/*! 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 28 At steady state, a power cycle h... [FREE SOLUTION] | 91Ó°ÊÓ

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At steady state, a power cycle having a thermal efficiency of \(38 \%\) generates \(100 \mathrm{MW}\) of electricity while discharging energy by heat transfer to cooling water at an average temperature of \(70^{\circ} \mathrm{F}\). The average temperature of the steam passing through the boiler is \(900^{\circ} \mathrm{F}\). Determine (a) the rate at which energy is discharged to the cooling water, in Btu/h. (b) the minimum theoretical rate at which energy could be discharged to the cooling water, in Btu/h. Compare with the actual rate and discuss.

Short Answer

Expert verified
Rate of energy discharged: 556,191,589.47 Btu/h. Minimum theoretical rate: 217,192,524.14 Btu/h.

Step by step solution

01

Determine electrical power output in Btu/h

Convert the electrical power output from MW to Btu/h. The conversion factor is 1 MW = 3,412,142 Btu/h. Thus, the electrical power generated is \( 100 \text{ MW} \times 3,412,142 \text{ Btu/h per MW} = 341,214,200 \text{ Btu/h} \).
02

Calculate the total energy input into the power cycle

Using the thermal efficiency formula \( \text{Efficiency} = \frac{\text{Electrical Power Output}}{\text{Total Energy Input}} \), the total energy input can be found: \( \frac{341,214,200 \text{ Btu/h}}{0.38} = 897,405,789.47 \text{ Btu/h} \).
03

Determine the rate of energy discharged to cooling water

The difference between the total energy input and the electrical power output is the energy discharged to the cooling water: \( 897,405,789.47 \text{ Btu/h} - 341,214,200 \text{ Btu/h} = 556,191,589.47 \text{ Btu/h} \). This is the rate at which energy is discharged to the cooling water.
04

Calculate the minimum theoretical energy discharge using Carnot Efficiency

For the Carnot efficiency, use the absolute temperatures: \( T_H = 900^{\text{°}F} + 459.67 = 1359.67^{\text{°}R} \)\( T_C = 70^{\text{°}F} + 459.67 = 529.67^{\text{°}R} \)\( \text{Carnot Efficiency} = 1 - \frac{T_C}{T_H} = 1 - \frac{529.67}{1359.67} = 0.611 \).
05

Calculate the minimum theoretical heat discharge rate

Using Carnot efficiency, find the minimum theoretical energy input, then the discharge: \( \frac{341,214,200 \text{ Btu/h}}{0.611} = 558,406,724.14 \text{ Btu/h} \). The heat discharged is \( 558,406,724.14 \text{ Btu/h} - 341,214,200 \text{ Btu/h} = 217,192,524.14 \text{ Btu/h} \).

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Key Concepts

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

steady state operation
In a power cycle operating at steady state, the conditions do not change over time. This means the amount of energy entering the system equals the amount of energy leaving over any given time period. Understanding steady state operation helps simplify calculations because we can assume consistent values for energy inputs and outputs.
In our problem, the power cycle operates at steady state, generating consistent electrical output while discharging a steady amount of energy to the cooling water.
  • Energy in = Energy out
  • Power cycles are common examples, like in power plants
  • Consistency helps with predicting performance
By maintaining steady state operation, we can calculate the energy conversion and efficiency more accurately.
thermal efficiency calculation
Thermal efficiency is a measure of how well a power cycle converts heat into work (or useful energy). It is given by the formula:
\[ \text{Efficiency} = \frac{\text{Electrical Power Output}}{\text{Total Energy Input}} \times 100 \text{ (to express as percentage)} \]
In this exercise, the thermal efficiency is 38%, meaning 38% of the energy input is converted into electrical power, while the rest is likely wasted, usually as heat to the environment.
Using the provided 38% efficiency, we calculated:
\[ \frac{341,214,200 \text{ Btu/h}}{0.38} = 897,405,789.47 \text{ Btu/h} \text{ (Total Energy Input)} \]
This Total Energy Input helps track how efficiently our power cycle is performing and also helps predict energy losses.
energy conversion
Energy conversion is the process of changing one form of energy into another. In a power cycle, we're primarily concerned with converting thermal energy (from fuel, like steam) into electrical energy.
However, not all of the thermal energy gets converted. Some of it is discharged as waste heat, often to cooling water. The calculations in our problem explored this:
  • Total energy input = 897,405,789.47 Btu/h
  • Electrical power output = 341,214,200 Btu/h
  • Energy discharged to cooling water = 556,191,589.47 Btu/h
This demonstrates the balance of energy in a power cycle: the total energy input minus useful output equals the energy that needs to be managed as waste.
To minimize waste, analyzing the minimum theoretical discharge using Carnot efficiency provides insight into what is achievable under ideal conditions.

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Most popular questions from this chapter

A hot thermal reservoir is separated from a cold thermal reservoir by a cylindrical rod insulated on its lateral surface. Energy transfer by conduction between the two reservoirs takes place through the rod, which remains at steady state. Using the Kelvin-Planck statement of the second law, demonstrate that such a process is irreversible.

For each \(\mathrm{kW}\) of power input to an ice maker at steady state, determine the maximum rate that ice can be produced, in \(\mathrm{kg} / \mathrm{h}\), from liquid water at \(0^{\circ} \mathrm{C}\). Assume that \(333 \mathrm{~kJ} / \mathrm{kg}\) of energy must be removed by heat transfer to freeze water at \(0^{\circ} \mathrm{C}\), and that the surroundings are at \(20^{\circ} \mathrm{C}\).

To increase the thermal efficiency of a reversible power cycle operating between reservoirs at \(T_{\mathrm{H}}\) and \(T_{C}\), would you increase \(T_{\mathrm{H}}\) while keeping \(T_{\mathrm{C}}\) constant, or decrease \(T_{\mathrm{C}}\) while keeping \(T_{\mathrm{H}}\) constant? Are there any natural limits on the increase in thermal efficiency that might be achieved by such means?

A certain reversible power cycle has the same thermal efficiency for hot and cold reservoirs at 1000 and \(500 \mathrm{~K}\), respectively, as for hot and cold reservoirs at temperature \(T\) and \(1000 \mathrm{~K}\). Determine \(T\), in \(\mathrm{K}\).

A reversible power cycle receives \(Q_{H}\) from a hot reservoir at temperature \(T_{\mathrm{H}}\) and rejects energy by heat transfer to the surroundings at temperature \(T_{0}\). The work developed by the power cycle is used to drive a refrigeration cycle that removes \(Q_{\mathrm{C}}\) from a cold reservoir at temperature \(T_{\mathrm{C}}\) and discharges energy by heat transfer to the same surroundings at \(T_{0}\). (a) Develop an expression for the ratio \(Q_{\mathrm{C}} / Q_{\mathrm{H}}\) in terms of the temperature ratios \(T_{\mathrm{H}} / T_{0}\) and \(T_{\mathrm{C}} / T_{0}\). (b) Plot \(Q_{\mathrm{C}} / Q_{\mathrm{H}}\) versus \(T_{\mathrm{H}} / T_{0}\) for \(T_{\mathrm{C}} / T_{0}=0.85,0.9\), and \(0.95\), and versus \(T_{C} / T_{0}\) for \(T_{H} / T_{0}=2,3\), and 4.

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