Engineering Thermodynamics Tutorial Sheet 1 Non Flow Process & Working Substance-Air QA perfect gas for which the ratio of specific heats is 14 occupies a volume of 0.06 m? at 10 bar and 500 K. The gas undergoes an expansion to 0.3 m’, Assuming R=0.25 ki/kg K for gas, find the heat absorbed or rejected by the gas for each of the following methods of expansion. (i) Constant Pressure (ji) Isothermal (iii) Hyperbolic (iv) According to law Pv! 1? =¢ (v) Adiabatic ‘oan Porgest gan, reams edsel G7 Py = MRT | Iwo 0°06 = an x 02S X SOD ? : 2 M=OUe ky 2 a ak Jeo G-GaR > & Om = OE gy 0-eoe BIT YeK v Get = 0e7e kal ky k S Te (a) Sto Li- fe lobor= loo ela Veo-obm TreSoPK i 4s Ser tr Rake lobar, Vy =O. d™ Wo = PCY) = {oor x (93 — 0°06) 2 = 240 kt y [ - ac 2b 2 vw, = mGlh-D 77 he = 0eUs ¢ 0-625 (25¥0~ $00) O = wr My = boo +240 er tuo kT "AH = madT dh = Mop (TT) = ONE KORTE x C2ew~Suy) = eyoke () Tsotrounat Sate ti. = (wba, Ye oem, T= SW k cera Dat Corbheh) Vyeo-tm? 0- Ww, = PY dl #)= foow x006 bn 9) We Fes7 ke OB. = Wh = 96ST bs @) Hypebwbe (pv=c) Gur on Detal go | Augporote pce bo Tre dard a, wootheunal poten } ” Ge) Revstding 1 Law pyle (pvc) Ti eso0 K Sto = tbm, Yr owosm, as ie 3 Stet 2i- AV = AY, , Vs oam rer oot Egg rk vy gee) . . % = % CB) - yore (2%)ons LN = hy o-asm Be (EY = CR ~ loonn aos — [68:5 x 03 loon oe — [68'S XO Os” = 27-4 RI V ma bt) = OME x 0-620 x (uok-9 svn) Yon 273s ke @. = wr wy = 27-32 + 747 Bo = 60rlY RI ), Adiabatic Prowns The head tramperd dump adinbahe’ pown & Yeo Myer = me (ET) 0-4 _ 0:06 = 262-65 wu" = ss) * c Oo owes (ape gm) = Te 8 Actkernaiwely \" sy orb) = los-0d kg P= Ck) = tox ($e eet) o- wo RY cBY, _ lomo xorot = jos-06 10:3 ve Wo oY = 2 kFQ2 Air undergoes two processes in series Process 1-2: expansion from 300 kPa, 0.019 mi/kg to 150 kPa, during which the pressure-volume relation is P v=const, Process 2-3: Constant pressure process to v3=v1. Sketch the process on p-v diagram and determine the net work per unit mass of air in kW/kg, assuming air as an ideal gas, Solukwr, P tate Hyperbolic CTsetexmol ) expamou Pia hey 2-B ie Osu Prenue hah ghch Li Re zerkla, % = 0-019 "ig Sto 2 PBA) P= Iso 4ePa, = 5 saz 31 a = Isola p= Bj= 001m, Prown 1-2: = A® 30 * 0-019 orem’ Be = tb a AKER = 003 ~ Py so /rp : 002g We = Wey (%) = 800 K O19 Dn (oe = 39s bol Procom 2-3 Wy, = ROKR) = Ix Cor0lg ~0-03¢') =z — US bo | Ky War = Wt = Br (22) = Py RT [hpQ:3 Air initially at 0.75 bar, 1000 K occupying a volume of 0.12 m* undergoes two processes. The air is compressed isothermally until the volume is halved, then it undergoes a constant pressure process until the volume is halved again. Assuming the ideal gas behaviour, sketch the process on p- v diagram and determine (@) the total work for the two processes (ii) the total heat transfer for the two processes Lotutun Proce |-2 1 Egetnewmal comronsum 2-8 Goutant Prue wolry Stale |: P= OTS ban V, = 012m, T= lovo . 5 = 0-0b%m Sia 2- = (or RMR Y) Vr=$\e = 0.03" Slate 3:~ Pa = Pa, ; Vy = LM = by = Proous (2 ? ae) i 4 2 Wa = AY, nC ; 0-06 = 7 KONL In a = 608e ko * 2maCb-T) = et Pe CET) e0 (y i tanitipians) vo | =-6 kt 6.2 Ma = 6.28e doal & epuol To | [nete- change of; utteunal enw fon on ial fe, | ger focan varourel perce , pecadine | sntousl enoyy of omperaiare 01y-- ) | Proeen 2-3 1 Pve = AV) pe PME Trl = Ist k fe ve tel s ® W, = BCR-Kw) = Id (OB - 0:06) Tee Hseb, = malh-B) For on Ga ON Rabe» Re 024k T ep, Sn cruse. of; adlaal goo Wh our, Atal et “ta y= MRT, TS 8 Ore = om 01287 RIOD me 0103186 be yo = oo3ise xaos (TT) 23 Promina Proce we (For Cota a Ly, 5 2. = 4 HHS k lood L U = 0.03136 x O7IE X Com pave Jas? &7 237 iWon cc $, = Yt 2: = ease + Cvs) = - lew kT Woo, Ave Gaatent phone poo _ —h a) CTs ) 0: 03136 #005 ¥ (se -1oe2) go = 23 = HICSS Rt Co) fatal wee for fr pe = Wot ay ©= 6-228 4 (Us) = 0-728 kg ii) fetal haat trope. fur tue pune, = Gr Ss = 76238 + C07) > -21996 bg QA A perfect gas undergoes a cycle comprised of three processes, It is first compressed isothermally from 1 bar, 27°C to one-eighth of its initial volume, Energy is the added at constant pressure increasing the temperature of the gas, and the cycle is completed by a reversible adiabatic expansion to the original conditions, The isobaric specific heat capacity is 1.25 kW/kg K and the characteristic gas constant is 0.5 kJ/kg K. Sketch the cycle on p-v diagram and calculate the following: | (@ maximum cycle temperature (ii) the net work per unit mass Setutun Proum |-2' — Thotramal osmpenaur a-3° Heat addubum al chnatonh pontine. a-|: Adiobahe eepansu Gp = bes I leg k R= Of BI [hy G-a = B pele = OS Oe = Or KT Ikp |}ye Gems 1. | a os 3 Ler ws Amira | runt man uf gan, Pv mR T, Ixy = 4x OX 36d}smS = too kPa, M vi Sta | Pye Shaw omen eT (& Bray), W=by, PVs J hv. = hy ae loo %V) Ye \ yall av om (4% ) = W, = 29 Me Te 7 smiz aie & = Provan |-21— Po v T= = pom = hom, TE2TC H Ber k, = ovlerc mS Tap? wen k Pe jv elicln (4h) = ~Biloae bs , Vy ad Us) ad1E oe Ve ad ) m4 “; ‘aye yi Ve Y ( = g.usoe 4 8 tL " too NE or) QiNsok al 689-3 kKMarymum Jemperarue of wa cycle = Tray = Th = 6¢9.2 k Mowcmum porwr 4, wre tyle Prop = Pah = 20 f Py WW, = CAM.) = Yoo (0°Ua0e — Olea) = 194 ORT Bs = MG(G—h) = Ie bern (6898 -30) Suge 6 ko Proun 27) , —— — fy, Y, = AMe ) I — BW x OUZog — [U2 KIS = werowor si x $1) Wi = 291-96 RF Ld 64 = 9 Net 7 M 4 OE. yk Var = Ne Cc os tl TNS FONG Y FPF HS = IU 7ke Nek sak bran is Beagie = Sear = Ot at 8) = 39 + 486-6 + O => \ITWYURS Thue, [fd = SEMQ.5 A quantity of gas occupying a volume of 0.3 m’ at a pressure of | bar, and temperature of 20°C. The gas is compressed isothermally to a pressure of 5 bar and then expanded adiabatically to its initial volume, Assuming the ideal gas behaviour, determine for this quantity of gas (i) the mass of gas (ii) the heat received or rejected during the compression (iii) the change of internal energy during the expansion . Assume y=1.4 and Cy=1 kJ/kg K Setubion The gas is amumad To behove Whe an Baal gay Procom \-2 t Tootnermal Compr ur 2-3! Adiobali erwponne q = | erly & yey Co t= @ e = J = 0-713 2 o Iv ° lek Ra GGe £70743 ¢ 0 -2es7 Ro) kek Sto Li- Pi = took fa, yeosm>, 7 228K gnie2in Pas km, hat (orbw= Au) = 4 ? ; Vg = V, = 0-37 Sie B!- Py Yq = Proce Ind Ue = AY a Vi = loxdd _ o-0g m3 ee. 0:06 = PYin( = tones In(gee = —4e-28 RT PV = MRT) | lox ose ™ KO-2ST HK 2A ma 0-368 kp6. eee Pawan 2-3 Be [20D By = ate “ ab - &¥ | vw, = V- er | Pa Yel = by! B (oa t= sw x (0-06)!% S28 KP | a = smn (9:06)! = ryt ce ) ) = ) = - a bkQ6 A volume of 0.36 m? of a gas at 288 K and 1.03 bar is compressed reversibly and adiabatically to 10 bar. It is then cooled at constant pressure to its original temperature after which it expands isothermally. Find the heat transfer at constant pressure, the heat transfer during expansion and the work transfer during eycle. Take y=1.41, R=0.287 ki/kg K Solutwn Rvp = MRT, 103 PORE = MRO WTR 2E m= 0-449 be ' Proeen I-21 Adtabahe compro us 2-3 Const Promur covlry g-| +7 Tsotheunal expan $ Ta star (:- P, = loa kPa, Vy 20:36, T)=X€K gta 2!- Pa lodokla , ayt= Ay = state b!- Py=hr=tovo kfa , Pv, = AY, (B=T) - #£ -9- qwiek 2» Ge wa F o-7 Gaon = 407 = 0-9F7 wer) pk 1 Proc I-21 - Ayla RV 1. M 036 = o-oTle ms Va= (7 OM) bh. - 298 Tay = w( By" (“38 bu bh = $se-76 k | We = Pil “Ble ak (1,-T.) © Yo Y-] i = OUT 0-287 % Coep-cxe-o) eg “4/Procans2-3 plea Ws = BOM ve) QB, =m O(B-TL) = ONG K 0-987 x (2H - <5B-E) = Nig-iz kg if Now 2-) PM = hl | Jos x0-26 = loo Wh | V, = 0 102708 nt | pees w= p.CYy-Ve) = love (0 -0370e — 0 -071e) (2 = -ay. zie fees Qe y= piv, on (Be = [03x0 > tn tem) > Sy. bs Net wn Me + ow, + W, = uw + (Ruue) C372) = —3S-Lkege Nek har eanufer = ‘So + 8,¢+ &, = 0 +(air) F S428 = —te kgQ.7 An ideal gas with a molecular weight 6.5 kg/kmole is compressed in a reversible manner from 690 kPa and 277 K to a final specific volume of 0.49 m'/kg according to the relation, P=561+200v+100v", where P is pressure in kPa and v is specific volume in m'/kg. The specific heat at constant volume is 0.887 ki/kg K. Determine (i) final temperature (ii) initial specific volume (iii) ‘work done on system (iv) The heat transfer be Ry, 2313 Tenstck Sube Pw aRT Re ee belt Ra 1-279 kI/, 690 yas, = ].279 6277 Bre B= OS1BS M?/ Ky Stl 200%" + loors,2 = Stl +460 Cowen) +1a0(0¥7 s a 5 61705 kPa Pv = RTL e179 ONT = OPTS Te ae Ue EK Ya Cou + 200 + wo) Fe Wa = S tae - -¥)) 4 wr) +12 (w?- ve SH CHEM) feo wow o-siss Ol (07-0135) +1 (o hosts 4! ay = 20-7 Rely ; g-8-277 au, = GCE T) = been CHEE O77) = wot kel a Uy ee = meas HOY) = —SU-TY ky |pQ 8A fluid system undergoes @ non-flow process following the pressure volume relation as P=4.5/V#2, where P js pressure in bar and V is volume in m'. During the process, the volume changes from 0.12 m? to 0.04m' and the system rejected 40 kJ heat, Determine the following (i) change in internal energy ({i) change in enthalpy S ni 29S baw Re Wyre LS 42 \ - [YS bar Pho AS yr = S42 = o-04 Fa Non-ftoo foam fi = faut (sw 4 Sa = favs (pay WwW = (ees SG ape yetee av dw = es sa) + acy) | * . is (53) bg a@oo-on) |* 1037 #F de At fw yo = dt (-s10°37) aus "70-37 ks Change om Rntadhy Methos T —— R ur dy + s (Pv) sor) J dit = - w) + (ay) WH, = W037 (usp x 0:04 ~3850 x01) HoH = 4S 87 ke ©E He utPv dts d+ Pav-+vaP L 2 au a fan = fav J Pave J va? 1 » > R-2 Ap 4S bn 27 fre {rey a 7 whl Uys 2 2.) = 494-37 bo we) Py omd fy mem bar ea er Py = MUS baw ua au 2 fan = fat J Parr sve \ ' f vat, = uauaT e (-s10%4) tity Werk, = usysT RSEngineering Thermodynamics Tutorial Sheet 2 Non Flow Process & Working Substance-Steam QI. One kg of steam at a pressure of 10 bar exists at followin; fraction, 0.8 (b) dry saturated (c) der 18 conditions: (a) wet with dryness temperature of steam is 200 °C, Determine specific volume, ity, enthalpy, intemal energy and entropy in each case @) P= bor, yao. From Steam Tablas ‘ = 0°tym/) = WORRY YR ay = 0:8(0-194) Vy ‘ye 9 x Or ISSA? key Keg = Deis ko] ry = 2139 kT Ne hy tO Rey 4 Ihe = 762-8 + 0-#( 2015-3) = 2375-04 ka lky us By = 44a halk k Ape = ZBTS-OU — [O00 ¥O-ISTTZ = 2219 &Y dea | ky A= bt Ay = 2189 + oF LY 4UE) = $6974 balty , a _ ! fot. if ~ 6 4u2d Ay fimd Onse 7 O) P= lobar, x= 4 om US By = O14 mY ky tha hy = 2778-1 ka] kp [ steom taute Us A-pw = 2778] —loronolay a rsey | kr )iy A> Ay = 6: S87 ka |p k f>4= iw = SHISGe kp fond C) P=lobm , T= 200% From Cieorm Tablas weo20eom' lip A= 28279 by 95 =0:2.060 ™"/ Ky ae 6690 ka Thy k = 2927.9 ht | by A= 6-694 kt Thy k Ua h-pur = 2827-9 -lovond206 = 2621-9 ba |hp @ fod Ju = Yo gsyy Replomnd foe 0.2060Q2 Steam at 10 bar and 240°C is cooled under constant pressure until the steam becomes 0.8 dry. Determine the initial and final specific volume and heat transfer per kg of steam, Sour Sta = P= lobar Ty = 20°C 0-2060) Cayo-200) From Steam Tolslbs .2397— 0° 2840 WB 2 012060 + (cae7 =) ?) 00%, | 250% so werl)| o-rve0 | 0:2397 A(eabtgy | -27°9 2qya-6 A a = 0-22736 ™* Thy, fy = 2ear.g (ners PTE) uoted) (as -200) = Mig. 6b kIT hy soe PraPalobn, A= ok From Skom “Totes eX Ne Bgr Ae lo baw 5) = 0: (98 = o¢(oey) vy 08 ie = 01S 2m Ip, Ap = eae "y Reg = 2015-3 holy AL > Ags + UC hye) = 762.8 + 0-8 (2015-3) = 2375. o4 Raley. ° AL- Ay 2237S OY — 2G 19-66 S oy, 62 ba] bp Moll dt a Ctate 4 amd Stale 2 are Pacated ut Pa (0 bar an State Lin obtamad by rowed of TY , ce Bw. TE Tazo w mot Shown on dioyram, is Ti _ bf . ne duraung Ta200¢ 9 T2200 pron or be dou Was Sepunh pad -Sta 2 in obtauned by antewedun 4 Pra loba and %=0-k A, = 2920 ka |ip A, 2 2380 by [ay 4 a hp, = 23% ~ 2920 = S10 ke ye Q.3 A pressure cooker contains 2 kg of steam at pressure of 5 bar and dryness fraction, 0.9. Find the Quantity of heat which must be rejected so asthe quality of steam becomes 0.5 dry Salut’ The beding of unter (food sbem una ppunure Coke, Aakwo plae ot coutant. volume. 7 V=C amd othe man trophont Da proton & Comtanr drove = State (.- Pech | %=09 Fasm Stem Tales At 5-0 ban By = 0-375 onl ty hy = 650-25 hy Ayy = Doms kal np Ay) %0-9 (0-275) OBB nb) ky A= Gur +0: 9( PS) = 2037 8F bhp (= us = 236q 1 RIN 205 She I- BLS, Ta A pet; 2 Beare = sue O-RTEyy, = 0837S mt rome [Real 8) Ve [0-63 40-5 w7 [069 | S163 0:337S & os (qr) W, = 06S nt | hy. Bay son viterp toh po Reg CT ae) (O67E-08) _ Rersh = ON (ovs69 0-693) id &, (or 2.675bax) Stop (MSH SHOP C2ETE- 2-H) (27-26) = UY 4s kal iy iv (ab a67s bor) = 2tI7B + (ity 2 277° D eae 1-29) = ajisel ka) bp Aye SIS poe (27D = lean kay ase ec hte = ORS 7s x 0°33 a tsuarar Roly GB, = MCR )= 2% (826-08) = |653-76 kyQ4 One kg of steam undergoes a reversible isothermal process from 20 bar, 250°C to a pressure of 30 bar. Calculate the heat-flow and the work done. Show the process on T-s diagram. Solyasr ‘= Stat |:- P= 2obar, Te290°c From Srtam Totes % = oll ons hy f= 290ns kolky B= 6SUS kok pe herp, = 2402-6 ~dwoV KOI) = depos I] hy saz 2: Pye Bobo Ta = 250°C From Seon. tables no = oroTee mp Ane 298 ka)ky - foe 69287 kalhe k Up = Map, = DBS S — Bos x 010706 = 2644 ko 2 > Map = q, = The-&) = (Q0+238) C6:287 - 6545) Ves 13498 Ro hag Uy = dey —2eeo Ss = —3ES AT) hep aU = Wye = (UU) + Ma ae (a yt ¥, 2 1 Wy 2 — FMS koh =13y9y = Bb (ML 2 My Moto D 7 Stale Ly bowted by mntoutctu® of f, 220 bar wa Maan euty Qeor 2 le bocateet by sncterscctun of P= 20ban T= 2v'C cwwe From Mothur diagrann 5 V=orlimj, hy = 2900 ka |hg A = 6S +9 Tega a) Da 0TH y Ay = Des holy Are 628 43 Map 4Q, = Tlbandi) = sora oem 67 = ~ poy 45 lig = APY, = 240 — rowox oI! = 26s ko | up Vy = ure Arh = 285y — 200d x 0:07 = 2c ae lig 4, = (Mey) + Me W421 = Caeyc-ree) + Wy Wy = 106-24 RT| hp Ue = Q5 A quantity of dry saturated steam occupies 0.396 m? at 15 bar. Determine the final condition of steam if itis compressed until the volume is halved, if the compression is carried out (a) isothermally (b) according to law pv=c. In case (a) determine the heat rejected and the work done during the ‘compression and in case (b) change in enthalpy, work transfer and heat transfer. = Solum (a) Togthoumal Sate 1!- PpalSban, Mab Fyom Geom Tables CTar = l9e-s%) IShar en i 0-13. m? / kp teste t= fy = aaa: keh 2 of Am Ay = 6-44 halby K | w= Ay PB, = 2792-2 —I800 x0 +132 4 = ray. 2 kale | 3 | a 1396 ™ = 3k ™= Te = Corsa Thy + | _T | seme BEM , BE m| . _ 9066 m?/Rg = 4 132) = 0r0 +T= Te 143°C ppowo|-2 2 Berea a2 will do wsbaue Prom Steam Fables t= 1S bow AT 1S-0 baw BX Bey Vy = O° 12m [hy D 606m ex, (0132) Ae = Puyea balay. Ay SOT Ay = (3978 kal Bp = sis kalhy k er) aos (19973) + we gg = WI8° Ih k = leg sckslap Bra Vi POS (U0) = ye kale tre Apt = Nees — 157 * 0,066 = iigess RITA 4g. = Tlk) = es 173) Car b4ve) ; = —973-25 kal ry wu, = Mg So ~ 2842 = — B7y 165 kaIpy = (4) ae egres = HoT fF Mo =) MW = —Se-se ao) he : T (6) Hypeuroe As the puta) phe w uncoryed , Bobar A. valuna Of puspecnes waged St Rernoris ba Adame State 2 Bal, hm= hy B® (4%) = xe, Ppa Se bar oe 4 (o- 182) = 0-066 m8 ug a0. 2 Py From. Steam Totes 2 fe Bo bow 0-06 ~ % ( 0-06677 ve, = 00667 mS/kip mu 0-499 v Ay = Jove. ¥ ka| hy hg loe-4 # 0-890 (1795-7) Reg = NM 786-7 daly = 27e0-25 o|by Us Ap Pte = 279¢s 1 - BoB w0-066 = ose7. ws AT | kp UW e WseT wr — 2942 = — 6-90 Ro |p Vee bh (Be). = [sex 0182 Mn ( 4) — 137-24 balay Ww y = (uy) + Wa = bas ¢ (187324) = —luyeg ko) by Apoh, = 27e5-1s- — 2792-2, =~ 695 dylhy.Q.6 A closed system containing steam undergoes an isentropic expansion from an initial pressure of 13 bar to 1.3 bar. If the steam is initially dry saturated, evaluate the index of expansion, work done and heat supplied Soluhior PA Bewhopu ows (42.=41) may be heppwoonted dy qquabin PUM = ennai adam behuetn ond T, mihi R= bm =) From Steam Tables We 1:0 bo aN ‘By = Orist m4] Ry fg = LIBS ba | ep r\ fy = 49S BI] k a - %, = Os) m3] —___________, Ay = 2187-6 kal by Ayo eas baby k ae uy = Ay Pye, = 27876 —120 ROIS) = 2891-3 alg Ste AI- Pa bSba, Asam 4y pet tn Tatty = 1.325%" Aa = 64k iy ee nol shy we ARCO AST Apg = 2238 10 +2 hy bp = ORT ba) ep le Bgg = S804 Ba Iip Ar= Apt He Agr) Gus = bae7 + 1 CS-8e4) A> 0 96s] Aye UNG + ogee ( 2238) = 239 44 kT |b Une Apo PV SB 2391-94 — 190K) OD = 2242-4] ka | bp Wm Ore6e CI) X I rms ay e&)nn pwr= RO Idx Cosi)” = 3 Chior Totany 1p om aida we 5h ne in ( lo) An (7-b1793) yn Wy = }3on % O-IS] ~ 130 -@ (> (302. Visy =) = 344. 0S ka) kp YW, = 2242p — 209)-3 = — BNE 89 tay Wd, % = & 4) te Ve aug. ag -rauq0r = OTe tole By Mellie daagwm. Sion com be, homed by & eer vitesecliin oleh andl potunatum pe (4 >17- 7 A Sta) Lime “Prom Gate £, o eepe (sg) slen a pl intent Qa | Bbor cue - From Mother Duyror Is a =! Die miley Me Lamy wer Ay = 2700 bo) by A, > 23¢5:0 kal Be 6S kalye Amos boltyr ® , PES Ge 26 7 aa = i OW = Tvey Maso wegeh N= Jo) Tatory 499 of 4 | wy Pech jamxo-le | \ ae PEG = BT Yr ka lig w= ARB = [30 x /12 by —J 2790 — [3u0y 0-16 = AED Ah Ups Ap hwe = 28 [30K PDS (wo )= 2229 hol hy — 353 ha bay = (Uw) + y (ve y) 2 asa ¢ 87142 Ie ya ao} ey Q7 In a steam jacketed cylinder, steam from 5 bar to 1.2 bar expands according to law Pv''!=c. Assuming that the initial dryness fraction 0.9, calculate the work done, heat supplied and change of entropy during the expansion. Sobukur State |1- Pas bar, BH) ® 0°9 (03%) m= or3a7s ™m8/ hp. A, = bYor2+ 09 (2108-5) = 2597 es RI ky hy bs) 237 Fe — SON 0-337T Us = 286-1 ko], Ave Ir Bot4 0-904 96) = 62259 ba] k, my sod From Steam tables AT S-0 bow a 0-375m3/ sa hy = bYor2 bal yy al Doss ly bee Velho} k, Byy? 961k T Sbar Ibo. @s%aim pth a= pall yf apebay 1:2 (ws) = 48H + O-teuG (ary 2) = 24%-Y ba V hy tee Aacpathe = 2380-4 —I20 x 2287) = 2232-2 holy doc. \:361 4 0-8649 (0:93%) 6 4se ka lie Wo = Pity = Pee n-) SHR 0-387 — 120 RK} 2387 ea eee, ee lu-) = 2os-3K ko \ty 28\ (usu, ) = 2132-2 ~23@-) i —\36-9 kg Vy a = Gy) + = 136-9 + 2.4 = bur ka |ep M-A= 6-99 bt259 = ONT Rage Q8 Steam at 14 bar is passed through a throttle valve with a pressure drop of 10 bar, The temperature after throttling is 150 °C. Determine (i) dryness fraction before throttling (ji) specific internal energy of steam before throttling. Selutin stare (i Pye ben, Rim ha Stotiz- Pra4ba, Thea Ive | Becaun Ata 2 3 daguned Complatey (Ly Hoe dndapendient Joroportia) | Hwe andl plank be quastin AMM | prte 2. Gok 2:- Rekba B=!sve From Steam Urey T abe ta = 0-471 an [by Ayam 2ISr-& kanye q we Slob dy~ ebay, Ay adr Z Aya 282-8 49 hyhia Ant % (Assy) 2752-¢= &30:3 + % (1904 D) my = 9-4) BX % Vz) ~ o-ge (oln) = O19e32m Ty ws Ay pw, = = as-is koly By Mots. Dapston Statc 2 con ve ated by sndescctum of, Pre Uborn wre ond Tee Isvr'e . Cte Thon deus a Aoiryontal Hove pom Shas: 2 by duck rob unterresd> Po IW bar awwe - Fm Met dapamn - a= OFesr A, = 2790 nolty Ue Ay - be, = = Lesy holy From. Steam Totes Ar 14-0 bay Ay = 1 Aye = 1aste7 ka bhp Lys Ot m5 | ep 30-3 kal ke 27582-% — [Yor % 013832 A ag B= OY (ty 2980 — yuoK OLY 309 A rigid vessel of volume |_m? contains steam at 19 bar and 300 °C, The vessel is cooled until the steam is just dry and saturated, Caleulate the mass of steam in the vessel, the final pressure of the ‘steam and the heat removed during the process, Satis '- P= bar, T= 300°C. From Steam Tele of 19bar =O iY = se otsay milty hye B026-Yeohy B= O-132u mee. AS 026-4 holy S30ve. we ahh = Borb-4 —[9und-192y¥ ae = 2774-84 fa [te 3 lma¥ = lo, = TTS bp | - + 7 | % O-way Pike | Sta 21- B= B, % =) Wy = 01324 in, = Vpn Sn Atom textos fer natwaked Ottom, Be value cf My 70am, v te of |S bow From Shon Table (abba) PL = IS bar Ag = 2792-2 olny Aye 2792-2 haley ws hh he = LJ9v2 —Isevnois2 = 94-2 ka | yy T S WAU, = 2342-27748 = = 130-64 ko] ky 8, = m4, )= 7-sv3 (-l8-64) | SS _18by 87 Rg “Prem _Metdu2s Dupom:- Sa L & boo by wtesectur of Re ISbox ET Hew amyFon Stab 4, drow wee bn whic *& unreureds paturclin Lume (1,= 1) Ay = Bolo ksfly Ar= 2790 kolhy Baw, = 0-12 dig Ps ISbow A, PB, = Bolo [9veK0 1d = 2763 bol 2730 —ISow0-13 Vy ups Meo bY = esas ka ly y \o 5 7-692 by BONS = T1692 (ase 2763) = —128%" ed O= ™ Cu-41) Q10 Wet steam at 4bar is enclosed in a rigid chamber and heated. When the pressure reaches 10 bar, the temperature is found to be 200 °C. Estimate the initial dryness fraction. The steam is now 5 4 bar, Estimate the final conditions. Also expanded according to law pv'? = C till the pressure ii compute the work done, the heat transferred and change in entropy in both the process. Solum State (2- Pa4baw | got 2'- Pe = lobo , T= 200°C gan 3 i= Ppa dbor, — Pata = BON Prot \-2! Cone volume haakmp atin Pv oe Becoun Atate 2 DB dokmast 5 Coenpulely ( dy wo Adupendent prepertis ) wwe hove BD Atak pom dete 2 rT P= lobar Tr= 2ove Stose 21- wo = 0-2060 08 | kp Bat went * Ape 2227-9 Ko Ip. At lobar & 20v%€ Ar = 6 694 ha] by Wy = 0/2060 MS) Ke AL= 2927.9 bull uae Aah fy = 6-694 ae i & 2927-4 — ura x 02060 = 2621-9 kg ly| State yp PraY bor, Bie, WH = B.=0-20607™S BR Hy, From Stam table bt 40 baw 02060 x 7%4(0-48S) By = O° Ue am Bl | % =o-vuyg he = bou-7 balip Arg = 2163-8 kyl fia bow zy oruung ( 2403-4) & = ee | = Isyog oy, Ag 2S N9 Ralep k Ave 1-777 + o-uyus (e119) = Yrosy ra} hy Us AyrR®, = 1554-08 — YoOR 0.2060 = U7), 6g ka Thy Gate 8-4 bm, mS Pal 400% %!7 = loon x (0.2060)? ss (12)"¥= or2060 = onan), Wy Vay O-Wyr & %y (O-UES) =) % XO GSue hye 604-7 + 0 -45ure (2133'&) = 2by27. 0 RI Tey use tahdg = 2b¥2-0 —UOORO-HY2 = dosa ely 4s » \ 777 + 0-98uTe CSI) = 6. OY ko Bye Proms (ot =P 34 2 yr b-by = 6 IY —4ose = Proven 2-3 pi Be ws > © Bes Wa = Ite bo | kp f° = sta) + as 2 -loT7 aaltp by by = bboy — b6IY = Uw, 2626S — Fury ep = 11S0-22 KI] py | aby ka hy k p= Hemme | tooo %_0:206 (ar) : byes = 2601-9) ++ Ib a 0:03 RI few QIIA vessel having a capacity of 0.6 m’ contains steam at 15 bar and 250 °C. Steam is blown off until the pressure drops to 4 bar, after which the valve is closed and steam is allowed to cool until the pressure is 3 bar. Assuming that the expansion of the steam remaining in the vessel is isentropic during blown-off period, calculate (i) The mass of steam blown off. (Gi) The dryness of steam after cooling. (iii) The heat transferred during cooling process. Stuhon = Ta = IS Sto 21- Pp = 4b , State 3!- Py = Bbor T22sve Arad) Vy Ur Procos 1-2 Blolmp off procera (Baerdacbe) 2-3! Stat i P= Sh Tre 20°C Pum Stow alles at ISbar & 2sn°e ~ ®, = Oviszond [Ap A, ama. 3 ho | ky a. = 6-709 Rs/by K Comatant volume. cooly ON 2 a 3 34Sur 2 Rs Abor, 42>) bo = 6708 Ro |e =A) Prom sreamtattts of 4:0 bar GIos—= N77 -4ACONG) | ye = 9.463 ntl Ae = O-9GNF bp =P TIT hating beg = SN kal iy BS Lepr hy = Cow karly % 0963 47( O'W) Ay = 188 hal XT O'We am day Ar = boW-7 +4 0-962u7 (21338) = 260-S5 baliey | Ur ahbhwh > 2AbbosD — Yoox Owes | = augat) Raley Bye Ba = Oe mB/ kp Stan 2: ee SbmM, a | From Steam. fblts of 3 -Obta | | | | VX wVya Vy = 0-606 m4 ily ONG % % (0406) Ae SS ta lky Ay = O-7389F hypo 2N6a-8 arly ye SUS 0-787 62-9) = Dsu-0 ka ly Uy = 2STy — SV KOUUS = 2020°2 ka) iy(a) mars ok beam —ewn off > Ma _ Ob = 394 Mm, = 5 - olse 7 ’ ‘t vob 134s by ™M > oS Owe Hy -M, B97 — 1NS = 2-602 ky (8) dynam oh Atenrn afin, Corley = Mz a 0 'T8SBF un = Q €) pas wonsponed dating ore Pe ~ 4, = Ug Ua * = p20 2. — WNED- 1) ~eeoey Ip blty = mG )= 18 CHM == 62), 2789Engineering Thermodynamics Tutorial Sheet 3 First Law-Steady Flow & Unsteady Flow QI. An air compressor is designed to compress atmospheric air( assumed to be at 100 kPa, 293 K) to ‘a pressure of 1 MPa. Heat losses to the environment are anticipated to be equal to 10% of the power ‘input to the compressor. The air enters at 50 m/s where the inlet area is 90 cm? and leaves at 120 m/s through an area of 5 cm?, Determine the following, a) Exit air temperature. B)Power input to the ‘compressor. Selurun At Tait i= P= look Pa T= 2k CLs sams AL= 400 = Fox joctm™ ac Bribie Pe = tooo kPa Ce= f20™/s Be = Some -- Ww. yao te Aw eR 2 Joo RU, = OTK 2B ca a, = o-suog mI hy : “4 ~ meohe&e = Foxe KSO Oss] ee Be 0 84053 mo hee - osisis BSR KID0 Be > Be. Wo oll m Ikp Pe We=RTe =) Jove KON = 0-278 Te Te = M0466 KW 2 =e 2 Ce! Plane SPER py G + eee ter Bt ee ¢ P-8 charges ww Be amd PE Negackup ” rot uw » Ayr hed + ne +¥ » GOT Te) + an Gigs lw_ 24% —390'66 sp? [20 Wo }oos (2 ) + aoa Ta (oe Nw = - 104-07 an Wa 946% thy Wa anw = 0-537 * (794-63) =D bY kahee Q2 Ina steady flow apparatus, 135 kI of work is done by each kg of fluid. The specific volume, Pressure and velocity at the inlet are 0.37m'/kg , 600 kPa and l6m/s, It leaves the system at a specific volume of 0.62 m'rkg, pressure of 100kPa, a velocity of 270mv/s. The inlet is 32m above the floor, and the discharge pipe is at floor level, The heat loss between the inlet and discharge is ki/kg of fluid, Determine the change in specific internal energy. Syutur Ae Tet P= Goo ka B= O37 mS/ky Ge long pr ourer(Be’) = fe lov kla Vem osim/iy Caz bron, w= 12 kale 2,-Ze = 32% a= —gkslhy PEE. 2 gt Am tee Ss ft taylt v at qze 4W = pet te t Sat (eo - ww a yea) 440 eth = (p.m -k¥) 4 CoS 4 ie a, 9 8l ay ( Gooeo 37 — 100” a) + Ibert dete = ecad~ 09) Se, ye 0 bol 38Q3 At inlet to a certain nozzle, the fluid parameters are: Specific enthalpy = 2850 ki/kg, Velocity Omn/s, Area = 0.1m’ Specific volume =0.18m'/kg. ‘At the discharge end, the specific enthalpy is 2650 kJ/kg and specific volume is 0.49m"/kg, The ‘nozzle is horizontal and there is negligible, heat loss from it. Calculate 4) Velocity of fluid at exiét from the nozzle ¥) Mass flow rate of fluid.j) Exit area of the nozzle. Sotuitaon “tars C= Somls Aye OTM ™ he resvtly = orlemly / 29 | etd - Cee 2 & he = 2650 bo ly Sree fet rage Be =0 saan iy ua 220 4 SD Lu Co = Um + Fes co GSH HS M/S Gi) oh Ane E O:] «SD 27-78 wi Be Orly . : 2778 = Aer bi Gi) me =) 0 +49 Ae = d-ouus me 39Q4 A steam turbine operates under steady flow conditions receiving steam at following state:- Pressure =I Sbar, specific internal energy = 2700ki/kg, specific volume = 0.17m*/kg Velocity =100m/s ‘The exhaust of steam from the turbine is at 0.1 bar, with specific internal energy ~ 2175 ki/kg,, specific volume =15m°/kg and velocity = 300 mis. The intake is 3m above the exhaust, The turbine develops 35kW and heat loss over the surface of turbine is 20 ki/kg. determine the steam flow rate through the turbine. At uilek P= Iso kPa UL_= 210d Aa |hp DLS Lo Fly Cee ms Ar er fos tok Pa Yes HIS pally we = Ismify Ce Some | as bhee 3 RS | Zrte= om We Th Rpfpee RC ISDORONT + 2262 + FE [ood see ar et waa ft. aS, pigs the GER AMT Tee Des + yee 4 WS | {ose | yor yp 9 81KS +(e) | 200 3S | ot ap IE = = lors 42198 S80 tm -W) oo SF Crores +0 2s 4 2IO® + as SW FIT TIT Fae om = 010614 ky [sew¥ Q5 A centrifugal air compressor delivers 15 kg of air/min. ‘The inlet and outlet conditions of air a mis, P=I bar, vj=0.5 m’/kg, C.=80 m/s, P.=7 bar, ve" enthalpy of air passing through the compressor is 160 kI/kg and heat loss to the surrounding is 720 ki/min, Assuming that inlet and outlet are at same level, find(.) motor power required to drive the compressor (_) ratio of inlet and outlet pipe diameters Sotuhon Drs P= tow kha Ee i- fee Too koe, Cu=lom/s Coz Bom/y, herby = 60 roly / B, =0.-0mS), Bea O-is mS hy, £3 /mn! = ome = ek y 1S by fama ‘e hy chomp Ww PE ww Qo on antler amd earch are ar Ao devel 2 a we Arrke )+ ose meh ws 2 2 = —|b60 +(e = 90 eto we —2lnis ke hy . . (mun HOR DRY wea 2s AL x ES x Ghee = 7 Ae Ce Gi) ome Aue Be By go 0's? Ce Ke a a 7 a ee Jo no-1S e we a Yyae 26 667- AL 26-66 F re Be. + ae \ obs ale.) alQ 6 The mass rate of flow into a steam turbine is 1.5 kg/s, and the heat transfer from the turbine is 8.5 kW.The following data are known for the steam entering and leaving the turbine. Determine the power output of the turbine. Inlet Conditions Outlet Conditions Pressure 2.0 MPa 0.1. MPa ‘Temperature 350°C | Qual TO0% ‘Velocity 50 mis 200 mis Elevation from reference plane 6m. 3m. Selukun ft unot:- p= Oba TL = BSE Ce sons Zpa om From — Akeomtalt oF fl ee. ALS BIBT 0 RT Wy [1mm = obo] ye Bek Po = Abow, ‘pee t Co = 26v™/§ Ze = 3™ Prom leon toile, of dba 2nd he = MIT SPER Ce” pZe WH hee Ge + fe += fhe Soe * 2600 (get ai bo [hp ABS Thee Stet ly, bs hee = 2 7 -3 S667 9, obo aeix (6 - BIBT 0 + SOT 1 = uses + 42We STIL RT| hy We Wr ye! = 655°7 kW Q.7 Steam at a pressure of 1.4 MPa, 300 ° C is flowing in a pipe. Connected to this pipe through a valve is an evacuated tank. The valve is opened and the tank fills with steam until the pressure is 1.4 MPa, and then valve is closed. The process takes place adiabatically and kinetic energies and potential energies are negligible. Determine the final temperature of the steam. aus takomit Cm, rm) | 4 Me - BM, 0 "0 nor atonce F zo Ce Qey em, Chet er qa) = SMe Chet Seed & 4 WwW, + [mala + oa 0925) — miu 7D ey ‘ © = = ~™ => Qu =o, Wey =8, meso ont (MM). =O Cranypo we KE Omd PE are magpie May Mhalomce egualim Fe? _ trong O me le Enoxpy Botener quoi gee fan 7@) any, OM Cao¥o4 —4,) = VY (2040-4 ~ 2549) By os2y ow (20¥0-u —Ur) = 37s+) fore) = Ot (QomH—ay) =2761 5 Boo"C aeo°< — we onle23, : . My D- 2003 O 217g | h Boyo-y Syg.¢ 3257-6 sly uw 2952-6 Why few 24-8 bk. Fiery, some. dee eo volun of FOU) tek the dete uF % dear bear the Vous of Fombrrotua (any behveon 20% and 499°C 45fiom unoar prterpdtoruss (ar (tbar & 340%) = ONMGT Ply A= 3127-66 Koly we 205203 Roly onc £Cyv) = 740 D Temperature fais Sehweon YORE and BSVC vt Iba & 242°e ay = Orla? |y Ap sev ksly we WSS br thy pra Tempore = Tha 342°C, M2 = Jos tara. = 2026 mon g nkeorntfusws wis fhe tom = Ma = My -M) = 2-024— O-WUE = [262 ky Q9 Ina steady flow system, a substance flows at the rate of 300 kg/min. It enters at a pressure of 6 bar, a velocity of 300 mvs, specific internal energy 2000 ki/kg and specific volume 0.4 m'/kg. It leaves the system at a pressure of 0.1 MPa, a velocity of 150 m/s, specific internal energy 1600 ki/kg, and specific volume 1.2 m’vkg. The inlet is 10m above the outlet. During its passage through the system, the substance has work transfer of 3 MW to the surroundings. Determine the heat transfer in kI/see. Solus) pt Tre - = SoD Rha , C, = Bovm/s Us 20k} P= ous | hy = OIMPe = Iovkla, , Ce= Iso m/s At Brit . Me = les ko \ey oe =homsly Zp-Ze = 1m om 2 20° K/min = © “bee jre 2éMw. 2ooo Pilbee gos kily Spee Shp faee Spee \- a! ic Zp Qe UR + Set 4yv 4 Ww 7 Ze = Pete te + 4 ee we 2 = (sp2—2 q= (Goxr2 — borrow) ae Clb —2050.) + pia + tein Cle) +, 400 \oso Yo wWrls wale 46= Yous ks s 7 eo = 230-76 kW Q10. Air flows steadily at the rate of 0.4 kg/see through an air compressor entering at 6 m/s with a pressure of | bar, specific volume 0.85 m'/kg. and leaving at 4.5 m/s, pressure of 7 bar and speci volume 0.16 m'/kg. The internal energy of the air leaving is 90 greater than that of air entering. Cooling water in a jacket surrounding the cylinder absorbs heat from the air at the rate of 60 ki/see. Calculate the following(a)Power required to drive the compressor(b) Inlet and outlet pipe cross- sectional area, Sotutus -- At Tater P= wore , me ote hy C.= mis at ct fee Tokla, Ye= ouemily Ce= WS mis ah = OUR fee Ue Me = 90 bay “to kslsec a Id kT [ty ou kpfsec A Pee Negle chap De one / i Wy W a Pe Up Ce Ga het © Fei + 2 ae ate w- Ge feise)) 4 Cu-Me) + as tL 2 L Gus C1) we [er vor ermrorid) + (90) + OSES we -267 kg % Wa eee de ot be aec —lob6-F kW 47YY. Q.11 An insulated tank of volume 1 m° contains dry saturated steam at 0.2 MPa. The tank is connected to a line carrying steam at 350 °C and 3 MPa and is filled with steam till the pressure reaches 3 MPa, Calculate the mass of steam in the tank at the end of filling operation and its temperature. Sdubm'- Man ens (m,-™, )t Sme- Em =0 @ Raton L Q, + em(he Se g2u) = me Che + &', 92e) Lu 8 [m (us Cog G22) — mC + 492) thy, Me=o | Qy =O, Wyo Chomgea om kG. and PE. Qe mee cred Eqpotws () gut mM, -M = Me ® Ruan ©) peo mA, = mum 4, o Cony = MUM, ms Chp—Ur) =m (An -U,)D ® Flom Steon fetos oF 2M (2bar) orol %, o/ ascieoatee blip bys 27067 rol y Tes che, = 2706-7 — 200% 0-886 - = pore ASN ty Ae} SMP (Bdbar) ond Igoe 48