Parallel & Counter Flow Heat Exchangers (LMTD): Heat Transfer for Mechanical Engineers

In this problem, we design a shell and tube heat exchanger. Specifically, we look at the difference in heat transfer area required by using a parallel or counter flow heat exchangers.

Heat exchanger problems typically have one of two potential main goals: (a) find the heat transferred between two fluids, OR (b) find the area needed to achieve a required target heat transfer between two fluids. In this example, we go through a systematic process to find the area required for shell and tube heat exchangers using the log mean temperature different (LMTD) method. I believe that understanding this process will allow you to solve any heat exchanger problem.

[davidholmes]

thanks a lot, great explanation! Maybe you should consider using a pop filter over your mic. Just trying to help and not trying to be rude.

[nawafaleid]

Your just awesome, very clear logical explaining.. Can I ask you for a copy of the matrix ?

[mahirkarim]

When calculating the area of the parallel or counter flow heat exchanger, I noticed that the units in the denominator do not align. You left the log-mean temperature in units of Celsius but the overall HT coefficient as W/m^2 K. In this case, wouldn't we have to convert the log-mean temperature to Kelvin in order to get the correct area for the heat exchangers?

[sevenmaster]

I don't have effective heat transfer co efficient value! And not even hi and ho value