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Understanding Thermodynamics: Ideal Gas Principles and

School
University of British Columbia**We aren't endorsed by this school
Course
CHBE 346
Subject
Chemistry
Date
Dec 11, 2024
Pages
12
Uploaded by aj1023
cbete ot e e N el A Introductory Chemical : Engineering Thermodynamics CHAPTER 2 {2.01) Three moles of an ideal gas.. Solution: The mass of the piston doesn't matter (1) nAU = Q- PAV = nAH = CpAT = Q 3(29,1)(275 = 375) = -8730) (B) AU = Q = 3(20.8)(275 - 375) ~ 6240 (2.02) One mole of an ideal gas (T' = 100 K, P' = 4) AU=AH = 0 since isothermal, Q= =W = RTIn(P/Py) = 8.314(100)In(0.1/0.7) = 1618 J/mol D W=0,T Ty(Py/P)) = 100*7700 K, AU = Q =Cw(T T)) = 20.79(700-100) = 12471 Vmol AH = Cp(T Ty) = 29.1(700-100) = 17459 Vmol €) Q=0, Ty = Ty(PyP)" " = 100(7)" = 174.4K AU = W =Cy(T; - Ty) = 20.79(174.4-100) = 1546 V/mol AH = Cp(T Ty) = 29.1(174.4-100) = 2164 V/mol 1MPa)... (2.03) One mole of an ideal gas (T = 298 K, P' = 1 MPa)... a) AU=AH = 0 since isothermal, o Q= -W = RTIn(P)/P:) = 8.314(298)In(0.1/0.25) = -2270 J/mol D W=0,T Ti(Py/P)) = 298*2.5-745 K. AU = Q =Cy(Ts - T)) = 12.47(745-298) = $574 Jmal AH = Cp(T; - T)) = 20.79(745-298) = 9291 J/mol ©) Q=0, Ty = Ty(PyP)*" = 2082.5)"" = 430K AU = W =Cy(T: -~ T;) = 12.47(430-298) = 1645 V/mol AH = Cp(Ts - Ty) = 20.79(430-298) = 2742 Jmol (2.04) One mole of an ideal gas (T' = 700 K, P' .75 MPa). ) AU=AH = 0 since isothermal, Q= -W = RTIn(P/P3) = $.314(700)In(0.75/0.1) = 11,726 J/mol Do ot post this copyright protected material on public or file shar jelliot@uakron,cdu or liraia e sharing to Chapter 2 - 1 = rmsu.edu Chapter 1
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b W-o T ;ll Q=CV(T> - T) ~20.79(93 3700 e 1=Cp(T> - 7)) = 29.1(93.3-700) ¢ :l 0, T, ‘m'; P = 700(0.1/0.75) \} Cv(T —T)) = 20.79(393.6-700) AH = Cp(T; - T,) = 29.1( T\(P2/Py) = 700/7.5-93, 3K, 2610 J/mol L ; ) AU=AH = 0 since isothermal, Q=-W = RTIn(P,/P,) = 8.314(500)In(0.6/0.1) = 7448 J/mol b) W=0,T,=Ty 500/6=83.3 K, AU = Q =CV(T, - T)) = 12.47(83.3-500) = -5196 J/mol AH = Cp(T; - T)) = 20.79(83.3-500) = -8663 J/mol ©) Q=0, Ty = T)(Py/P) = 500(0.1/0.6)°° = 244.2K AU = W =Cv(T; - T,) = 12.47(244.2-500) 190 J/mol AH = Cp(T; - T)) = 20.79(244.2-500) = -5318 J/mol (2.06) (a) What is the enthalpy change needed to change 3 kg... a) H" at 0°C ~ 0 and since at 150°C, P = 0.4762MPa > 0.1 MPa, so we use superheated steam table to get H' for steam = 2776.4 kJ/kg AH = 3%(2776.6 - 0) = 8329.8kJ b) Use linear interpolation or the relation H = H3™ + VSATH(P-p™), H'=0+04=04k kg. So AH =3%(2776.6 - 0.4) = 8328.6 kJ ¢) Use interpolation to get H" at 0.4MPa and 4°C and correct for pressure, H' = H™" +v>! Ap~ 16.8 + 0.4 =172 kl/k AH =(2752.8-17.2) = 2736 kJ d) Look up saturated steam table. Since H = H™" + gAH"; only q changed AH = [(qz -qi)(AH™) = 1(0.8-0.6) * (2113.75) = 422.75 kJ ¢) S n is in superheated condition H' = 2950 kl/kg at 0.8MPa, H" at 0.8 MPa from saturated steam table = 720.86 ki/kg AH = 721- 2950 = -2229 kJ/kg f) H' = 2950 ki/kg at 0.8MPa; H™" at AH = 1085.8 -2950 = -1864 kl/kg 2) H" at 230°C = 2802.9 kl/kg, H" at 100°C and 0.05MPa = 2682.4 k) kg VY at 230°C = 0.0715 m¥kg, V" at 100°C, 0. a=3.4187 m'/kg enthalpy decreases, volume increases. h) At0.20MPa, 120.21°C steam is saturated; At 0.5MPa, 0.5MPa, 153°C steam is superheated i) For 1kg of water at | MPa, V = V" +qAV'™® 0.15=0.001127 + q; (0.1944 - 0.001127) => qi=77% For 5 kg of water, 0.15 = 5*[0.001127 +q5(0.1944 -0.001127)] q 5= 14.85% to be in the tank. We can have 5 kg of water in the tank ) V¥t 6.8 MPa and 93°C = 0.001034m kg from compressed liquid table by doube interpolation, 50°C = 1085.8kJ/kg 51.83°C steam is saturated:; At I)uvnol post this copyright protected material on public or file sharing sites Repot” Jelliott@uakron.edu or lira@egr.msu.edu
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AH=Cp(T, - Ty) = 29,1 y 1. 29.1(93.3-700) ©)Q ‘ 0, T2 = Ty(Py/P X = 700(0.1/0.75) AU =W =Cv(T, —T)) = 20.79(393.6-700) AH = (T2~ Ty) =29, 1(393.6-700) b) \\’sU 0, T2 = Ty(Po/Py) = 700/7.5-93.3 K. ( M= Q=CU(Ty — ) = 20.79(93.3-700) : (2.05) One mole of an ideal gas (T = 500 K, a) AU=AH = 0 since isothermal, Q= -W = RTIn(Py/P,) = 8.314(500)In(0.6/0.1) = 7448 J/mol b) W =0, T, = Ty(P./P,) = 500/6=83.3 K, - AU = Q =CW(T, - Ty) = 12.47(8. AH = Cp(T; — Ty) = 20.79( €) Q=0, T = Ty(Py/P,)V = AU = W =Cv(T> — T)) = 12.47(244.2-500 AH = Cp(T, - T)) = 20.79(244.2- /mol (2.06) (a) What is the enthalpy change needed to change 3 kg... a) HY at 0°C ~ 0 and since at 150°C, P$T = 0.4762MPa > 0.1 MPa, so we use superheated steam table to get H' for steam = 2776.4 kl/kg AH = 3%(2776.6 - 0) = 8329.8kJ o l linear interpolation or the relation H = H3™ + VSATH(P-P™), b) U H'=0+0.4=04kJ/kg, SoAH =3*%2776.6 - 0.4) = 8328.6 k] ¢) Use interpolation to get H" at 0.4MPa and 4°C and correct for pressure, H" = H** +V*Ap~ 16.8 + 0. AH = (275 team table. Since H = H*" + gAH"™; only q changed JAH™) = 1(0.8- 0.6) * (2113.75) =422.75 kJ ¢) Steam is in superheated condition H* = 2950 kJ/kg at 0.8MPa, H" at 0.8 MPa from saturated steam table = 720.86 kl/kg AH = 721- 2950 =-2229 kJ/kg £) H' = 2950 kJ/kg at 0.8MPa; H™" at 250°C = AH = 1085.8 -295 1864 kl/kg 2) H" at 230°C = 2802.9 ki/kg, H" at 100°C and 0.05MPa = 2682.4 kJ kg V¥ at230°C = 0.0715 mkg, V" at 100°C, 0.05MPa = 3.4187 m'/kg enthalpy decreases, volume increases. h) At0.20MPa, 120.21°C steam is saturated; At 0.5MPa, 151.83°C steam is saturated; At 0.5MPa, 153°C steam is superheated i) For Ikg of water at | MPa, V = V' +qAV'# 0.15=0.001127 + q; (0.1944 - 0.001127) => q,=77% For 5 kg of water, 0.15 = 5*[0.001127 +q5(0.1944 -0.001 127)] it = 14.85% to be in the tank. We can have 5 kg of water in the tank 1) V"at 6.8 MPa and 93°C = 0.001034m’/kg from compressed liquid table by double interpolation, d) Look up 1085.8kJ/kg Do not post this copyright protected material on public or file sharing sites. Report e e Jelliott@uakron.edu or lira@egr.msu.edu =
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€ - gaadey) 01 Bueys ojy Hoday “snts Fuwivys ayy 10 dtjqnd o (e P Sur IR 1o npa @Ol 1w pasatosd 1y3usdos sugy 1sod 108 o (ueisuod 7y 10y 6ge, 15102 7 40) 6£0°01-) jowBypy g9z 10WS/f 6SL°€5€8 [OW/W [$0020°0 = [OW/E, WD 1400 = £0/¢ ) 10wBY/ Y 65.°¢ W+ Av=nv 801 -= "HV+'HV = HY ssoo0ud 304 ouw/f 68 10U/ 6L0°01- = HV 1241 Jo dey Yuq A woy 140t = /do anfea Suisn Ud) = (udl-"Hv="nv 10w 60,8801 -=.22°D[="HV [enwousjod do Suisp amssarg weisuo) :fj doig 0="2V 0="HV ammaduwia | umsuo) ¢ doig Latl ¥ T AEpouL 18 d ™ bbb = 1910601453 =NV S-= 0'8EEE-TIILT=HV BYNY0'1LST = BaN £0°D 081 ),0 BV T'19L2 = (PN £0°D 051 ) 1 BV 1°910€ = (RaIN § BYPE0'REEE = (BaN $°€ D 05k ) 1 " BdW §°€ pue 9 06 wouy abueyd ajess e sa06 Jepun weas (20'2) Y0S"TL = 99S0TAE 680L 000TT) = g HV WA JHw-f) =bos 9980 = 4o HV * BN €6'80L = H * 110002 HG W T9E°E = (FEO100°0 - TS9'1)eT = AV B/, 789'1 ~uonwodiony affuss &G aiqe pawatRns WOy 5,057 U TN 910 10 A 8£0100°0 $Z0100°0 | 6£0100°0 1101000 001 FE0100°0 €6 ST0100°0 L20100°0 08 89| <
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g/kgmol ~ AH =-602.709 kl/kg ( 59 for Cp constant) AU =-464.098 ki/kg (421 for Cp constant) (2.08) Five grams of the specified pure solvent. For parts a-d neglect effect of pressure on H's. H =RHAY(T-298.15)+B"*(T°-298.15%)2+C*(T° - 208.1 where superscripts L denote liquid Cp parameters HY= [ CtaT + AH™ +[ crar R¥*(A(T-298.15)+ BT, 298.1 Since we only know AH'™ at boiling point, use ?)/2+CL¥(T,* - 298.1 and T is A(T- ToHB*(T- Ty )2+CH(T - Ty )3+ DX(T - Ty )4 superscri| 303.150r323.15K moles of benzene, n moles of ethanol, n (5 g/MW MW pis L denote liquid Cp parameters, where T, is the nors 5/78.15 = 6.4E-2 moles 5/46.07 = 0.1085 moles 57 30 or323.15K + AH™ + moles of water, n = (5 g)/ MW = 5/18.02 = 0.2775 moles 14,83 /g from all values read. To 3 & ac! 3 J/g 77 For part d). to change the reference state to 25C, subtract ll‘ at 30°C, H™- = (125.73- obtain enthalpy in J/mol, multiply by 18.02 g/mol. Example at : 104.83)*18.02 = 377 J/mol d water sant a tor water 2 roeve ehanol ethanol woler water i = T i 03,15 32315 30315 m; Sre0l 278601 A 31666 assee 8712 B s ATIEOT 47301 125E03 1IED: a 3.49E-04 3.49E-04 1.80E-07 -1.80E-07 5 9014 9014 3224 3224 8 2.14E-01 2.14E-01 1.92E-03 1.92E-03 :: -8.39E-05 -8.39E-05 1.06E-05 1.06E-05 o 1.37€-09 137E-09 -3.60E-09 -3.60E-09 ™ wies s Fafs 1S s6270 200202 37710 188789 37662 188327 655186 655186 568120 568129 38580 38580 40856 40656 040223 204480 237725 470181 4172963 4308706 43060.04 4463538 441610744 44806.0092 AH= 334842 329172 4116693 4018504 4358203 4274749 4378536 4zezate 1740056 1967047 2114616 2299454 2216857 2326163, 2226030 2334454 111364 125891 2294.36 249491 6151.78 6455.10 6179.73 547814 75% vaporized o)) 576267 2776840 14 E M RS TRR MY W% wwn sses s swsw = 584 534 942071 9217.34 94559 neglect the effect of pressure on e temperature, 1 bar = 750 mm Hg). Po not post this co Jelliott@uakron, ed, Vi e ateri B 0 Pyright protected materia] on public or file sharing sites. Report file sharit? W Chapter 2 uor lira@ egr.msu.edu mal boilir poir atin K, and T is nthalpies (however, note tha pressure does affect the satura!i
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Up to the saturation temperature, the cn}halpy of liquid w ill be Hf’ R¥(AX(T-298.15)+B" *(T2-298.152)/2+C"4(T° - 298.15)3 The enthalpy of saturated vapor will be H - l[l’ CrdT+AH™ +[ CpdT — R*(AX(T;-298.15)+B"*(T,-298.1 +CI¥(T,* - 298.15%)3 + AH™ + A(T- ToyHB*(T%- T, ) 2+C*(T° - T, )3+ DX(T* - Ty )4 where Ty is the boiling point at 760 mm He. Above the normal boiling point, also use the last equation, except T>Th. The saturation temperature at 750 mm Hg can be found using the Antoine cocfficients by Jogu(750)= A - B(T=+C) > T = B/(A-logi(750) P For liquids, U=H, for vapors, U = H- PV, using idcal gas law, U=H-RT. For part d), to change the reference state to =%, subtract 104.83 J/g from all values © cad. To obtain enthalpy in Jimol, multiply by 18.02 g/mol. Example at 99.6 1°C, H™" = (417.5- 104.83)%18.02 = 5634 J/mol, H*™ = (2257.5 ~ 104.8)*18.02= 46314 J/mol Note: U and H are not identically 0 at ref state, but difference is so small that it is not tabulated. U=H-PV part a b ¢ d benzene ethanol water water AL 0747 33866 8712 BL 6.80E-02 -1.73E-01 1.25E-03 cL 3.78E-05 3.49E-04 -1.80E-07 A -3.39E+01 0014 3224 B 474E-01 244E-01 192E-03 c 302E-04 -8.39E-05 1.06E-05 D 7.13E-08 1.37E-09 -3.60E-09 Tb 35325 35165 37315 AH™ 30756 38580 40656 AntoineA 6.87987 81122 807131 AntoineB 1196.76 1592.864 1730.63 AntoineC 219161 226184 233426 Tsat(K) 352.81975 351.1138 3727777 37278 25C H(J/mol) 0.00 0.00 0.00 0.00 U(J/mol) 0.00 0.00 0.00 0.00 Tsat HL(J/mol) 7891.05 6479.06 565296 5634.3134 HV(J/mol) 38670.80 45092 21 46324.54 46313.562 UL(J/mol) 789105 6479.06 5652.96 5632.5114 UV(JImol) 35737.452 42173.05 43225.27 43260.974 110C HV(J/mol) 41799.35 47538.79 46680.11 46603.865 UV(Jimol) 38613.84 44353.28 4349460 43479.377 (2.10) One kg of methane is contained... AU = Q-RTIn(V>/V,), AU=0 since ideal gas, Q = RTIn(Py/P)=8.314(523.15)In(0.8/0.3) = 4266 J/mol n=m/MW = 1000g/(16g/mol) = 62.5 mol, Q = 4266*62.5 = 267 kJ Do not post this copyright protected material on fi ing si SCDOME ublic or file sharing sites. Rej e sharing R T e Il e - oy
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W‘j‘ I /V/ A ice... 0.6958m’/kg V' =0.5524mkg, W {2.11) One kg of steam in a piston/cylinder dev a) AU=Q-PAV, U'=2727. TkI/kg. U’ 67.5 ki/kg, V' -PAV 8)kJ/k; 50.2 kikg Q= (2567.5 2727.7)-502 = 210.4 Kk . ) b) need to find P where V' = 0.6958 at 150°C, between 0.2 and 0.3 MP: 0.281, at this pressure U ke, W=0AU=25722-2 a, interpolating, P* i jum is isotherm (2.12) In one stroke of a reciprocating compressor, helium is isothermally compressed ... Assume Ideal Gas 1 mol AU+ -+ dQ+dW, +dW,, where isothermal (4U = 0) = - 5704.8204 J/gmol 5704.8204 J/gmol 7 = 7 5704.8204 J/gmol Heat Removed = ~0 (2.13) Air at 30 C and 2 MPa enters a throttle with a velocity of 25m/s and exits at w 0.3MPa. The pipe diameter does not change. Compute T°, u°"", = Et E-balance AH'+ /g = 0, put enthalpy on a mass basis, Cp=T*RI(20.0288) = 1010 1 Cp(T™ - ") + (™ -0 10107 - 306182 + (' 1) M-balance V=ud, i : = _ R RIS (2.1 o pel 2] o). iz T 1819, = ) W =T1.819 (2) Solving (1) and (2) sim & 2) simultaneously with Exc, - temperature drop is abou |2 xeel, 7™ =29]1 K,y = T he temperatire rop g o2 C: T Problemn used very high veloe T b) s negligibl fo an gl . Y locities. Normally the Do 10t post this copyrig D B Jelliot
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e— . {2.14) Argon at 400K and 50 bar is adiabatically... Cp=52R RICp=2/5 Ty = Ty(Py/P)" " = 400(1/50) 83.7K Wy = AH = Cp(T; - T)) = (5*8.314/2)(83.7-400) = -6574 kJ/mol _ (2.15) Steam at 500 bar and 500 C undergoes a throttling expansion ... < For Joule-Thomson expansion: ~ AH = Hf - Hj=0 Pj =500 bar = 50 MPa; Tj = 500°C; H' (50 MPa; 500°C) 6ki/kg (Steam Tables) H'=H'=2722.6 kl/kg, P'= | bar=0.1 MPa T' = (H"- 2675.8)/(2776.6 - 2675.8)(150-100) + 100 = 123.2°C For an ideal gas case: Enthalpy is a function of Temperature only H(T),Py)=H(T5,P2) becomes H(T()=H(T2) Which implies T;=T-=500 C Steam is very non-ideal under the given inlet conditions. (2.16) An adiabatic turbine operates... mAH = Wy = 750 kW = s (2865.9 - 3451.6) ki/kg., i = 1.28 ki (2.17) A steam turbine expands steam from 500 C, 3.5 MPa to 200°C and 0.3MPa. W=1750kW and Q = 100kW, what is mass flow? Ebal:0 = m" H" — ™ H™ +Q + W, W, =-750 kW 0 =-100kW = =0 H™at 200 °C, 0.3 MPa = 2865.6 ki/kg 3 H"at 500 °C, 3.5 MPa = 3451.6 ki/kg u -850 = = a [H™ —H") o (2.18) (a) P"" = 80bar steam is throttled to 1 bar and 400°C. Compute T". E (b) P = Thar (satLiq) is throttled to 1bar. Compute ¢°“. 2 o b a @) 0=H +O+W, ;e where 9=, =0 e S =g = :Illyih pressurc - H*(at 1 bar, 400 °C) = 3278.6 ki/kg 4 T"(at H" = H"and § MPa) = 452.1 °C valve b) _Du not post this copyright protected material on public or file sharing sites. Report file sharing to Jelliott@uakron.edu or lira@egr.msu.edu Chapter2 - 7 n— A
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{2.14) Argon at 400K and 50 bar is adiabatically... R Cp RICp .4 p Ty = Ty(Po/Py)¥P = 400(1/50) = 83.7K Wa = AH = Cp(T: - T;) = (5*8.314/2)(83.7-400) = -6574 kJ mol (2.15) Steam at 500 bar and 500 C undergoes a throttling expansion ... For Joule-Thomson expansion: AH = Hf - Hi =0 Pj =500 bar = 50 MPa; Tj = 500°C; H' (50 MPa; 500°C) =2722.6 ki/kg (Steam Tables) H=H .6 kI/kg, P= 1 bar = 0.1 MPa T' = (H' 2675.8)/(2776.6 - 2675.8)(150-100) + 100 = 123.2°C thalpy is a function of Temperature only. becomes H(T;)=H(T:) Which implies T=T,= 500 C. For an ideal H(T,.P)=H( Steam is very non-ideal under the given inlet conditions. (2.16) An adiabatic turbine operates... mAH=Ws = 750 kW = /i (2865.9 - 3451.6) ki/kg, i = 1.28 kg/sec (2.17) A steam turbine expands steam from 500 C, 3.5 MPa to 200°C and 0.3MPa. W=750kW and @ = 100kW, what is mass flow? Ebal:0= " H" — ™ H"™ + 0 + W. - 750 kW - 100 kW H*"at 200 °C, 0.3 MPa = 2865.6 kl/kg > H"at500°C, 3.5 MPa = 3451.6 ki/kg u -850 ) s —————=1.45 kg/sec 3 =7 - (2.18) (a) P = 80bar steam is throttled to 1 bar and 400°C. Compute T". (b) P" = Tbar (satLiq) is throttled to 1bar. Compute ™. a) 0=H"-H"+0+W. . whee Q=W =0 £ e~ Highpressre H"(at 1 bar, 400 °C) = 3278.6 ki/kg - T"(at H" = H*"and 8 MPa)=452.1 °C b) Do ot post this copyright protected material on public or file sharing sites. Report file sharing to jelliott@uakron.edu or lira@egr.msu.edu Chapter2 -7
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i i and two (2.19) An overall balance “around part of a plant involves three inlets | outlets which only contains water Balance dn a10.15 MPa, g = 0.9) H,= 461.1 +(0.9%2229.5) = 2467.6 ki/kg .8 MPa) #,=2768.3 klkg 3072.1 ki’kg 943.1) - 30(2467.6) —120 =163015 KJ/min =2716.9 kW s SoL {2.20) Steam at 550 kPa and 200C....a) En, A H®=H™ p s interpolating between 0.5 and 0.6 MPa e H® = 0.5(2855.8 + 2850.6) v Froin: At p 853 at 0.2 MPa t P T revert interpolation formula at 0.2 MPa ! 2769.1 + x(2870.7-2769.1), x = 0.83, T= 150+ 0.83(50)=191°C A e = b) Energ U +0.83(2654.6 1) =2641 kl/kg Ul U™=0.5(2642.9 + 2638.9) = 2641 ki’kg V0. 1mp AU =0 even though AT = -9°C. —— DG ot po jellion@u, . . “nort file Domot post this copyright protected material on public or file sharing sites. Report 00 Jelliott@uakron.edu or lira@ egr.msu.edu
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It leaks to 1bar. S37) A 0.1 m® cylinder is initially at 10 bar and 300K. : =4.01 mols @2 0 300K, n, = PY/RT = 0. 1MPa(0.1E6cm’)/8.314/300 Ty(P/PY P =300(1/10) = 1554K 7,) = (7*8.314/2)(155 4-300) = 4208 kJ. mol 2.22) As part of a supercritical extraction of coal ... ba U-n H (™ — m™); w'=0=m"™ = ol=m", U=H",PF 20 MPa, s 20MPa, 400°C, H” = 2816.9 kJ/mol s s between 450 and 500°C. Interpolating, 6.9 - 2807.2)/(2945.3 - 2807.2) * 50 = 453 y interpolation, m = ¥/ = 1/0.01286 his V2 Hdn™ - H™dn™; ndU = (H"-H)(dn/dt)dt, G G (), dt ) [2.24) An adiabatic tank of negligible heat capacity ... pen Unsteady System: AUm = HAm+ 0+ =t he = . 0 m' H" = U/ =H" Table. H"(1 MPa, 200°CY =1 MPa and U=2828.3 kI/k LIc 684m’ / kg kikg 73 kg balance: A(Um)=HAm +Q+W - U'm'= H5(m' - m') VA0.1MP=.150°C) = 1.9367 m'/kg and U' 0t post this copyright protected material on public or file sha e: i uskron.edu or lira@egr.msu.edu 4 SRR fi('?hiha""wg “:) apter 2 -
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rt a) me as pal ‘. which in principle canlt:»cqfiifi_‘: ;.l e )’ and V', Howcwr._n s quickey H")(U-H )r‘i mula for U an ngv‘i"c function, v ‘J( U on for ing as an obj cvare e i l’t‘:‘l o “gglr:er. the temperature ca g " Using G al Seek or S 0 g Goal ). Using Note: w, If w, enging, intg v, dellioe 544
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