A water turbine has a velocity of 6 m/s at the entrance of the draft tube and a velocity of 1.2 m/s at the exit. The tailrace is 5 m below the entrance of the draft tube.

Calculate the pressure head at the entrance of the draft tube, if the friction losses in the draft tube are 0.1 m.

Given:

Velocity at inlet, V1= 6m/s

Velocity at outlet,V2=1.2 m/s

Friction loss $h_f$ =0.1m

Vertical height between tail race and inlet of draft tube = 5m

Let y= vertical height between tailrace and outlet of draft tube.

Applying Bernoulli's equation at the inlet and outlet of draft-tube and taking reference line passing through section (2-2) , we get

$\frac{p_1}{ρg }$+$\frac {V1^2}{2g}$+$Z1$ = $\frac{p2}{ρg}$+$\frac{V2^2}{ρg}$+$Z2$ +$ h_f$

where Z1=5+y; V1= 6 m/s; V2 = 1.2 m/s, $h_f$=0.1

$\frac{p_2}{ρg }$= Atmospheric pressure head + y = $\frac{p_a}{ρg }+y$

Z2=0

Substituting the values we get

$\frac{p_1}{ρg }$+$\frac {6^2}{2×9.81}$+$(5+y)$ = $\left(\frac{p_a}{ρg}+y\right)$+$\frac{1.2^2}{2×9.81}$+0 +0.1

$\frac{p_1}{ρg }$+1.835+5+y = $\frac{p_a}{ρg}+y$+0.0734 +0.1

$\frac{p_1}{ρg }$+6.835 = $\frac{p_a}{ρg}$+0.1734 → 1 } $$

If $\frac{p_a}{ρg }$ is taken 0 then we will get $\frac{p_1}{ρg }$ as vacuum pressure head at inlet of draft tube.

But if $\frac{p_a}{ρg }$ = 10.3m of water ,then we will get $\frac{p_1}{ρg }$ as absolute pressure head at inlet of draft tube.

Taking $\frac{p_a}{ρg }$ =0 and substituting this in eq 1 we get,

$\frac{p_1}{ρg }$+ 6.835 = 0+ 0.1734

$\frac{p_1}{ρg }$ = -6.835 + 0.1734 = - 6.6616 m