Question

In a concentric tube counter flow heat exchanger, hot oil enters at $102\text{ }^{\circ}\text{C}$ and leaves at $65\text{ }^{\circ}\text{C}$. Cold water at $25\text{ }^{\circ}\text{C}$ and leaves at $42\text{ }^{\circ}\text{C}$. The log mean temperature difference (LTMD) is _______}

• A. $65.25\text{ }^{\circ}\text{C}$
• B. $49.32\text{ }^{\circ}\text{C}$
• C. $74.12\text{ }^{\circ}\text{C}$
• D. $102.6\text{ }^{\circ}\text{C}$

Hint:

For counter-flow, always draw a simple schematic showing fluid directions to avoid mixing up the inlet and outlet temperatures when computing $\Delta T_1$ and $\Delta T_2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks for the Logarithmic Mean Temperature Difference (LMTD) of a counter-flow concentric tube heat exchanger with given inlet and outlet temperatures.

Step 2: Key Formula or Approach:
For a counter-flow heat exchanger, the temperature differences at the two ends are defined as:
\[ \Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} \]
\[ \Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} \]
The Log Mean Temperature Difference (LMTD) is calculated using:
\[ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \]

Step 3: Detailed Explanation:

• Identify the given temperatures:
- Hot fluid inlet temperature, $T_{h,\text{in}} = 102\text{ }^{\circ}\text{C}$
- Hot fluid outlet temperature, $T_{h,\text{out}} = 65\text{ }^{\circ}\text{C}$
- Cold fluid inlet temperature, $T_{c,\text{in}} = 25\text{ }^{\circ}\text{C}$
- Cold fluid outlet temperature, $T_{c,\text{out}} = 42\text{ }^{\circ}\text{C}$

• Calculate the temperature differences at the two ends of the counter-flow heat exchanger:
\[ \Delta T_1 = 102 - 42 = 60\text{ }^{\circ}\text{C} \]
\[ \Delta T_2 = 65 - 25 = 40\text{ }^{\circ}\text{C} \]

• Calculate the LMTD using the formula:
\[ \text{LMTD} = \frac{60 - 40}{\ln\left(\frac{60}{40}\right)} = \frac{20}{\ln(1.5)} \]

• Find the value of $\ln(1.5)$:
\[ \ln(1.5) \approx 0.405465 \]

• Perform the division:
\[ \text{LMTD} = \frac{20}{0.405465} \approx 49.32\text{ }^{\circ}\text{C} \]

Step 4: Final Answer:
The log mean temperature difference is $49.32\text{ }^{\circ}\text{C}$.