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One dimensional steady state heat transfer occurs from a flat vertical wall of length 0.1 m into the adjacent fluid. The heat flux into this fluid is 21 W/m3. The wall thermal conductivity is 1.73 W/mK. If the heat transfer coefficient is 30 W/m2K and the Nusselt number based on the wall length is 20, then the magnitude of the temperature gradient at the wall on the fluid side (in K /m) is
Correct answer is '140'. Can you explain this answer?
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Most Upvoted Answer
One dimensional steady state heat transfer occurs from a flat vertical...
From (1) and (3) ,
Given data,
Putting values, we get
= 140 K/ m
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One dimensional steady state heat transfer occurs from a flat vertical...
Given:
- Length of the wall (L) = 0.1 m
- Heat flux into the fluid (q) = 21 W/m3
- Wall thermal conductivity (k) = 1.73 W/mK
- Heat transfer coefficient (h) = 30 W/m2K
- Nusselt number based on wall length (NuL) = 20
Calculating Temperature Gradient:
The heat flux (q) can be related to the temperature gradient (∆T/∆x) using the following equation:
q = -k * (∆T/∆x)
Here, the negative sign indicates that heat is transferring from the wall to the fluid. Rearranging the equation, we have:
∆T/∆x = -q/k
Calculating the Temperature Gradient at the Wall:
The temperature gradient at the wall (∆T/∆x) can be found by substituting the given values:
∆T/∆x = -21/1.73
∆T/∆x = -12.14 K/m
Calculating the Temperature Gradient at the Fluid Side:
The Nusselt number (NuL) can be related to the heat transfer coefficient (h) and the thermal conductivity (k) using the following equation:
NuL = h * L / k
Rearranging the equation, we can solve for the heat transfer coefficient (h):
h = NuL * k / L
Now, substituting the given values:
h = 20 * 1.73 / 0.1
h = 346 W/m2K
Applying Newton's Law of Cooling:
According to Newton's Law of Cooling, the heat flux (q) can be related to the temperature difference (∆T) and the heat transfer coefficient (h) using the following equation:
q = h * ∆T
Rearranging the equation, we can solve for the temperature difference (∆T):
∆T = q / h
Substituting the given values:
∆T = 21 / 346
∆T = 0.06 K
Calculating the Temperature Gradient at the Fluid Side:
The temperature gradient at the fluid side (∆T/∆x) can be found by dividing the temperature difference (∆T) by the length of the wall (L):
∆T/∆x = ∆T / L
Substituting the values:
∆T/∆x = 0.06 / 0.1
∆T/∆x = 0.6 K/m
Therefore, the magnitude of the temperature gradient at the wall on the fluid side is 0.6 K/m.
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One dimensional steady state heat transfer occurs from a flat vertical wall of length 0.1 m into the adjacent fluid. The heat flux into this fluid is 21 W/m3. The wall thermal conductivity is 1.73 W/mK. If the heat transfer coefficient is 30 W/m2K and the Nusselt number based on the wall length is 20, then the magnitude of the temperature gradient at the wall on the fluid side (in K /m) isCorrect answer is '140'. Can you explain this answer?
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