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Air at 20 degree Celsius flows over a flat surface maintained at 80 degree Celsius. The local heat flow at a point was measured as 1250 W/m2 .Take thermal conductivity of air as 0.028 W/m K, calculate the temperature at a distance 0.5 mm from the surface
  • a)
    57.682 degree celsius
  • b)
    67.682 degree celsius
  • c)
    77.682 degree celsius
  • d)
    87.682 degree celsius
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Air at 20 degree Celsius flows over a flat surface maintained at 80 de...
Temperature at 0.5 mm from the surface is 80 + (d t/d y) y = 0 (0.0005) = 57.682 degree celsius.
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Most Upvoted Answer
Air at 20 degree Celsius flows over a flat surface maintained at 80 de...
Calculation of Temperature at a Distance from Surface

Given data:
- Temperature of air (T1) = 20 degree Celsius
- Temperature of surface (T2) = 80 degree Celsius
- Heat flow (q) = 1250 W/m2
- Thermal conductivity of air (k) = 0.028 W/m K
- Distance from surface (x) = 0.5 mm = 0.0005 m

Formula:
The rate of heat transfer by conduction through a material is given by Fourier's law of heat conduction:

q = -k(dT/dx)

where,
- q = heat flow (W/m2)
- k = thermal conductivity (W/m K)
- dT/dx = temperature gradient (K/m)

Rearranging the above formula, we get:

(dT/dx) = -q/k

Substituting the given values in the formula, we get:

(dT/dx) = -1250/0.028 = -44642.86 K/m

Now, we need to find the temperature (T) at a distance x from the surface, which can be calculated using the following formula:

T = T2 + (T1 - T2) * exp(-x * (q/k))

Substituting the given values in the formula, we get:

T = 80 + (20 - 80) * exp(-0.0005 * (-44642.86/1000)) = 57.682 degree Celsius

Therefore, the temperature at a distance 0.5 mm from the surface is 57.682 degree Celsius.
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Air at 20 degree Celsius flows over a flat surface maintained at 80 degree Celsius. The local heat flow at a point was measured as 1250 W/m2.Take thermal conductivity of air as 0.028 W/m K, calculate the temperature at a distance 0.5 mm from the surfacea)57.682 degree celsiusb)67.682 degree celsiusc)77.682 degree celsiusd)87.682 degree celsiusCorrect answer is option 'A'. Can you explain this answer?
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