4 Example Problems with Solutions on Thermodynamics | ENSC 2213, Study notes of Thermodynamics

Download 4 Example Problems with Solutions on Thermodynamics | ENSC 2213 and more Study notes Thermodynamics in PDF only on Docsity! ENGSC 2213 THERMODYNAMICS Example Problem 1 A control volume operating at steady state has one inlet and two exits, as shown in Fig. P4.20. Using data on the figure, detetmine (a) the mass flow rate at the inlet, in ke/s, (b) the mass flow rate at 2, in kg/s, and (e) the velocity at 3, in m/s. pees eee = 12 | f> ig =? Steam nd weet Ts mess Ay =06n2 | e > Sat. liquid Y= 50ms be==a=e= 4 pg = 1.5 bars py = 10bars . Ag = 0.018 m2 T, = 400°C thy = 50 kgs  FROGLEM 4.20 Known: A cowtrel volume operates at steady slate. Dak ave known at - the Inlet aud ot each of two exits, Euue: = Determine (ay the ranss flow vate at yha thlet,(b) dhe macs Hownte oubat 2 audcey the velocity at, . SCHEMATIC é GAYE DATA p=, 1 4 ‘ fig =? Steam in fantrol i 2 Ay = 0.6m? = | I Sat. liquid Wy 50m/s Peeeeoa + py = 1.5 bars PB, = 10 bars Ag = 0.018.m2 Ty = to0°c tng = 50 kes ASSUMPTION: The control volume is at steady hele, ANAINSIS:@)For the ass flow wate in ati, use Be Ub ANT - my vd From Table A-4, Uy = 06,3066 we/kg thus ~ 0.6 m2)( 50 m/s) \ "0, 3066 wi) eg . HALES kglsg m cb) Using the mass vate lochance o a . ¥ . : ss geet my He- m3 => ms mw wa Thess Wa,> 47.85 -50 = 4785 kl sg vas (2) Using Beg Hl Wy US Via As From Tolle A-3 ok 1.5 bars , Uy 2, = 1.0528 X)0) Me] ey. Thus _ C80 kgls) (i.o528 4103 wi/kg) = ore se Ki mits) (0.018 wy = 2.92 wis oy m3 Na 4-24 RHE PROGLEM 42°77 .. F ‘ious: Air flows through @ nozzle with known conditions atte mlet and exit. Heat transfer eceurs from the aim fo the surround wigs. Fiala: Determine the exit velocity. ‘Sesemanc £ Given Dama’ 2 . ee = fi —> nozwle-+—> 72 =S70% Wee a1 | 60! @) ie = eulm 2 Te Shas lb ASSUMOTLONS (1) The control voli.me is at steady state. (2) For the corte! E volume; We, *0.(3) The air bebiives as an ideal gas, (4) The inlet kinetic F enérgy and potential energy effects Are negligible. ANALYSIS: To determine the éxit velocity, besin with the steady-state energy balance e - gO. . 2 e O= Qy “he +m {ch-ha\ + LEN) 59 03/7203] Solving for Vid ves falthhs) ¢ Se] " From Table A-22& , laj= 140.81 Btusle and hy = 136.26 @tu/l, Thus el =| ave = 186.26) Bia/lb + (L10 Btu ](178 Ht (32.2 Hb /s 1 Ib =/Su tts . Ve 4-28  ENGSC 2213 THERMODYNAMICS Example Problem 3 A well-insulated turbine operating at steady state is sketched in Fig. P4.40. Steam enters at 3 MPa, 400°C, _ With a volumetric flow rate of 85 m3/min. Some steam is extracted from the turbine at a pressure of 0.5 MPa anda ‘temperature of 180°C. The rest expands to a pressure of 6 kPa and exits with a mass flow rate of 40,000 kg/h and a quality of 90%. Kinetic and potential energy effects can be neglected. Determine Q , {a} the diameter, in m, of the duct through which steam is extracted, if the velocity there is 20 m/s, (b) the power developed by the turbine, in kW and Btu/h. Power —pD ou [a Pi = 3MPa T, = 400°C p> (AW), = 85 m3/min , P3. = 6kPa V2 = 20 mis 3 = 90% po = 0.5MPa rig = 40,000 kg/h Ty = 180°C PROBLEM 4. 40 kNows: Stenm passes +hrough « well-insulated extraction, turbine With kugum condition at the inlet aud exik. wb: Determine (4) the diameter of the duct where steam is extracted aud (6) the power developed hy Ha durbrhe, SCHEMATIC ¢@ e1VEN DATAL 2 re oe OL eee Dy = 30 bars 4 Turbine LE f= Treaoorc Hh é b ASTUMPTIONS: (1) The cantrel volume is at (AM = 85 nmin EE Stendy Staté, (2) Heat transfer is w BEmf 5 3 eqligi He. a no. G) Kinetic aud polewbial energy effects com V2 = 20 mis 2 oloon Pa = & bars ma = 40,000 kg/h be neglected . T; = 1g0°c 3 . ANALYSIS: (a) To find the diameter Dy, beaiw with the. Wass vate bafence eg 4 Soe : = = Wh, - Mantas or Woe = ms Evaluating va, with ax zo.ngg 4 mike frm Table A4 ~ AT) | (85 w/min) Ge) mF er (0.0994 meg) 60S FURS gis “ Ts vay =(14.25 keals) ~Ho,cos ky, Lh) #34139 kgls aud NZ A= Tha = 234 kgis) (0, 4045 mii) og gage mt . > (ao ms) where the value of vu, is from Table A-4, Finally HA, 2 Dy ap = 0.284 m » 2 (b) To determine the pour developed’ use Ha energy vate balance oO 40. . Oo 4a oa a a =f cv T™, Ciey gf \- imal ered vag leap if where the (udviated tevms- ave deleted based wpon the assamphions. T Wey > wy, -vahs- ms hs From Table A-4 5 hh, = 2230.9 KTlky aid bh, = 2812.0 keTlkeg . the Specific ewtaloy at 3 is determmed usihg dain From Table Abs Wig? Wigg+ X3 Wag, 2151534 (A) aus4 = 2525.8 hk wo = 2gi7u e 4-42 PROBLEM 4.74 kNowal: Air flows through two turbine Stages and an late r-connecting heat exchanger. A Separate hot qir'stream passes im counter tow through the heat exchaager. Data are known at various locations. Nb. Determine the temperature of the main air sheam exiting the heat exchanger and the power output af the Second Hirbine. SeHemaric £ Gwen DéATh: . = : 1 . We 718000 bw . We 2? ASoumpTioas :(1) The tontral i = } bra a | j Volumes are at steady state. @) Heat Fausfer fotke Sur rounidings can ue vieglected. 100K @) Kinetic and potential | . peti may energy effects are negligible. ae analy : (4) The air behaves as an i Z i ; ideal gas.G) for the heat Teleak TAMA Ty2 780K y = aL 4 = exchanger, Wey 29. Aeiibors | u Fame F Re Tbar 771200 K : i, (200 A. Tg Hak Az t bar a 22 iba Wage 1260 kgf/min Avatysis: First, find the air flow pate at 4. Begin with steadsy-stecte energy amd mass balances for turbine ° 9 . . . ° 02 We + mr, [Oh he) + CEs ge5Fa.] a ee a a Selning form, we * Mm th * an Chi-ha) From TableA-22 4 h,= 1515.42 btlleg ama hy= Hol.07 kolky. Thus . _ (70,200 kw) \ sas \ _ ™* Cisisaa = 161.07) ig 1 ew 28.22 k5/s Turn ny next do the heat exchanger wa, 2 a = Ws = ty : WMs=mi , ‘ap . aD oy BeZ ‘O . : 2 oO O= Br? yal? saa hth chy 2 +4g¢ede tngL Cas he) HES +gey-ag]| D = tH, Cha-hg) +s Chg he) ec hae hit "AE (ag-he) Agatis, From Table 8-22 5-lg = Votlng kT Vey amc hy 212-77.79 let hy . Thus 4-33 PROGLEM 4.74 CCgat'd) _ C220 kglnuis ). hg NV6l.o7 lee #2 gl skeen) (1611.19 - 1277.74) 213490.D kT Iheg Interpoleting iin Table A=22) Tz * = 1245 Ke wT Now, writthg the steadey- ~ stele energy balance for furbive 2 02 AE Wat ig [ths he) GPRS, ¢ gepZ2] “Wa = ig Chg hy) From Table 4-22; hy # (023.25 ETTky , and We, =@8,22 kg/s)(134¢0.0- 1023.25) ET thy na) or = 10.350 kw Wate 4-34