shouldn't the overall heat transfer coefficient be U = (1/h_oil + (radius_oil / radius_water) * 1/h_water) ^ -1? Since we're in cylindrical coordinates there has to be the ratio of pipe radii to account for a constant heat flux in the varying pipes diameters?
You're correct about needing the pipe radii ratio, but your equation is back to front (done w.r.t. to h_inner, not h_outer); U = (1/h_oil * [r_water / r_oil] + 1/h_water) ^ -1
Only convection is needed; pipe wall is described as thin (r_pipe_inner-wall ≈ r_pipe_outer-wall), therefore resistance through pipe wall is negligible (but yes, presenter didn't include pipe ratios).
I like the way you explain is very clear, thank you for your input!!, I am trying to calculate the length of a inmerse evaporator to produce 1 ton of ice, like iceblock through a brine but would like to know the dimensions to use, I would appreciate if you can give me an idea if the same formula applies
Good video for understanding concepts for first year Chem engg students but bad video for actual engineers sizing heat exchangers. Often times people are yelling and screaming not knowing the heat transfer coefficient. That is never known. No one tells you either. You have to work it out using the dimensions number equations. This video basically shows you the last step in designing the exchanger (if you don't include pressure drop calculations) when all values are known.
shouldn't the overall heat transfer coefficient be
U = (1/h_oil + (radius_oil / radius_water) * 1/h_water) ^ -1? Since we're in cylindrical coordinates there has to be the ratio of pipe radii to account for a constant heat flux in the varying pipes diameters?
You're correct about needing the pipe radii ratio, but your equation is back to front (done w.r.t. to h_inner, not h_outer); U = (1/h_oil * [r_water / r_oil] + 1/h_water) ^ -1
No. That is why in the beginning of the problem she assumes "thin walled" so she can throw out that term.
Can you tell h values of water and oil,how it take
Why in the design equation there is no conductivity coefficient since using different material for heat transfer will have different conductivity
Only convection is needed; pipe wall is described as thin (r_pipe_inner-wall ≈ r_pipe_outer-wall), therefore resistance through pipe wall is negligible (but yes, presenter didn't include pipe ratios).
I like the way you explain is very clear, thank you for your input!!, I am trying to calculate the length of a inmerse evaporator to produce 1 ton of ice, like iceblock through a brine but would like to know the dimensions to use, I would appreciate if you can give me an idea if the same formula applies
Thanks a lot.. That was very helpful
please can you tell me what kind of software did you use to do this educational vedio? i love it, thank you.
thanks alot u have help
i love you
Good video for understanding concepts for first year Chem engg students but bad video for actual engineers sizing heat exchangers. Often times people are yelling and screaming not knowing the heat transfer coefficient. That is never known. No one tells you either. You have to work it out using the dimensions number equations. This video basically shows you the last step in designing the exchanger (if you don't include pressure drop calculations) when all values are known.