Question

During open-heart surgery, a patient’s blood is cooled down to 25°C from 37°C using a concentric tube counter-flow heat exchanger. Water enters the heat exchanger at 4°C and leaves at 18°C. Blood flow rate during the surgery is 5 L/minute.
Use the following fluid properties: \[ \begin{array}{|c|c|c|} \hline \text{Fluid} & \text{Density (kg/m}^3\text{)} & \text{Specific heat (J/kg-K)}
\hline \text{Blood} & 1050 & 3740
\text{Water} & 1000 & 4200
\hline \end{array} \] Effectiveness of the heat exchanger is ________________ (round off to 2 decimal places).

Hint:

The effectiveness of a heat exchanger can be calculated by dividing the actual heat transfer rate by the maximum possible heat transfer rate.

Answer 1

Correct Answer: 0.4

We use the effectiveness equation for a heat exchanger in the counterflow configuration: \[ \epsilon = \frac{\dot{Q}_{\text{actual}}}{\dot{Q}_{\text{max}}} \] Where: - \( \dot{Q}_{\text{actual}} \) is the actual heat transferred, - \( \dot{Q}_{\text{max}} \) is the maximum possible heat transfer. The actual heat transfer rate \( \dot{Q}_{\text{actual}} \) can be calculated using the blood: \[ \dot{Q}_{\text{actual}} = \dot{m}_\text{blood} \cdot c_\text{blood} \cdot (T_\text{inlet} - T_\text{outlet}) \] Where:
- \( \dot{m}_\text{blood} = \rho_\text{blood} \cdot V_\text{blood} \),
- \( c_\text{blood} = 3740 \, \text{J/kg-K} \) is the specific heat of blood,
- \( T_\text{inlet} = 37^\circ \text{C} \), and \( T_\text{outlet} = 25^\circ \text{C} \) are the inlet and outlet temperatures of the blood.
First, convert the blood flow rate to kg/s: \[ \dot{V}_\text{blood} = 5 \, \text{L/min} = \frac{5}{60} \, \text{L/s} = \frac{5}{60} \times 10^{-3} \, \text{m}^3/\text{s} \] Now, calculate the mass flow rate of blood: \[ \dot{m}_\text{blood} = \rho_\text{blood} \cdot \dot{V}_\text{blood} = 1050 \, \text{kg/m}^3 \times \frac{5}{60} \times 10^{-3} \, \text{m}^3/\text{s} = 0.0875 \, \text{kg/s}. \] Now, calculate the actual heat transfer: \[ \dot{Q}_{\text{actual}} = 0.0875 \, \text{kg/s} \times 3740 \, \text{J/kg-K} \times (37 - 25) = 0.0875 \times 3740 \times 12 = 3927 \, \text{W}. \] Next, calculate the maximum heat transfer rate \( \dot{Q}_{\text{max}} \) using the water: \[ \dot{Q}_{\text{max}} = \dot{m}_\text{water} \cdot c_\text{water} \cdot (T_\text{outlet} - T_\text{inlet}) \] Where:
- \( \dot{m}_\text{water} = \rho_\text{water} \cdot \dot{V}_\text{water} \),
- \( c_\text{water} = 4200 \, \text{J/kg-K} \) is the specific heat of water,
- \( T_\text{outlet} = 18^\circ \text{C} \), and \( T_\text{inlet} = 4^\circ \text{C} \).
First, calculate the mass flow rate of water: \[ \dot{V}_\text{water} = \dot{V}_\text{blood} = \frac{5}{60} \times 10^{-3} \, \text{m}^3/\text{s}, \] \[ \dot{m}_\text{water} = 1000 \, \text{kg/m}^3 \times \frac{5}{60} \times 10^{-3} = 0.0833 \, \text{kg/s}. \] Now, calculate the maximum heat transfer: \[ \dot{Q}_{\text{max}} = 0.0833 \, \text{kg/s} \times 4200 \, \text{J/kg-K} \times (18 - 4) = 0.0833 \times 4200 \times 14 = 4884.12 \, \text{W}. \] Finally, calculate the effectiveness of the heat exchanger: \[ \epsilon = \frac{\dot{Q}_{\text{actual}}}{\dot{Q}_{\text{max}}} = \frac{3927}{4884.12} \approx 0.40. \] Thus, the effectiveness of the heat exchanger is \( \boxed{0.40} \).