mech 230 thermodynamics 1 final workbook solutions
TRANSCRIPT
MECH 230 โ Thermodynamics 1
Final Workbook Solutions
CREATED BY JUSTIN BONAL
MECH 230 Final Exam Workbook
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Contents 1.0 General Knowledge ........................................................................................................................... 1
1.1 Unit Analysis .................................................................................................................................. 1
1.2 Pressure ......................................................................................................................................... 1
2.0 Energies ............................................................................................................................................ 2
2.1 Work .............................................................................................................................................. 2
2.2 Internal Energy .............................................................................................................................. 3
2.3 Heat ............................................................................................................................................... 3
2.4 First Law of Thermodynamics ....................................................................................................... 3
3.0 Ideal gas ............................................................................................................................................ 4
3.1 Universal Gas Constant .................................................................................................................. 4
3.2 Polytropic Work Using Ideal Gas .................................................................................................... 4
4.0 Specific Heat ..................................................................................................................................... 5
4.1 Constant Volume Heat Addition .................................................................................................... 5
4.2 Constant Pressure Heat Addition .................................................................................................. 6
4.3 Relating cv and cp for an Ideal Gas................................................................................................... 6
5.0 Control Volume (CV) Analysis ............................................................................................................ 8
5.1 Mass Flow Rate .............................................................................................................................. 8
5.2 Conservation of Energy with CV .............................................................................................. 8
6.0 Applications of CV ....................................................................................................................... 9
6.1 Nozzles and Diffusers .............................................................................................................. 9
6.2 Turbine .................................................................................................................................... 9
6.3 Compressor/Pump ................................................................................................................. 10
6.4 Throttling Device ................................................................................................................... 11
6.5 Heat Exchangers .................................................................................................................... 11
7.0 Second Law of Thermodynamics ..................................................................................................... 12
7.1 Efficiencies ................................................................................................................................... 12
7.2 Reversible and Irreversible Processes........................................................................................... 12
7.3 Entropy ........................................................................................................................................ 13
7.3.1 Entropy Change for Ideal Gas ................................................................................................ 13
7.3.2 Isentropic Processes for Ideal Gas ......................................................................................... 14
8.5 Control Volume Entropy Balance ................................................................................................. 15
8.5.1 Isentropic Efficiencies of Turbines and Compressors ............................................................. 15
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8.5.2 Internally Reversible Steady-State Flow Work .......................................................................... 16
9.0 The Rankine Cycle ........................................................................................................................... 18
9.1 Steps ........................................................................................................................................... 18
9.1 Thermal Efficiency ................................................................................................................. 19
9.2 Back Work Ratio .................................................................................................................... 20
9.3 Ideal Rankine Cycle ................................................................................................................ 20
9.4 Increasing Thermal Efficiency ................................................................................................ 21
10.0 Gas Powered Cycles .................................................................................................................. 22
10.1 General Engine Knowledge ........................................................................................................ 23
10.2 Air Standard Otto Cycle ......................................................................................................... 23
10.2.1 Otto Cycle Thermal Efficiency ............................................................................................. 24
10.2.2 Cold Air Standard Analysis Efficiency for Otto Cycle ........................................................... 25
10.2 Air Standard Diesel Cycle ....................................................................................................... 27
10.2.1 Diesel Cycle Thermal Efficiency ..................................................................................... 28
10.2.2 Cold Air-Standard Analysis for Diesel Cycle ................................................................... 29
11.0 Gas Turbine Power Plants .............................................................................................................. 31
11.1 Air Standard Brayton Cycle ........................................................................................................ 31
11.1.1 Ideal Air Standard Brayton Cycle ......................................................................................... 32
11.1.2 Ideal Cold Air-Standard Brayton Cycle................................................................................. 32
11.1.3 Increasing Cycle Efficiency ................................................................................................... 33
11.2 Aircraft Gas Turbines ................................................................................................................. 35
References ............................................................................................................................................ 39
List of Figures Figure 1: A useful way to conceptualize measuring pressures[1] ............................................................. 1
Figure 2: Example of a manometer [https://www.researchgate.net/figure/Fig3-9-Simple-U-tube-
manometer_fig2_318378486] ................................................................................................................. 2
Figure 3: A schematic of a nozzle and diffuser ......................................................................................... 9
Figure 4: A schematic of a turbine ......................................................................................................... 10
Figure 5: A schematic of a pump ........................................................................................................... 10
Figure 6: A schematic of a throttling device .......................................................................................... 11
Figure 7: A schematic of a tube-in-tube heat exchanger ........................................................................ 11
Figure 8: Ts and hs diagrams of an isentropic expansion versus real expansion ..................................... 15
Figure 9: Ts and hs diagrams of an isentropic compression versus real compression ............................. 16
Figure 10: Schematic of the Basic Rankine Cycle ................................................................................... 19
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Figure 11: Schematic for the ideal Rankine Cycle. Since the cycle is ideal, processes 1->2 and 3->4 are
vertical (no change in entropy) .............................................................................................................. 20
Figure 12: Schematic of the Rankine Cycle with reheat, along with the Ts diagram of the ideal cycle ... 21
Figure 13: Pv and Ts diagrams for the air standard diesel cycle ............................................................. 28
Figure 14: Schematic of the Brayton cycle ............................................................................................ 31
Figure 15: Ts and Pv diagrams for the ideal Brayton cycle ..................................................................... 32
Figure 16: Two stage turbine reheat ...................................................................................................... 33
Figure 17: Ts diagram of the Brayton cycle with reheat ......................................................................... 34
Figure 18: Compression with Intercooling ............................................................................................. 34
Figure 19: Ts and Pv diagram for compression with intercooling ........................................................... 35
Figure 20: Jet propulsion cycle .............................................................................................................. 35
List of Tables Table 1: Useful Derived Units for Thermodynamics................................................................................. 1
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1.0 General Knowledge Thermodynamics! The class everyone has seen memes about being terrible and horrendous. Englinks is
here to ease the process! If you master the basics, and learn from the ground up, this class will NOT be
as bad as everyone says.
1.1 Unit Analysis Units are a core part of Thermodynamics, and an easy place to lose marks. They can be tricky, so make
sure to have the derived units found in table 1 on your formula sheet
Table 1: Useful Derived Units for Thermodynamics
Name Formula Sign Composition
Force m*a Newton (N) kg*m/s2
Pressure F*A Pascal (Pa) kg*m3/s2
Energy F*d Joule (J) kg*m2/s2
Density m/V rho (๐) kg/m3
Specific Volume V/m || 1/๐ Upsilon (๐) m3/kg
Often, you will need to change from a given unit into fundamental unit (ie atm to Pa). Always change
your values into fundamental units before you perform a calculation. Useful conversions can be found
below.
1 ๐๐ก๐ = 101.326 ๐๐ = 760๐๐๐ป๐|| 1 ๐๐๐ = 100,000 ๐๐ = 100๐๐๐ = 1๐๐๐ || 1 ๐3 = 1000๐ฟ
1.2 Pressure Pressure is defined as force over area and has units of Pascals. There are two ways to measure
pressures:
Absolute pressure (Pabs): Measured relative to 0kPa (a perfect vacuum)
Gauge pressure (Pg): Measured relative to the atmospheric pressure
Atmospheric pressure (Patm ) is the local pressure of the measurement. Figure 1 and equation [1] are
useful to conceptualize these definitions and should be on your formula sheet.
๐ท๐ = ๐ท๐๐๐ โ ๐ท๐๐๐
[1]
Figure 1: A useful way to conceptualize measuring pressures[1]
There are various ways to measure pressures, one of which is a manometer. An example of a
manometer can be found in Figure 2. In this case, the pressure of the gas is calculated using equation
[2].
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๐๐ = ๐๐๐๐ฌ โ ๐๐๐ญ๐ฆ = ๐๐ฅ๐ข๐ช๐ฎ๐ข๐ โ ๐ โ ๐
[2]
Figure 2: Example of a manometer [https://www.researchgate.net/figure/Fig3-9-Simple-U-tube-manometer_fig2_318378486]
2.0 Energies Energy has units of Joules [J]. It is a property that tells you how the system reacts in process. It is
essential to know where the energy in a system is going during a process or cycle to understand how
the system can be used practically.
There are three types of energy that you are going to be working with in Thermodynamics: Work,
Heat, and Internal.
2.1 Work Although energy is a property of a system, work is not a property of the system. This is because work
depends on the path that it takes from one point in a process to another.
In general, the work covered in this course will be due to a pressure force. An example of this is a piston-
cylinder assembly. The pressure force will cause a change in volume so,
๐พ๐โ๐ = โซ ๐ท๐๐ ๐ฝ
๐
๐
[3]
Where Pg is the pressure of a gas and dV is a differential change in volume. From equation [3] it is easy
to see that a constant volume process will have no work.
โ๐ฝ = ๐ โด โ๐พ = ๐
[4]
When you are told that a system undergoes a polytropic process, that means it behaves according to
equation [5].
๐ท๐ฝ๐ = ๐
[5]
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Where P is pressure, V is volume, c is a constant, and n is the polytropic index. From this, it is easy to
see the following relation: ๐1๐1๐ = ๐2๐2
๐, since c is a constant.
From equation [5], three different ways to calculate work can be derived. Equation [6] shows the work
for when ๐ โ 1, equation [7] for when n = 1, and equation [8] for when n = 0. When n = 0, this is called a
constant pressure process.
๐พ๐โ๐ =
๐ท๐๐ฝ๐ โ ๐ท๐๐ฝ๐
๐ โ ๐ ๐๐๐ ๐ โ ๐
[6]
๐พ๐โ๐ = ๐ท๐๐ฝ๐ ๐ฅ๐ง (
๐ฝ๐
๐ฝ๐) ๐๐๐ ๐ = ๐
[7]
๐พ๐โ๐ = ๐ทโ๐ฝ ๐๐๐ ๐ = ๐ [8] Something essential to note about calculating the work in a process is the sign convention. Work done
by the system is positive, and work done on the system is negative.
2.2 Internal Energy Internal energy can be thought of the energy that is contained within the fluid in a system, and is
denoted by U. You can think back to the kinetic molecular theory learned in first year Chemistry.
If a process has no change in temperature, โ๐ = 0, it is called an isothermal process. For an ideal gas,
internal energy is a function of temperature. So, if there is no change in temperature for an ideal gas,
there is no change in internal energy.
โ๐ป = ๐ โด โ๐ผ = ๐ โผ (๐ฐ๐ซ๐ฌ๐จ๐ณ ๐ฎ๐จ๐บ)โผ
[9]
2.3 Heat Energy transferred by a difference in temperature between the system and its surroundings is called
heat, and is denoted by Q. Contrary to work, heat transferred to the system is positive and heat
transferred from the system is negative.
Equation [10] shows the general formula for heat transfer.
๐1โ2 = โซ ๐ฟ๐
2
1
[10]
An adiabatic process is a process where there is no heat exchange from the system to the
surroundings. Therefore โ๐ธ = ๐.
2.4 First Law of Thermodynamics The first law is something that will follow you everywhere in this course and is essential to almost every
question you will do. Equation [11] states the first law.
๐ธ2 โ ๐ธ1 = ๐ โ ๐
[11]
Where E2 is the energy at the end of a process and E1 is the energy at the beginning. The change in
energy of a system can also be dictated by equation [12].
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๐ธ2 โ ๐ธ1 = โ๐พ๐ธ + โ๐ + โ๐๐ธ
[12]
Where KE is kinetic energy and PE is potential energy. Unless otherwise stated in a problem, the
changes in potential and kinetic energy can be neglected. So, combining equations [11] and [12] yields
equation [13].
โ๐ผ๐โ๐ = ๐ธ๐โ๐ โ ๐พ๐โ๐
[13]
3.0 Ideal gas You most likely have some experience working with the ideal gas law from previous courses. In
thermodynamics, you are going to take the same concepts and apply them on a broader scale.
3.1 Universal Gas Constant
The universal gas constant, ๏ฟฝ๏ฟฝ = 8314๐ฝ
๐๐๐๐โ๐พ= 8.314
๐๐ฝ
๐๐๐๐ ๐พ, is something that is consistent for all
gases. However, in thermodynamics it is often more convenient to use the gas constant, R, specific to a
substance. To find R, apply the following
๐น =
๏ฟฝ๏ฟฝ
๐
[14]
After dividing by the molar mass, the gas constant has units of ๐ฝ
๐๐โ๐พ. So, your calculations will be in
terms of mass, rather than mols.
The general ideal gas law is given as equation [15], where P is pressure, V is volume, n is mols, and T is
temperature. Equation [16] shows the ideal gas law in terms of the gas constant, R, where M is mass.
๐๐ = ๐๏ฟฝ๏ฟฝ๐
[15]
๐ท๐ฝ = ๐ด๐น๐ป
[16]
Equation [16] can be rearranged to be calculated on a per mass basis by dividing through my M or V.
Equation [17] shows the equation in terms of specific volume (divide by M), and equation [18] in terms
of density (divide by V).
๐๐ = ๐ ๐
[17]
๐ = ๐๐ ๐
[18]
3.2 Polytropic Work Using Ideal Gas After being told that a gas is ideal, multiple equations become โunlockedโ to solve a process.
Remembering back to a polytropic process, the work of the process for n=/ 1 is given by equation [6].
However, was have just learned that PV=MRT from equation [16]. So, we can make a substitution to
yield equation [18].
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๐พ๐โ๐ =๐ด๐น(๐ป๐ โ ๐ป๐)
๐ โ ๐ ๐๐๐ ๐ โ ๐
[19]
For an ideal gas going though a polytropic process with n=1 means it is an isothermal process (โ๐ป =
๐). Therefore equation [7] can be modified to the following.
๐พ๐โ๐ = ๐ด๐น๐ป ๐ฅ๐ง (
๐ฝ๐
๐ฝ๐) ๐๐๐ ๐ = ๐
[20]
Equation [17] can be rearranged as equation [21]. We can use equations [3] (polytropic) and [21] to
make relationships from one point to another.
๐ =๐๐
๐
[21]
These relationships simplify to equations [22] and [23], where n is the polytropic index. These formulas
are essential for any ideal polytropic process. DO NOT USE UNLESS YOU ARE TOLD THAT THE
SUBSTANCE IS AN IDEAL GAS.
๐ป๐
๐ป๐= (
๐ท๐
๐ท๐)
๐โ๐๐
[22]
๐ป๐
๐ป๐= (
๐๐
๐๐)
๐โ๐
[23]
4.0 Specific Heat Specific heat, c, is a property that represents the amount of heat required to raise the temperature of 1
kg of a substance by 1-degree Kelvin, as seen in equation [24]. The units of specific heat are ๐๐ฝ
๐๐ยฐ๐พ.
๐ =โ๐
๐ โ โ๐
[24]
4.1 Constant Volume Heat Addition For a constant volume, there is no work being done. Therefore, first law gets reduced to
โ๐ = โ๐
Rearranging equation [23], dividing through by M, and plugging it into the above yields
โ๐ = ๐๐โ๐ป
[25]
Noting that ๐๐ฃ depends on both P and T, equation [25] gets generalized to
๐ข2 โ ๐ข1 = โซ ๐๐ข
2
1
= โซ ๐๐ฃ(๐, ๐)๐๐2
1
[26]
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4.2 Constant Pressure Heat Addition For a constant pressure heat addition, enthalpy is calculated using
๐ = ๐๐โ๐ป
[27]
Like ๐๐ฃ, ๐๐ is dependent on P and T. So, equation [27] generalizes to
โ2 โ โ1 = โซ ๐โ
2
1
= โซ ๐๐(๐, ๐)๐๐2
1
[28]
4.3 Relating cv and cp for an Ideal Gas A commonly used term is the specific heat ratio, k. For an ideal gas
๐(๐ป) =
๐๐
๐๐
[29]
The following equations are useful for solving specific heats, given one another
๐๐(๐ป) =
๐น
๐ฒ(๐ป) โ ๐
[30]
๐๐(๐ป) =๐ฒ(๐ป)๐น
๐ฒ(๐ป) โ ๐
[31]
2013 FINAL QUESTION 1
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5.0 Control Volume (CV) Analysis Control volume analysis is used when mass is flowing in and out of a system. The way this is measured
is by seeing how much mass flows through the control volume of the system being analyzed.
5.1 Mass Flow Rate The rate at which mass flows through a control volume is called the mass flow rate, denoted by ๏ฟฝ๏ฟฝ, and
has unites of kg/s. The change in mass of the control volume is given by equation [31]
๐๐๐ถ๐
๐๐ก = โ๏ฟฝ๏ฟฝ๐ โ โ๏ฟฝ๏ฟฝ๐ [32]
Where the subscript i denotes inlet, and e denotes exits For a steady state system, there is no change in
mass of the control volume. Therefore, equation [31] reduces to
โ๏ฟฝ๏ฟฝ๐ = โ๏ฟฝ๏ฟฝ๐ [33] There are a few different equations that come from mass flow rate. Mass flow rate can also be denoted
by
๏ฟฝ๏ฟฝ = ๐๐จ๐ฝ [34] Where ๐ is density, A is the cross-sectional area, and V is the velocity perpendicular to the control
surface. The mass flux is defined as the mass flow per unit area [๐๐
๐2๐ ]
๏ฟฝ๏ฟฝ
๐จ= ๐๐ฝ =
๐ฝ
๐ [35]
The volumetric flow rate is defined as [๐3
๐ ]
๐
๐= ๐ฝ๐จ
[36]
5.2 Conservation of Energy with CV When we look at heat and work in a CV analysis, the typical Q and W have a dot above them, ๏ฟฝ๏ฟฝand ๏ฟฝ๏ฟฝ.
The dot signifies the rate at which either heat or work is being transferred to or from the system, and
has units of kJ/s. Like with a closed system, the capitals can be exchanged for lower case letters to
represent energy transfer on a per mass basis, ๏ฟฝ๏ฟฝ and ๏ฟฝ๏ฟฝ. This is done by dividing ๏ฟฝ๏ฟฝand ๏ฟฝ๏ฟฝ by the mass
flow rate, ๏ฟฝ๏ฟฝ, and has units of J/kg. This means that heat and work are being added into the CV on a per
mass basis.
The general formula for conservation of energy with CV is
๐๐ธ๐๐ฃ
๐๐ก= ๏ฟฝ๏ฟฝ โ ๏ฟฝ๏ฟฝ๐ โ๐๐๐ก + โ๏ฟฝ๏ฟฝ๐ (โ๐ +
๐๐2
2+ ๐๐๐) โ โ๏ฟฝ๏ฟฝ๐ (โ๐ +
๐๐2
2+ ๐๐๐) [37]
For this course, we assume that ๐๐ธ๐๐ฃ
๐๐ก= 0, that there is only one inlet and one outlet, and that the
systems has steady flow, ๏ฟฝ๏ฟฝ๐ = ๏ฟฝ๏ฟฝ๐ = ๏ฟฝ๏ฟฝ. Therefore, dividing through by ๏ฟฝ๏ฟฝ reduces the equation to
๐ = ๏ฟฝ๏ฟฝ โ ๏ฟฝ๏ฟฝ + (๐๐ +๐ฝ๐
๐
๐+ ๐๐๐) โ (๐๐ +
๐ฝ๐๐
๐+ ๐๐๐) [38]
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Equation [37] is the formula that you will be using when solving CV problems in MECH 230. Often
changes in KE and PE are neglected.
6.0 Applications of CV 6.1 Nozzles and Diffusers Nozzles and diffusers are devices that either increase or decrease the flow velocity by changing the
cross-sectional area through which the fluid is flowing. Combining equations [32] and [34] and
assuming that ๐1 = ๐2 for subsonic flow yields the following
๐จ๐๐ฝ๐ = ๐จ๐๐ฝ๐ [39]
Figure 3: A schematic of a nozzle and diffuser
A diffuser decreases the flow rate of the fluid, therefore ๐ด2 > ๐ด1 โ ๐2 < ๐1.
A nozzle increases the flow rate of the fluid, therefore ๐ด1 > ๐ด2 โ ๐1 < ๐2.
Applying the CV energy equation, assuming no heat transfer, work, or change in height, yields the
following
๐ฝ๐๐ = ๐ฝ๐
๐ + ๐(๐๐ โ ๐๐) [40]
For a rocket nozzle, ๐2 โซ ๐1, and the above equation simplifies to
๐22 = 2(โ1 โ โ2) [41]
6.2 Turbine A turbine is a mechanical device through which shaft work is developed.
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Figure 4: A schematic of a turbine
Applying first law for CV, assuming steady state, no change in PE, and no heat transfer, yields
๏ฟฝ๏ฟฝ = (โ1 โ โ2) + (๐1
2
2โ
๐22
2) [42]
Assuming โ1 โ โ2 โซ๐1
2
2โ
๐22
2 simplifies the above to the following
๏ฟฝ๏ฟฝ = (๐๐ โ ๐๐) [43]
Note that since a turbine develops work, โ1 > โ2. The power is the work output per unit time
๏ฟฝ๏ฟฝ = ๏ฟฝ๏ฟฝ โ ๏ฟฝ๏ฟฝ = (โ1 โ โ2) [44]
6.3 Compressor/Pump A pump is a mechanical device that uses shaft work input to raise the pressure of the flowing fluid.
Figure 5: A schematic of a pump
Like before, we apply first law with no change in PE, heat transfer, and steady state, which yields the
following
๏ฟฝ๏ฟฝ = (โ1 โ โ2) + (๐1
2
2โ
๐22
2) [45]
Assuming the change in KE is small compared to the change in h
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๏ฟฝ๏ฟฝ = (๐๐ โ ๐๐) [46]
Rather than work output like a turbine, work is inputted into a pump. Therefore โ2 > โ1.
6.4 Throttling Device A throttling device is a device that creates a pressure drop by restricting the flow.
Figure 6: A schematic of a throttling device
Applying first law, no heat transfer, steady state, and no change in PE yields
(โ1 โ โ2) = (๐1
2
2โ
๐22
2) [47]
Looking downstream from each side of the device, we can assume that ๐1 = ๐2. This assumption yields
๐๐ = ๐๐ [48]
6.5 Heat Exchangers A heat exchanger is a device that transfers energy between fluids at different temperatures to heat or
cool a fluid.
Figure 7: A schematic of a tube-in-tube heat exchanger
Applying first law to the above system with no heat loss, no steady flow, and no change in KE and PE
yields
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0 = ๏ฟฝ๏ฟฝ1โ1+๏ฟฝ๏ฟฝ3โ3 โ ๏ฟฝ๏ฟฝ2โ2 โ ๏ฟฝ๏ฟฝ4โ4 [49]
Steady flow โด ๏ฟฝ๏ฟฝ1 = ๏ฟฝ๏ฟฝ2 ๐๐๐ ๏ฟฝ๏ฟฝ3 = ๏ฟฝ๏ฟฝ4
0 = ๏ฟฝ๏ฟฝ1(โ1 โ โ2) + ๏ฟฝ๏ฟฝ3(โ3 โ โ4) [50]
Which further simplifies to
๏ฟฝ๏ฟฝ๐
๏ฟฝ๏ฟฝ๐=
๐๐ โ ๐๐
๐๐ โ ๐๐ [51]
7.0 Second Law of Thermodynamics The second law of thermodynamics can be explained by the Clausius Statement. Clausius states that it
is impossible for a system to operate in such a way that the sole result is the transfer of heat from a cold
to a hot body.
The Second law encompasses two things โ the direction in which a spontaneous process will go, and
the maximum work that a process can develop. From having the maximum work of a process, the
efficiency of the process can be determined. These two properties can be described by a new
thermodynamic property โ entropy (S).
7.1 Efficiencies The thermal efficiency of a cycle is given by the work performed during the cycle, divided by the
maximum work of the cycle
๐๐๐ฆ๐๐๐ =๐ค๐๐๐ ๐๐๐๐
๐๐๐ฅ๐๐๐ข๐ ๐ค๐๐๐ =
๐
๐๐๐ด๐ [52]
Applying first law to a cycle, we can see the following
0 = ๐๐๐ โ ๐๐๐ข๐ก โ ๐
Where Qin is the heat transferred into the system to create work, and Qout is the heat loss of the cycle.
This can be due to many factors, including friction. If we have an ideal system, the Qout=0, and no
energy would be lost to heat. In this case, Wmax =Qin, simplifying the efficiency equation to the following
๐๐๐ฆ๐๐๐ =
๐
๐๐๐
[53]
Or further rearranged to the following
๐๐๐ฆ๐๐๐ =๐๐๐ โ ๐๐๐ข๐ก
๐๐๐= 1 โ
๐๐๐ข๐ก
๐๐๐ [54]
For internal combustion (IC) engines, there could be other factors that lower their efficiencies, such as
mechanical inefficiencies. So, the overall efficiency of an engine is
๐๐๐ฃ๐๐๐๐๐ = ๐๐๐ฆ๐๐๐๐๐๐๐โ๐๐๐๐๐๐ [55]
7.2 Reversible and Irreversible Processes It is impossible to attain an engine with 100% efficiency. However, the maximum efficiency of a cycle
will be the cycle with a series of ideal reversible processes. A reversible process is a process in which
MECH 230 Final Exam Workbook
13
both the system and its surroundings can be returned to their original state after the process has been
completed. Quasi-equilibrium (or slow) expansion or compression processes are reversible.
An irreversible process is a process in which the system and surroundings cannot be returned to their
original state. This process includes irreversibilityโs that make it impossible to return to its original
state. Rapid compression and expansion processes are irreversible.
An internally reversible process is a process in which the system can be returned to its original state,
but irreversibilityโs can occur in its surroundings.
7.3 Entropy The definition of entropy, S, is the amount of disorder in a system. It has units of J/K, and is given by the
following
The most general equation for an entropy balance for a close system is given by the following
๐2 โ ๐1 = โซ๐ฟ๐
๐+ ๐๐๐๐ [56]
Where S2 -S1 is the change in entropy of the system, โซ๐ฟ๐
๐is the entropy transfer to the system by heat,
and Sgen is the entropy generated in the system due to irreversibilityโs. The following table shows the
possible values of Sgen given a process:
This is a good way to check your work, as if you have a Sgen value that is less than 0, there has been a
calculation error somewhere.
A process that is both adiabatic and reversible is called an isentropic process. In this case, โ๐ = 0
(adiabatic) and Sgen=0 (reversible), so entropy is constant.
โด ๐1 โ ๐2 = 0, ๐1 = ๐2.
7.3.1 Entropy Change for Ideal Gas We know that for an ideal gas ๐๐ข = ๐๐ฃ(๐)๐๐, ๐โ = ๐๐(๐)๐๐, ๐๐ = ๐ ๐.
Plugging these equations into the above yields the following
๐ 2(๐2, ๐2) โ ๐ 1(๐1, ๐1) = โซ๐๐(๐)๐๐
๐
2
1
+ ๐ ๐๐(๐2
๐1) [57]
And
๐ 2(๐2, ๐2) โ ๐ 1(๐1, ๐1) = โซ๐๐(๐)๐๐
๐
2
1
โ ๐ ๐๐(๐2
๐1) [58]
The bounds of โซ๐๐(๐)๐๐
๐
2
1 are tabulated as a function of T in air in Table A-22 under the label ๐ ยฐ. So,
โซ๐๐(๐)๐๐
๐
2
1= ๐ ยฐ2 โ ๐ ยฐ1. Therefore equation [57] can be written as
MECH 230 Final Exam Workbook
14
๐๐ โ ๐๐ = (๐ยฐ๐ โ ๐ยฐ๐) โ ๐น๐๐(๐ท๐
๐ท๐) [59]
If cv and cp are taken as constant, changes in entropy are calculated by the following
๐๐ โ ๐๐ = ๐๐ ๐ฅ๐ง (๐ป๐
๐ป๐) + ๐น๐๐ (
๐๐
๐๐) [60]
And
๐๐ โ ๐๐ = ๐๐ ๐ฅ๐ง (๐ป๐
๐ป๐) โ ๐น๐๐ (
๐ท๐
๐ท๐) [61]
7.3.2 Isentropic Processes for Ideal Gas A process that is both adiabatic and reversible is called an isentropic process. In this case, โ๐ = 0
(adiabatic) and Sgen=0 (reversible), so entropy is constant.
โด ๐1 โ ๐2 = 0, ๐1 = ๐2.
Plugging S1=S2 into equation [27]
๐ ยฐ2 โ ๐ ยฐ1
๐ = ln (
๐2
๐1)
๐2
๐1=
exp (๐ ยฐ2๐ )
exp (๐ ยฐ1๐ )
Now we define relative pressure, ๐ท๐ = ๐๐๐(๐ยฐ
๐น). Therefore
(๐ท๐
๐ท๐) =
๐ท๐๐
๐ท๐๐ (๐ = ๐๐๐๐๐)
[62]
From the ideal gas law,
๐2
๐1=
๐2
๐1(
๐1
๐2) =
(๐2๐๐2
)
(๐1Pr1
)
Define vr =T/Pr, therefore
(๐๐
๐๐) =
๐๐๐
๐๐๐ [63]
The values of vr and Pr can be found as a function of T for air in table A-22.
7.3.2.1 Isentropic for Ideal Gas with constant cv and cp
We know that cv=R/(k-1). Plugging into equation [41], with s2=s1 (isentropic), and simplifying yields
(๐ป๐
๐ป๐) = (
๐๐
๐๐)
๐โ๐
, ๐ = ๐๐๐๐๐, ๐๐ = ๐๐๐๐๐ [64]
Substituting in ๐2
๐1=
๐2๐2
๐1๐1,
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(๐ท๐
๐ท๐) = (
๐๐
๐๐)
๐
, ๐ = ๐๐๐๐๐, ๐๐ = ๐๐๐๐๐ [65]
For a polytropic process, ๐๐๐ = ๐๐๐๐ ๐ก. For an isentropic process, n=k. Combining equations yields
(๐1
๐2) = (
๐2
๐1)
1๐
= (๐2
๐1)
1๐โ1
(๐ป๐
๐ป๐) = (
๐ท๐
๐ท๐)
๐โ๐
๐, ๐ = ๐๐๐๐๐, ๐๐ = ๐๐๐๐๐ [66]
8.5 Control Volume Entropy Balance The rate of entropy change in a control volume is given by
๐๐๐ถ๐
๐๐ก =
๏ฟฝ๏ฟฝ๐
๐๐โ๏ฟฝ๏ฟฝ๐๐ ๐ โ โ๏ฟฝ๏ฟฝ๐๐ ๐ [67]
Assuming steady state, one inlet one outlet, and isothermal CV, the above simplifies to
๐ =๐
๏ฟฝ๏ฟฝ(
๏ฟฝ๏ฟฝ๐๐
๐ป) + ๐๐๐ โ ๐๐๐๐ +
๏ฟฝ๏ฟฝ๐๐๐
๏ฟฝ๏ฟฝ
[68]
8.5.1 Isentropic Efficiencies of Turbines and Compressors
Recall for a turbine ๏ฟฝ๏ฟฝ๐๐ฃ
๏ฟฝ๏ฟฝ= โ1 โ โ2 > 0, since turbines create work. For a real turbine it is impossible to
get rid of all irreversibilityโs, so s2>s1. In Figure 8, 2s is only achievable with no irreversibilityโs (Sgen=0)
and s2 =s1. The maximum theoretical work from a turbine is through an isentropic expansion.
Figure 8: Ts and hs diagrams of an isentropic expansion versus real expansion
We know that the efficiency of a turbine is given as ๐ =๐๐๐๐๐
๐๐๐๐ฅ. Now we know that the maximum work of
a turbine is given as ๏ฟฝ๏ฟฝ๐๐ฃ
๏ฟฝ๏ฟฝ= โ1 โ โ2๐ .
Plugging in, we get the isentropic turbine efficiency
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16
๐ผ๐ =๐๐ โ ๐๐
๐๐ โ ๐๐๐ [69]
For a compressor, the same thought process follows. Figure 9 shows the isentropic and real paths taken
on a T-s and h-s diagram for a compressor. The minimum theoretical compressor work required is the
isentropic compressor work.
Figure 9: Ts and hs diagrams of an isentropic compression versus real compression
We know that the compressor efficiency is given by ๐๐ =๐๐๐๐ฅ
๐ . We know that ๐๐๐๐ฅ = โ1 โ โ2๐ . From
this, the isentropic compressor efficiency is given by
๐ผ๐ =๐๐ โ ๐๐๐
๐๐ โ ๐๐ [70]
8.5.2 Internally Reversible Steady-State Flow Work Assuming a single inlet and exit CV, steady state, and no change in KE and PE. For pumps, turbines, and
compressors, work done is
๏ฟฝ๏ฟฝ๐๐ฃ
๏ฟฝ๏ฟฝ= โ โซ ๐๐๐
2
1
[71]
Liquids are incompressible, so there is no change in the specific volume ๐1 = ๐2 = ๐. โด
๏ฟฝ๏ฟฝ๐๐
๏ฟฝ๏ฟฝ= โ๐(๐ท๐ โ ๐ท๐) [72]
For a gas going through a polytropic process, with ๐๐ = ๐ ๐
๏ฟฝ๏ฟฝ๐๐
๏ฟฝ๏ฟฝ= โ
๐๐น๐ป๐
๐ โ ๐(
๐ป๐
๐ป๐โ ๐) , ๐ โ ๐ [73]
OR substituting (๐2
๐1) = (
๐2
๐1)
๐โ1
๐ we get
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17
๏ฟฝ๏ฟฝ๐๐ฃ
๏ฟฝ๏ฟฝ= โ
๐๐ ๐1
๐ โ 1((
๐2
๐1)
๐โ1๐
โ 1) , ๐ โ 1 [74]
If a process is isothermal (T1 = T2)
๏ฟฝ๏ฟฝ๐๐
๏ฟฝ๏ฟฝ= โ๐น๐ป๐๐(
๐ท๐
๐ท๐) [75]
2015 FINAL QUESTION 2
MECH 230 Final Exam Workbook
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9.0 The Rankine Cycle The Rankine cycle is a series of processes used to model the performance of steam turbine systems. It is
an idealized cycle of a heat engine that converts heat into mechanical work.
9.1 Steps Figure 10 shows the cycle that describes the basic Rankine cycle.
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19
Figure 10: Schematic of the Basic Rankine Cycle
Each process is analyzed using conservation of energy. Assuming steady state, no change in KE and PE
gives
0 =๏ฟฝ๏ฟฝ๐๐ฃ
๏ฟฝ๏ฟฝโ
๏ฟฝ๏ฟฝ๐๐ฃ
๏ฟฝ๏ฟฝ+ (โ๐๐ โ โ๐๐ข๐ก)
Now we analyze each step individually.
1 โ 2 ๐ด๐๐๐๐๐๐ก๐๐ ๐๐ข๐๐๐๐๐ ๐ธ๐ฅ๐๐๐๐ ๐๐๐
Adiabatic so ๏ฟฝ๏ฟฝ = 0.
โด ๐ค๐๐ข๐ก = โ1 โ โ2
2 โ 3 ๐ถ๐๐๐๐๐๐ ๐๐ (๐๐ ๐ค๐๐๐)
No work so ๏ฟฝ๏ฟฝ = 0.
โด ๐๐๐ข๐ก = โ2 โ โ3
3 โ 4 ๐ด๐๐๐๐๐๐ก๐๐ ๐๐ข๐๐ ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐
Adiabatic so ๏ฟฝ๏ฟฝ = 0
โด ๐ค๐๐ = โ4 โ โ3
4 โ 1 ๐๐ก๐๐๐ ๐บ๐๐๐๐๐๐ก๐๐ (๐๐ ๐ค๐๐๐)
No work so ๏ฟฝ๏ฟฝ = 0
โด ๐๐๐ = โ1 โ โ4
9.1 Thermal Efficiency The thermal efficiency of a heat engine is defined by
๐ =๐๐๐ก ๐ค๐๐๐ ๐๐ข๐ก
โ๐๐๐ก ๐๐๐๐ข๐ก
Net work out is the work output minus the work input. This reduces the above to the following
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20
๐๐๐๐๐๐๐๐ =๐ค๐๐ข๐ก โ ๐ค๐๐
๐๐๐=
(โ1 โ โ2) โ (โ4 โ โ3)
โ1 โ โ4 [76]
9.2 Back Work Ratio The Back-Work Ratio (bwr) is a measure of performance sometimes used in power plants. It is the
fraction of work that the turbine produces that is consumed by the pump. So, the formula for bwr is
given by the following
๐๐๐๐๐๐๐๐๐๐ =๐๐๐(๐๐๐๐)
๐๐๐๐(๐๐๐๐๐๐๐)=
๐๐ โ ๐๐
๐๐ โ ๐๐ [77]
9.3 Ideal Rankine Cycle In the ideal Rankine Cycle, no irreversibilityโs occur. The resulting cycle is the following
1 โ 2 ๐ผ๐ ๐๐๐ก๐๐๐๐๐ ๐ธ๐ฅ๐๐๐๐ ๐๐๐ ๐๐ก ๐๐ข๐๐๐๐๐
2 โ 3 ๐ถ๐๐๐๐๐๐ ๐๐ก๐๐๐ ๐๐ก ๐ถ๐๐๐ ๐ก๐๐๐ก ๐๐๐๐ ๐ ๐ข๐๐
3 โ 4 ๐ผ๐ ๐๐๐ก๐๐๐๐๐ ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐ ๐๐ก ๐๐ข๐๐
4 โ 1 ๐๐ก๐๐๐ ๐บ๐๐๐๐๐๐ก๐๐๐ ๐๐ก ๐ถ๐๐๐ ๐ก๐๐๐ก ๐๐๐๐ ๐ ๐ข๐๐
The T-s diagram for a Rankine Cycle can be found in Figure 11. Something to note is at position 3, the
process ends when the working fluid is in a fully saturated liquid state.
Figure 11: Schematic for the ideal Rankine Cycle. Since the cycle is ideal, processes 1->2 and 3->4 are vertical (no change in entropy)
For an internally reversable pump, the pump work can be evaluated by
๐ค๐๐ = โ โซ ๐๐๐4
3
If the working fluid is a pure liquid then the specific volume can be assumed to be constant, yielding the
following
๐ค๐๐ = ๐3(๐4 โ ๐3)
Like the thermal efficiency of a Carnot Cycle, the ideal Rankine Efficiency is given by
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21
๐ผ๐๐ ๐๐๐ ๐๐๐๐๐๐๐ = ๐ โ๐ป๐๐๐
๐ป๐๐ [78]
Where Tin is the mean temperature for the process 4->1 and Tout the temperature from 2->3.
The bwr for an ideal Rankine cycle is given by
๐๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐ =๐๐๐
๐๐๐๐=
๐๐(๐ท๐ โ ๐ท๐)
(๐๐ โ ๐๐๐) [79]
9.4 Increasing Thermal Efficiency As can be seen in equation [77], the most efficient way to increase the efficiency of the cycle is by
increasing Tin. The most widely used method is to use a two-stage turbine, with a reheat in the middle.
Figure 12 shows both the schematic for a Rankine Cycle with reheat, and the corresponding T-s
diagram.
Figure 12: Schematic of the Rankine Cycle with reheat, along with the Ts diagram of the ideal cycle
In this case, the efficiency of the cycle would be given by
๐ผ๐๐๐๐๐๐๐ ๐๐๐๐๐๐ =
(๐๐โ๐ + ๐๐โ๐) โ ๐๐โ๐
(๐๐โ๐ + ๐๐โ๐)
=(๐๐ โ ๐๐) + (๐๐ โ ๐๐) โ (๐๐ โ ๐๐)
(๐๐ โ ๐๐) + (๐๐ โ ๐๐)
[80]
2015 FINAL QUESTION 3
MECH 230 Final Exam Workbook
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10.0 Gas Powered Cycles A gas-powered cycle is one that produces power while the working fluid is always gas and has no
change in phase. There are two types of Internal Combustion (IC) engines that we will be working with.
The first is spark ignition (Otto), and the second is compression (Diesel).
MECH 230 Final Exam Workbook
23
For the purpose of this course, we will be using Air-Standard Analysis in order to greatly simplify the
problems. The simplifications are
1) Fixed amount of ideal gas for the working fluid
2) Combustion is replaced by constant volume heat addition
3) Intake and exhaust not considered. Cycle completed with constant volume heat removal
4) All processes are internally reversible
10.1 General Engine Knowledge The engine power, ๏ฟฝ๏ฟฝ, (or horsepower) is the amount of power outputted by an engine. It is defined by
๏ฟฝ๏ฟฝ = ๐พ๐๐๐๐๐ โ๐ต
๐ [81]
Where N is the crank shaft speed, in revolutions per second. A useful conversion to have on your
formula sheet is
๐ ๐๐๐๐๐๐๐๐๐๐ = ๐๐๐ ๐พ๐๐๐๐
Another useful parameter used in thermodynamic analysis is mean effective pressure, mep. This is a
theoretical constant pressure such that if acted onto the piston during a power stroke, it would develop
the same amount of net work for one cycle. The formula for mep is
๐๐๐ =๐๐๐ ๐๐๐๐ ๐๐๐ ๐๐๐๐๐
๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐=
๐พ๐๐๐๐๐
๐ฝ๐ โ ๐ฝ๐ [82]
10.2 Air Standard Otto Cycle A term used for Otto cycles is the compression ratio. This is a ratio of the volume of the cylinder at
different parts of the process, and is given by
๐ =๐1
๐2=
๐4
๐3 [83]
Or, since one of the assumptions is fixed mass
๐ =๐๐
๐๐=
๐๐
๐๐ [84]
The air standard Otto Cycle goes through the following processes, assuming no change in KE and PE. P-
v and T-s diagrams for the cycle can be found in Figure 4.
1 โ 2 ๐ผ๐ ๐๐๐ก๐๐๐๐๐ ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐
Isentropic, Q=0. Compression, so work is negative. Applying first law,
๐ค๐๐ = ๐ข2 โ ๐ข1
Isentropic, so we can use relative specific volumes found in table A-22 and the compression ratio
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๐ฃ๐2
๐๐1=
๐2
๐1=
1
๐
And since the working fluid is an ideal gas
๐2๐2
๐2=
๐1๐1
๐1โ
๐2
๐1=
๐2
๐1โ
๐1
๐2=
๐2
๐1โ ๐
2 โ 3 ๐ถ๐๐๐ ๐ก๐๐๐ก ๐ฃ๐๐๐ข๐๐ โ๐๐๐ก ๐๐๐๐๐ก๐๐๐
Constant volume so W = 0. Applying first law,
๐๐๐ = ๐ข3 โ ๐ข2
Ideal gas with constant volume, so
๐3
๐2=
๐3
๐2
3 โ 4 ๐ผ๐ ๐๐๐ก๐๐๐๐๐ ๐ธ๐ฅ๐๐๐๐ ๐๐๐
Isentropic, Q = 0. Expansion, work is positive. Applying first law,
๐ค๐๐ข๐ก = ๐ข3 โ ๐ข4
Reversible process with compression ratio
๐๐4
๐๐3=
๐4
๐3= ๐
Ideal gas, so
๐4
๐3=
๐4
๐3โ
๐3
๐4=
๐4
๐3โ
1
๐
4 โ 1 ๐ถ๐๐๐ ๐ก๐๐๐ก ๐ฃ๐๐๐ข๐๐ โ๐๐๐ก ๐๐๐๐๐ฃ๐๐
Constant volume, W = 0. Heat removal, Q is negative. Applying first law,
๐๐๐ข๐ก = ๐ข4 โ ๐ข1
Ideal gas with constant volume, so
๐4
๐4=
๐1
๐1
10.2.1 Otto Cycle Thermal Efficiency Like other heat engines, the efficiency of the cycle is given by
๐๐๐ฆ๐๐๐ =๐ค๐๐ข๐ก โ ๐ค๐๐
๐๐๐
So, for an Otto Cycle the efficiency is
๐ผ๐๐๐๐ =(๐๐ โ ๐๐) โ (๐๐ โ ๐๐)
(๐๐ โ ๐๐)= ๐ โ
๐๐ โ ๐๐
๐๐ โ ๐๐ [85]
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10.2.2 Cold Air Standard Analysis Efficiency for Otto Cycle For cold air standard analysis, specific heats are assumed to be constant. This unlocks multiple
equations for use.
For process 1->2 we know
๐2
๐1= (
๐1
๐2)
๐โ1 = ๐๐โ1 and
๐2
๐1= (
๐2
๐1)
๐โ1
๐
For process 3->4 we know
๐4
๐3= (
๐3
๐4)
๐โ1 = (
1
๐)๐โ1 and
๐4
๐3= (
๐4
๐3)
๐โ1
๐
From equation [84], and using the fact that โ๐ข = ๐๐โ๐ for constant cv gives us
๐ผ๐๐๐๐ ๐๐๐๐๐ ๐๐ฝ= ๐ โ
๐ป๐
๐ป๐= ๐ โ
๐
๐๐โ๐ [86]
2015 FINAL QUESTION 4
MECH 230 Final Exam Workbook
26
MECH 230 Final Exam Workbook
27
10.2 Air Standard Diesel Cycle
The diesel cycle is like the Otto cycle in many ways. It has the same compression ratio, ๐ =๐1
๐2. However,
it has an additional ratio, the cut-off ratio. The cut-off ratio is the ratio of the volume after combustion
to volume before combustion, defined by
๐๐ =๐ฝ๐
๐ฝ๐=
๐๐
๐๐ [87]
The cycle goes through four different processes
1 โ 2 ๐ผ๐ ๐๐๐ก๐๐๐๐๐ ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐
Isentropic, Q=0. Compression, W is negative. Applying first law,
๐ค๐๐ = ๐ข2 โ ๐ข1
Isentropic, so we can use relative specific volumes found in table A-22 and the compression ratio
๐ฃ๐2
๐๐1=
๐2
๐1=
1
๐
And since the working fluid is an ideal gas
๐2๐2
๐2=
๐1๐1
๐1โ
๐2
๐1=
๐2
๐1โ
๐1
๐2=
๐2
๐1โ ๐
2 โ 3 ๐ถ๐๐๐ ๐ก๐๐๐ก ๐๐๐๐ ๐ ๐ข๐๐ ๐ป๐๐๐ก ๐ด๐๐๐๐ก๐๐๐
Heat addition, Q is positive. Constant pressure, so ๐ = ๐2(๐3 โ ๐2). Applying first law,
๐ข3 โ ๐ข2 = ๐๐๐ โ ๐2(๐3 โ ๐2).
Rearranging
๐๐๐ = (๐ข3 + ๐3๐3) โ (๐ข2 + ๐2๐2)
Using definition h = u + Pv
๐๐๐ = โ3 โ โ2
Ideal gas with constant pressure
๐3
๐2=
๐3
๐2= ๐๐
3 โ 4 ๐ผ๐ ๐๐๐ก๐๐๐๐๐ ๐ธ๐ฅ๐๐๐๐ ๐๐๐
Isentropic, Q = 0. Expansion, work is positive. Applying first law,
๐ค๐๐ข๐ก = ๐ข3 โ ๐ข4
Using relative specific volumes
๐๐4
๐๐3=
๐4
๐3
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We know that ๐4 = ๐1
๐4
๐3=
๐4
๐2โ
๐2
๐3=
๐1
๐2โ
๐2
๐3=
๐
๐๐
๐๐4
๐๐3=
๐4
๐3=
๐
๐๐
Applying ideal gas law
๐4
๐3=
๐4
๐3โ
๐๐
๐
4 โ 1 ๐ถ๐๐๐ ๐ก๐๐๐ก ๐๐๐๐ข๐๐ ๐ป๐๐๐ก ๐ ๐๐๐๐ฃ๐๐
Constant volume, W = 0. Heat removal, Q is negative. First law
๐๐๐ข๐ก = ๐ข4 โ ๐ข1
Ideal gas with constant volume, so
๐4
๐4=
๐1
๐1
The schematic of the cycle on P-v and T-s diagrams can be found in Figure 13
Figure 13: P-v and T-s diagrams for the air standard diesel cycle
10.2.1 Diesel Cycle Thermal Efficiency
Like all other cycles, the efficiency is defined by ๐ =๐ค๐๐ฆ๐๐๐
๐๐๐. Plugging in for the equations solved above
yields the diesel cycle thermal efficiency
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๐ผ๐ ๐๐๐๐๐ = ๐ โ๐๐ โ ๐๐
๐๐ โ ๐๐ [88]
10.2.2 Cold Air-Standard Analysis for Diesel Cycle For a cold air analysis, specific heats are constant. For the two isentropic processes with constant
specific heat,
For process 1->2 we know
๐2
๐1= (
๐1
๐2)
๐โ1 = ๐๐โ1 and
๐2
๐1= (
๐2
๐1)
๐โ1
๐
For process 3->4 we know
๐4
๐3= (
๐3
๐4)
๐โ1 = (
1
๐)๐โ1 and
๐4
๐3= (
๐4
๐3)
๐โ1
๐
Using the definitions โ๐ข = ๐๐ฃโ๐ and โโ = ๐๐โ๐, and applying it to equation 13 yields the efficiency for
a diesel cycle with constant specific heats to be
๐ผ๐ ๐๐๐๐๐ ๐๐๐๐๐ ๐๐ฝ
= ๐ โ๐
๐๐โ๐[๐
๐โ
(๐๐๐ โ ๐)
(๐๐ โ ๐)]
[89]
2013 FINAL QUESTION 3
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11.0 Gas Turbine Power Plants Gas Turbine Power Plants are better suited for transportation. This is due to their higher power output
to weight ratio compared to vapor power plants.
11.1 Air Standard Brayton Cycle The Brayton cycle is an example of a gas turbine power plant. Figure 14 depicts the cycle.
Figure 14: Schematic of the Brayton cycle
Assuming steady state and applying first law 0 = ๏ฟฝ๏ฟฝ โ ๏ฟฝ๏ฟฝ + โ๐๐ โ โ๐๐ข๐ก.
For each step of the process
1 โ 2 ๐ด๐๐๐๐๐๐ก๐๐ ๐ถ๐๐๐๐๐๐ ๐ ๐๐๐
Adiabatic so Q = 0. Compression so W is negative
๏ฟฝ๏ฟฝ๐๐ = โ2 โ โ1
2 โ 3 ๐ป๐๐๐ก ๐ด๐๐๐๐ก๐๐๐ (๐๐ ๐ค๐๐๐)
W=0
๏ฟฝ๏ฟฝ๐๐ = โ3 โ โ2
3 โ 4๐ด๐๐๐๐๐๐ก๐๐ ๐ธ๐ฅ๐๐๐๐ ๐๐๐
Adiabatic so Q = 0. Expansion so W is positive
๏ฟฝ๏ฟฝ๐๐ข๐ก = โ3 โ โ4
4 โ 1 ๐ป๐๐๐ก ๐ ๐๐๐๐ฃ๐๐ (๐๐ ๐ค๐๐๐)
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W=0
๏ฟฝ๏ฟฝ๐๐ข๐ก = โ4 โ โ1
The thermal efficiency for a cycle is given by ๐๐๐ฆ๐๐๐ = 1 โ๏ฟฝ๏ฟฝ๐๐ข๐ก
๏ฟฝ๏ฟฝ๐๐. Plugging in gives the Brayton cycle
thermal efficiency
๐ผ๐๐๐๐๐๐๐ = ๐ โ๐๐ โ ๐๐
๐๐ โ ๐๐ [90]
The back-work ratio for the Brayton cycle is given by
๐๐๐๐๐๐๐๐๐๐ =๐๐ โ ๐๐
๐๐ โ ๐๐ [91]
11.1.1 Ideal Air Standard Brayton Cycle A few more equations get unlocked when the process is ideal. Processes 1->2 and 3->4 are isentropic if
the cycle is ideal.
1 โ 2 ๐ผ๐ ๐๐๐ก๐๐๐๐๐
๐๐2 = ๐๐1 (๐2
๐1)
3 โ 4 ๐ผ๐ ๐๐๐ก๐๐๐๐๐
๐๐4 = ๐๐3 (๐4
๐3)
The T-s and P-v diagrams for the ideal Brayton cycle are found in Figure 6.
Figure 15: Ts and Pv diagrams for the ideal Brayton cycle
11.1.2 Ideal Cold Air-Standard Brayton Cycle Cold air standard analysis means that specific heats are constant. Processes 1->2 and 3->4 are
isentropic if the cycle is ideal.
1 โ 2 ๐ผ๐ ๐๐๐ก๐๐๐๐๐, ๐๐๐๐ ๐ก ๐๐ฃ
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๐2
๐1= (
๐2
๐1)
๐โ1๐
3 โ 4 ๐ผ๐ ๐๐๐ก๐๐๐๐๐, ๐๐๐๐ ๐ก ๐๐ฃ
๐4
๐3= (
๐4
๐3)
๐โ1๐
However, P3=P2 and P1=P4, thus ๐2
๐1=
๐3
๐4. Combining the two above equations, we get
๐2
๐1=
๐3
๐4 [92]
The thermal efficiency for a Brayton Cycle with constant cv is given by
๐ผ๐๐๐๐๐๐๐ ๐๐๐๐๐ ๐๐ฝ
= ๐ โ๐ป๐
๐ป๐= ๐ โ
๐
(๐ท๐๐ท๐
)
๐โ๐๐
[93]
The bwr for a Brayton Cycle with constant cv is given by
๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐ ๐๐=
๐ป๐ โ ๐ป๐
๐ป๐ โ ๐ป๐ [94]
11.1.3 Increasing Cycle Efficiency There are two ways to increase the efficiency of a gas turbine. One is compression with intercooling,
and the other is reheat between the turbines.
11.1.3.1 Brayton Cycle with Reheat
Similar to the Rankine cycle, reheating the working fluid between two turbines increased the efficiency.
Figure 16 depicts what this would look like. Figure 17 shows the T-s diagram of the Brayton cycle with
reheat.
Figure 16: Two stage turbine reheat
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Figure 17: Ts diagram of the Brayton cycle with reheat
11.1.3.2 Brayton Cycle with Intercooling
The amount of power that the compressor needs can be reduced by compressing with cooling between
stages. Figure 18 shows what this would look like. Figure 19 shows the P-v and T-s diagrams for the
Brayton cycle with intercooling.
Figure 18: Compression with Intercooling
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Figure 19: Ts and Pv diagram for compression with intercooling
11.2 Aircraft Gas Turbines The ideal air-standard jet propulsion cycle can be seen in Figure 20.
Figure 20: Jet propulsion cycle
The turbine is used to power the compressor, so the net work of the cycle is equal to zero ๏ฟฝ๏ฟฝ๐๐ฆ๐๐๐ = 0.
Therefore,
๏ฟฝ๏ฟฝ๐๐๐๐๐๐๐ ๐ ๐๐ = ๏ฟฝ๏ฟฝ๐ก๐ข๐๐๐๐๐
๐๐ โ ๐๐ = ๐๐ โ ๐๐ [95]
Generally, the process through the diffuser and the nozzle are taken as isentropic. Applying
conservation of energy to the Diffuser and Nozzle yields
0 = โ๐๐ +๐๐๐
2
2โ โ๐๐ข๐ก โ
๐๐๐ข๐ก2
2
The diffuser is used to slow the working fluid to zero velocity, V1=0
๐ โ 1
๐๐ = ๐๐ +๐ฝ๐
๐
๐
[96]
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The nozzle accelerates the gas to such a speed that the velocity before the process is negligible, V4=0
4 โ 5
๐ฝ๐ = โ๐(๐๐ โ ๐๐)
[97]
OR with constant cv
๐ฝ๐ = โ๐ โ ๐๐(๐ป๐ โ ๐ป๐)
[98]
The thrust of the engine is given by the difference in the speed of the working fluid at the inlet and
outlet, multiplied by the mass flow rate.
๐ญ๐ป = ๏ฟฝ๏ฟฝ(๐ฝ๐ โ ๐ฝ๐) [99]
2013 FINAL QUESTION 4
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References [1] All schematics are taken from Prof. Ciccarelliโs notes. Exam questions come from Prof. Ciccarelli.