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LMTD of a counter flow heat exchanger is 20°C and cold fluid enters at 20°c & hot fluid at 100°C, mass flow rate of cold fluid is twice that of mass flow rate of hot fluid. $C_p$ of hot fluid is twice that of cold fluid's. How to find the exit temp. of Cold fluid?

I did it in this way.

Its 40°C

$$\begin{align} LMTD &= \dfrac{\Delta T_1 - \Delta T_2}{\ln\left(\dfrac{\Delta T_1}{\Delta T_2}\right)} \\ \Delta T_2 &= \Delta T_1 \\ \therefore LMTD &= \dfrac{0}{0} \end{align}$$

Let $\Delta T_2 = x\cdot \Delta T_1$

$$\begin{align} LMTD &= \dfrac{\Delta T_1 - x\cdot \Delta T_1}{\ln\left(\dfrac{\Delta T_1}{x\cdot \Delta T_1}\right)} \\ &= \dfrac{(1-x)\Delta T_1}{\ln\left(\dfrac{1}{x}\right)} \\ &= \dfrac{(1-x)\Delta T_1}{-\ln x} \end{align}$$

Applying L'Hopital Rule:

$$\begin{align} LMTD &= \lim_{x\rightarrow1} x \cdot \Delta T_1 \\ &= \Delta T_1 = \Delta T_2 \end{align}$$

20 = Cold outlet Temp - (Cold inlet Temp = 20)

Cold outlet Temp = 40

Similarly, The Output Temp. of Hot Fluid is 80°C

Is this approach correct?

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  • $\begingroup$ This site is not a homework answering service. If you would like some help, provide evidence of prior working & state what aspect you are having difficult with. $\endgroup$ – Fred Oct 19 '19 at 5:58
  • $\begingroup$ Hi Fred, This is Knowledge sharing Site, where you can post your queries, especially when things get trickier. PS: This is not Homework. People are giving wrong answers in other forums, hence wanted to resolve it. And one cannot provide evidences for getting a doubt clarified. Please feel free to answer it. $\endgroup$ – Hari kumar Oct 19 '19 at 6:03
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    $\begingroup$ This looks like a homework question (even if it's not actual homework). In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$ – Wasabi Oct 22 '19 at 13:52
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Your approach is one method to solve a balanced heat exchanger. I might instead have simply divided the equation by $\Delta T_2$ to obtain

$$ \Delta T_{LMTD} = \Delta T_2 \left[\frac{R - 1}{\ln(R)} \right]$$

Finding the limit as $R \rightarrow 1$ gives the same answer.

An alternative approach with the above is to graph the numerator $R - 1$ and denominator $\ln(R)$.

Graph of numerator (red) and denominator (blue) of function

The graph shows in the limit as $R \rightarrow 1$ that the ratio of numerator to denominator approaches unity.

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