Question

Consider a large 3-cm-thick stainless steel plate in which heat is generated uniformly at a rate of \(5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming the plate is losing heat from both sides, determine the heat flux on the surface of the plate during steady operation. Answer: $75 \mathrm{~kW} / \mathrm{m}^{2}$

Short answer

Answer: The heat flux on the surface of the stainless steel plate during steady operation is 75 kW/m².

Step-by-step solution

Identify the given values

The problem gives us the following information: - Thickness of the stainless steel plate (t): 3 cm = 0.03 m (converted to meters) - Heat generation rate (q_gen): \(5 \times 10^{6}\) W/m³ - Heat loss occurs from both sides of the plate Step 2: Calculate the heat loss from one side of the plate

Calculate the heat loss from one side of the plate

Since the plate is losing heat from both sides, we can calculate the heat loss from one side and then double the result. We know that the heat generation rate inside the plate is uniform. So, the heat generated within a unit volume of the plate is given by: q_gen (W/m³) = Heat generated in unit volume (W) Step 3: Find the total volume of the steel plate

Find the total volume of the steel plate

To find the heat generated by the steel plate, we need to multiply the heat generation rate (q_gen) by the volume of the plate. Assuming the width and height of the plate are 1 meter each, the volume of the plate can be calculated as: Volume (V) = Thickness × Width × Height = 0.03 m × 1 m × 1 m = 0.03 m³ Step 4: Calculate the total heat generated in the plate

Calculate the total heat generated in the plate

Now, multiplying the heat generation rate by the volume of the plate, we have: Total heat generated (Q) = q_gen × V = \(5 \times 10^{6}\) W/m³ × 0.03 m³ = \(1.5 \times 10^{5}\) W Step 5: Calculate the heat flux lost through one side of the plate

Calculate the heat flux lost through one side of the plate

Since the plate loses heat from both sides, the heat flux (q) lost through one side of the plate will be half of the total heat generated: q = Q / 2 = \(1.5 \times 10^{5}\) W / 2 = \(7.5 \times 10^{4}\) W To convert this value to kW/m², we have: q = \(7.5 \times 10^{4}\) W × (1 kW / 1000 W) = 75 kW/m² The heat flux on the surface of the plate during steady operation is 75 kW/m², which matches the given answer.