# Confused by basic hydropower concepts

1. Jul 17, 2017

### Micmic85

Hi everybody,

Consider the situation illustrated in the attached figure . A large water tank at elevation z1 is connected to a penstock with diameter D and outlet at elevation z2. The flow in the penstock is Q. The outlet of the pipe is connected to a valve and nozzle that direct a jet of water to an impulse turbine. Yes, the most basic stuff....

Let's assume that we remove the nozzle and valve at the outlet of the pipe. Q is in that case the natural flow in the pipe, i.e. the maximum flow that a pipe of a given diameter can carry under the sole action of gravity. From the book of Jones, Gravity Driven Water Flows in Networks, the power generated by the turbine in that case would be 0...I cannot understand this as the water coming out of the pipe has some velocity and carries then kinetic energy that could be used to rotate the turbine. Can someone clarify?

Thanks

2. Jul 17, 2017

### Baluncore

Welcome to PF.

The natural flow will be limited when the hydrostatic pressure of the water is lost as the water flows down the penstock. There will be no pressure available at the turbine input. The nozzle is needed to speed up the water velocity and narrow the jet to match the impulse turbine dimensions. Without the jet, the turbine will be flooded with low velocity water that has insufficient speed to spin the turbine fast enough to generate power.

3. Jul 17, 2017

### Staff: Mentor

Forger the hydropower and just think of a garden hose with a nozzle. The stream in the air has no more pressure, but it does have velocity. The nozzle gives it more velocity but less flow, until the nozzles opens far enough that flow is limited by friction in the hose. After that, opening the nozzle more has no effect.

Any real life turbine has a some fraction of impulse (velocity change) and reaction (pressure change). But the ratios vary dramatically in different situations.

Condider the pelton wheel turbine, the Francis turbine, and the propeller turbine. That is only three designs, but there are an infinite number of combinations of head and flow rates. So most real life cases would be optimum with a compromise design combining all three.

4. Jul 17, 2017

### jim hardy

I agree with you there's minimal velocity. So an old fashioned paddlewheel turning slowly could extract some energy.

But a hydro plant turbine is not a low speed paddlewheel
and if we constrain it to operation near synchronous speed
i expect its blades would be moving faster than the water
so it would deliver power to the water instead of the other way round, just churning like an eggbeater .

If the author is a hydro-power guy he may simply have forgotten that once upon a time he didn't know that, so neglected to explain it.
Haste makes waste. That's the price of "Publish or Perish" mentality in academia

Make any sense ?

old jim

5. Jul 17, 2017

### jartsa

With a nozzle at the bottom: All potential energy of water becomes kinetic energy of water.

With no nozzle at the bottom: Well, same as above, I guess.

At the place where the pipe starts, there's a 'nozzle', above the 'nozzle' there's the pressure caused by the water above, below the 'nozzle' there's the under pressure caused by the water in the pipe.

I would guess a nozzle must be carefully designed, which the top 'nozzle' is not. So maybe the pressure energy is converted mostly to heat energy of the water at that point?

6. Jul 17, 2017

### Staff: Mentor

Specifically, the basic scenario assumes the velocities in the reservoir and tailings pond to both be zero. No flow velocity, no spinning turbine. There are some other assumption in there that are hidden, but might be visible in the context in the textbook we aren't seeing.

7. Jul 17, 2017

### Staff: Mentor

I agree, the problem must be mis-stated. Unless there is something nonzero somewhere in the problem, there's nothing to solve.

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