Question

Saturated steam at 100 \(^\circ\)C condenses on the outside of a tube. Cold fluid enters the tube at 20 \(^\circ\)C and exits at 50 \(^\circ\)C. The value of the Log Mean Temperature Difference (in \(^\circ\)C) is:

• A. 54.89
• B. 63.83
• C. 72.39
• D. 89.36

Hint:

For phase-change heat exchangers (condensers/evaporators), the LMTD is completely identical for both parallel and counter-flow arrangements because the hot or cold fluid temperature remains absolutely flat. As a quick estimation check, the LMTD must always lie between the values of \(\Delta T_1\) and \(\Delta T_2\). Here, \(50 < 63.83 < 80\), verifying that our answer falls within the correct physical bound.

Answer 1

The Correct Option is B

Concept: The Log Mean Temperature Difference (LMTD) is utilized to analyze heat transfer driving forces in heat exchangers. It represents a logarithmic average of the temperature differences between the hot and cold streams at both ends of the exchanger. The general formula for LMTD is defined as: \[ \text{LMTD} = \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \] where:
• \(\Delta T_1\) is the temperature difference between the two fluids at one end of the exchanger.
• \(\Delta T_2\) is the temperature difference between the two fluids at the alternate end of the exchanger. In phase-change scenarios such as condensation or boiling, one of the fluids maintains a constant temperature throughout the unit regardless of whether the configuration is parallel-flow or counter-flow.

Step 1: Identify fluid temperature parameters
Let the hot fluid be the condensing saturated steam. Since it undergoes a phase change (condensation) at constant pressure, its temperature remains uniform across the entire length of the tube: \[ T_{h, \text{in}} = T_{h, \text{out}} = 100\ ^\circ\text{C} \] Let the cold fluid be the liquid flowing inside the tube. Its inlet and outlet temperatures are provided as: \[ T_{c, \text{in}} = 20\ ^\circ\text{C} \] \[ T_{c, \text{out}} = 50\ ^\circ\text{C} \]

Step 2: Compute terminal temperature differences
Let us establish the temperature differences at both operational boundaries of the heat exchanger unit: At the fluid entrance side: \[ \Delta T_1 = T_{h, \text{in}} - T_{c, \text{in}} = 100\ ^\circ\text{C} - 20\ ^\circ\text{C} = 80\ ^\circ\text{C} \] At the fluid exit side: \[ \Delta T_2 = T_{h, \text{out}} - T_{c, \text{out}} = 100\ ^\circ\text{C} - 50\ ^\circ\text{C} = 50\ ^\circ\text{C} \]

Step 3: Calculate the logarithmic mean difference
Substitute the calculated values of \(\Delta T_1 = 80\) and \(\Delta T_2 = 50\) into the fundamental LMTD relation: \[ \Delta T_{lm} = \frac{80 - 50}{\ln\left(\frac{80}{50}\right)} \] Simplifying the numerator: \[ 80 - 50 = 30 \] Simplifying the argument within the natural logarithm: \[ \frac{80}{50} = 1.6 \] Now evaluate the natural log value: \[ \ln(1.6) \approx 0.4700036 \] Dividing the numerator by this logarithmic value yields: \[ \Delta T_{lm} = \frac{30}{0.4700036} \approx 63.8293\ ^\circ\text{C} \] Rounding to two decimal places, we get \(63.83\ ^\circ\text{C}\), which corresponds to Option (2).