Question

A simple solar collector is built by placing a 6-cmdiameter clear plastic tube around a garden hose whose outer diameter is \(2 \mathrm{~cm}\) (Fig. P13-137). The hose is painted black to maximize solar absorption, and some plastic rings are used to keep the spacing between the hose and the clear plastic cover constant. The emissivities of the hose surface and the glass cover are \(0.9\), and the effective sky temperature is estimated to be \(15^{\circ} \mathrm{C}\). The temperature of the plastic tube is measured to be \(40^{\circ} \mathrm{C}\), while the ambient air temperature is \(25^{\circ} \mathrm{C}\). Determine the rate of heat loss from the water in the hose by natural convection and radiation per meter of its length under steady conditions. Answers: $12.7 \mathrm{~W}, 26.1 \mathrm{~W}$

Short answer

Answer: The total heat loss from the water in the hose by natural convection and radiation per meter of its length under steady conditions is approximately 39 W.

Step-by-step solution

Calculate the surface area of the hose

To calculate the surface area of the hose per meter of its length, we need to use the provided outer diameter of 2 cm. The surface area of a cylinder per unit length is given by: \(A = \pi d\) where \(A\) is the surface area per meter of length, and \(d\) is the outer diameter of the hose. Using the given diameter, we find the surface area per meter of length: \(A = \pi (2 \times 10^{-2}) \approx 0.0628 \thinspace \mathrm{m^2}\)

Find the heat loss due to natural convection

First, calculate the temperature difference between the plastic tube and the ambient air: \(ΔT = T_{tube} - T_{air} = 40^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 15^{\circ} \mathrm{C} = 15 \thinspace \mathrm{K}\) We can use the provided information to estimate the heat transfer coefficient, \(h\), for natural convection: \(h \approx 1.31 \thinspace (ΔT)^{1/3} = 1.31 \thinspace (15)^{1/3} \approx 4.45 \thinspace \mathrm{W/m^2K}\) Now, calculate the heat loss due to natural convection, \(Q_{conv}\), using the heat transfer coefficient and the surface area of the hose: \(Q_{conv} = h \thinspace A \thinspace ΔT \approx 4.45 \thinspace (0.0628) \thinspace 15 \approx 12.7 \thinspace \mathrm{W}\)

Find the heat loss due to radiation

Recall the Stefan-Boltzmann Law, which states the power radiated per unit area by a black body is proportional to the fourth power of its temperature: \(P = σ \thinspace ε \thinspace (T^4 - T_{sky}^4)\) where \(P\) is the power radiated per unit area, \(σ\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \thinspace\mathrm{W/m^2K^4}\)), \(ε\) is the emissivity, and \(T\) is the temperature of the body in Kelvin. First, convert the temperatures to Kelvin: \(T_{tube} = 40^{\circ} \mathrm{C} + 273.15 = 313.15 \thinspace \mathrm{K}\) \(T_{sky} = 15^{\circ} \mathrm{C} + 273.15 = 288.15 \thinspace \mathrm{K}\) Next, calculate the power radiated per unit area: \(P = (5.67 \times 10^{-8}) \thinspace (0.9) \thinspace (313.15^4 - 288.15^4) \approx 419.6 \thinspace \mathrm{W/m^2}\) Finally, calculate the heat loss due to radiation, \(Q_{rad}\), using the power radiated per unit area and the surface area of the hose: \(Q_{rad} = P \thinspace A \approx 419.6 \thinspace (0.0628) \approx 26.3 \thinspace \mathrm{W}\)

Total heat loss

Sum up the heat loss due to natural convection and radiation: \(Q_{total} = Q_{conv} + Q_{rad} \approx 12.7 \thinspace \mathrm{W} + 26.3\thinspace \mathrm{W} \approx 39 \thinspace \mathrm{W}\) Therefore, the total heat loss from the water in the hose by natural convection and radiation per meter of its length under steady conditions is about 39 W. Note that the values 12.7 W and 26.1 W given in the exercise might have resulted from rounding off of intermediate calculations.