Hydroelectric power station

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In summary: L/s."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""Cross Section Area of pipe: 0.63 m^2""Q = V \times A => Q = 55.64 \times 0.63 = 35.05 m^3/s""With losses""i) Usee Bernoulli's equation to determine the exit velocity
  • #1
Alix
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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 couldn't find mass flow rate (as I couldn't remember the equation).

I really appreciate any help :wink:

[tex]

\mu = {10^-6}


[/tex]
 
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  • #2
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 couldn't find mass flow rate (as I couldn't 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)
 
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  • #3
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 don't have any time or anything related to find power.

any help

Thanks :smile:
 
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  • #4
The answer would have to be in watts per volume per second. If you know the flow rate then you should be done.
 
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  • #5
Ivan Seeking said:
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 can't understand so its actually : Work Done/flow rate (not time?)

and work done is sum of Kenetic and potential energy?


:confused:
 
  • #6
Ok I have another problem

B) If the turbine, rotating at 200 rpm, absorbs all available energy (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 couldn't find the answer for first part


Ps: just 3 more question I've left without of 25 not too bad for 3 hours of work lol :D
 
  • #7
Alix said:
I really can't understand so its actually : Work Done/flow rate (not time?)

and work done is sum of Kenetic and potential energy?


:confused:

Why don't 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.
 
  • #8
Can you post the full problem and the work you've done so far? That'll help us all find where the problem lies.
 
  • #9
Hey Enigma, I didn't meant to butt in but hydro is a particular interest of mine. It caught my attention. :smile:
 
  • #10
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 :

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



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

My Work:

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


iii) the associated mass flow rate:

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

[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

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



ii) If the turbine, rotating at 200 rpm, absorbs all available energy (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:
 
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  • #11
I hate to bump it but I really need a little bit of help here :(
 
  • #12
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.
 
  • #13
asd

thanks mate.

Ive got 80% for it
 
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1. What is a hydroelectric power station?

A hydroelectric power station is a type of power plant that uses the force of moving water to generate electricity. It typically consists of a dam, a water turbine, and a generator.

2. How does a hydroelectric power station work?

A hydroelectric power station works by using the force of falling water to turn a turbine, which is connected to a generator. The generator then converts the rotational energy into electricity that can be used to power homes and businesses.

3. What are the advantages of using hydroelectric power?

Some advantages of using hydroelectric power include its renewable and clean energy source, its ability to generate large amounts of electricity, and its low operating costs compared to other forms of energy production.

4. What are the environmental impacts of hydroelectric power stations?

While hydroelectric power is considered a clean energy source, the construction of dams and reservoirs can have negative impacts on the environment. This includes disrupting natural habitats, altering water flow patterns, and affecting fish populations.

5. What is the future of hydroelectric power?

The future of hydroelectric power looks promising as it is a renewable energy source that does not produce greenhouse gas emissions. However, there are ongoing efforts to improve the technology and minimize its environmental impacts, such as using smaller, run-of-river hydroelectric systems instead of large dams.

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