Tarik Al-Shemmeri. Engineering Thermodynamics (776123), страница 3
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Thermodynamics working fluidsEngineering Thermodynamics1000179.910.001120.1944762.812,7782.13876.58652000212.420.001170.0996908.792,8002.44746.340910000311.060.001450.018031407.562,7253.35965.614120000365.810.002360.00581,8262,4104.01394.926922120374.150.003170.003172,0842,0844.434.43Table 2.2 Saturated Steam table at selected pressuresPlease 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.com222.
Thermodynamics working fluidsEngineering ThermodynamicsEnthalpy of Superheated Steam (kJ/kg)SaturationPressure(bar) Temperature (oC) SaturationTemperature (oC)20025030035040045027962925303931483256336415198.3279220212.427992904302531383248335730233.828092858299531173231334340250.328012963309432143330Table 2.3 Superheated Steam table at selected pressuresWorked Example 2.1Self ignition would occur in the engine using certain brand of petrol if the temperature due tocompression reached 350°C.Calculate the highest ratio of compression that may be used to avoid pre-ignition if the law ofcompression isPV 1.3 = cPV 1.4 = cCalculate the final pressure in each case.
Assume inlet condition of 27°C and 1 bar.Solution(a)(b)V1 T2 = V 2 T1 1n −11 V1 V2 349 + 273 0.3 = = 11.36 i 300 V1 V2 349 + 273 0.4 = = 6.19 ii 300 VP2 = P1 1 V21nDownload free ebooks at bookboon.com232. Thermodynamics working fluidsEngineering ThermodynamicsP2i = 1(11.36)1.3 = 23.5 barP2ii = 1(6.19)1.4 = 12.8 barWorked Example 2.2Calculate the density of Ethane at 171 bar and 458K; assume for Ethane:Tc=305 KPc=48.80 barR = 319.3 J/kgKa) assuming it behaves as a perfect gasb) using the compressibility chart.Solution:a) for a perfect gasρ=P171x10 5== 117 kg / m 3R.T 319.3 x 458b) using the com[pressibility chart:TR = T / TCR = 458 / 305.4= 1.5PR = P / PCR = 171 / 48.8 = 3.52.7READ Z from the chart ( Z = 0.8 )ie 80% less dense compared with the perfect gas behaviour.Or density = 146 kg/m3Download free ebooks at bookboon.com242.
Thermodynamics working fluidsEngineering ThermodynamicsWorked Example 2.3Find the specific volume for H2O at 10 MPa and 500°C using:a) Compressibility chart;b) Steam tables (below)Tp = 10.0 MPa (311.06 deg-C)vuHsSat.0.0180262544.42724.75.61413250.0198612610.42809.15.75683500.022422699.22923.45.94434000.026412832.43096.56.21204500.029752943.43240.96.41905000.032793045.83373.76.59665500.035643144.63500.96.75616000.038373241.73625.36.9029Source: http://www.sfsb.hr/Please 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.com252.
Thermodynamics working fluidsEngineering ThermodynamicsSolution:a) Compressibility ChartPR=100P== 0.47P c 221.2T R=500 + 273T== 1.19T c 374.15 + 273but R = 8.3144/18.015 = 0.4615kJ/kgKUsing Figure 2.2, Z = 0.9v=∴PV= 0.9RTRxTxZ 461.5 x 773x 0.9 = 0.032m 3 / kg=5P100 x10b) From Steam Tables:The steam is superheatedAt 100 bar and 500°C , v = 0.03279 m3 /kgBoth results are similar within to the 3rd decimalplace.Worked Example 2.4Determine the pressure of water vapour at 300°C, if its specific volume is 0.2579 m3/kg, using thefollowing methods:a) Ideal gas equationb) Van-der-Waals equationsSolution:a) Pv = RTP=RT 461.5 x573== 1.025MPa0.2579vDownload free ebooks at bookboon.com262.
Thermodynamics working fluidsEngineering Thermodynamicsb) a =27 R 2Tc 2 27 x 461.5 2 x647.3 2== 170464 Pc64 x 22.09 x10 6b=RTc 461.5 x647.3== 1.69 x10 −368Pc 8 x 22.09 x10P=RTa− 2v−b v=461.5 x5731704−−30.2579 − 1.69 x100.2579 2= 1032120.1 – 25619= 1.006 MPaWorked Example 2.5An unkown gas has a mass of 1.5 kg contained in a bottle of volume 1.17 m3 while at a temperatureof 300 K, and a pressure of 200 kPa.
Determine the ideal gas constant and deduce the gas?Solution:Assuming perfect gas behaviour:PV = mRT200 x 103 x 1.17 = 1.5 x R x 300∴R = 520 J/kgKbut R =RoMhence M = Ro/R = 8314.3/520 = 15.99The nearest gas with such a molar mass is Methane, for which M=16.02 kg/Kmol.The small difference may be attributed to measurements errors.Download free ebooks at bookboon.com272. Thermodynamics working fluidsEngineering ThermodynamicsWorked Example 2.6A 6 m3 tank contains helium at 400K is evacuated form atmospheric pressure of 100kPa to a finalpressure of 2.5kPa.Determinea) the mass of helium remaining in the tank;b) the mass of helium pumped out;c) if the temperature of the remaining helium falls to 10oC, what is the pressure in kPa?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.com282.
Thermodynamics working fluidsEngineering ThermodynamicsSolution:a) P2V2 = m2 RT2R=Ro 8314.3== 2077 J / kgKM4.003The mass remaining is:m2 =2500 x 6= 0.018 kg2077 x 400b) initial mass is:m1 =100000 x 6= 0.722 kg2077 x 400∴mass pumped out = m1 - m2 = 0.704 kgsince m3 = m2, thenc)P3 =m3 .R.T3 0.018 x 2077 x 283== 1763 PaV36Worked Example 2.7A motorist equips his automobile tyres with a relief-type valve so that the pressure inside the tyrewill never exceed 220 kPa (gauge). He starts the trip with a pressure of 200 kPa (gauge) and atemperature of 23oC in the tyres. During the long drive the temperature of the air in the tyresreaches 83oC. Each tyre contains 0.11 kg of air.
Determine:a) the mass of air escaping each tyre,b) the pressure of the air inside the tyre when the temperature returns to 23oC.Download free ebooks at bookboon.com292. Thermodynamics working fluidsEngineering ThermodynamicsSolution:a) P1V1 = m1RT1P1 V1200 x 10 3 x V1m1 === 0.11 kgRT1 287 x (273 + 23)V1 =0.11 x 287 x 296= 0.04672m 33200 x 10V1 = V2 = constantP2V2 = m2 RT2220 x 103 x 0.04672 = m2 x 287 x (273 + 83)∴m2 = 0.1006 kg∴dm = m1 - m2 = 0.11 - 0.1006 = 0.0094 kgb) V3 = V2 = V1P3 =andm3 = m2m.R.T3 0.1006 x 287 x 296== 183 kPaV40.04672Worked Example 2.8300 kg/minute of steam at 3 MPa and 400oC is supplied to a steam turbine. determine the potentialheat released from steam if it is condensed at constant pressure.
Can you deduce the specific heat ofthe steam under this condition?p = 3.00 MPa (233.90 C)TvuhsDownload free ebooks at bookboon.com302. Thermodynamics working fluidsEngineering ThermodynamicsSat.0.066682604.12804.26.18692500.070582644.02855.86.28723000.081142750.12993.56.53903500.090532843.73115.36.74284000.099362932.83230.96.92124500.107873020.43344.07.08345000.116193108.03456.57.2338225Source: http://www.sfsb.hr/Solution:a) Constant pressure processh1= 3230.9h2 = 2804.2Thermal energy available Q =m x (h2 – h1 ) = (300/60)*(3230.9 – 2804.2) = 2133.5 kWPlease 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.com312. Thermodynamics working fluidsEngineering Thermodynamicsb) It can be seen from the table that the temperature at saturation is 233.90 C, so if the equationfor heat exchange is used, this is the same heat found above:Q = m.Cp.(Τ2 − Τ1)Hence Cp = Q / (m x (Τ2 − Τ1)) = 2133.5 /( 2 x ( 500 – 233.9) = 4.009 kJ/kg K.Which is lower than that at lower pressures, at 1 bar Cp for water is about 4.18 kJ/kgK.Worked Example 2.9Self-ignition would occur in an engine using certain brand of petrol if the temperature due tocompression reaches 350 oC; when the inlet condition is 1 bar, 27oC.Calculate the highest compression ratio possible in order to avoif self-ignition, if the compressionis according toa) adiabatic, with index of 1.4; andb) polytropic, with index of 1.3Solution:PressurevolumeThe compression ratio is calculated as follows:When n = 1.4, the volume ratio is :V1 T2 = V2 T1 1 /( n −1) 349.9 + 273 = 27 + 273 1 / 0.4= 6.2Download free ebooks at bookboon.com322.
Thermodynamics working fluidsEngineering Thermodynamicsand when n = 1.3, the volume ratio is :V1 T2 = V2 T1 1 /( n −1) 349.9 + 273 = 27 + 273 1 / 0.3= 11.4Worked Example 2.10The gas in an internal combustion engine, initially at a temperature of 1270 oC; expandspolytropically to five times its initial volume and one-eights its initial pressure. Calculate:a) the index of expansion, n, andb) the final temperature.Solution:PressurevolumeP2 V1 = P1 V2 a) Sincenn can be found by taking log of both sides, then rearranging the above equationp2) Ln( 1 )p18 = 1.292=n=V11Ln( ) Ln( )V25Ln(b) the final temperature is now evaluated:Download free ebooks at bookboon.com332. Thermodynamics working fluidsEngineering ThermodynamicsPT2 = T1 2 P1( n −1) / n1= (1270 + 273) 8= 698K0.2921 / 1.292Worked Example 2.11Determine using Steam Tables, the volume occupied by 2 kg of steam at 500 kPa, under thefollowing conditions and specify the state of steam.a)b)c)d)pure liquid statewhen it is in a pure vapour state20% moisture content20% dry.Please 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.com342.
Thermodynamics working fluidsEngineering Thermodynamicsptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)500151.860.001090.3749640.232,7491.86076.8213Solution:From Steam Table:- vf = 0.00109 m3/kg andvg = 0.3749 m3/kga) when pure liquid∴v = 0.00109 m3/kgV = 2 x 0.00109 = 0.00218 m3b) when it is saturated vapour∴v = vg = 0.3749 m3/kgV = 2 x 0.3749 = 0.7498 m3c) The steam is obviously in its wet phase. X=0.8∴v∴V= (1 – x) vf + xvg = 0.2 x 0.00109 + 0.8 x 0.3749= 0.2785 m3/kg= 2 x 0.2785 = 0.557 m3d) The steam is obviously in its wet phase.
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