Increasing the Pressure by Decreasing the Pipe Size

I will try to explain in this way, the calculations are a bit tedious and involves a bit of complex equations/

- reservoir with a constant level of H above the discharge point - It has a bit of Potential Energy available based on Height H.

- This energy is converted to KE at the end of the penstock/ Turbine.

- While carrying the water from source to turvine (through the pipes), there is an Energy loss due to friction (pipe), direction change (elbows, pipe bends), Pipe size changes etc (in fact anything on that pipe including pipe itself will have a head loss) say it is ΔH

- Hence the Head available to be converted to KE is (H-ΔH)

- With the efficiency of the Turbine η_t it is converted to rotational motion

- With the η_g - Efficiency of Generator it is converted to Electrical Energy.

Thus the Electricity generated

E α η_tη_g(H-ΔH)

Out of these obviously

η_t dependes on the velocity and quantity of the water available (Q) ie the KE available Vs it is designed for,

Same for η_g

H is the starting point.

ΔH depends on Q, Pipe Diameter and other factors as explained above and must be minimized for maximum energy available at the turbine stage.

To minimize ΔH -

The velocity should be low (lower friction loss), The lower velocity also reduces the Re (Reynolds No) and improves the flow

Bends, etc should be avoided.

Thus if you want the maximum energy available at the Turbine, pipe dia should be more and not less.

People misinterpret the Pressure as the Energy.

If the flow is stopped, the pressure available (what ever may be the pips dia) will be the Head.

If pipe dia is reduced, the pressure will sharply drop as the flow increases so nett energy available will reduce.

What is usually done (and what is causing the confuson) is bring the water through larger pipe. In the end put a nozzle to generate the velocity (for impulse turbines) for reaction turbines you need not do anything. The flow will be controlled by it.

An example may be in order

Put a polythene (I thing the one we use for aquariums. garden watering the flexible transparent one) pipe of 1/2" say some 4-5 ft on your tap and in the end close the flow by fingers as we do to spray water on plants - and see the jet available .

Now you just put a 6mm pipe of same length and do the same thing, you don't close the end now - what is the throw of the spray ?

Am I out in left field or is the following the case? I can't see how you would increase the pressure at the inlet of the turbine by decreasing the pipe size. Velocity yes, pressure no.

Considering this form of Bernoulli's eqn. if the top of the reservoir is point 1, then you can say that P1 is atmospheric and V1 is zero and leaves the equation. Z1 is your total reservoir height, which is static. V2 is a function of the flow rate since the flow is incompressible, m = rho * V2 * A. Z2 is zero relative to the height of the reservoir. Hf is the head loss due to friction which I believe is a function of pipe length and diameter (4fL/D?) (this is also where the bends loss goes), solve for P2. Note that the V2 term will increase with a decrease in area (pipe size) as will the Hf term, making the pressure P2 drop as these two increase.

If you have a constant level source the flow will be roughly proportional to the pipe cross area. Velocity at the nozzle will be almost the same if pipe dia is not changed too much (as already mentioned). If you connect a pipe to a PUMP then things get changed and here is the misunderstanding! Why? Let assume you have a pipe with D1 at the end the flow velocity will depend on the characteristic of the pump and if it is a free jet you will see a parabola. Now if you reduce the pipe( nozzle) diameter the velocity will increase because you move on the pump characteristic to a higher net resistance and the flow will go down but the velocity at the outlet will be higher and the parabola of the jet flatter.

This is NOT valid for a constant height supply.

I have the feeling here is the misinterpretation.