Question

Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.1\)-m-long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at \(5 \mathrm{~cm}\) from the leading edge and \((b)\) the average convection heat transfer coefficient over the entire plate.

Short answer

In this problem, we were asked to find (a) the local convection heat transfer coefficient at a specific point (5 cm from the leading edge) and (b) the average convection heat transfer coefficient over the entire plate. After a series of calculations, we found that (a) the local convection heat transfer coefficient at 5 cm from the leading edge is approximately 19738.23 W/(m²·K), and (b) the average convection heat transfer coefficient over the entire plate is approximately 18163.36 W/(m²·K).

Step-by-step solution

Calculate Reynolds number

First, we need to calculate the Reynolds number (\(Re_x\)) to determine the flow regime. The Reynolds number is given by the formula: \[Re_x = \frac{u_x \cdot x}{v}\] Where \(u_x\) is the velocity of the fluid, \(x\) is the distance from the leading edge, and \(v\) is the kinematic viscosity of the fluid. For liquid mercury, the kinematic viscosity is given as \(8.33 \times 10^{-7}\mathrm{~m^2/s}\). For this exercise, we need to calculate the Reynolds number at \(x = 0.05 \mathrm{~m}\). \[Re_{0.05} = \frac{0.3 \mathrm{~m/s} \cdot 0.05 \mathrm{~m}}{8.33 \times 10^{-7}\mathrm{~m^2/s}} = 1.8 \times 10^4\] Since \(Re_{0.05} < 10^5\), the flow can be considered laminar.

Calculate Prandtl number

Next, we need to calculate the Prandtl number (\(Pr\)) using the following formula: \[Pr = \frac{c_p \cdot \mu}{k}\] Where \(c_p\) is the specific heat of liquid mercury at the film temperature, \(\mu\) is the dynamic viscosity, and \(k\) is the thermal conductivity. For liquid mercury, \(c_p = 138 \mathrm{~J/(kg\cdot K)}\), \(\mu = 1.523 \times 10^{-3} \mathrm{~kg/(m\cdot s)}\), and \(k = 8.61 \mathrm{~W/(m\cdot K)}\). \[Pr = \frac{138 \mathrm{~J/(kg\cdot K)} \cdot 1.523 \times 10^{-3} \mathrm{~kg/(m\cdot s)}}{8.61 \mathrm{~W/(m\cdot K)}} = 0.02359\]

Calculate Nusselt number

Now, we will calculate the local Nusselt number (\(Nu_x\)) for laminar flow over a flat plate with constant surface temperature using the following expression: \[Nu_x = 0.332 \cdot Re_x^{1/2} \cdot Pr^{1/3}\] Substitute the values we calculated for \(Re_{0.05}\) and \(Pr\): \[Nu_{0.05} = 0.332 \cdot (1.8 \times 10^4)^{1/2} \cdot (0.02359)^{1/3} = 121.93\]

Calculate local convection heat transfer coefficient

Now, we can calculate the local convection heat transfer coefficient (\(h_x\)) using the relationship between the local Nusselt number and local convection heat transfer coefficient: \[h_x = \frac{Nu_x \cdot k}{x}\] Substitute the values for \(Nu_{0.05}\), \(k\), and \(x\): \[h_{0.05} = \frac{121.93 \cdot 8.61 \mathrm{~W/(m\cdot K)}}{0.05 \mathrm{~m}} = 19738.23 \mathrm{~W/(m^2\cdot K)}\] Hence, the local convection heat transfer coefficient at 5 cm from the leading edge is approximately \(19738.23 \mathrm{~W/(m^2\cdot K)}\).

Calculate average convection heat transfer coefficient

Finally, we will calculate the average convection heat transfer coefficient (\(\overline{h}\)) using the following relationship: \[\overline{h} = \frac{Nu_L \cdot k}{L}\] Where \(Nu_L\) is the average Nusselt number and \(L\) is the length of the flat plate (\(0.1\mathrm{~m}\)). To calculate \(Nu_L\), we will use the same expression as before, but this time we will use the Reynolds number at \(x = L\): \[Re_L = \frac{0.3 \mathrm{~m/s} \cdot 0.1 \mathrm{~m}}{8.33 \times 10^{-7}\mathrm{~m^2/s}} = 3.6 \times 10^4\] As the flow is still laminar, the average Nusselt number is given by: \[Nu_L = 0.664 \cdot Re_L^{1/2} \cdot Pr^{1/3}\] \[Nu_L = 0.664 \cdot (3.6 \times 10^4)^{1/2} \cdot (0.02359)^{1/3} = 210.92\] Now we can calculate the average convection heat transfer coefficient: \[\overline{h} = \frac{210.92 \cdot 8.61 \mathrm{~W/(m\cdot K)}}{0.1 \mathrm{~m}} = 18163.36 \mathrm{~W/(m^2\cdot K)}\] Hence, the average convection heat transfer coefficient over the entire plate is approximately \(18163.36 \mathrm{~W/(m^2\cdot K)}\).