Question

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at $500 \mathrm{~W} / \mathrm{m}^{2}\( with a surrounding temperature of \)0^{\circ} \mathrm{C}$. The convection heat transfer coefficient at the absorber surface is $5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as $T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\(, and determine net heat flux \)\dot{q}_{0}$ absorbed by the solar collector.

Short answer

Question: Determine the temperature distribution T(x) of the absorber plate as a function of position and calculate the net heat flux absorbed by the solar collector. Solution: From our analysis and calculations, the temperature distribution of the absorber plate can be expressed as: \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\) Moreover, the net heat flux absorbed by the solar collector, \(\dot{q}_{0}\), can be found by solving the energy balance equation: \(q_{in} = q_conv + q_rad + \int \dot{q}(x) dx\) Using the given values for absorptivity, emissivity, and solar radiation, and converting temperatures to Kelvin, we can calculate the individual heat transfer terms and ultimately find the value of \(\dot{q}_{0}\).

Step-by-step solution

Identify the relevant energy balance equation

For the absorber plate, the energy balance equation can be written as: Incident solar radiation = Convective heat transfer + Radiative heat transfer + rate of change of internal energy.

Find the expressions for individual heat transfer terms

We are given the values for absorptivity (α = 0.9), emissivity (ε = 0.9), the temperature of the absorber plate surface (T0 = 35°C), and the incident solar radiation (500 W/m²). Let's put these values in the appropriate terms. Incident solar radiation (q_in): \(q_{in} = \alpha \cdot G = 0.9 \cdot 500 \, \mathrm{W/m^2}\) Convective heat transfer (q_conv): \(q_{conv} = h \cdot (T_0 - T_{ambient}) = 5 \, \mathrm{W/m^2\cdot K} \cdot (35\,^\circ\mathrm{C} - 25\,^\circ\mathrm{C})\) Radiative heat transfer (q_rad): \(q_{rad} = \epsilon \cdot \sigma \cdot (T_0^4 - T_{surrounding}^4)\), where \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \, \mathrm{W/m^2\cdot K^4}\). Convert the temperatures to Kelvin. T0(K) = 35°C + 273.15 = 308.15 K Tsurrounding(K) = 0°C + 273.15 = 273.15 K Now calculate \(q_{rad}\): \(q_{rad} = 0.9 \cdot (5.67 \times 10^{-8}) \cdot (308.15^4 - 273.15^4) \, \mathrm{W/m^2}\)

Calculate net heat flux absorbed by solar collector

Now, using the energy balance equation: q_in = q_conv + q_rad + rate of change of internal energy. The rate of change of internal energy is given by the integration of the heat flux along the x-axis \(\int \dot{q}(x) dx\) Since the convective and radiative heat transfers are given in terms of T0(K), we can write the equation as: \(q_{in} = q_conv + q_rad + \int \dot{q}(x) dx\) Calculate \(q_{in}\), \(q_{conv}\), and \(q_{rad}\) using values from Steps 2. Solve for the net heat flux absorbed by the solar collector, \(\dot{q}_{0}\), which is the value of \(\dot{q}(x)\) at x=0. Rearrange and integrate the energy balance equation and obtain an expression for T(x). \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\) Conclude that the variation of temperature in the absorber plate can be expressed as this equation, and report the value of \(\dot{q}_{0}\).