**1 Answer**

written 5.4 years ago by |

**If the side clearance angle is 15 degree and discharge through nozzle is $0.1 m^3/3$**

**Find :**

Power available at the nozzle.

Hydraulic efficiency of turbine.

**Solutions :**

**Given :**

1] Diameter of wheel, D = 1.0 m

2] Speed of wheel, N = 1000 r.p.m

$\therefore$ Tangential velocity of the wheel,

$u = \frac{\pi DN}{60}$

$= \frac{\pi \times 1 \times 1000}{60}$

u = 52.36 m/s

3] Net Head, H = 700m

4] Side clearance angle, $\theta = 15$ degree

5] Discharge, $q = 0.1m^3/s$

Velocity of jet at inlet,

$v_1 = cv \sqrt{2gH}$

$= 1 \times \sqrt{2 \times 9.81 \times 700}$ [cv = 1.0]

$v_1 = 117.19 m/s$

Step No. (1) Power available at the nozzle is given by

$W.P = \frac{W \times H}{1000}$

$= \frac{P \times g \times Q \times H}{1000}$

$= \frac{1000 \times 9.81 \times 1.0 \times 700}{1000}$

= 686.7 kw.

Step No (2) Hydraulic efficiency is given by

$n_h = \frac{2 (v_1 - u) (1 + coz \phi ) u}{v_1^2}$

$= \frac{ 2(117.19 - 52.36) (1 + coz 15˚) \times 52.36}{117.19 \times 117.19}$

$= \frac{2 \times 64.83 \times 1.966 \times 52.36}{117.19 \times 117.19}$

= 0.9718

$n_h$ = 97.18%