Tarik Al-Shemmeri. Engineering Thermodynamics (776123), страница 6
Текст из файла (страница 6)
The thirdlaw of thermodynamics codifies this observation and setsfor all elements and compounds in their most stable and perfect crystalline state at absolute zeroand one atmosphere pressure. (All except for helium, which is a liquid at the lowest observabletemperatures at one atmosphere.)The advantage of this law is that it allows us to use experimental data to compute the absoluteentropy of a substance. For example, suppose we want to calculate the absolute entropy of liquidwater at 25o C. We would need to know the Cp of ice from 0 K to 273.15 K and the Cp of liquidwater from 273.15 K to 298.15 K. The value of the Entropy is determined by the followingequation:(52)Download free ebooks at bookboon.com553.
Laws of ThermodynamicsEngineering ThermodynamicsProcessLawConstantVolumeConstantPressureConstantTemperaturePolytropicP= constTV= constTPV = constPVn = constReversibleAdiabatic orIsentropicPVY = constP1V1n = P2V 2nP, V, T.RelationChange inInternalEnergy ΔUWorkTransferW=∫ pdvP1 P2=T1 T2mCv(T2-T1)0V1 V 2=T1 T2mCv(T2 –T1 )P(V2 –V1 )OrmR(T2 –T1 )Change inEntropyΔS=S2 –S1T= 1 T20mCv (T1 –T2 )T2mCv1nT1orP2mCv1nP1mCp (T1 –T2OrV2mRT 1nV1OrmRT 1nV2V1V2mR1nV1ororV2V1mR1nn n −1P1V1 − P2V2(n − 1)mR(T1 − T2(n − 1)P1 V2 = P2 V1 yy T y −1= 1 T2 mCv (T2 –T1 )P1V1 − P2V2(γ − 1)mR(T1 − T2(γ − 1)mCv (T1 –T2 )V2V1T2mCp1nT1mCp1nnmCv (T2 –T1 )V2PV 1nV1PV 1nHeatTransferQP 1V 1 = P 2 V 2P1 V2 = P2 V1 P1V1 y = P2V 2yP1P2W + (U2 – U1 )0VT m R1n 2 + C v 1n 2 V1T1 PT m R1n 1 + C p 1n 2 P2T1 0VPm C p1n 2 + Cv1n 2 V1P1 Table 3.3: Perfect Gas Process-RelationDownload free ebooks at bookboon.com563.
Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.1A closed rigid container has a volume of 1 m3 and holds air at 345 kPa and 20°C. Heat is addeduntil the temperature is 327°C. Determine the change in Internal Energy:a) Using an average value of the specific heat.b) Taking into account the variation of specific heat with temperature.Solution:a) ΔU = mCv ΔTCv =m=764 + 718= 741 J / kgK2PV 345 x10 3 x1== 4.1026 kgRT287 x 293Please click the advertdongenergy.com/jobMay we offer you oneof the world’s greatestchallenges?In all humility.We are looking for highlyeducated engineers andfinance students.Join us at dongenergy.com/jobDownload free ebooks at bookboon.com573.
Laws of ThermodynamicsEngineering ThermodynamicsThereforeΔU = 4.1026 x 741 (327 – 20) = 932 kJ∫b) ∆U = m.T2T1Cv .T∆u = ∫ TT 12 CvT = Cv 2T2 − Cv 1T1= 764 x 600 – 718 x 293= 248,319 J/kgm=PV 345 x10 3 x1== 4.1026kgRT287 x 293Therefore,ΔU = m x Δu = 1018.7 kJDownload free ebooks at bookboon.com583. Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.2An adiabatic steam turbine expands steam from a pressure of 6 MPa and a temperature of 500°C toa pressure of 10 KPa The isentropic efficiency of the turbine is 0.82 and changes in kinetic andpotential energy may be neglected.
Determine the state of the steam at exit from the turbine and thespecific work transfer.p = 6.0 MPa (257.64 deg-C)TvuhsSat.0.032442589.72784.35.88925000.056653082.23422.26.8803ptsvfvghfhgsfsg(kPa)10(oC)45.81(m3/kg)0.001(m3/kg)14.674(kJ/kg)191.83(kJ/kg)2,585(kJ/kg.K)0.6493(kJ/kg.K)8.1502Solution:From steam tables at 6 MPa and 500°C, h1 = 3422.2 kJ/kg, and s1 = 6.8803 kJ/kg KAt 10 KPa, Sf = 0.6493, and Sg = 8.1502S2’ = s1 and x2’ is found usingThen6.8803 = 0.6493+ x2’ (8.1502-0.6493), from which x2’ = 0.917This is the state of steam at exit from the turbine ( mixed phase with dryness fraction, x=0.917)Thush2 = hf + x hfg= 191.83 + 0.917 x (2585 – 191.83) = 2387 kJ/kgFromη it =h1 − h2h1 − h2'3422.2 − h23422.2 − 2387ie,0.82 =from whichh2 = 2573 kJ/kgDownload free ebooks at bookboon.com593. Laws of ThermodynamicsEngineering ThermodynamicsThe turbine specific work (m=1 kg/s) is =W = m *(h2 – h1) = 1 x ( 3422.2 – 2573) = 849 kJ/kg.Worked Example 3.3Show that for a polytropic process, the change in entropy may be expressed as :S 2 − S1 =γ −nn −1.m.Cv .LnT1T2Solution:1st Law: Q = U + WwithU = Cv.dTandW = P.dVfrom the 2nd law: Q = T.dShenceT.dS = mCv.dT + P.dVWith PV=RT, Then the above equation becomes:T.dS = mCv.dT + (mRT/V).dVDivide by TdS = mCv.
dT/T + mR.dV/Vintegrate between states 1 and 2S 2 − S1 = m[Cv .LnT2V+ R.Ln 2 ]T1V1Download free ebooks at bookboon.com603. Laws of ThermodynamicsEngineering Thermodynamicsfor a polytropic processhenceV2 T1 = V1 T2 S 2 − S1 = m[Cv .Ln1n −1T2TTRR−.Ln 2 ] = m.[Cv [1 −].Ln 2 ]T1 n − 1T1n −1T1but Cp = Cv +Rand Cp = γ. Cvhence R = Cv.( γ -1)and thusS 2 − S1 = m.Cv [γ −nn −1].LnT1T2QEDPlease 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.com613.
Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.4Steam at 1.0 MPa, 0.95 dry is throttled to 10 kPa. What is the quality of the steam after throttling?ptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)1045.810.00114.674191.832,5850.64938.15021000179.910.001120.1944762.812,7782.13876.5865Solution:SFEEQ - W = m (∆h = ∆ke + ∆Pe)The SFEE, reduces to: h2 = h1at 1.0 MPa , hf = 762.81 kJ/kg and hg = 2,778 kJ/kgh1 = 762.81+ 0.95 x (2778-762.81) = 2677.24 kJ/kgat 10 kPa with hg = 2,585 kJ/kgsince h2 = h1 = 2677.24 kJ/kg , which is higher then hg at 10 kPa, the steam will be in a superheatedcondition.Download free ebooks at bookboon.com623. Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.5A steam turbine receives steam at 2 MPa and 250 oC, and exhausts at 0.1 MPa, 0.85 dry.a) Neglecting heat losses and changes in ke and Pe, estimate the work output per kg steam.b) If, when allowance is made for friction, radiation, and leakage losses, the actual workobtained is 80% of that estimated in (a), calculate the power output of the turbine whenconsuming 600 kg of steam per minute.Saturated table extract for P = 100 KPaptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)10099.630.001041.694417.462,6761.30267.3594T (C )Superheated table extract at P = 2 MPaV (m3/kg)h (kJ/kg)2400.10842876S (kJ/kgK)6.495Solution:a) Q - W = m (∆h + ∆ke + ∆Pe)Q = 0 , ∆ke = 0 , ∆Pe = 0 are all neglected.∴ W = m(h1 - h 2)h1 = 2876 kJ/kgh2 = 417.46 + 0.85 x (2676 – 417.46) = 2337.2 kJ/kg∴ Wideal = 1x(2876 - 2337)= 539 kJ/kgDownload free ebooks at bookboon.com633.
Laws of ThermodynamicsEngineering Thermodynamicsb) Wactual= 0.8 x 539 = 431 kJ/kgPower = m x W = (600/60)x 431 = 4.31 MWWorked Example 3.6A boiler receives feed water at 40oC and delivers steam at 2 MPa and 500oC. If the furnace is oilfired, the calorific value of oil being 42000 kJ/kg and 4000 kg oil are burned while 45000 kg ofsteam are produced, determine :a) the heat supplied in the boiler.b) the efficiency of the boiler.Assume the values of enthalpies at the two state points as :h1 = hf@40oC = 169.33 kJ/kgat 2 MPa , 500 C, h2 = 3467.6 kJ/kgPlease 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.com643. Laws of ThermodynamicsEngineering ThermodynamicsSolution:a) Constant pressure process.h1 = hf@40oC = 169.33 kJ/kgh2 = 3467.6 kJ/kgSFEE ignoring W, ∆ke and ∆Pe:Qs= ms (h2 - h1)= 45000 (3467.6 - 169.33)= 1.484 x 108kJb) The heat generated by burning oil in the furnace is= mass of oil burned x calorific value= 4000 x 42000 = 1.68 x 108 kJ∴ Boiler efficiency =Energy Output 1.484 x10 8== 88%Energy Input1.68 x10 8Worked Example 3.7An air compressor receives air at 27oC and delivers it to a receiver at the rate of 0.5 kg/s. It isdriven by an electric motor which absorbs 10 kW and the efficiency of the drive is 80%.Water jacket cooling is used at the rate of 6 kg/min while its temperature rises from 10oC to 20oC.Estimate the temperature of the air delivered.Data : Cpw = 4.186, and Cpa = 1.005 kJ/kgKDownload free ebooks at bookboon.com653.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
