/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A proposed method for generating... [FREE SOLUTION] | 91Ó°ÊÓ

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A proposed method for generating electricity from solar irradiation is to concentrate the irradiation into a cavity that is placed within a large container of a salt with a high melting temperature. If all heat losses are neglected, part of the solar irradiation entering the cavity is used to melt the salt while the remainder is used to power a Rankine cycle. (The salt is melted during the day and is resolidified at night in order to generate electricity around the clock.) Consider conditions for which the solar power entering the cavity is \(q_{\mathrm{sal}}=7.50 \mathrm{MW}\) and the time rate of change of energy stored in the salt is \(\dot{E}_{\mathrm{st}}=3.45 \mathrm{MW}\). For a cavity opening of diameter \(D_{s}=1 \mathrm{~m}\), determine the heat transfer to the Rankine cycle, \(q_{R}\). The temperature of the salt is maintained at its melting point, \(T_{\text {salt }}=T_{\text {m }}=1000^{\circ} \mathrm{C}\). Neglect heat loss by convection and irradiation from the surroundings.

Short Answer

Expert verified
Using the energy balance equation \(q_{sal} = q_{R} + \dot{E}_{st}\), we can solve for the heat transfer to the Rankine cycle, \(q_{R}\), as follows: \(q_{R} = q_{sal} - \dot{E}_{st}\). Plugging in the given values, \(q_{sal} = 7.50\text{ MW}\) and \(\dot{E}_{st} = 3.45\text{ MW}\), we have: \(q_{R} = 7.50\text{ MW} - 3.45\text{ MW}\), which means the heat transfer to the Rankine cycle, \(q_{R}\), is \(4.05\text{ MW}\).

Step by step solution

01

Recognize the energy balance relationship

We are given that the solar power entering the cavity (\(q_{sal}\)) is divided between the heat transfer to the Rankine cycle (\(q_{R}\)) and the energy used to melt the salt (\(\dot{E}_{st}\)). This can be expressed through the following equation: \[ q_{sal} = q_{R} + \dot{E}_{st} \]
02

Solve for the heat transfer to the Rankine cycle (\(q_{R}\))

Using the energy balance equation from Step 1, we can solve for \(q_{R}\) as follows: \[ q_{R} = q_{sal} - \dot{E}_{st} \] Plugging in the given values, \(q_{sal} = 7.50\text{ MW}\) and \(\dot{E}_{st} = 3.45\text{ MW}\), we have: \[ q_{R} = 7.50\text{ MW} - 3.45\text{ MW} \]
03

Calculate the heat transfer to the Rankine cycle (\(q_{R}\))

Now, we can compute the value of \(q_{R}\): \[ q_{R} = 4.05\text{ MW} \] The heat transfer to the Rankine cycle, \(q_{R}\), is 4.05 MW.

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Key Concepts

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

Heat Transfer
Heat transfer is an essential process where energy is moved from a hotter body to a colder one. In solar power generation, heat transfer plays a crucial role in converting sunlight into usable electricity.
Here's how it works:
  • Solar irradiation heats a cavity in the system.
  • The heat is transferred to a salt with a high melting temperature inside the cavity.
  • This leads to two potential outcomes: part of the heat melts the salt, and another portion thermally energizes a Rankine cycle.
Understanding heat transfer allows engineers to balance energy used for melting salt and powering the Rankine cycle effectively, thereby maximizing overall efficiency. By concentrating heat, systems can achieve high temperatures needed for efficient electricity production.
Rankine Cycle
The Rankine cycle is a fundamental principle in thermodynamics used primarily for power generation. It's a cycle that converts heat into mechanical work, often implemented in power plants.
The cycle includes these critical steps:
  • Heat is absorbed from a high-energy source, like solar-heated salt.
  • A working fluid (often water or steam) is used to carry this heat.
  • The fluid expands, doing work on a turbine to generate electricity.
  • The fluid is then cooled and compressed back into a liquid state to restart the cycle.
In a solar power setup, the specifications of the Rankine cycle can be optimized to use heat from molten salt efficiently, leading to sustainable energy production even when sunlight is not available.
Energy Balance
Energy balance is crucial for determining how effectively a system uses energy. It involves accounting for all energy entering and leaving the system. For the solar cavity system:
  • Total energy input is provided by solar irradiation, denoted as \( q_{sal} \).
  • Energy is divided: some melts the salt (\( \dot{E}_{st} \)) and some transfers to the Rankine cycle (\( q_R \)).
  • The balance is expressed by the equation \( q_{sal} = q_R + \dot{E}_{st} \), ensuring no excess waste if heat losses are ignored.
Accurate energy balance calculations are vital to designing systems where energy is effectively harnessed, ensuring maximum conversion of solar energy to electrical output.
Melting Temperature
Melting temperature is the point at which a solid becomes a liquid. In this solar power system, the salt's melting point is pivotal for operation efficiency.

Here’s why it matters:
  • The salt inside the cavity must maintain a consistent temperature near its melting point (\( 1000^{\circ} \mathrm{C} \)) to efficiently absorb and release heat.
  • As the sun shines, the salt absorbs energy, melts, and stores this heat.
  • At night, when the solar input ceases, the solidifying salt releases stored heat to power a Rankine cycle continuously.
Choosing salt with an appropriate melting temperature allows the system to store and use solar energy efficiently across varying conditions.

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Most popular questions from this chapter

A procedure for measuring the thermal conductivity of solids at elevated temperatures involves placement of a sample at the bottom of a large furnace. The sample is of thickness \(L\) and is placed in a square container of width \(W\) on a side. The sides are well insulated. The walls of the cavity are maintained at \(T_{w}\), while the bottom surface of the sample is maintained at a much lower temperature \(T_{e}\) by circulating coolant through the sample container. The sample surface is diffuse and gray with an emissivity \(\varepsilon_{s}\). Its temperature \(T_{s}\) is measured optically. (a) Neglecting convection effects, obtain an expression from which the sample thermal conductivity may be evaluated in terms of measured and known quantities \(\left(T_{w}, T_{s}, T_{c}, \varepsilon_{s}, L\right)\). The measurements are made under steady-state conditions. If \(T_{w}=1400 \mathrm{~K}, T_{s}=1000 \mathrm{~K}, \varepsilon_{s}=0.85, L=\) \(0.015 \mathrm{~m}\), and \(T_{c}=300 \mathrm{~K}\), what is the sample thermal conductivity? (b) If \(W=0.10 \mathrm{~m}\) and the coolant is water with a flow rate of \(\dot{m}_{c}=0.1 \mathrm{~kg} / \mathrm{s}\), is it reasonable to assume a uniform bottom surface temperature \(T_{c}\) ?

Consider the metallic surface of Example 12.7. Additional measurements of the spectral, hemispherical emissivity yield a spectral distribution which may be approximated as follows: (a) Determine corresponding values of the total, hemispherical emissivity \(\varepsilon\) and the total emissive power \(E\) at \(2000 \mathrm{~K}\). (b) Plot the emissivity as a function of temperature for \(500 \leq T \leq 3000 \mathrm{~K}\). Explain the variation.

The spectral emissivity of unoxidized titanium at room temperature is well described by the expression \(\varepsilon_{\lambda}=0.52 \lambda^{-0.5}\) for \(0.3 \mu \mathrm{m} \leq \lambda \leq 30 \mu \mathrm{m}\). (a) Determine the emissive power associated with an unoxidized titanium surface at \(T=300 \mathrm{~K}\). Assume the spectral emissivity is \(\varepsilon_{\mathrm{\lambda}}=0.1\) for \(\lambda>30 \mu \mathrm{m}\). (b) Determine the value of \(\lambda_{\max }\) for the emissive power of the surface in part (a).

Neglecting the effects of radiation absorption, emission, and scattering within their atmospheres, calculate the average temperature of Earth, Venus, and Mars assuming diffuse, gray behavior. The average distance from the sun of each of the three planets, \(L_{s p}\), along with their measured average temperatures, \(\bar{T}_{p}\), are shown in the table below. Based upon a comparison of the calculated and measured average temperatures, which planet is most affected by radiation transfer in its atmosphere? \begin{tabular}{lcc} \hline Planet & \(\boldsymbol{L}_{s-p}(\mathbf{m})\) & \(\bar{T}_{p}(\mathbf{K})\) \\\ \hline Venus & \(1.08 \times 10^{11}\) & 735 \\ Earth & \(1.50 \times 10^{11}\) & 287 \\ Mars & \(2.30 \times 10^{11}\) & 227 \\ \hline \end{tabular}

Isothermal furnaces with small apertures approximating a blackbody are frequently used to calibrate heat flux gages, radiation thermometers, and other radiometric devices. In such applications, it is necessary to control power to the furnace such that the variation of temperature and the spectral intensity of the aperture are within desired limits. (a) By considering the Planck spectral distribution, Equation \(12.30\), show that the ratio of the fractional change in the spectral intensity to the fractional change in the temperature of the furnace has the form $$ \frac{d I_{\lambda} / I_{\lambda}}{d T / T}=\frac{C_{2}}{\lambda T} \frac{1}{1-\exp \left(-C_{2} / \lambda T\right)} $$ (b) Using this relation, determine the allowable variation in temperature of the furnace operating at \(2000 \mathrm{~K}\) to ensure that the spectral intensity at \(0.65 \mu \mathrm{m}\) will not vary by more than \(0.5 \%\). What is the allowable variation at \(10 \mu \mathrm{m}\) ?

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