =  652.82  m/s

(iii)The static temperature and pressure at nozzle exit assuming a nozzle efficiency of  0.96  1016.30
p2 = 1.99  bar

and mass flow =  39.10 kg/s

(iv)The rotor blade root stress assuming the blade is tapered with a stress taper factor K of 2/3 and
the blade material density is  8000.00  = 243.74  MPa

(v) The approximate average mean blade temperature is Tb =  1062.56  K