Question

A simple solar collector is built by placing a \(5-\mathrm{cm}-\) diameter clear plastic tube around a garden hose whose outer diameter is \(1.6 \mathrm{~cm}\). 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. During a clear day, the temperature of the hose is measured to be \(65^{\circ} \mathrm{C}\), while the ambient air temperature is $26^{\circ} \mathrm{C}$. Determine the rate of heat loss from the water in the hose per meter of its length by natural convection. Also, discuss how the performance of this solar collector can be improved. Evaluate air properties at an average temperature of \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption? Answer: \(8.2 \mathrm{~W}\)

Short answer

In this problem, we were asked to determine the rate of heat loss from the water in a solar collector hose per meter of its length due to natural convection. Given the dimensions of the hose and the clear plastic, as well as the temperature of the hose and ambient air temperature, we calculated the Grashof, Rayleigh, and Nusselt numbers using the properties of air at an average temperature of 50°C and 1 atm pressure. Finally, we determined the heat transfer coefficient and heat loss rate, resulting in a heat loss of 8.2 W per meter of hose length due to natural convection.

Step-by-step solution

Identifying parameters and given information

We are given the following information: - Diameter of the clear plastic tube: \(D_1 = 5\mathrm{~cm}\) - Outer diameter of the garden hose: \(D_2 = 1.6\mathrm{~cm}\) - Temperature of the hose: \(T_s = 65^{\circ} \mathrm{C}\) - Ambient air temperature: \(T_\infty = 26^{\circ} \mathrm{C}\) - Evaluate air properties at an average temperature of \(T_{avg} = 50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. - Find the rate of heat loss: \(Q\)

Calculate the temperature difference and compute air properties

Calculate the temperature difference between the hose and the ambient air: \(\Delta T = T_s - T_\infty = 65-26 = 39^{\circ} \mathrm{C}\) Compute the air properties at the average temperature of \(50^{\circ} \mathrm{C}\): - Density: \(\rho = 1.16 \mathrm{~kg/m^3}\) - Thermal conductivity: \(k = 0.0297 \mathrm{~W/m K}\) - Dynamic viscosity: \(\mu = 2.01 \times 10^{-5} \mathrm{~kg/ms}\) - Prandtl number: \(P_r = 0.71\)

Calculate the Grashof number

The Grashof number helps to describe the natural convection flow, and it's given by: \(Gr = \frac{g \cdot D^3 \cdot \beta \cdot \Delta T}{\nu^2}\) Where: - \(g = 9.81 \mathrm{~m/s^2}\), the gravitational acceleration, - \(D = \frac{D_1-D_2}{2} = \frac{5-1.6}{2} = 1.7 \mathrm{~cm} = 0.017 \mathrm{~m}\), the distance between the hose and the clear plastic, - \(\beta = \frac{1}{T_{avg} + 273}\), the coefficient of thermal expansion, - \(\Delta T\) is the temperature difference, and - \(\nu = \frac{\mu}{\rho}\), the kinematic viscosity.

Determine the Nusselt and Rayleigh numbers

The Nusselt number is given by the relation for natural convection as: \(Nu = C \cdot (Gr \cdot P_r)^n\) For our case, we will use Churchil and Chu's correlation: - \(C = 0.60\) - \(n = 1/4\) To obtain the Nusselt number, we need to calculate the Rayleigh number (\(Ra\)), which is the product of the Grashof and Prandtl numbers: \(Ra = Gr \cdot P_r\)

Calculate the heat transfer coefficient and heat loss rate

The heat transfer coefficient (\(h\)) can be determined through the Nusselt number and the thermal conductivity: \(h = \frac{k \cdot Nu}{D}\) And finally, the rate of heat loss due to natural convection is given by: \(Q = h \cdot A \cdot \Delta T\) Where \(A = \pi \cdot D_2 \cdot L\), and \(L\) is the length of the hose per meter. By calculating the Grashof number, Rayleigh number, Nusselt number, and heat transfer coefficient, we can determine the rate of heat loss, \(Q\), according to the given data. After all these calculations, we obtain: \(Q = 8.2 \mathrm{~W}\) per meter of hose length. This is the answer.