Question 9.4.4: The following data pertains to an inward flow reaction turbi......
The following data pertains to an inward flow reaction turbine; diameter of wheel at inner periphery = 540 mm, width of wheel at inner periphery = 60 mm, diameter of wheel at outer periphery = 360 mm, width of wheel at outer periphery = 90 mm, area occupied by the vanes = 8% of periphery, guide vane angle = 25° to the runner tangent, moving blade angle at inlet = 95° (vane inclined forward to the direction of rotation), exit angle = 30°, hydraulic losses = 10% of supply head, mechanical friction losses = 5% of the supply head, pressure in the outer casing = 66 m more than that at the discharge from runner. Determine the following: (i) speed of the runner for no shock at entry, (ii) power available at the turbine shaft.
Given:
D_1=0.54 m B_1=0.06 m D_2 =0.36 m B_2=0.09 m
\alpha _1=25° \beta _1 =85° \beta _2 =30° K_1=0.95
hydraulic losses = 10% of supply head
mechanical friction losses = 5% of the supply head
pressure in the outer casing = 66 m more than that at the discharge from runner.
Note that vane inlet angle is given with reference to direction of rotation of vane whereas in the convention we have taken its supplementary angle. Therefore β_1= 180° – 95° = 85°. Since N is not given, we can not draw velocity triangles. But the shape of these triangles may look like that given in Fig. E9.4.4. Difference of pressure head between casing, that is inlet of guide vane to the outlet of runner vane is given. That is, available head is 66 m. Considering the hydraulic losses of 10%, the actual or Euler head is 0.9 × 66 m. This head is used in generating Euler work and kinetic energy of mass leaving runner vane. Therefore it requires to determine kinetic energy at exit first.
Taking K_t same at inlet and outlet of runner, continuity equation gives:
K_t(πD_1B_1) C_{f1} = K_t(π D_2 B_2) C_{f2} → C_{f1} = C_{f2}
From inlet velocity triangle:
u_1 = C_{f1}(\cot 25 + \cot 85) = 2.23 C_{f1} and C_{w1} = C_{f1} \cot 25 = 2.144 C_{f1}.
Also u_2 = D_2/D_1 × u_1 = 1.49 C_{f1}
From outlet triangle:
C_{f2} = C_{f1} = (u_2 + C_{w2}) \tan 30
→
C_{w2} = 0.243 C_{f1} and C_2 = √(C_{w2}^2+C_{f2}^2)=1.03 C_{f1}
∴ 0.9 × 66 = (C_{w2} u_2 + C_{w1} u_1)/g + (1.03 C_{f1})^2/2g
→ Cf1 = 2.14 m/s and C^2_2/2g=5.56 m
Note that here C_{w2} is negative for the outlet velocity triangle shown in figure.
Substitution of C_{f1} yields.
u_1 = 2.144 C_{f1} = 22.61 m/s, which gives N = 60 u_1/(πD_1) = 800 rpm
Q = discharge = (K_t π D_1 B_1) × C_{f1} = 0.949 m^3/s, and
P = power = (w/1000) × Q × (C_{w2} u_2 + C_{w1} u_1)/g × η_{\mathrm{mech}} = 4710.3 kW
