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. 
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,
$$\Delta T _{AM}= \frac{\Delta T_i-\Delta T_e}{2} $$
Best Answer
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 $$\dot{Q}_{total}=UPL\frac{(\Delta T_{in}+\Delta T_{out})}{2}$$