Question

A solar collector consists of a horizontal copper tube of outer diameter $5 \mathrm{~cm}\( enclosed in a concentric thin glass tube of diameter \)12 \mathrm{~cm}$. Water is heated as it flows through the tube, and the annular space between the copper and the glass tubes is filled with air at $1 \mathrm{~atm}$ pressure. The emissivities of the tube surface and the glass cover are \(0.85\) and \(0.9\), respectively. During a clear day, the temperatures of the tube surface and the glass cover are measured to be $60^{\circ} \mathrm{C}\( and \)40^{\circ} \mathrm{C}$, respectively. Determine the rate of heat loss from the collector by natural convection and radiation per meter length of the tube.

Short answer

The rate of heat loss from the solar collector by natural convection and radiation per meter length of the tube is approximately 68.23 W/m.

Step-by-step solution

Calculate Convective Heat Transfer Coefficient

Using the Churchil-Chu correlation for natural convection in annular enclosures filled with air, we can find the convective heat transfer coefficient (h_c) by the following equation: $$h_c = \frac{1.32 * \Delta T^{1/3}}{(D_i/D_o)^{1/22} * (D_o - D_i)^{1/6}}$$ where, \(\Delta T\) - temperature difference between the inner and outer surface \((60^{\circ}C - 40^{\circ}C)\) \(D_i\) - inner diameter of the annular space (outer diameter of the copper tube) \(D_o\) - outer diameter of the annular space (inner diameter of the glass tube) Plug in the values: $$h_c = \frac{1.32 * (20)^{1/3}}{((0.05/0.12)^{1/22}) * (0.07)^{1/6}}$$ Calculate \(h_c\): $$h_c \approx 4.41 \, W/(m^2K)$$

Calculate Radiative Heat Transfer

Using the Stefan-Boltzmann law, we can find the radiative heat transfer (Q_r) as follows: $$Q_r = F_G * A * \sigma * (T_i^4 - T_o^4)$$ where, \(F_G\) - radiative shape factor for parallel tubes \(A\) - surface area per meter length of tube \(\sigma\) - Stefan-Boltzmann constant \((5.67 * 10^{-8} \, W/(m^2 K^4))\) \(T_i\) - inner surface temperature in Kelvin \(T_o\) - outer surface temperature in Kelvin First, convert the temperatures to Kelvin: \(T_i = 60^{\circ}C + 273.15 = 333.15K\) \(T_o = 40^{\circ}C + 273.15 = 313.15K\) As the radii are much greater than the plate separation, we can use the following approximation for the radiative shape factor: $$F_G \approx \frac{1}{1/\epsilon_i + 1/\epsilon_o - 1}$$ $$F_G \approx \frac{1}{1/0.85 + 1/0.9 - 1} \approx 0.4312$$ Now, find the surface area of the copper tube per meter length: $$A = \pi * D_i$$ $$A = \pi * 0.05 \, m^2$$ Calculate the radiative heat transfer: $$Q_r = 0.4312 * (\pi * 0.05) * (5.67 * 10^{-8}) * (333.15^4 - 313.15^4)$$ $$Q_r \approx 54.4 \, W/m$$

Calculate Total Heat Loss

To find the total heat loss, we need to add the heat losses due to convection and radiation. First, let's calculate the convective heat transfer (Q_c): $$Q_c = h_c * A * \Delta T$$ $$Q_c = 4.41 * (\pi * 0.05) * 20$$ $$Q_c \approx 13.83\, W/m$$ Now, find the total heat loss (Q_t): $$Q_t = Q_c + Q_r$$ $$Q_t = 13.83 + 54.4$$ $$Q_t \approx 68.23\, W/m$$ Hence, the rate of heat loss from the collector by natural convection and radiation per meter length of the tube is about 68.23 W/m.