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A coolant fluid at 30°C flows over a heated flat plate maintained at a constant temperature of 100°C. The boundary layer temperature distribution at a given location on the plate may be approximated as T= 30 + 70 expt(-y) where y(in m) is the distance normal to the plate and fis in °C. If thermal conductivity of the fluid is 1.0 W/mK, the local convective heat transfer coefficient (in W/m2K) at that location will be
[2009]
  • a)
    0.2
  • b)
    1
  • c)
    5
  • d)
    10
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A coolant fluid at 30°C flows over a heated flat plate maintained ...
Under steady state conditionsHeat transfer by conduction = Heat transfer by convection
∴ 
or 

But y = 0

or h(100–30) = k x (100 - 30)
or h = k = 1 W/m2K
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Most Upvoted Answer
A coolant fluid at 30°C flows over a heated flat plate maintained ...
Understanding the Problem
In this scenario, we have a coolant fluid flowing over a heated flat plate. The temperature distribution in the boundary layer is given by the equation:
T = 30 + 70 * exp(-y)
where T is the temperature in °C and y is the distance from the plate in meters. The plate is maintained at a constant temperature of 100°C.
Boundary Layer Temperature Analysis
- At y = 0 (on the plate), T = 100°C.
- As y increases, T approaches 30°C, indicating that the fluid cools down as it moves away from the plate.
Local Convective Heat Transfer Coefficient
To find the local convective heat transfer coefficient (h), we can use Newton's law of cooling:
q = h * (T_plate - T_infinity)
Here:
- q is the heat transfer rate.
- T_plate = 100°C (temperature of the plate).
- T_infinity = 30°C (temperature of the fluid far from the plate).
Heat Transfer Rate Calculation
The heat transfer rate (q) can also be expressed using thermal conductivity (k) and temperature gradient (dT/dy):
q = -k * (dT/dy)
Given:
- k = 1.0 W/mK
Now, we need to determine dT/dy from the temperature distribution equation:
dT/dy = d(30 + 70 * exp(-y))/dy = -70 * exp(-y)
At y = 0 (the surface of the plate):
dT/dy = -70 * exp(0) = -70
Substituting values into the heat transfer equation:
q = -1.0 * (-70) = 70 W/m²
Now, substituting into Newton’s law of cooling:
70 = h * (100 - 30)
This simplifies to:
70 = h * 70
Thus, h = 1 W/m²K.
Conclusion
Therefore, the local convective heat transfer coefficient at that location is 1 W/m²K, which matches option 'B'.
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A coolant fluid at 30°C flows over a heated flat plate maintained at a constant temperature of 100°C. The boundary layer temperature distribution at a given location on the plate may be approximated as T= 30 + 70 expt(-y) where y(in m) is the distance normal to the plate and fis in °C. If thermal conductivity of the fluid is 1.0 W/mK, the local convective heat transfer coefficient (in W/m2K) at that location will be[2009]a)0.2b)1c)5d)10Correct answer is option 'B'. Can you explain this answer?
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