The peripheral velocity = 0.26$\sqrt{29H}$ and the radial velocity of flow at inlet is 0.96$\sqrt{29H}$ The wheel runs at 150 r.p.m and the hydraulic losses in the turbine are 22% of the available energy, assuming radial discharge

Determine.

1] The guide blade angle.

2] The wheel vane angle at inlet.

3] Diameter of the wheel at inlet.

4] Width of the wheel at inlet.

Given:

Overall efficiency $n_o$ = 75% = 0.75

power produced S.P. = 148.25 kw

Head H = 7.62 m

Peripheral velocity $u_1$ = 0.26 $\sqrt{29H}$

= 0.26 $\sqrt{2 \times 9.81 \times 7.62}$

= 3.179 m/s

Velocity of flow at inlet, vf1 = 0.96 $\sqrt{29H}$

= 0.96 $\sqrt{2 \times 9.81 \times 7.62}$

= 11.738 m/s

Speed, N = 150 r.p.m.

hydraulic losses = 22% of available energy

$v_{w2} = 0, v_{f2} = v^2$

Step No - (1) Hydraulic efficiency

$n_h = \frac{H - 0.22 H}{H}$

$\frac{0.78H}{H}$

= 0.78

But, $n_h = \frac{v_{w1} u_1}{9H}$

$\therefore$ $0.78 = \frac{v_{w1} u_1}{9H}$

$\therefore$ $v_{w1} = \frac{0.78 \times 9 \times H}{u_1}$

$= \frac{0.78 \times 9.81 \times 7.62}{3.179}$

= 18.34 m/s

Step No (2) The guide blade angle,

$tan a = \frac{v_{f1}}{v_{w1}}$

$= \frac{11.738}{18.34}$

tan a = 0.64

$a = tan^-1 (0.64)$

a = $32.619^0$

Step No (3) The wheel vane at inlet.

$tan \theta = \frac{v_{f1}}{v_{w1} - u_1}$

$= \frac{11.738}{18.34 - 3.179}$

$tan \theta = 0.774$

$\theta = tan^{-1} (0.774)$

$\theta = 37.74^0$

Step No (4) Diameter of wheel at inlet

$u_1 = \frac{\pi D_1 N}{60}$

$D_1 = \frac{60 \times u_1}{\pi \times N}$

$= \frac{60 \times 3.179}{\pi \times 50}$

$D_1 = 0.4047m$

Step No (5) Width of the wheel at inlet (B1)

$n_o = \frac{S.P.}{W.P.}$

$\frac{148.25}{w.p}$

But $w.p. = \frac{WH}{1000}$

$w.p = \frac{p \times 9 \times Q \times H}{1000}$

$w.p = \frac{1000 \times 9.81 \times Q \times 7.62}{1000}$

$\therefore$ $n_o = \frac{148.25}{\frac{1000 \times 9.81 \times Q \times 7.62}{1000}}$

$\therefore$ $Q = \frac{148.25 \times 1000}{1000 \times 9.81 \times n_o \times 7.62}$

$= \frac{148.25 \times 1000}{1000 \times 9.81 \times 0.75 \times 7.62}$

Q = 2.644 $m^3/s$

$\therefore$ Using equation $Q = \pi D_1 \times B_1 \times v_{f1}$

$2.644 = \pi \times 0.4047 \times B_1 \times 11.738$

$\therefore$ $B_1 = \frac{2.644}{\pi \times 0.4047 \times 11.738}$

$\therefore$ $B_1 = 0.177 m$