Question

The local convection heat transfer coefficient for air flowing parallel over a 1 -m-long plate with irregular surface topology is experimentally determined to be \(h_{x}=0.5+12 x-0.7 x^{3}\), where \(h_{x}\) is in $\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the plate surface is coated with water, determine the corresponding average mass convection coefficient over the entire plate. Assume properties can be evaluated at \(298 \mathrm{~K}\) and $1 \mathrm{~atm}$.

Short answer

Answer: The average mass convection coefficient over the entire plate is 0.00532 kg/m²s.

Step-by-step solution

Calculate the average heat transfer coefficient over the plate

To find the average value of the heat transfer coefficient, we can integrate the function \(h_x\) along the length of the plate (1 meter) and then divide by the length of the plate. Thus, $$\bar{h} = \frac{1}{1-0}\int_0^1 (0.5+12 x-0.7 x^3) dx$$ Evaluate the integral to find the average heat transfer coefficient.

Evaluate the integral

To evaluate the integral, we need to find the antiderivative of the function and then apply the limits of integration: $$\int (0.5 + 12 x - 0.7x^3) dx = \left(\frac{1}{2}x^2 + 6x^2 - \frac{1}{5}x^4 \right)$$ Now, apply the limits of integration: $$\bar{h} = \left(\frac{1}{2}(1)^2 + 6(1)^2 - \frac{1}{5}(1)^4 \right) - \left(\frac{1}{2}(0)^2 + 6(0)^2 - \frac{1}{5}(0)^4 \right) = 6.3 W/m^2K$$ So the average value of the heat transfer coefficient is \(6.3 W/m^2K\).

Use heat and mass transfer relationship to find average mass convection coefficient

The relationship between heat and mass transfer is given as: $$\bar{h} = \rho C_p \bar{G_m}$$ where \(\bar{G_m}\) is the average mass convection coefficient, \(\rho\) is the air density, and \(C_p\) is the specific heat at constant pressure. We are given that the properties can be evaluated at 298 K and 1 atm. Using typical values for air at these conditions: - Air density, \(\rho = 1.184 kg/m^3\) - Specific heat at constant pressure, \(C_p = 1005 J/kgK\) We'll plug these values into the equation and solve for the average mass convection coefficient. $$6.3 = (1.184)(1005)\bar{G_m}$$ Solving for \(\bar{G_m}\): $$\bar{G_m} = \frac{6.3}{(1.184)(1005)} = 0.00532 kg/m^2s$$ So, the corresponding average mass convection coefficient over the entire plate is \(0.00532 kg/m^2s\).