Question

In a counter flow heat exchanger, cold fluid enters at 30\(^\circ\)C and leaves at 50\(^\circ\)C, whereas the hot fluid enters at 150\(^\circ\)C and leaves at 130\(^\circ\)C. What is the mean temperature difference?

• A. 50\(^\circ\)C
• B. 120\(^\circ\)C
• C. 80\(^\circ\)C
• D. 100\(^\circ\)C

Hint:

For any counter-flow heat exchanger, if the temperature drop of the hot fluid equals the temperature rise of the cold fluid, the temperature differences at both ends are identical (\(\Delta T_1 = \Delta T_2\)). In this case, you do not need to use logarithms; the LMTD is simply equal to this common temperature difference.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The mean temperature difference in a heat exchanger is calculated as the Logarithmic Mean Temperature Difference (LMTD).
This value accounts for the non-linear temperature profiles of both the hot and cold fluids as they flow through the system.
Key Formula or Approach:
For a counter-flow heat exchanger, the temperature differences at the two ends are:
\[ \Delta T_1 = T_{\text{h,in}} - T_{\text{c,out}} \]
\[ \Delta T_2 = T_{\text{h,out}} - T_{\text{c,in}} \]
The Logarithmic Mean Temperature Difference (LMTD) is defined as:
\[ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \]
In cases where \(\Delta T_1 = \Delta T_2\), the mathematical limit of the LMTD expression equals this common value:
\[ \text{LMTD} = \Delta T_1 = \Delta T_2 \]

Step 2: Detailed Explanation:
Let us identify the given temperatures:
- Hot fluid inlet temperature, \(T_{\text{h,in}} = 150\ ^\circ\text{C}\)
- Hot fluid outlet temperature, \(T_{\text{h,out}} = 130\ ^\circ\text{C}\)
- Cold fluid inlet temperature, \(T_{\text{c,in}} = 30\ ^\circ\text{C}\)
- Cold fluid outlet temperature, \(T_{\text{c,out}} = 50\ ^\circ\text{C}\)
Now, calculate the temperature differences at the two ends of the counter-flow heat exchanger:
\[ \Delta T_1 = 150\ ^\circ\text{C} - 50\ ^\circ\text{C} = 100\ ^\circ\text{C} \]
\[ \Delta T_2 = 130\ ^\circ\text{C} - 30\ ^\circ\text{C} = 100\ ^\circ\text{C} \]
Since the temperature differences at both ends are identical (\(\Delta T_1 = \Delta T_2 = 100\ ^\circ\text{C}\)), the temperature profiles of the two fluids are parallel lines.
The Logarithmic Mean Temperature Difference is equal to this constant temperature difference:
\[ \text{LMTD} = 100\ ^\circ\text{C} \]

Step 3: Final Answer:
The mean temperature difference of the heat exchanger is 100\(^\circ\)C. Hence, the correct option is (D).