Power Plant Engineering

368 POWER PLANT ENGINEERING

Substituting the value of Q in the above equation, we get

P ∝

75

ρ

3/ 2

2

()H

N







H ∝

5/2

2

()H

N

∴ N^2 ∝

()H5/2

P

∴ N ∝

()H5/4

P

= C

()H5/4

P

where C is knour as constant depending upon the type of the turbine.

If the turbine develops 1 B.H.P. under one metre head then

C = N = N.

where Ns is the specific speed as per the definition.

Substituting the value of C in the above equation, we get

Ns = 5/4

()

NP

H

when P is in H.P.

= 5/4

1.165

()

NKW

H

when the power is in kW. ...(1)

By definition, the specific speed is number of revolutions per minute at which a given runner

would revolve if it were so reduced in proportions that it would develop one H.P. under one metre-head.

Sometimes the power developed is given in kilowatts instead of metric H.P., the head being in

metre as before.

The specific speed of a single jet Pelton wheel in terms of diameter of runner and diameter of jet

in metric units is given by

Ns (single jet petrol) = 244.75

d

D

...(2)

In a multi-jet pelton wheel, the H.P. is directly proportional to the number of jet if the head
remains constant. The specific speed of multi jet Pelton wheel is given by

Ns ∝ n as Ns ∝ P and P ∝ n.
Therefore, the specific speed of multi jet unit can be calculated by multiplying the specific speed
of single jet unit with a factor n where n is number of jets used.
It is necessary to know a characteristic of an imaginary machine identical in shape for comparing
the characteristics of machines of different types. The imaginary turbine is called a specific turbine. The
specific speed provides a means of comparing the speed of all types of hydraulic turbines on the same
basis of head and horse power capacity.

The overall cost of installation (runner + generator + power house and auxiliary equipments) is
lower if a runner of high specific speed is used for a given head and H.P. output. The selection of too