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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

A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.

A freezer compartment is covered with a 2 -mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

A transmission case measures \(W=0.30 \mathrm{~m}\) on a side and receives a power input of \(P_{i}=150 \mathrm{hp}\) from the engine. The switch is set to open at \(70^{\circ} \mathrm{C}\), the maximum dryer air temperature. To operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches \(70^{\circ} \mathrm{C}\left(T_{\text {set }}\right)\) when the air temperature \(T\) is less than \(T_{\text {set. }}\). If the convection heat transfer coefficient between the air and the exposed switch surface of \(30 \mathrm{~mm}^{2}\) is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how much heater power \(P_{e}\) is required when the desired dryer air temperature is \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

Chips of width \(L=15 \mathrm{~mm}\) on a side are mounted to a substrate that is installed in an enclosure whose walls and air are maintained at a temperature of \(T_{\text {sur }}=25^{\circ} \mathrm{C}\). The chips have an emissivity of \(\varepsilon=0.60\) and a maximum allowable temperature of \(T_{s}=85^{\circ} \mathrm{C}\). (a) If heat is rejected from the chips by radiation and natural convection, what is the maximum operating power of each chip? The convection coefficient depends on the chip-to-air temperature difference and may be approximated as \(h=C\left(T_{s}-T_{\infty}\right)^{1 / 4}\), where \(C=4.2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{5 / 4}\). (b) If a fan is used to maintain airflow through the enclosure and heat transfer is by forced convection, with \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum operating power?
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