# Is it wrong to derive Log mean temperature difference (LMTD) of a Heat Exchanger, this way?

I tried deriving the LMTD of a parallel flow double pipe heat exchanger by first finding the temperature profiles of the hot and cold fluid and then averaging them over the entire length of the Heat exchanger. Then I subtracted these average values hoping that it would give me the relation for LMTD, but I'm stuck. The relations for temperature variation and average temperature are in the picture.

EDIT: I have come one step closer to make the relation look like the one for LMTD given in books.

The relation would match with the relation of LMTD given in books if we assume the wall surface temperatures to be equal. Do books take that assumption while deriving LMTD?

Please Help.

• Welcome to stack exchange! To get better reception of your questions, please write the equations in mathjax. Mar 20, 2021 at 9:57
• Oh okay. Thank You. Mar 20, 2021 at 10:44

## 1 Answer

This is not analyzed correctly. $$T_{s1}$$ and $$T_{s2}$$ are not constant, and they need to be eliminated from the analysis all-together. If $$\dot{Q}$$ is the rate of heat flow per unit length along the exchanger, then $$\dot{Q}=\rho_ih_i(T_h-T_{s1})=\rho_oh_o(T_{s2}-T_c)=\frac{(\rho_i+\rho_o)}{2}\frac{k_W}{\delta}(T_{s1}-T_{s2})$$where $$\delta$$ is the wall thickness and $$k_w$$ is its thermal conductivity. If we eliminate $$T_{s1}$$ and $$T_{s2}$$ from these equations, we obtain: $$\dot{Q}=U\rho (T_h-T_c)$$Where $$\frac{1}{U\rho}=\frac{1}{h_1\rho_i}+\frac{1}{\frac{(\rho_i+\rho_o)}{2}\frac{k_W}{\delta}}+\frac{1}{\rho_oh_o}$$ Based on this, the heat balance equations for the heat exchanger become $$\dot{m}_hC_h\frac{dT_h}{dx}=-U\rho(T_h-T_c)$$and$$\dot{m}_cC_c\frac{dT_c}{dx}=+U\rho(T_h-T_c)$$ I leave it to you to complete the analysis and show that the overall heat load Q is given by:$$Q=(U\rho x_f)\Delta T_{LM}$$

• Thank You for your response. I was able to get to the relation of LMTD following your approach. However I was wondering what goes wrong in my approach? The temperature difference between the hot and cold fluid varies along the HX, the average of this temperature difference variation is LMTD (correct me if I'm wrong). So I just tried going the other way round I first calculated the average value of the temperature variations of the hot and cold fluid and then took the difference. What goes wrong in my approach? Please help if you could. Struggling on this for a week now. Mar 19, 2021 at 11:44
• The LMTD is a kind of average of the temperature difference. It is the value that gives the correct heat load, or Q/UA. Plus, the hot and cold temperatures vary in tandem, so you can/t take them individually. Your calculated variations of the temperatures with x are incorrect. Plus, the wall temperature varies axially, and this needs to be accounted for. Sorry to say that, in my judgment, every single part of the analysis was done incorrectly. Mar 19, 2021 at 13:09
• Ohh I'm sorry I forgot to mention, that any heat transfer in the axial direction is neglected. I'm doing my graduation, they have taken this simplified approach at the preliminary level. I have calculated the temperature variations from the concepts of internal flow in pipes with heat transfer. If we have a pipe inside which a heated fluid flows the temperature variation along the pipe will be exponential. That's what I learnt in the previous chapters. Wanted to ask, are you pretty much sure that the temperature variations are wrong, because if they are then I ll search for what is right. Mar 19, 2021 at 13:36
• Yes, the axial heat transfer is certainly negligible. But your analysis neglects the axial temperature variations of the wall temperatures, even though the heat transfer in the radial direction changes with axial position. The wall temperatures are simply not independent of axial position. If you analyzed the problem the way I indicated (which is what you said you did), then you should have found axiall temperature variations for the fluids that differ from the relationships you originally used. That is because the fluid temperatures are correlated with one another. The temperate Mar 19, 2021 at 13:45
• variations of each of the fluids depends on $\dot{m}C$ for both the fluids. Mar 19, 2021 at 13:46