Homework 3 Solutions - Chemical Engineering Thermodynamics | CHE 2164, Study notes of Chemistry

Download Homework 3 Solutions - Chemical Engineering Thermodynamics | CHE 2164 and more Study notes Chemistry in PDF only on Docsity! Chemical Engineering – Thermodynamics 2164 Homework 3 Solutions 1. (a) U2 = 200 kJ,  U= 20 kJ (b) U1 = 10 kJ,  U= 50 kJ (c) under specified (d) Q = 10 kJ, U2= 40 kJ (e) W = 209 kJ,  U = 169 kJ 2.   16 1313 1 Jkg10189.0 )kgm 001101.0kgm3157.0( MPa 60.0 )Pconstant (kJkg 2.2086           if VVP PdVw hq wqu u=(2086.2–189)Jkg-1 = 1.90 Mkg-1 OR: P)(constant vPu PvvPuh Pvuh    3a) Method: 1. Mass is conserved and can be calculated from the initial conditions 2. Constant P and initial and final volumes known, so can determine work. 3. (i)initial and final conditions are specified so we can calculate U and therefore Q=U –W or (ii) note that heat at constant pressure is the enthalpy so we can look up enthalpy in steam tables. Initial mass ignore volume of liquid Mass of gas kg 082.1 kg0.462m m 50.0 1-3 3  Mass of liquid kg 271.0gas mass25.0  Total mass = 1.352 kg     by water done work kJ 279 m50.0kgm88934.0kg352.1kPa 400 31-3      if VVp pdVW if UUU WUQ   kJ 4230 kJ 352.13129 kJkg 3129)C500 kPa, 400( 1     f f U u    kJ 1309 kJ 2921 2553kJ/kgkg 08.1kJ/kg 604kg 271.0 ))C143(sat kPa, 400(     U umumU gglli  At constant 400 kPa,   MJ 6.1~ kJ 1588 kJ 2791309    WUQ OR Q = HF - HI 3b) Constant volume:   MJ 1.3 kJ 304 kJ 2921kJ/kg 3125kg 352.1 kJ/kg 3129 i.e. problem theofpart first in the asenergy internal same theusecan weso pressure theoffunction weak a isgasot a ofenergy internal specific they,Alternatel kJ/kg 3125 is Û se weion,interpolat Using occurs. V̂ f value this whereÛ find toC 500at tablessteam thuplook can We kgm372.0 kg51 m50.0 5.V V̂ problem thefromV̂get can T, have Yes, variablesstate intensivet independen twoneed specified? st te final theis kJ 2921 0 13 3 3 final F F       Q U v M V U UQ W f F F final in ti Check: less heat is required to reach same temp if no work is done. 4a)   kJ 2493 K 200mol100.1kJmol 31.8 2 3 constant) is C so gas,perfect ( 311 v       ifV V TTC dTCU U is 2.5 MJ 4b) f water f air f water f air f air f airf air final air air air init air VV TT P nRT V P n RT pV n P            3 11 3 m 02.0 kPa 150 (final) mol 325.0 k 373kJmol 31.8 m 01.0kPa 101 kPa 100 Need equation  fwaterfwater TfV  , solve simultaneously. Need mass of water. Ignore volume of liquid water /kgm 00104.0 3fv At 100 kPa, 100°C: 13kgm69.1 gv   kg 00786.0 waterof mass total kg 00194.0 kg 00592.0 7.24100 7.24 water liquid of mass kg 00592.0 kgm 69.1 m01.0 or water vapof mass 13 3         Don’t have analytical form for V(T) or V(P) so do trial and error: 1. All water is gas at 100°C  still 100 kPa 2. Sat steam at 150 kPa, T = 111°C If T = 111°C Volume of air,   3 3 3 1 m0131.0 m00693.002.0 m00693.0 kPa150 K384JK31.8325.0       f water f air V P nRT V Is there enough gas (water vapor) to fill this volume? 009112.0 15933.100786.0 13    kgmkg mvVolume f Not enough volume, need greater T Raise T: superheated steam, 200°C Volume of air,   3 3 3 1 m0115.0 m00851.002.0 m00851.0 kPa150 K473JK31.8325.0       f water f air V P nRT V At 200 °C and 150 kPa, vwater = 0.00786 kg ×1.445 m3kg-1 = 0.0113 m3 approx. correct Final temp ~ 200°C ΔUUair:     J 675 mol 325.0K 100KJmol 31.8 2 5 111       air ifV Vair U TTC dTCU ΔUUwater: Too difficult to obtain from q and w. State function, get from initial, final     kJ3.5 kJ64.15 kg00592.0kJ/kg2506kg00194.0kJ/kg419,100 kJ 9.20 kg 00786.0kJ/kg 2656kPa 150,C 200      water initial final U satCU U   Note, Uwater has risen by much more than Uair because of the energy to vaporize the water, and the greater heat capacity of water vapor than air.