What in case if you have to determine the pressure in that tank that is closed, and the water is flowing to an open reservoir with the volume per second given? I tried to solve this question with Bernoulli's equation multiple times but I don't get the same answer that is given.
If the pipe was not open to the environment, say it was a closed pipe, would we still be able to apply Bernoulli's equation to solve for pressure at Point 2. With point 1 still being at the top of the reservoir and a given flowrate
Absolutely correct, I did not make it clear in this old video. I now have an updated video that gives more full explanation about atmospheric pressure in this example: ua-cam.com/video/9vQR5hpNHJ8/v-deo.html
Thank you for all you great videos. I watched "Calculating the power a turbine can generate on a hydroelectric scheme using Bernoulli’s equation" and I can't reach the same velocity of 1.5 m/s at 40m head. Are these videos completely separate concepts? In my problem, I need to calculate Flow when I only have Head and pipe diameter. I'm attempting to calculate potential power for a Micro Hydro system. Is it possible without Flow rate?
Thanks for the feedback! You would either need to flow-rate, or an estimation of the losses from the turbine do solve, else there would be too many unknowns.
This is a very good question. If you increase the diameter at the outlet, the velocity would stay the same and the discharge would increase, as the velocity at the outlet is set purely by the difference in head between the tank and the outlet and a larger area with the same velocity gives larger discharge. However, if you increased the diameter at some local point along the pipe, that is not open the atmosphere, this would lead to a reduction velocity, to conserve mass with respect to the discharge at the outlet, but the same overall discharge. Hope that makes sense?
How the pressure can be energy ...there is no flow in the closed tank..so no flow work is there...there is no flow work or pressure energy..plz explain me sir..
Pressure = density ×gh Density = m/v So, Pressure = (m/v)×gh = mgh/v (mgh) is potential energy so...pressure is also potential energy per cubic meter. So pressure has dimensions of energy in it.
I am just having a sit back while sipping my afternoon green tea as I watch fluids mechanic concepts series streaming in my brain reservoir.
thanks for the clear and concise explanation! liked
Thanks for the comment! I've now remade this video with a real model if interested:
ua-cam.com/video/9vQR5hpNHJ8/v-deo.html
What in case if you have to determine the pressure in that tank that is closed, and the water is flowing to an open reservoir with the volume per second given? I tried to solve this question with Bernoulli's equation multiple times but I don't get the same answer that is given.
Thanks
Please show workdone in this siphoning in joules ?
How much temperature increased since volume and pressure also changing at two tanks ?
If the pipe was not open to the environment, say it was a closed pipe, would we still be able to apply Bernoulli's equation to solve for pressure at Point 2. With point 1 still being at the top of the reservoir and a given flowrate
P1 is not zero is atmospheric same at the orifice, so they cancel out (not because they are zero but because they are the same (P1 =P2))
Absolutely correct, I did not make it clear in this old video. I now have an updated video that gives more full explanation about atmospheric pressure in this example:
ua-cam.com/video/9vQR5hpNHJ8/v-deo.html
@@fluidsexplained1901 thanks,
Thank you for all you great videos.
I watched "Calculating the power a turbine can generate on a hydroelectric scheme using Bernoulli’s equation"
and I can't reach the same velocity of 1.5 m/s at 40m head. Are these videos completely separate concepts?
In my problem, I need to calculate Flow when I only have Head and pipe diameter. I'm attempting to calculate potential power for a Micro Hydro system. Is it possible without Flow rate?
Thanks for the feedback! You would either need to flow-rate, or an estimation of the losses from the turbine do solve, else there would be too many unknowns.
When we increase the section area of the pipe then the discharge will increase?.
This is a very good question. If you increase the diameter at the outlet, the velocity would stay the same and the discharge would increase, as the velocity at the outlet is set purely by the difference in head between the tank and the outlet and a larger area with the same velocity gives larger discharge. However, if you increased the diameter at some local point along the pipe, that is not open the atmosphere, this would lead to a reduction velocity, to conserve mass with respect to the discharge at the outlet, but the same overall discharge. Hope that makes sense?
How the pressure can be energy ...there is no flow in the closed tank..so no flow work is there...there is no flow work or pressure energy..plz explain me sir..
It’s the equivalent to potential energy. Like a car being stationary at the top of a hill.
Pressure = density ×gh
Density = m/v
So,
Pressure = (m/v)×gh
= mgh/v
(mgh) is potential energy so...pressure is also potential energy per cubic meter. So pressure has dimensions of energy in it.