Tarik Al-Shemmeri. Engineering Thermodynamics (776123), страница 9
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We are not.Stop talking and make a careermoving energy forward.Ambitious engineers andfinance students go todongenergy.com/jobDownload free ebooks at bookboon.com883. Laws of ThermodynamicsEngineering Thermodynamicsb) The SFEE can now be used to determine the mass flow rateV 2 2 − V 21 − W = m C p (T2 − T1 ) +2000 55 2 − 0 − 2 = m 1.005(164.7 − 293) +2000 ∴m =−2= +0.0157 kg / s− 128.88 + 1.51Worked Example 3.25Steam at a pressure of 2 MPa and a temperature of 240ºC enters a nozzle with a velocity of 15m/s.The steam expands reversibly and adiabatically in the nozzle to a pressure of 100 kPa and a drynessfraction of 0.9. Calculate the velocity of the steam at exit from the nozzle,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 )240Superheated table extract at P = 2 MPaV (m3/kg)h (kJ/kg)S (kJ/kgK)0.108428766.495Solution:a) SFEE applies,V 2 2 − V 21+ g (Z 2 − Z1 )Q − W = m (h2 − h1 ) +2Q = 0 adiabatic, W=0, No work transfer, and the Potential Energy change is negligibleDownload free ebooks at bookboon.com893.
Laws of ThermodynamicsEngineering ThermodynamicsHence the energy equation can now be rearranged as follows:2V2 = V1 + 2m(h1 − h2 )from the steam tables, h1 = 2876 kJ/kgh2 = 417.46 + 0.9 x (2676-417.46) = 2450 kJ/kg2V2 = V1 + 2m(h1 − h2 ) = 15 2 + 2 x 0.6 x ( 2876 − 2450)x10 3= 715m / sWorked Example 3.26A one pass steam generator receives saturated water at 20 bar, and converting it into steam at400ºC. The mass flow rate of steam is 1200 kg/h, calculate the heat transfer in the generator.Enthalpy of Superheated Steam (kJ/kg)Pressure(bar)Saturation Temperature(oC)20212.4Temperature (oC)Saturation 200 25027993003504004502904 3025 3138 3248 3357Solution:SFEE applies,V 2 2 − V 21+ g ( Z 2 − Z 1 )Q − W = m (h2 − h1 ) +2Neglecting the changes in Kinetic and Potential energies,also No work done W = 0hence the SFEE reduces to:Q = m (h2 – h1)from the steam table, we find:h1 = 2799Download free ebooks at bookboon.com903.
Laws of ThermodynamicsEngineering Thermodynamicsh2 = 3248 kJ/kghence the boiler capacity ( heat gained by the steam):Q=1200(3248 − 2799) = 149.667 kW3600Worked Example 3.271kg of gas occupies a volume of 0.4m3 at a pressure of 100 kN/m2. The gas is compressedisothermally to a pressure of 450 kN/m2. Determine the work of compression, and the change inentropy of the gas during the compression. Assume for the gas R = 300 J/kgKSolution:a) For an isothermal process, the work is given byW = PVnV2V1Since isothermal, PV= c∴PressureV2 P1=V1 P2W = PVnPV = CP1P2volume∴W = 100 x10 3 x0.4n100= - 60 kJ450(-ve indicates compression)b) the change of Entropy is given by:∆s = m.R.nP1100= 1x 0.300 x ln= −0.451 kJ / kgKP2450Download free ebooks at bookboon.com913.
Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.28A gas contained in a closed system at a pressure of 1 bar and temperature of 15ºC. A mass of 0.9kgof the gas is heated at constant pressure to raise its temperature from 15ºC to 250ºC. Determine thework done, and during the process. R = 0.185 kJ/kgKSolution:Using the ideal gas equation:V1 =mRT1 0.9 x0.185 x 288 x10 3== 0.4795mP11x10 5For a constant pressure:V2 = V1T2 250 + 273 3= 0.4795 = 0.871mT115273+Please 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.com923. Laws of ThermodynamicsEngineering Thermodynamicsthe work during an isobaric process is given by:W = P(V2 − V1 )W = 10 5 (0.871 − 0.4795) = 39.1kJ+ ve implies that the gas is expanding.Worked Example 3.29The cylinder of an engine has a stroke of 300mm and a bore of 250mm.
The volume ratio ofcompression is 14:1. Air in the cylinder at the beginning of compression has a pressure of 96kN/m2 and a temperature of 93 ºC. The air is compressed for the full stroke according to the lawPV1.3 = C. Determine the work transfer per unit mass of air. Assume air R = 287 J/kgK.Solution:compression work is given by:2300 ∧ 250 3V1 =x = 0.0147 m1000 4 1000 PressureV2 = V1 / 14 = 0.00105m 3P1 = 96 x10 3 N / m 2V P2 = P1 1 V2 W =volume1.3= 96(14)1.3 = 2966.45kN / m 2P1V1 − P2V2n −196 x10 3 x 0.0147 − 2966.45 x10 3 x 0.00105∴W == −5.678 kJ / stroke1. 3 − 1p.v 96 x10 3 x 0.00147∴m === 0.0134 kg / strokeR.T287 x ( 273 + 93)Download free ebooks at bookboon.com933.
Laws of ThermodynamicsEngineering ThermodynamicshenceW = ( - 5.6785 kJ/stroke ) / (0.0134 kg/stroke) = -424 kJ/kg-ve sign indicates compression.Worked Example 3.30A mass of air at 330ºC, contained in a cylinder expanded polytropically to five times its initialvolume and 1/8th its initial pressure which is 1 bar.
Calculate:a) the value of the expansion index,b) the work transfer per unit mass.Solution:a) for a polytropic process,P1 V2 = P2 V1 Pressurenvolumeln(P2 / P1 ) ln(8)== 1.292n=ln(V 2/ V1 ) ln(5)b) The work done for a polytropic process is,W =P1V1 − P2V2n −1V1 = mRT1 / P1= 1x 287 x 603 / 10 5= 1.7306m 3V2 = 5V1 = 8.653m 3Download free ebooks at bookboon.com943. Laws of ThermodynamicsEngineering Thermodynamics551P1V1 − P2V2 1x10 x1.7306 − 8 x10 x 8.653Therefore W === +222 kJ1.292 − 1n −1Worked Example 3.31Steam at a pressure of 10 bar and dryness fraction of 0.96 expands adiabatically to a pressure of 2bar according to PV1.12 = constant.
Determine the work done during expansion per unit mass ofsteam.ptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)1000179.910.001120.1944762.812,7782.13876.5865Please 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.com953. Laws of ThermodynamicsEngineering ThermodynamicsSolution:At 10 bar saturated steam, the specific volume is:v = vf + x (vg -vf)= 0.00112 + 0.96 x (0.1944 - 0.00112)= 0.18667 m3/kgfor the processPv 2 = v 1 1 P2(1 / n ) 10 V2 = 0.18667 x 2Work done is∴W =W =(1 / 1.12 )= 0.7855 m 3 / kgP1V1 − P2V2n −110 x10 5 x 0.18667 − 2 x10 5 x 0.7855= 246 kJ / kg1.12 − 1Worked Example 3.32A nuclear reactor generates 3000 MW of heat.
The heat is transferred in a heat exchanger ofenergy transfer efficiency 75% into steam which is expanded in a turbine in order to produce apower output. The steam is condensed in a condenser, releasing 1800 MW of heat, and pumpedback through the heat exchanger by a feed pump which requires 3% of the power output from theturbine. Determine:a) The net power output from the plant.b) The power output from the turbine.c) The overall thermal efficiency of the plant.Download free ebooks at bookboon.com963. Laws of ThermodynamicsEngineering ThermodynamicsSolution:a) Consider the first law of thermodynamics for a cycle:∑Q − ∑ w = 0(3000 x0.75 − 1800) − (Wt − 0.03Wt ) = 0450 − 0.97Wt = 0450Wt == 463.9 MW0.97HeatExchangerTurbineW net = W t − W p = 0.97W t= 450 MWNuclearReactorThis is the net power outputCondenserThe cycle efficiency is:η=Pump450Wnet=x100 = 15%3000QsWorked Example 3.33Milk initially at 30ºC is to be kept in a chilled tank at 5ºC.
If the total volume of milk is 100 litres,its density is 1100kg/m3 and the specific heat capacity of 4.2kJ/kgK.a) Determine the heat extraction rate assuming the chiller to be perfectly insulatedb) What would be the chiller consumption if heat transfer through the chiller body is?i.+ 5kW gain in summer, ii) -5kW loss in winterDownload free ebooks at bookboon.com973. Laws of ThermodynamicsEngineering ThermodynamicsSolution:a) The Energy balance of the system implies that heat absorbed by the refrigerant is takenaway from the milk in order to keep it cool.Q = mC p ∆Tbut m = ρVi.e. Q = ρVC p ∆T=100= x1100 x 4.2 x(30 − 5) = 11.55kW1000b) In this case there is another source of heat exchange, hence;Qchiller = mC p ∆T (+or −)∆QPlease 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.com983.
Laws of ThermodynamicsEngineering ThermodynamicsThe sign (+ or -) depends on whether heat is lost as in winter or gained as in summer.hence;Qchiller = 11.5 + 5 = 16.55kW in summerandQchiller = 11.55 − 5 = 6.55kW in winterWorked Example 3.34You have a 200 gram cup of coffee at 100 C, too hot to drink.a) How much will you cool it by adding 50 gm of water at 0 C?b) How much will you cool it by adding 50 gm of ice at 0 C?for ice assume hi = -333.5and hf = 417 kJ/kgKSolution:(a)Heat lost by coffee = heat gained by cold watermc x Cpc x ( 100- tc2 ) = mw x Cpw x ( tc2- 0 )0.200 x 4200 x (100 – tc2) = 0.050 x 4200 x( tc2 – 0)solve to get tc2 = 80oC(b)Heat lost by coffee = heat gained by icemc x Cpc x ( 100- tc2 ) = mw x hig + mw x Cpw x ( tc2- 0 )0.200 x 4200 x (100 – tc2) =0.05 x (417-333.5)+ 0.050 x 4200 x( tc2 – 0)solve to get tc2 = 64oCDownload free ebooks at bookboon.com993.
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