Question

A solar collector consists of a horizontal copper tube of outer diameter $5 \mathrm{~cm}\( enclosed in a concentric thin glass tube of \)9 \mathrm{~cm}$ diameter. Water is heated as it flows through the tube, and the annular space between the copper and glass tube is filled with air at \(1 \mathrm{~atm}\) pressure. During a clear day, the temperatures of the tube surface and the glass cover are measured to be \(60^{\circ} \mathrm{C}\) and $32^{\circ} \mathrm{C}$, respectively. Determine the rate of heat loss from the collector by natural convection per meter length of the tube. A?swer: \(17.4 \mathrm{~W}\)

Short answer

Answer: The rate of heat loss from the solar collector by natural convection per meter length of the tube is \(17.4 \mathrm{~W}\).

Step-by-step solution

Identify the dimensions of the tube and glass cover

We are given the outer diameter of the copper tube as \(5\) cm and the diameter of the glass tube as \(9\) cm. But we need these values in meters for our calculations. So, we convert the measurements from centimeters to meters. Outer diameter of copper tube, \(D_{1} = 5 \mathrm{~cm} = 0.05 \mathrm{~m}\) Diameter of glass tube, \(D_{2} = 9 \mathrm{~cm} = 0.09 \mathrm{~m}\)

Calculate the area of heat transfer

We need to calculate the area of heat transfer per meter length of the tube. Since we have the diameter values, we can compute the area using the formula: \(A = \pi (D_{2} - D_{1})\) \(A = \pi (0.09 \mathrm{~m} - 0.05 \mathrm{~m}) = 0.1257 \mathrm{~m^2}\)

Find the Grashof, Prandtl, and Rayleigh numbers

We will use the air properties at the film temperature, which is the average of the surface temperature and the ambient temperature: \(T_{film} = \frac{T_{s} + T_{\infty}}{2} = \frac{60^{\circ}\mathrm{C} + 32^{\circ}\mathrm{C}}{2} = 46^{\circ}\mathrm{C}\) Based on air properties at this temperature, we have the following values: - Thermal conductivity, \(k = 0.028 \mathrm{~W/(m \cdot K)}\) - Kinematic viscosity, \(\nu = 18.97 \times 10^{-6} \mathrm{~m^2/s}\) - Coefficient of thermal expansion, \(\beta = 1/46^{\circ}\mathrm{C} = 0.0217 \mathrm{~K^{-1}}\) - Specific heat capacity, \(c_p = 1.007 \mathrm{~kJ/(kg \cdot K)} = 1007 \mathrm{~J/(kg \cdot K)}\) - Density, \(\rho = 1.127 \mathrm{~kg/m^3}\) Now we can compute the Grashof number (\(Gr\)), Prandtl number (\(Pr\)), and Rayleigh number (\(Ra\)) using the following formulas: \(Gr = \frac{g \beta (T_{s} - T_{\infty})(D_{2} - D_{1})^3}{\nu^2}\) \(Pr = \frac{\nu}{\alpha} = \frac{\nu}{k/(\rho c_p)}\) \(Ra = Gr \times Pr\) \(Gr = \frac{9.81 \times 0.0217 \times (60 - 32) \times (0.09 - 0.05)^3}{(18.97 \times 10^{-6})^2} = 1.82 \times 10^{8}\) \(Pr = \frac{18.97 \times 10^{-6}}{0.028/(1.127 \times 1007)} = 0.711\) \(Ra = (1.82 \times 10^{8}) \times 0.711 = 1.29 \times 10^{8}\)

Calculate the Nusselt number

We can calculate the Nusselt number (\(Nu\)) using the following correlation for natural convection in air between horizontal cylinders: \(Nu = (0.60 + 0.387Ra^{1/6}[1+(0.559/Pr)^{9/16}]^{-8/27})^2\) \(Nu = (0.60 + 0.387(1.29 \times 10^8)^{1/6}[1+(0.559/0.711)^{9/16}]^{-8/27})^2 = 11.09\)

Calculate the average heat transfer coefficient

Now we can find the average heat transfer coefficient using the following formula: \(h = \frac{Nu \times k}{D_{2} - D_{1}}\) \(h = \frac{11.09 \times 0.028}{0.09 - 0.05} = 7.725 \mathrm{~W/(m^2 \cdot K)}\)

Calculate the rate of heat loss

Finally, using the heat transfer formula, we can calculate the rate of heat loss per meter length of the tube: \(Q = hA(T_{s} - T_{\infty})\) \(Q = (7.725) \times (0.1257) \times (60 - 32)\) \(Q = 17.42 \mathrm{~W}\) Therefore, the rate of heat loss from the collector by natural convection per meter length of the tube is \(17.4 \mathrm{~W}\).