Flat-plate solar collectors are often tilted up toward the sun in order to intercept a greater amount of direct solar radiation. The tilt angle from the horizontal also affects the rate of heat loss from the collector. Consider a \(1.5-\mathrm{m}\)-high and 3-m-wide solar collector that is tilted at an angle \(\theta\) from the horizontal. The back side of the absorber is heavily insulated. The absorber plate and the glass cover, which are spaced \(2.5 \mathrm{~cm}\) from each other, are maintained at temperatures of \(80^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss from the absorber plate by natural convection for \(\theta=0^{\circ}, 30^{\circ}\), and \(90^{\circ}\).

Let's write down the given information: - \(h = 1.5\,\text{m}\) (height of the solar collector) - \(w = 3\,\text{m}\) (width of the solar collector) - \(d = 2.5\,\text{cm} = 0.025\,\text{m}\) (distance between absorber plate and glass cover) - \(T_p = 80^{\circ}\text{C} = 353\,\text{K}\) (temperature of absorber plate) - \(T_c = 40^{\circ}\text{C} = 313\,\text{K}\) (temperature of glass cover)

We will now calculate the Grashof number using the formula: \(Gr = \dfrac{g\beta(T_p - T_c)d^3}{\nu^2}\) The gravitational acceleration \(g = 9.807\,\text{m}/\text{s}^2\), the coefficient of thermal expansion \(\beta = 1/T_f\), where \(T_f = \dfrac{T_p + T_c}{2} = 333\,\text{K}\) is the average temperature, and the dynamic kinematic viscosity \(\nu\) at \(T_f\) can be obtained from tables.

Calculate the coefficient of thermal expansion (\(\beta\)) and the kinematic viscosity (\(\nu\)): \(\beta = \dfrac{1}{T_f} = \dfrac{1}{333\,\text{K}} = 3.003\,x\,10^{-3}\,\text{K}^{-1}\) Look up the kinematic viscosity from a table with properties for air: \(\nu = 1.8\,x\,10^{-5}\,\text{m}^{2}\,\text{s}^{-1}\)

Now, we can calculate the Grashof number: \(Gr = \dfrac{9.807\,\text{m} \cdot\text{s}^{-2}\,x\,3.003\,x\,10^{-3}\,\text{K}^{-1}(353\,\text{K} - 313\,\text{K})(0.025\,\text{m})^3}{(1.8\,x\,10^{-5}\,\text{m}^2 \cdot\text{s}^{-1})^2} \approx 1.775\,x\,10^{7}\)

Now we'll calculate the Nusselt number using the formula: \(Nu = 0.098\,Gr^{1/3} \approx 7.389\)

We can now calculate the heat transfer coefficient (\(h_n\)) using the formula: \(h_n = \dfrac{Nu\,k}{d}\) where \(k\) is the thermal conductivity of the fluid (air). Look up the thermal conductivity \(k = 0.03\,\text{W} \cdot\text{m}^{-1} \cdot\text{K}^{-1}\) in a table with properties for air. \(h_n = \dfrac{7.389(0.03\,\text{W} \cdot\text{m}^{-1} \cdot\text{K}^{-1})}{0.025\,\text{m}} \approx 8.867\,\text{W} \cdot\text{m}^{-2}\cdot\text{K}^{-1}\)

Now, we can calculate the rate of heat loss (\(q_n\)) for each tilt angle \(\theta\) using the formula: \(q_n = h_nA(T_p - T_c) = h_n(hw)(T_p - T_c)\) For \(\theta=0^{\circ}\) (horizontal position): \(q_n = 8.867\,\text{W} \cdot\text{m}^{-2}\cdot\text{K}^{-1}\,x\,(1.5\,\text{m})(3\,\text{m})(353\,\text{K} - 313\,\text{K}) \approx 1.688\,x\,10^{3}\,\text{W}\) For \(\theta=30^{\circ}\) and \(\theta=90^{\circ}\), we don't have to consider the tilt angle since there is no change in heat transfer. Thus, the heat loss remains the same as the horizontal case. In conclusion, the rate of heat loss from the absorber plate by natural convection is approximately \(1.688\,x\,10^{3}\,\text{W}\) for all tilt angles \(\theta = 0^{\circ}, 30^{\circ}\), and \(90^{\circ}\).