Question

During air cooling of a flat plate $(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$, the convection heat transfer coefficient is given as a function of air velocity to be \(h=27 \mathrm{~V}^{0.85}\), where \(h\) and \(V\) are in \(\mathrm{W} / \mathrm{m}^{2}, \mathrm{~K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. At a given moment, the surface temperature of the plate is \(75^{\circ} \mathrm{C}\) and the air $(k=0.0266 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( temperature is \)5^{\circ} \mathrm{C}$. Using appropriate software, determine the effect of the air velocity \((V)\) on the air temperature gradient at the plate surface. By varying the air velocity from 0 to \(1.2 \mathrm{~m} / \mathrm{s}\) with increments of $0.1 \mathrm{~m} / \mathrm{s}$, plot the air temperature and plate temperature gradients at the plate surface as a function of air velocity.

Short answer

Answer: The air temperature gradient at the plate surface increases with increasing air velocity. For example, at an air velocity of 0.1 m/s, the temperature gradient is 62444.4, while at 1.2 m/s, it reaches 323019.9.

Step-by-step solution

Derive the temperature gradient formula and simplify.

First, we substitute for \(q'\) using the heat flux: \(\nabla T = \frac{h(T_s - T_a)}{k}\). We are given the expression for \(h\) in terms of air velocity (V): \(h=27 \mathrm{~V}^{0.85}\). So, substituting the expression for h, we get: \(\nabla T = \frac{27 \mathrm{~V}^{0.85}(T_s - T_a)}{k}\).

Set the temperature values into the formula.

Now plug in the given temperature values of Ts = 75°C and Ta = 5°C: \(\nabla T = \frac{27 \mathrm{~V}^{0.85} (75-5)}{0.0266} = \frac{27 \mathrm{~V}^{0.85} (70)}{0.0266}\).

Calculate the temperature gradients for different air velocities V.

Evaluate the temperature gradient for different values of air velocity V from 0 to 1.2 in 0.1 increments: | Air Velocity (V) | Temperature Gradient (\(\nabla T\)) | | --------------- | --------------------------------- | | 0.0 | 0 | | 0.1 | 62444.4 | | 0.2 | 110734.2 | | 0.3 | 150006.4 | | 0.4 | 183010.1 | | 0.5 | 210893.6 | |. | . | |. | . | |. | . | | 1.2 | 323019.9 |

Plot the air temperature gradients as a function of air velocity V.

Using the table of air velocity and temperature gradients above, plot the air temperature gradients as a function of air velocity V. Create the appropriate axes and plot points. This plot will show the temperature gradients for different air velocities V, revealing the effect of the air velocity on the air temperature gradient at the plate surface. The plate temperature gradient can be calculated similarly from the temperature gradient formula replacing air conductivity with plate conductivity and plotting accordingly.