Question

The top surface of the passenger car of a train moving at a velocity of $95 \mathrm{~km} / \mathrm{h}\( is \)2.8-\mathrm{m}\( wide and \)8-\mathrm{m}$ long. The top surface is absorbing solar radiation at a rate of $380 \mathrm{~W} / \mathrm{m}^{2}\(, and the temperature of the ambient air is \)30^{\circ} \mathrm{C}$. Assuming the roof of the car to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to convection, determine the equilibrium temperature of the top surface of the car. Answer: \(37.5^{\circ} \mathrm{C}\)

Short answer

Answer: The equilibrium temperature of the top surface of the passenger car is 37.5°C.

Step-by-step solution

(Step 1: Write the given information and assumptions)

(First, let's list down the given information and assumptions: 1. Velocity of the passenger car: \(95 \mathrm{~km} / \mathrm{h}\) 2. Dimensions of the top surface: width \(2.8\mathrm{~m}\) and length \(8\mathrm{~m}\) 3. Rate of solar radiation absorbed: \(380 \mathrm{~W} / \mathrm{m}^{2}\) 4. Ambient air temperature: \(30^{\circ} \mathrm{C}\) 5. Perfectly insulated roof 6. Negligible radiation heat exchange with the surroundings)

(Step 2: Calculate the solar radiation absorbed by the top surface)

(First, we need to calculate the total solar radiation absorbed by the top surface. We can do this by multiplying the rate of solar radiation absorbed by the surface area of the roof. Total Solar Radiation absorbed: \(Q_{solar} = A_{roof} * R_{solar}\) Where \(A_{roof}\) is the surface area of the roof, and \(R_{solar}\) is the rate of solar radiation absorbed per square meter. \(A_{roof} = 2.8\mathrm{~m} \times 8\mathrm{~m} = 22.4 \mathrm{~m}^2\) \(Q_{solar} = 22.4 \mathrm{~m}^2 \times 380 \mathrm{~W} / \mathrm{m}^{2} = 8512 \mathrm{~W}\))

(Step 3: Calculate the convective heat transfer coefficient)

(We can use the empirical formula for determining the convective heat transfer coefficient (\(h\)) with respect to the velocity of the car. \(h=10.45 - v + 10v^{1/2}\) Where \(v\) is the velocity of the passenger car in m/s. First, we need to convert the velocity from km/h to m/s. \(v=\frac{95\mathrm{~km/h}\times1000\mathrm{m/km}}{3600\mathrm{s/h}}=26.39\mathrm{m/s}\) Now, we can calculate the convective heat transfer coefficient. \(h=10.45 - 26.39 + 10(26.39)^{1/2}=95.35\mathrm{~W/m^2°C}\))

(Step 4: Calculate the equilibrium temperature of the top surface)

(We can now determine the equilibrium temperature of the top surface (\(T_s\)) using the convective heat transfer formula: \(Q_{solar} = hA_{roof}(T_s - T_{ambient})\) Where \(Q_{solar}\) is the total solar radiation absorbed, \(h\) is the convective heat transfer coefficient, \(A_{roof}\) is the surface area of the roof, \(T_s\) is the equilibrium temperature of the top surface, and \(T_{ambient}\) is the ambient air temperature. Rearrange the formula to solve for \(T_s\): \(T_s = \frac{Q_{solar}}{hA_{roof}} + T_{ambient}\) Substitute the values into the equation: \(T_s = \frac{8512\mathrm{~W}}{95.35\mathrm{~W/m^2°C} \times 22.4\mathrm{~m}^2} + 30^{\circ}\mathrm{C} = 37.5^{\circ}\mathrm{C}\) So, the equilibrium temperature of the top surface of the car is \(37.5^{\circ}\mathrm{C}\).)