Question

Mercury at \(25^{\circ} \mathrm{C}\) flows over a 3 -m-long and 2 -m-wide flat plate maintained at \(75^{\circ} \mathrm{C}\) with a velocity of $0.01 \mathrm{~m} / \mathrm{s}$. Determine the rate of heat transfer from the entire plate.

Short answer

Answer: The rate of heat transfer from the entire plate is approximately 2345.7 W.

Step-by-step solution

Calculate the Reynolds number

In this step, we'll calculate the Reynolds number to find the flow type on the plate. We will use the formula: $$Re = \frac{\rho VD}{\mu}$$, where \(\rho\) is the density of Mercury, \(V\) is the velocity, \(D\) is the characteristic length (for a flat plate, it's the length of the plate), and \(\mu\) is the dynamic viscosity of Mercury. At 25°C, Mercury properties are: \(\rho = 13400 ~kg/m^3\), \(\mu = 15.9 \times 10^{-4} ~Pa \cdot s\). The Reynolds number is: $$Re = \frac{(13400 ~kg/m^3)(0.01 ~m/s)(3 ~m)}{15.9 \times 10^{-4} ~Pa \cdot s} = 251.57$$ Since the Reynolds number is less than 5000, the flow is laminar.

Calculate the Nusselt number

For a flat plate with a laminar flow, the Nusselt number (Nu) can be calculated using the formula: $$Nu = 0.664 ~Re^{1/2} ~Pr^{1/3}$$, where Pr is the Prandtl number. The Prandtl number for Mercury is 0.0252. We already calculated the Reynolds number, Re=251.57. With these parameters, the Nusselt number is: $$Nu = 0.664(251.57)^{1/2}(0.0252)^{1/3} = 2.728$$

Calculate the heat transfer coefficient

By definition, the Nusselt number is the ratio of convective to the conductive heat transfer, which can be expressed as: $$Nu = \frac{hD}{k}$$, where \(k\) is the thermal conductivity of Mercury, which is \(8.55 ~W/(m \cdot K)\). We can solve for the heat transfer coefficient (h) using the values we calculated: $$h = \frac{k \cdot Nu}{D} = \frac{(8.55 ~W/(m \cdot K))(2.728)}{3 ~m} = 7.819 ~W/(m^2 \cdot K)$$

Calculate the rate of heat transfer

Now, we can calculate the rate of heat transfer using the formula: $$q = hA\Delta T$$, where \(A\) is the area (3 m length x 2 m width) and \(\Delta T\) is the temperature difference between the plate and Mercury (\(75^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 50^{\circ} \mathrm{C}\)). The rate of heat transfer is: $$q = (7.819 ~W/(m^2 \cdot K))(6 ~m^2)(50 ~K) = 2345.7 ~W$$ Thus, the rate of heat transfer from the entire plate is approximately 2345.7 W.