I really like the lectures content, the clarity and the graphics. Thank you very much. I think you are helping a lot of aspiring professional Engineers. [Q=731,850 W, Re=232,401, NU=750, A=27sq.m, L=4.30m]
05:46 At T=323K, the dynamic viscosity (mu) should be 548*10^-6 Ns/m^2 (typo with the units on the slide). The density (1/specific volume) should be 987 kg/m^3 (not 0.987 kg/^3). 12:13 With these, I computed the Reynolds Number to be 2.32*10^5. Similar to Eduardo Suarez's result in the other comment. Thanks again for putting these videos up on UA-cam.
In problems like these, we need to (1) define fluid properties and (2) pick a good equation to find the Re. You can see where we get properties at around 5:00. We get properties for water at around 6:05. We put that information into the Re equation at ~ 11:55. This equation might look a bit different than you are used to, but you could also use (/rho)(U)(D)/(\mu) = (U)(D)/( u). Note that in the heat exchanger videos, I seem to have often made the mistake of taking the specific volume (v_f) instead of the kinematic viscosity (/nu). But in this problem, the equation with mass flow rate works better because (m\dot) is given in the problem, as is the tube diameter (remember to convert to m). So the only property we needed to look up for Re is the dynamic viscosity (\mu). Others have pointed out that there may be a clerical error in the math here...I didn't go back to check if they are correct. The general method is correct, but you might want to double check the algebra.
At first glance this looks like it would work. But we need to remember that the Delta(T) for convection is the temperature difference between the fluid and the tube surface. And the temperature of the tube surface varies along the length of the tube.
I really like the lectures content, the clarity and the graphics. Thank you very much. I think you are helping a lot of aspiring professional Engineers. [Q=731,850 W, Re=232,401, NU=750, A=27sq.m, L=4.30m]
That is very kind of you Eduardo. I'm glad that you find the videos useful!
05:46 At T=323K, the dynamic viscosity (mu) should be 548*10^-6 Ns/m^2 (typo with the units on the slide). The density (1/specific volume) should be 987 kg/m^3 (not 0.987 kg/^3).
12:13 With these, I computed the Reynolds Number to be 2.32*10^5. Similar to Eduardo Suarez's result in the other comment.
Thanks again for putting these videos up on UA-cam.
Good catch. Looks like I was doing these too quickly. I appreciate you pointing out these errors.
At 8:31 you say Cmin is W/K but at 13:12 Cmin is W/kgK, I am a little confused by this?
can i have a document on thermophysics properties of saturated please? thank!
hi .. thanks for the video
can i ask you where are the ten tubes ? I know the 8 passes but can't figure out where are the 10 tubes?
The tubes aren't shown in the picture. The sketch shows a single cross section. There would be ten tubes lined up in parallel (into the page).
Amazing video thank you
Can anyone how explain he got the Reynolds number please
In problems like these, we need to (1) define fluid properties and (2) pick a good equation to find the Re. You can see where we get properties at around 5:00. We get properties for water at around 6:05. We put that information into the Re equation at ~ 11:55. This equation might look a bit different than you are used to, but you could also use (/rho)(U)(D)/(\mu) = (U)(D)/(
u). Note that in the heat exchanger videos, I seem to have often made the mistake of taking the specific volume (v_f) instead of the kinematic viscosity (/nu). But in this problem, the equation with mass flow rate works better because (m\dot) is given in the problem, as is the tube diameter (remember to convert to m). So the only property we needed to look up for Re is the dynamic viscosity (\mu). Others have pointed out that there may be a clerical error in the math here...I didn't go back to check if they are correct. The general method is correct, but you might want to double check the algebra.
Was it better to use:
Q= {mCp(T2-T1)}c = {hA(T1-T2)}h , and calculate HT area directly?
At first glance this looks like it would work. But we need to remember that the Delta(T) for convection is the temperature difference between the fluid and the tube surface. And the temperature of the tube surface varies along the length of the tube.
Very helpful