Glad it helped! There are some other videos on the playlist if you are interested (ruclips.net/p/PL_eoqBWpA6iLOjQaEiG3hQD0hGQyWFlGj). I also developed a course with some of this information, and it's free for all to view on the internet (www.e-education.psu.edu/emsc297/).
Just a minor error in your discharge calculation. You used 0.06309 l/s or Kg/s from 1 gpm conversion in stead of 3.15 l/s or kg/s from 50 gpm. Good work overall!
This might be a very basic doubt. 8:48 while quoting the example of niagra falls, discharge is given. In every tutorials, the discharge or velocity of the water exiting is given. So my question is, during the design stages of a hydroelectric project, how do they calculate power when the discharge rate or velocity is unknown. Can you or anybody explain how discharge or velocity for a given pipe size from a given height can be calculated, please? Feel free to abuse me if I'm being stupid. Thanking in advance, :)
Hi, Thanks for the great video. At 17.30 you use 0.06309l/s instead of 3.1545l/s in the power output formula. Is this a mistake or am I missing something? Thanks!
Sorry, have not checked this for a while. You are correct - I definitely messed that up when I applied the power equation. I wish I could do annotations to update it. Thanks for letting me know.
Please can you solve this question .the reserover area of a hydro-electric generating plant is spread over an area of 4000m^2 with a storage capacity of 8000000m^3 . the net head of water available to the turbine is 70m . Assuming an efficiency of 0.78 and 0.93 for water turbine and generator respectively, calculate the electrical energy generated by plant. Estimate the difference in water level if a load of 30MW,is continuously supplied by the generator for 6 hours
This is not my expertise, but the equation for output is P (kW) = net head (m) X system efficiency X flow (m3/s). You can get all of that easily except for flow. I don't know how to determine the flow, but I assume that there is an equation that will tell you that based on the depth of the water (which you can figure out by using the area and the storage capacity) and the net head, but I am not sure. This paper might help (cdn.intechopen.com/pdfs/40550/InTech-Hydro_power.pdf). A version of that equation is on p. 98. You can figure out the volume of water that went through the turbine after 30 MW of generation for 6 hours by using the same equation and solving for flow when output is 30 MW. Then you can use that the figure out how much the level would drop by using the area of the reservoir. Hope this helps! I honestly don't know how to figure out the first question, but the second one should work if you run those numbers.
Hi Dan, I have a quick question regards a calculation I have - I have calculated the power of the following - FLow = 2 litres/s, G = 9.81, Head = 15M, Efficiency = 75%, Head efficiency = 10%. P as per my calculations = 199Watts (Im not as concerned about the calculations and more what the P in my instance means). If the Flow is constant, is it right to say that my setup would produce 199 Watts per second, and 11919 Watts per minute, 6.264705240 Giga Watts in a year? Is this the right way to look at it or what am I missing? I want to understand what a hydro setup like the one outlined above would produce in a year in KwH so I can compare it to the average house consumption.
Sorry, Jaleel. I have no way of doing that. I just searched for "3d animation hydroelectric power plant" on RUclips and a bunch of videos came up, though.
Dear sir, The equation you applied for hydro power calculation does not include ρ, to the best of my knowledge, the equation is: P=ρQgHη. Can you please explain a little bit? Thanks for your time. Regards Leroy
Sorry for the delay. You are correct, but my formula used liters/s for Q instead of cubic meters/s. Since there are 1, 000 liters in a cubic meter, and water has a density of 1,000 kg/cubic meter, you can skip the density part if you use l/s. Hopefully that makes sense - it's a bit difficult to explain without writing on a board!
This has come timely for me. Next week I will be participating in the determination of Pelton Turbine Unit efficiency. Keep it up Dan.
Thank you for your video 😊 greatly appreciated, more power to you sir!
GOD bless!!!
Thanks very much 👍. This is very interesting. Keep it up 🎉
Appreciated sir. Such a Nice and easy Explanation.
it is helpful but you have forgotten density of water when you calculate the power . 13:00
I appreciated this video👍
Glad it helped! There are some other videos on the playlist if you are interested (ruclips.net/p/PL_eoqBWpA6iLOjQaEiG3hQD0hGQyWFlGj). I also developed a course with some of this information, and it's free for all to view on the internet (www.e-education.psu.edu/emsc297/).
this really helped me a lot
Just a minor error in your discharge calculation. You used 0.06309 l/s or Kg/s from 1 gpm conversion in stead of 3.15 l/s or kg/s from 50 gpm. Good work overall!
This might be a very basic doubt.
8:48 while quoting the example of niagra falls, discharge is given.
In every tutorials, the discharge or velocity of the water exiting is given.
So my question is, during the design stages of a hydroelectric project, how do they calculate power when the discharge rate or velocity is unknown.
Can you or anybody explain how discharge or velocity for a given pipe size from a given height can be calculated, please?
Feel free to abuse me if I'm being stupid.
Thanking in advance,
:)
Thinking the same thing right now
What if we want to fill an artificial reservoir; to pump water back up.....will it need the same amount of energy?
Thank you for your video and Can you give the presentation link
Hi,
Thanks for the great video.
At 17.30 you use 0.06309l/s instead of 3.1545l/s in the power output formula. Is this a mistake or am I missing something?
Thanks!
Hi thanks for the video
There was a mistake
U put the value of 1gpm instead of 50gpm
The result 2.25 should be multiplied by 50
Stephen Burke u r right
Sorry, have not checked this for a while. You are correct - I definitely messed that up when I applied the power equation. I wish I could do annotations to update it. Thanks for letting me know.
Although I just noticed that the answer is correct, but the number in the formula is wrong...
You are right he missed it the answer is 2.228KW
Hi, you have mentioned that there is a wind power calculation class (or video), can you please share the link?
If we have a pressure of 16 Bar and a flow Q of 160 liters per second, what is the capacity of the station and how to calculate that
In case of niagara fall the total gained power is 1.2GW. What is the total of energy in Joules? And how to convert the 1.2GW in GWh please?
Please can you solve this question
.the reserover area of a hydro-electric generating plant is spread over an area of 4000m^2 with a storage capacity of 8000000m^3 . the net head of water available to the turbine is 70m . Assuming an efficiency of 0.78 and 0.93 for water turbine and generator respectively, calculate the electrical energy generated by plant. Estimate the difference in water level if a load of 30MW,is continuously supplied by the generator for 6 hours
This is not my expertise, but the equation for output is P (kW) = net head (m) X system efficiency X flow (m3/s). You can get all of that easily except for flow. I don't know how to determine the flow, but I assume that there is an equation that will tell you that based on the depth of the water (which you can figure out by using the area and the storage capacity) and the net head, but I am not sure. This paper might help (cdn.intechopen.com/pdfs/40550/InTech-Hydro_power.pdf). A version of that equation is on p. 98.
You can figure out the volume of water that went through the turbine after 30 MW of generation for 6 hours by using the same equation and solving for flow when output is 30 MW. Then you can use that the figure out how much the level would drop by using the area of the reservoir.
Hope this helps! I honestly don't know how to figure out the first question, but the second one should work if you run those numbers.
18:00 a mistake: you wrote the value of 1L but calculated the actual 50gpm. Messes the mind before one understands it is not bad maths but a mistake
Hi Dan, I have a quick question regards a calculation I have - I have calculated the power of the following - FLow = 2 litres/s, G = 9.81, Head = 15M, Efficiency = 75%, Head efficiency = 10%. P as per my calculations = 199Watts (Im not as concerned about the calculations and more what the P in my instance means). If the Flow is constant, is it right to say that my setup would produce 199 Watts per second, and 11919 Watts per minute, 6.264705240 Giga Watts in a year? Is this the right way to look at it or what am I missing? I want to understand what a hydro setup like the one outlined above would produce in a year in KwH so I can compare it to the average house consumption.
Its been a year lol, i don't think he is answering😂😂😂
12:20 "...it's Greek, basically" And non basically it might be non-Greek or any different from Greek? ;)
Whats the difd btw kw and kwh?
Which software areuysing to teach?
Hi, can you upload 3d animation video of these chapter.
Sorry, Jaleel. I have no way of doing that. I just searched for "3d animation hydroelectric power plant" on RUclips and a bunch of videos came up, though.
hi sir I need some calculation about flywheel can u help me
Sorry, Sajjad. I am not familiar with that calculation.
The falls had enough energy to send Marty back home. 1.2 Giga watts.
Dear sir,
The equation you applied for hydro power calculation does not include ρ, to the best of my knowledge, the equation is: P=ρQgHη. Can you please explain a little bit? Thanks for your time.
Regards
Leroy
Sorry for the delay. You are correct, but my formula used liters/s for Q instead of cubic meters/s. Since there are 1, 000 liters in a cubic meter, and water has a density of 1,000 kg/cubic meter, you can skip the density part if you use l/s. Hopefully that makes sense - it's a bit difficult to explain without writing on a board!