91Ó°ÊÓ

In this problem, we have a few given values: - Outer diameter of copper tube = 5 cm - Diameter of glass tube = 9 cm - Temperature of copper tube surface = \(60^{\circ} \mathrm{C}\) - Temperature of glass cover = \(32^{\circ} \mathrm{C}\) We can see that the difference in temperature between the copper tube and the glass cover is what causes heat to be transferred by natural convection. The overall temperature difference is: \(\Delta T = T_{copper} - T_{glass} = 60 - 32 = 28^{\circ} \mathrm{C}\)

Before calculating the natural convection heat transfer coefficient, we need to find the annular space between the copper and glass tubes. The radius of the glass tube and copper tube are 4.5 cm and 2.5 cm, respectively, so the annular space can be calculated as: Annular space = \(R_{glass} - R_{copper} = 4.5 - 2.5 = 2\) cm

The Grashof number, Gr, is a dimensionless number that helps determine if natural convection will occur. To calculate Gr, we will use the following formula: \(Gr = \frac{g \cdot \beta \cdot \Delta T \cdot L^3}{\nu^2}\) where: - \(g\) is the acceleration due to gravity (\(9.81 \frac{m}{s^2}\)) - \(\beta\) is the coefficient of thermal expansion of air (we'll use \(\beta \approx \frac{1}{T_{avg}}\), where \(T_{avg}\) is the average temperature in Kelvin: \(T_{avg} = \frac{T_{copper}+T_{glass}}{2} = \frac{60+32}{2} + 273.15 = 333.15 K\)) - \(\Delta T\) is the overall temperature difference (28°C) - \(L\) is the annular space between the copper and glass tubes (0.02 m) - \(\nu\) is the kinematic viscosity of air (approximately \(15.11 \times 10^{-6} \frac{m^2}{s}\) at 333.15 K) Now, calculate the Grashof number: \(Gr = \frac{9.81 \cdot \frac{1}{333.15} \cdot 28 \cdot (0.02)^3}{(15.11 \times 10^{-6})^2} \approx 4.18 \times 10^4\)

The Nusselt number, Nu, is a dimensionless number used to determine the natural convection heat transfer coefficient. For this problem, we can use the following correlation for an annular space: \(Nu = 1 + 0.14 \cdot Gr^{1/3} \cdot Pr^{0.5}\) where: - \(Gr\) is the Grashof number - \(Pr\) is the Prandtl number, another dimensionless number, which is approximately 0.703 for air at 333.15 K Calculate the Nusselt number: \(Nu = 1 + 0.14 \cdot (4.18 \times 10^4)^{1/3} \cdot (0.703)^{0.5} \approx 5.76\)

Now, use the Nusselt number to find the natural convection heat transfer coefficient, \(h\), using the following formula: \(h = \frac{Nu \cdot k}{L}\) where: - \(Nu\) is the Nusselt number - \(k\) is the thermal conductivity of air (approximately \(0.026 \frac{W}{m \cdot K}\) at 333.15 K) - \(L\) is the annular space (0.02 m) Calculate the natural convection heat transfer coefficient: \(h = \frac{5.76 \cdot 0.026}{0.02} \approx 7.47 \frac{W}{m \cdot K}\)

Finally, calculate the rate of heat loss, \(Q\), per meter length of the tube using the formula: \(Q = h \cdot A \cdot \Delta T\) where: - \(h\) is the natural convection heat transfer coefficient - \(A\) is the surface area per meter length of the copper tube (\(A = 2 \pi R_{copper} \cdot L\), with \(L\) now representing 1 meter length of the tube) - \(\Delta T\) is the overall temperature difference (28°C) Calculate the rate of heat loss: \(Q = 7.47 \cdot (2 \pi (0.025) \cdot 1) \cdot 28 \approx 65.6 W\) The rate of heat loss from the collector by natural convection per meter length of the tube is approximately 65.6 W.