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Steam at a given state enters a turbine operating at steady state and expands adiabatically to a specified lower pressure. Would you expect the power output to be greater in an internally reversible expansion or an actual expansion?

Short Answer

Expert verified
The power output is greater in an internally reversible expansion than in an actual expansion.

Step by step solution

01

Understand the Problem

Identify that the problem involves a turbine where steam expands adiabatically from a high pressure to a low pressure state. The goal is to compare power output between an internally reversible expansion and an actual expansion.
02

Internally Reversible Expansion

In an internally reversible process, the expansion occurs in a manner where no entropy is generated, and the process is isentropic (entropy remains constant). The efficiency of this expansion is maximized because there are no dissipative effects.
03

Actual Expansion

An actual expansion in a turbine is not perfectly efficient and involves irreversible processes. These include friction, turbulence, and heat losses, leading to an increase in entropy. Thus, the work output is less than that of the internally reversible process.
04

Compare Work Outputs

Use the first law of thermodynamics for both cases. For the internally reversible process: \[ W_{\text{rev}} = \text{h1} - \text{h2s} \] where \( h1 \) is the initial specific enthalpy and \( h2s \) is the specific enthalpy in the isentropic state after expansion. For the actual process: \[ W_{\text{act}} = \text{h1} - \text{h2a} \] where \( h2a \) is the specific enthalpy in the actual state after expansion. Generally, \( h2a > h2s \), so \( W_{\text{act}} < W_{\text{rev}} \).
05

Conclusion

Since the internally reversible process maximizes efficiency and minimizes entropy generation, the work output for an internally reversible expansion is greater than that of an actual expansion.

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Key Concepts

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

Isentropic Process
In thermodynamics, an isentropic process is a particular type of process where entropy remains constant. This means there is no entropy generation within the system. An isentropic process is ideal because it represents a fully reversible process, free from any dissipative effects like friction or turbulence.
For a turbine, an isentropic expansion means that the steam expands from a higher pressure to a lower pressure without any loss of energy due to internal friction or heat transfer.
This implies that the efficiency of an isentropic expansion is maximized. Mathematically, we describe an isentropic expansion by ensuring the entropy at the start and end of the process remains constant:
\( s1 = s2 \)
where \( s \) is the entropy at those states.
Utilizing the First Law of Thermodynamics for an isentropic process, the work output, \( W_{\text{rev}} \), is determined as follows:
\[ W_{\text{rev}} = h1 - h2s \]
Here, \( h1 \) is the initial specific enthalpy and \( h2s \) is the specific enthalpy after isentropic expansion.
Entropy Generation
Entropy generation occurs in any real process where irreversibilities like friction, turbulence, or heat loss are present. These factors cause the process to deviate from the ideal behavior observed in an isentropic process. In practical turbine operation, actual expansion is not entirely efficient due to these effects.
As a result, the actual expansion process generates entropy, denoted by \( \Delta S > 0 \). A higher entropy state means that some potential work is lost, and the process is less efficient.
In the formula for work output for an actual process, the specific enthalpy in the post-expansion state, \( h2a \), is higher due to entropy generation:
\[ W_{\text{act}} = h1 - h2a \]
Generally, \( h2a \) is greater than \( h2s \), indicating that \( W_{\text{act}} \) is less than \( W_{\text{rev}} \). The increase in enthalpy post-expansion reflects energy losses due to irreversibilities.
First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle that describes the conservation of energy. It states that the energy within a closed system remains constant. For any process happening in a system, the change in internal energy is equal to the heat added to the system minus the work done by the system.
Mathematically, the First Law is expressed as:
\[ \Delta U = Q - W \]
For an adiabatic turbine where heat transfer, \( Q \), is zero, the First Law simplifies to:
\[ \Delta U = -W \]
With the turbine operating under steady-state conditions, the work done becomes directly related to change in specific enthalpy, \( h \):
\[ W = h1 - h2 \]
For an internally reversible (isentropic) process, this modifies to:
\[ W_{\text{rev}} = h1 - h2s \]
And for an actual process plagued by irreversibilities:
\[ W_{\text{act}} = h1 - h2a \]
By comparing the two, it is evident that \( W_{\text{rev}} \) exceeds \( W_{\text{act}} \) because \( h2a > h2s \. \). This underlines the importance of minimizing entropy generation to maximize the power output of turbines.

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Most popular questions from this chapter

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.

What are some of the principal irreversibilities present during operation of (a) an automobile engine, (b) a household refrigerator, (c) a gas-fired water heater, (d) an electric water heater?

A refrigeration cycle having a coefficient of performance of 3 maintains a computer laboratory at \(18^{\circ} \mathrm{C}\) on a day when the outside temperature is \(30^{\circ} \mathrm{C}\). The thermal load at steady state consists of energy entering through the walls and windows at a rate of \(30,000 \mathrm{~kJ} / \mathrm{h}\) and from the occupants, computers, and lighting at a rate of \(6000 \mathrm{~kJ} / \mathrm{h}\). Determine the power required by this cycle and compare with the minimum theoretical power required for any refrigeration cycle operating under these conditions, each in \(\mathrm{kW}\).

A building for which the heat transfer rate through the walls and roof is \(400 \mathrm{~W}\) per degree temperature difference between the inside and outside is to be maintained at \(20^{\circ} \mathrm{C}\). For a day when the outside temperature is \(4^{\circ} \mathrm{C}\), determine the power required at steady state, \(\mathrm{kW}\), to heat the building using electrical resistance elements and compare with the minimum theoretical power that would be required by a heat pump. Repeat the comparison using typical manufacturer's data for the heat pump coefficient of performance.

A heat pump receives energy by heat transfer from the outside air at \(0^{\circ} \mathrm{C}\) and discharges energy by heat transfer to a dwelling at \(20^{\circ} \mathrm{C}\). Is this in violation of the Clausius statement of the second law of thermodynamics? Explain.

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