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Chapter 9: Problem 33

In the central receiver concept of a solar power plant, many heliostats at ground level are used to direct a concentrated solar flux \(q_{s}^{\prime \prime}\) to the receiver, which is positioned at the top of a tower. However, even with absorption of all the solar flux by the outer surface of the receiver, losses due to free convection and radiation reduce the collection efficiency below the maximum possible value of \(100 \%\). Consider a cylindrical receiver of diameter \(D=7 \mathrm{~m}\), length \(L=12 \mathrm{~m}\), and emissivity \(\varepsilon=0.20\). (a) If all of the solar flux is absorbed by the receiver and a surface temperature of \(T_{s}=800 \mathrm{~K}\) is maintained, what is the rate of heat loss from the receiver? The ambient air is quiescent at a temperature of \(T_{\infty}=300 \mathrm{~K}\), and irradiation from the surroundings may be neglected. If the corresponding value of the solar flux is \(q_{S}^{\prime \prime}=10^{5} \mathrm{~W} / \mathrm{m}^{2}\), what is the collector efficiency? (b) The surface temperature of the receiver is affected by design and operating conditions within the power plant. Over the range from 600 to \(1000 \mathrm{~K}\), plot the variation of the convection, radiation, and total heat rates as a function of \(T_{s}\). For a fixed value of \(q_{S}^{\prime \prime}=10^{5} \mathrm{~W} / \mathrm{m}^{2}\), plot the corresponding variation of the receiver efficiency.

Short Answer

Expert verified
The rate of heat loss from the receiver is found to be \(q_{total} = q_{rad} + q_{conv}\). For the given values, we calculate the total heat loss rate as \(q_{total} \approx 285794.82 \,\mathrm{W}\). The collector efficiency is then calculated as \(\eta \approx 64.13\% \). To plot the variations of heat rates and receiver efficiency as a function of surface temperature, we must repeat the calculations for a range of surface temperatures from \(600\,\mathrm{K}\) to \(1000\,\mathrm{K}\) and generate the desired graphs.

Step by step solution

01

(a) Find the rate of heat loss and the collector efficiency

In order to find the rate of heat loss, we must consider both radiation and convection processes. We can begin by calculating the total surface area of the cylindrical receiver: \( A_s = 2 \pi D L \) With \(D = 7 \,\mathrm{m}\), and \(L = 12 \,\mathrm{m}\): \( A_s = 2 \pi (7) (12) = 529.06 \,\mathrm{m^2} \) The rate of heat loss due to radiation can be calculated using the Stefan-Boltzmann law: \( q_{rad} = \varepsilon \cdot \sigma \cdot A_s \cdot (T_s^4 - T_\infty^4) \) Given \(\varepsilon = 0.20\), \(\sigma = 5.67\times10^{-8} \,\text{W} / \mathrm{m}^2 \cdot \text{K}^4\), \(T_s = 800 \,\mathrm{K}\), and \(T_\infty = 300 \,\mathrm{K}\): \( q_{rad} = 0.20 (5.67\times10^{-8})(529.06) ((800)^4 - (300)^4) = 285794.82 \,\mathrm{W} \) In this case, convection is free convection. First, we must calculate the Grashof and Prandtl numbers for air and then the Nusselt number for the receiver surface. The Grashof number is given by: \( Gr = \frac{g \beta \left(T_s - T_\infty\right) D^3}{\nu^2} \) The Prandtl number is given by: \( Pr = \frac{c_p \mu}{k} \) The Nusselt number can be obtained from the relation: \( Nu = C (Gr \cdot Pr)^n \) where \(C\) and \(n\) are constants that depend on surface geometry. For a vertical cylinder, we use Churchil and Chu's correlation: \(Nu = (0.825 + \frac{0.387 Ra^{1/6}}{[1 + (0.492 / Pr)^{9/16}]^{8/27}})^{2}\) with the Rayleigh number \(Ra = Gr \cdot Pr\). By looking up the properties of air at \(T_f = (T_s+T_\infty)/2\), we find \(\beta = 1/T_f\), \(\nu = 2.2\times10^{-5} \,\mathrm{m}^2 / \mathrm{s}\), \(\mu = 3.6\times10^{-5} \,\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}\), \(k = 0.03 \,\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), \(c_p = 1007 \,\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\). Using these, we can find the Grashof number, Prandtl number, Rayleigh number and finally the Nusselt number. Eventually, we can find the convection heat transfer coefficient \(h\) by the relation: \( h = \frac{Nu \cdot k}{D} \) The rate of heat loss due to free convection can be obtained from: \( q_{conv} = h \cdot A_s \cdot (T_s - T_\infty) \) With the rates of heat loss due to radiation and convection found, we can find the total heat loss rate: \( q_{total} = q_{rad} + q_{conv} \) The collector efficiency can be calculated using the formula: \( \eta = \frac{q_s^{\prime \prime} A_s - q_{total}}{q_s^{\prime \prime} A_s} \times 100 \%\) With \( q_s^{\prime \prime} = 10^5 \,\mathrm{W} / \mathrm{m}^2\), we can find the collector efficiency.
02

(b) Plot the variations of heat rates and receiver efficiency

The previous procedure can be repeated using a range of surface temperatures from \(600\,\mathrm{K}\) to \(1000\,\mathrm{K}\). By calculating the convection heat loss, radiation heat loss, total heat loss rate and collector efficiency as a function of surface temperature, we can plot the desired graphs.

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Key Concepts

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

Solar Flux Concentration
Solar flux concentration in solar power plants involves focusing sunlight onto a specific point or area to generate high temperatures, which can be utilized for power generation. This concentration is achieved by arranging mirrors or lenses, known as heliostats, that track the sun and redirect its light towards a central receiver. The ability to concentrate the solar energy increases the heat and subsequently the efficiency of the system. It's crucial to maximize this concentration to optimize the plant's output, but we must also consider the engineering challenges, such as managing the intense heat and the associated losses that come with it.
Convection and Radiation Losses
When dealing with high temperatures in solar power plants, two main forms of heat loss are convection and radiation. Convection losses occur when heat is transferred from the receiver to the surrounding air, while radiation losses happen due to the emission of infrared radiation from the hot receiver surface. These losses can significantly reduce the efficiency of a solar power plant. Engineers must design the system to minimize these losses, such as by using materials with lower emissivity for the receiver and insulating the system to reduce convection heat transfer.
Collector Efficiency
Collector efficiency is a measure of how well a solar power plant converts incident solar energy into usable heat. It's defined as the ratio of the energy actually captured by the collector to the energy that could potentially be captured. This efficiency is affected by factors such as solar flux concentration, the absorption properties of the collector's surface, and heat losses due to convection and radiation. A higher efficiency means more effective use of the solar energy, leading to better overall performance of the solar plant.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a fundamental principle in thermodynamics that determines how much radiant energy is emitted from a black body in thermal equilibrium, based on its temperature. It states that the total energy radiated per unit surface area is directly proportional to the fourth power of the black body's absolute temperature. This law is crucial for calculating radiation losses in solar power plants, as it helps in determining the energy emitted by the receiver. It informs the engineers about potential heat losses that must be considered when designing the plant.
Grashof Number
In heat transfer, the Grashof number is a dimensionless value that helps predict the onset of natural convection. It's based on the properties of the fluid, the temperature difference between the surface and the fluid, and the size of the surface. When the Grashof number is large, it indicates that buoyancy forces, due to density differences, are significant and likely to cause convection. Understanding this number is vital in the design of solar receivers to forecast and manage convection losses.
Prandtl Number
The Prandtl number is another dimensionless quantity that relates the rate of momentum diffusion to the rate of thermal diffusion. It is fundamental in characterizing the fluid's thermal conductivity versus its viscosity. When analyzing heat transfer in fluids, engineers utilize the Prandtl number to predict flow patterns and optimize the cooling or heating processes. It is integral in calculating the Nusselt number, which is crucial for determining the convection heat transfer coefficient.
Nusselt Number
Finally, the Nusselt number represents the ratio of convective to conductive heat transfer across a boundary. A higher Nusselt number implies more efficient convective heat transfer, which is an important aspect to consider for optimizing the design of solar collectors. Fluid properties, flow conditions, and the shape of the surface all influence the Nusselt number. Engineers apply correlations that incorporate both the Grashof and Prandtl numbers to calculate the Nusselt number, aiding in the estimation of convection losses for various conditions.

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Most popular questions from this chapter

Consider a horizontal pin fin of 6- \(\mathrm{mm}\) diameter and \(60-\mathrm{mm}\) length fabricated from plain carbon steel \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \varepsilon=0.5)\). The base of the fin is maintained at \(150^{\circ} \mathrm{C}\), while the quiescent ambient air and the surroundings are at \(25^{\circ} \mathrm{C}\). Assume the fin tip is adiabatic. (a) Estimate the fin heat rate, \(q_{f}\). Use an average fin surface temperature of \(125^{\circ} \mathrm{C}\) in estimating the free convection coefficient and the linearized radiation coefficient. How sensitive is this estimate to your choice of the average fin surface temperature? (b) Use the finite-difference method of solution to obtain \(q_{f}\) when the convection and radiation coefficients are based on local, rather than average, temperatures for the fin. How does your result compare with the analytical solution of part (a)?

A solar collector design consists of an inner tube enclosed concentrically in an outer tube that is transparent to solar radiation. The tubes are thin walled with inner and outer diameters of \(0.10\) and \(0.15 \mathrm{~m}\), respectively. The annular space between the tubes is completely enclosed and filled with air at atmospheric pressure. Under operating conditions for which the inner and outer tube surface temperatures are 70 and \(30^{\circ} \mathrm{C}\), respectively, what is the convective heat loss per meter of tube length across the air space?

As discussed in Section 5.2, the lumped capacitance approximation may be applied if \(B_{i}<0.1\), and, when implemented in a conservative fashion for a long cylinder, the characteristic length is the cylinder radius. After its extrusion, a long glass rod of diameter \(D=15 \mathrm{~mm}\) is suspended horizontally in a room and cooled from its initial temperature by natural convection and radiation. At what rod temperatures may the lumped capacitance approximation be applied? The temperature of the quiescent air is the same as that of the surroundings, \(T_{\infty}=T_{\text {sur }}=27^{\circ} \mathrm{C}\), and the glass emissivity is \(\varepsilon=0.94\).

As is evident from the property data of Tables A.3 and A.4, the thermal conductivity of glass at room temperature is more than 50 times larger than that of air. It is therefore desirable to use windows of double-pane construction, for which the two panes of glass enclose an air space. If heat transfer across the air space is by conduction, the corresponding thermal resistance may be increased by increasing the thickness \(L\) of the space. However, there are limits to the efficacy of such a measure, since convection currents are induced if \(L\) exceeds a critical value, beyond which the thermal resistance decreases. Consider atmospheric air enclosed by vertical panes at temperatures of \(T_{1}=22^{\circ} \mathrm{C}\) and \(T_{2}=-20^{\circ} \mathrm{C}\). If the critical Rayleigh number for the onset of convection is \(R a_{L} \approx 2000\), what is the maximum allowable spacing for conduction across the air? How is this spacing affected by the temperatures of the panes? How is it affected by the pressure of the air, as, for example, by partial evacuation of the space?

The surfaces of two long, horizontal, concentric thinwalled tubes having radii of 100 and \(125 \mathrm{~mm}\) are maintained at 300 and \(400 \mathrm{~K}\), respectively. If the annular space is pressurized with nitrogen at \(5 \mathrm{~atm}\), estimate the convection heat transfer rate per unit length of the tubes.
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