MECH 230 Thermodynamics 1 Final Workbook Solutions

MECH 230 – 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 (𝜌) kg/m3

Specific Volume V/m || 1/𝜌 Upsilon (𝜐) 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 𝑎𝑡𝑚 = 101.326 𝑃𝑎 = 760𝑚𝑚𝐻𝑔|| 1 𝑏𝑎𝑟 = 100,000 𝑃𝑎 = 100𝑘𝑃𝑎 = 1𝑀𝑃𝑎 || 1 𝑚3 = 1000𝐿

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.

𝑷𝒈 = 𝑷𝒂𝒃𝒔 − 𝑷𝒂𝒕𝒎

[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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𝐏𝐠 = 𝐏𝐚𝐛𝐬 – 𝐏𝐚𝐭𝐦 = 𝛒𝐥𝐢𝐪𝐮𝐢𝐝 ∗ 𝒈 ∗ 𝒉

[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,

𝑾𝟏→𝟐 = ∫ 𝑷𝒈𝒅𝑽

𝟐

𝟏

[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.

∆𝑽 = 𝟎 ∴ ∆𝑾 = 𝟎

[4]

When you are told that a system undergoes a polytropic process, that means it behaves according to

equation [5].

𝑷𝑽𝒏 = 𝒄

[5]

https://www.researchgate.net/figure/Fig3-9-Simple-U-tube-manometer_fig2_318378486


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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: 𝑃1𝑉1𝑛 = 𝑃2𝑉2

𝑛, since c is a constant.

From equation [5], three different ways to calculate work can be derived. Equation [6] shows the work

for when 𝑛 ≠ 1, equation [7] for when n = 1, and equation [8] for when n = 0. When n = 0, this is called a

constant pressure process.

𝑾𝟏→𝟐 =

𝑷𝟐𝑽𝟐 − 𝑷𝟏𝑽𝟏

𝟏 − 𝒏 𝒇𝒐𝒓 𝒏 ≠ 𝟏

[6]

𝑾𝟏→𝟐 = 𝑷𝟏𝑽𝟏 𝐥𝐧 (

𝑽𝟐

𝑽𝟏) 𝒇𝒐𝒓 𝒏 = 𝟏

[7]

𝑾𝟏→𝟐 = 𝑷∆𝑽 𝒇𝒐𝒓 𝒏 = 𝟎 [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, ∆𝑇 = 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.

∆𝑻 = 𝟎 ∴ ∆𝑼 = 𝟎 ‼ (𝑰𝑫𝑬𝑨𝑳 𝑮𝑨𝑺)‼

[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.

𝑄1→2 = ∫ 𝛿𝑄

2

1

[10]

An adiabatic process is a process where there is no heat exchange from the system to the

surroundings. Therefore ∆𝑸 = 𝟎.

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.

𝐸2 − 𝐸1 = 𝑄 − 𝑊

[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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𝐸2 − 𝐸1 = ∆𝐾𝐸 + ∆𝑈 + ∆𝑃𝐸

[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].

∆𝑼𝟏→𝟐 = 𝑸𝟏→𝟐 − 𝑾𝟏→𝟐

[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, �� = 8314𝐽

𝑘𝑚𝑜𝑙∗𝐾= 8.314

𝑘𝐽

𝑘𝑚𝑜𝑙 𝐾, 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

𝑹 =

��

𝝁

[14]

After dividing by the molar mass, the gas constant has units of 𝐽

𝑘𝑔∗𝐾. 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.

𝑃𝑉 = 𝑛��𝑇

[15]

𝑷𝑽 = 𝑴𝑹𝑻

[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).

𝑃𝜐 = 𝑅𝑇

[17]

𝑃 = 𝜌𝑅𝑇

[18]

3.2 Polytropic Work Using Ideal Gas After being told that a gas is ideal, multiple equations become “unlocked” 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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𝑾𝟏→𝟐 =𝑴𝑹(𝑻𝟐 − 𝑻𝟏)

𝟏 − 𝒏 𝒇𝒐𝒓 𝒏 ≠ 𝟏

[19]

For an ideal gas going though a polytropic process with n=1 means it is an isothermal process (∆𝑻 =

𝟎). Therefore equation [7] can be modified to the following.

𝑾𝟏→𝟐 = 𝑴𝑹𝑻 𝐥𝐧 (

𝑽𝟐

𝑽𝟏) 𝒇𝒐𝒓 𝒏 = 𝟏

[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.

𝑅 =𝑃𝜐

𝑇

[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.

𝑻𝟐

𝑻𝟏= (

𝑷𝟐

𝑷𝟏)

𝒏−𝟏𝒏

[22]

𝑻𝟐

𝑻𝟏= (

𝝊𝟐

𝝊𝟏)

𝒏−𝟏

[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 𝑘𝐽

𝑘𝑔°𝐾.

𝑐 =∆𝑄

𝑀 ∗ ∆𝑇

[24]

4.1 Constant Volume Heat Addition For a constant volume, there is no work being done. Therefore, first law gets reduced to

∆𝑈 = ∆𝑄

Rearranging equation [23], dividing through by M, and plugging it into the above yields

∆𝒖 = 𝒄𝒗∆𝑻

[25]

Noting that 𝑐𝑣 depends on both P and T, equation [25] gets generalized to

𝑢2 − 𝑢1 = ∫ 𝑑𝑢

2

1

= ∫ 𝑐𝑣(𝑃, 𝑇)𝑑𝑇2

1

[26]


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4.2 Constant Pressure Heat Addition For a constant pressure heat addition, enthalpy is calculated using

𝒉 = 𝒄𝒑∆𝑻

[27]

Like 𝑐𝑣, 𝑐𝑝 is dependent on P and T. So, equation [27] generalizes to

ℎ2 − ℎ1 = ∫ 𝑑ℎ

2

1

= ∫ 𝑐𝑝(𝑃, 𝑇)𝑑𝑇2

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

𝒌(𝑻) =

𝒄𝒑

𝒄𝒗

[29]

The following equations are useful for solving specific heats, given one another

𝒄𝒗(𝑻) =

𝑹

𝑲(𝑻) − 𝟏

[30]

𝒄𝒑(𝑻) =𝑲(𝑻)𝑹

𝑲(𝑻) − 𝟏

[31]

2013 FINAL QUESTION 1


7


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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 ��, and

has unites of kg/s. The change in mass of the control volume is given by equation [31]

𝑑𝑀𝐶𝑉

𝑑𝑡 = ∑��𝑖 − ∑��𝑒 [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

∑��𝑒 = ∑��𝑖 [33] There are a few different equations that come from mass flow rate. Mass flow rate can also be denoted

by

�� = 𝝆𝑨𝑽 [34] Where 𝜌 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 [𝑘𝑔

𝑚2𝑠]

��

𝑨= 𝝆𝑽 =

𝑽

𝒗 [35]

The volumetric flow rate is defined as [𝑚3

𝑠]

𝒎

𝝆= 𝑽𝑨

[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, ��and ��.

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, �� and ��. This is done by dividing ��and �� by the mass

flow rate, ��, 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

𝑑𝐸𝑐𝑣

𝑑𝑡= �� − ��𝑠ℎ𝑎𝑓𝑡 + ∑��𝑖 (ℎ𝑖 +

𝑉𝑖2

2+ 𝑔𝑍𝑖) − ∑��𝑒 (ℎ𝑒 +

𝑉𝑒2

2+ 𝑔𝑍𝑒) [37]

For this course, we assume that 𝑑𝐸𝑐𝑣

𝑑𝑡= 0, that there is only one inlet and one outlet, and that the

systems has steady flow, ��𝑖 = ��𝑒 = ��. Therefore, dividing through by �� reduces the equation to

𝟎 = �� − �� + (𝒉𝒊 +𝑽𝒊

𝟐

𝟐+ 𝒈𝒁𝒊) − (𝒉𝒆 +

𝑽𝒆𝟐

𝟐+ 𝒈𝒁𝒆) [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 𝜌1 = 𝜌2 for subsonic flow yields the following

𝑨𝟏𝑽𝟏 = 𝑨𝟐𝑽𝟐 [39]

Figure 3: A schematic of a nozzle and diffuser

A diffuser decreases the flow rate of the fluid, therefore 𝐴2 > 𝐴1 → 𝑉2 < 𝑉1.

A nozzle increases the flow rate of the fluid, therefore 𝐴1 > 𝐴2 → 𝑉1 < 𝑉2.

Applying the CV energy equation, assuming no heat transfer, work, or change in height, yields the

following

𝑽𝟐𝟐 = 𝑽𝟏

𝟐 + 𝟐(𝒉𝟏 − 𝒉𝟐) [40]

For a rocket nozzle, 𝑉2 ≫ 𝑉1, and the above equation simplifies to

𝑉22 = 2(ℎ1 − ℎ2) [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

�� = (ℎ1 − ℎ2) + (𝑉1

2

2−

𝑉22

2) [42]

Assuming ℎ1 − ℎ2 ≫𝑉1

2

2−

𝑉22

2 simplifies the above to the following

�� = (𝒉𝟏 − 𝒉𝟐) [43]

Note that since a turbine develops work, ℎ1 > ℎ2. The power is the work output per unit time

�� = �� ∗ �� = (ℎ1 − ℎ2) [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

�� = (ℎ1 − ℎ2) + (𝑉1

2

2−

𝑉22

2) [45]

Assuming the change in KE is small compared to the change in h


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�� = (𝒉𝟏 − 𝒉𝟐) [46]

Rather than work output like a turbine, work is inputted into a pump. Therefore ℎ2 > ℎ1.

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

(ℎ1 − ℎ2) = (𝑉1

2

2−

𝑉22

2) [47]

Looking downstream from each side of the device, we can assume that 𝑉1 = 𝑉2. This assumption yields

𝒉𝟏 = 𝒉𝟐 [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 = ��1ℎ1+��3ℎ3 − ��2ℎ2 − ��4ℎ4 [49]

Steady flow ∴ ��1 = ��2 𝑎𝑛𝑑 ��3 = ��4

0 = ��1(ℎ1 − ℎ2) + ��3(ℎ3 − ℎ4) [50]

Which further simplifies to

��𝟏

��𝟑=

𝒉𝟒 − 𝒉𝟑

𝒉𝟏 − 𝒉𝟐 [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 – 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 – 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

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒

𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑤𝑜𝑟𝑘 =

𝑊

𝑊𝑀𝐴𝑋 [52]

Applying first law to a cycle, we can see the following

0 = 𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 − 𝑊

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

𝜂𝑐𝑦𝑐𝑙𝑒 =

𝑊

𝑄𝑖𝑛

[53]

Or further rearranged to the following

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡

𝑄𝑖𝑛= 1 −

𝑄𝑜𝑢𝑡

𝑄𝑖𝑛 [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

𝜂𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = 𝜂𝑐𝑦𝑐𝑙𝑒𝜂𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 [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


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 irreversibility’s 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 irreversibility’s 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

𝑆2 − 𝑆1 = ∫𝛿𝑄

𝑇+ 𝑆𝑔𝑒𝑛 [56]

Where S2 -S1 is the change in entropy of the system, ∫𝛿𝑄

𝑇is the entropy transfer to the system by heat,

and Sgen is the entropy generated in the system due to irreversibility’s. 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, ∆𝑄 = 0

(adiabatic) and Sgen=0 (reversible), so entropy is constant.

∴ 𝑆1 − 𝑆2 = 0, 𝑆1 = 𝑆2.

7.3.1 Entropy Change for Ideal Gas We know that for an ideal gas 𝑑𝑢 = 𝑐𝑣(𝑇)𝑑𝑇, 𝑑ℎ = 𝑐𝑝(𝑇)𝑑𝑇, 𝑃𝜈 = 𝑅𝑇.

Plugging these equations into the above yields the following

𝑠2(𝑇2, 𝜈2) − 𝑠1(𝑇1, 𝜈1) = ∫𝑐𝑉(𝑇)𝑑𝑇

𝑇

2

1

+ 𝑅𝑙𝑛(𝜈2

𝜈1) [57]

And

𝑠2(𝑇2, 𝑃2) − 𝑠1(𝑇1, 𝑃1) = ∫𝑐𝑃(𝑇)𝑑𝑇

𝑇

2

1

− 𝑅𝑙𝑛(𝑃2

𝑃1) [58]

The bounds of ∫𝑐𝑃(𝑇)𝑑𝑇

𝑇

2

1 are tabulated as a function of T in air in Table A-22 under the label 𝑠°. So,

∫𝑐𝑃(𝑇)𝑑𝑇

𝑇

2

1= 𝑠°2 − 𝑠°1. Therefore equation [57] can be written as


14

𝒔𝟐 − 𝒔𝟏 = (𝒔°𝟐 − 𝒔°𝟏) − 𝑹𝒍𝒏(𝑷𝟐

𝑷𝟏) [59]

If cv and cp are taken as constant, changes in entropy are calculated by the following

𝒔𝟐 − 𝒔𝟏 = 𝒄𝒗 𝐥𝐧 (𝑻𝟐

𝑻𝟏) + 𝑹𝒍𝒏 (

𝝂𝟐

𝝂𝟏) [60]

And

𝒔𝟐 − 𝒔𝟏 = 𝒄𝒗 𝐥𝐧 (𝑻𝟐

𝑻𝟏) − 𝑹𝒍𝒏 (

𝑷𝟐

𝑷𝟏) [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, ∆𝑄 = 0

(adiabatic) and Sgen=0 (reversible), so entropy is constant.

∴ 𝑆1 − 𝑆2 = 0, 𝑆1 = 𝑆2.

Plugging S1=S2 into equation [27]

𝑠°2 − 𝑠°1

𝑅= ln (

𝑃2

𝑃1)

𝑃2

𝑃1=

exp (𝑠°2𝑅 )

exp (𝑠°1𝑅 )

Now we define relative pressure, 𝑷𝒓 = 𝒆𝒙𝒑(𝒔°

𝑹). Therefore

(𝑷𝟐

𝑷𝟏) =

𝑷𝒓𝟐

𝑷𝒓𝟏 (𝒔 = 𝒄𝒐𝒏𝒔𝒕)

[62]

From the ideal gas law,

𝜈2

𝜈1=

𝑇2

𝑇1(

𝑃1

𝑃2) =

(𝑇2𝑃𝑟2

)

(𝑇1Pr1

)

Define vr =T/Pr, therefore

(𝝂𝟐

𝝂𝟏) =

𝒗𝒓𝟐

𝝂𝒓𝟏 [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

(𝑻𝟐

𝑻𝟏) = (

𝝂𝟏

𝝂𝟐)

𝒌−𝟏

, 𝒔 = 𝒄𝒐𝒏𝒔𝒕, 𝒄𝒑 = 𝒄𝒐𝒏𝒔𝒕 [64]

Substituting in 𝑇2

𝑇1=

𝑃2𝜈2

𝑃1𝜈1,


15

(𝑷𝟐

𝑷𝟏) = (

𝝂𝟏

𝝂𝟐)

𝒌

, 𝒔 = 𝒄𝒐𝒏𝒔𝒕, 𝒄𝒑 = 𝒄𝒐𝒏𝒔𝒕 [65]

For a polytropic process, 𝑃𝜈𝑛 = 𝑐𝑜𝑛𝑠𝑡. For an isentropic process, n=k. Combining equations yields

(𝜈1

𝜈2) = (

𝑃2

𝑃1)

1𝑘

= (𝑇2

𝑇1)

1𝑘−1

(𝑻𝟐

𝑻𝟏) = (

𝑷𝟐

𝑷𝟏)

𝒌−𝟏

𝒌, 𝒔 = 𝒄𝒐𝒏𝒔𝒕, 𝒄𝒑 = 𝒄𝒐𝒏𝒔𝒕 [66]

8.5 Control Volume Entropy Balance The rate of entropy change in a control volume is given by

𝑑𝑆𝐶𝑉

𝑑𝑡 =

��𝑗

𝑇𝑗∑��𝑖𝑠𝑖 − ∑��𝑒𝑠𝑒 [67]

Assuming steady state, one inlet one outlet, and isothermal CV, the above simplifies to

𝟎 =𝟏

��(

��𝒄𝒗

𝑻) + 𝒔𝒊𝒏 − 𝒔𝒐𝒖𝒕 +

��𝒈𝒆𝒏

��

[68]

8.5.1 Isentropic Efficiencies of Turbines and Compressors

Recall for a turbine ��𝑐𝑣

��= ℎ1 − ℎ2 > 0, since turbines create work. For a real turbine it is impossible to

get rid of all irreversibility’s, so s2>s1. In Figure 8, 2s is only achievable with no irreversibility’s (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 𝜂 =𝑊𝑟𝑒𝑎𝑙

𝑊𝑚𝑎𝑥. Now we know that the maximum work of

a turbine is given as ��𝑐𝑣

��= ℎ1 − ℎ2𝑠.

Plugging in, we get the isentropic turbine efficiency


16

𝜼𝒕 =𝒉𝟏 − 𝒉𝟐

𝒉𝟏 − 𝒉𝟐𝒔 [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 𝜂𝑐 =𝑊𝑚𝑎𝑥

𝑊 . We know that 𝑊𝑚𝑎𝑥 = ℎ1 − ℎ2𝑠. From

this, the isentropic compressor efficiency is given by

𝜼𝒄 =𝒉𝟏 − 𝒉𝟐𝒔

𝒉𝟏 − 𝒉𝟐 [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

��𝑐𝑣

��= − ∫ 𝜈𝑑𝑃

2

1

[71]

Liquids are incompressible, so there is no change in the specific volume 𝜈1 = 𝜈2 = 𝜈. ∴

��𝒄𝒗

��= −𝝂(𝑷𝟐 − 𝑷𝟏) [72]

For a gas going through a polytropic process, with 𝑃𝜈 = 𝑅𝑇

��𝒄𝒗

��= −

𝒏𝑹𝑻𝟏

𝒏 − 𝟏(

𝑻𝟐

𝑻𝟏− 𝟏) , 𝒏 ≠ 𝟏 [73]

OR substituting (𝑇2

𝑇1) = (

𝑃2

𝑃1)

𝑛−1

𝑛 we get


17

��𝑐𝑣

��= −

𝑛𝑅𝑇1

𝑛 − 1((

𝑃2

𝑃1)

𝑛−1𝑛

− 1) , 𝑛 ≠ 1 [74]

If a process is isothermal (T1 = T2)

��𝒄𝒗

��= −𝑹𝑻𝒍𝒏(

𝑷𝟐

𝑷𝟏) [75]



18

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.


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 =��𝑐𝑣

��−

��𝑐𝑣

��+ (ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡)

Now we analyze each step individually.

1 → 2 𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛

Adiabatic so �� = 0.

∴ 𝑤𝑜𝑢𝑡 = ℎ1 − ℎ2

2 → 3 𝐶𝑜𝑛𝑑𝑒𝑛𝑠𝑜𝑟 (𝑛𝑜 𝑤𝑜𝑟𝑘)

No work so �� = 0.

∴ 𝑞𝑜𝑢𝑡 = ℎ2 − ℎ3

3 → 4 𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑃𝑢𝑚𝑝 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛

Adiabatic so �� = 0

∴ 𝑤𝑖𝑛 = ℎ4 − ℎ3

4 → 1 𝑆𝑡𝑒𝑎𝑚 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 (𝑛𝑜 𝑤𝑜𝑟𝑘)

No work so �� = 0

∴ 𝑞𝑖𝑛 = ℎ1 − ℎ4

9.1 Thermal Efficiency The thermal efficiency of a heat engine is defined by

𝜂 =𝑛𝑒𝑡 𝑤𝑜𝑟𝑘 𝑜𝑢𝑡

ℎ𝑒𝑎𝑡 𝑖𝑛𝑝𝑢𝑡

Net work out is the work output minus the work input. This reduces the above to the following


20

𝜂𝑟𝑎𝑛𝑘𝑖𝑛𝑒 =𝑤𝑜𝑢𝑡 − 𝑤𝑖𝑛

𝑞𝑖𝑛=

(ℎ1 − ℎ2) − (ℎ4 − ℎ3)

ℎ1 − ℎ4 [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

𝒃𝒘𝒓𝒓𝒂𝒏𝒌𝒊𝒏𝒆 =𝒘𝒊𝒏(𝒑𝒖𝒎𝒑)

𝒘𝒐𝒖𝒕(𝒕𝒖𝒓𝒃𝒊𝒏𝒆)=

𝒉𝟒 − 𝒉𝟑

𝒉𝟏 − 𝒉𝟐 [77]

9.3 Ideal Rankine Cycle In the ideal Rankine Cycle, no irreversibility’s occur. The resulting cycle is the following

1 → 2 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑎𝑡 𝑇𝑢𝑟𝑏𝑖𝑛𝑒

2 → 3 𝐶𝑜𝑛𝑑𝑒𝑛𝑠𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒

3 → 4 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑎𝑡 𝑃𝑢𝑚𝑝

4 → 1 𝑆𝑡𝑒𝑎𝑚 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒

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

𝑤𝑖𝑛 = − ∫ 𝜈𝑑𝑃4

3

If the working fluid is a pure liquid then the specific volume can be assumed to be constant, yielding the

following

𝑤𝑖𝑛 = 𝜈3(𝑃4 − 𝑃3)

Like the thermal efficiency of a Carnot Cycle, the ideal Rankine Efficiency is given by


21

𝜼𝒊𝒅𝒆𝒂𝒍 𝒓𝒂𝒏𝒌𝒊𝒏𝒆 = 𝟏 −𝑻𝒐𝒖𝒕

𝑻𝒊𝒏 [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

𝒃𝒘𝒓𝒊𝒅𝒆𝒂𝒍 𝒓𝒂𝒏𝒌𝒊𝒏𝒆 =𝒘𝒊𝒏

𝒘𝒐𝒖𝒕=

𝝂𝟑(𝑷𝟒 − 𝑷𝟑)

(𝒉𝟏 − 𝒉𝟐𝒔) [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

𝜼𝒓𝒂𝒏𝒌𝒊𝒏𝒆 𝒓𝒆𝒉𝒆𝒂𝒕 =

(𝒘𝟏→𝟐 + 𝒘𝟐→𝟑) − 𝒘𝟓→𝟔

(𝒒𝟔−𝟏 + 𝒒𝟐→𝟑)

=(𝒉𝟏 − 𝒉𝟐) + (𝒉𝟑 − 𝒉𝟒) − (𝒉𝟔 − 𝒉𝟓)

(𝒉𝟏 − 𝒉𝟔) + (𝒉𝟑 − 𝒉𝟐)

[80]



22

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).


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, ��, (or horsepower) is the amount of power outputted by an engine. It is defined by

�� = 𝑾𝒄𝒚𝒄𝒍𝒆 ∗𝑵

𝟐 [81]

Where N is the crank shaft speed, in revolutions per second. A useful conversion to have on your

formula sheet is

𝟏 𝒉𝒐𝒓𝒔𝒆𝒑𝒐𝒘𝒆𝒓 = 𝟕𝟒𝟔 𝑾𝒂𝒕𝒕𝒔

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

𝒎𝒆𝒑 =𝒏𝒆𝒕 𝒘𝒐𝒓𝒌 𝒑𝒆𝒓 𝒄𝒚𝒄𝒍𝒆

𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒅 𝒗𝒐𝒍𝒖𝒎𝒆=

𝑾𝒄𝒚𝒄𝒍𝒆

𝑽𝟏 − 𝑽𝟐 [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

𝑟 =𝑉1

𝑉2=

𝑉4

𝑉3 [83]

Or, since one of the assumptions is fixed mass

𝒓 =𝝂𝟏

𝝂𝟐=

𝝂𝟒

𝝂𝟑 [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 → 2 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛

Isentropic, Q=0. Compression, so work is negative. Applying first law,

𝑤𝑖𝑛 = 𝑢2 − 𝑢1

Isentropic, so we can use relative specific volumes found in table A-22 and the compression ratio


24

𝑣𝑟2

𝜈𝑟1=

𝜈2

𝜈1=

1

𝑟

And since the working fluid is an ideal gas

𝑃2𝜈2

𝑇2=

𝑃1𝜈1

𝑇1→

𝑃2

𝑃1=

𝑇2

𝑇1∗

𝜈1

𝜈2=

𝑇2

𝑇1∗ 𝑟

2 → 3 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 ℎ𝑒𝑎𝑡 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛

Constant volume so W = 0. Applying first law,

𝑞𝑖𝑛 = 𝑢3 − 𝑢2

Ideal gas with constant volume, so

𝑃3

𝑃2=

𝑇3

𝑇2

3 → 4 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛

Isentropic, Q = 0. Expansion, work is positive. Applying first law,

𝑤𝑜𝑢𝑡 = 𝑢3 − 𝑢4

Reversible process with compression ratio

𝜈𝑟4

𝜈𝑟3=

𝜈4

𝜈3= 𝑟

Ideal gas, so

𝑃4

𝑃3=

𝑇4

𝑇3∗

𝜈3

𝜈4=

𝑇4

𝑇3∗

1

𝑟

4 → 1 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 ℎ𝑒𝑎𝑡 𝑟𝑒𝑚𝑜𝑣𝑎𝑙

Constant volume, W = 0. Heat removal, Q is negative. Applying first law,

𝑞𝑜𝑢𝑡 = 𝑢4 − 𝑢1


𝑃4

𝑇4=

𝑃1

𝑇1

10.2.1 Otto Cycle Thermal Efficiency Like other heat engines, the efficiency of the cycle is given by

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑤𝑜𝑢𝑡 − 𝑤𝑖𝑛

𝑞𝑖𝑛

So, for an Otto Cycle the efficiency is

𝜼𝒐𝒕𝒕𝒐 =(𝒖𝟑 − 𝒖𝟒) − (𝒖𝟐 − 𝒖𝟏)

(𝒖𝟑 − 𝒖𝟐)= 𝟏 −

𝒖𝟒 − 𝒖𝟏

𝒖𝟑 − 𝒖𝟐 [85]


25

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

𝑇2

𝑇1= (

𝜈1

𝜈2)

𝑘−1 = 𝑟𝑘−1 and

𝑇2

𝑇1= (

𝑃2

𝑃1)

𝑘−1

𝑘


𝑇4

𝑇3= (

𝜈3

𝜈4)

𝑘−1 = (

1

𝑟)𝑘−1 and

𝑇4

𝑇3= (

𝑃4

𝑃3)

𝑘−1

𝑘

From equation [84], and using the fact that ∆𝑢 = 𝑐𝑉∆𝑇 for constant cv gives us

𝜼𝒐𝒕𝒕𝒐 𝒄𝒐𝒏𝒔𝒕 𝒄𝑽= 𝟏 −

𝑻𝟏

𝑻𝟐= 𝟏 −

𝟏

𝒓𝒌−𝟏 [86]



26


27

10.2 Air Standard Diesel Cycle

The diesel cycle is like the Otto cycle in many ways. It has the same compression ratio, 𝑟 =𝜈1

𝜈2. 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

𝒓𝒄 =𝑽𝟑

𝑽𝟐=

𝝂𝟑

𝝂𝟐 [87]

The cycle goes through four different processes

1 → 2 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛

Isentropic, Q=0. Compression, W is negative. Applying first law,

𝑤𝑖𝑛 = 𝑢2 − 𝑢1

Isentropic, so we can use relative specific volumes found in table A-22 and the compression ratio

𝑣𝑟2

𝜈𝑟1=

𝜈2

𝜈1=

1

𝑟

And since the working fluid is an ideal gas

𝑃2𝜈2

𝑇2=

𝑃1𝜈1

𝑇1→

𝑃2

𝑃1=

𝑇2

𝑇1∗

𝜈1

𝜈2=

𝑇2

𝑇1∗ 𝑟

2 → 3 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐻𝑒𝑎𝑡 𝐴𝑑𝑑𝑖𝑡𝑖𝑜𝑛

Heat addition, Q is positive. Constant pressure, so 𝑊 = 𝑃2(𝑉3 − 𝑉2). Applying first law,

𝑢3 − 𝑢2 = 𝑞𝑖𝑛 − 𝑃2(𝑉3 − 𝑉2).

Rearranging

𝑞𝑖𝑛 = (𝑢3 + 𝑃3𝜈3) − (𝑢2 + 𝑃2𝜈2)

Using definition h = u + Pv

𝑞𝑖𝑛 = ℎ3 − ℎ2

Ideal gas with constant pressure

𝑇3

𝑇2=

𝜈3

𝜈2= 𝑟𝑐

3 → 4 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛

Isentropic, Q = 0. Expansion, work is positive. Applying first law,

𝑤𝑜𝑢𝑡 = 𝑢3 − 𝑢4

Using relative specific volumes

𝜈𝑟4

𝜈𝑟3=

𝜈4

𝜈3


28

We know that 𝜈4 = 𝜈1

𝜈4

𝜈3=

𝜈4

𝜈2∗

𝜈2

𝜈3=

𝜈1

𝜈2∗

𝜈2

𝜈3=

𝑟

𝑟𝑐

𝜈𝑟4

𝜈𝑟3=

𝜈4

𝜈3=

𝑟

𝑟𝑐

Applying ideal gas law

𝑃4

𝑃3=

𝑇4

𝑇3∗

𝑟𝑐

𝑟

4 → 1 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉𝑜𝑙𝑢𝑚𝑒 𝐻𝑒𝑎𝑡 𝑅𝑒𝑚𝑜𝑣𝑎𝑙

Constant volume, W = 0. Heat removal, Q is negative. First law

𝑞𝑜𝑢𝑡 = 𝑢4 − 𝑢1


𝑃4

𝑇4=

𝑃1

𝑇1

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 𝜂 =𝑤𝑐𝑦𝑐𝑙𝑒

𝑞𝑖𝑛. Plugging in for the equations solved above

yields the diesel cycle thermal efficiency


29

𝜼𝒅𝒊𝒆𝒔𝒆𝒍 = 𝟏 −𝒖𝟒 − 𝒖𝟏

𝒉𝟑 − 𝒉𝟐 [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,


𝑇2

𝑇1= (

𝜈1

𝜈2)

𝑘−1 = 𝑟𝑘−1 and

𝑇2

𝑇1= (

𝑃2

𝑃1)

𝑘−1

𝑘


𝑇4

𝑇3= (

𝜈3

𝜈4)

𝑘−1 = (

1

𝑟)𝑘−1 and

𝑇4

𝑇3= (

𝑃4

𝑃3)

𝑘−1

𝑘

Using the definitions ∆𝑢 = 𝑐𝑣∆𝑇 and ∆ℎ = 𝑐𝑝∆𝑇, and applying it to equation 13 yields the efficiency for

a diesel cycle with constant specific heats to be

𝜼𝒅𝒊𝒆𝒔𝒆𝒍 𝒄𝒐𝒏𝒔𝒕 𝒄𝑽

= 𝟏 −𝟏

𝒓𝒌−𝟏[𝟏

𝒌∗

(𝒓𝒄𝒌 − 𝟏)

(𝒓𝒄 − 𝟏)]

[89]



30


31

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 = �� − �� + ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡.

For each step of the process

1 → 2 𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛

Adiabatic so Q = 0. Compression so W is negative

��𝑖𝑛 = ℎ2 − ℎ1

2 → 3 𝐻𝑒𝑎𝑡 𝐴𝑑𝑑𝑖𝑡𝑖𝑜𝑛 (𝑛𝑜 𝑤𝑜𝑟𝑘)

W=0

��𝑖𝑛 = ℎ3 − ℎ2

3 → 4𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛

Adiabatic so Q = 0. Expansion so W is positive

��𝑜𝑢𝑡 = ℎ3 − ℎ4

4 → 1 𝐻𝑒𝑎𝑡 𝑅𝑒𝑚𝑜𝑣𝑎𝑙 (𝑛𝑜 𝑤𝑜𝑟𝑘)


32

W=0

��𝑜𝑢𝑡 = ℎ4 − ℎ1

The thermal efficiency for a cycle is given by 𝜂𝑐𝑦𝑐𝑙𝑒 = 1 −��𝑜𝑢𝑡

��𝑖𝑛. Plugging in gives the Brayton cycle

thermal efficiency

𝜼𝒃𝒓𝒂𝒚𝒕𝒐𝒏 = 𝟏 −𝒉𝟒 − 𝒉𝟏

𝒉𝟑 − 𝒉𝟐 [90]

The back-work ratio for the Brayton cycle is given by

𝒃𝒘𝒓𝒃𝒓𝒂𝒚𝒕𝒐𝒏 =𝒉𝟐 − 𝒉𝟏

𝒉𝟑 − 𝒉𝟒 [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 → 2 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐

𝑃𝑟2 = 𝑃𝑟1 (𝑃2

𝑃1)

3 → 4 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐

𝑃𝑟4 = 𝑃𝑟3 (𝑃4

𝑃3)

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 → 2 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐, 𝑐𝑜𝑛𝑠𝑡 𝑐𝑣


33

𝑇2

𝑇1= (

𝑃2

𝑃1)

𝑘−1𝑘

3 → 4 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐, 𝑐𝑜𝑛𝑠𝑡 𝑐𝑣

𝑇4

𝑇3= (

𝑃4

𝑃3)

𝑘−1𝑘

However, P3=P2 and P1=P4, thus 𝑃2

𝑃1=

𝑃3

𝑃4. Combining the two above equations, we get

𝑇2

𝑇1=

𝑇3

𝑇4 [92]

The thermal efficiency for a Brayton Cycle with constant cv is given by

𝜼𝒃𝒓𝒂𝒚𝒕𝒐𝒏 𝒄𝒐𝒏𝒔𝒕 𝒄𝑽

= 𝟏 −𝑻𝟏

𝑻𝟐= 𝟏 −

𝟏

(𝑷𝟐𝑷𝟏

)

𝒌−𝟏𝒌

[93]

The bwr for a Brayton Cycle with constant cv is given by

𝒃𝒘𝒓𝒃𝒓𝒂𝒚𝒕𝒐𝒏 𝒄𝒐𝒏𝒔𝒕 𝒄𝒗=

𝑻𝟐 − 𝑻𝟏

𝑻𝟑 − 𝑻𝟒 [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


34

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


35

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 ��𝑐𝑦𝑐𝑙𝑒 = 0.

Therefore,

��𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 = ��𝑡𝑢𝑟𝑏𝑖𝑛𝑒

𝒉𝟐 − 𝒉𝟏 = 𝒉𝟑 − 𝒉𝟒 [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 = ℎ𝑖𝑛 +𝑉𝑖𝑛

2

2− ℎ𝑜𝑢𝑡 −

𝑉𝑜𝑢𝑡2

2

The diffuser is used to slow the working fluid to zero velocity, V1=0

𝑎 → 1

𝒉𝟏 = 𝒉𝒂 +𝑽𝒂

𝟐

𝟐

[96]


36

The nozzle accelerates the gas to such a speed that the velocity before the process is negligible, V4=0

4 → 5

𝑽𝟓 = √𝟐(𝒉𝟒 − 𝒉𝟓)

[97]

OR with constant cv

𝑽𝟓 = √𝟐 ∗ 𝒄𝒑(𝑻𝟒 − 𝑻𝟓)

[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.

𝑭𝑻 = ��(𝑽𝟓 − 𝑽𝒂) [99]



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39

References [1] All schematics are taken from Prof. Ciccarelli’s notes. Exam questions come from Prof. Ciccarelli.

Documents

MECH 230 Thermodynamics 1 Final Workbook Solutions