Tarik Al-Shemmeri. Engineering Thermodynamics (776123), страница 4

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X=0.2∴v∴V= (1 – x) vf + xvg = 0.8 x 0.00109 + 0.2 x 0.3749= 0.0758 m3/kg= 2 x 0.0758 = 0.152 m3Worked Example 2.12The model '6SE-TCA3 Perkins' diesel engine have a stroke of 190 mm and a bore of 160 mm. If itsclearance volume is 5% of the swept volume, determine the pressure and temperature at the end ofcompression when the inlet condition is 1 bar, 27oC.Download free ebooks at bookboon.com352. Thermodynamics working fluidsEngineering ThermodynamicsAssume n = 1.38PressurevolumeSolution:The swept volume isV1 = V2 =∧∧ 22D L = (0.160 ) x 0.190 = 0.00382 m 344Please click the advertdongenergy.com/jobAmbitiousand curious?We are looking for young andhighly educated engineersand finance students.Join us at dongenergy.com/jobDownload free ebooks at bookboon.com362.

Thermodynamics working fluidsEngineering ThermodynamicsThe clearance volume isVc =5x 0.00382 = 0.00019 m 3100or 0.19 litrehence V1 = VS + VC= 3.82 + 0.19= 4.01 litresThe final pressure is:VP2 = P1  1 V2n 4.01  = 1 0.19 1.38= 67.2 barThe final temperature is:V T2 = T1  1  V2 n −14.01= 300 0.19 0 .38= 956KDownload free ebooks at bookboon.com373. Laws of ThermodynamicsEngineering Thermodynamics3.

Laws of ThermodynamicsThere are four laws which relates the thermodynamics of substances.3.1 Zeroth Law of ThermodynamicsIf an object with a higher temperature comes in contact with a lower temperature object, it willtransfer heat to the lower temperature object. The objects will approach the same temperature, andin the absence of loss to other objects, they will maintain a single constant temperature. Therefore,thermal equilibrium is attained.The "zeroth law" states that if two systems are at the same time in thermal equilibrium with a thirdsystem, they are in thermal equilibrium with each other.Figure 3.1: Analogy of the Zeroth Law of Thermodynamics.If A and C are in thermal equilibrium with B, then A is in thermal equilibrium with B.

Practicallythis means that all three are at the same temperature , and it forms the basis for comparison oftemperatures. The Zeroth Law states that:“two systems which are equal in temperature to a third system are equal in temperature to eachother”.3.1.1 Methods of Measuring TemperatureThe Zeroth Law means that a thermometer can be used to assign a label to any system, the labelbeing the value shown on the thermometer when it is thermal equilibrium with the system. We callthe label ``temperature''. Temperature is a function of state which determines whether a system willbe in equilibrium with another.To put this into practice we need a thermometer, which is a device that has some easily measuredproperty, , that varies with temperature.

This might be the length of the mercury column in amercury-in-glass thermometer for instance, or the pressure of a constant volume gas thermometer.We then require easily reproduced calibration temperatures. For instance, the CentigradeDownload free ebooks at bookboon.com383. Laws of ThermodynamicsEngineering Thermodynamicstemperature scale assigns temperatures of 0 and 100, to the temperature of ice in equilibrium withwater (known as the ice point) and to the temperature of boiling water (the steam point).

Letting thevalues ofat these two points be X0 and X100, then the temperature in Centigrade,Therefore, measuring temperatures may be based on one of the following properties:a) Expansion of materials due temperature variations, eg gas thermometer, liquid-in-glassthermometer, bi-metal strip.b) Electrical resistance of materials.c) Electro-motive force induced within a circuit made up of two dissimilar materials.d) Radiative properties of surfaces.3.1.2 International Temperature ScaleThis scale is used to calibrate temperature measuring devices. It consists of a number of fixedpoints of known temperature which can be reproduced accurately (Table 3.1).Please click the advertdongenergy.com/jobEverybodyis talking......about future energy supply. We are not.Stop talking and make a careermoving energy forward.Ambitious engineers andfinance students go todongenergy.com/jobDownload free ebooks at bookboon.com393.

Laws of ThermodynamicsEngineering ThermodynamicsUnits of TemperatureThe Kelvin (SI units) is the fraction(1/273.16) of the thermodynamic temperature of the triple pointof water. Generally, conversion of Celsius to Kelvin : T(K) = T(°C) + 273.15Fixed PointIce PointTemperature (deg C)0.01Steam Point100.00Solidification of Antimony630.74Solidification of Gold1064.43Table 3.1 Example of Thermodynamic fixed temperatures3.2 First Law of ThermodynamicsThe first law of thermodynamics is the application of the conservation of energy principle.3.2.1 First Law of Thermodynamics Applied to closed Systemsconsider a closed system where there is no flow into or out of the system, and the fluid massremains constant.

For such system, the first law statement is known as the Non-Flow EnergyEquation, or NFEE abbreviated, it can be summerised as follows:∆U = Q − W(25)The first law makes use of the key concepts of internal energy (∆U), heat (Q), and system work (W).3.2.2 Internal EnergyInternal energy is defined as the energy associated with the random, disordered motion ofmolecules. It is separated in scale from the macroscopic ordered energy associated with movingobjects; it refers to the invisible microscopic energy on the atomic and molecular scale. Forexample, a room temperature glass of water sitting on a table has no apparent energy, eitherpotential or kinetic .

But on the microscopic scale it is a seething mass of high speed moleculestraveling at hundreds of meters per second. If the water were tossed across the room, thismicroscopic energy would not necessarily be changed when we superimpose an ordered large scalemotion on the water as a whole.During a non flow process the change in internal energy is calculated assuming the closed’s systemvolume remains constant, the following equation is usedDownload free ebooks at bookboon.com403. Laws of ThermodynamicsEngineering ThermodynamicsU  m.Cv .T(26)Where Cv is the specific heat capacity of the fluid, and T is the temperature difference during theprocess3.2.3 Specific HeatHeat may be defined as energy in transit from a high temperature object to a lower temperatureobject.

An object does not possess "heat"; the appropriate term for the microscopic energy in anobject is internal energy. The internal energy may be increased by transferring energy to the objectfrom a higher temperature (hotter) object - this is properly called heating.In order to heat or cool a given quantity of a gas in a given time, the following equation is used:Quantity of Heat (Q)= mass(m) x specific heat capacity(C ) x temperature differenceSince this heat exchange may take placeEither at constant pressure:Or at constant volume: Cp (T2 - T1)Q= m Cv (T2 - T1)Q= m(27)(28)Where:CpCvspecific heat at constant pressure (kJ/kg K), see Table 3.2specific heat at constant volume (kJ/kg K) , see Table 3.2Note that for a perfect gasCp = Cv + RandnCp / Cv(29)Specific Heat at Constant Volume CvConsider a closed system of unit mass, the first law of thermodynamics applied to this system is:q – w = duIf the volume is constant, then w = 0,ButHenceOrit follows thatq = duq = Cv dTdu = CvdTCv = du/dTThis is known as Joule’s Law of internal energy which states that “the internal energy of a perfectgas depends upon its temperature only”.Specific Heat at Constant Pressure CpDownload free ebooks at bookboon.com413.

Laws of ThermodynamicsEngineering ThermodynamicsSpecific Heat at Constant Pressure CpConsider a constant pressure non-flow process, the first law:q – w = duFor a constant pressure processW = p(v2 –v1)= (p2 v2 –p1 v1)hence q – (p2 v2 -p1 v1 ) = u2 -u1orq = (u2 + p2 v2 ) – (u1 + p1 v1 )= h 2 – h1butq = Cp dThence h2 - h1 = Cp (T2 - T1)orCp = dh/dTPlease click the advertdongenergy.com/jobdDear highly educated engineeringand finance students,if you are driven, ambitious, open-mindedand focused - we have a challenge for you.Actually, the greatest challenge in theworld.

Curious? Visit dongenergy.com/jobBest wishesDONG EnergyDownload free ebooks at bookboon.com423. Laws of ThermodynamicsEngineering ThermodynamicsRelationship Between Specific HeatsSinceH = U + PVdH = dU + d(PV);d(PV) = d(mRT) = mRdTm Cp dT = m Cv dT + m R dTTherefore,Cp = Cv + RIeCp > CvThe ratioCp/Cv = n is called the adiabatic index.The reason for the differences between Cp and Cv is that in the constant pressure process part ofthe heat transferred is used in doing work against the moving system boundary and all heattherefore is not available for raising the gas temperature.TemperatureKCpJ/kgKCvJ/kgKnCp / Cv25010037161.40130010057181.40035010087211.39840010137261.39545010207331.39150010297421.38755010407531.38160010517641.37665010637761.37070010757881.36475010878001.35980010998121.35490011218341.344100011428551.336Table 3.2 Specific heat capacities for air at standard atmosphereDownload free ebooks at bookboon.com433.

Laws of ThermodynamicsEngineering Thermodynamics3.2.4 System WorkWork performed on or by the working fluid within a syste’s boundary is defined as the summation(or integration) of the product of pressure and volume of the fluid during a process.W = ∫ P.dV(30)In calculating the process work, it is important to point out that for each process, the work donewill be different, since there are four distinctly different processes, in the following sections, anexpression for the work done will be evaluated for each process.a.

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