Question

The preliminary design of a space station calls for a power cycle that at steady state receives energy by heat transfer at \(T_{\mathrm{H}}=600 \mathrm{~K}\) from a nuclear source and rejects energy to space by thermal radiation according to Eq. 2.33. For the radiative surface, the temperature is \(T_{\mathrm{C}}\), the emissivity is \(0.6\), and the surface receives no radiation from any source. The thermal efficiency of the power cycle is one- half that of a reversible power cycle operating between reservoirs at \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\) - (a) For \(T_{\mathrm{C}}=400 \mathrm{~K}\), determine \(\dot{W}_{\text {cycle }} / \mathrm{A}\), the net power developed per unit of radiator surface area, in \(\mathrm{kW} / \mathrm{m}^{2}\), and the thermal efficiency. (b) Plot \(\dot{W}_{\text {cycle }} / \mathrm{A}\) and the thermal efficiency versus \(T_{\mathrm{C}}\), and determine the maximum value of \(\dot{W}_{\text {cycle }} / \mathrm{A}\). (c) Determine the range of temperatures \(T_{\mathrm{C}}\), in \(\mathrm{K}\), for which \(\dot{W}_{\text {cycle }} / \mathrm{A}\) is within 2 percent of the maximum value obtained in part (b). The Stefan-Boltzmann constant is \(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\).

Short answer

(a) For \( T_{\mathrm{C}} = 400 \mathrm{~K} \), \( \dot{W}_{\text{cycle}} / A \approx 0.015 \mathrm{kW} / \mathrm{m}^{2} \) and \( \eta = \frac{1}{6} \). (b) Plot \( \dot{W}_{\text{cycle}} / A \) and efficiency versus \( T_{\mathrm{C}} \). (c) Determine \( T_{\mathrm{C}} \) range for 2% of the max value.

Step-by-step solution

- Determine the reversible efficiency

The efficiency of a reversible power cycle operating between two thermal reservoirs is given by \[ \eta_{\text{rev}} = 1 - \frac{T_{\mathrm{C}}}{T_{\mathrm{H}}} \]Given: \( T_{\mathrm{H}} = 600\, \mathrm{K} \) and \( T_{\mathrm{C}} = 400\, \mathrm{K} \),\[ \eta_{\text{rev}} = 1 - \frac{400}{600} = 1 - \frac{2}{3} = \frac{1}{3} \]

- Determine the actual efficiency

The actual efficiency is half of the reversible efficiency,\[ \eta = \frac{1}{2} \eta_{\text{rev}} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \]

- Determine the heat transfer rate per unit area

The power cycle rejects energy to space by thermal radiation, using the Stefan-Boltzmann Law:\[ q = \epsilon \sigma T_{\mathrm{C}}^4 \]where \( \epsilon = 0.6 \) and \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4} \).Given \( T_{\mathrm{C}} = 400 \, \mathrm{K} \),\[ q = 0.6 \times 5.67 \times 10^{-8} \times (400)^4 \text{ W/m}^2 \]Calculate \( q \):\[ q = 0.6 \times 5.67 \times 10^{-8} \times 2.56 \times 10^9 = 87.55 \text{ W/m}^2 \approx 88 \text{ W/m}^2 \]

- Determine the power per unit area

The net power developed per unit of surface area is given by the product of the actual efficiency and the heat rejected per unit area:\[ \dot{W}_{\text{cycle}} / A = \eta \times q \]Substitute \( \eta = 1/6 \) and \( q = 88\, \mathrm{W}/\mathrm{m}^2 \):\[ \dot{W}_{\text{cycle}} / A = \frac{1}{6} \times 88 = 14.67 \text{ W/m}^2 \approx 0.015 \text{ kW/m}^2 \]

- Plot \( \dot{W}_{\text{cycle}} / A \) and efficiency versus \( T_{\mathrm{C}} \)

To plot \( \dot{W}_{\text{cycle}} / A \) and efficiency as functions of \( T_{\mathrm{C}} \), repeat Steps 1 through 4 over a range of \( T_{\mathrm{C}} \) values (e.g., from 200 K to 500 K), and record the results for each step.

- Determine the maximum \( \dot{W}_{\text{cycle}} / A \)

From the plotted data, observe the temperature \( T_{\mathrm{C}} \) at which \( \dot{W}_{\text{cycle}} / A \) reaches its maximum value.

- Find the temperature range for 2% of maximum

Using the maximum value determined, calculate \( 2\text{%} \) of this value and find the range of \( T_{\mathrm{C}} \) for which \( \dot{W}_{\text{cycle}} / A \) remains within \( 2\text{%} \) of the maximum.

Key concepts

Reversible Power Cycles

A reversible power cycle is an idealized thermodynamic cycle that operates with maximum efficiency. It follows the principles of Carnot's theorem, which states that no engine operating between two heat reservoirs can be more efficient than a Carnot engine. The efficiency of a reversible power cycle is given by \[ \eta_{\text{rev}} = 1 - \frac{T_{\mathrm{C}}}{T_{\mathrm{H}}} \]where \( T_{\mathrm{H}} \) is the higher temperature reservoir and \( T_{\mathrm{C}} \) is the lower temperature reservoir. Because this is an ideal scenario, real-world engines cannot achieve this efficiency due to irreversibilities and losses. Yet, it serves as a benchmark for understanding how efficient an engine can theoretically be.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law relates the power radiated per unit area of a black body to its temperature. Formulated as \[ q = \epsilon \sigma T_{\mathrm{C}}^{4} \],this law states that the emitted radiation power depends on the emissivity \( \epsilon \) of the surface, the Stefan-Boltzmann constant \( \sigma \) (given as \( 5.67 \times 10^{-8} \) \( \, W/m^{2} \cdot \, K^{4} \)), and the temperature of the radiative surface in Kelvin. This equation is critical when calculating heat transfer by radiation, especially in environments like space where convection and conduction are minimal.

Heat Transfer

Heat transfer is the process by which heat energy is exchanged between physical systems. It can occur via conduction, convection, or radiation. In the context of the given exercise, heat transfer happens by radiation, as the space station radiates energy into the vacuum of space. The efficiency of this process is governed by the temperatures of the heat reservoirs and the properties of the radiative surface. By using the Stefan-Boltzmann Law, one can determine the heat flux \( q \), allowing for the calculation of the net power produced per unit area.

Radiative Surface

A radiative surface is an area through which energy is emitted as electromagnetic radiation. Key parameters include:

• Emissivity (\( \epsilon \)): Measures how effectively a surface emits radiation compared to a perfect black body.
• Surface area (\( A \)): Larger areas can radiate more energy.
• Temperature (\( T_{\mathrm{C}} \)): Higher temperatures lead to higher radiative power.

In applications like a space station, the cooling mechanism relies on the radiative surface to reject excess heat. By understanding these parameters, engineers can design systems that effectively manage thermal loads, ensuring the operational stability of the space station.