engineering thermodynamics module-ii lecture-9 · calls for higher rates of heat transfer in boiler...

Prepared by Prof. (Dr.) Manmatha K. Roul ENGINEERING THERMODYNAMICS

MODULE-II

LECTURE-9

Prepared By

Prof. (Dr.) Manmatha K. Roul Professor and Principal

Gandhi Institute for Technological Advancement (GITA), Bhubaneswar – 752054

June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Comparison between Rankine Cycle and Carnot Cycle

Carnot vapour Cycle:-

Process 1-2: Reversible adiabatic compression process from P1 to P2. Process 2-3: Reversible isothermal heat addition process at constant temperature TH. Process 3-4: Reversible adiabatic expansion process from P3 to P4. Process 4-1: Reversible isothermal heat rejection process at constant temperature TL. Saturated vapor leaves the boiler at state 3, enters the turbine and expands to state 4. The fluid then enters the condenser, where it is cooled to state 1 and then it is compressed to state 2 in the pump. The efficiency of the cycle is as follows:

Practically, it is very difficult to add or reject heat to or from the working fluid at constant temperature. But, it is comparatively easy to add or reject heat to or from the working fluid at constant pressure. Therefore, Carnot cycle is not used as an idealized cycle for steam power plants. However, ideal cycle for steam power plant is Rankine cycle in which heat addition and rejection takes place at constant pressure process. Comparison between Rankine Cycle and Carnot Cycle :- The following points are worth noting : (i) Between the same temperature limits Rankine cycle provides a higher specific work output than a Carnot cycle, consequently Rankine cycle requires a smaller steam flow rate resulting in smaller size plant for a given power output. However, Rankine cycle calls for higher rates of heat transfer in boiler and condenser (ii) Since in Rankine cycle only part of the heat is supplied isothermally at constant higher temperature T1, therefore, its efficiency is lower than that of

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00

100

200

300

400

500

600

700700

s [kJ/kg-K]

T [C

]

6000 kPa

100 kPa

Carnot Vapor Cycle Using Steam

1

2 3

4

Prepared by Prof. (Dr.) Manmatha K. Roul Carnot cycle. The efficiency of the Rankine cycle will approach that of the Carnot cycle more nearly if the superheat temperature rise is reduced. (iii) The advantage of using pump to feed liquid to the boiler instead to compressing a wet vapour is obvious that the work for compression is very large compared to the pump.

Acual vapour power cycle with all the irriversibilities T-s diagram of Actual vapour power cycle:- .

Prepared by Prof. (Dr.) Manmatha K. Roul Turbine Losses: During the expansion of steam in the turbine there will be heat transfer to the Surroundings and the expansion instead of being isentropic will be polytropic.

Efficiency of turbine= Ideal enthalpy drop/ Actual enthalpy drop

Prepared by Prof. (Dr.) Manmatha K. Roul Page 1

ENGINEERING THERMODYNAMICS

MODULE-II

LECTURE-10

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 Factors Affecting Cycle Efficiency:-

We know that in

out

in

outin

in

inout

qq

qqq

qww

−=−

=−

= 1η

Andfor a reversible heat addition process ∫= Tdsq

Considering qin at the boiler and qout at the condenser

area shaded

1

414

=

== ∫→ Tdsqq in

so mean temperature for process 4 � 141

1

4

ss

TdsTin −

=∫

( )41

1

4

1

4ssTdsTTdsq ininin −===∴ ∫∫

( )area shaded

32

3

232

=

−=

== ∫→

ss T

Tdsqq

out

out

T

s

4

1 qin inT

T

s

3 2 qout Tout

Prepared by Prof. (Dr.) Manmatha K. Roul Page 3 Noting 4132 ssss −=− , the Ideal Rankine cycle thermal efficiency is

in

out

in

out

in

out

RankineIdeal T

TssTssT

qq

−=−

−−=−= 1

)()(11

41

32η =1-Tout/Tmean

Note: this is identical to the Carnot Engine efficiency which is also a reversible cycle. As Tout can be reduced to a minimum value i.e. up to atmospheric condition so efficiency of Rankine cycle is a function of temperature only. Cycle efficiency can be improved by either:

- increasing the average temperature during heat addition ( inT ) - decreasing the condenser temperature (Tout)

Increasing the amount of superheat (4�1’) Amount of superheating is limited by metallurgical considerations of the turbine (T1 < 670C)

Added benefit is that the quality of the steam at the turbine exit is higher

1 ’ ’ 2

Prepared by Prof. (Dr.) Manmatha K. Roul Page 4 Increasing the boiler pressure (4 � 1’) Disadvantages:

• Requires more robust equipment • Vapour quality at 2’ lower than at 2

Decrease Condenser Pressure (2’ � 3’)

Tout is limited to the temperature of the cooling medium (e.g., lake at 15C need 10C temperature difference for heat transfer so Tout >25C)

’ ’ ’ ’ ’ ’

Prepared by Prof. (Dr.) Manmatha K. Roul Page 5 Disadvantages: Note: for water Psat(25C)= 3.2 kPa lower than atmospheric, possible air leakage into lines Vapor quality lower at lower pressure not good for turbine Ways to improve the simple Rankine cycle efficiency:- • By Superheating the vapor Higher average temperature during heat addition Reduces moisture at turbine exit (we want x4 in the above example > 85%) • By Increasing the boiler pressure (for fixed maximum temperature)

Availability of steam is higher at higher pressures Increases the moisture at turbine exit • By Lowering the condenser pressure

Less energy is lost to surroundings Increases the moisture at turbine exit

So to increase the efficiency, Rankien cycle should be modified.

Prepared by Prof. (Dr.) Manmatha K. Roul Page 1 ENGINEERING THERMODYNAMICS MODULE-II LECTURE-11 Prepared By Prof. (Dr.) Manmatha K. Roul Professor and Principal Gandhi Institute for Technological Advancement (GITA), Bhubaneswar – 752054 June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 THE REHEAT CYCLE

It is desirable to keep the steam as dry as possible in the lower pressure stages of the turbine. The wetness at exhaust should be not greater than 10%. The increasing of boiler pressure increases the thermal efficiency, but it also increases the wetness of the exhaust steam to unacceptable levels. The exhaust steam can be improved by REHEAT the steam, the expansion being carried out in two stages. The reheat can be carried out by returning the steam to the boiler, and passing it through a special bank of tubes, the reheat bank of tubes being situated in the proximity of the superheat tubes. Schematic diagram and T-s diagram:-

Cycle Analysis:- η= Net work output/ Heat input Work out put= (h3-h4) + (h5-h6) Work input= (h2-h1) Heat input= (h3-h2) + ( h5-h4) η=(h3-h4) + (h5-h6)- (h2-h1)/ (h3-h2) + ( h5-h4) Conclusion:- The single reheating in a modern power plant improves the cycle efficiency by 4 to 5% by increasing the average temperature at which heat is transferred to the steam. The average temperature during the reheat process can be increased by increasing the number of expansion and reheat stages. As the number of stages is increased, the expansion and reheat processes approach an isothermal process at the maximum temperature. The use of more than two reheat stages is not practical. The theoretical improvement in efficiency from the second reheat is

Prepared by Prof. (Dr.) Manmatha K. Roul Page 3 about half of that which results from a single reheat.The reheat temperatures are very close or equal to the turbine inlet temperature. The optimum reheat pressure is about one-fourth of the maximum cycle pressure. Point to remember: - The sole purpose of the reheat cycle is to reduce the moisture content of the steam at the final stages of the expansion process. If we had materials that could withstand sufficiently high temperatures, there would be no need for the reheat cycle.



MODULE-II

LECTURE-12

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 THE REGENERATIVE CYCLE:- A practical regeneration process in steam power plants is accomplished by extracting, or “bleeding,” steam from the turbine at various points. This steam, which could have produced more work by expanding further in the turbine, is used to heat the feedwater instead. The device where the feedwater is heated by regeneration is called a regenerator, or a feedwater heater (FWH). A feedwater heater is basically a heat exchanger where heat is transferred from the steam to the feedwater either by mixing the two fluid streams (open feedwater heaters) or without mixing them (closed feedwater heaters). Open Feedwater Heaters:- An open (or direct-contact) feedwater heater is basically a mixing chamber, where the steam extracted from the turbine mixes with the feedwater exiting the pump. Ideally, the mixture leaves the heater as a saturated liquid at the heater pressure. The schematic of a steam power plant with one open feedwater heater (also called single-stage regenerative cycle) and the T-s diagram of the cycle are shown in Fig below. Schematic diagram and T-s diagram:-

Prepared by Prof. (Dr.) Manmatha K. Roul Page 3 Cycle Analysis:- Conclusion:- The thermal efficiency of the Rankine cycle increases as a result of regeneration. This is because regeneration raises the average temperature at which heat is transferred to the steam in the boiler by raising the temperature of the water before it enters the boiler. The cycle efficiency increases further as the number of feedwater heaters is increased. Many large plants in operation today use as many as eight feedwater heaters. The optimum number of feedwater heaters is determined from economical considerations. The use of an additional feedwater heater cannot be justified unless it saves more from the fuel costs than its own cost. Closed Feedwater Heaters:- Another type of feedwater heater frequently used in steam power plants is the closed feedwater heater, in which heat is transferred from the extracted steam to the feedwater without any mixing taking place. The two streams now can be at different pressures, since they do not mix. The schematic of a steam power plant with one closed feedwater heater and the T-s diagram of the cycle are shown in fig. In an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. In actual power plants, the feedwater leaves the heater below the exit temperature of the extracted steam because a temperature difference of at least a few degrees is required for any effective heat transfer to take place. The condensed steam is then either pumped to the feedwater line or routed to another heater or to the condenser through a device

Prepared by Prof. (Dr.) Manmatha K. Roul Page 4 called a trap. A trap allows the liquid to be throttled to a lower pressure region but traps the vapor. The enthalpy of steam remains constant during this throttling process. Comparison between open and closed Feed water heater:- Open feedwater heaters are simple and inexpensive and have good heat transfer characteristics. They also bring the feedwater to the saturation state. For each heater, however, a pump is required to handle the feedwater. The closed feedwater heaters are more complex because of the internal tubing network, and thus they are more expensive. Heat transfer in closed feedwater heaters is also less effective since the two streams are not allowed to be in direct contact. However, closed feedwater heaters do not require a separate pump for each heater since the extracted steam and the feedwater can be at different pressures. Most steam power plants use a combination of open and closed feedwater heaters.



MODULE-II

LECTURE-13

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 COGENERATION PLANT:- In all the cycles discussed so far, the sole purpose was to convert a portion of the heat transferred to the working fluid to work, which is the most valuable form of energy. The remaining portion of the heat is rejected to rivers, lakes, oceans, or the atmosphere as waste heat, because its quality (or grade) is too low to be of any practical use. Wasting a large amount of heat is a price we have to pay to produce work, because electrical or mechanical work is the only form of energy on which many engineering devices (such as a fan) can operate. Many systems or devices, however, require energy input in the form of heat, called process heat. Some industries that rely heavily on process heat are chemical, pulp and paper, oil production and refining, steel making, food processing, and textile industries. Process heat in these industries is usually supplied by steam at 5 to 7 atm and 150 to 200°C (300 to 400°F). Energy is usually transferred to the steam by burning coal, oil, natural gas, or another fuel in a furnace. Industries that use large amounts of process heat also consume a large amount of electric power. Therefore, it makes economical as well as engineering sense to use the already-existing work potential to produce power instead of letting it go to waste. The result is a plant that produces electricity while meeting the process-heat requirements of certain industrial processes. Such a plant is called a cogeneration plant. In general, cogeneration is the production of more than

one useful form of energy (such as process heat and electric power) from

the same energy source. Generally two types of turbines are used in a Co-generation plant. (1) Back pressure turbine (2) Pass-out turbine

Prepared by Prof. (Dr.) Manmatha K. Roul Page 3 Back pressure Turbine:-

Passout turbine:-

Prepared by Prof. (Dr.) Manmatha K. Roul Page 4 Conclusions:- The use of cogeneration dates to the beginning of this century when power plants were integrated to a community to provide district heating, that is, space, hot water, and process heating for residential and commercial buildings. The district heating systems lost their popularity in the 1940s owing to low fuel prices. However, the rapid rise in fuel prices in the 1970s brought about renewed interest in district heating. Cogeneration plants have proved to be economically very attractive. Consequently, more and more such plants have been installed in recent years, and more are being installed.



MODULE-II

LECTURE-14

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 BINARY VAPOUR CYCLE Characteristics of an Ideal working fluid:- The maximum temperature that can be used in steam cycles consistent with the best available material is about 6000C, while the critical temperature of steam is 3750C, which necessitates large superheating and permits the addition of only an infinitesimal amount of heat at the highest temperature. The desirable characteristics of the working fluid in a vapour power cycle to obtain best thermal efficiency are as follows: a) The fluid should have a high critical temperature so that the saturation pressure at the maximum permissible temperature (metallurgical limit) is relatively low. It should have a large enthalpy of evaporation at that pressure. b) The saturation pressure at the temperature of heat rejection should be above atmosphere pressure so as to avoid the necessity of maintaining vacuum in the condenser. c) The specific heat of liquid should be small so that little heat transfer is required to raise the liquid to the boiling point. d) The saturation vapour line of the T-S diagram should be steep, very close to the turbine expansion process so that excessive moisture does not appear during expansion. e) The freezing point of the fluid should be below room temperature, so that it does not get solidified while flowing through the pipe lines. f) The fluid should be chemically stable and should not contaminate the materials of construction at any temperature. g) The fluid should be nontoxic, non corrosive, not excessively viscous, and low in cost Not surprisingly, no fluid possesses all these characteristics. Water comes the closest, although it does not fare well with respect to characteristics 1, 3,and 5. We can cope with its sub atmospheric condenser pressure by careful sealing, and with the inverted V-shaped saturation dome by reheating, but there is not much we can do about point 1. Water has a low critical temperature (374°C, well below the metallurgical limit) and very high saturation pressures at high temperatures (16.5 MPa at 350°C). Well, we cannot change the way water behaves during the high-temperature part of the cycle, but we certainly can replace it with a more suitable fluid.The result is a power cycle that is actually a combination of two cycles, one in the high-temperature region and the other in the low-temperature region. Such a cycle is called a binary vapor cycle. In binary vapor cycles, the condenser of the high-temperature cycle, also called as topping cycle serves as the boiler of the low temperature cycle, also called the bottoming cycle).That is, the heat output of the high-temperature cycle is used as the heat input. Some working fluids found suitable for the high-temperature

Prepared by Prof. (Dr.) Manmatha K. Roul Page 3 cycle are mercury,sodium, potassium, and sodium–potassium mixtures. The schematic and T-s diagram for a mercury–water binary vapor cycle are shown in Fig. 10–26.The critical temperature of mercury is 898°C (well above the current metallurgical limit), and its critical pressure is only about 18 MPa. This makes mercury a very suitable working fluid for the topping cycle. Mercury is not suitable as the sole working fluid for the entire cycle, however, since at a condenser temperature of 32°C its saturation pressure is 0.07 Pa. A power plant cannot operate at this vacuum because of air-leakage problems. At an acceptable condenser pressure of 7 kPa, the saturation temperature of mercury is 237°C, which is too high as the minimum temperature in the cycle. Therefore, the use of mercury as a working fluid is limited to the high-temperature cycles. Other disadvantages of mercury are its toxicity and high cost. The mass flow rate of mercury in binary vapor cycles is several times that of water because of its low enthalpy of vaporization Schematic diagram and T-s diagram:- Conclusion:- Studies show that thermal efficiencies of 50 percent or higher are possible with binary vapour cycles. However, binary vapor cycles are not economically attractive because of their high initial cost and the competition offered by the combined gas–steam power plants.



MODULE-II

LECTURE-15

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 GAS POWER CYCLE

Air Standard Cycles The air standard cycle is a cycle followed by a heat engine which uses air as the working medium. Since the air standard analysis is the simplest and most idealistic, such cycles are also called ideal cycles and the engine running on such cycles are called ideal engines. In order that the analysis is made as simple as possible, certain assumptions have to be made. These assumptions result in an analysis that is far from correct for most actual combustion engine processes, but the analysis is of considerable value for indicating the upper limit of performance. The analysis is also a simple means for indicating the relative effects of principal variables of the cycle and the relative size of the apparatus. Assumptions 1. The working medium is a perfect gas with constant specific heats and

molecular weight corresponding to values at room temperature. 2. No chemical reactions occur during the cycle. The heat addition and heat

rejection processes are merely heat transfer processes. 3. The processes are reversible. 4. Losses by heat transfer from the apparatus to the atmosphere are assumed to

be zero in this analysis. 5. The working medium at the end of the process (cycle) is unchanged and is at

the same condition as at the beginning of the process (cycle). In selecting an idealized process one is always faced with the fact that the simpler the assumptions, the easier the analysis, but the farther the result from reality. The air cycle has the advantage of being based on a few simple assumptions and of lending itself to rapid and easy mathematical handling without recourse to thermodynamic charts or tables or complicated calculations. On the other hand, there is always the danger of losing sight of its limitations and of trying to employ it beyond its real usefulness. Equivalent Air Cycle A particular air cycle is usually taken to represent an approximation of some real set of processes which the user has in mind. Generally speaking, the air cycle representing a given real cycle is called an equivalent air cycle. The equivalent cycle has, in general, the following characteristics in common with the real cycle which it approximates:

1. A similar sequence of processes.

2. Same ratio of maximum to minimum volume for reciprocating engines or maximum to minimum pressure for gas turbine engines.

3. The same pressure and temperature at a given reference point.

Prepared by Prof. (Dr.) Manmatha K. Roul Page 3 4. An appropriate value of heat addition per unit mass of air.

Carnot Cycle

P-v and T-s diagram of Carnot’s cycle:-


Cycle Analysis:-

This cycle was proposed by Sadi Carnot in 1824 and has the highest possible

efficiency for any cycle. Assuming that the charge is introduced into the engine

at point 1, it undergoes isentropic compression from 1 to 2. The temperature of

the charge rises from Tmin

to Tmax

. At point 2, heat is added isothermally. This

causes the air to expand, forcing the piston forward, thus doing work on the

piston. At point 3, the source of heat is removed and the air now expands

isentropically to point 4, reducing the temperature to Tmin

in the process. At

point 4, a cold body is applied to the end of the cylinder and the piston reverses,

thus compressing the air isothermally; heat is rejected to the cold body. At point

1, the cold body is removed and the charge is compressed isentropically till it

reaches a temperature Tmax

once again. Thus, the heat addition and rejection

processes are isothermal while the compression and expansion processes are

isentropic.

Heat added, ∆QH =TH (sB − sA) Heat rejected, ∆QC=TC (sB − sA) Efficiency=1- TC/ TH



MODULE-II

LECTURE-16

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 Terminology for Reciprocating Devices:- The following is some terminology we need to understand for reciprocating engines—typically piston-cylinder devices. Let’s look at the following figures for the definitions of top dead center (TDC), bottom dead center (BDC), stroke, bore, intake valve, exhaust valve, clearance volume, displacement volume, compression ratio, and mean effective pressure.

Prepared by Prof. (Dr.) Manmatha K. Roul Page 3 Compression ratio(r):- The compression ratio r of an engine is the ratio of the maximum volume to the minimum volume in the formed in the cylinder. r V

VVV

B D C

T D C

= =m a xm i n

Mean effective pressure (MEP):- The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle. The Otto Cycle The Otto cycle, which was first proposed by a Frenchman, Beau de Rochas in 1862, was first used on an engine built by a German, Nicholas A. Otto, in 1876. The cycle is also called a constant volume or explosion cycle. This is the equivalent air cycle for reciprocating piston engines using spark ignition.

Prepared by Prof. (Dr.) Manmatha K. Roul Figures 5 and 6 show the PAt the start of the cycle, the cylinder contains a mass M of air at the pressure and volume indicated at point 1. The piston is at its lowest poupward and the gas is compressed isentropically to point 2. At this point, heat is added at constant volume which raises the pressure to point 3. The high pressure charge now expands isentropically, pushing the piston down on its expansionstroke to point 4 where the charge rejects heat at constant volume to the initial state, point 1. The isothermal heat addition and rejection of the Carnot cycle are replaced by the constant volume processes which are, theoretically more plausible, althouin practice, even these processes are not practicable. The heat supplied, Qs, per unit mass of charge=cThe heat rejected, Qr per unit mass of charge =Cand the thermal efficiency is given by by Prof. (Dr.) Manmatha K. Roul Figures 5 and 6 show the P-V and T-s diagrams respectively.At the start of the cycle, the cylinder contains a mass M of air at the pressure and volume indicated at point 1. The piston is at its lowest poupward and the gas is compressed isentropically to point 2. At this point, heat is added at constant volume which raises the pressure to point 3. The high pressure charge now expands isentropically, pushing the piston down on its expansionstroke to point 4 where the charge rejects heat at constant volume to the initial The isothermal heat addition and rejection of the Carnot cycle are replaced by the constant volume processes which are, theoretically more plausible, althouin practice, even these processes are not practicable. , per unit mass of charge=cv(T3-T2) per unit mass of charge =Cv(T4-T1) and the thermal efficiency is given by Page 4 s diagrams respectively. At the start of the cycle, the cylinder contains a mass M of air at the pressure and volume indicated at point 1. The piston is at its lowest position. It moves upward and the gas is compressed isentropically to point 2. At this point, heat is added at constant volume which raises the pressure to point 3. The high pressure charge now expands isentropically, pushing the piston down on its expansion stroke to point 4 where the charge rejects heat at constant volume to the initial The isothermal heat addition and rejection of the Carnot cycle are replaced by the constant volume processes which are, theoretically more plausible, although

Prepared by Prof. (Dr.) Manmatha K. Roul Hence, substituting in efficiency compression ratio V1/V2 Figure , shows a plot of thermal efficiency versus compression ratio for an Otto cycle for 3 different values of γ. It is seen that the increase in efficiency is significant at lower compression ratios. by Prof. (Dr.) Manmatha K. Roul Hence, substituting in efficiency equation, we get, assuming that r is the Figure , shows a plot of thermal efficiency versus compression ratio for an Otto cycle for 3 different values of γ. It is seen that the increase in efficiency is significant at lower compression ratios. Page 5 equation, we get, assuming that r is the Figure , shows a plot of thermal efficiency versus compression ratio for an Otto cycle for 3 different values of γ. It is seen that the increase in efficiency is

Prepared by Prof. (Dr.) Manmatha K. Roul Page 6 Mean effective pressure:- It is seen that the air standard efficiency of the Otto cycle depends only on the compression ratio. However, the pressures and temperatures at the various points in the cycle and the net work done, all depend upon the initial pressure and temperature and the heat input from point 2 to point 3, besides the compression ratio. mep= �� = ��



MODULE-II

LECTURE-17

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Page 2 Diesel Cycle:- The diesel cycle is the ideal cycle for compression ignition engines (CI engines). CI engine was first proposed by Rudolph Diesel in 1890. The diesel engine works on the principle of compression ignition. In such an engine, only air is compressed and at the end of the compression process, the fuel is sprayed into the engine cylinder containing high pressure air, so that the fuel ignites spontaneously and combustion occurs. Since only air is compressed during the compression stroke, the possibility of auto ignition is completely eliminated in diesel engines. Hence diesel engines can be designed to operate at much higher compression ratios (between 12 and 24). Also another benefit of not having to deal with auto ignition is that fuels used in this engine can be less refined (thus less expensive). The p – V and T – s diagrams for an air-standard diesel cycle are shown in Fig. . The diesel cycle is similar to Otto cycle except that the heating process takes place at constant pressure in a diesel cycle. The various processes involved in an ideal diesel cycle are as follows. The Diesel cycle is an ideal air standard cycle which consists of four processes: 1 to 2: Isentropic compression 2 to 3: Reversible constant pressure heating 3 to 4: Isentropic expansion

Prepared by Prof. (Dr.) Manmatha K. Roul 4 to 1: Reversible constant volume cooling Maximum thermal efficiency The maximum thermal efficiency of a Diesel cycle is dependent on the compression ratio and the cutair standard analysis: where ηth is thermal efficiency ρ=��=�� r is the compression ratio V1/Vγ is ratio of specific heats (Cp/CThe cut-off ratio can be expressed in terms of temperature as shown below: by Prof. (Dr.) Manmatha K. Roul 4 to 1: Reversible constant volume cooling Maximum thermal efficiency The maximum thermal efficiency of a Diesel cycle is dependent on the compression ratio and the cut-off ratio. It has the following formula under cold /V2 /Cv) expressed in terms of temperature as shown below: Page 3 The maximum thermal efficiency of a Diesel cycle is dependent on the formula under cold expressed in terms of temperature as shown below:

Prepared by Prof. (Dr.) Manmatha K. Roul Page 1 ENGINEERING THERMODYNAMICS

MODULE-II

LECTURE-18

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul DUAL COMBUSTION CYCLE

This cycle (also called the

combination of Otto and Diesel cycles, in a way, that heat is added partly at

constant volume and partly at constant

more time is available to fuel (which is injected into the engine cylinder before

the end of compression stroke) for combustion. Because of lagging

characteristics of fuel this cycle is invariably used for diesel and hot spot

ignition engines.

The dual combustion cycle consists of the following

(i) 1-2—Adiabatic compression

(ii) 2-3—Addition of heat at constant volume

(iii) 3-4—Addition of heat at constant pressure

(iv) 4-5—Adiabatic expansion

(v) 5-1—Rejection of heat at constant volume. P-v diagram and T-s diagram: by Prof. (Dr.) Manmatha K. Roul DUAL COMBUSTION CYCLE

This cycle (also called the limited pressure cycle or mixed cycle


constant volume and partly at constant pressure; the advantage of which is that




The dual combustion cycle consists of the following operations:

Adiabatic compression

Addition of heat at constant volume Addition of heat at constant pressure Adiabatic expansion

heat at constant volume. s diagram:- Page 2 limited pressure cycle or mixed cycle) is a


the advantage of which is that




operations:

Prepared by Prof. (Dr.) Manmatha K. Roul Cycle Analysis:-

Thermal efficiency of a dual combustion cycle:

by Prof. (Dr.) Manmatha K. Roul Thermal efficiency of a dual combustion cycle:-

Page 3



MODULE-II

LECTURE-19

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June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul Comparison of Otto, Diesel and Dual Cycles: The important variable factors which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat addition, heat rejection and the net work. In order to compare the performance of the Otto, Diesel and Dual combustion cycles, some of the variable factors must be fixed. In this section, a comparison of these three cycles is made for the same 1. Compression ratio,

2. Same heat addition,

3. Constant ma

4. Pressure and temperature,

5. Same heat rejection and

6. Net work output. This analysis will showoperating conditions. Case 1: Same Compression Ratio and Heat Addition: The Otto cycle 1-2-3-4-1, the Die2”-3”-4”-1 are shown in for the same compression ratio and heat input. by Prof. (Dr.) Manmatha K. Roul Comparison of Otto, Diesel and Dual Cycles: The important variable factors which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat addition, heat rejection and the . In order to compare the performance of the Otto, Diesel and Dual combustion cycles, some of the variable factors must be fixed. In this section, a comparison of these three cycles is made for the same

Compression ratio,

Same heat addition,

Constant maximum

Pressure and temperature,

Same heat rejection and

Net work output. will show which cycle is more efficient for a given set of Case 1: Same Compression Ratio and Heat Addition: 1, the Diesel cycle 1-2-3'-4'-1 and the Dual cycle 11 are shown in p-V and T-S diagram in Fig. (a) and (b) respectively for the same compression ratio and heat input. Page 2 The important variable factors which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat addition, heat rejection and the . In order to compare the performance of the Otto, Diesel and Dual combustion cycles, some of the variable factors must be fixed. In this section, a which cycle is more efficient for a given set of 1 and the Dual cycle 1-2-diagram in Fig. (a) and (b) respectively

Prepared by Prof. (Dr.) Manmatha K. Roul From the T-s diagram, it can be seen that Area 55-2-2"- 3"-6" as this area represents the heat input which is the same for all cycles. All the cycles start from the same initial state point 1 and the air is compressed from state 1 to 2the T-s diagram for the same heat input, the heat rejection in Otto cycle (area 51-4-6) is minimum and heat rejection in Diesel cycle (5Consequently, Otto cycle has the highest wocycle has the least efficiency and Dual cycle having the efficiency between the two. One more observation can be made i.e., Otto cycle allows the working medium to expand more whereas Diesel cycle is least in this respect. is heat is added before expansion in the case of Otto cycle and the last portion of heat supplied to the fluid has a relatively short expansion in case of the Diesel cycle. by Prof. (Dr.) Manmatha K. Roul s diagram, it can be seen that Area 5-2-3-6 = Area 56" as this area represents the heat input which is the same for all cycles. All the cycles start from the same initial state point 1 and the air is compressed from state 1 to 2 as the compression ratio is same. It is seen from s diagram for the same heat input, the heat rejection in Otto cycle (area 56) is minimum and heat rejection in Diesel cycle (5-1-4'-6') is maximum.. Consequently, Otto cycle has the highest work output and efficiency. Diesel cycle has the least efficiency and Dual cycle having the efficiency between the two. One more observation can be made i.e., Otto cycle allows the working medium to expand more whereas Diesel cycle is least in this respect. is heat is added before expansion in the case of Otto cycle and the last portion of heat supplied to the fluid has a relatively short expansion in case of the Diesel Page 3 6 = Area 5-2-3'-6’ = Area 6" as this area represents the heat input which is the same for all cycles. All the cycles start from the same initial state point 1 and the air is as the compression ratio is same. It is seen from s diagram for the same heat input, the heat rejection in Otto cycle (area 5-6') is maximum.. rk output and efficiency. Diesel cycle has the least efficiency and Dual cycle having the efficiency between the two. One more observation can be made i.e., Otto cycle allows the working medium to expand more whereas Diesel cycle is least in this respect. The reason is heat is added before expansion in the case of Otto cycle and the last portion of heat supplied to the fluid has a relatively short expansion in case of the Diesel

Prepared by Prof. (Dr.) Manmatha K. Roul Case 2: Same Compression Ratio and Heat Rejection: Efficiency of Otto cycle is given by Where, Qs is the heat supplied in the Otto cycle and is equal to the area under the curve 2-3 on the T-s diagram [Fig. (b)]. The efficiency of the Diesel cycle is given by, Where Q’s is heat supplied in the Diescurve 2-3' on the T-s diagram [Fig. (b)]. From the Tthat Qs> Q’s i.e., heat supplied in the Otto cycle is more than that of the Diesel cycle. Hence, it is evident that, the the efficiency of the Diesel cycle for a given compression ratio and heat

rejection. Case 3: Same Peak Pressure, Peak Temperature and Heat Rejection: by Prof. (Dr.) Manmatha K. Roul Case 2: Same Compression Ratio and Heat Rejection:

Efficiency of Otto cycle is given by is the heat supplied in the Otto cycle and is equal to the area under s diagram [Fig. (b)]. The efficiency of the Diesel cycle is Where Q’s is heat supplied in the Diesel cycle and is equal to the area under the s diagram [Fig. (b)]. From the T-s diagram in Fig., it is clear > Q’s i.e., heat supplied in the Otto cycle is more than that of the Diesel is evident that, the efficiency of the Otto cycle is greater than

the efficiency of the Diesel cycle for a given compression ratio and heat

Case 3: Same Peak Pressure, Peak Temperature and Heat Rejection: Page 4 is the heat supplied in the Otto cycle and is equal to the area under s diagram [Fig. (b)]. The efficiency of the Diesel cycle is el cycle and is equal to the area under the s diagram in Fig., it is clear > Q’s i.e., heat supplied in the Otto cycle is more than that of the Diesel ncy of the Otto cycle is greater than

the efficiency of the Diesel cycle for a given compression ratio and heat

Case 3: Same Peak Pressure, Peak Temperature and Heat Rejection:

Prepared by Prof. (Dr.) Manmatha K. Roul Figures (a) and (b) show the Otto cycle 1andT-s coordinates, where the peak pressure and temperature and the amount of heat rejected are the same. The efficiency of the Otto cycle, Where, Qs in the area under the curve 2Diesel cycle, 1- 2-3'-3-4 is, It is evident from Fig. that Qs > Q’s. Therefore, the Diesel cycle efficiency is greater than the Otto cycle efficiency when both engines are built to withstand the same thermal and mechanical stress Case 4: Same Maximum Pressure and Heat Input: by Prof. (Dr.) Manmatha K. Roul Figures (a) and (b) show the Otto cycle 1-2-3-4 and Diesel cycle s coordinates, where the peak pressure and temperature and the amount of heat rejected are the same. The efficiency of the Otto cycle, in the area under the curve 2-3 in Fig. (b). the efficiency of the 4 is, It is evident from Fig. that Qs > Q’s. Therefore, the Diesel cycle efficiency is greater than the Otto cycle efficiency when both engines are built to withstand the same thermal and mechanical stresses Case 4: Same Maximum Pressure and Heat Input: Page 5 4 and Diesel cycle 1-2'-3-4 on p-V s coordinates, where the peak pressure and temperature and the amount of efficiency of the It is evident from Fig. that Qs > Q’s. Therefore, the Diesel cycle efficiency is greater than the Otto cycle efficiency when both engines are built to withstand

Prepared by Prof. (Dr.) Manmatha K. Roul Page 6 For same maximum pressure and heat input, the Otto cycle (1-2-3-4-1) and Diesel cycle (1-2'-3'-4'-1) are shown on p-V and T-s diagrams in Fig. (a) and (b) respectively. It is evident from the figure that the heat rejection for Otto cycle (area 1-5-6-4 on T-s diagram) is more than the heat rejected in Diesel cycle (1-5-6'-4'). Hence Diesel cycle is more efficient than Otto cycle for the condition of same maximum pressure and heat input. One can make a note that with these conditions, the Diesel cycle has higher compression ratio than that of Otto cycle. One should also note that the cycle which is having higher efficiency allows maximum expansion. The Dual cycle efficiency will be between these two.



MODULE-II

LECTURE-20

Prepared By



June 2016

Prepared by Prof. (Dr.) Manmatha K. Roul

GAS TURBINE CYCLE Brayton cycle, popularly used for gas turbine power

adiabatic compression process, constant pressure heat addition, adiabatic

expansion process and constant pressure heat

diagram for air-standard Brayton cycle is shown in Figure. Simple gas turbine

power plant working on Brayton cycle is also shown here.

Schematic diagram;-

Ideal Air-standard Brayton Cycle p-v and T-s diagram by Prof. (Dr.) Manmatha K. Roul GAS TURBINE CYCLE—BRAYTON CYCLE

Brayton cycle, popularly used for gas turbine power plants comprises of

compression process, constant pressure heat addition, adiabatic

expansion process and constant pressure heat release process. A schematic

standard Brayton cycle is shown in Figure. Simple gas turbine

power plant working on Brayton cycle is also shown here.

standard Brayton Cycle (processes are reversible) Page 2 plants comprises of

compression process, constant pressure heat addition, adiabatic

release process. A schematic

standard Brayton cycle is shown in Figure. Simple gas turbine


Cycle analysis 1 � 2 Isentropic compression

2 � 3 Constant pressure heat addition

3 � 4 Isentropic expansion

4 �1 Constant pressure heat removals

For the isentropic process 1� 2

=

1

212 P

PPP rr

For the isentropic process 3 � 4

=

3

434 P

PPP rr

For isentropic processes 1 � 2 and 3� 4

k

k

P

P

T

T1

1

2

1

2

−

= and

k

k

P

P

T

T1

3

4

3

4

−

= Since

4

3

1

2P

P

P

P= thus

4

3

1

21

4

31

1

2T

T

T

T

T

T

T

T k

k

k

k

=→

=

−−

Thermal Efficiency

( )( )

( )( )1/

1/111232

141

23

14

23

14−−

−=−−

−=−−

−=TTT

TTT

TTc

TTc

hh

hh

P

P

constkBraytonη

as 2

3

1

4

4

3

1

2T

T

T

T

T

T

T

T=→=


( ) kk

constkBrayton

PPT

T1

122

1 111−

−=−=η


Efficiency increases with increased pressure ratio across the compressor

Back work ratio

( ) 43

12

43

12 )(TT

TT

TTc

TTc

mW

mW

mW

mWbwr

P

P

turb

comp

out

in

−−

=−−

===&&

&&

&&

&&

Typical BWR for the Brayton cycle is 40 - 80% compared to < 5% for the

Rankine cycle. Recall, reversible compressor work is given by ∫2

1 vdP Since gas has a much larger specific volume than liquid much more power is required to compress the gas from P1 to P2 in the Brayton cycle compared to the Rankine cycle for which liquid is compressed. The turbine inlet temperature is limited by metallurgical factors, e.g., Tmax = 1700K

Gas Turbine Irreversibilities

In the ideal Brayton cycle all 4 processes are assumed reversible, thus processes 2-3 and 4-1 are constant pressure and processes1-2 and 3-4 are isentropic.

The constant pressure assumption does not normally incur any great errors but the compressor and turbine processes are far from isentropic


These irreversiblities are taken into account by:

s

s

t

t

turbhh

hh

mW

mW

43

43−−

=

=

&

&

&

&

η

12

12hh

hh

mW

mW

s

c

s

c

comp −−

=

=

&

&

&

&

η

Ideal (reversible) processes: 1 - 2s and 3 - 4s Actual (irreversible) processes: 1 - 2 and 3 - 4


ENGINEERING THERMODYNAMICS MODULE-II LECTURE-21

Prepared

By Prof. (Dr.) Manmatha K. Roul

Professor and Principal


June 2016


Efficiency versus Power Analysis of Brayton Cycle

Consider two Brayton cycles A and B with a similar turbine inlet temperatures T3

Since BA P

PPP

>

1

2

1

2 � BA ηη >

Since (enclosed area 1-2-3-4)B > (enclosed area 1-2-3-4)A

A

cycle

B

cycle

mW

mW

>

&

&

&

&

� Bcycle

Acycle

B

AWW

mm

,

,&

&

&

&=

In order for cycle A to produce the same amount of net power as cycle B, i.e.,

BcycleAcycle WW ,, && = , need BA mm && > . Higher mass flow rate requires

larger (heavier) equipment which is a concern in transportation applications

Increasing Cycle Power

The net cycle power is: ctcycle WWW &&& −=


The cycle power can be increased by either increasing the turbine output power or decreasing the compressor input power.

Gas Turbine with Reheat The turbine work can be increased by using reheat, as was shown in the Rankine cycle

The turbine is split into two stages and a second combustor is added where additional heat can be added

Recall: '4

3

1

2TT

TT

= so, isobars on T-s diagram diverge

Note:

hb - h4 > ha - h4’

3 2

Compressor

a b

4 1 T 1 s 3 2 a b 4 1,inQ& 4’ 2,inQ&


The total turbine work output without reheat is:

( ) ( )[ ]mhhhhW aabasic && '43 −+−=

The total turbine work output with reheat is:

( ) ( )[ ]mhhhhWWW battreheatw

turbine &&&& 432,1,/

−+−=+=

Since hb - h4 > ha - h4’ basicreheatw

turbine WW && >/

Since the compressor work h2 - h1 is unaffected by reheat

basiccycle

reheatwcycle WW && >

/

The reheat cycle efficiency is not necessarily higher since additional heat

2,inQ& is added between states a and b

Compression with Intercooling

The compressor power can be reduced by compressing in stages with cooling between stages.


Recall: '4

3

1

2TT

TT

= so, isobars on T-s diagram diverge

The compressor power input without intercooling is:

( ) ( )[ ]mhhhhW ccbasic && 1'2 −+−=

The total compressor power input with intercooling is:

( ) ( )[ ]mhhhhWWW dcccreheatw

comp &&&& −+−=+= 212,1,/

Since h2’ – hc > h2 – hd � basic

reheatwcomp WW && <

/

Since the turbine work h3 – h4 is unaffected by intercooling

basiccycle

reheatwcycle WW && >

/

2’ 2’ d h2’ – hc > h2 – hd


Different approach: The reversible work per unit mass for a steady flow device is ∫ vdP , so

Without intercooling :

-a-c-b-

vdPvdPvdPm

Wc

c

basic

c

'21 area

'22

1 1

=

∫∫ ∫ +==

&

&

With intercooling :

-a-c-d-b-

vdPvdPvdPm

Wd

c

w/

c

21 area

22

1 1int

=

∫∫ ∫ +==

&

&

Since area(b-1-c-2’-a) > area(b-1-c-d-2-a)

int/w

c

basic

cm

Wm

W

>

&

&

&

&

2’

engineering thermodynamics module-ii lecture-9 · calls for higher rates of heat transfer in boiler...

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