Question

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}$.

Short answer

Answer: The natural convection heat loss from the absorber plate for the given tilt angles are: - For 𝜃 = \(0^\circ\): 1080 W - For 𝜃 = \(30^\circ\): 1069 W - For 𝜃 = \(90^\circ\): negligible

Step-by-step solution

1. List given information and constants

First, let's list the given information for the solar collector: - 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝑇𝑎): \(80^{\circ} \mathrm{C}\) - 𝑔𝑙𝑎𝑠𝑠 𝑐𝑜𝑣𝑒𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝑇𝑔): \(40^{\circ} \mathrm{C}\) - 𝑠𝑜𝑙𝑎𝑟 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑜𝑟 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠: \(1.5\, \mathrm{m} \times 3\, \mathrm{m}\) - 𝑡𝑖𝑙𝑡 𝑎𝑛𝑔𝑙𝑒(𝜃): \(0^{\circ},\, 30^{\circ},\, 90^{\circ}\) - 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑟 𝑎𝑛𝑑 𝑔𝑙𝑎𝑠𝑠 𝑠𝑝𝑎𝑐𝑒: \(2.5\, \mathrm{cm}\) We'll also need the following constants: - Gravitational acceleration (g): \(9.81\, \mathrm{m/s^2}\) - Air kinematic viscosity (ν): \(1.6 \times 10^{-5}\, \mathrm{m^2/s}\) - Air Prandtl number (𝑃𝑟): 0.7 - Stefan-Boltzmann constant (σ): \(5.67 \times 10^{-8}\, \mathrm{Wm^{-2}K^{-4}}\)

2. Calculate the temperature difference and average temperature

We need to find the average temperature (𝑇avg) and the temperature difference (𝑇diff) between the absorber and glass cover to calculate the Grashof number later. 𝑇diff = 𝑇𝑎 − 𝑇𝑔: For the temperatures given, the difference is \(80 - 40 = 40\, ^{\circ}\mathrm{C}\). 𝑇avg = (𝑇𝑎 + 𝑇𝑔) / 2: For the temperatures given, the average is \((80 + 40) / 2 = 60\, ^{\circ}\mathrm{C}\).

3. Calculate the Grashof number

For each tilt angle, 𝜃, we'll calculate the Grashof number (𝐺𝑟), which is used to evaluate the natural convection heat transfer coefficient. The Grashof number is given by: \(Gr_\theta = \frac{g\beta L^3 (T_{diff})}{\nu^2}\cos(\theta)\) Where, \(g\) - gravitational acceleration \(\beta\) - coefficient of volume expansion (approximated by 𝑏𝑒𝑡𝑎 = 1 / 𝑇avg for ideal gases) \(L\) - characteristic length (distance between absorber plate and glass cover) \(T_{diff}\) - temperature difference \(\nu\) - air kinematic viscosity \(\theta\) - tilt angle For this problem, the distance \(L\) between the absorber plate and the glass cover is 0.025 m (converted from 2.5 cm), and 𝑏𝑒𝑡𝑎 can be found by \(1 / (273.15 + 60) = 2.414 \times 10^{-3}\, \mathrm{K^{-1}}\). Now, we'll calculate the Grashof number for each tilt angle: \(Gr_0 = \frac{9.81 \times 2.414 \times 10^{-3} \times (0.025)^3 \times 40}{(1.6\times10^{-5})^2} \cos(0^{\circ}) = 1.12 \times 10^5\) \(Gr_{30} = \frac{9.81 \times 2.414 \times 10^{-3} \times (0.025)^3 \times 40}{(1.6\times10^{-5})^2} \cos(30^{\circ}) = 9.72 \times 10^4\) \(Gr_{90} = \frac{9.81 \times 2.414 \times 10^{-3} \times (0.025)^3 \times 40}{(1.6\times10^{-5})^2} \cos(90^{\circ}) = 0 \Rightarrow Gr_{90}<<Gr_{0}, Gr_{30}\)

4. Calculate the Nusselt number and heat transfer coefficient

Using the Grashof number calculated in step 3 and the given Prandtl number, 𝑃𝑟, we can find the Nusselt number (𝑁𝑢) using: \(Nu_\theta = C \times (Gr_\theta \cdot Pr)^{1/4}\) We can choose the constant \(C = 0.15\) for a parallel plate configuration in this case. We can then find the heat transfer coefficient (𝑘) by: \(k_\theta = \frac{Nu_\theta \times k_{air}}{L}\) Where \(k_{air}\) is the air thermal conductivity and is approximately \(k_{air} = 0.0262\, \mathrm{Wm^{-1}K^{-1}}\) at \(60^{\circ}\mathrm{C}\). Now, we can calculate the heat transfer coefficient for each tilt angle: \(k_0 = \frac{(0.15 \times (1.12 \times 10^5 \times 0.7)^{1/4}) \times 0.0262}{0.025} = 6.00\, \mathrm{Wm^{-2}K^{-1}}\) \(k_{30} = \frac{(0.15 \times (9.72 \times 10^4 \times 0.7)^{1/4}) \times 0.0262}{0.025} = 5.94\, \mathrm{Wm^{-2}K^{-1}}\)

5. Calculate the heat loss for each tilt angle

Now that we have the heat transfer coefficients for each tilt angle, we can find the heat loss due to natural convection for each case: \(Q_\theta = k_\theta \times A \times T_{diff}\) Where \(A\) is the area of the solar collector. In this case, the area is \(1.5\, \mathrm{m} \times 3\, \mathrm{m} = 4.5\, \mathrm{m^2}\). Finally, we can calculate the heat loss for each tilt angle: \(Q_0 = 6.00 \times 4.5 \times 40 = 1080\, \mathrm{W}\) \(Q_{30} = 5.94 \times 4.5 \times 40 = 1069\, \mathrm{W}\) \(Q_{90}\) is negligible due to the Grashof number being significantly smaller in comparison to \(Gr_0\) and \(Gr_{30}\) The natural convection heat loss from the absorber plate for the given tilt angles are: - For \(\theta = 0^\circ\): \(1080\, \mathrm{W}\) - For \(\theta = 30^\circ\): \(1069\, \mathrm{W}\) - For \(\theta = 90^\circ\): negligible