Question

Water at \(43.3^{\circ} \mathrm{C}\) flows over a large plate at a velocity of \(30.0 \mathrm{~cm} / \mathrm{s}\). The plate is \(1.0-\mathrm{m}\) long (in the flow direction), and its surface is maintained at a uniform temperature of \(10.0^{\circ} \mathrm{C}\). Calculate the steady rate of heat transfer per unit width of the plate.

Short answer

Answer: The steady rate of heat transfer per unit width of the plate is approximately 116,324.7 W/m².

Step-by-step solution

Identify the given information

Here, we are given: - Water temperature: \(T_1 = 43.3^{\circ} \mathrm{C}\) - Plate surface temperature: \(T_2 = 10.0^{\circ} \mathrm{C}\) - Flow velocity: \(v = 30.0 \mathrm{~cm} / \mathrm{s}\) - Plate length: \(L = 1.0 \mathrm{~m}\)

Calculate the temperature difference

We need to find the temperature difference between the water and the plate surface: \(\Delta T = T_1 - T_2 = 43.3^{\circ} \mathrm{C} - 10.0^{\circ} \mathrm{C} = 33.3^{\circ} \mathrm{C}\)

Convert velocity and plate length into meters

Since the flow velocity is given in cm/s, we need to convert it into m/s: \(v = 30.0 \mathrm{~cm} / \mathrm{s} * \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.3 \mathrm{~m}/\mathrm{s}\)

Calculate the Reynolds number

For the water flow over the plate, we can calculate the Reynolds number (\(Re\)) using the following formula: \(Re = \frac{v L}{\nu}\) Here \(\nu\) is the kinematic viscosity of water, which can be derived from the temperature \(T_1 = 43.3^{\circ} \mathrm{C}\). For this temperature, we find \(\nu \approx 6.63 \times 10^{-7} \mathrm{m^2/s}\) from a fluid properties table or using an online calculator. Now, let's calculate the Reynolds number: \(Re = \frac{0.3 \mathrm{~m}/\mathrm{s} \times 1.0\mathrm{~m}}{6.63 \times 10^{-7} \mathrm{m^2/s}} \approx 452,344\)

Calculate the Prandtl number

We also need to determine the Prandtl number (\(Pr\)) for water at \(T_1 = 43.3^{\circ} \mathrm{C}\). From a fluid properties table or online calculator, we can find \(Pr \approx 4.38\).

Calculate the heat transfer coefficient using Sieder-Tate correlation

We can use the Sieder-Tate correlation to determine the heat transfer coefficient (\(h\)): \(h = 0.023 \times Re^{0.8} \times Pr^{0.4} \times k_f / L\) Here, \(k_f\) is the thermal conductivity of water at the given temperature. For \(T_1 = 43.3^{\circ} \mathrm{C}\), we find \(k_f \approx 0.627 \mathrm{W/m\cdot K}\). Now, let's calculate the heat transfer coefficient: \(h = 0.023 \times (452,344)^{0.8} \times (4.38)^{0.4} \times 0.627 \mathrm{W/m\cdot K} / 1.0\mathrm{~m} \approx 3492.3 \mathrm{W/m^2 \cdot K}\)

Calculate the steady rate of heat transfer per unit width

Finally, we can use Fourier's law of heat conduction to find the steady rate of heat transfer per unit width (\(q''\)): \(q'' = h \Delta T\) \(q'' = 3492.3 \mathrm{W/m^2 \cdot K} \times 33.3^{\circ} \mathrm{C} \approx 116,324.7 \mathrm{W/m^2}\) Therefore, the steady rate of heat transfer per unit width of the plate is approximately 116,324.7 W/m².