Consider a capital draft-tube as shown,

Applying Bernoulli's equation to inlet [(1)-(1)]

and outlet [(2){2)] take section [(2)(2)] as datum line,

$\frac{P_1}{pg} + \frac{v1^2}{2g} + (Hs + y) = \frac{P_2}{P_g} + \frac{v_2^2}{2g} + 0 + hf$ -----(1)

$h_f$ = losses of energy between (1) - (1) and (2) - (2)

But, $\frac{P_2}{P_g}$ Atmospheric pressure head + y = $\frac{P_a}{P_g} + y$

Substituting this value of $\frac{P_a}{P_g}$ in equation (1)

$\frac{P_1}{P_g} + \frac{v_1^2}{2g} + (Hs + y) = \frac{P_a}{P_g} + y + \frac{v_2^2}{2g} + hf$

$\therefore$ $\frac{P_1}{P_g}+ \frac{v_1^2}{2g} + (Hs) = \frac{P_a}{P_g} + \frac{v_2^2}{2g} + hf$

$\therefore$ $\frac{P_a}{P_g} + \frac{P_a}{P_g} + \frac{v_2^2}{2g} + hf - \frac{v_1^2}{2g} - Hs$

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

Efficiency of Draft - $tube^L$

The ratio of actual conversion of kinetic head into pressure head in the draft-tube to the kinetic head at the inlet of the draft-tube.

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

where,

$v_1$ = Velocity of water at inlet of draft-tube.

$v_2$ = Velocity of water at outlet of draft-tube.

$h_f$ = Loss of head in draft - tube.