Hydroelectric power station

[Alix]

I have a question that I have to answer and it seems I really stuck its for my mechanical engineering course (I'm actually studying Aerospace) I'd appreciate any help.

Q: Water driving a hydroelectric power station falls a total distance of 158 m.
Calculate the total power available assuming all energy is harnessed.

Ive already found the answer for exit velocity of water and volume flow rate couldnt find mass flow rate (as I couldnt remember the equation).

I really appreciate any help :wink:



\mu = {10^-6}

[enigma]

Welcome to the forums, Alix!

Always good to see another aerohead here!

Couldn't you calculate the specific power? Basically ignore the mass term and find an answer in W/kg.

EDIT
Ive already found the answer for exit velocity of water and volume flow rate couldnt find mass flow rate (as I couldnt remember the equation).

Ignore what I just said. Mass is volume times density (keep the units in the calculation... you'll see they cancel out and leave mass)

[Alix]

Thanks for warm welcome enigma :smile: .
If I´m not mistaken, I need time in order to find power as " Power = Work Done / time " and we dont have any time or anything related to find power.

any help

Thanks :smile:

[Ivan Seeking]

The answer would have to be in watts per volume per second. If you know the flow rate then you should be done.

[Alix]

The answer would have to be in watts per volume per second. If you know the flow rate then you should be done.

I really cant understand so its actually : Work Done/flow rate (not time?)

and work done is sum of Kenetic and potential energy?


:confused:

[Alix]

Ok I have another problem

B) If the turbine, rotating at 200 rpm, absorbs all available enery (assuming there is no losses) determine the torque transmitted by the turbine shaft.

Ive converted 200 rpm to 20.94 rad/sec .
What i think is I should use Power = Torque x the thing looks like w [is it omega or something]

But i have no idea how to convert energy to power and i still couldnt find the answer for first part


Ps: just 3 more question ive left with out of 25 not too bad for 3 hours of work lol :D

[Ivan Seeking]

I really cant understand so its actually : Work Done/flow rate (not time?)

and work done is sum of Kenetic and potential energy?


:confused:

Why dont you show what you have worked out first? I was saying that if you have the flow, then you need the watts per volumetric flow, multiply by the flow, and you're done.

Bottom line is that you need the answer in watts - in units of energy per unit time; joules per second.

[enigma]

Can you post the full problem and the work you've done so far? That'll help us all find where the problem lies.

[Ivan Seeking]

Hey Enigma, I didn't meant to butt in but hydro is a particular interest of mine. It caught my attention. :smile:

[Alix]

sorry guys I should of told you the whole problem: Ok,

Q: Water Driving a hydropower station falls a total distance of 158 m

A. Assuming no losses

i) The exit velocity (m/s) of the water assuming all potential energy is converted to kenetic energy

My Work :

mgh = 1/2mv^2 => 2gh=v^2 => V = 55 .64 m/s



ii) The volume flow rate (m^3/s) if the pipe bore is 900 mm.

My Work:

Cross Section Area of pipe: 0.63 m^2
Q = V \times A => Q = 55.64 \times 0.63 = 35.05 m^3/s [Volume Flow Rate]


iii) the associated mass flow rate:

m= \rho \times V \times A => 1000 \times 55.64 \times 0.63 = 35053.2 Kg/s

[NOT SURE IF THE ANSWER IS RIGHT]


iv) The total power availabe assuming all energy is harnessed

B.
i) If the turbine accelerates to an angular velocity of 158 rpm from rest, in 15 secs, determine the rate of acceleration.

My Work:
158 rpm -> 16.54 rad/s

\omega_2 = \omega_1 + \alpha t => 16.54 = 0 + \alpha \times 15 => \alpha= 1.10 rad/s^2



ii) If the turbine, rotating at 200 rpm, absorbs all available enery (assuming there is no losses) determine the torque transmitted by the turbine shaft.


c. [Haven't started the work on this part yet] With losses

i) Usee Bernoulli's equation to determine the exit velocity of the water if the outlet pressure is 0.58 bar above atmospheric and lossed are equivalent to fluid head of 12.5 m.

ii) Deduce the power generated if the turbine/generator has an efficieny of 63% I believe I need the answer for part (A. iv) to find this ansswer.

iii) Determine the overall efficiency of the complete energu conversion systm.


Thanks guys
I will appreciate all your helps :smile:

[Alix]

I hate to bump it but I really need a little bit of help here :(

[Clausius2]

I am not going to do the calculations, but in order to obtain the power (W) exerted in the turbine, you only have to seek for the Bernoulli equation:

W=m(ho2-ho1)

where m=mass flow;
hoi=stagnation entalphy or total entalphy if z=cte.
If we suppose a perfect liquid:

ho2-ho1=(Po2-Po1)/rho

You can aproximate this difference with the hidrostatic height:
Po2-Po1=rho*g*H

where H is the difference of heights.

The total torque exerted in the turbine rotor is:

C=W/(k*w)

where k=power coefficient: it takes into account stator losses (k aprox. 1)
w=angular velocity

good luck.

[Alix]

thanks mate.

Ive got 80% for it