Question

A solar collector consists of a horizontal aluminum tube having an outer diameter of \(2.5\) in enclosed in a concentric thin glass tube of diameter $5 \mathrm{in}$. Water is heated as it flows through the tube, and the annular space between the aluminum and the glass tube is filled with air at $0.5 \mathrm{~atm}$ pressure. The pump circulating the water fails during a clear day, and the water temperature in the tube starts rising. The aluminum tube absorbs solar radiation at a rate of \(30 \mathrm{Btu} / \mathrm{h}\) per foot length, and the temperature of the ambient air outside is $75^{\circ} \mathrm{F}$. The emissivities of the tube and the glass cover are 0.9. Taking the effective sky temperature to be \(60^{\circ} \mathrm{F}\), determine the temperature of the aluminum tube when thermal equilibrium is established (i.e., when the rate of heat loss from the tube equals the amount of solar energy gained by the tube).

Short answer

Answer: To find the equilibrium temperature of the aluminum tube, we need to solve the equation: $$T_{al} = \frac{q_{loss} - k_{air}\frac{A_{al}(T_{al} - T_{ambient})}{L_{gap}}}{\epsilon_{al} \sigma A_{al}(2T_{al}^4 - T_{ambient}^4 - T_{sky}^4)}$$ Use the given information and a numerical solver or try various temperatures to find the equilibrium temperature of the aluminum tube, denoted as \(T_{al}\).

Step-by-step solution

List out given information

We have been given the following information: - Outer diameter of the aluminum tube: \(d_{al} = 2.5\,\text{in}\) - Diameter of the thin glass tube: \(d_{g} = 5\,\text{in}\) - Air pressure between the tubes: \(P_{air} = 0.5\,\text{atm}\) - Absorbed solar radiation by the aluminum tube: \(q''_{solar} = 30\,\frac{\text{Btu}}{\text{h}\cdot\text{ft}}\) - Ambient air temperature: \(T_{ambient} = 75^{\circ}\text{F}\) - Effective sky temperature: \(T_{sky} = 60^{\circ}\text{F}\) - Emissivities of the aluminum tube and glass tube: \(\epsilon_{al} = \epsilon_{g} = 0.9\)

Determine heat loss rate in Btu/h

We need to convert the absorbed solar radiation per unit length to a rate. The area of heat absorption per foot is the lateral surface of the aluminum tube: $$A_{al} = \pi d_{al}L = \pi (2.5\,\text{in})(1\,\text{ft})$$ Now, the rate of heat loss is simply the product of the area and the absorbed solar radiation per unit area: $$q_{loss} = q''_{solar} A_{al} = 30\,\frac{\text{Btu}}{\text{h}\cdot\text{ft}} \times \pi (2.5\,\text{in})(1\,\text{ft})$$

Calculate the heat loss by radiation and conduction

Next, we want to find the heat loss by radiation and conduction. Total heat loss can be expressed as the sum of both contributions: $$q_{loss} = q_{rad} + q_{cond}$$ Heat loss by radiation can be found using the following formula: $$q_{rad} = \epsilon_{al} \sigma A_{al}(T_{al}^4 - T_{env}^4)$$ Where \(\sigma = 5.67 \times 10^{-8}\, \frac{\text{W}}{\text{m}^2 \cdot \text{K}^4}\) is the Stefan-Boltzmann constant and \(T_{env}^4\) is the environmental temperature (\(T_{ambient}^4\) or \(T_{sky}^4\)). We will assume that radiation heat loss is mainly to the ambient and the sky, so we have: $$q_{rad} = \epsilon_{al} \sigma A_{al}(T_{al}^4 - T_{ambient}^4) + \epsilon_{al} \sigma A_{al}(T_{al}^4 - T_{sky}^4)$$ Heat loss by conduction through the air gap is found using the following formula: $$q_{cond} = k_{air}\frac{A_{al}(T_{al} - T_{g})}{L_{gap}}$$ Where \(k_{air}\) is the air thermal conductivity, \(T_g\) is the glass tube temperature, and \(L_{gap}\) is the thickness of the air gap between the tubes. We will assume that the temperature of the glass tube is constant and equal to the ambient temperature; therefore, $$q_{cond} = k_{air}\frac{A_{al}(T_{al} - T_{ambient})}{L_{gap}}$$ Substituting in the equations for \(q_{rad}\) and \(q_{cond}\) into the equation for \(q_{loss}\), we get: $$q_{loss} = \epsilon_{al} \sigma A_{al}(T_{al}^4 - T_{ambient}^4) + \epsilon_{al} \sigma A_{al}(T_{al}^4 - T_{sky}^4) + k_{air}\frac{A_{al}(T_{al} - T_{ambient})}{L_{gap}}$$

Solve for aluminum tube temperature

Now, we have only one unknown in the equation, the aluminum tube temperature \(T_{al}\). We can solve for it by rearranging the equation: $$q_{loss} = \epsilon_{al} \sigma A_{al}(2T_{al}^4 - T_{ambient}^4 - T_{sky}^4) + k_{air}\frac{A_{al}(T_{al} - T_{ambient})}{L_{gap}}$$ $$T_{al} = \frac{q_{loss} - k_{air}\frac{A_{al}(T_{al} - T_{ambient})}{L_{gap}}}{\epsilon_{al} \sigma A_{al}(2T_{al}^4 - T_{ambient}^4 - T_{sky}^4)}$$ To find the temperature \(T_{al}\), you can use a numerical solver or try various temperatures until you find one which satisfies the equation. This will give you the equilibrium temperature of the aluminum tube.