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# THERMODYNAMICS ENSC 2213

OK State

GPA 3.78

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This 0 page Class Notes was uploaded by Ms. Abe Bernier on Sunday November 1, 2015. The Class Notes belongs to ENSC 2213 at Oklahoma State University taught by Afshin Ghajar in Fall. Since its upload, it has received 13 views. For similar materials see /class/232953/ensc-2213-oklahoma-state-university in Engineering & Applied Science at Oklahoma State University.

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Date Created: 11/01/15

EN GSC 2213 THERMOD YNAMICS Example Problem 1 A control volume operating at steady state has one inlet and two exits as shown in Fig P420 Using data on the gure determine a the mass ow rate at the inlet in kgs b the mass ow rate at 2 in kgs and c the velocity at 3 in ms 39 2 Steam in 3 06 m2 gt Sat hquld v1 50 ms L p3 v15 bars 171 10 bars 0 A3 0018 m2 T1 400 G0 n33 50 kgs Control39 volume mam LlD KMowN 39 A to halvchaw 6 ks a1 s rm sk a Ufanave lawown cs 39 m lA QJ 6753A GEEM a Wyatt75 I Ezmz 39Ddtmmha m an mas ew ratquot2 aft4amp1 MMAb ap was 5H owm Mod 1414 c 44M velad 113 swim 1g gym M 2 1 Hgt 7 Control 3 Sta m volume J 3 A1 06 mg D Sat llqmd Vl 50 ms p3 15 bars A 10 bars A3 0018 m2 T1 400 C quot3913 50 kgs hggmenow 39T ni camml volsz is 55 shah w mmsar we Mass Haw minim 41 1152 39ll b ANT 7 w u l Fm Ta ole A39 transom wake quotMas 06 Mz50 Mls 39 asceIawa Kj 4185 Ices 5 us39m Jr39Ae MASS 3923 bcQma M dA V Y VALVA3 MLVI Y3 Thus max9185 50 H185 53154 4y 5 Reina Haul 39 V39V s u39z a T3 w From 39TaMe A a 9A Ls bars v3 17 19528 X10 1K5 Thu so ksls 10528 Xio 33n3K3 008 n13 39 V 2A1 mls I 39 3 3 ENGSC 2213 THERMOD YNAMICS Example Problem 2 I Air enters a nozzle operating at steady state at 800 R With negligible velocity and exits the nozzle at 570 R Heat transfer occurs from the air to the surround ings at a rate of 10 Btu per 1b of air owing Assuming ideal gas behavior and neglecting potential energy e ects determine the velocity39at the exit in fts 890 TABLES IN ENGLISH UNITS Tau AZZE Ideal Gas Properties of Air T R h and uBtulb 37130116 11 when As 0 when A 0 i5039 460 056235 1080 26097 18693 076964 480 057255 1120 27103 19425 077880 500 058233 1160 28114 20163 078767 520 059172 1200 29130 20905 079628 537 059945 1240 30152 71653 080466 540 060078 39 397 081280 560 060950139 quot 580 quot0161793 K 600 0626071 quot 2 620 06339 640 064159 1480 36389 26244 085062 660 1520 37447 27026 085767 680 1560 38508 27813 086456 700 1600 39574 78606 087130 720 067002 1650 40913 29603 087954 067665 9 83 14398 4734 1950 092504 070747 6573 49088 35720 1571 4598 860 20646 14750 071323 7149 4457 2000 50471 36761 093205 1740 42758 880 21135 15102 071886 7761 4201 2050 51861 37808 093891 1923 3949 900 21626 15457 072438 8411 3964 2100 53255 38860 094564 2121 3667 920 22118 15812 072979 9102 3744 2150 54654 39917 095222 2335 3410 1 p and 0 data for use with Eqs 643 and 644 respectively FRDQLEM 4 1 KM Afr lows Hirauqh a mule wH k known wnd 39lims ake 1415 and exit Heuf n Yan s fer occurs ram 59 air a Hne SumunAIA s am Dekmshe 142 exH Velocffy V WAnc eweu 39 39 1 ecbog gt U720 v 0739 gV39 I r nozzle gt E 57092 I lt2 I WhA ro 15th V Wm Thecbnffvl volume is mquot shade awe 2 ForHe mam volume 39 Way 0 3 The mm Lehaves a an ideal 513 4 The u nlei lakeHg s energy MIA fdkn lIhl W733 aged are he uylue W113 de ermfne he xi velocHy begin he sMsJm energy balance v f 0quot Rev o UM ha i3cz 56ng 4390 V1quot 2 ilk1M 4quot From Table A zza by me 941mb Whitman eatu Thus V z 7 77s44l Anzt LILs 11 J 2msx lasagnau Q BMILJH6TW 5us 39 39 r v 39 WE 425 ENGSC 2213 THERMOD YNAMICS Example Problem 3 V A wellinsulated turbine operating at steady state is sketched in Fig P440 Steam enters at 3 MPa 400 C 4 with a volumetric39 ow rate of 85 m3min Some steam is extracted from the turbine at a pressure of 05 MPa and a temperature of 180 C The rest expands to a pressure of 6 kPa and exits with a mass ow rate of 40000 lggh and a quality of 90 Kinetic andpotential energy effects can be neglected Determine 39 a the diameter in m of the duct through which steam is extracted if the velocity there is 20 ms 1 the power developed by the turbine in kW and Btuh A Po ver D 3 cu P1 3MPG T1 400 C Saba Av1 85 m3min t 39 p3 5kPa V2 20 ms x3 90 p2 05MPa 13 40000 kgh T2 180 C Hammad 4 30 93ng 54PM passes 44ka a wleM msulafed Mcan Whig WE HA known cami um ed M vale339 anal aids Eight Bakumine M Hue Ahmadarm Hag dud where sham is exmcu1 1M b Hm Power developsA blj Ha Wbuhe SeHEMMm g ast Mm 15 I 30 bars 39 1 0 C Z esgum Pnous l The vohfahg 5 a AV 85 m3min sieud S hexz He39d muskr Is he Ifsiue 3 Kmdu xv poisum mixqu 2444 mm A V2 20 MY 353 2 p3 006 bar a P2 5 bars be a hc ki T2 mac MALXSIS 03 To 4th quotHA9 diama eer D3 bani X3 901 1513 40000 kgh w39rHa ke ms nuke ba amca d w 39 4 WIl mz ma or 3 7 k y 3 EvahuAkg n n w d k 4n 0 u l i 4 3149 WT b eA39 39 Ar 85M3Mim mm W Tr mom m T 515 l4ng MVHWZ39 M 39 3 W k I s m Qbms ksls 39aoo 9 Sims 3139 kjls MAI A1 jaw 3m kals 4045M3i 3 00335 r 70 mls when name a 39U39i is 39VDVM TilHe F kau 1 7 028 Wt D H To d ermMe 39HAE power devalopeal use Hm envy53 rade ba auce 9 0 D a n 0 a h 39quot v TW I Hg I M1kzg m3k39 s ni iga Hm uciLinL veA HM are deKeM baseAW musuwam V90quot 9 quot 39W lkl W s me39TcMe kH lm 323a lekj M hzaslao Erika The Seecl39Hc ewkhmkpas M 5 is Ademmd v5le denim Fm Table A3 L13 hi3 x5 M433 151534 mamas 23258 5amp3 9quot 25579 3 4 42 PRAQUEM 4140 Cem d I serh uj Vaku s r k F H7 13 5 Iocoo kw Wu L a s Iszsoqua 31333l2b Whaziglbms I 1137 lewd VOW Cemva MB 4 Bulk SZOMLIJ v1 311 kw 352mquot ahah 1 WW W 39 BM COMMENT This ProHem Nahmks quotHm use 04 mass awi awry rake bauxite 4w Volum es LUin mu i ipla Mlds wad0r wigs 44 3 EN GSC 2213 THERMOD YNAMICS Example Problem 4 Air ideal gas ows through the turbine and heat exchanger system shown below Data for the two ow streams are shown on the gure Estimate the temperature of the turbineside air leaving the heat exohanger T3 K and the power output of the second turbine We in Wu 10000 kW 39gt TE 1100 K p2 4 bars in 1 T1 1400 K 171 12 bars T5 1480 K 5 p5 1 bar ni5 1200 kgmin 6 Heat exchanger 7 T5 1200 K P 17 6 5 Air PRCELEM LL14 Snowy Air Hows Hawnuah huo39 Wayne sings and an der conneq rbg heaf39 exchanger A Separaia h f am 171247 passes in cauufer ow Hm ugh Hag had exchanger Dam ara known a varzbus Ia mans ND De rmlhe Hi hwm xn ex 12 main air SWM EKHIhg 39HAE hear exgbmnger amt re power om p m 514 he Secand r r pe SC EMIiC 5 111121 2m 39 39 r 39 1 gggmgngwgzonhe Con39l39r39ul 39 Volumes an m seadg Shae 2 Hemquot mns r Jo te Stur 510000 LW b mundahgs can be waglecfzal 32K 3 may cud po wh39al I quot5 emu35 eqec39 are Hegtiglbla if W 1 The air behaves as an 1 z T dead gasfs For H e hea f 77Hoa1lt AAA 5 39 7Bolt fzbars 4 L l bar exchangeq w 0 p aaamger 7100 Ilt 2 1 Ear m g bur gt 745 1200 Igmin g qgls Fd S l39 nd We air How rah ad 1 Begin 15an shangV kda ememg GUM mass balmIces h urbu39nel D 0112 V Vt 39 quot f OhIna L z 3c7 1 Md 0 mpnaz gt Se v na for 14 Wt A l klk1 From TablaAzzg 14 1515le Mlle 5 and 3941 Mam7150 lug i i EELS m Sls q2ll l07 Tkamp H ZE39ZZkaS Turn n3 nexi 5 HA2 head xchmm er Wx z A WA W W5TAL 9 o 7 20 9 W7 0 O cu39wum139quotfk3 7392 WE3Mskks l e VB 7 393tg D W1M139M3MSCASl 53 39V of 13 lnzi V chsLJ Again from mag A225 Ins Mumq lef mm kG27739z Iallrj ws 4 33 FKDGLEM 4m emu lnquot Hvlo1 7 3 s 1512 kgs os mm mm 771 39 3qdo my 39 gt39 a BaniSK I gt T3 mg mma 142 k sgaqa39wg balance T39Wgge z V0 QCVquot Maiquot 6 3 340 t l 394qu 4 W ii 339C 13h3945 V w 39 Fram Table 422 h 102325 ml5 I 39 1228ZZLglsnla oo10152ETc5w Or l icd39ls 0550 Lu v We 4 24 ENSC 2213 T HERM 0D YNAMI CS ENSC 2213 THERMOD YNAMICS Chapter 4 THE FIRST LAW FOR OPEN SYSTEMS AJ Johannes PhD PE Professor of Chemical Engineering School of Chemical Engineering Oklahoma State University Copies of Class Lecture Notes Portions of these notes were originally compiled by Professor Robert L Robinson Jr ENSC 2213 T HERJW 0D YNAMI CS THE FIRST LAW FOR OPEN SYSTEMS A Topics for Discussion lDevelop the Open System Mass Balance 2 Develop the First Law for Open Systems B The Open System Mass Balance Consider mass owing into and out of a system at various points on the system control volume CV boundary control surface cs ENSC 2213 T HERM 0D YNAMI CS Conservation of mass requires Final mass Initial massCV 2 Mass in Mass outcs 01 m2 m1Zme e Where sum over all inlets quotiquot 1 In rate form we get dm Zmi Zme dt i e Where n1i i ow rate of mass into the CV Closed System Zmi 22me 0 no mass crosses boundary so m2 m1 3 ENSC 2213 T HERM 0D YNAMI CS C The First Law for Open Systems Let s see What the energy effects are for mass owing into and out of a system Consider the following situation Here an element of mass 5m flows into the system control volume What is the effect on the energy of the system ENSC 2213 T HERM 0D YNAMI CS Let s focus only on the mass flow across the boundary disregarding other heat and work effects Control Volume Closed System Elclosed ElCV E15m E2closed E2CV SO E2 E1cv 2 E2 Elclosed E15m ENSC 2213 T HERM 0D YNAMI CS For the closed system we know how to write the rst law for the process shown E2 E1closed 1W2 the boundary moves SO E2 E1closed PidVi but dVi ViESmi 0139 E2 E1closed Pivismi SO E2 E1cv E15m PiV18mi but E18m 2 U1 gzi V12 51111 SO E2 E1cv ui P1V1 gzi 7128mi hi gzi 7128mi ENSC 2213 T HERM 0D YNAMI CS Thus the mass coming into the control volume increases the energy in the CV by the amount of its ke its pe and it39s hi ENTHALPY pronounced N THALPY The fact that it brings its speci c energy plus a pm term arises from the fact that it compresses the uid already in the system in order to get in I39 39z A WF V Wk o quot u mm mm m 22gt 41 M I I i OWL a Ml ENSC 2213 T HERM 0D YNAMI CS A similar but opposite effect occurs when mass ows out When we account for all effects 1Q2 and other 1W2 effects the 1st law becomes d UmgzEVZJ5Q 5W Differential Form 2 ofthe 1st Law hi gzi Vf5mi Zhe gze ve28me This can be put in rate form by dividing by dt i Umgz72jQ lV dt Rate Form 1 A2 ofthe lSt Law hi gzi 3Vi mi Zhe gze 515 me where Q 2 heat transfer rate W 2 work rate m 2 mass ow rate ENSC 2213 T HERM 0D YNAMI CS Eccnditicns T p etc at the inlets and outlets do not change With time we can integrate the differential form to get Integral Form of 1st Law Recall U2 n12gz2 722 U1m1gzl712 1Q2 1W2 v2 1 l Zmihigzi2 1 l 2 2 Zme he gze V6 6 ENSC 2213 T HERM 0D YNAMI CS D SteadvState SteadvFlow SVstems SSSF In many practical situations systems operate for long periods Where no changes in conditions occur Within the process e g process temperature pressure and ow rates are constant EJPB gt y3 gj 53th 53 gt195 W mi me are constant and E2 E Then 1Q2 mi hi gzi 712 1W2 me he gze 7 2 So the SSSF 1st Law in rate form is QIhi gzi Vi2 WIhe he gze Ve2 10 ENSC 2213 T HERM 0D YNAMI CS E Some Useful Flow Relations Q 2 VA W Here Q i heat 1me 11 ENSC 2213 T HERM 0D YNAMI CS F Summary of First Law Relations CV gt Control volume system CS 2 Control surface system boundary For any process dECV 5Emcs 39 5Eoutcs Accumulation f Flow of f Flow of of energy energy in energy out inside CV across CS across CS 111 m dECV dUCV gZCV 2 VCV C C aEmCS5mihi ziiyjj 8Q gc Zea cs anutCS5mehe zeiyjj 8W gc ZgC CS 12 ENSC 2213 T HERM 0D YNAMI CS 1 General First Law Differential Form dUz vzj 5Q5w gc 2gc CV 25mihi zi712 i gc 2gc CS 2 5mehe ze LV62 e g 2g CS This is the general form of the rst law and all other forms can be derived from it KNOW What each term means and how and at What location to evaluate it 2 Integral Form of First LaW U2 mzg Z2 m2 V22 go 2gc CV U1Z1n1 v12 1Q21W2 go go cv Zmi hi zi LV12 i go 2gc CS 1 A Zmehe ze V3 6 go 2gc CS 13 ENSC 2213 T HERM 0D YNAMI CS 3 Rate Form of First Law iUz 72 Q W dt gc 2gc CV Zmihi zi73 i gc 2gc CS Zmehe zei7 2 e gc 2gc CS G Summarv of Mass Material Balance Relations Differential form lt6mgtcv amacs lt6 macs Integral form m2 m1gtcv macs g macs Rate form d 3011M mi CS g me CS 14 ENSC 2213 T HERM 0D YNAMI CS APPLICATIONS OF FIRST LAW TO OPEN SYSTEMS A Topics for Discussion 1 Develop applicable forms of 1st laW for speci c processes 2 Solve example problems B General Forms of 1St Law 1 Closed Systems Mass Balance m me m2 m1 1St Law g 2g CV g 2g CV 1Q21W2 15 ENSC 2213 T HERM 0D YNAMI CS 2 SteadyState Open Systems SSSF MB d 39 a mCV O gt 2 mi CS 2 me CS EB 1 A 1ECVO QZmihi zi V12 dt g0 2gC CS WZmehe zei7 2 gc 2gc CS 16 ENSC 2213 ITHHUWODYNAAHCS C What Can We Calculate Using 1gt Law Closed Systems 0 1Q2 only from 1St law until we study heat transfer 0 1W2 from 1St law g de nition of work 0 Initial state of system Final state of system Open SSSF Systems Q only from 1st law 0 W from 1st law or de nition of work Inlet conditions Outlet conditions Calculation Tools lst Law 0 P V T u h relations 0 Description of process 0 Mass balance mpA 17 ENSC 2213 T HERM 0D YNAMI CS D Some Typical Open System Applications 1 Turbine 39 ke and pe often negligible 4 o Q20 often the case check 39 rhiz le 39 pe lt pi Wmhi hegt0 2 Pump or Compressor 39 ke and pe often negligible ml W 0M Q20 often the case check 1rhi rhe for single inlet and outlet 39 pe gtpi Wmhi helt0 quot quotnumber 18 ENSC 2213 T HERM 0D YNAMI CS 3 Heat Exchangers W39ul 39 i I 39 Q Svstem Overall Exchangg o Q20 Often the case insulated 0 W20 mu mle and H121 mZe M o ke and pe effects Often negligible Then mu hm m21 h2i mle hle Ih2e h2e OI39 m1he hi1 m2 he hi2 Svstem One of the Streams 39 h h 91 11116 1 Q1Q2 Q2m2 he hi2 So Heat loss by one stream is gained by the other 19 ENSC 2213 T HERM 0D YNAMI CS 4 Throttling Process This applies to a throttling valve or diffuser Valve Diffuser 39 2 g I w a 2 r 1quot 0 W20 0 QOcften the case insulated o keamppe effects are often negligble Then mi hi 2 me he but Ini rne mass balance SO hi 2 he Note for a Diffuser the ke change may nct be negligible 20 ENSC 2213 T HERM 0D YNAMI CS E Comparison of Thermal Kinetic Energv and Potential Energv Effects Example Heating one lbm of liquid water 10 F increases its internal energy or enthalpy about 10 BTU Comparable PE effect is 10 BTU AZ go gt AZ i 7780 ft Comparable KE effect is 10 BTU 230722 4712 L 2 V2 2708 ftsec This illustrates Why KE amp PE effects are often neglected 21 ENSC 2213 T HERM 0D YNAMI CS F Behavior of Properties for Compressed Liquids Consider the following data for water T 50 F p psia V ft3lbIII u BTUlbIII h BTUlbIII 018 psat 001602 1806 1806 500 001600 1802 1950 Difference 500 psat 000002 004 144 For v effect of p const T is negligible For 11 effect of p const T is negligible Forh hupV AhAuApV AupAvVAp Ah z VAp T constant 22 ENSC 2213 T HERM 0D YNAMI CS Check the results for Ah Ah h 500 psi h 018 psi Does Ah VAp i Vpr 0016ft3500 0181bf144in2 BTU 1b in2 ft2 778ftlbf m 148 vs 144 from table gt OK Thus for compressed liquids at pressures not too far removed from the saturation pressure you can use u p19 T1 i u psat9 T1 V p19 T1 i V psat9 T1 h pi T1 i h psat9 T1 Vf 131 psat 23 ENSC 2213 T HERM 0D YNAMI CS Example 1 A control volume operating at steady state has one inlet and two outlets as shown in the gure below Using data on the gure determine 1 the mass ow rate at the inlet in kgs 2 the mass ow rate at 2 in kgs 3 the velocity at 3 in ms I w quotl 12 1 II I Hgt 5 Steam in 33713 393 2 A 0 6 2 I lH gt Sat liquid 1 quot m l v1 50 rule P3 15 bars p1 10 bars A3 0018 m2 T1 400 C 3913 50 W5 24 ENSC 2213 T HERM 0D YNAMI CS Example 2 Air at 800 R enters a nozzle operating at steady state with a very low velocity and exits the nozzle at 570 R Heat transfer occurs from the air to the surroundings at a rate of 10 Btulbm of air Assuming ideal gas behavior and neglecting potential energy effects determine the velocity at the exit in fts 25 ENSC 2213 T HERM 0D YNAMI CS Example 3 A wellinsulated turbine Operating at steady state is sketched in Fig P440 Steam enters at 3 MP3 400 C with a volumetric ow rate of 85 HIPmin Some steam is extracted from the turbine at a pressure of 05 MPa and a temperature of 180 C The rest expands to a pressure of 6 kPa and exits with a mass ow rateef 40000 kgh and a quality of90 Kinetic and39pdtential energy e 39ects can be neglected Determine quot a the diameter in m of the duct throngh which steam is extracted if the velocity there is 20 1115 53 the power developed by the whine in kW and Btuh c Power 39D Gut 2 73 P1 3MPa T1 400 C Lam up Awl 85 m3fmin d W 15 3 I p3 Gthz V2 20 11115 x3 90 p2 05 we m3 40000 kgh 2 180 C 39 26 ENSC 2213 Thermodynamics NAM PRACTICE FINAL EXAM 7 7 7 7 Problem 1 A rigid container contains 5 kg of a gas R 04 kJkg K The gas in the container is heated from 25 C to 75 C The gas can be considered ideal Calcuiate the total change in internal energy in k if the heat capacity is represented by 36 000 LT amt 72 L 04 7 CVC ZK39LB Oo IT 3 MM 4000quot 7 775 MM 000 1 T 017 2 8 598 2 WT moz r A 0442493 wm NSF gt mt i983 90 7 3 Numg V 4 4 Wind kn VIJ V WA 3 665 gt Au 9644 A Answer 3277 kJ EN SC 2213 Thermodynamics NAME PRACTICE FINAL EXAM Problem 2 Air assume ideal gas 2 10 is contained in a pistoncylinder apparatus as shown below The cylinder contains st ops which limit the travel of the piston between the upper and lower volumes of 06 and 02 m respectively The piston may be considered frictionless and 5 9 8 0 r a quotU 23 no if a vr the 0287 072kJkgK 2 M 93wa 194 a What is the initial pressure kPa of the air V Tquot fv m QT M AT a ozxLLeo 39 O T 06m v P 3917 Tb m VO m3 4 E39me A b What is the nal pressure kPa of the all a o z lg loggingquot 7 CV 1 09 c What IS the final temperature K of the air rev eevt 4 lt21 59004 7 quot7 V77 3quot 400 900 d HOW much work k1 is done by the air during the process v am clan 71 3 10 cmnlaml quot we V pdv 7 VV VL V39 gt 00 V c How much heat kl is transferred to the air during the process Ct Tm TI5 amp Q Q Ma m3l66 299000 Answers a 700 kPa b 500 kPa c 143 K a 150 M e 8272 kJ 30 W 2 NAME ENSC 2213 Thermodynamics PRACTICE FINAL EXAM Problem 3 Two Carnot engines operate in series The rst cycle receives energy by heat transfer QH 500 H from a high temperature reservoir at TH 727 C and rejects energy QCJ to a reservoir at an intermediate temperature Tl 427 C The heat rejected from the rst cycle qu is the heat input to the second cycle QHg The second cycle produces half the work of the rst cycle and rejects heat Qc z by heat transfer to a cold temperature reservoir at To Determine the temperature of the cold temperature reservoir Tc C 7 A 7r quot TM MMK M 04 9 991 quot 3 I 139 0quot 0c 5 7 an 6MP met on 39 Kan 7 94 0a WP 50 I W w I Is 771 J in i 025 514 QC iO KL T W f 7 quot 7 if TH TMWH Z 72 WIN 1245 Answer 2772 quotC ENSC 2213 Thermodynamics NAME PRACTICE FINAL EXAM 40 Problem 4 Air an ideal gas at p 100 kPa and T1 l7 C enters a well insulated compressor operating at stead state and exists at p 3373 kPa and T2 147 C Kinetic and potential energy effects for the compressor are negligible Determine the isentropic compressor ef ciency 11c in two ways 3 Assuming variable specz c heals use data from Table AZZ X 5742491 S39 b at 1 Assuming consrant speci c heats use k 14 244 P 100W lt FL 9953 591 z 9 a g T13 T mm M vizze Frquot I12quot 0 wu hr 03 7m 44 quot94quot Lug w I 1 l 4239 tw Z 39 14 UW Xw I 39 J I gt 7 mapa 1 fl r f 7F n397 106 446 L 0 5 9277 LIL3916 gt 7909 03 hz A1k I39LI mu 393 73 I 1Wquot IMf K 40 40 A III12 Kg 15 mo agilz WIl 4mm 39hlg blM 42539 mm 41 57 z md o Z t 1 77 41 lE 06 0426 42 A Answers a 923 b 927 ENSC 2213 Thermodynamics NAME PRACTICE FINAL EXAM 5 40 Problem 5 Consider a Rankin cycle with water as the working uid Superheated vapor enters the turbine at 480 C and 10 MPa and exits at 6 kPa The turbine has an isentropic ef ciency of 80 Determine for the cycle a The actual turbine work klkg b The actual rate of heat transfer from the working uid passing through the condenser to the cooling water klkg J T g V w I t wok 17 L339 2 W WWW 1 P fu no aw ALE 10 95132 611 25b0HI529 7 39 90394 a zl 141 200923 1 77f 0 30 J an A 3 hquot 7M in quot453 quot1 571m 8 2HIH1quot 939 3 a 39 k 39 A3 lelss k li li397Zzw rAgfs a 723 397 4111 a Gmg39 nlwzooms at 4 5 w z A quot73 7 gumWM 0 7 Hr 39 V WWW9 EWI JLMV Answers 1 10493 kJkg 0 21206 kJkg ENSC 2213 Thermodynamics NAME PRACTICE FINAL EXAM 40 Problem 6 An ideal cold air stanclard constant speci c heat analysis Diesel cycle operates under the following conditions Vi V2 l7V3Vg2 13195 kPaTL725 C k14 29 K For this cycle determine a The thermal efficiency b The temperature K and pressure kPa of the air process c The temperature K and pressure kPa of the air at the end of the heat addition process 1 The temperature K and pressure kPa ofthe air at the end of the expansion process 3 2 I flu I a Melff L 39 01 igt Q l5 zquotquot 1 bzagt4ml 7 n Wqu at the end of the compression in L4 1 L l J r gt 717 T e 1 am qTZ 7 2 593 7 N39 P3 1 golSVQq q J L l 4 H 1 v a 39 quota L z Ti7 v TqMr gtW ltvl II394 f2 PM gtsowi Fq Z WWW Answers a 623 b 9255 K amp 5016 kPa c 1851 K amp 5016 kPa 1 7864 K amp 2507 kPa ENSC 2213 Thermodynamics NAME EXAM 3 Problem 1 A Carnot power cycle operates between two reservoirs at 800 C and 20 C One half of the work output of the power cycle is used to drive a Carnot heat pump that removes heat from the cold surroundings at 2 C and transfers it to a house maintained at 22 C if the house is losing heat at a rate of 62000 kJh determine the rate of heat supply to the power cycle in kJh required to keep the house at 22 C Ti 0006 1073K Tu az c 245 K 39 H GZ oa Eb 3 l4 5 Q zo c273lt 122 c 575K 62 1 614 pump 31 2 quotm 2 75 72 7 eaZR erxx QHHP LiH Qbup We WW t Lia 71 teaser Tiq 0 MA quot 15 r H 7347th 2755 1amp4 a W antease more Krh 37 on 1 z I V 7quot 073K Q M 2 were 35 21125 5 7 vtX 0397Z 7 H 2w EN SC 2213 Thermodynamics NAME I f EXAM 3 gas expands from an initial state of 527 C and 2 MPa Problem 2 Nitrogen N2 as an ideal he change in the 929119 entropy of Nitrogen to a nal state of 27 C and i MPa Determine 1 during this expansion process in kJkgK in two ways a Assume Constant speci c hears use data from Table A 20 at 277 C 3 b Assume variablg speci c heats use data from Table A 23 LOW aw 622 E T7 l 7 A5 5 Cpih 4 R in l 39 71271 2 550K when 15mm 755 14 20 CF 5 Nae 39 CP KTKS K R g 83 KIvmolK 0249 KjKJ39K 39 z 01 KJKwi M Lgt From Tidal A i 7 KT MPa 5 1 4l 27379K OZ391W 1 A 065 K6Jlt H 5211133K KJ K ZMPR gm i639HS 6 O205 13 As 0413 KTKA H FYO M a6 0 205 o o 7 A551551 53I elem5 4mm W525 AvZB 700quot M2 JVLuff o 775270C 00K 5 2220407 KIKmeiK I a 72 27 9a Sack 54 2 KIKhwIK a K I 332 me A5 LIQIVQZ 220 0m 3 LIKMK n 2 57ezlts w261225 23LiQZL KSKWK A5 A5 40507379 Agni M Kg Kmoi ENSC 2213 Thermodynamics NAME quot EXAM 3 Problem 3 Steam initially at 10 MPa and 200 C undergoes an isothermal reversible compression process in a closed piston cylinder assembly The process stops when the steam in the cylinder is saturated vapor Determine the heat transfer and the work for this 5 compression process in kJkg Enm bisAna Jar a cMtt 5353 33 P 4114 m m 30fcfma FWaJ l MW AImm varr M Q 77 T5ZquotltIgt EZG23 Y Q 7 5 5 o m l 2 Fn 735 A f 7L0 MP1 439 7 Zoo 6 0E 14293 5 23 5 Mf KJIK 75644 2 2 7200 C j 557 anyr Ff M KT 5 15 KJ Ma 975 a 52 3 23 W Ttszrsl 2 M 20DZ73lt a 323 4M a E 39p Kj K L3quot slzs gk 244733 3 2575 3 2627 f ENSC 2213 Thermodynamics NAME EXAM 3 Problem 4 Air an ideal gas at p 449 bar and T1 277 C enters an insulated turbine operating t steady state and exits at p2 144 bar The isentropic turbine ef ciency is 85 Kineatl ic and potential energy effects are negligible Assuming variable specific heats use data from Table A22 determine the actual work developed by the turbine in kJkg 39l l WVra Me 7t 52 9 Z Vial Z 4 52 Mm V435 mm 44 MW 7 27774273 550K 7 S i thl KT KO Fa get 7 7 B ltF Y 5 z 5 F u lq Z39v J a l 7 WM l l l quotPas 1 kg of nxao 5 330 FrOM Wick A412 yak at 3404 his Lioo ie KTle W ZL7l125 0 555174 9100452 K47 lBDj KITKg

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#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.