Thermodynamics of Open Systems: Energy Equations and Problems, Lecture notes of Law

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( ) same thebemust msenergy ter theall of units the )365(W) 2 V(hmQ Wgz) 2 Vpv(umQ W)gz 2 Vvp(um)gz 2 Vvp(umQ gz 2 Vu(T)PEKEu(T)E vp mvp mWWWWW vp mVpVVppdVW LawFirstW∆EQ shaft 2 shaft 2 shaft1 2 1 1111 2 2 222 2 2211shaftout flowin flowshaft 1111final 1initial 11in flow −+++∆= ++++∆×= ++++×−+++×= ++=++= −+=++= ==−== += ∫ gz FIRST LAW IN OPEN SYSTEMS Steady Flow Energy Equation Open, steady flow thermodynamic system - a region in space Q shaftW p1 v2 1V 2V 1z 2z p2 Steady Flow Processes Devices EquationEnergy FlowSteady Wgz) 2 V(hmQ shaft 2 +++∆= Turbine, Compressor, Pump ( )outin hhmW −= ∆== =∆∆ hm∆HW 0Q Elevation, Velocity, ( )outin hhmQ −= == =≅≅ h m∆∆HQ 0 Work0,∆Elevation 0,∆Velocity Boiler, Condenser, Heat Exchanger Diffuser, Nozzle Valve - throttling process outin hh = = = ==== outin HH 0∆H 0W0,Q0,∆Elevation0,∆Velocity 2 Vh 2 Vh 2 2 2 2 1 1 +=+ ==≅ 0W0,Q0,∆Elevation 4.233 An adiabatic air compressor is to be powered by a direct coupled adiabatic steam turbine that is also driving a generator. Steam enters the turbine at 12.5 MPa and 500 C at a rate of 25 kg/sec and exits at 10 kPa and a quality of .92. Air enters the compressor at 98 kpa and 295 K at a rate of 10 kg/sec and exits at 1 MPa. Determine the net power delivered to the generator by the turbine. airsteam v v g2393.2kJ/k2392.8.92191.83h @10kPaxh@10kPahh 3341.812.5)p500,(Tsuperheat@h 628.1kJ/kg620)Tairtable@(h 210.5kJ/kg295)Tairtable@(h t2 fgft2 t1 c2 c1 =×+= += ==== === === 12.5 kPa 500 C 10 kPa x=.92 netW compressor turbinegenerator p1 Mpa 620 K p 98 kPa 295 K ( ) ( ) ( ) ( ) kJ/sec 19539W 210.5628.1 sec kg102393.23341.8 sec kg 25W hhmhhmWWW net net c2c1ct2t1tcompressorturbinenet = −−−×= −−−=−= 4.65 Steam at 3Mpa and 400 C enters an adiabatic nozzle steadily with a velocity of 40 m/sec and leaves at 2.5 MPa and 300 m/sec. Determine (a) the exit temperature and (b) the ratio of inlet to exit area. Open thermodynamic system - a region in space 1 3 MPa 400C 40m/sec 2.5MPa 300m/sec p 2 v /kgm 09936.v kJ/kg 3203.9h 3.)P400.,(T @superheat v&h 3 1 1 11 = = === ( ) ( ) .1517 1.1757 .1783 A A 273.15364.78 300 A 2.5 273.15400 40 A 3 V A RT pV A RT p AV RT pρAVm b) C 364.78T 2.5)p3159.7,(h @superheat T a) 1 2 21 21 2 2 11 1 1 2 2 == + = +       =           == = === kJ/kg 3159.744.23203.9h /secm 1000 kJ/kg 1 sec m 2 300 2 40kJ/kgm 3203.9h 2 V 2 Vhh equationenergy flowsteady 2 Vh 2 Vh 2 222 222 2 2 2 2 1 12 2 2 2 2 1 1 =−=             −+=       −+= +=+ Steam at 3Mpa and 400 C enters an adiabatic nozzle steadily with a velocity of 40 m/sec and leaves at 2.5 MPa and 300 m/sec. Determine (a) the exit temperature and (b) the ratio of inlet to exit area. Open thermodynamic system - a region in space 1 3 MPa 400C 40m/sec 2.5MPa 300m/sec p 2 v .1872 74.535 13.951 A A 11626. 300 A 2.5 09938. 40 A 3 V A V A ρAVm b) 11626. v 2.5)p3187.5,(h @ v C, 376.5T 2.5)p3187.5,(h @ T a) 1 2 21 2 21 1 11 2 2 2 2 == = = = = == = == vv /kgm 09938.v kJ/kg 3231.7h Mpa) 3.P400.,(T @superheat v&h 3 1 1 11 = = === kJ/kg 3187.544.23231.71h /secm 1000 kJ/kg 1 sec m 2 300 2 40kJ/kgm 3231.71.9h 2 V 2 Vhh equationenergy flowsteady 2 Vh 2 Vh 2 222 222 2 2 2 2 1 12 2 2 2 2 1 1 =−=             −+=       −+= +=+ 4.65 im A 200 cubic ft tank contains 2. lbm carbon dioxide and .1 mole helium at an initial temperature of 70 F. 3 lbm of air at 14.7 and 70 F are admitted to the tank. What is the final temperature of the tank? ( )fi1 1 1 1 mmm T u h −= 1p ( )( ) ( ) ( ) ( ) ( ) ( ) F167T R627T 0381.6323.831.125TQ 381.670460.243hmm 323.8370460.745.4.15652um 1.125TT.745.4.15652.1743um hmmumumQ o f o f f 1fi ii ffff iifiiff = = =−−= =+××=− =+××+×= =×+×+×= −−−= 4-153 kJ/kg 3167.7h C350 MPa, @.5 kJ/kg 2292.51h 2163.8.8561.47h hxhh kJ/kg 2725.3vh C 133.55 @300kPaTT kPa 300 @ o o 1 1 fgf1 g2 saturation2 = = ×+= ×+= == == ( ) ( ) kg 9.78kg 10kg 19.78mm kg 19.78 kJ/kg 2725.3kJ/kg 3167.7 kJ/kg 2292.51kJ/kg 3167.710kgm )h(h )h(hmm )h(hm)h(hm0 pv,usince vpmumhmvpmumhm0 vpmvpmumumhmhm0 umumh )m(m)vpmvp(m0 ,for W ngsubstituti vpmvpm W umumh )m(mWQ m cylinder,piston in thefinally mass theis system the 12 2 1o 1o 12 1o1o22 11111o122222o2 1112221122o1o2 1122o12111222 boundary 111222boundary 1122o12boundary 2 =−=− = − − = − − = −+−= += −−+++−= −+−++−= −=−+−− −= −=−+− boundaryW8 kg liquid water and 2 kg vapor at 300 kPa are contained in an insulated piston cylinder. Steam at .5 MPa and 350 C are admitted until the piston cylinder contains only vapor. Determine the final temperature and the amount of steam admitted. oh steam ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 hhmHH W0,Qfor hhmHHQ 0,for W EEWWQQ WWW 0 233) (page UUEEWWQQ W∆EQ WUUQ UUWWQQ 0 173) (page UUEEWWQQ 32)-(2 UUEEWWQQ UKE, PE, W,Q, forms, all is E 72) (page ∆EEE conservedEnergy , definedEnergy 212net 1212net in massout massoutinoutin flowshaft 12out massin massoutinoutin 12 12outinoutin 12out massin massoutinoutin 12massoutmassinoutinoutin outin −=−== −=−== −=−+− += −=−+−+− += +−= −=−+− −=−+−+− −=−+−+− =− space in region a SYSTEM OPEN mass of quantity contained a SYSTEM CLOSED LawFirst