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Chapter 1: Problem 67

A photovoltaic panel of dimension \(2 \mathrm{~m} \times 4 \mathrm{~m}\) is installed on the roof of a home. The panel is irradiated with a solar flux of \(G_{S}=700 \mathrm{~W} / \mathrm{m}^{2}\), oriented normal to the top panel surface. The absorptivity of the panel to the solar irradiation is \(\alpha_{S}=0.83\), and the efficiency of conversion of the absorbed flux to electrical power is \(\eta=P / \alpha_{S} G_{S} A=0.553-0.001 \mathrm{~K}^{-1} T_{p}\), where \(T_{p}\) is the panel temperature expressed in kelvins and \(A\) is the solar panel area. Determine the electrical power generated for (a) a still summer day, in which \(T_{\text {sur }}=T_{\infty}=35^{\circ} \mathrm{C}\), \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (b) a breezy winter day, for which \(T_{\text {sur }}=T_{\infty}=-15^{\circ} \mathrm{C}, h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The panel emissivity is \(\varepsilon=0.90\).

Short Answer

Expert verified
Q_abs = 23156 W #tag_title#Step 2: Calculate the electrical power generated#tag_content#To calculate the electrical power generated, we need to multiply the total absorbed solar flux by the efficiency of the panel (畏). The efficiency of the panel depends on the temperature. 畏 = \(0.553 - 0.001T_{p}\) where \(T_{p}\) is the panel temperature in kelvins. (a) For a still summer day, we have: \(T_{\infty}\) = 35掳C + 273.15 = 308.15 K (convert to kelvins) Now substitute this value into the efficiency equation: 畏 = \(0.553 - 0.001 * 308.15\) 畏 = 0.246 Now, calculate the electrical power generated (P_summer): P_summer = 畏 * Q_abs P_summer = \(0.246 * 23156\) P_summer = 5684 W (b) For a breezy winter day, we have: \(T_{\infty}\) = -15掳C + 273.15 = 258.15 K (convert to kelvins) Now substitute this value into the efficiency equation: 畏 = \(0.553 - 0.001 * 258.15\) 畏 = 0.295 Now, calculate the electrical power generated (P_winter): P_winter = 畏 * Q_abs P_winter = \(0.295 * 23156\) P_winter = 6821 W Therefore, the electrical power generated on a still summer day is 5684 W, and on a breezy winter day, it is 6821 W.

Step by step solution

01

Calculate the absorbed solar flux

Assuming the panel is completely irradiated, the solar energy absorbed per area can be determined by the product of the solar flux and absorption coefficient. Absorbed solar flux per area (S_abs) = \(\alpha_{S} G_{S}\) Now, we can find the total absorbed solar flux by multiplying it by the total area of the photovoltaic panel (A=Area=L*W). Total absorbed solar flux (Q_abs) = \(\alpha_{S} G_{S} * A\) Let's plug in the values: \(\alpha_{S}\) = 0.83,\ \(G_{S}\) = 700 W/m虏,\ Area, A = 2m x 4m = 8 m虏. Now, Calculate total absorbed solar flux (Q_abs): Q_abs = \(0.83 * 700 * 8\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solar Irradiance
Solar irradiance is the power per unit area received from the Sun in the form of electromagnetic radiation. It is a key concept in photovoltaic power generation, as it sets the amount of solar energy available to be converted into electricity. This energy is measured in watts per square meter (W/m虏). A typical photovoltaic panel's efficiency heavily depends on the solar irradiance, as it defines how much sunlight is available to generate electricity. For example, on highly sunny days, the irradiance can be very high, leading to more potential energy being absorbed by solar panels.
  • The sun's angle and weather conditions affect solar irradiance.
  • Clear skies increase irradiance, while clouds or pollution decrease it.
Understanding solar irradiance helps in assessing the performance potential of solar panels.
Thermal Efficiency
Thermal efficiency in the context of photovoltaic panels refers to the efficiency with which absorbed solar energy is converted into electrical energy. The formula given for the panel efficiency in the exercise is \[\eta = 0.553 - 0.001\, \text{K}^{-1} T_p,\]where \(T_p\) represents the panel temperature in kelvins. This equation shows how the panel's efficiency decreases as its temperature increases, due to the reduction in the efficiency of converting solar energy into electricity.
  • Lower temperatures generally improve the efficiency.
  • This means panels work better during cooler parts of the day or year.
Thermal efficiency is important for optimizing the performance of solar panels and maximizing energy production.
Panel Temperature
Panel temperature is essential when considering the performance of solar panels because it directly influences efficiency. The temperature of the panel increases as it absorbs solar flux but can be affected by external factors like wind speed and air temperature. For instance:
  • During a hot day, the panel might reach higher temperatures, affecting its efficiency negatively.
  • Conversely, wind can help cool the panels down, maintaining a more efficient operating temperature.
Panel temperature is a crucial factor in power generation calculations since its rise can diminish electrical conversion efficiency.
Absorptivity
Absorptivity, denoted as \(\alpha\), is a measure of how well a material absorbs solar radiation. In photovoltaic panels, absorptivity indicates the fraction of solar energy that is absorbed by the panel. In our scenario, the value of \(\alpha_S = 0.83\) means that 83% of the incoming solar energy is absorbed by the panel.
  • Higher absorptivity values mean more energy is captured for conversion.
  • Materials with high absorptivity are critical in designing efficient solar panels.
Understanding absorptivity is vital to evaluate how effective a panel is at harnessing solar energy.
Emissivity
Emissivity, represented by \(\varepsilon\), indicates how effectively a surface emits infrared radiation compared to an ideal black body. In solar panels, emissivity affects how heat is dissipated.The given emissivity value for the panels, \(\varepsilon = 0.90\), suggests that the panel is very efficient in emitting heat as infrared radiation, which helps in managing temperature.
  • Higher emissivity helps panels cool down by releasing more heat.
  • Efficient thermal management sustains better performance by maintaining optimal operating temperatures.
Balancing absorptivity and emissivity is pivotal in photovoltaic design to ensure panels don鈥檛 overheat and maintain their efficiency.

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Most popular questions from this chapter

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The oven air and wall are at temperatures \(T_{\infty}=T_{\text {sur }}=180^{\circ} \mathrm{C}\). Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of \(T_{i}=24^{\circ} \mathrm{C}\). If the convection feature is activated, the forced convection heat transfer coefficient is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of \(0.97\) for the emissivity of the cake batter and pan.

The heat flux that is applied to one face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The opposite face is exposed to air at temperature \(30^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the wall exposed to air is measured and found to be \(50^{\circ} \mathrm{C}\). Do steady-state conditions exist? If not, is the temperature of the wall increasing or decreasing with time?

A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

An instrumentation package has a spherical outer surface of diameter \(D=100 \mathrm{~mm}\) and emissivity \(\varepsilon=0.25\). The package is placed in a large space simulation chamber whose walls are maintained at \(77 \mathrm{~K}\). If operation of the electronic components is restricted to the temperature range \(40 \leq T \leq 85^{\circ} \mathrm{C}\), what is the range of acceptable power dissipation for the package? Display your results graphically, showing also the effect of variations in the emissivity by considering values of \(0.20\) and \(0.30\).

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.
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