Unit Quantities:

1] Unit speed.

2] Unit discharge.

3] Unit Power.

1] Unit speed.

It is defined as the speed of a turbine working under a unit head (i.e. under a head of 1m) it is denoted by 'Nu'

N = Speed of a turbine under a head 'H'

H = Head under which a turbine is working.

u = Tangential velocity.

The tangential velocity, absolute velocity of water and head on the turbine are related as,

$u \alpha v$ (where $V \alpha \sqrt H$)

$\alpha \sqrt H$ ----(1)

Also tangential velocity (u) is given by

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

For a given turbine, the diameter (D) is constant.

$\therefore$ $(u \alpha N)$ or $(N \alpha u)$ or $(N \alpha \sqrt H)$

$\therefore$ $N = k_1 \sqrt H$ ---- (2)

Where $k_1$ is a constant of proportionality.

If head on the turbine becomes unity, the speed becomes unit speed or when

(H = 1), (N = Nu)

Substituting these values in equation (2), we get…

$N u = k_1 \sqrt 1.0$

$= K_1$

substituting the value of $k_1$ in equation (2)

$N = Nu \sqrt H$ or Nu = $\frac{N}{\sqrt H}$

2] Unit Discharge.

It is defined as, the discharge passing through a turbine, which is working under a unit head (i.e. lm). It is denoted by the symbol 'Qu'

H = Head of water on the turbine.

Q = Discharge passing through turbine when head is 'H' on the turbine.

a = Area of flow of water.

The discharge passing through a given turbine under a head 'H' is given by

Q = Area of flow x Velocity

But for a turbine, area of flow is constant and velocity is proportional to $\sqrt H$

$Q$ $\alpha$ velocity $\alpha \sqrt H$ $\therefore$ $Q = K_2 \sqrt H$ ------- (3)

where $K_2$ is constant of proportionality.

If, H = 1 , Q = Qu

substituting these values in equation (3), we get

$qu = $K_2$ \sqrt 1.0$

= $K_2$

substituting the value of k2 in equation (3), we get

$Q = Qu \sqrt H$

$\therefore$ $Qu = \frac{Q}{\sqrt H}$

3] Unit power.

It is defined as the power developed by a turbine, working under a unit head. it is denoted by 'Pu' The expression is as follows.

H = Head of water on the turbine.

P = Power developed by turbine.

Q = Discharge through turbine under head 'H'

The overall efficiency ($n_o$) is given by

$n_o = \frac{\text{power developed}}{\text{water power}}$

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

$\therefore$ $p = n_o \times \frac{p \times \ h \times Q \times H}{1000}$

$\alpha Q \times H$

$\alpha \sqrt H \times H$ ($\because Q \alpha \sqrt H$)

$\alpha H^3/2$

$\therefore$ $p = k_3 H^3/2$

where $k_3$ is constant of proportionality

when, H = 1m , p = pu

$\therefore$ $pu = k_3 (1)^3/2$

= $k_3$

Substituting the value of $k_3$ in equation (4)

we get,

$P = P_u H^3/2$

$\therefore$ $ p_u = \frac{p}{H^3/2}$

use of unit quantities $(N_u, Q_u, P_u)$

$N_u = \frac{N_1}{\sqrt H_1} = \frac{N_2}{\sqrt{H_2}}$

$Q_u = \frac{Q_1}{\sqrt H_1} = \frac{Q_2}{\sqrt{H_2}}$

$P_u = \frac{P_1}{H_1^3/2} = \frac{P_2}{H_2^3/2}$