Question

Air enters a 7-cm-diameter and 4-m-long tube at \(65^{\circ} \mathrm{C}\) and leaves at \(15^{\circ} \mathrm{C}\). The tube is observed to be nearly isothermal at \(5^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2},{ }^{\circ} \mathrm{C}\), the rate of heat transfer from the air is (a) \(491 \mathrm{~W}\) (b) \(616 \mathrm{~W}\) (c) \(810 \mathrm{~W}\) (d) \(907 \mathrm{~W}\) (e) \(975 \mathrm{~W}\)

Short answer

Answer: The calculated rate of heat transfer from the air is closest to 308 W. However, this value is not listed among the provided answer choices. The closest answer is (b) 616 W, but this represents a significant difference from the calculation. There may be an error in the provided answer choices, or the question is ill-posed.

Step-by-step solution

Calculate the Surface Area of the Tube

We can calculate the surface area of the tube using the formula for the surface area of a cylinder: \(A = 2\pi r L\). The radius (r) of the tube can be calculated from the given diameter: \(r = \frac{d}{2} = \frac{0.07 \, \mathrm{m}}{2} = 0.035 \, \mathrm{m}\). The length (L) is given as 4 m. Plugging these values into the formula gives: $$A = 2\pi(0.035\,\mathrm{m})(4\,\mathrm{m}) = 0.44\,\mathrm{m^2}$$.

Calculate the Temperature Difference

The temperature difference (ΔT) between the air and the tube's surface can be determined from the given temperatures. The average air temperature as it passes through the tube is the average of the initial (65°C) and final (15°C) temperatures: $$T_{avg} = \frac{65^{\circ} \mathrm{C} + 15^{\circ} \mathrm{C}}{2} = 40^{\circ} \mathrm{C}$$ We are given that the tube is nearly isothermal at 5°C. Therefore, the temperature difference is: $$\Delta{T} = T_{avg} - T_{tube} = 40^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 35^{\circ} \mathrm{C}$$

Calculate the Rate of Heat Transfer

Now we can calculate the rate of heat transfer (Q) using the formula for convection heat transfer: \(Q = hA\Delta{T}\). We are given the average convection heat transfer coefficient (h) as 20 W/m²°C. Plugging in the known values, we get: $$Q = (20\,\mathrm{W/m^2°C})(0.44\,\mathrm{m^2})(35^{\circ} \mathrm{C}) = 308\,\mathrm{W}$$ The calculated rate of heat transfer from the air is closest to 308 W, which is not listed among the answer choices. There may be an error in the provided answer choices, or the question is ill-posed. The closest answer to the calculated value is (b) 616 W, but this represents a significant difference from the calculation.