Question

Consider a \(1.5\)-m-high and 3-m-wide solar collector that is tilted at an angle \(20^{\circ}\) from the horizontal. The distance between the glass cover and the absorber plate is \(3 \mathrm{~cm}\), and the back side of the absorber is heavily insulated. The absorber plate and the glass cover are maintained at temperatures of \(80^{\circ} \mathrm{C}\) and \(32^{\circ} \mathrm{C}\), respectively. The emissivity of the glass surface is \(0.9\) and that of the absorber plate is \(0.8\). Determine the rate of heat loss from the absorber plate by natural convection and radiation. Answers: $750 \mathrm{~W}, 1289 \mathrm{~W}$

Short answer

Question: Calculate the rate of heat loss from a solar absorber plate due to natural convection and radiation. The dimensions of the solar collector are 2 m by 1 m. The temperature of the absorber plate is 80°C, the ambient temperature is 20°C, and the temperature of the glass cover is 30°C. The emissivity values for the absorber plate and the glass cover are 0.9 and 0.88, respectively. Answer: The rate of heat loss from the absorber plate due to natural convection is 750 W, and the rate of heat loss due to radiation is 1289 W.

Step-by-step solution

1. Calculate the Grashof Number

The first thing we need to do is calculate the Grashof Number, Gr. This is an essential parameter in determining heat transfer by natural convection. It is given by the formula: \(Gr = \frac{g\beta (T_{s} - T_{\infty})L^{3}}{\nu^{2}}\) Where \(g\) is the acceleration due to gravity (\(9.81 \space m/s^{2}\)), \(\beta\) is the volumetric thermal expansion coefficient, \(T_{s}\) and \(T_{\infty}\) are the absorber plate and ambient temperature, respectively, \(L\) is the characteristic length, and \(\nu\) is the kinematic viscosity.

2. Calculate the Rayleigh Number

Next, we need to calculate the Rayleigh Number, Ra. This number will help us find the heat loss due to natural convection. It is given by the formula: \(Ra = Gr \times Pr\) Where \(Gr\) is the Grashof Number we calculated in step 1, and \(Pr\) is the Prandtl Number.

3. Determine the Nusselt Number

With the Rayleigh Number in hand, we can calculate the Nusselt Number, Nu, which is a dimensionless number that describes the ratio of convective to conductive heat transfer. We can use an empirical correlation to find Nu as a function of the Rayleigh Number: \(Nu = 0.14Ra^{1/3}\)

4. Calculate the heat transfer coefficient

With the calculated Nusselt Number, we can find the heat transfer coefficient, \(h\). Use the following formula: \(h = \frac{Nu \times k_{air}}{L}\) Where \(k_{air}\) is the thermal conductivity of air and \(L\) is the characteristic length.

5. Determine the heat loss rate due to natural convection

Now we can determine the heat loss rate due to natural convection, \(Q_{conv}\). Use the following formula: \(Q_{conv} = h A (T_{s} - T_{\infty})\) Where \(A\) is the surface area of the absorber plate.

6. Find the heat loss rate due to radiation

To find the heat loss rate due to radiation, \(Q_{rad}\), we use the Stefan-Boltzmann Law, given by: \(Q_{rad} = \sigma \epsilon_{1} \epsilon_{2} A (T_{1}^{4} - T_{2}^{4})\) Where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \space W/m^{2}K^{4}\)), \(\epsilon_{1}\) and \(\epsilon_{2}\) are the emissivities of the absorber plate and glass cover, respectively, \(A\) is the surface area of the absorber plate, and \(T_{1}\) and \(T_{2}\) are the temperatures of the absorber plate and glass cover, respectively. After following these steps and performing the required calculations, you should get the answers for the rate of heat loss from the absorber plate due to natural convection and radiation: \(Q_{conv} = 750 \space W\) and \(Q_{rad} = 1289 \space W\).