The goal of this task is to obtain a set of angles that will fulfill the conditions of the
engine design and give satisfactory performance. To find it, the variables will be the
turbine flow coefficient  and the flow air angle 3.
We have to check the requirement following:
 Nozzle pressure ratio should not exceed chocking with more than 1O%
 The relative Mach number in the root of the rotor should not exceed 0.75
 A positive reaction in the root assuming by designing the blade using the free vortex
condition
To model the performances of the engine, I used MATLAB.
Hypothesis and Assumptions:
Some hypothesis had been taken for this model:
 First, we consider that transformations are isentropic, as the compression and
expansion processes are suppose to be reversible and adiabatic.
In this way we can use the isentropic efficiency and its formulas.
 We also consider that the working fluid has the same composition throughout the
cycle and that is a perfect with constant specific heat.
 We consider the specific heat constant, as the temperature does not exceed 1500K.
We also consider γ =1,33 and Cpg constant for the combustion gas product in the
turbine.
 We consider that the mass flow is constant throughout the cycle so that the losses of
mass flow are nearly equal to the fuel injection.
 In our single stage turbine, C1 will be axial so that 1=0 and C1 = Ca1.
 The axial flow velocity Ca is constant through the rotor. So that Ca1=Ca2=Ca3.
And C1=C3.
To apply the Free Vortex Design Theory, we assume:
 The stagnation enthalpy ho is constant over the annulus.
 The axial velocity is constant over the annulus.
 The whirl velocity is inversely proportional to the radius.
The fallowing MATLAB code had been computed to design the turbine of the W1 engine.
The values of the outlet swirl angle 3 and the nozzle flow coefficient N had been taken as
variables to meat the requirement R=2*N with the task-1 turbine efficiency value.
MATLAB Code :
%%%%%%%%% TASK-2%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% CONSTANT PARAMETERS :
clear all ;
m=26*0.45359237; % Flux in kg/sec
Cpa=1005;
Cpg=1148;
% calorific coeff, J/(K.kg)
gammaG=1.333;
% For GAZ
N=17750;
h= 2.4*0.0254 ;
Dm= 14*0.0254;
Dt=Dm+h ;
Dr=Dm-h ;
% rpm
% Blade Lenght
% Axial Turbine medium diameter
% Tip Diameter
% Root Diameter
n=72;
% number of Blades
eta_t=0.84;
% Isentropic Turbine efficiency
T01=1050;
T03=889;
%Inlet turbine Temperature (from Task_1)
%Outlet turbine Temperature (from Task_1)
P01=3.8;
P03=1.6;
Pa=1.01325;
%Inlet turbine Pressure (from Task_1)
%Outlet turbine Pressure (from Task_1)
% Atmospheric pressure
R=0.287;
% Constant of the gaz [kJ/K.Kg]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% DEFINITION OF THE TURBINE-GEOMETRY :
%% ASSUMPTION : C1=C3 & Ca2=Ca3
% Coeff of Temperature Drop (Psi):
Um=N*pi*Dm/60 ;
Delta_T0=T01-T03 ;
Psi = (Cpg*2*Delta_T0)/Um^2 ;
%STARTING ASSUMPTIONS :
% Flow Coeff (Phi)
Phi= 0.8 ;
%Phi=[0.8 ; 1]
alpha_3 = 0 ;
%No Swirl
% Psi=[3;5]
%%%%%%%
beta_3=atan(1/Phi+tan(alpha_3));
LAMBDA=1/2*(tan(beta_3)*2*Phi-1/2*Psi);
% Degree of Reaction Coefficient
%%%%%%%
%% Computation of the Root and Tip's blade Degree of reaction
%% (Which don't have to be negative)
Ur=N*pi*Dr/60 ;
Psi_r = (Cpg*2*Delta_T0)/Ur^2 ;
% Psi=[3;5]
beta_3_r=atan(1/Phi);
LAMBDA_r=1/2*(tan(beta_3_r)*2*Phi-1/2*Psi_r);
% Degree of Reaction Coefficient
%%If negative value of the Degree of Reaction Coefficient (LAMBDA)
if LAMBDA_r<0
alpha_3deg = 20 ;
% Introduction of Swirl
alpha_3 = alpha_3deg*pi/180; % Angle values in radian
beta_3=atan(1/Phi+tan(alpha_3));
LAMBDA=1/2*(tan(beta_3)*2*Phi-1/2*Psi);
end
% Degree of Reaction Coefficient
beta_2 = atan(1/(2*Phi)*(1/2*Psi-2*LAMBDA)) ;
alpha_2 = atan(tan(beta_2)+1/Phi) ;
% Values Converted in degrees :
beta_2deg = beta_2*180/pi ;
alpha_2deg = alpha_2*180/pi ;
beta_3deg = beta_3*180/pi ;
alpha_3deg = alpha_3*180/pi ;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Estimate the annulus aera :
%%% We Start with zone 2 values, to check the convergence nozzle,
%%% if it is chocked or unchocked :
Ca_2 = Um*Phi ;
% axial velocity in 2
C_2 = Ca_2/cos(alpha_2); % absolute velocity in 2
Cw_2 = sqrt(C_2^2-Ca_2^2); % Component of momentum
V_2 = Ca_2/(cos(beta_2));
% Relative Velocityin 2
T02 = T01;
% No energetic variation throught the nozzle
T2 = T02-C_2^2/(2*Cpg) ; % Dedinition of the stagnation temperature
%%% First estimation of the Nozzle Flow Coefficient :
Lambda_N = 0.125;
T22 = T2-Lambda_N*C_2^2/(2*Cpg);
r = (T01/T22)^(gammaG/(gammaG-1)) ; % Pressure ratio r = P01/P2
P2 = P01/r ;
%%% As a First Approximation, we Ignore the effect of friction on the critical pressure ratio :
rc1 = ((gammaG+1)/2)^(gammaG/(gammaG-1)); % rc = P01/PC
%%% Display the Nozzle state :
if r<rc1
disp('Nozzle is uncchocked')
else
disp('Nozzle is chocked')
end
if r>1.1*rc1
disp('Nozzle chocked not acceptable')
end
rho_2 = 100*P2/(R*T2);
A2 = m/(rho_2*Ca_2);
A2N = A2*cos(alpha_2);
% Density of the outlet Nozzle
% Annulus aera
% Throat area Nozzles required
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%% We now do the same with area 1 and 3 :
%%%% AREA-1 %%%
%%%% The assumption iss that Ca_1 = C_1, C1 = C3 and Ca_3 = Ca_2 :
Ca_3 = Ca_2;
C_3 = Ca_3/cos(alpha_3);
C_1 = C_3;
Ca_1 = C_1;
T1 = T01-C_1^2/(2*Cpg);
P1 = P01*(T1/T01)^(gammaG/(gammaG-1));
rho_1 = 100*P1/(R*T1); % Density of the inlet Nozzle
A1 = m/(rho_1*Ca_1);
% Annulus aera
%%%% AREA-1 %%%
T3 = T03-C_3^2/(2*Cpg);
P3 = P03*(T3/T03)^(gammaG/(gammaG-1));
rho_3 = 100*P3/(R*T3); % Density of the inlet Nozzle
A3 = m/(rho_3*Ca_3);
% Annulus aera
V_3 = Ca_3/(cos(beta_3));
% Relative Velocity in 3
C_3 = Ca_3/cos(alpha_3);
% absolute velocity in 3
Cw_3 = sqrt(C_3^2-Ca_3^2); % Component of momentum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% We assume that the Rotor Flow Coefficient is the double of the Nozzle Flow Coefficient
%% Lambda_R = 2*Lambda_N ;
%% So we have to check this assumption
%% Considering a Isotropic Reaction between area 2 and area 3, we have :
T333 = T2*(P3/P2)^((gammaG-1)/gammaG);
%% We can compute the Rotor Flow Coefficient :
Lambda_R = (T3-T333)/(V_3^2/(2*Cpg));
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Application of the Free Vortex Theory
%We now get the dimensions of the blades of the turbine :
h1 = A1*N/(Um*60);
h2 = A2*N/(Um*60);
h3 = A3*N/(Um*60);
hd1 = 1/2*(h1+h2);
hd2 = 1/2*(h2+h3);
w1 = hd1/3 ;
w2 = hd2/3 ;
ws = 0.5*w2;
H = ws+w1+w2;
alpha = 0.5*(h3-h1)/H;
alpha_deg = alpha*180/pi;
rm = Dm/2 ;
rt1 = 1/2*(Dm + h1) ;
rt2 = 1/2*(Dm + h2) ;
rt3 = 1/2*(Dm + h3) ;
rr1 = 1/2*(Dm - h1) ;
rr2 = 1/2*(Dm - h2) ;
rr3 = 1/2*(Dm - h3) ;
%% Define the Flow angles values at the Tip and at the Root :
alpha_2r = atan((rm/rr2)*tan(alpha_2));
alpha_2t = atan((rm/rt2)*tan(alpha_2));
alpha_2r_deg = alpha_2r*180/pi ;
alpha_2t_deg = alpha_2t*180/pi ;
beta_2r = atan((rm/rr2)*tan(alpha_2)-(rr2/rm)*Um/Ca_2);
beta_2t = atan((rm/rt2)*tan(alpha_2)-(rt2/rm)*Um/Ca_2);
beta_2r_deg = beta_2r*180/pi ;
beta_2t_deg = beta_2t*180/pi ;
beta_3r = atan((rm/rr3)*tan(alpha_3)+(rr3/rm)*Um/Ca_3);
beta_3t = atan((rm/rt3)*tan(alpha_3)+(rt3/rm)*Um/Ca_3);
beta_3r_deg = beta_3r*180/pi ;
beta_3t_deg = beta_3t*180/pi ;
%%%%%% Check the Root Velocity (because the greatest velocity is at the root):
V_2r = Ca_2/cos(beta_2r);
C_2r = Ca_2/cos(alpha_2r);
T2r = T02 - C_2r^2/(2*Cpg) ;
Mv_2r = V_2r/(sqrt(gammaG*1000*R*T2r));
if Mv_2r > 0.75
disp('Nozzle root Mach number not acceptable')
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Computation of the Nozzle isentropic efficiency etaj_N :
%% (to get more accurate value of the critical ratio)
etaj_N = (T01-T2)/(T01-T22);
%% CRITICAL PRESSURE COMPUTATION :
rc2=1/[1-1/etaj_N*(gammaG-1)/(gammaG+1)]^(gammaG/(gammaG-1)); %New critical ratio
RESULTS :

I’ve meet the assumption of R=2*N with N = 12.5
The angle 3 had been taken at 20°
And the flow coefficient:  = 0.8

The turbine nozzle operate unchoked
Nozzle isentropic efficiency:
j_N = 89%
With this result, the critical pressure ratio can be calculated with more accuracy:
P01/P2 = 2.01

The space between the stator and the rotor blades has been taken of about 50% of the
blade with to limit the risk of flow separation in the divergent.
I found the following values
- H = 4.4cm
-  = 25°
- h1 = 3,3 cm; h2 = 5cm; h3 = 7,2cm
- ws = 0,5cm; w1 = 1.4cm; w2 = 2cm
- hd1 = 4,1cm; hd2 = 6cm
Turbine stage schematic:
w2
ws
w1
 = 25°
h1

h2
h3
Applying the Free Vortex Theory, we get the values of the angles in the root of the
rotor:
root
root 
root 
root 
63.8°
43.7°
24.5°
55.5°