in water, if the atmospheric pressure head is 10.3m of water and loss of head due to friction in the draft tube is equal to 0.2 x velocity head at outlet of the tube.

Find.

1] Pressure head at inlet.

2] Efficiency of the draft tube.

Given:

Diameter at inlet, $D_1$ = 1.0m

Diameter at outlet, $D_2$ = 1.5m

Velocity at outlet, $V_2$ = 2.5 m/s

Total length of tube, $H_s$ + y = 6.0m

Length of tube in water, y = 1.20 m

Hs = 6.0 - 1.20

= 4.80 m

Atmospheric pressure head, $\frac{P_a}{P_g} = 10.3m$

Loss of head due to friction,

$hf = 0.2 \times \ velocity head at outlet$

$=0.2 \times \frac{v_2^2}{2g}$

Discharge through tube,

Q = $A_2 V_2$

$= \frac{\pi }{4} \times (D_2)^2 \times 2.5$

$= \frac{\pi }{4} \times (1.5)^2 \times 2.5$

$= 4.4178 \ m^2/s$

Velocity at inlet,

$V_1 = \frac{Q}{A_1}$

= $\frac{4.4178}{\frac{\pi }{4} \times (1)^2}$

= 5.625 m/s

Step No (1) Pressure head at inlet $(\frac{P_1}{P_g})$

$\frac{P_1}{P_g} = \frac{P_1}{P_g} - Hs - (\frac{v_1^2}{2g} - \frac{v_2^2}{2g} - Hf)$

$= 10.3 - 4.8 - (\frac{5.625^2}{2 \times 9.81} - \frac{2.5^2}{2 \times 9.81} - 0.2 \times \frac{v_2^2}{2g})$

$= 10.3 - 4.8 - ( 1.6126 - 0.3185 - \frac{0.2 \times 2.5^2}{2 \times 9.81})$

= 10.3 - 4.8 - (1.6126 - 0.3185 - 0.0637)

= 5.5 - (1.2304)

= 4.269

= 4.27 (abs)

Step No (2) Efficiency of draft tube (nd)

$n_d = \frac{(\frac{v_1^2}{2g} - \frac{v_2^2}{2g}) - hf}{\frac{v_1^2}{2g}}$

$= \frac{ \frac{v_1^2}{2g} - \frac{v_2^2}{2g} - \frac{0.2v_2^2}{2g}} {\frac{v_1^2}{2g}}$

$= \frac{v_1^2 - 1.2v_2^2}{v_1^2}$

$= 1 - 1.2 (\frac{v^2}{v^1})^2$

$= 1-1.2 (\frac{2.2}{5.625})^2$

= 1- 0.237

[ $n_d$ = 0.763 or 76.3%]