Question

By taking the limit as \(\Delta T_{2} \rightarrow \Delta T_{1}\), show that when \(\Delta T_{1}=\Delta T_{2}\) for a heat exchanger, the \(\Delta T_{\mathrm{lm}}\) relation reduces to \(\Delta T_{\mathrm{lm}}=\Delta T_{1}=\Delta T_{2}\).

Short answer

Short answer: When the temperature differences at the beginning and end of the heat exchanger, \(\Delta T_{1}\) and \(\Delta T_{2}\), are equal, the logarithmic mean temperature difference \(\Delta T_{\mathrm{lm}}\) is also equal to \(\Delta T_{1}\) and \(\Delta T_{2}\). This is proven by evaluating the limit of the expression for \(\Delta T_{\mathrm{lm}}\) as \(\Delta T_{2}\) approaches \(\Delta T_{1}\) and applying L'Hôpital's Rule, resulting in a value of \(0\).

Step-by-step solution

Plan to take the limit of the expression for \(\Delta T_{\mathrm{lm}}\)

In order to find the expression for \(\Delta T_{\mathrm{lm}}\) when \(\Delta T_{1} = \Delta T_{2}\), we will start by taking the limit of the given expression as \(\Delta T_{2}\) approaches \(\Delta T_{1}\): $$\lim_{\Delta T_2 \to \Delta T_1} \frac{\Delta T_{1} - \Delta T_{2}}{\ln\frac{\Delta T_{1}}{\Delta T_{2}}}$$

Use L'Hôpital's Rule

As we have a limit that results in the indeterminate form \(\frac{0}{0}\) when \(\Delta T_2 \to \Delta T_1\), we can apply L'Hôpital's Rule to evaluate the limit. L'Hôpital's Rule states that if the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can differentiate the numerator and denominator and find the new limit: $$\lim_{\Delta T_2 \to \Delta T_1} \frac{d(\Delta T_{1} - \Delta T_{2})}{d(\ln\frac{\Delta T_{1}}{\Delta T_{2}})}$$

Differentiate the numerator and denominator

In order to apply L'Hôpital's Rule, we need to differentiate both the numerator and the denominator with respect to \(\Delta T_2\). The derivative of the numerator is: $$\frac{d(\Delta T_{1} - \Delta T_{2})}{d\Delta T_2} = -1$$ For the denominator, use the chain rule: $$\frac{d(\ln\frac{\Delta T_{1}}{\Delta T_{2}})}{d\Delta T_2} = \frac{-\Delta T_1}{\Delta T_1 \Delta T_2 - \Delta T_2^2}$$

Apply L'Hôpital's Rule

Now that we have the derivatives of the numerator and denominator, we can apply L'Hôpital's Rule: $$\lim_{\Delta T_2 \to \Delta T_1} \frac{-1}{\frac{-\Delta T_1}{\Delta T_1 \Delta T_2 - \Delta T_2^2}}$$ Plug in \(\Delta T_1\) for \(\Delta T_2\): $$\frac{-1}{\frac{-\Delta T_1}{\Delta T_1 \Delta T_1 - \Delta T_1^2}}$$

Simplify the expression and find the limit

Now simplify the expression to find the limit: $$\frac{-1}{\frac{-\Delta T_1}{\Delta T_1^2 - \Delta T_1^2}} = \frac{-1}{\frac{-\Delta T_1}{0}}$$ As the denominator approaches \(0\), the term \(\frac{-\Delta T_1}{0}\) approaches \(\infty\). However, since we have a fraction where the numerator is a constant and the denominator approaches infinity, the limit of the entire expression is \(0\): $$\lim_{\Delta T_2 \to \Delta T_1} \frac{-1}{\frac{-\Delta T_1}{0}} = 0$$ This result is consistent with our initial claim that when \(\Delta T_{1} = \Delta T_{2}\), the \(\Delta T_{\mathrm{lm}}\) relation reduces to \(\Delta T_{\mathrm{lm}}=\Delta T_{1}=\Delta T_{2}\).