Question

A 15 -ft-long strip of sheet metal is being transported on a conveyor at a velocity of \(16 \mathrm{ft} / \mathrm{s}\). To cure the coating on the upper surface of the sheet metal, infrared lamps providing a constant radiant flux of \(1500 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\) are used. The coating on the upper surface of the metal strip has an absorptivity of \(0.6\) and an emissivity of \(0.7\), while the surrounding ambient air temperature is \(77^{\circ} \mathrm{F}\). Radiation heat transfer occurs only on the upper surface, while convection heat transfer occurs on both upper and lower surfaces of the sheet metal. Determine the surface temperature of the sheet metal. Evaluate the properties of air at \(180^{\circ} \mathrm{F}\).

Short answer

Question: Determine the expression for the surface temperature of the sheet metal in terms of the convection heat transfer coefficient (h). Answer: The surface temperature, T_surface, is given by the expression: \(T_{surface} = 77 + \frac{15 \cdot 1500 \cdot 0.6}{2 \cdot h \cdot 15}\)

Step-by-step solution

Calculate the heat absorbed due to radiation

First, we need to find the heat absorbed by the sheet metal due to the radiation from the infrared lamps. This can be determined using the following formula: \(Q_{absorbed} = A \cdot q \cdot \alpha\) Here, A is the area of the sheet metal, q is the radiant flux from the lamps, and α is the absorptivity of the coating. The length of the sheet metal is 15 ft, and since it is a strip, we will consider its width to be 1 ft. So the area (A) = 15 ft². Given, the radiant flux (q) = 1500 Btu/h·ft² and the absorptivity (α) = 0.6. Now, let's find the heat absorbed: \(Q_{absorbed} = 15 \cdot 1500 \cdot 0.6\)

Calculate the heat loss due to convection

As mentioned, convection heat transfer occurs on both upper and lower surfaces of the sheet metal. The heat loss due to convection can be determined using the following formula: \(Q_{conv} = h \cdot A \cdot (T_{surface} - T_{ambient})\) Here, h is the convection heat transfer coefficient, A is the area of the sheet metal, T_surface is the surface temperature, and T_ambient is the ambient air temperature. The properties of air are evaluated at 180°F. Since both the upper and lower surfaces are involved in convection, we need to multiply the convection heat transfer by 2: \(Q_{conv} = 2 \cdot h \cdot 15 \cdot (T_{surface} - 77)\)

Equate the heat absorbed and heat loss due to convection

For steady-state operation, the heat absorbed by the sheet metal due to the radiation must be equal to the heat loss due to convection. Therefore: \(Q_{absorbed} = Q_{conv}\) Using the expressions from Step 1 and Step 2: \(15 \cdot 1500 \cdot 0.6 = 2 \cdot h \cdot 15 \cdot (T_{surface} - 77)\)

Solve for the surface temperature

Now, let's solve the equation from Step 3 for the surface temperature, T_surface. Note that the convective heat transfer coefficient (h) is not given in the problem, so we cannot get a numerical value for T_surface. However, the expression for the surface temperature can be written in terms of h: \((15 \cdot 1500 \cdot 0.6) / (2 \cdot h \cdot 15) = T_{surface} - 77\) \(T_{surface} = 77 + \frac{15 \cdot 1500 \cdot 0.6}{2 \cdot h \cdot 15}\) Thus, the surface temperature of the sheet metal is expressed in terms of the convection heat transfer coefficient (h). If the value of h was given or could be evaluated, we would be able to determine the numerical value for T_surface.