MECH 230 – Thermodynamics 1 Final Workbook Solutions

CREATED BY JUSTIN BONAL

MECH 230 Final Exam Workbook

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

MECH 230 Final Exam Workbook

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

MECH 230 Final Exam Workbook

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

iii

MECH 230 Final Exam Workbook

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MECH 230 Final Exam Workbook

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 (휌) kg/m3 Specific Volume V/m || 1/휌 Upsilon (휐) 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 푎푡푚 = 101.326 푃푎 = 760푚푚퐻𝑔|| 1 푏푎푟 = 100,000 푃푎 = 100푘푃푎 = 1푀푃푎 || 1 푚3 = 1000퐿 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.

푷 = 푷 − 푷 품 풂풃풔 풂풕풎 [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].

MECH 230 Final Exam Workbook

퐏 = 퐏 – 퐏 = 훒 ∗ 품 ∗ 풉 퐠 퐚퐛퐬 퐚퐭퐦 퐥퐢퐪퐮퐢퐝 [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,

ퟐ 푾ퟏ→ퟐ = ∫ 푷품풅푽 ퟏ [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.

∆푽 = ퟎ ∴ ∆푾 = ퟎ [4]

When you are told that a system undergoes a polytropic process, that means it behaves according to equation [5].

푷푽풏 = 풄 [5]

MECH 230 Final Exam Workbook

Where P is pressure, V is volume, c is a constant, and n is the polytropic index. From this, it is easy to 푛 푛 see the following relation: 푃1푉1 = 푃2푉2 , since c is a constant. From equation [5], three different ways to calculate work can be derived. Equation [6] shows the work for when 푛 ≠ 1, equation [7] for when n = 1, and equation [8] for when n = 0. When n = 0, this is called a constant pressure process.

푷ퟐ푽ퟐ − 푷ퟏ푽ퟏ 푾 = 풇풐풓 풏 ≠ ퟏ ퟏ→ퟐ ퟏ − 풏 [6]

푽ퟐ 푾ퟏ→ퟐ = 푷ퟏ푽ퟏ 퐥퐧 ( ) 풇풐풓 풏 = ퟏ 푽ퟏ [7]

푾ퟏ→ퟐ = 푷∆푽 풇풐풓 풏 = ퟎ [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, ∆푇 = 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.

∆푻 = ퟎ ∴ ∆푼 = ퟎ ‼ (푰푫푬푨푳 푮푨푺)‼ [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.

2 푄1→2 = ∫ 훿푄 1 [10]

An adiabatic process is a process where there is no heat exchange from the system to the surroundings. Therefore ∆푸 = ퟎ. 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.

퐸 − 퐸 = 푄 − 푊 2 1 [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].

MECH 230 Final Exam Workbook

퐸 − 퐸 = ∆퐾퐸 + ∆푈 + ∆푃퐸 2 1 [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].

∆푼 = 푸 − 푾 ퟏ→ퟐ ퟏ→ퟐ ퟏ→ퟐ [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 퐽 푘퐽 The universal gas constant, 푅̅ = 8314 = 8.314 , is something that is consistent for all 푘푚표푙∗퐾 푘푚표푙 퐾 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

푹̅ 푹 = 흁 [14]

퐽 After dividing by the molar mass, the gas constant has units of . So, your calculations will be in 푘푔∗퐾 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.

푃푉 = 푛푅̅푇 [15]

푷푽 = 푴푹푻 [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).

푃휐 = 푅푇 [17]

푃 = 휌푅푇 [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].

MECH 230 Final Exam Workbook

푴푹(푻ퟐ − 푻ퟏ) 푾 = 풇풐풓 풏 ≠ ퟏ ퟏ→ퟐ ퟏ − 풏 [19]

For an ideal gas going though a polytropic process with n=1 means it is an isothermal process (∆푻 = ퟎ). Therefore equation [7] can be modified to the following.

푽ퟐ 푾ퟏ→ퟐ = 푴푹푻 퐥퐧 ( ) 풇풐풓 풏 = ퟏ 푽ퟏ [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. 푃휐 푅 = 푇 [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.

풏−ퟏ

푻ퟐ 푷ퟐ 풏 = ( ) [22] 푻ퟏ 푷ퟏ

풏−ퟏ 푻ퟐ 흊ퟐ = ( ) [23] 푻ퟏ 흊ퟏ 4.0 Specific Heat Specific heat, c, is a property that represents the amount of heat required to raise the temperature of 1 푘퐽 kg of a substance by 1-degree Kelvin, as seen in equation [24]. The units of specific heat are . 푘푔°퐾 ∆푄 푐 = 푀 ∗ ∆푇 [24]

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

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

∆풖 = 풄 ∆푻 풗 [25]

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

2 2 푢2 − 푢1 = ∫ 푑푢 = ∫ 푐푣(푃, 푇)푑푇 1 1 [26]

MECH 230 Final Exam Workbook

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

풉 = 풄 ∆푻 풑 [27]

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

2 2 ℎ2 − ℎ1 = ∫ 푑ℎ = ∫ 푐푝(푃, 푇)푑푇 1 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

풄풑 풌(푻) = 풄풗 [29]

The following equations are useful for solving specific heats, given one another 푹 풄 (푻) = 풗 푲(푻) − ퟏ [30]

푲(푻)푹 풄 (푻) = 풑 푲(푻) − ퟏ [31]

2013 FINAL QUESTION 1

MECH 230 Final Exam Workbook

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

푑푀퐶푉 = ∑푚̇ − ∑푚̇ [32] 푑푡 푖 푒

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

∑푚̇ 푒 = ∑푚̇ 푖 [33] There are a few different equations that come from mass flow rate. Mass flow rate can also be denoted by

풎̇ = 흆푨푽 [34] Where 휌 is density, A is the cross-sectional area, and V is the velocity perpendicular to the control 푘푔 surface. The mass flux is defined as the mass flow per unit area [ ] 푚2푠 풎̇ 푽 = 흆푽 = [35] 푨 풗 푚3 The volumetric flow rate is defined as [ ] 푠 풎 ̇ = 푽푨 [36] 흆 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, 푄̇ and 푊̇ . 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, 푞̇ and 푤̇ . This is done by dividing 푄̇ and 푊̇ by the mass flow rate, 푚̇ , 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

2 2 푑퐸푐푣 푉 푉푒 = 푄̇ − 푊̇ + ∑푚̇ (ℎ + 푖 + 𝑔푍 ) − ∑푚̇ (ℎ + + 𝑔푍 ) [37] 푑푡 푠ℎ푎푓푡 푖 푖 2 푖 푒 푒 2 푒 푑퐸 For this course, we assume that 푐푣 = 0, that there is only one inlet and one outlet, and that the 푑푡 systems has steady flow, 푚̇ 푖 = 푚̇ 푒 = 푚̇ . Therefore, dividing through by 푚̇ reduces the equation to

ퟐ ퟐ 푽 푽풆 ퟎ = 풒̇ − 풘̇ + (풉 + 풊 + 품풁 ) − (풉 + + 품풁 ) [38] 풊 ퟐ 풊 풆 ퟐ 풆

MECH 230 Final Exam Workbook

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 휌1 = 휌2 for subsonic flow yields the following

푨ퟏ푽ퟏ = 푨ퟐ푽ퟐ [39]

Figure 3: A schematic of a nozzle and diffuser

A diffuser decreases the flow rate of the fluid, therefore 퐴2 > 퐴1 → 푉2 < 푉1.

A nozzle increases the flow rate of the fluid, therefore 퐴1 > 퐴2 → 푉1 < 푉2. Applying the CV energy equation, assuming no heat transfer, work, or change in height, yields the following

ퟐ ퟐ 푽ퟐ = 푽ퟏ + ퟐ(풉ퟏ − 풉ퟐ) [40]

For a rocket nozzle, 푉2 ≫ 푉1, and the above equation simplifies to

2 푉2 = 2(ℎ1 − ℎ2) [41] 6.2 Turbine A turbine is a mechanical device through which shaft work is developed.

MECH 230 Final Exam Workbook

Figure 4: A schematic of a turbine

Applying first law for CV, assuming steady state, no change in PE, and no heat transfer, yields

2 2 푉1 푉2 푤̇ = (ℎ − ℎ ) + ( − ) [42] 1 2 2 2 푉2 푉2 Assuming ℎ − ℎ ≫ 1 − 2 simplifies the above to the following 1 2 2 2

풘̇ = (풉ퟏ − 풉ퟐ) [43]

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

[44] 푊̇ = 푚̇ ∗ 푤̇ = (ℎ1 − ℎ2) 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

2 2 푉1 푉2 푤̇ = (ℎ − ℎ ) + ( − ) [45] 1 2 2 2 Assuming the change in KE is small compared to the change in h

MECH 230 Final Exam Workbook

풘̇ = (풉ퟏ − 풉ퟐ) [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

2 2 푉1 푉2 (ℎ − ℎ ) = ( − ) [47] 1 2 2 2 Looking downstream from each side of the device, we can assume that 푉1 = 푉2. This assumption yields

풉ퟏ = 풉ퟐ [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

MECH 230 Final Exam Workbook

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

Steady flow ∴ 푚̇ 1 = 푚̇ 2 푎푛푑 푚̇ 3 = 푚̇ 4

0 = 푚̇ 1(ℎ1 − ℎ2) + 푚̇ 3(ℎ3 − ℎ4) [50] Which further simplifies to

풎̇ ퟏ 풉ퟒ − 풉ퟑ = [51] 풎̇ ퟑ 풉ퟏ − 풉ퟐ 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 푤표푟푘 푑표푛푒 푊 휂푐푦푐푙푒 = = [52] 푚푎푥𝑖푚푢푚 푤표푟푘 푊푀퐴푋 Applying first law to a cycle, we can see the following

0 = 푄푖푛 − 푄표푢푡 − 푊

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 푊 휂푐푦푐푙푒 = 푄푖푛 [53]

Or further rearranged to the following

푄푖푛 − 푄표푢푡 푄표푢푡 휂푐푦푐푙푒 = = 1 − [54] 푄푖푛 푄푖푛 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

휂표푣푒푟푎푙푙 = 휂푐푦푐푙푒휂푚푒푐ℎ푎푛푖푐푎푙 [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 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 훿푄 푆 − 푆 = ∫ + 푆 [56] 2 1 푇 푔푒푛 훿푄 Where S2 -S1 is the change in entropy of the system, ∫ is the entropy transfer to the system by heat, 푇 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, ∆푄 = 0

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

∴ 푆1 − 푆2 = 0, 푆1 = 푆2. 7.3.1 Entropy Change for Ideal Gas We know that for an ideal gas 푑푢 = 푐푣(푇)푑푇, 푑ℎ = 푐푝(푇)푑푇, 푃휈 = 푅푇. Plugging these equations into the above yields the following

2 푐푉(푇)푑푇 휈2 푠2(푇2, 휈2) − 푠1(푇1, 휈1) = ∫ + 푅푙푛( ) [57] 1 푇 휈1 And

2 푐푃(푇)푑푇 푃2 푠2(푇2, 푃2) − 푠1(푇1, 푃1) = ∫ − 푅푙푛( ) [58] 1 푇 푃1 2 푐 (푇)푑푇 The bounds of ∫ 푃 are tabulated as a function of T in air in Table A-22 under the label 푠°. So, 1 푇 2 푐 (푇)푑푇 ∫ 푃 = 푠° − 푠° . Therefore equation [57] can be written as 1 푇 2 1

MECH 230 Final Exam Workbook

푷ퟐ 풔ퟐ − 풔ퟏ = (풔°ퟐ − 풔°ퟏ) − 푹풍풏( ) [59] 푷ퟏ If cv and cp are taken as constant, changes in entropy are calculated by the following

푻ퟐ 흂ퟐ 풔ퟐ − 풔ퟏ = 풄풗 퐥퐧 ( ) + 푹풍풏 ( ) [60] 푻ퟏ 흂ퟏ And

푻ퟐ 푷ퟐ 풔ퟐ − 풔ퟏ = 풄풗 퐥퐧 ( ) − 푹풍풏 ( ) [61] 푻ퟏ 푷ퟏ

7.3.2 Isentropic Processes for Ideal Gas A process that is both adiabatic and reversible is called an isentropic process. In this case, ∆푄 = 0

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

∴ 푆1 − 푆2 = 0, 푆1 = 푆2.

Plugging S1=S2 into equation [27]

푠°2 − 푠°1 푃2 = ln ( ) 푅 푃1 푠° exp ( 2) 푃2 푅 = 푠° 푃1 exp ( 1) 푅 풔° Now we define relative pressure, 푷 = 풆풙풑( ). Therefore 풓 푹 푷 푷 ( ퟐ) = 풓ퟐ (풔 = 풄풐풏풔풕) 푷ퟏ 푷풓ퟏ [62]

From the ideal gas law, 푇 ( 2 ) 휈2 푇2 푃1 푃 = ( ) = 푟2 푇 휈1 푇1 푃2 ( 1 ) Pr1

Define vr =T/Pr, therefore

흂ퟐ 풗풓ퟐ ( ) = [63] 흂ퟏ 흂풓ퟏ 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

풌−ퟏ 푻ퟐ 흂ퟏ ( ) = ( ) , 풔 = 풄풐풏풔풕, 풄풑 = 풄풐풏풔풕 [64] 푻ퟏ 흂ퟐ 푇 푃 휈 Substituting in 2 = 2 2, 푇1 푃1휈1

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풌 푷ퟐ 흂ퟏ ( ) = ( ) , 풔 = 풄풐풏풔풕, 풄풑 = 풄풐풏풔풕 [65] 푷ퟏ 흂ퟐ For a polytropic process, 푃휈푛 = 푐표푛푠푡. For an isentropic process, n=k. Combining equations yields

1 1 휈1 푃2 푘 푇2 푘−1 ( ) = ( ) = ( ) 휈2 푃1 푇1

풌−ퟏ 푻ퟐ 푷ퟐ 풌 [66] ( ) = ( ) , 풔 = 풄풐풏풔풕, 풄풑 = 풄풐풏풔풕 푻ퟏ 푷ퟏ 8.5 Control Volume Entropy Balance The rate of entropy change in a control volume is given by ̇ 푑푆퐶푉 푄푗 = ∑푚̇ 푖푠푖 − ∑푚̇ 푒푠푒 [67] 푑푡 푇푗 Assuming steady state, one inlet one outlet, and isothermal CV, the above simplifies to ̇ ퟏ 푸̇ 풄풗 푺품풆풏 [68] ퟎ = ( ) + 풔 − 풔 + 풎̇ 푻 풊풏 풐풖풕 풎̇ 8.5.1 Isentropic Efficiencies of Turbines and Compressors 푊̇ Recall for a turbine 푐푣 = ℎ − ℎ > 0, since turbines create work. For a real turbine it is impossible to 푚̇ 1 2 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

푊 We know that the efficiency of a turbine is given as 휂 = 푟푒푎푙. Now we know that the maximum work of 푊푚푎푥 푊̇ a turbine is given as 푐푣 = ℎ − ℎ . 푚̇ 1 2푠 Plugging in, we get the isentropic turbine efficiency

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풉ퟏ − 풉ퟐ 휼풕 = [69] 풉ퟏ − 풉ퟐ풔 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

푊 We know that the compressor efficiency is given by 휂 = 푚푎푥 . We know that 푊 = ℎ − ℎ . From 푐 푊 푚푎푥 1 2푠 this, the isentropic compressor efficiency is given by

풉ퟏ − 풉ퟐ풔 휼풄 = [70] 풉ퟏ − 풉ퟐ 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

2 푊̇푐푣 = − ∫ 휈푑푃 [71] 푚̇ 1 Liquids are incompressible, so there is no change in the specific volume 휈1 = 휈2 = 휈. ∴

푾̇ 풄풗 = −흂(푷 − 푷 ) [72] 풎̇ ퟐ ퟏ For a gas going through a polytropic process, with 푃휈 = 푅푇

푾̇ 풄풗 풏푹푻ퟏ 푻ퟐ = − ( − ퟏ) , 풏 ≠ ퟏ [73] 풎̇ 풏 − ퟏ 푻ퟏ 푛−1 푇 푃 OR substituting ( 2) = ( 2) 푛 we get 푇1 푃1

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푛−1 푊̇푐푣 푛푅푇1 푃2 푛 = − (( ) − 1) , 푛 ≠ 1 [74] 푚̇ 푛 − 1 푃1

If a process is isothermal (T1 = T2)

푾̇ 풄풗 푷ퟐ = −푹푻풍풏( ) [75] 풎̇ 푷ퟏ

2015 FINAL QUESTION 2

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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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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

푄̇푐푣 푊̇푐푣 0 = − + (ℎ − ℎ ) 푚̇ 푚̇ 푖푛 표푢푡 Now we analyze each step individually.

1 → 2 퐴푑𝑖푎푏푎푡𝑖푐 푇푢푟푏𝑖푛푒 퐸푥푝푎푛푠𝑖표푛

Adiabatic so 푄̇ = 0.

∴ 푤표푢푡 = ℎ1 − ℎ2 2 → 3 퐶표푛푑푒푛푠표푟 (푛표 푤표푟푘)

No work so 푊̇ = 0.

∴ 푞표푢푡 = ℎ2 − ℎ3 3 → 4 퐴푑𝑖푎푏푎푡𝑖푐 푃푢푚푝 퐶표푚푝푟푒푠푠𝑖표푛

Adiabatic so 푄̇ = 0

∴ 푤푖푛 = ℎ4 − ℎ3 4 → 1 푆푡푒푎푚 퐺푒푛푒푟푎푡표푟 (푛표 푤표푟푘)

No work so 푊̇ = 0

∴ 푞푖푛 = ℎ1 − ℎ4 9.1 Thermal Efficiency The thermal efficiency of a heat engine is defined by 푛푒푡 푤표푟푘 표푢푡 휂 = ℎ푒푎푡 𝑖푛푝푢푡

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

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푤표푢푡 − 푤푖푛 (ℎ1 − ℎ2) − (ℎ4 − ℎ3) 휂푟푎푛푘푖푛푒 = = [76] 푞푖푛 ℎ1 − ℎ4 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

풘풊풏(풑풖풎풑) 풉ퟒ − 풉ퟑ 풃풘풓풓풂풏풌풊풏풆 = = [77] 풘풐풖풕(풕풖풓풃풊풏풆) 풉ퟏ − 풉ퟐ 9.3 Ideal Rankine Cycle In the ideal Rankine Cycle, no irreversibility’s occur. The resulting cycle is the following

1 → 2 퐼푠푒푛푡푟표푝𝑖푐 퐸푥푝푎푛푠𝑖표푛 푎푡 푇푢푟푏𝑖푛푒 2 → 3 퐶표푛푑푒푛푠푎푡𝑖표푛 푎푡 퐶표푛푠푡푎푛푡 푃푟푒푠푠푢푟푒 3 → 4 퐼푠푒푛푡푟표푝𝑖푐 퐶표푚푝푟푒푠푠𝑖표푛 푎푡 푃푢푚푝 4 → 1 푆푡푒푎푚 퐺푒푛푒푟푎푡𝑖표푛 푎푡 퐶표푛푠푡푎푛푡 푃푟푒푠푠푢푟푒 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

4 푤푖푛 = − ∫ 휈푑푃 3 If the working fluid is a pure liquid then the specific volume can be assumed to be constant, yielding the following

푤푖푛 = 휈3(푃4 − 푃3) Like the thermal efficiency of a Carnot Cycle, the ideal Rankine Efficiency is given by

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푻풐풖풕 휼풊풅풆풂풍 풓풂풏풌풊풏풆 = ퟏ − [78] 푻풊풏 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

풘풊풏 흂ퟑ(푷ퟒ − 푷ퟑ) 풃풘풓풊풅풆풂풍 풓풂풏풌풊풏풆 = = [79] 풘풐풖풕 (풉ퟏ − 풉ퟐ풔) 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 (풘 + 풘 ) − 풘 휼 = ퟏ→ퟐ ퟐ→ퟑ ퟓ→ퟔ 풓풂풏풌풊풏풆 풓풆풉풆풂풕 (풒 + 풒 ) ퟔ−ퟏ ퟐ→ퟑ [80] (풉ퟏ − 풉ퟐ) + (풉ퟑ − 풉ퟒ) − (풉ퟔ − 풉ퟓ) = (풉ퟏ − 풉ퟔ) + (풉ퟑ − 풉ퟐ) 2015 FINAL QUESTION 3

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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).

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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, 푾̇ , (or horsepower) is the amount of power outputted by an engine. It is defined by 푵 푾̇ = 푾 ∗ [81] 풄풚풄풍풆 ퟐ Where N is the crank shaft speed, in revolutions per second. A useful conversion to have on your formula sheet is

ퟏ 풉풐풓풔풆풑풐풘풆풓 = ퟕퟒퟔ 푾풂풕풕풔 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

풏풆풕 풘풐풓풌 풑풆풓 풄풚풄풍풆 푾풄풚풄풍풆 풎풆풑 = = [82] 풅풊풔풑풍풂풄풆풅 풗풐풍풖풎풆 푽ퟏ − 푽ퟐ

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

푉1 푉4 푟 = = [83] 푉2 푉3

Or, since one of the assumptions is fixed mass

흂ퟏ 흂ퟒ 풓 = = [84] 흂ퟐ 흂ퟑ

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

푤푖푛 = 푢2 − 푢1 Isentropic, so we can use relative specific volumes found in table A-22 and the compression ratio

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푣푟2 휈2 1 = = 휈푟1 휈1 푟 And since the working fluid is an ideal gas

푃2휈2 푃1휈1 푃2 푇2 휈1 푇2 = → = ∗ = ∗ 푟 푇2 푇1 푃1 푇1 휈2 푇1 2 → 3 퐶표푛푠푡푎푛푡 푣표푙푢푚푒 ℎ푒푎푡 푎푑푑𝑖푡𝑖표푛 Constant volume so W = 0. Applying first law,

푞푖푛 = 푢3 − 푢2 Ideal gas with constant volume, so

푃3 푇3 = 푃2 푇2 3 → 4 퐼푠푒푛푡푟표푝𝑖푐 퐸푥푝푎푛푠𝑖표푛 Isentropic, Q = 0. Expansion, work is positive. Applying first law,

푤표푢푡 = 푢3 − 푢4 Reversible process with compression ratio

휈푟4 휈4 = = 푟 휈푟3 휈3 Ideal gas, so

푃4 푇4 휈3 푇4 1 = ∗ = ∗ 푃3 푇3 휈4 푇3 푟 4 → 1 퐶표푛푠푡푎푛푡 푣표푙푢푚푒 ℎ푒푎푡 푟푒푚표푣푎푙 Constant volume, W = 0. Heat removal, Q is negative. Applying first law,

푞표푢푡 = 푢4 − 푢1 Ideal gas with constant volume, so

푃4 푃1 = 푇4 푇1 10.2.1 Otto Cycle Thermal Efficiency Like other heat engines, the efficiency of the cycle is given by

푤표푢푡 − 푤푖푛 휂푐푦푐푙푒 = 푞푖푛 So, for an Otto Cycle the efficiency is

(풖ퟑ − 풖ퟒ) − (풖ퟐ − 풖ퟏ) 풖ퟒ − 풖ퟏ 휼풐풕풕풐 = = ퟏ − [85] (풖ퟑ − 풖ퟐ) 풖ퟑ − 풖ퟐ 24

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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

푘−1 푇 휈 푘−1 푇 푃 2 = ( 1) = 푟푘−1 and 2 = ( 2) 푘 푇1 휈2 푇1 푃1 For process 3->4 we know

푘−1 푇 휈 푘−1 1 푇 푃 4 = ( 3) = ( )푘−1 and 4 = ( 4) 푘 푇3 휈4 푟 푇3 푃3

From equation [84], and using the fact that ∆푢 = 푐푉∆푇 for constant cv gives us 푻 ퟏ 휼 = ퟏ − ퟏ = ퟏ − 풐풕풕풐 풄풐풏풔풕 풄푽 풌−ퟏ [86] 푻ퟐ 풓 2015 FINAL QUESTION 4

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10.2 Air Standard Diesel Cycle 휈 The diesel cycle is like the Otto cycle in many ways. It has the same compression ratio, 푟 = 1. However, 휈2 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

푽ퟑ 흂ퟑ 풓풄 = = [87] 푽ퟐ 흂ퟐ The cycle goes through four different processes

1 → 2 퐼푠푒푛푡푟표푝𝑖푐 퐶표푚푝푟푒푠푠𝑖표푛 Isentropic, Q=0. Compression, W is negative. Applying first law,

푤푖푛 = 푢2 − 푢1 Isentropic, so we can use relative specific volumes found in table A-22 and the compression ratio

푣푟2 휈2 1 = = 휈푟1 휈1 푟 And since the working fluid is an ideal gas

푃2휈2 푃1휈1 푃2 푇2 휈1 푇2 = → = ∗ = ∗ 푟 푇2 푇1 푃1 푇1 휈2 푇1 2 → 3 퐶표푛푠푡푎푛푡 푃푟푒푠푠푢푟푒 퐻푒푎푡 퐴푑푑𝑖푡𝑖표푛

Heat addition, Q is positive. Constant pressure, so 푊 = 푃2(푉3 − 푉2). Applying first law,

푢3 − 푢2 = 푞푖푛 − 푃2(푉3 − 푉2). Rearranging

푞푖푛 = (푢3 + 푃3휈3) − (푢2 + 푃2휈2) Using definition h = u + Pv

푞푖푛 = ℎ3 − ℎ2 Ideal gas with constant pressure

푇3 휈3 = = 푟푐 푇2 휈2 3 → 4 퐼푠푒푛푡푟표푝𝑖푐 퐸푥푝푎푛푠𝑖표푛 Isentropic, Q = 0. Expansion, work is positive. Applying first law,

푤표푢푡 = 푢3 − 푢4 Using relative specific volumes

휈푟4 휈4 = 휈푟3 휈3

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We know that 휈4 = 휈1

휈4 휈4 휈2 휈1 휈2 푟 = ∗ = ∗ = 휈3 휈2 휈3 휈2 휈3 푟푐

휈푟4 휈4 푟 = = 휈푟3 휈3 푟푐 Applying ideal gas law

푃4 푇4 푟푐 = ∗ 푃3 푇3 푟 4 → 1 퐶표푛푠푡푎푛푡 푉표푙푢푚푒 퐻푒푎푡 푅푒푚표푣푎푙 Constant volume, W = 0. Heat removal, Q is negative. First law

푞표푢푡 = 푢4 − 푢1 Ideal gas with constant volume, so

푃4 푃1 = 푇4 푇1 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 푤 Like all other cycles, the efficiency is defined by 휂 = 푐푦푐푙푒. Plugging in for the equations solved above 푞푖푛 yields the diesel cycle thermal efficiency

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풖ퟒ − 풖ퟏ 휼풅풊풆풔풆풍 = ퟏ − [88] 풉ퟑ − 풉ퟐ 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

푘−1 푇 휈 푘−1 푇 푃 2 = ( 1) = 푟푘−1 and 2 = ( 2) 푘 푇1 휈2 푇1 푃1 For process 3->4 we know

푘−1 푇 휈 푘−1 1 푇 푃 4 = ( 3) = ( )푘−1 and 4 = ( 4) 푘 푇3 휈4 푟 푇3 푃3

Using the definitions ∆푢 = 푐푣∆푇 and ∆ℎ = 푐푝∆푇, and applying it to equation 13 yields the efficiency for a diesel cycle with constant specific heats to be

ퟏ ퟏ (풓풌 − ퟏ) 휼 = ퟏ − [ ∗ 풄 ] 풅풊풆풔풆풍 풄풐풏풔풕 풄푽 풌−ퟏ 풓 풌 (풓풄 − ퟏ) [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 = 푞̇ − 푤̇ + ℎ푖푛 − ℎ표푢푡. For each step of the process

1 → 2 퐴푑𝑖푎푏푎푡𝑖푐 퐶표푚푝푟푒푠푠𝑖표푛 Adiabatic so Q = 0. Compression so W is negative

푤̇ 푖푛 = ℎ2 − ℎ1 2 → 3 퐻푒푎푡 퐴푑푑𝑖푡𝑖표푛 (푛표 푤표푟푘) W=0

푞̇푖푛 = ℎ3 − ℎ2 3 → 4퐴푑𝑖푎푏푎푡𝑖푐 퐸푥푝푎푛푠𝑖표푛 Adiabatic so Q = 0. Expansion so W is positive

푤̇ 표푢푡 = ℎ3 − ℎ4 4 → 1 퐻푒푎푡 푅푒푚표푣푎푙 (푛표 푤표푟푘)

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W=0

푞̇표푢푡 = ℎ4 − ℎ1

푞̇표푢푡 The thermal efficiency for a cycle is given by 휂푐푦푐푙푒 = 1 − . Plugging in gives the Brayton cycle 푞̇푖푛 thermal efficiency

풉ퟒ − 풉ퟏ 휼풃풓풂풚풕풐풏 = ퟏ − [90] 풉ퟑ − 풉ퟐ The back-work ratio for the Brayton cycle is given by

풉ퟐ − 풉ퟏ 풃풘풓풃풓풂풚풕풐풏 = [91] 풉ퟑ − 풉ퟒ 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 퐼푠푒푛푡푟표푝𝑖푐

푃2 푃푟2 = 푃푟1 ( ) 푃1 3 → 4 퐼푠푒푛푡푟표푝𝑖푐

푃4 푃푟4 = 푃푟3 ( ) 푃3 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 퐼푠푒푛푡푟표푝𝑖푐, 푐표푛푠푡 푐푣

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푘−1 푇2 푃2 푘 = ( ) 푇1 푃1

3 → 4 퐼푠푒푛푡푟표푝𝑖푐, 푐표푛푠푡 푐푣

푘−1 푇4 푃4 푘 = ( ) 푇3 푃3

푃2 푃3 However, P3=P2 and P1=P4, thus = . Combining the two above equations, we get 푃1 푃4

푇2 푇3 = [92] 푇1 푇4

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

푻ퟏ ퟏ 휼 = ퟏ − = ퟏ − 풃풓풂풚풕풐풏 풄풐풏풔풕 풄푽 푻 풌−ퟏ ퟐ 푷 풌 [93] ( ퟐ) 푷ퟏ The bwr for a Brayton Cycle with constant cv is given by

푻ퟐ − 푻ퟏ 풃풘풓풃풓풂풚풕풐풏 풄풐풏풔풕 풄풗 = [94] 푻ퟑ − 푻ퟒ 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

MECH 230 Final Exam Workbook

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 푊̇ 푐푦푐푙푒 = 0. Therefore,

푤̇ 푐표푚푝푟푒푠푠표푟 = 푤̇ 푡푢푟푏푖푛푒

풉ퟐ − 풉ퟏ = 풉ퟑ − 풉ퟒ [95] Generally, the process through the diffuser and the nozzle are taken as isentropic. Applying conservation of energy to the Diffuser and Nozzle yields

2 2 푉 푉표푢푡 0 = ℎ + 푖푛 − ℎ − 푖푛 2 표푢푡 2

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

푎 → 1

ퟐ 푽풂 풉 = 풉 + ퟏ 풂 ퟐ [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 → 5

푽 = √ퟐ(풉 − 풉 ) ퟓ ퟒ ퟓ [97]

OR with constant cv

푽 = √ퟐ ∗ 풄 (푻 − 푻 ) ퟓ 풑 ퟒ ퟓ [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.

푭푻 = 풎̇ (푽ퟓ − 푽풂) [99]

2013 FINAL QUESTION 4

MECH 230 Final Exam Workbook

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