Measuring energy capture of river turbine

[gloo68]

I understand that work done if we have (510,000 Kg or 5,000,000 N) of water falling 30 meters (like in a dam) would just be 5,000,000N * 30m= 150,000,000 Newton meters.

But how would one measure the work captured by a turbine for a river that say travels 50km/h. I guess assuming the same mass of water passing by it , 510,000kg). I am not sure what acceleration I am suppose to use...is it 9.81 because the water is falling and thus the speed of 50km/h is irrelevant?

 

[mfb]

It depends on the turbine. As an upper estimate, you can assume the whole kinetic energy of the water gets used, and you can work out the kinetic energy of 510,000 kg of water moving at 50km/h.
No real turbine will reach this upper limit as the water has to keep some velocity to leave the turbine, but the real efficiency depends on the turbine design then.

(50km/h is really fast for a river, by the way).

 

[gloo68]

Yeah, i was kind of ignoring turbine efficiency in this case or just assuming 100 percent. I was wondering if it was correct to use the same equation of force times distance. I understand in the dam situation it falls 30m before turning the turbine, but what is the distance used in a river flow? I don't want to employ the rate of power equation (Watts)... Just want to understand how to calculate total work done if 510,000 kg of water flows and moves past the turbine. Doesn't seem clear to me? What variables am I missing to calculate this?

 

[mfb]

See post #2, you can calculate the kinetic energy of the moving water. There are no missing variables.

 

[gloo68]

hmmm...so w=f*d

f=510,000kg*9.81m/s
= 5,000,000 N

w=5,000,000 *50km
= =250,000,000 J

So in one hour the turbine will capture the entire 510,000Kg of water moving at 50km/h for an equivalent of 250 million joules??

Forgive me, but I am not a physics major but a finance guy. Havent' done this stuff in over 30 years.

 

[mfb]

Does your water fall from a height of 50km? Certainly not.
50km/h is a velocity, 50km is a distance. They are completely different things.

$E=\frac{1}{2} m v^2$. If you plug in the mass in kg and the velocity v in m/s, you get the result in Joule.

 

[gloo68]

50km/h = 0.014km/s

E=0.5 *510,000kg * 0.014km/s *0.014km/s
=49.98 Joules??

Does that look right?

 

[mfb]

0.014km/s = 14m/s
You have to multiply your result by 1 million. 1J = 1kg m²/s²

14m/s is approximately the velocity of water falling down by 10 meters, or 1/3 the distance of the other problem. It is not a coincidence that the energy is also 1/3.

 

[gloo68]

So 50,000,000 Joules give or take. That seems more right than 50.

So if it was a really long river, it would be viable to put as many of these in as possible? I guess the last one may theoretically receive less because the water may be slowed down by the previous ones? I wonder why they don't do that with hydroelectric dams before they fall down the penstocks?? Put a series of turbines before it reaches the dam and the water falls.

 

[russ_watters]

So 50,000,000 Joules give or take. That seems more right than 50.

So if it was a really long river, it would be viable to put as many of these in as possible? I guess the last one may theoretically receive less because the water may be slowed down by the previous ones? I wonder why they don't do that with hydroelectric dams before they fall down the penstocks?? Put a series of turbines before it reaches the dam and the water falls.

Two reasons:
1. The input speed you provided is wildly unrealistic, so your power/energy calculation is about a million times too high.
2. Extracting energy blocks the flow of a river. So one way or another, it's a dam and one big dam with big turbines is more practical than thousands of small ones.

Also, since energy generation is continuous, it is a lot more useful to calculate the rate of capture in watts instead of the total capture in joules: you won't know the total until the day you close the plant!

 

[mfb]

So 50,000,000 Joules give or take. That seems more right than 50.

So if it was a really long river, it would be viable to put as many of these in as possible? I guess the last one may theoretically receive less because the water may be slowed down by the previous ones? I wonder why they don't do that with hydroelectric dams before they fall down the penstocks?? Put a series of turbines before it reaches the dam and the water falls.

The calculation assumes the water gets stopped. You cannot stop water multiple times, unless there is something in between to accelerate it again - like a height difference.

While multiple turbines along a river are possible, they are expensive and often inefficient. A single dam, "collecting" the height difference of a large section of the river, and a few big turbines are more efficient.

 

[gloo68]

eah, i know river don't flow that fast. I was just using examples for calculations. Was just thinking about the Verdant power company putting in all those small turbines in the Hudson River to catch tidal flow. Curious about their economics cause i heard it was so tough to put in the turbines at river bed...they were so strong it snapped the turbine blade.

Thanks to both of you (Russ and mfb)

 

[russ_watters]

You're welcome, but I really don't understand: the question seems real/legitimate so it makes no sense to me to try to answer it with a calculation known to be seriously flawed. Seems like wasted effort to me.

 

[gloo68]

I know my questions are strange. To be honest Russ, i have a patent that I am trying to market which has relevance for my strange questions. And no, it's not some kind of perpetual motion wheel in the water...most people think that right away.

I appreciate your help as always Russ.