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Consider a double pipe parallel flow heat exchanger. For such a heat exchanger the LMTD is given as

$$\Delta T _{LM}= \frac{\Delta T_i-\Delta T_e}{ln(\frac{\Delta T_i}{\Delta T_e})} $$

The relation above can be derived by using the following approach - https://www.ques10.com/p/15892/derive-the-expression-for-log-mean-temperature-dif/

This is the relation that we get for mean temperature difference when the temperature variations of the hot and cold fluids are exponential.

However, there is also something known as arithmetic temperature difference which if used instead of LMTD will give us incorrect results. I was wondering if the temperature variations of the hot and cold fluid were linear then would the arithmetic temperature difference would have given correct results?

If yes, then if we follow the same method used for deriving the mean temperature difference when the temperature profiles were exponential, in a case where the temperature profiles are linear, then we should get the arithmetic temperature difference. However, by following similar approach and assuming the temperature profiles to be linear I still get a logarithmic mean. enter image description here

Whether I go by taking an exponential temperature profile or linear temperature profile I get the same result, which is - a logarithmic mean temperature difference. Shouldn't the mean temperature difference when having a linear temperature profile be,

$$\Delta T _{AM}= \frac{\Delta T_i-\Delta T_e}{2} $$

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  • $\begingroup$ The temperature profiles can't be linear if the streams are exchanging heat with one another and the heat transfer coefficient is constant. A linear profile is inconsistent with what is happening physically. So, if you assume a linear profile, one of your equations will not be satisfied. $\endgroup$
    – Chet Miller
    Commented Aug 14, 2021 at 11:36
  • $\begingroup$ Yes Sir, the temperature profiles can't be linear in real. What I'm trying to figure out is that when the temperature profiles are exponential then LMTD gives us correct results, but the use of AMTD will give rise to errors. I read somewhere that had the temperature profiles been linear then the use of AMTD would have given correct results. Hence, I was trying to start with a case of linear temperature profiles and was expecting to get AMTD relation. $\endgroup$
    – Harshit Rajput
    Commented Aug 14, 2021 at 11:53
  • $\begingroup$ I was able to come up with the AMTD relation by assuming linear temperature profile by averaging the linear temperature function (which becomes very easy to determine by just a knowledge of inlet and outlet temperature and length) over the entire length of heat exchanger but by using the method above (as given in the picture) i wasnt able to $\endgroup$
    – Harshit Rajput
    Commented Aug 14, 2021 at 11:55

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If the temperature difference is linear with x, then $$\Delta T=\Delta T_{in}+(\Delta T_{out}-\Delta T_{in})\frac{x}{L}$$So, $$d\dot{Q}_x=UP\left[\Delta T_{in}+(\Delta T_{out}-\Delta T_{in})\frac{x}{L}\right]dx$$If you integrate this between x = 0 and x = L, you get $$\dot{Q}_{total}=UPL\frac{(\Delta T_{in}+\Delta T_{out})}{2}$$

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