Question

Cold air at 10°C is forced to flow over a plate maintained at 40°C. The mean heat transfer coefficient is (\(h_m\)= 30 W/m²·°C). The heat flow rate from plate to air through a plate area of A=2 m² will be

• A. 0.9 kW
• B. 1.8 kW
• C. 2.7 kW
• D. 3.6 kW

Hint:

Always check the units of the final answer required by the options. In this case, the calculation yields Watts (W), but the options are in kilowatts (kW). A simple conversion is often the final step to getting the correct answer.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem deals with convective heat transfer. Convection is the mode of heat transfer between a solid surface and a moving fluid (in this case, the plate and the air). The rate of heat transfer is governed by Newton's Law of Cooling.

Step 2: Key Formula or Approach:
Newton's Law of Cooling formula for convective heat transfer is:
\[ Q = h \cdot A \cdot \Delta T \] Where:
Q = Heat flow rate (in Watts, W)
h = Heat transfer coefficient (in W/m²·°C or W/m²·K)
A = Surface area through which heat is transferred (in m²)
\(\Delta T\) = Temperature difference between the surface and the fluid (in °C or K)
\(\Delta T = T_{surface} - T_{fluid}\)

Step 3: Detailed Explanation:
First, identify the given values from the problem statement:

• Fluid (air) temperature, \(T_{fluid}\) = 10°C
• Surface (plate) temperature, \(T_{surface}\) = 40°C
• Mean heat transfer coefficient, \(h_m\) = 30 W/m²·°C
• Plate area, A = 2 m² Calculate the temperature difference, \(\Delta T\):
\[ \Delta T = T_{surface} - T_{fluid} = 40°C - 10°C = 30°C \] Now, substitute the values into Newton's Law of Cooling:
\[ Q = h_m \cdot A \cdot \Delta T \] \[ Q = (30 \text{ W/m²·°C}) \cdot (2 \text{ m²}) \cdot (30°C) \] \[ Q = 60 \cdot 30 \text{ W} \] \[ Q = 1800 \text{ W} \] The options are given in kilowatts (kW). To convert Watts to kilowatts, divide by 1000.
\[ Q \text{ (in kW)} = \frac{1800}{1000} = 1.8 \text{ kW} \]

Step 4: Final Answer:
The heat flow rate from the plate to the air is 1.8 kW.