Question

A counter flow, shell and tube type heat exchanger is to be used to cool water from TH, in = 22 \(^\circ\text{C}\) to TH, out = 6 \(^\circ\text{C}\), using brine entering at TC, in = -2 \(^\circ\text{C}\) and leaving at TC, out = 3 \(^\circ\text{C}\). The logarithmic mean temperature difference (\(\Delta T_m\)) for such system will be

• A. \(\frac{19-8}{\ln(19/8)}\)
• B. \(\frac{16-5}{\ln(16/5)}\)
• C. \(\frac{22-6}{\ln(8/3)}\)
• D. \(\frac{24-9}{\ln(24/9)}\)

Hint:

For counter-flow, remember that fluids travel in opposite directions. Thus, the inlet of one fluid meets the outlet of the other:
\(\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}}\) and \(\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Logarithmic Mean Temperature Difference (LMTD) is used to analyze temperature-change profiles in heat exchangers.

Step 2: Key Formula or Approach:
For a counter-flow heat exchanger, LMTD is defined as:
\[ \Delta T_m = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} \] where:
\(\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}}\) (temperature difference at one end)
\(\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}}\) (temperature difference at the other end)

Step 3: Detailed Explanation:
Given values for hot fluid (water):
\[ T_{h,\text{in}} = 22 ^\circ\text{C}, \quad T_{h,\text{out}} = 6 ^\circ\text{C} \] Given values for cold fluid (brine):
\[ T_{c,\text{in}} = -2 ^\circ\text{C}, \quad T_{c,\text{out}} = 3 ^\circ\text{C} \] Now, calculate the terminal temperature differences:
\[ \Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 22 - 3 = 19 ^\circ\text{C} \] \[ \Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 6 - (-2) = 8 ^\circ\text{C} \] Substitute \(\Delta T_1\) and \(\Delta T_2\) into the LMTD formula:
\[ \Delta T_m = \frac{19 - 8}{\ln(19/8)} \]

Step 4: Final Answer:
The correct option is 1, which corresponds to \(\frac{19-8}{\ln(19/8)}\).