Technology & Engineering
Isentropic Efficiency
Isentropic efficiency is a measure of how effectively a process or device can convert input energy into useful work without incurring losses. It is often used in thermodynamics and fluid mechanics to assess the performance of compressors, turbines, and nozzles. A higher isentropic efficiency indicates a more efficient conversion of energy.
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9 Key excerpts on "Isentropic Efficiency"
eBook - PDFRadial Flow Turbocompressors
Design, Analysis, and Applications
- Michael Casey, Chris Robinson(Authors)
- 2021(Publication Date)
- Cambridge University Press(Publisher)
Compressor efficiency is best defined as the ratio of the minimum effort that would be required to compress the flow in an ideal reversible compression process between two pressure levels relative to the effort actually expended to achieve this pressure rise. It is expressed in such a way that the efficiency has the value of unity (or 100%) for a perfect machine. This statement looks very straightforward but is actually highly complex. It is necessary to consider slowly and methodically all the issues that may cause difficulties and require clarification, and this discussion is started in this section and continued throughout the chapter. The ideal thermodynamic process in the reversible machine which is being used as a basis to define the minimum work has to be defined and, in practice, this reference process could be isentropic, polytropic or isothermal. The isentropic adiabatic process might naturally be considered to be the most useful definition of the reference process for this perfect machine as it has no losses and no heat transfer. This leads to the definition of the Isentropic Efficiency. But as shown in this chapter, it is sometimes more appropriate to consider a reversible machine with a reference process that is the same as the polytropic process which actually occurs in the real machine. This leads to the definition of the polytropic efficiency, which in compressors is larger than the Isentropic Efficiency. In intercooled compressors, a reversible isothermal process may be more useful as reference, and this leads to the definition of the isothermal efficiency. Because of the divergence of the constant pressure lines in an h-s diagram towards high temperature, an isothermal process defines the minimum possible work between two pressure levels.
eBook - PDFThermodynamics
A Smart Approach
- Ibrahim Dinçer(Author)
- 2020(Publication Date)
- Wiley(Publisher)
However, the story differs in actual cases, and the performance is quanti-fied by so-called: Isentropic Efficiency, which needs to be specifically written for each device. The following subsections cover these in detail. (a) Turbine A fluid or gas is expanded in a turbine to produce power. There are of course both steam (i.e. using water) and gas (i.e. using air) turbines; these are considered for analysis. Turbines are normally assumed to be adiabatic for analysis purposes. The performance of an adiabatic turbine is usually expressed by Isentropic Efficiency. Consider a turbine with inlet state 1 with temperature T 1 and pressure P 1 and an exit state 2 with temperature T 2 and pressure P 2 as shown in Figure 4.17. The power output from this State 2 State 1 Turbine W out Figure 4.17 Schematic diagram of a turbine. 4.9 Isentropic Efficiencies for Ideal Gases 193 compressor would be maximum if the fluid is expanded reversibly and adiabatically (i.e. isentropically) between the given initial state and given exit pressure. The isentropic effi-ciency is then the ratio of actual work rate to the isentropic work rate: η is , t = W a W s = m h 1 -h 2 a m h 1 -h 2 s = mC p T 1 -T 2 a mC p T 1 -T 2 s = T 1 -T 2 a T 1 -T 2 s 4 12 where m is the mass flow rate of gas ( kg/s ), C p is the constant pressure specific heat of gas ( kJ/kgK ), h 1 is inlet enthalpy of the gas ( kJ/kg ), h 2s is the enthalpy of the gas at the turbine outlet ( kJ/kg ) if the process was isentropic while h 2a is the actual enthalpy of the gas ( kJ/kg ), and the subscript t refers to turbine. T 1 , T 2a , and T 2s are the inlet, actual exit and isentropic exit temperatures of the gas. A few reminders are listed for the above outlined methodology for turbine as follows: • The relationship h = C p T is used for ideal gases to further simplify the equation. • The isentropic enthalpy at the exit may directly be obtained from exit pressure and exit entropy (since s 1 equals s 2 s ).
- John Reisel(Author)
- 2021(Publication Date)
- Cengage Learning EMEA(Publisher)
All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 219 6.5 Isentropic Efficiencies in order to either maximize the power output (for turbines) or minimize power consumption (for compressors and pumps). Therefore, for these devices we are seeking to compare the actual performance to that which would be seen for a reversible and adiabatic device. Consider Eq. (6.35): S ? gen 5 m ? ss 2 2 s 1 d 2 o n j 51 Q ? j T b, j (6.35) For a reversible process, S ? gen 5 0, and for an adiabatic process, Q ? 5 0. Substituting these into Eq. (6.35) yields 0 5 m ? ss 2 2 s 1 d 2 0 which logically reduces to s 2 5 s 1 The entropy is constant for a reversible and adiabatic process, and a constant entropy process is also called an isentropic process. Therefore, because a reversible and adiabatic process implies that the process is isentropic, we call the type of comparison reflecting this an Isentropic Efficiency although it is more correctly termed a reversible and adiabatic efficiency. Be aware, though, that an isentropic process does not need to be reversible and adiabatic. We could have a constant entropy process where heat is removed from a system and entropy is generated but whose contributions from each balance out, which causes the process to be isentropic. 6.5.1 Isentropic Processes in Ideal Gases To solve problems involving isentropic processes for fluids such as water and refrigerants, we generally find the appropriate property values using the initial entropy and the final pressure, temperature, or specific volume—depending on what is known.
eBook - PDF- Kaufui Vincent Wong(Author)
- 2011(Publication Date)
- CRC Press(Publisher)
End (b) Use definitions of isentropic efficiencies to determine the actual states, and thereby the actual efficiency. Start The logic diagram for solving the above problem may be represented as shown. It is similar to that for Example 4.7. The mass conservation equation is not used explicitly; a unit mass of the working fluid is considered as undergoing the cycle. The energy conser-vation equation (first law) is used to determine the unknown quantity, which is thermal efficiency of the ideal cycle in this case. Then, the definitions of isentropic efficiencies (second law) are used to find the actual states and thereby the thermal efficiency of the actual cycle. Second Law of Thermodynamics and Entropy 4 -45 Example 4.16 Problem A device operates on an ideal vapor-compression refrigeration cycle between 0.16 and 1 MPa. Refrigerant-134a is the working fluid. Determine (i) the coefficient of performance if the device is operating as a refrigerator, as well as when operating as a heat pump, (ii) if the Isentropic Efficiency of the compressor is 85%, evaluate the COP’s of the correspond-ing refrigerator and heat pump. 1 2 s T 3 4 4s Solution Assumptions : 1. K.E. and P.E. changes are negligible. Analysis : From P 1 = 0.16 MPa, x 1 = 1.0, h 1 = 388.7 kJ/kg, s 1 = 1.736 kJ/(kg K) From P 2 = 1 MPa, s 2 = s 1 , h 2 = 427.0 kJ/kg, From P 3 = 1 MPa, x 3 = 0.0, h 3 = 255.6 kJ/kg, s 3 = 1.188 kJ/(kg K) From P 4 = 0.16 MPa, h 4 = h 3 , s 4 = 1.219 kJ/(kg K) 1. When the device operates as a refrigerator, the heat transfer quantity of concern is the heat transferred to the evaporator. When the device operates as a heat pump, the heat transfer quantity of interest is the heat transferred out at the con-denser.
eBook - PDFThermodynamics
From Concepts to Applications, Second Edition
- Arthur Shavit, Chaim Gutfinger(Authors)
- 2008(Publication Date)
- CRC Press(Publisher)
a. Find the power of the turbine. b. Find the heat interaction in the heat exchanger. c. What is the Isentropic Efficiency of the turbine? 266 Thermodynamics: From Concepts to Applications d. Find the entropy change across the turbine. e. Find the maximum work and the irreversibility of the turbine. f. Find the change in entropy of everything involved in the process. g. What is the irreversibility of the total process? 10.51 Air at 150 kPa and 200°C enters an adiabatic diffuser at 240 m/s, at a rate of 2 kg/s and leaves at 60 m/s. The Isentropic Efficiency of the device is ε c = 0.8 and the environment is at 100 kPa and 300 K. Find a. The exit temperature. b. The exit pressure. c. The cross-sectional areas at the inlet and outlet of the device. d. The change in entropy of everything involved in the process. e. The irreversibility and maximum work of this process. 10.52 Steam at 150 kPa and 200°C enters an adiabatic diffuser at 240 m/s at a rate of 2 kg/s and leaves at 60 m/s. The Isentropic Efficiency of the device is ε c = 0.8 and the environment is at 100 kPa and 300 K. Find a. The exit temperature. b. The exit pressure. c. The cross-sectional areas at the inlet and outlet of the device. d. The change in entropy of everything involved in the process. e. The irreversibility and maximum work of this process. 10.53 Steam at 3 MPa and 400°C enters an adiabatic nozzle at 100 m/s and a rate of 2 kg/ s. It leaves the nozzle at 400 kPa and 200°C. The environment is at 100 kPa and 300 K. Find a. The exit velocity. b. The Isentropic Efficiency. c. The cross-sectional areas at the inlet and outlet of the device. d. The change in entropy of everything involved in the process. e. The irreversibility and maximum work of this process. 10.54 Air at 2.5 MPa and 600 K enters an adiabatic nozzle at 100 m/s and a rate of 3 kg/s. It leaves the nozzle at 500 kPa and 430 K. The environment is at 100 kPa and 300 K. Find a. The exit velocity. b. The Isentropic Efficiency.
No longer available |Learn more- Irving Granet, Maurice Bluestein(Authors)
- 2014(Publication Date)
- CRC Press(Publisher)
Using the energy equation for a reversible, nonflow process enabled us to derive some general rela-tions among entropy, enthalpy, and internal energy. Key Terms Terms used for the first time in this chapter are as follows: absolute thermodynamic temperature: An arbitrary temperature function used with the Carnot cycle. For practical purposes, it is identical to the absolute temperature scales of Kelvin or Rankine. Carnot cycle: A reversible cycle proposed by Sadi Carnot in 1824 that consists of two constant-temperature and two adiabatic processes. cycle: A series of thermodynamic processes during which the working fluid can be made to undergo changes involving energy transitions and is subsequently returned to its original state. 174 Thermodynamics and Heat Power entropy: A property of a substance; also a measure of the unavailability that occurs in an irreversible process. heat engine: A continuously operating system across whose boundaries flow only heat and work. isentropic: A process carried out at constant entropy. A reversible adiabatic process is isentropic. isothermal: A process carried out at constant temperature. principle of the increase of entropy: The entropy of an isolated system increases or in the limit remains the same. reversible process: Any process performed so that the system and all its surroundings can be restored to their initial states by performing the process in reverse. second law of thermodynamics: Heat cannot, of itself, pass from a lower temperature to a higher temperature. thermal efficiency: The ratio of the net work of a cycle to the heat added to the cycle.
- Allan D. Kraus, James R. Welty, Abdul Aziz(Authors)
- 2011(Publication Date)
- CRC Press(Publisher)
Determine (a) the specific work output of the turbine, (b) the change in specific entropy of the steam, and (c) the Isentropic Efficiency of the turbine. 7.44: A well-insulated steam turbine operating at steady state develops 8 MW of power for a steam flow of 10 kg/s. The steam enters at 4 MPa and leaves at 8 kPa with a quality of 0.92. Kinetic and potential energy changes are negligible. Determine the inlet temperature of the steam. 7.45: A gas turbine receives hot gases at 600 kPa and 1400 K and exhausts them to the atmosphere at 100 kPa. The Isentropic Efficiency is 0.88. Assume the hot gases to Entropy 217 behave like an ideal gas with c p = 0 . 9861 kJ/kg-K and c v = 0 . 7472 kJ/kg-K and determine the temperature at the turbine exhaust. 7.46: A compressor draws in nitrogen at 100 kPa and 300 K and delivers it at 300 kPa and 500 K. Determine the Isentropic Efficiency assuming that nitrogen is an ideal gas with constant specific heats. 7.47: Air at 100 kPa and 27 ◦ C enters a compressor at the rate of 2 kg/s and leaves at 600 kPa. Assuming air to be an ideal gas with constant specific heats, plot (a) the temperature of the air leaving the compressor and (b) the power input to the compressor as a function of the Isentropic Efficiency of the compressor for values ranging from 0.50 to 1.00 in increments of 0.10. What conclusions can you draw from the plot? 7.48: A compressor for a gas turbine system draws air at 100 kPa and 298 K and delivers it at 700 kPa and 600 K. Determine (a) the polytropic exponent, n , if the compression follows PV n = constant and (b) the efficiency of the compressor. 7.49: A feedwater pump draws saturated liquid condensate at 10 kPa and delivers it to a boiler at 7.5 MPa. The Isentropic Efficiency of the pump is 0.87, and kinetic and potential energy changes are negligible. Determine the outlet enthalpy. 7.50: A compressor draws in refrigerant R-134a at 0.60 bar and − 20 ◦ C and delivers it at 10 bar.
No longer available |Learn more- (Author)
- 2014(Publication Date)
- Academic Studio(Publisher)
During this time it was usually called availability or available work . As a simple example of exergy, air at atmospheric conditions of temperature, pressure, and composition contains energy but no exergy when it is chosen as the thermodynamic reference state known as ambient. Individual processes on Earth like combustion in a power plant often eventually result in products that are incorporated into a large atmosphere, so defining this reference state for exergy is useful even though the atmosphere itself is not at equilibrium and is full of long and short term variations. If standard ambient conditions are used for calculations during plant operation when the actual weather is very cold or hot, then certain parts of a chemical plant might seem to have an exergy efficiency of greater t han 100% and appear on paper to be a perpetual motion machine! Using actual conditions will give actual values, but standard ambient conditions are useful for initial design calculations. ________________________ WORLD TECHNOLOGIES ________________________ One goal of energy and exergy methods in engineering is to compute what comes into and out of several possible designs before a factory is built. Energy input and output will always balance according to the First Law of Thermodynamics or the energy conservation principle. Exergy output will not balance the exergy input for real processes since a part of the exergy input is always destroyed according to the Second Law of Thermodynamics for real processes. After the input and output are completed, the engineer will often want to select the most efficient process. An energy efficiency or first law efficiency will determine the most efficient process based on wasting as little energy as possible relative to energy inputs. An exergy efficiency or second-law efficiency will determine the most efficient process based on wasting and destroying as little available work as possible from a given input of available work.
eBook - ePubEssentials of Energy Technology
Sources, Transport, Storage, Conservation
- Jochen Fricke, Walter L. Borst(Authors)
- 2013(Publication Date)
- Wiley-VCH(Publisher)
Chapter 3
Thermodynamic Energy Efficiency
The conversion of most forms of primary energy to useful energy is still highly inefficient today. Only about one-third of the primary energy is actually put to use (Figure 3.1 ). There are several reasons for this: relatively low energy costs (especially in the United States), marginal incentives for higher efficiency, and fundamental thermodynamic limitations.(Source: Adapted from Ref. [1].)Figure 3.1Typical flow of energy in an industrialized country, here Germany. Most primary energy comes from coal, natural gas, oil, and uranium. The end energies are electricity, gasoline, and diesel fuel, for industry, trade, heating and air conditioning, transportation, and so on.3.1 Carnot's Law
The proper application of thermodynamics is the key for higher energy efficiency. The upper limit for the conversion of heat into useful work is given by Carnot's law:3.1where ηC is the energy efficiency of the Carnot engine (Figure 3.2 ) operating between two reservoirs at absolute temperatures Th (hot) and Tc (cold). First described by Sadi Carnot in 1824, this law shows that the higher the temperature Th is for a fixed Tc , the higher is the efficiency. In the Carnot engine, all processes are assumed to be reversible and infinitely slow so that the system is in equilibrium at all times.Figure 3.2(a) Energy flow chart for the Carnot process. (b) Pressure-Volume (pV) diagram. The shaded area of the closed loop is the extracted work W. (c) Temperature–Entropy (TS) diagram. Vertical transitions represent isentropic (adiabatic) processes.Two adiabatic and two isothermal processes make up the closed cycle, in which heat Q34 at high temperature Th is used to produce work W, with heat Q12 discharged at low temperature Tc . The efficiency ηC in Eq. (3.1 ) is defined as the work W divided by the provided heat Q34 .Applying the first law of thermodynamics (conservation of energy), we know thatand we obtain3.23.3Problem 3.1
Derive Eq. (3.1 ) from the TS diagram in Figure 3.2
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