MECH 310 Homework Assignment 2, Fall 2013

Problem 2 ; Steam is woled iv a closed rigid container froma known initial state fo a known final pressure, FinD: Determine thetenperatureat which condensation first occurs, the fraction oF the total mass tHrat condenses, aud the volume occupied by saturated liquid at the fthal state. SCHEM, Vv : P P,2 7 bar T,+S00°C ASSUMPTIONS: (1) The steam is a closed system. (2) The volume is constant, ANALYSIS: By assumphonscn aul (2), the specific vatume is constaut . Thus, using data from Table A-4 w= Vat Ny = 0.50710 wm Skg Then Leon Table A ca 5 To Ta =Tiat@ Ya = 140°C <—————— Next, the fruction of the total mass that condenses is Sractin Mee condensed Sw FI Xe Ur- Vg, 05010 -1.0300 x10" KF = = = O.4kIO > 1G ,-U 3.420 -(,03 00¥ (073 Thus fraction condonsed = |- 0.1430 =0,8S20 5 ° 3 The total moss of tre systems is mal = Ce = (972 kg Then , the volume of the liquid ak state QV is Vex =~ Meg = (4712 kg) (1.0300 x (07? wP/ keg) 3 Ver Ve, = 0.00203 we ess Problem 3 ‘ ASSo MPTIONS + The closed systtin shown by adashed Irue ontha aboye achemahc. 2- The gas Geheves aa au rdeaR gas, 3. forthe process, Q=W=0 ) aad emehc and potatos energy changes are neg b ble, Vv Aye kT RV Te epf{i kal nN: w RT > RY "oT > iS ale] Vi The finek temperatine u asgurved, Applying an enuersy loalance anti géeraus gut + aU=o Since whvnat euansy of an ident gas de pends a™ HRuperatuye eLone, Tr = Ty. Thus p= (Sbar){t][ seas |= 2 bar Problem 4 ScHtEMmATCC AGEN DATA* Pam = 100 kPa ny Diameter = 10 cm | is quality Mass = 40) kg Xy = 25% ASSUMPTIONS: 1) The water is a Closed tySteun. 2. The pressure is constaut vuntl +e pisten bits +he Steps. 3. friction behveen te piston = aad eq linda weil cor ler ignered. + For the syskm, APE » SKE can be ignored, S. Go constant. To begin, fan +ha three sStoter located he dots on the pry miagram, Stele | & fixed ey Ka 25% and Pp, » WUCCH ts found frown « force balance on on the Lower face of the piston \qveale tan pisten weds We -plus +h force averted on ths top face by to a tre sy Were: RA= Mpist 9 + Patim Ay A= ® Dose /+ > bar Lear _ ihe 2 BOG ABM (SD) [LN |= Goo xt A> Manet Pate zx fern kg-mels* | * rt a = 15 bare Stote a is Lred by Pye PpehS bar and re speeste Volume , UR. Frow the olen arometry , Net ASV or S24, , whee 10828. o.2¢(sisq- 1.0828) = 0,2q0S¢ Milky 104 103 Vy = Veit Xi Vgi- YG) = D> Vt 4.5(0.2908#) = | 3074 mil kG Sma Wy Va (is bar) , Stake Patte wi tee superhsetad vey or Ak yim , a0 Shown wo ttre prov Kagram, SkUBG fixed by Vy=Va and py> 3 bar, The fotet vrase of) wale we ae m: = 74 (0.1 wa )(0.01 m) = 2-703 xo t «9 yy 0.270% iF] FF Since dieu ts ne werk assonated with tee Constant volume porkiaw yt (roeest, ths totat work obinined 20 the water under goes the oer stent pw ssure ercpans torr frm 1 to as We if pav = plVa-Vi} = PLV-W) = mpl — Vi) ! That wo , with 42 4555 r nae SO] (sNl0-e905 et IS | 24.23 7 _ Ws @.Fo2K«0"4 9) (1-Sber) Anw aneray balance feads Aut OFF +2P1 >= Q-W > Q:- oUtW => mM (uz-u,) +W tus prstuy; The force exerted by the water where Uy, = Ug, + x, C Ug - utr) 2 FCCGE + OUST — 466494) = 990.13 CNes To find ur, interpolate at Boar usrng vz! Up = 32G3.73 TT aa. os Q= (a. Tes o¥ 3) ( 3203.53 - ean) | Ka | 4 46237 > eS FYIT —— =