Heat Exchangers

Question

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?

Answer 1

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 $\ln(R)$.

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