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Chapter 5: Problem 18

A reversible power cycle operating between hot and cold reservoirs at \(900 \mathrm{~K}\) and \(200 \mathrm{~K}\), respectively, receives \(120 \mathrm{~kJ}\) by heat transfer from the hot reservoir for each cycle of operation. Determine the net work developed in 12 cycles of operation, in \(\mathrm{kJ}\).

Short Answer

Expert verified
1120 kJ

Step by step solution

01

Identify given values

Given values are: Hot reservoir temperature, \( T_H = 900 \mathrm{~K} \) Cold reservoir temperature, \( T_C = 200 \mathrm{~K} \) Heat received from hot reservoir per cycle, \( Q_H = 120 \mathrm{~kJ} \).
02

Calculate efficiency of the Carnot cycle

The efficiency \( \eta \) of a Carnot cycle operating between two reservoirs is given by: \[ \eta = 1 - \frac{T_C}{T_H} \] Substitute the given temperatures: \[ \eta = 1 - \frac{200}{900} = 1 - \frac{2}{9} = \frac{7}{9} \]
03

Calculate net work per cycle

The net work developed per cycle \( W_{net, cycle} \) can be determined using: \[ W_{net, cycle} = \eta Q_H \] Substituting the values: \[ W_{net, cycle} = \left( \frac{7}{9} \right) \times 120 \mathrm{~kJ} = 93.33 \mathrm{~kJ} \]
04

Calculate net work for 12 cycles

The net work for 12 cycles \( W_{net, 12 cycles} \) is: \[ W_{net, 12 cycles} = 12 \times W_{net, cycle} \] Substitute the value of \( W_{net, cycle} \): \[ W_{net, 12 cycles} = 12 \times 93.33 \mathrm{~kJ} = 1120 \mathrm{~kJ} \]

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Key Concepts

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

Carnot cycle efficiency
The Carnot cycle is an idealized thermodynamic cycle that provides a benchmark for the efficiency of real engines.
It operates between two reservoirs: a hot reservoir at temperature \( T_H \) and a cold reservoir at temperature \( T_C \).
The efficiency \( \eta \) of any Carnot engine depends solely on these temperatures and is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
Higher \( \eta \) means the engine does more useful work.
For example, with temperatures \( T_H = 900 \K \) and \( T_C = 200 \K \), we calculate:
\[ \eta = 1 - \frac{200}{900} = \frac{7}{9} \]
This means 7/9 of the heat taken in gets converted into useful work and the rest is expelled.
thermodynamic reservoirs
Thermodynamic reservoirs are large bodies that can absorb or supply finite amounts of heat without undergoing any change in temperature.
The hot reservoir provides heat energy to the cycle, whereas the cold reservoir absorbs the expelled heat.
These reservoirs are assumed to have infinite thermal capacity.
In the Carnot cycle, the heat transfer between the system and the reservoirs occurs at constant temperatures \( T_H \) and \( T_C \).
This simplifies calculations and helps us focus on the efficiency of the cycle.
net work calculation
To determine the net work \( W_{net} \) produced by a reversible power cycle, we multiply the efficiency \( \eta \) by the total heat input \( Q_H \).
Using our previous efficiency calculation \( \eta = \frac{7}{9} \):
\[ W_{net, cycle} = \eta Q_H \]
Substituting the given \( Q_H = 120 \mathrm{~kJ} \):
\[ W_{net, cycle} = \left( \frac{7}{9} \right) \times 120 \mathrm{~kJ} = 93.33 \mathrm{~kJ} \]
For multiple cycles, multiply this value by the number of cycles:
\[ W_{net, 12 \; cycles} = 12 \times 93.33 \mathrm{~kJ} = 1120 \mathrm{~kJ} \]
This means the total net work output for 12 cycles is 1120 kJ.

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Most popular questions from this chapter

If a window air conditioner were placed on a table in a room and operated, would the room temperature increase, decrease, or remain the same? Explain.

An inventor has developed a refrigerator capable of maintaining its freezer compartment at \(-5^{\circ} \mathrm{C}\) while operating in a kitchen at \(25^{\circ} \mathrm{C}\), and claims the device has a coefficient of performance of (a) 7, (b) \(8.93\), (c) 10. Evaluate the claim in each of the three cases.

To maintain a dwelling at a temperature of \(20^{\circ} \mathrm{C}, 600 \mathrm{MJ}\) energy is required per day when the temperature outside the dwelling is \(4^{\circ} \mathrm{C}\). A heat pump is used to supply this energy. Determine the minimum theoretical cost to operate the heat pump, in \(\$ /\) day, if the cost of electricity is \(\$ 0.10\) per \(\mathrm{kW} \cdot \mathrm{h}\).

A carnot engine absorbs \(250 \mathrm{~J}\) of heat from reservoir at \(100^{\circ} \mathrm{C}\) and rejects heat to a reservoir at \(20^{\circ} \mathrm{C}\). Find (a) the heat rejected, (b) work done by the engine, and (c) thermal efficiency.

The freezer compartment of a refrigerator is maintained at \(-6^{\circ} \mathrm{C}\). The temperature of the surrounding air is \(25^{\circ} \mathrm{C}\). The refrigerant absorbs heat from the freezer compartment at the rate of \(10,000 \mathrm{~kJ} / \mathrm{h}\). The power input required to operate the refrigerator is \(3500 \mathrm{~kJ} / \mathrm{h}\). Determine the coefficient of performance of the refrigerator, and determine the coefficient of performance of a reversible refrigeration cycle operating between the same given temperatures.
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