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A coolant fluid at 30°C flows over a heated flat plate maintained at aconstant temperature of 100°C. The boundary layer temperature distribution at a given location on the plate may be approximated asT = 30 + 70exp (–y) where y (in m) is the distance normal to the plate andT is 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: 
  • a)
    0.2
  • b)
    1
  • c)
    5
  • d)
    10
Correct answer is option 'B'. Can you explain this answer?
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A coolant fluid at 30°C flows over a heated flat plate maintained...
Ans. (b)
Given T = 30 + 70 e-y
or

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A coolant fluid at 30°C flows over a heated flat plate maintained...
Given data:
- Coolant fluid temperature, Tcoolant = 30°C
- Plate temperature, Tplate = 100°C
- Boundary layer temperature distribution equation, T = 30 + 70exp(y)
- Thermal conductivity of the fluid, k = 1.0 W/mK

Objective:
To determine the local convective heat transfer coefficient at a given location on the plate.

Solution:
The heat transfer coefficient can be determined using Newton's Law of Cooling, which states that the rate of heat transfer between a solid surface and a fluid is directly proportional to the temperature difference between the two and the heat transfer coefficient.

The heat transfer rate can be expressed as:
q = hA(Tplate - Tcoolant)

Where:
q = heat transfer rate
h = convective heat transfer coefficient
A = surface area
Tplate = plate temperature
Tcoolant = coolant temperature

To determine the convective heat transfer coefficient, we need to evaluate the surface temperature gradient (∂T/∂y) at the given location.

Given that the boundary layer temperature distribution is given by:
T = 30 + 70exp(y)

Differentiating the above equation with respect to y, we get:
∂T/∂y = 70exp(y)

Calculation:
At the given location, let's assume the surface temperature gradient (∂T/∂y) is equal to ∂T/∂y = 70exp(0) = 70.

Using the formula for the convective heat transfer coefficient, we can rearrange the equation as follows:
h = q / (A(Tplate - Tcoolant))

Since the area is not given, we can assume a unit area, i.e., A = 1 m².

Substituting the given values, we get:
h = q / (1 * (100 - 30))
h = q / 70

Since the thermal conductivity of the fluid is given as 1.0 W/mK, the heat transfer rate (q) can be calculated using Fourier's Law of Heat Conduction:
q = kA(∂T/∂y)

Substituting the given values, we get:
q = 1.0 * 1 * 70

Now substituting the value of q in the heat transfer coefficient equation, we get:
h = (1.0 * 1 * 70) / 70
h = 1.0 W/m²K

Therefore, the local convective heat transfer coefficient at the given location is 1.0 W/m²K. Hence, the correct answer is option 'B'.
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A coolant fluid at 30°C flows over a heated flat plate maintained at aconstant temperature of 100°C. The boundary layer temperature distribution at a given location on the plate may be approximated asT = 30 + 70exp (–y) where y (in m) is the distance normal to the plate andT is 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:a)0.2b)1c)5d)10Correct answer is option 'B'. Can you explain this answer?
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