Question

Water (\(C_{\text{p}} = 4.18\text{ kJ/kg}\cdot\text{K}\)) at \(80^\circ\text{C}\) enters a counter-flow heat exchanger with a mass flow rate of \(0.5\text{ kg/s}\). Air (\(C_{\text{p}} = 1\text{ kJ/kg}\cdot\text{K}\)) enters at \(30^\circ\text{C}\) with a mass flow rate of \(2.09\text{ kg/s}\). If the effectiveness of the heat exchanger is 0.8, the LMTD (\(^\circ\text{C}\)) is

• A. 40
• B. 20
• C. 10
• D. 5

Hint:

For any balanced counter-flow heat exchanger ($C_{\text{h}} = C_{\text{c}}$), the temperature difference is uniform along the length.
The LMTD is equal to this constant temperature difference, which can be computed directly as $\Delta T = (1 - \epsilon)(T_{\text{h,in}} - T_{\text{c,in}})$.
Applying this formula here: $(1 - 0.8)(80 - 30) = 0.2 \times 50 = 10^\circ\text{C}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to calculate the Log Mean Temperature Difference (LMTD) for a counter-flow heat exchanger with known fluid capacities, inlet conditions, and heat exchanger effectiveness.

Step 2: Key Formula or Approach:
The heat capacity rate \(C\) is calculated as:
\[ C = \dot{m} C_{\text{p}} \]
We must identify \(C_{\text{h}}\) for the hot fluid (water) and \(C_{\text{c}}\) for the cold fluid (air).
The actual heat transfer rate is:
\[ q = \epsilon C_{\text{min}} (T_{\text{h,in}} - T_{\text{c,in}}) \]
The outlet temperatures can be calculated from energy balances:
\[ q = C_{\text{h}} (T_{\text{h,in}} - T_{\text{h,out}}) = C_{\text{c}} (T_{\text{c,out}} - T_{\text{c,in}}) \]

Step 3: Detailed Explanation:

•
Step 3.1: Calculate Heat Capacity Rates:
For the hot fluid (water):
\[ C_{\text{h}} = \dot{m}_{\text{h}} C_{\text{p,h}} = 0.5 \times 4.18 = 2.09\text{ kW/K} \]
For the cold fluid (air):
\[ C_{\text{c}} = \dot{m}_{\text{c}} C_{\text{p,c}} = 2.09 \times 1 = 2.09\text{ kW/K} \]
Since both heat capacity rates are identical (\(C_{\text{h}} = C_{\text{c}} = 2.09\text{ kW/K}\)), this is a special case of balanced flow where the temperature difference between the hot and cold fluids remains constant throughout the length of the exchanger.

•
Step 3.2: Calculate Heat Transfer:
The minimum heat capacity rate \(C_{\text{min}} = 2.09\text{ kW/K}\).
The maximum possible heat transfer is:
\[ q_{\text{max}} = C_{\text{min}} (T_{\text{h,in}} - T_{\text{c,in}}) = 2.09 \times (80 - 30) = 104.5\text{ kW} \]
The actual heat transfer rate is:
\[ q = \epsilon q_{\text{max}} = 0.8 \times 104.5 = 83.6\text{ kW} \]

•
Step 3.3: Determine Exit Temperatures:
For the hot fluid:
\[ 83.6 = 2.09 \times (80 - T_{\text{h,out}}) \implies T_{\text{h,out}} = 80 - 40 = 40^\circ\text{C} \]
For the cold fluid:
\[ 83.6 = 2.09 \times (T_{\text{c,out}} - 30) \implies T_{\text{c,out}} = 30 + 40 = 70^\circ\text{C} \]

•
Step 3.4: Compute LMTD:
The temperature difference at the hot fluid inlet end:
\[ \Delta T_{1} = T_{\text{h,in}} - T_{\text{c,out}} = 80 - 70 = 10^\circ\text{C} \]
The temperature difference at the hot fluid exit end:
\[ \Delta T_{2} = T_{\text{h,out}} - T_{\text{c,in}} = 40 - 30 = 10^\circ\text{C} \]
Since \(\Delta T_{1} = \Delta T_{2}\), the log mean temperature difference simplifies directly to:
\[ \text{LMTD} = \Delta T_{1} = \Delta T_{2} = 10^\circ\text{C} \]

Step 4: Final Answer:
The log mean temperature difference of the heat exchanger is \(10^\circ\text{C}\).