SOLUTION THERMODYNAMICSCHAPTER 6 6.1 Starting with Eq.(6.8), show that isobars in the vapor region of a Mollier (H S) diagram must have positive slope and positive curvature. SOLUTION: {∂∂HS } =T Proses Isobar dan memiliki slope bernilai P positif Membedakan persamaan di atas: { } { } 2 ∂ H ∂T = 2 ∂S P ∂S P Jika di substitusikan dengan contoh 6.17 { } ∂2 H T = 2 ∂ S P Cp Proses isobar dan memiliki kelengkungan positif 6.2 (a) Making use of the fact that Eq. (6.20) is an exact differensial expression, show that : (∂ Cp/∂ P) T = -T ( ∂ V/ ∂ T2)P 2 What is the result of application of this equation to an ideal gas? (b) Heat capacities Cv and Cp are defined as temperature derivatives respectively of U and H. Because these properties are related, one expects the heat capacities also to be related. Show that the general expression connecting Cp to Cv is : Cp=Cv+ T ( ∂P ) ∂T ∂V ¿ V( ∂T P SOLUTION : (a) Aplikasikan persamaan 6.12 ke persamaan 6.20 : ∂V ∂ V −T ( )P ∂T ∂Cp = ∂P T ∂T P [ ( ) { } ] ∂2 V ∂V − 2 ∂T P ∂T ( ) ( ) ( ) ( ) ∂Cp ∂V = ∂ P T ∂T −T P P Sehingga , ∂Cp ∂2 V =−T ∂P T ∂T2 ( ) ( ) P Untuk gas ideal : ∂V ∂T ( ) 2 = P ( ) R ∂ V dan =0 2 P ∂T P (b) Persamaan 6.21 dan 6.33 adalah persamaan umum untuk dS dan untuk mengubah kedua persamaan harus diberikan nilai dS yang sama. Oleh karena itu persamaan menjadi: Cp−Cv ∂P ∂V =( dV + ( dP ( ∂V ) ) ∂T ∂T ∂T ) V P Dalam kondisi P konstan, menjadi : ∂ P ∂V Cp=Cv+ T ∂T V ∂T P ( )( ) Dengan contoh persamaan 3.2 dan 6.34 : ( ∂V∂T ) =βV dan( ∂∂ PT ) = Kβ P P Substitusikan dengan persamaan pada P konstan : Cp−Cv=βTV ( Kβ ) 6.3 If U is considered a function of T and P, the natural heat capacity is neither Cv nor Cp, but rather the derivate (∂ U /∂ T ) p . Develop the following connections between (∂ U /∂ T ) p , Cp, and Cv : ∂V (∂ U /∂ T )P = Cp−P ∂T ( ) =Cp−βPV P [ ( ) ]( ) Cv+ T = ∂P ∂V β v−P p=Cv+ ( βT −kP ) V ∂T ∂T k To what do these equations reduce for an ideal gas? For an incompressible liquid? SOLUTION : Untuk definisi H,U=H=PV, turunannya: ∂H ∂V =( −P ( ( ∂U ) ) ∂T ∂T ∂T ) P P P or ( ∂U ∂T ) P = Cp−P ( ∂∂ VT ) P Substitusi turunan akhir dari persamaan 3-2 definisi dari (∂ V /∂ T )P =Cp−βPV β 32.4 The PVT behavior of a certain gas is described by the equation of state: P(V-b) = RT Where b is constant.34 dan 3.Pembagian persamaan 6. SOLUTION : a Berdasarkan persamaan 6. P(V-b)ɣ = const.2 substitusinya adalah : β p=Cv+ ( β T −kP ) V ( ∂U ) ∂T k 6.32: dU = [ C v dT + T ( ∂P ) −P ∂T V ] oleh persamaan keadaan: P = RT (V −b) ∂P ∂T dimana ( R P ¿ ¿V = = V −b T dengan mensubstitusikan persamaan keadaan ke persamaan 6. If in addition Cv is constant. show that: a U is a function of T only b ɣ = constant c For a mechanically reversible process.32 dengan dT dan terbatas untuk P konstan hasilnya adalah ∂P ∂V p=Cv+ T ( v −P ( ( ∂U ) ) [ ∂T ] ∂T ) p ∂T Pemecahan untuk turunan kedua dari persamaan 6. maka diperoleh: dU = [ C v dT + T ( R RT ) − V −b V V −b persamaan akhir menjadi: ] dU = fungsi T) b dH = dU + d(PV) d(PV) = R d T + b d P …… (1) …… (2) ⟹ C v dT dU = [ C v dT + ( RT RT ) − V −b V V −b ] . (terbukti bahwa U hanya merupakan . sesuai dengan persamaan keadaan: dU = C v dT =−P dV = . What is the model ? SOLUTION : It follows immediately from Eq. These results are consistent with those for an important model of gas-phase behavior. H. dU = dW . substitukan dengan persamaan keadaan: P ( V −b ) (V −b)γ−1 =konstan . where Γ (T) is a substance-specific function of temperature.( γ −1 ) d ln(V-b) ⟹ hubungkan dengan persamaan: T γ−1 d lnT + d ln (V −b) =0 (V −b) γ−1 = konstan. and Cv. ( 6. CV CV maka: d lnT = . Cp. diperoleh: C (¿¿ v dT ) + (R d T + b d P) dH = ¿ ⟹ C ¿ ¿ ¿ integralkan persamaan (3). U. maka: P (V −b) γ = const (TERBUKTI) R 6. maka persamaan dU = dW berubah menjadi: d ( V −b ) −RT dV =−RT V −b V −b Disederhanakan menjadi: dari jawaban (b). gunakan persamaan + R) dT + b d P …… (3) ∂H =C V + R ∂T hasil akhir menjadi: Cp = Cv + R.kombinasikan persamaan (1) dan (2) dengan persamaan akhir dari jawaban (a).10 ) that : . S. Determine for such a fluid expressions for V. karena Cv konstan. maka: C γ ≡ P = konstan CV c Untuk persamaan mekanik proses adiabatic reversibel.5 A pure fluids is described by the canonical equation of state : G = Γ (T) + RT ln P. C v dT γ≡ CP CV dT −R = d ln ( V −b ) T CV = konstan. diubah menjadi R C P−C V = =γ−1 . R = Cp – R The equation for V gives the ideal-gas value.( ∂T P Differentation of the give equation of state yields: V= RT P and S=- d Γ(T) dT . we can apply the equations : H = G + TS and U = H-PV = H – RT These become : H= Γ (T ) -T dΓ (T ) ` dT and U= Γ (T ) -T dΓ (T ) dT – RT By Eq. Determine for such a fluid expressions for V.10 : . and Cv. The equation for S shows its relation to P to be that of an ideal gas. Cp.∂G ¿ V=( ∂P T and ∂G ¿ S = . The equations for CP and CV show these properties to be functions of T only. which conforms to ideal-gas behavior.R ln P Once V and S ( as well as G ) are known. ∂H ¿ Cp = ( ∂ T Because Γ P and ∂U ¿ Cv = ( ∂ T T is a function of temperature only. S. which conforms to ideal-gas behavior. 6. as does the result. these become: d2 T Cp = -T dT 2 and d2 T Cv = -T dT 2 . These results are consistent with those for an important model of liquid phase behavior. H. described by the canonical equation of state : G = F(T) + KP where F(T) is a substance specific function of temperature and K is a substance specific constant. We conclude that the given equation of state is consistent with the model of ideal-gas behavior. U. What is the model ? SOLUTION : Dilihat dari persamaan 6. CP = CV + R.6 A pure fluid. The equations for H and U show these properties to be functions of T only. U=F ( T ) + KP−T U=F ( T )−T dF (T ) −PK dT dF(T ) dT Sedangkan dari persamaan (2. V =K . H=F ( T ) + KP−T dF (T ) dT Sehingga.16) dan (2. Cv=−T V d2 F dT2 F(T) dianggap konstan Maka kesimpulannya adalah Cp = Cv Persamaan untuk V menunjukkan hal itu akan konstan dari kedua T dan P.G = F (T) + K P Diferensiasi dari persamaan yang diberikan : Dalam V V= ( ∂G ∂T ) T .20) : Cp= ( ∂∂HT ) Cp=−T P d2 F d T2 ( ∂∂UT ) . Cv= . S= Dalam S ( ∂∂GT ) S=− P −dF (T ) dT Maka V dan S telah diketahui dalam persamaan : H =G+TS dan U = H – PV dan U= H – PK Maka persamaan menjadi. Ini adalah definisi dari cairan termampatkan H dipandang sebagai fungsi dari kedua T dan P . 282 Cp = 2. V . SOLUTION : Isobutane: Tc = 408.551 x 10-3 m3 kg-1. T . Estimate the temperature change and the entropy change of the isobutane. SOLUTION : At constant temperature Eqs. Dp) For an estimate.63). Estimates of V and β may be found from Eq.123 gm / mol . Vt = 1. Semua ini konsisten dengan model. dan Cv adalah fungsi T saja.β.sedangkan U.26) can be written : dS = −β .095 x 10 K −1 ∆S = .095 x 10 K −1 .8 Liquid isobutane is throttled through a valve from an initial state of 360 K and 4000 kPa to a final pressure of 2000 kPa.T. For saturated liquid ammonia at 270K.1K Zc = 0. (6. cairan mampat. The specific heat of liquid isobutane at 360 K is 2.V dP and dH = 1−( β . Kami juga memiliki hasil yang Cp = Cv. S.7 J / kg 6.78 J g-1 ○c-1.V (P2 – P1) = -2.661 J / KG.V (P2 – P1) ∆H = 1 . 6. K = 551.7 Estimate the changes in enthalpy and entropy when liquid ammonia at 270K is compressed from its saturation pressure of 381 kPa. 551 x 10-3 m3 kg-1 −3 β=2. Cp.25) and (6. assume properties independent of pressure T = 270 K P1 = 381 kPa P2 = 1200 kPa V = 1.β. and −3 β=2. (3.78 J / gm K P1 = 4000 kPa P2 = 2000 kPa mol (wt) = 58. T = ( 359 . 132.662 cm3/mol V = 262.31 x 662 x 10−4 ) m3 .105 cm3/mol (V1 ) (V / T ) → 4. ( 1.662) cm3/mol = (361 – 359) = 2K = 1.5457 = 131.662 .7 cm3/mol x (0.282)0.8) is integrated to yield: H = TS V P S Then at 360 K. 0. 0.767 cm3/mol Assume that changes in T and V are negligible during throtling.098755 x 103 K-1 =( 1 131. Zc ( 1 – Tr )0.63) for volume of a saturated liquid may be used for the volume of a compressed liquid if the effect of pressure on liquid volume is neglected.113 .7 cm3/mol x (0. 2.105 cm3 / mol T1 → T2 K = (132.5391 = 132.282)0. (−2 x 106 ) Pa 359 k 263. 360 .885 ) (The elements are denoted by subscripts 1. & 3) V = Vc .7 cm3 / mol Eq.662 cm 3 mol−1 3 ).5430 = 132. We use the additional values of T and V to estimate the volume expansivity: V = 1.2857 V = (131. mol−1 .88 .32 x 4 359 Pa m3/mol K = 0. 132.Vc = 262. = V1P2P1T1 = = S but H = 0 ( −1.767 – 131.7 cm3/mol x (0.733 J / mol K. (3.882 .282)0.767 ) cm3 / mol V = 262. 361 ) K Tr = T / Tc Tr = ( 0.(6.113 cm3/mol V = 262. Then Eq.105 cm mol 2K −1 ) . 29) may be integrated to give = Cp .−1 x 10 Pa mol 58.125429 J/mol K Whence T = (V / Cp ) .105 131.662 . Determine Q.78 j/gmk .123 gm .mol−1 = (129.14) . ΔH.123 )K 6. W.78 J/gm K) . ( - 1.05429 + 1.324 K = 4. (6. ΔU. -2x106 Pa = (0. SOLUTION : V2 = V1.= 1.55x10-3 – 4. 1.2 K = 1.196351 x 10 K .11) J/ mol K = 1. 5. j −3 −1 −4 3 −1 6 K + 4. Eq. 1. 196351 x 10-3 K-1 Assuming properties independent of pressure. EXP[(-45x10-6 bar-1)( 1500 bar-1 bar )] .105 58.196351x10-3 K-1 .9 One kilogram of water ( V1=1003 cm3kg-1 ) in a piston / cylinder device at 25 0C and 1 bar is compressed in a mechanically reversible.32181x10-4 m3/mol .EXP[-k(P2-P1)] = 1003 c m3 kg .31662 x 10 m mol . ( ∂ T / T ) – ( ∂ V/ ∂ P) S = 2.125429 (359 K/2.105 263. isothermal process to 1500 bar. and ΔS given that β = 250 x 10 -6K-1 and k = 45x10-6 bar1 . [ ( ∂ S ∂ V1 ∂ P) / mol wt ] = 1. 194 ¿ = 134606.T) .298.(1499 bar) ¯ c m3 . 0.067455) 3 . (1 – β.(1500 bar – 1 bar) = 970.= 1003 = 1003 c m3 kg = 1003 cm kg = .698 ¿ ¯ c m3 . (0.302 c m3 kg (1 – 0. EXP[(-45x10-6 bar-1)(1499 bar)] . ¿ kg = 1346069. EXP( -0. ¿ kg = 1454482.9348 c m3 kg = 937.302 ΔH c m3 kg = Vave .9254625) .9194 J kg .604 Vave 3 cm kg V 1+V 2 2 = 1003 cm kg 3 + 937.604 cm kg 3 2 = 970.15K)). (P2 – P1) = 970.302 c m3 kg ( 1 – (250 x 10-6K-1.0745375). kJ kg = 134,6069194 = 134,607 ΔU ΔS kJ kg = ΔH – (P2.V2 – P1.P1) = 134,607 kJ kg = 134,607 kJ kg = 134,607 kJ kg – (140,6406 = 134,607 kJ kg – 140,5403 = -5,9333 kJ kg – [(1500 bar . 937,604 3 ¯ cm . ¿ kg – (1406406 ¿ – 1003 – 0,1003 c m3 kg ) kJ kg ) kJ kg = -β . Vave (P2 – P1) -6 -1 = (-250 x 10 K )(970,302 3 = -363,6207 cm . = -0,03636207 = -0,0364 = T. ΔS c m3 kg )(1500 bar – 1 bar) c m3 kg . K )(1499 bar) = (-0,2425755 Q kJ kg c m3 kg ) – (1 bar . 1003 ¯¿ kg . K ¿ kJ kg . K kJ kg . K c m3 kg )] = 298,15K.( -0,0364 ) kJ kg = 4,91936 W kJ kg . K = ΔU – Q = (-5,9333 kJ kg ) – (-10,85266 = 4,91936 kJ kg kJ kg ) 6.10 Liquid water at 250C and 1 bar fills a rigid vessel. If heat is added to the water until its temperature reaches 500C. What pressure is developed? The avarege β value of between 25 and 500C is 36.2 x 10-5 K-1. The value of ĸ at 1 bar and 500C is 4.42 x 10-5 bar-1 and may be assumed is independent of P. The specific volume of liquid water at 250C is 1.0030 cm3g-1. SOLUTION : Berdasarkan persamaan 3.5 (perubahan volume konstan) β.∆T – ĸ .∆ P = 0 β (T – T ) – ĸ 2 1 β (T – T ) = ĸ 2 1 P2 (P2 – P1) = 0 (P2 – P1) = β ĸ = 4.42 X 10−5 ¿¯−1 36.2 x 10−5 K−1 ¿ (T2 – T1) + P1 = 205.75 bar (323.15 K – 298.15 K) + 1 bar 6.11 Determine expressions for GR,HR,and SR implied by the three-term virial equation in volume,Eq (3.39) SOLUTION : From Eq. 3.39: Z= PV RT B C + = 1+ V V 2 Vig = gas ideal,dimana : P.V = n.R.T Vig = Z= RT P PV RT V= ZRT P Sehingga, Vr = V- Vig =V- = Vr = RT P ZRT RT − P P RT P ( Z - 1) Volum Residu : R V = RT P (Z-1) Dimana : dG = VdP - … (1) ∫ dT T konstan sehingga : dG = V.Dp Integrasi dari nol menuju P > P GR = ∫ VR dP 0 … (4) the answer is R P G dP =∫ ( Z−1 ) RT 0 P Substitusi persamaan (3) ke persamaan (5) GR RT P ∫ ( Z−1 ) dP P = … (6) 0 Substitusi persamaan (2) ke persamaan (6) GR RT P dZ + ∫ ( Z−1 ) dP P Z = 0 (Z-1) Diferensiasi kondisi 2 menjadi : GR RT ¿ P ∫ ( Z−1 ) = 0 dP +¿ Z-1) – ln Z … (7) P Persamaan (4) dapat ditulis d G RT = 1 RT dG - G RT 2 Dt dimana dG disubstitusi dengan persamaan (4) dan G dengan G = H – TS menjadi (dalam keadaan T konstan) . GR RT P P = ∫ 0 R dP 0 RT ( Z−1) P RT dP P = ∫ 1P ( Z −1 ) dP 0 Finally.Dibagi RT GR RT P = dP ∫ VR RT = ∫ VRT … (5) 0 Maka. TSR Dan entropinya adalah SR R = HR RT - GR RT …3 …2 .GR d RT ) = ¿ VR RT dP - HR dT = (Z-1) RT 2 HR RT 2 dT dP P GR – d ( RT ) … (8) Persamaan (8) dibagi dT dan dalam P konstan. Substitusinya menghasilkan : HR RT = -T ∂Z ∂T ¿ P ∫¿ dP P )P + Z-1 … (9) 0 Dan dari persamaan 3.39 didapatkan = V = 1/ρ Z-1 = Bρ + Cρ2 Pensubstitusian persamaaan (7) dan persamaan (9) akan menghasilkan : GR RT HR RT 3 2 2 Cρ – ln Z = 2Bρ + = B T[( T dB C )]ρ + ( dT T - …1 - 1 2 dC 2 dT )ρ Persamaan Gibbs umum : G = H – TS Untuk residual GR = HR . maka : HR RT 2 GR ∂( ) ∂P RT ] ( ∂T )P – [ P ∂T Z−1 P = Diferensiasi persamaan (1) menyediakan hasil di kanan dan diferensiasi persamaan (7) menyediakan hasil kedua. dan P pada fase vapor dan pada fase liquid denganσ = € = 0. 6. Maka persamaan van der wall menjadi β= Pr 8Tr dan q = 27 8Tr maka dapat disubstitusikan pada persamaan tersebut menjadi Hr RT =Z−1− qβ Z dan Sr RT = ln(Z − β) 6. ql) R.6.63b to eq 6. Determine expressions for GR. V. eq.12 Nilai parameter dari persamaan van der walls telah ditunjukkan pada baris pertama tabel 3.56 and eq. parameters a and b are functions of composition only. 6.T GR = ( z−1−ln ( z −β ) . ql) RT Substitusi eq. SOLUTION : According to eq6. halaman 99. HR.13 .56 P = ZρRT I.1. and SR implied by the Dieterici equation: Here. Maka persamaan menjadi : Gr RT = Z − 1 − ln(Z − β) – qβ/Z Telah diberikan T.13. Pada bagian bawah halaman 214 ditunjukkan pada persamaan I = β/Z. 6.63 GR =( z−1−ln ( z−β ) . 13 R ZρRT = −S VRT ) RT exp ¿ R v−S ZρRT = d ln α ( Tr ) −1 qlR d ln Tr VRT RT exp ¿ d ln α ( Tr ) v−z−1+ −1 qlR d ln Tr −z−1+ ( ( ) ) . ql ) 1 exp ( ) V v−( z−1−ln ( z− β ) .65 R d ln α ( Tr ) S =z−1+ −1 ql R d ln Tr ( S R =z−1+ ) ( ddlnαln Tr(Tr ) −1) qlR Substitusi eq. ql) RT VRT RT exp ¿ v−( z−1−ln ( z− β ) . Eq. ql ) ) ρRT II. 6.56 and eq. ql) RT ZρRT = −( z−1−ln ( z− β ) .R ZρRT = RT −G exp ( ) R VRT V −G ZρRT = −( z−1−ln ( z −β ) . ql ) 1 exp ( ) V (v −( z−1−ln ( z −β ) . 6. 6.65 to eq 6. ql) Z= −( z−1−ln ( z −β ) . 13 ZρRT = −H R VRT ) RT exp ¿ R v−H . )) d ln α ( Tr ) v−z−1+ −1 ql ρR d ln Tr ( exp ¿ Eq.56 and eq.−z−1+ ZρRT = ( d dlnlnαTr( Tr ) −1) ql VT T exp ¿ d ln α ( Tr ) v−z−1+ −1 ql d ln Tr ( −z−1+ Z= ) ( ddlnlnαTr( Tr ) −1) ql VT T exp ¿ d ln α ( Tr ) (v −z−1+ −1 ql) ρRT d ln Tr ( ) −z−1+ ( d dlnlnαTr( Tr ) −1) ql VT Z= 1 ( III.64 to eq 6. 6. 6.64 d ln α ( Tr ) HR =Z−1+ −1 ql RT d ln Tr ( H = Z −1+ R ( ) d ln α ( Tr ) −1 qlRT d ln Tr ) Substitusi eq. 6. Acetylene at 300K and 40 bar b.H and R by the Redlich/Kwong equation for one of the following and compare result with values found from suitable generalized correlations: a.Hydrogen sulfide at 400 K and 70 bar k.Argon at 175 K and 75 bar c.n-Octane at 575 K and 15 bar m.( ddlnlnα Tr(Tr ) −1) qlRT −Z−1+ ZρRT = VRT ) RT exp ¿ d ln α ( Tr ) v−Z−1+ −1 qlRT d ln Tr ( Z= ) 1 exp ( d ln α ( Tr ) v−Z−1+ −1 qlRTρ d ln Tr ( ( d dlnlnα Tr( Tr ) −1) ql ) −Z−1+ ) VRTρ 14.Carbon monoxide at 175 K and 60 bar g.Carbon dioxide at 325 K and 60 bar f.52) .53) Guess Z 1 Given Z=1 Eq(3.Calculate Z.n-Butane at 500 K and 30 bar e.Propane at 375 K and 25 bar n.Cyclohexane at 650 K and 50 bar i.Propylene at 475 K and 75 bar SOLUTION : Redlich/kwong equation : Eq(3.Ethylene at 300 K and 35 bar j.Benzene at 575 K and 30 bar d.Nitrogen at 150 K and 50 bar l.Carbon tetrachloride at 575 K and 35 bar h. Z q I 114 HRi RTi Ziqi 1 1.744 -4.685 -6.5qili Eq (6.14 and compare results with values found from suitable generalized correlations.08664 = 0.542 -3.623·103 6.503·103 0.302·103 0.15 Calculate ZR.42748 C = (0.766 -4.523 -2.775 -3.767 HRi -3.257 -2. SOLUTION : Soave/Redlich/Kwong equation: = 0.480 + 1.39·103 0.648 -2.115 -1.67) The derivative in these SRi Rln Ziqi I 0.12 -5.6.772 -4.709 -5.75 -5.122·103 -8.026 -2. and SR by the Soave/Redlich/Kwong equation for the substance and conditions given by one of the parts of Pb.176.461 -2.706 -5.068·103 0.978 -4.68) Equations equal-0.659·103 0.729 -5.5qili Eq (6.0.764·103 0.939 -3. .362·103 0.316·103 0.381·103 0. 2) .346 -1.3·103 0.75 -5.663 -7.698 -1.319·103 0.5 Z i qI 0.771 -4.695 Sri 0.HR.605 -5.574.488·103 0.024 -4. (3.Ii] So.769 z−β 0.776.qi.090.103 -2.103 -3. Tri 0.722 0.51) Tr 0. Sri = R [ ln (Z) ( βi .103 -1. qi )−1−¿ [ci.(3. maka0..14 -6.Ii] Tri 0.-2.806.408 -ci.787 + β) E.1.(3.795 Tri ( ) 0.246 i=1.947 -4. -6..103 -1. qi+ βi -5. -8.585.618 Ii = ln Z¿ -4.673 Given: 0. qi -7.412 The derivative in the following equations equals.5 ∝ i ) +1].595.qi.5]2 0.5 ∝ i ) .749 z=1 0.q.849 -4.103 -4.103 α = [ 1+c.098 ¿ βi .751.451 βi .774 β = ( Tr Eq.974 -6.q. Z ( βi.741 Guess: 0.(z 0.807.715 0.776 Z = 1+ β – q.103 -2.816.103 -3.5 -7.821..323.103 -5..49) -6.qi )− βi [ci.691 0.406.Ti [ Z ( βi.103 -2.103 -4. Tr ).103 ¿ HRi= R. β z .Tr0..K ..581 Z (¿) -5.103 -4.244.741 α (Tr ) 0.768 q= E.774 0.02 dimana. qi ) = HRi=J/mol SRi = J/mol.527.606 Pr Pr ¿ β = (.50) 0. ∝ i -5.482 ¿ -6.857.103 -1.103 -2. (3.6.941 B0 = 0.422/Tr 1.69). For this we estimate the liquid volume Eq. & (3. B1) Pr/Tr) = 7.6 = -0. Eq. (3. (6.8858− t+2788.271 Pr = P/Pc By Eqs.621 V vap = R .375 P(50) = 36.15 K t= T K 273.72).172/Tr 4.095 x 10-3 K-1.151 cm3/mol Pr = 0.083-0.2 = -1.200 kPa.T / P (1 + (B0 +ω.66). and β = 2.61).210 Tc = 562. Zc (1 – Tr 2/7) = 93.375 kPa K a.79 ) P(t) 1.15 t = 50 The pressure is the vapor pressure given by the Antoine equation: P(t) = exp d dt (13. (3.17 Estimate the change in enthalpy and entropy when liquid ammonia at 270 K is compressed from its saturaration pressure of 381 kPa to 1.166 kPa dPdt =1.306 x 104 cm3/mol By Eq. For benzene: ω = 0.63) and the vapor volume by the generalized virial correlation.166 P = 36.63) B0 = 0. (3.55 x 10 -3m 3 kg-1.575 Tr = T / Tc Zc = 0. For saturated liquid ammonia at 270 K. Vt = 1.98 bar Vc = 295 cm3/mol Tr = 0. Vliq = Vc .(3.007 . SOLUTION : T = 323. The entropy change of vaporization is equal to the latent heat divided by the temperatur. we need the volume change of vaporization.51 220.65).139-0.2 K Pc = 48. For the Clapeyron equation. pers.72) for the latent heat and divide by T to get the entropy change of vaporization: ∆S = dPdt. (Vvap – V liq) = 100. 6. T / P x dPdt = 102.14 J / mol.34 J / mol.18 Let P1sat and P2sat be values of the saturation vapor pressure of a pure liquid at absolute temperature T1 and T2..Solve Eq.. K (a) Here for the entropy change of vaporization: ∆S = R. Justify the following interpolation formula for estimation of the vapor pressure Psat at intermediate temperature T : ln Psat = ln P1sat + T ₂ ( T −T ₁ ) T ( T ₂−T ₁ ) ln P 2sat P 1sat SOLUTION : Ln P2sat = A – B T2 B T Ln Psat = A – Ln P1sat = A – ………………A ………………….. T 2 )……………………….C Eliminasi C dari A B B − Ln P2 – ln P1 = A – A – ( T 2 T 1 ) sat sat Ln P 2sat P 1sat Ln P2 P 1sat Ln P 2sat P 1sat sat =0+ B B − T 1 T2 1 1 − = B( T 1 T 2 ) = B( Eleminasi C dari B T 2−T 1 T 1.B B T 1 ……………….1 . (6. .1 dan pers.......ln ln Psat = ln P1sat + = T ₂ ( T −T ₁ ) T ( T ₂−T ₁ ) P1 sat T −T 1 T 1... lau aplikasikan (A) pada area titik didih normal: ....70).70) dalam log10 sehinggga menjadi : log P sat = A – B/T..2 Ln P sat P 1sat Ln P sat P1 Ln P sat sat P1 : Ln P 2sat P 1sat : Ln P2 sat P1 sat Ln P sat sat = ..... derive Edmister’”s formula for estemation of the acentric factor: Where Ѳ ≡ Tn/Tc. Ln P 2sat P 1sat …………………(terbukti) 6....(C) jika P sat dalam (atm).......(B) masukan perbedaan T... T )…………………………pers.(A) Masukan titik kritis : log Pc = A – B/Tc.... log P sa = B(i/Tc – 1/T) = B ((Tr – 1)/T).19 Assumsing the validity of Eq...B B − Ln P – ln P1 = A – A – ( T T 1 ) sat sat Ln P sat P 1sat =0+ Ln P sat P 1sat 1 1 − = B( T 1 T ) Ln P sat P 1sat B B − T1 T T −T 1 T 1.. and Pc is an (atm) SOLUTION : Tuliskan persamaan (6. T 2 ) ln P 2sat sat P1 .2 = B( Bandingkan pers......... Tn is the normal boiling point.. Ln P 2sat sat P1 T ₂ ( T −T ₁ ) T ( T ₂−T ₁ ) T ₂ ( T −T ₁ ) T ( T ₂−T ₁ ) T 2−T 1 T 1.... T ) : B( = B( = T ₂ ( T −T ₁ ) T ( T ₂−T ₁ ) . (6.... Eq.54) ω = −1.4 : H1 = 1156. 6.4 BTU lbm . maka: Dari persamaan (3.7 Jadi.21 The state of 1 (lb m) of steam is changed from saturated vapor at 20 (psia) to superheated vapor at 50 (psia) and 1.log 1 = A – B/Tn or A = B/Tn dengan θ ≡ Tn/Tc. What are the enthalpy and entropy changes of the steam? What would the enthalpy and entropy changes be if steam were an ideal gas? SOLUTION : Tabel F. (B) sekarang dapat ditulis menjadi :: Dimana: Persamaan menjadi: Masukan Tr = 0.0 − log(Psatr )Tr=0.7.3 BTU lbm H2 = 1533.000 ( 0F). 96 + 459.470 + R 1.123+1 + 2 2 2. rankine Berdasarkan persamaan 4.BTU S1 = 1.9977 lbm .1 BTU lbm ∆ S=¿ S – S 2 1 BTU =0. rankine BTU S2 = 1. 105 ( ) 382.017) = 3.9 dan 5.215 J/molK .123 B C D T 0 ( τ +1 ) + T 20 ( τ 2 +τ +1 ) + 2 2 3 τT0 (Cp) H =3.450 .51392 x 8.928 K 382. rankine Sehingga.017 2. 10−3 0.7320 lbm .57)rankine P1 = 20 psi P2 = 50 psi T0 = 382. ∆ H=¿ H – H 2 1 = 377.51391 Cp ( ¿ ¿H = 3.51391 x R = 3.121 .123(382.928 K τ= T1 T0 = 810.017 K (Cp) H = A+ R = 2.18 (uap sebagai gas ideal) T0 = (227.314 J/molK = 29.017 K T1 = 810.266 lbm .67)rankine T1 = (1000+459. 470 + D 2 τ T0 2 () τ +12 )]( τ−1 ln τ ) 1.2076.1.350514 X 8. gm Sliquid = 3.22. 10−3 .314 = 19.753 ) = 4.121 .15m³.635 J ig (C p )s =A+ R [ ( [ B T 0+ C T 20 + = 3.000 kPa Vliquiid = 1.753) – 0.123 (382. .123−1 0.384.350514 ∆ S=2. J gm .450 . table F.2 pada 8.916 = 2. A two-phase system of liquid water and water vapor in equilibrium at 8.017) )( )]( 2.911 K) =12530.000 kPa consist of equal volumes of liquid and vapor.338 ∆S R ig = (C p )s T 1 P ln −ln 1 R T0 P0 = 4.Cp ¿ ¿ ∆ H=¿ = 29.382.215 J/molK(428.123+ 1 2 2 2 2.105 2.338(0.017+ ( 0. what is the total enthalpy H’ and what is total entropy s’? SOLUTION : Data.5422 J/K 6. If the total volume V’=0. K cm³ gm J Hliquid = 1317. 191 kg 2.15 .15 .9. . cm³ gm J Hvapor = 2759. K 1.96 KJ + 8798.15 .768 gr = 54191 gr = 54.7471 J gm .1 4.188 kg. 1317. 106 cm ³ 2 Vvapor = 0. Hliquid + m vapor.188 kg. gm Svapor = 5.191 kg.m liquid.15 . 106 cm ³ 2 =Vliquid 6 = 0. Mvapor 3. 2759 KJ kg = 71374.321 KJ/K = 192. 5. M liquid 0. 106 c m3 2 c m3 23.525.384 gr = 150000 2.7471 KJ kg/ K = 173.191 kg.56 KJ = 80173. If the .525 gr = 150000 47.23 A vessel contains 1kg og H2O as liquid and vapor and equilibrium 100 kPa. Hvap = 54.5 KJ KJ = 54. 10 cm ³ 2 cm ³ 1.Vvapor = 23.2076 kg/ K + 3.188 kg = Htotal = . Htotal =- 0. 3.05 gr = 3188 gr = 3.145 KJ/K 6.82 KJ/K + 18. Stotal KJ kg + 3. 2 at 1000 kPa: .vapor occupies 70% of the volume of the vessel. determine H dan S for the 1kg of H2O SOLUTION : Data dari tabel F. 1382 J/gm K + 0.3 ft lbm BTU lbm .3 pada 350 °F diketahui: Vliq = 0.194.1.Svap = (1 – 0.013 H = (1 – x) . 342 Hvap = 1192. 2776.013) 762.1 27 cm 3/ gm 30 X = 0.013 .29 cm3 /gm 70 = (1 – x) . Hvap = (1 – 0.24 Reaktor bertekanan mengandung liquid air dan uap air di keadaan setimbang pada suhu 350 °F total massa dari liquid dan uap adalah 3 lbm.2 J/gm Mencari x = fraksi masa dari vapor x .Vvap 70 = (1 – x) .5828J/gmK = 2.495 J/gm S = (1 – x) .Vliq 30 x . berapakah total entalpi dari reaktor? SOLUTION : Data dari tabel F.29 cm3/gm Hvap = 2776. Hliq + x.Vliq = 1.2 J/gm = 789.198 J/gm K 6.127 cm3/gm Hliq = 762. 6.76 ft lbm BTU lbm Vvap = 3. Sliq + x.605 J/gm + 0. 01799 Hliq = 321. Jika volum dari uap adalah 50 kali volum liquid.605 J/gm Huap =14.013 .013) 2. Vliq + = 3 lbm Vvap 3× lbm Vliq 1+50 Vuap mliq = 3 lbm ft lbm 1+50 ft 3.209 cm /gr 3 V vapor=71.1 BTU 6. H. Vliq mliq + 50. 636 lb Htotal = mliq×Hliq + mvap ×Hvap = (2. mliq . mliq.3 BTU lbm ) Htotal = 1519.01799 = 3 lbm = 2.342 lmb 0. 364 lb mvap = 3 lbm mvap .mliq + mvap = 3 lbm mvap ×Vvap = 50.1 pada suhu 230oC 3 V liquid =1. 76 BTU lbm ) + (0.45cm / gr .2. SOLUTION : ρ= m V ρ= 1 V V= 1 cm3 0.25 Wet steam at 2300C has a density of 0. and S.636 lb x 1192. 364 lb x 321. Determine x. 364 lb mvap = 0.025 gr Data pada Tabel F.025 g cm-3. 704 gr gr H=1990.241 x=0.3 H=443.552 .241 gr x= 38.45 −1.2107 J /gr K Mencari x V =( 1−x ) V liquid+ x uap x= V −V liquid V uap−V liquid 1 cm 3 cm3 −1.552 ) 990.025 gr x= cm3 70.025 gr gr x= 3 cm cm 3 71.61202 J /gr K S vapor =6.791 70.209 gr gr 3 1−0.0 J /gr K S liquid=2.3 J / gr H vapor=2802.209 0.552 Mencari H H=( 1−x ) H liquid + x H uap H=( 1−0.030225 cm 0.H liquid =990.3584 J gr Mencari S S=( 1−x ) S liquid + x S uap .0 gr gr J J + 1546.6554 J J +0. 2802. 15 m −¿ ¿¿ ¿ 4.004 . SOLUTION : Berdasarkan persamaan 6.61202 S=1.93 gr kg .3942m3 /kg mvap= = 0.4 o Pada suhu 150 C.73b.382 kg V tot −mtot .10 ¿ 3 0. V liq =1. 10−3 gr kg 3 ∆ V lv =32930 mtot = V tot 0.1. 3 o Pada suhu 30 C.4283 gr gr S=4.V liq ∆ V lv 0.Vlv Dari tabel F.15m3 mengandung uap saturated pada 150 oC didinginkan menjadi 30 oC.543 ×10−3 kg −3 3 m kg 3 cm m =32.Vliq + mvap.S=( 1−0.15 m3 = V vap 0.2107 gr gr J J +3.004 3 cm m =1.6.1 didapat: cm3 m3 =0.004 .382 kg .5983 J gr 6.3924 gr kg V vap =392. Tentukan volume akhir dan massa air liquid pada vessel.552 ) 2.552 .26 Sebuah vessel dengan volume 0. Vtot = mtot.17 J J +0. V liq ¿ 377.27 Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.15 K (105°C).7−781.1 J gm Interpolate pada 101.543 ×10 kg = 0.124 = 2779. 1100 kPa : Hliq Hvap H2 = 781.124 x = 0.72 gram V tot liquid =mliq . where its temperature is 378.mliq=mtot −mvap −3 ¿ 0. –H throttling : H2 = Hliq + x .124 2779.2.953 x J gm .1−781. (Hvap – Hliq) = H 2−H liq H vap−H liq x = 2686.23 cm3 6.382 kg−4. What is the quality of the steam in its initial state? SOLUTION : According from table F.72 gram.7 J gm = 2686.325 kPa & 105 degC: Const.004 cm gr 3 = 379.1 .325 kPa.37772 kg = 377. What is the temperature of the steam in its final state and what its entropy change? What would be the final temperature and entropy change for an ideal gas ? SOLUTION : Data pada table F.302 J gm.29 Steam at 300(psia) and 500( of) expands at constant enthalpy (as in a throttling process) to 20(psia).80 degC and S2 = 7.015 gm gm.6.100 kPa and 260 oC expands at constant enthalpy (as in a throttling process) to 125 kPa.K ∆S= 1. K 6.2 at 2100 kPa and 260 degC. there would be no temperature change and the entropy change would be given by : P1 = 2100 KPa P2 = 125 KPa ∆S = -R ln P2 molwt P1 ∆S=1.8316 J ∆S = S2 . Table F.5 J S1 = 6.S1 gm. the final temperature is 224.4.K mol 2923.5640 J gm H2 = mol wt=18.28 Steam at 2. K For steam as an ideal gas. untuk tekanan 300 psia dan 500OF dan enthalpy konstan . What is the temperature of the steam in its final state and what is its entropy change? What would be final temperature and entropy change for an ideal gas? SOLUTION : Data. by interpolation : H1 = 2923.268 J gm.5 J gm Final state is at this enthalpy and a pressure of 125 KPa By interpolation at these conditions. 87 BTU lbm rankin S2 = 1.5703 BTU lbm rankin = 0.7 BTU lbm H2= 1257.T = 500oF P1 = 300 psia BTU lbm H1 = 1257.8606 – 1.2903 ¿ −¿ ¿ 20 ) 300 .2903 Untuk steam pada gas ideal. tidak ada perubahan temperature maka −RT ln( P2 ) P1 mol ∆S = molwt = 18 lbmol 8.5703 Namun entalpi yang diperlukan pada tekanan akhir yaitu 20 psia Untuk itu diperlukan interpolasi dan diperoleh hasil interpolasi yaitu : P2 = 20 psia T = 438.314 x 500 x ln ( = = 0.8606 ∆S = S2 – S1 BTU lbm rankin =1.7 BTU lbm rankin S1 = 1. 0912 ) = = 0.564 J gm = 2599.6 J gm J gm + 0.30 Superheated steam at 500 kPa and 300 0C expands insentropically to 50 kPa.33 kPa At 50ºC Psat = 12.122 .564 J gm ) 6.166 kPa = 0.031 101.98.33 kPa ? SOLUTION : At 25ºC Psat = 3.0912 ) gm .6. ( 2646.34 kPa Xwater = Psat P = 12.33 kPa ? at 50ºC and 101. K ( 7.4614−1. K J ( 7.9 . ( Hvap – Hliq ) = 340.31 What is the mole fraction of water vapor in air that is saturated with water at 25ºC and 101.34 kPa = 0.5947−1 .98 = Hliq + x.x340. ( Svap – Sliq) x= H S 1−Sliq Svap−Sliq J gm .33 kPa Xwater = Psat P = 3. What is its final enthalpy? SOLUTION : S2 = S1 = Sliq + x.166 kPa P = 101. Heat is transferred to the vessel until one phase just disappears.014 + 0. and what are its temperature and pressure? How much heat is transferred in the process? SOLUTION : Berdasarkan tabel F.021 m3 Vliq = 1.Which phase (liquid or vapor) remains. vap mass .02011 gm + 8.101.3682 x 10-6 gm = 0. and single phase remains.035 m3 Mass = mliq + mvap = 0.044 cm3/gm Uliq = 41.02012 gm X = m.9 J/gm Vvapor = 1673.021 m3 1.014 m3 Volume liquid = 0.014 m3 of saturated-vapor in equilibrium with 0.021 m3 of saturated-liquid water at 1000C.33 kPa 6.021) m3 = 0.0 cm 3 / gm = 0.044 cm3 /gm + 0.32 A rigid vessel contains 0.5 J/gm Vtotal = Vliq + Vvap = (0.0 cm3/gm Uvap = 2506.014 m3 1673.1 pada suhu 1000C.maka: Volume vapor = 0. 035 0.75 since the total volume and mass don’t change.739 cm3/gm Mencapai saturated liquid pada suhu 349.3682 x 10 gm 0.5001 kPa U2 = 1641. 25% condenses making the quality 0.−6 = 8.(V(v)-V(l)) .8682 J/gm = 419.83 K ( fase liquid) P = 16.419.0 J/gm + 0.5 – 419.159 x 10-4 V2 = Vtotal mass = 0.33 A vessel of 0.02012 gm = 4.832 J/gm 6.0) J/gm = 419.8682 J/gm Q = U2 – U1 = 1641. We have for the final state: V2=V1=V(l)+X.159 x 10-4 ( 2506.8682 J/gm Q = 1221. how much heat is transferred and what is the final pressure? SOLUTION : Of this total mass.9 J/gm + 4. If the vessel is cooled until 25% f the steam has condensed.7 J/gm U1 = Uliq + X(Uvap – Uliq) = 41.7 J/gm .25 m3 capacity is filled with saturated steam at 1500 kPa.02012 cm3/gm = 1. 75 = 175.41 J/gm + 0.044 cm3/gr Vvap = 1673.9 gm (2134.9 J/gm =2134.09 J = -869.4 J/gm) = 1898.41 J/gm) = 782.we make a preliminary calculation to estimate: Vvap = V1 X = 131.66 cm3/gm/0.75 (2584.959 J/gr .75 (1802.3 J/gm -2592.9 gm ( -458.5 kPa and Vliq = 782.9 J/gm V2 = Vliq + X (Vvap-Vliq) = 782.75 for wet steam Since the liquid volume is much smaller than the vapor volume. Interpolate on X to find P = 1114.547 cm3/gm This Value Occurs at a pressure a bit above 1100 kPa.Sementara itu: X= V 1−V (l) V ( v )−V (l) Find P for which (A) yields the value X=0.how much heat is transferred and what is the final pressure? SOLUTION : Vliq = 1.34 A vessel of 0.41 J/gm Vvap = 2584.41 J/gm + 0.49 J/gm = 782. Evaluate X at 1100 and 1150 kPa by (A).25 m3 capacity is filled with saturated steam at 1500 kPa.31 J/gm Q = mass (V2-V1) = 1898.1 J/gm ) = -869886.9 KJ 6.9 J/gm) -782. If the vassel is cooled until 25 % of the steam has condensed.41 J/gm + 1351.0 cm3/gr Uliq = 418. 5 J/gr U vap = 2598.059 x 1671.645 cm3 /gr = 99.044 cm3 /gr + 98.184 x 103 gr m total = m liq + m vap = 20.689 cm3 /gr Berdasarkan tabel F.059 (1673.959 J / gr ) = 418.689 cm3 /gr U1 = Uliq x U vap x Uliq = 418.044 cm3/gr) = 1.044 cm3/gr + 0.341 x 103 gr x V1 = = = 0.157 x 103 gr = m vap = = = 1.0 cm3/gr .Uvap = 2506.059 (2506.5 J/gr m liq = = 19.5 J / gr .165 J / gr = 542.1 maka didapat T = 212 oC U liq = 904.418.1.124 J / gr V1 = V2 = 99.541 J / gr = 418.959 J / gr + 0.959 J / gr + 0.959 J / gr + 123.Vliq) = 1.058 = Vliq + x (Vvap .956 cm3/gr = 1.044 cm3 /gr + 0.059 x 2087.0 J/gr U2 = Uliq x U vap x U liq . 4 m³ volume is filled with steam at 800 kPa and 350 c. a.904.124 J / gr) = 20.5 J/gr + 0.044 x 103 J = 9369. A rigid vessel of 0.4.36 1 Kg of steam is contained in a piston / cylinder device at 800 kPa dan 200oC. isothermal expansion to 150kPa.058 (2598.4 m³ T = 2000c U2 = 2638.341 x 103 gr x 460.= 904. SOLUTION : Appendiks F2 : P = 800 kPa T = 350c V1 = 354.10 cm ³ cm ³ 354.341 x 103 gr (1002.058 x 1693.0 J/gr .044 kJ 6.723 J/gr Q = m total ( U2 – U1 ) = 20. How much heat does it absorb? . If it undergoes a mechanically reversible.5 J/gr ) = 904. How much heat must be transferred from the steam to bring its temperature to 200c.7 Q= Vtotal V1 J gr (U₂-U₁) 6 = 0.5 J/gr + 0.599 J / gr = 9369.34 Cm ³ gr U1 = 2878 J gr Vtotal = 0.542.723 J/gr .5 J/gr + 98.233 J/gr = 1002.5 J/gr = 904.34 gr 6.35. 3 gm J gm . K Uliq = 466.6439 T = 473 oK Q=1 kg .4336 J gm . If it undergoes reversible. K J gm .473 K .3−2629.2234 J gm .9 S1 = 6.b. K } J −392. Keadaan Isotermal. K Q = 392. Entropy konstant pada saat 150 kPa Sliq = 1.2 (Superheted Steam) pada 800 kPa dan 200 oCU1 = 2629.4 { W = 1 kg .9 ) J kg−392. Pada Saat 150 kPa dan 200 oC J Pada Tabel F. ∆ S Q=473 kg .1643 KJ gm B. ( 2656.8148 ) Q=m . K ( 0. adiabatic expansion to 150kPa. What is its final temperature and how much work is done ? SOLUTION : Dari Tabel F.8148 J gm . K S2 = 7.8291 ) J gm.1643 KJ gm W =−365.968 J gm Svap = 7.7643 K J gm .1643 KJ W =(m . K a.6439−6.2 U2 = 2656.T . ( 7. ∆ U)−Q W =26. 8148−1.367 x 103−2629. Table F. K X= J ( 7.4336 ) gm .2234−1. by interpolation: H2 = 3633. (575°C).367 x 103 J gm W =m. How much heat is required per kj ? SOLUTION : Data.4336 ) X =0.37 Steam at 2000 kPa containing 6% moisture is heated at constant pressure to 848.15 K. (3633.J gm Uvap = 2519. ( U 2−U 1) W =1 Kg ( 2.9 ) J gm W =−262.929 U 2=U liq + X ( U vap−U liq ) U 2=2.589 x J gm For superheated vapor at 2000 kPa and 575 degC.2 at 2000 kPa: H vap = 2797.527 KJ 6.684X103) J/gm .2 x J gm Hliq = 908. K ( 6.4 – 2.5 S vap−¿ S S −S X = 1 liq ¿ liq J gm .4 x Q J gm = mass H2 H1 = 1 kg . = 949.2244 gm .It is than heated at constant volume until it is saturated vapor.5924 gm . Determine Q and W for the process? SOLUTION : Kondisi pertama adiabatik ekspansi P1=2700kPa X1=0.52 kJ. maka: U23= Q23 + 0 U23= Q23 U3-U2= Q23 Q23 = U3 U2 Untuk proses keseluruhan Q = U3 U2 dan W = U2 U1 Untuk tekanan P1=2700kPa Dari tabel F2 halaman 710 J Uliq=977.9 .90 undergoes a reversible.90 P2=400 kPa Kondisi ke dua pemanasan dengan volume konstan 1 Q12 = 0 U12= Q12 + W12.adiabatic expansion in a nonflow process to 400 kPa.8 gm J Sliq=2.968 gm J Uvap= 2601.700 kPa and with a quality of 0.38 Steam at 2.. 6. maka: U12= 0 + W12 U12=W12 U2 –U1=W12 W12 = U2 U1 2 Kondisi ke dua pemanasan dengan volume konstan W23 = 0 U23= Q23 + W23 karena W23 = 0 . K J Svap=6. K X1=0.karena Q12 = 0 . 439 103 gm S1 Sliq x1Svap Sliq J S1 2.2244 J S1 2.8 gm J U1 977. K cm gm Since step 1 is isentropic.7 J gm cm3 Vliq = 1.5924 gm . K S1= 5.U1 Uliq x1Uvap Uliq J U1 977.92601.237 gm Uvap = 2552.96. K Untuk tekanan P2=400 kPa Dari tabel F2 halaman 700 J Sliq = 1.968 gm J gm J 977.084 gm 3 Vvap = 462. K .5924 gm . K J gm . K J Uliq =604. K 0.968 gm J U1= 2.968 gm J 0.7764 gm .8943 gm .5924 gm . K J Svap= 6.22 J gm . S2 = S1 = Sliq x2Svap Sliq J 2.861 J gm . K gm .7764 gm .237 gm 0. K 5.084 gm 0.2(X1) 2560. K x2= J J 6. K gm .159 103 gm V2 Vliq x2Vvap Vliq cm3 V2 1.22 cm3 V2 1. K 4. vapor with this specific volume.0846 x2 0.8943 −1.7982552.23 0C Dari tabel F1 halaman 691 Volume spesifik U spesifik 373. K x2= J 5.3 (Y1) 369.070528 cm gm cm3 gm cm3 1.Interpolate to find that this V occurs at T = 509.1 (X2) 2562.861 J gm.084 gm cm3 gm 3 V2=V3 (karena pemanasan dengan volume tetap/konstan) The final state is sat.1 (Y2) Interpolasi .1179 gm .S 1−Sliq x2= Svap−Sliq J J −1.7764 gm .798 U2 Uliq x2Uvap Uliq J U2 604.7 J J gm 604.798462.237 gm gm J U2= 2.070528 (X) / V3 ???? (Y)/ U3 355.237 gm J J U2 604.084 gm V2= 369. 439 103 gm J W= .0548 J/gm.2. yaitu : U2 = 2578.8 J/gm S1 = 7.2 ).2 J Y= 2560.4017 10 3 J gm Work U2 U1 J J W= 2. Determine Q and W for the process.0.5607 103 Q=0.7 gm Y= U3.K .yaitu : U1 = 2605.X −X 1 Y=Y1 + ( X 2−X 1 ) (Y2-Y1) 369.2 Y= 2560.6221 J/gm.39 Four kilogram of steam in a piston/cylinder device at 400 kPa and 175 0C undergoes a mechanically reversible. 6.159 103 gm .070528−373.3) 355.5607 103 J gm Q U3 U2 J J 3 2. maka J U3 =2560.K Untuk saturated steam pada suhu 1750C diperoleh lah nilai ( Dari tabel F. isothermal compression to a final pressure such that stem it just saturated.280 J gm Tanda minus menunjukkan bahwa sistem melakukan kerja.28 103 gm W= .8 J/gm S2 = 6.159 10 gm gm Q 2.1 -2560.1−373. SOLUTION : Untuk P = 400 kPa & T = 1750C diperolehlah nilai ( Dari tabel F.1 ).7 gm = 2.3+ ( ) (2562. W = m x (U2 . 0. 1.15 K x (6.638 ICPS(723.8 J/gm ) – (.0854 J gm. 3.15) K J gm J gm .2605.470.15 K P1 = 3000 kPa P2 = 235 kPa ICPH(723. hasilnya ialah : a. K T2 = (140 + 273.121×105) = -2.415901 .15.2.413.8 J/gm .66 kJ b.415901 ICPH = -1343. K Table F.5 J gm S 2=7. Determine ∆H and ∆S: (a) From steam-table data (b) By equations for an ideal gas (c) By appropriate generalized correlations (a) Table F.121×105) = -1343. 3.000kPa to a final state of 140oC and 235 kPa.775.1.450×10-3 .15) K T1 = 723. 3000 kPa and 450 oC SOLUTION : H 1=3344.K ) = .K . 0.2.7.450×10-3 .6221 J/gm.66 kJ ) = 667. K ∆ H=−600.15. 413.15.0548 J/gm.40.1 ∆ S=S 2−S1 ∆ S=0. 0. Steam undergoes a change from an initial state of 450oC and 3.15.638 K ICPS = -2.6 J gm S 1=7.0.775.470.2003 ∆ H=H 2−H 1 J gm .S1 ) = 4 kg x 448. Q = m x T x ( S2 .67 kJ 6.0. interpolate 235 kPa and 140 oC H 2=2744.0.U1 ) – Q = 4 kg x ( 2578.115 (b) T1 = (450 + 273.15 K T2 = 413.Jadi. K 6. It then expands.11752 Pr =0.04422 SRB(0.345) = -0.13602.01066 2 2 The generalized virial-coefficient correlation is suitable here HRB(1. to the initial temperature of 473. 0. 0.13341 HRB1 = -0. 0. 0.08779 SRB1 = -0. 0.15 K(200°C).Eqs. the steam is compressed in a mechanically reversible.05048 ∆ H=∆ H ig + ∆ S=∆ S ig + R .( HR B2−HR B1 ) molwt R .6 ∆ Sig = J gm ∆ Sig =0.13341 SRB(1. K w = 0.2 Superheat Steam pada 550 kPa dan 200OC: V1 = 385.41 A piston / cylinder device operating in a cycle with steam as the working fluid executes the following steps: Steam at 550 kPa and 473. ICPH ∆ H ig = molwt ∆ H ig =−620.(HR B 2−HR B1) molwt ∆ H=−593. 0.86) & (6.01066.87) for an ideal gas: molwt=18 R .08779 HRB(0. SOLUTION : Data table F. isothermal processto the initial pressure of 550 kPa.95 ∆ S=0.1 K Pc = 220.0605 (c) Tc = 647.63846.0. Finally.55 bar T r =1.04422 HRB2 = -0.15 K (200°C) is heated at constant volume to a pressureof 800 kPa.13602.345 T r =0.T c .345) = -0.11752.13602 1 gm mol 1 P2 ) P1 ( ) molwt J gm .345) = -0.(ICPS−ln R . reversibly and adiabatically.01066.078 J gm J gm.19 cm3/gm . (6.05048 SRB2 = -0.345) = -0.0.63846 Pr =0.11752.63846. reversibly and adiabatically.5782 J/gm.465 J/gm Untuk Siklus perubahan energy dalah adalah = 0 Wsiklus = -Qsiklus n = .5782 ) = -268.6 ) J/gm = 322.5782 J/gm. pada volume spesifik awal P. to the initial temperature of 417.1 - 2640.0108 J/gm. maka: Q23 = 0 S3 = S2 S3 = 7.K Sehingga.35ºF .4 J/gm Ekspansi ke T awal.1672 6.4 = 0.Q12 – Q31 1+ −268.U1 = ( 2963.465 322. Q12 = U2. V = Konstan U2 = 2963.7.42 A piston/ cylinder device operating in a cycle with steam as the working fluid executes the following steps : Saturated – vapor steam at 300(psia) is heated constant preasure to 900ºF In the expand.K Pemanasan Volume konstan pada tekanan 800 kPa.15 K Q31 = T ( S1 – S3 ) = 473.1 J/gm S2 = 7.74 oC.6 J/gm S1 = 7. interpolasi yang didapat t = 401.Wsikuls / Q12 n = 1+ Q 31 Q 12 = = .K Temperatur konstat dikompresi ke P awal: T = 473.0108 .U1 = 2640.15 (7. 300(psi) : T1 = (417. F. Vapor. Rankine Superheated steam at 300(psi) and 900ºF H2 = 1473. (S1-S3) Q31 = -218. Rankine T1 = 877. Berapakan tekanan yang dikeluarkan dalam bentuk aliran uap basah dengan nilai 0.6 BTU S2 = 1.67) rankine H1 = 1202.isothermal proses to the initial state What is the thermal efficiency of the cycle ? SOLUTION : Table. jadi kita mencarinya dengan cara interpolasi. P2 = 572. sat.02 rankine S1 = 1. Berapakah tekanan yang dikeluarkan dalam bentuk aliran uap jenuh? b.95? SOLUTION : Dari data tabel F. a. Pertama kita mencari tekanan dimana tekanan uap jenuhnya memiliki entropi.2 superheated steam pada 4000 kPa dan 400 °C: S1= 6. the steam is compressed in a mechanically reversible.7591 BTU lbm Q12 = H2-H1 Lbm.9 BTU Lbm. Finally.5105 BTU Lbm.35 + 459.7733 J gmK Dari kedua masalah didapatkan: S2 S1 a.43 Uap masuk ke dalam turbin pada tekanan 400 kPa dan suhu 400 °C terekspansi secara reversibel dan adiabatis.4. Hal ini terdapat pada tekanan di bawah 575 kPa.83kPa .027 BTU lbm 6. Rankine Q31 = T1. 8 kJ/kg .9 kJ/kg .6071 gmK Svap = 7. the final entropy would be S2 S1 Table F.0520 J gmK S2= Sliq x Svap Sliq J S2 = 1.317.0261kJ/kgK) x2= 0.0261kJ/kgK Svap = 7.8 kJ/kg If the turbine were to operate isentropically.3404 kJ/kgK H2 = 2683. Untuk uap basa.7 kJ/kg) / (2.3259. and for the final condition of 40 kPa and 100 degC: SOLUTION : H1 = 3259.2 for superheated vapor at the initial conditions.1.0520 J J ¿ 1.7733 Didapatkan dari interpolasi P2 = 250.95 H' Hliq x2Hvap Hliq H'= 317.7 kJ/kg S1 = 7. Tekanan ini terdapat jika dibawah tekanan 250 kPa. Pada tekanan 250 kPa nilainya: J Sliq = 1.b.522 103 kJ/kg H2 H1) / (H' H1) 2683.95 S2 = Sliq xSvap Sliq Jadi kita harus mendapatkan tekanan dimana persamaannya diketahui.6071 gmK gmK J gmK atau > 6.16 kJ/kg) H' =2.9 kJ/kg x2= (S2 Sliq) / (Svap Sliq) x2= (7.522 103 kJ/kg . 1300 kPa and 400 degC. liquid and vapor at 40 kPa: Sliq = 1.46 Table F.1.6709 kJ/kgK Hliq = 317.6071 gmK = 6.16 kPa 6.7798 + 0.95 ( 7.2 for sat. entropinya di dapatkan: x= 0.16 kJ/kg + 0.7 kJ/kg) .3259.3404 kJ/kgK .16 kJ/kg Hvap = 2636.6709 kJ/kgK .0261kJ/kgK) / (7.95(2636. 3 J / gm .7 J / gr Sig = 10.3500.15 cm3 / gr HR = H .0.Hig HR = 2856.7 J / gm HR = .15 K 1.5503 J / gr K Hig = 2928.6 atm V = 132.15 K Molwt = 18 gr / mol R molwt R V =V– T P 3 0.6 atm .85 cm3 / gr – 3633 cm3 / gr VR = .08206 atm VR = 132.47 From steam table data estimate values for the residual properties V R . HR .72.6 x 10 5 Pa = 1.78 6.85 cm3 / gr – 18 gr mol VR = 132.2928.0681 J / gr K T = 225 oC = 498.3 J / gr S = 6.85 cm3 / gr H = 2856.4 J / gm Δ Sig = −R molwt ln P Po dm mol K 498. SR for steam at 225 oC and 1600 kPa and compare with values found by a suitable generalized correlation? SOLUTION : P = 1600 kPa = 1. 172 0.558 = 0.345 Tc = 647.−o .1 K Pc = 220.3.139 – 0.0.6 ¿ = = 0.76982 = 0.36 SR = S – ( Sig + Δ Sig) J gr K = 6.0.6 Tr Bi = 0.377 ) .5503 .769824.0681 J gr K .0.( 10.55 bar Tr = T Tc = 498.083 – 0.36 J gr K J gr K = .1 K Pr = P Pr 220.00725 ¿ ¿ Bo = 0.139 – 0.172 Tr 4.2 = 0.083 – 0.422 1.76982 = .422 1.2298 Reduced condition ω = 0.55 ¯¿ ¯ 1.3.6 0.0336 dm3 mol K gr mol ln 1600 kPa 1 kPa atm dm3 gr K J gr K = .08206 atm = 18 = .2 = .0.15 K 647. 0076 J / gr K -0.15 K mol K gr 1.9.0.558 + 0.08206 atm HR = -0.17858 SRB = .76982 Z = 1 – 0.643 J / gr SR = R molwt SRB 0.167101 HR = R Tc molwt HRB dm 3 647.00648 Z = 0.345 x -0.00725 0.0.377 ) 0.1 K molK gr 18 mol 0.6 atm18 mol 0.17858 HR =-52.08206 atm VR = ( 0.19755 cm3 / gr HRB = .99352 VR = RT P molwt (Z–1) dm3 498.08206 atm SR = 18 dm 3 molK gr mol SR = .99352 – 1 ) VR =.167101 .Z = 1 + ( Bo + ω Bi ) Pr Tr Z = 1 + ( -0.0. 15 K = 4.15 K x 6.8 cm3 gr -1 – ( 8. T = 453. Sig = 8.15 K r = ∆Hlv / T = 2. and SR for saturated vapor at 1000 kPa. By interpolation in table F.1393 J gr-1 k-1 = 4. and SR for saturated vapor at 1000 kPa.453.4427 J gr-1 k-1 ∆Slv = Sv .1.Hl = 2776.2 at 1 kPa Hig = 2831.32 J gr-1 Gv = Hv – T.3 J gr -1 .314 J mol-1 k-1 x 453.0132 J gr -1 ∆Slv = Sv .3 J gr -1 .88 degC and 1 kPa is an ideal gas.5819 J gr-1 k-1 .672 cm3 gr -1 ∆Hlv = Hv .8 cm3 gr -1 . Does this result agree with the steamtable value? Apply appropriate generalized correlations for evaluation of VR.2.6.Sl = 6.48 From data in the steam tables: (a) Determine values for Gl and Gv for saturated liquid and vapor at 1000 kPa. ∆Vlv = Vv .29 J gr-1 b.128 cm3 gr -1 = 192.Sv = 2776.5819 J gr-1 k-1 .1393 J gr-1 k-1 = -206.0132 J gr -1 .875 cm3 gr-1 For enthalpy and entropy. VR = Vv – ( R.1393 J gr-1 k-1 = 4.2. ∆Hlv = 2. assume that steam at 179. Do these results agree with the values found in (c)? SOLUTION : So.1 J gr -1 = 2.15 K x 2. T / molwt. HR .763.4426 J gr-1 k-1 a.4426 J gr-1 k-1 c.Sl = 6. Should these be the same? (b) Determine values for ∆Hlv/ T and ∆Slv at 1000 kPa. P) = 193. P0 = 1 kPa .5819 J gr-1 k-1 = -206.015 gr mol-1 x 1000 = -14. Gl = Hl – T.1 J gr -1 – 453. (d) Estimate a value for d P sat /dT at 1000 kPa and apply the Clapeyron equation to evaluate ∆Slv at 1000 kPa.Vl = 193. HR . Sl = 763.0132 J gr -1 / 453. Should these be the same? (c) Find values for VR.2 J gr-1 .15 K / 18.7994 J gr-1 k-1 . 1126 J gr-1 k-1 d.15 / 647. slope = slope (x.139 – ( 0. 1050 ) kPa . molwt ) Z-1 = ( 8.93 cm3 gr-1 HR = ( R.63 .3 J gr -1 .6 ) = -0. Pc = 220.314 / 18.4 J gr-1 k-1 Reduced conditions : ώ = 0.664 B1 = 0.8. Tr) i = 1.15 / 10.664 + 0.61) + (3.Tc / molwt) (HRB.62) and (3.984 kPa k-1 = 4.1 = 0.188 J mol-1 k-1 SR = Sv – Sig + ∆Sig = 6.2 J gr-1 = -36.314 . 179. B1 .984 kPa k-1 ∆Slv = ∆Vlv .345 .79. Pr / Tr = 1 + -0.63 By Eqs (3.7001 K Pr = P / Pc = 10 / 220. Assume ln P vs 1/T linear and fit three data pts @ 975.2 ) = -0.63) along with Eq.1 K . Yi = ln (ppi / kPa) ..0453 bar The generalized virial-coefficient correlation is suitable here B0 = 0.T / P. 0. 1000.02 ) (degC) Xi = 1/ ti + 273.015) 0.943 – 1 = -11.9 J gr-1 ∆Sig = (-R / molwt ) x ln (P/P0) = ( -8. dPdT = 192.345 .55 = 0.422 / Tr 1. t: ( 178.943 VR = ( R.015) ln (1000/1) = -3. 1050 kPa Data : pp: ( 975.672 cm3 gr -1 x 22. the entropy does depend on P : HR = Hv – Hig = 2776.0453 / 0.88.7994 J gr-1 k-1 + 3.172 / Tr 4. 18.5819 J gr-1 k-1 . 1000.15 .083 – ( 0.7001 = 0. 182.The enthalpy of an ideal gas is independent of pressure .40) Z = 1 + B0 + ώ . 453.y) slope = -4717 dPdT = (-P / T2) ln slope K = 22.2831.188 J mol-1 k-1 = -0.55 bar Tr = T / Tc = 453. Tc = 647. -0. (6.3 . 1 R .65 btu/lbm .45 btu/lbm a.5695 btu/lbm rankine Vlv = Vv .Hl = (1194. Tr) = -0.65) btu/lbm = 863.4 didapat : Vl = 0. Should these be the same? b) Determine values for ΔHlv /T and ΔSlv at 150(psia). Gl=H l TSl - = 330. HR.(818. lv ∆ S =Sv - Sl .069 J gr-1 k-1 6.49 From data in the steam tables : a) Determine values for Gl and Gv for saturated liquid and vapor at 150(psia). 1.5141 btu/lbm Rankine) = -89.91 btu/lbm b.935 btu/lbm v G =H v TS - v = (1194. and SR for saturated vapor at 150(psia) d) Estimate a value for d Psat /dT at 150(psia) and apply the Clapeyron equation to evaluate ΔSlv at 150(psia). 0.1 R . Does this result agree with the steam-table value? Apply appropriate generalized correlations for evaluation of VR.1 btu/lbm) – (818.43+459.67) Rankine = 818.014 ft3/lbm Hl = 330. Should these be the same? c) Find values for VR.1 – 330.0181 ft3/lbm Vv = 3.1 btu/lbm Sl = 0.Vl = (3. HR.= -43.014 – 0. and SR for saturated vapor at 150(psia).5695 btu/lbm Rankine) = -89.1 Rankine SOLUTION : Dari tabel F.0181) ft3/lbm Hlv = Hv .18 J gr-1 SR = R / molwt ( SRB.5141 btu/lbm rankine Sv = 1. Do these results agree with the values found in (c)? T= (358.65 btu/lbm Hv = 1194. 43−350 = 400−350 2.1445 H ig =1222. P ¿ 3. didapat: T −T b H ig −H b = T a−T b H a−H b T −T b S ig −Sb = T a−T b Sa −S b H ig −1218.1 R .8−1218.43−350 = 400−350 1241. sehingga.5141) btu/lbm R = 1.0554 btu/lbm R ∆ H lv 863. tetapi entropi selalu tergantung pada tekanan. Dengan interpolasi pada tabel F.014 ft 3 /lbm− 10. V R =V V − = 1.7 358. c.4 pada 1 psi.015lbm /lbmol .5 btu/lbm .45 btu /lbm = T 818.150 psia 3 ¿−0.1 btu/¿ = -28.= (1.1 R 18.6 btu/ ¿ lbm ¿ 1194.1492 Btu lbm R Po = 1 psi Entalpi gas tidak bergantung pada tekanan.73 ft 3 psia/lbmol R .7 Sig −2.6 Btu lbm S ig =2.1722−2.2345 ft lbm Untuk entalpi dan entropi asumsikan sistem pada 358.43oF dan 1 psi adalah gas ideal. R v H =H −H ig lbm−1222. 818.1445 358.0554 btu/lbm R RT BM .5695 – 0. 0. . dan 155 psia).77 358. X 1= 1 =1. 150. t= 1 t i+ 459.552 btu /lbmol R S R =Sv −( S ig −∆ Sig ) ¿ 1.67 . 10−3 358.015 lbm/lbmol 1 = .67 .67 .0.67 .77+ 459. X i= 1 T linear untuk data (145.43 361. Asumsikan ln P vs [ ] 145 150 psi 155 Data : . 2.10−3 355.∆ Sig = ¿ −R P ln BM Po ( ) −1.218 .0339 [ ] 355.02+459.02 0 F i=1.987 btu /lbmol R 150 ln 18.0277 btu /lbmol R d.43+ 459.1492−0. y=ln ( Ppi Psi ) y 1=ln ( 145 150 ) = -0.10 361.552)btu/ lbmol R = . X 3= 1 −3 =1. X 2= 1 =1.23 .5695 btu/lbmol R−(2.22.3 . 1. y 3=ln ( 155 150 ) = 0.501 .987 btu/lbmol R lbm R ft 3 psi 10. slope 2 T 818.5 =0.905 psi/R .7024 647..73 lbmol R = 1. ∆ Slv =∆ V lv dPdT 2.996 = ft 3 psi .0469 P c 220. y 2=ln ( 150 150 ) =0 . 1.1 R ¿ ¿ ¿ 2 = ¿ −150 psia ¿ = 1.1 K P = 150 psi = 10.342 bar Pc = 220.905 .5 K Tc = 647.0328 3 Slope = -8.345 T = 358.43 0F = 454.057 btu/lbm R Kondisi Reduced W = 0. 10 dP dT = −P .44 T Tc = 454.55 bar Pr = Tr = P 10.342 = =0.1 . 1 R lbmol R (0.0336 . B o=0.2 Tr 0.7024 ¿ 1+ ¿ = 0. dan 3. 3.083− = 0.024 btu/lbm . Br R.7024)4 . 818.Korelasi keofisien virial yang sesuai disini : .0336 0.083− 0.63 dan persamaan 6.1884 ft3/lbm HR T r .7024)1. V R= RT (z−1) P BM 3 = ft psia .73 = -0.0469 0. Pr z=1+ Bo+ω B ( ) 1 .083− 0.Tc ¿ HRB ¿ ) BM = -19.015 lbm/lbmol 10.62 .61 . 2 = -0.139− = 0.172 4. B 1=0.62 Dengan persamaan 3.66+(¿) 0. Tr −0.6 = -0.942 .422 Tr 1.66 .62 x 0. ω .40 dari interpolasi Tr dan Pr didapat ω=0.6 0. 18.942−1) 150 psia.422 (0.172 (0. 1 Rankine P=150 psi Molwt= 18.T r .1 BTU/lbm-818.4 di 1 psi.6 BTU/lbm .055 BTU/lbm.Sl= (1.5695 BTU/lbm.5141 BTU/lbm.4 . Should these be the same? f) Determine values for ΔHlv /T and ΔSlv at 150(psia). Br R HR= SRB¿ ) BM = -0.67) Rankine = 818.1 rankine.0.rankinbe pada Po=1 psi .Rankine = 1.65 BTU/lbm .235 ft3/lbm Untuk enthalpy dan entropi. HR.Rankine r=ΔHlv /T = 863.1 rankine. diketahui : Vl= 0.94 BTU/lbm Gv= Hv –T.43+459. Should these be the same? g) Find values for VR.Rankine c.5141 BTU/lbm. Sv= 1.1 BTU/lbm Sl=0. Dengan interpolasi pada tabel F.rankine a. Hv=1194.996 ft3/lbm ΔHlv=Hv-Hl=863. and SR for saturated vapor at 150(psia) h) Estimate a value for d Psat /dT at 150(psia) and apply the Clapeyron equation to evaluate ΔSlv at 150(psia). Vv=3.Sl = 330.Rankine ΔVlv=Vv-Vl=2. ΔSlv= Sv .45 BTU/lbm .014 ft3/lbm Hl=330.43 OF dan 1 psi adalah suatu gas ideal. VR= Vv- = =-0. Gl=Hl-T.015 gr/mol Dari tabel F.45 BTU/lbm / 818.5141) BTU/lbm.5695 BTU/lbm. Do these results agree with the values found in (c)? T= (358.0168 btu /lbm rankin 6. Sv= 1194.Rankine = -89.rankine = -89.055 BTU/lbm. ω . asumsikan bahwa uap berda pada 358.49 From data in the steam tables : e) Determine values for Gl and Gv for saturated liquid and vapor at 150(psia).65 BTU/lbm-818.1492 BTU/lbm. and SR for saturated vapor at 150(psia).0181 ft3/lbm .91 BTU/lbm b. maka : Hig= 1222. 0.1 Rankine = 1. Sig=2.5695 . 1. Does this result agree with the steam-table value? SOLUTION : Apply appropriate generalized correlations for evaluation of VR. HR. 0469 Koefisien virial umum yang cocok disini adalah : Bo=0.B1.552) BTU/lbm.Pr/Tr) = 0.0274 BTU/lbm. dPdT= 2.55 bar Pr= P/Pc = 0.rankine d.056 .139-(0.5695 BTU/lbm.Slope.1 BTU/lbm .62 Z=1+ (Bo+ω.905 BTU/lbm.1.HR=Hv-Hig = 1194.172/Tr. Asumsikan ln P VS 1/T linier dan didapatkan tiga data.4.024 BTU/lbm SR= = -0.6 BTU/lbm = -28.02 Slope = -8501 dPdT=(-P.77 2 150 358.7024 Pc=220. yaitu : i 1 Pp (psi) 145 T (F) 355.rankine \ ω=0.942 VR= = -0.rankine)/T2 = 1.905 psi/rankine ΔSlv= ΔVlv.1492 +0.422/Tr.rankine–(2. 1.Rankin = -0.(0.5 BTU/lbm ΔSig= SR = =Sv-(Sig+ = -0.1 K Tr= T/Tc= 0.rankine psi/rankine = 1.083.Rankine ΔSig)=1.0168 BTU/lbm.1222.1894 ft3/lbm HR= = -19.2) = -0.6) = -0.66 Bl=0.996 ft3/lbm.345 Tc=647.43 3 155 361.552 BTU/lbm. 152 T= 195 + 273.28 = 3.08314 ) 369.496( 0.2 cm3 mol HRO =-2.674 X 103 mol HR1=-0.15) V= 135 V=184.15 369.639 V= ZRT P 0.178 = Gunakan korelasi antara lee atau kesler dengan interpolasi Zo=0.1636 Z= ZO + ω .6141 Z1=0. SOLUTION : Untuk propane : TC =369.639 ( 0.TC =-2.8 K PC =42.R.6141 + 0.50 Propane gas at 1 bar and 35 0C is compressed to a final state of 135 bar and 1950C.152 (0.15 K P=135 bar Tr= Pr = T TC P PC pO=1bar = 468.28 bar ω=0.8 = 1. propane may be assumed an ideal gas.Z1 Z=0.586 . In its initial state.8 K J HRO=-7.15 K = 468.496 .08314 ) (468.266 135 42.1636) Z=0.6. estimate the molar volume of the propane in the final state and the enthalpy and entropy changes for the process. 6.48 bar ω = 0.Estimate ∆H and ∆S for the process by suitable generalized correlations SOLUTION : Untuk Propana Tc = 369.139−0.422 T r 1.083−0.8 K Pc = 42.9275 B 1=0.172 0.9275 B 0=0.083− 1500 kPa 4248 kPa Pr=0.152 T = (70+273)K = 343 K P0 = 101.6 0.51 Propane at 70 oC and 101.139− 0.4549 0 B =−0.422 0.33 kPa is compressed isothermally to 1500 kPa.33 kPa P = 1500 kPa Tr = Tr= T Tc Pr= 343 K 369.083− Pr= 0.3719 B 1=0.2 Tr B 1=0.8 K Tr=0.3531 Diasumsikan propana merupakan gas ideal. maka : B =0.139− 0.9275 0 0 B =0.172 4.1854 P Pc . 675 =−0.152 (−0.3531 [−1.0464−0.0469+ (−0.178 ) ] ¿ 0.0464 0 d B 0.722 0.152 (−0.3068 SR d B0 d B1 =−Pr +ω R dTr dTr ( ) SR =−0.9275 d B1 0.351 (−0.6 ¿ 0.3531 [ −0.925 ¿−0.2525 )+ 0.675 = d T r T r 2.0994 R .2525 0.3719−0.9275 (−0.3531 (−0.2525+ 0.9275 (−0.203 ) ) R R S =−0.722 = d T r T r 5.203 ) ) ] ¿ 0.2 ¿ 0.2833 ) R R S =0.1 B =−0.203 R [ 0 ( 1 H dB dB 0 1 =Pr B −Tr +ω B −Tr Tc dTr dTr )] ¿ 0. jadi , dari keterangan rumus diatas , dapat dihitung nilai dari ∆ S dan ∆ H , yaitu: ∆ H=( Cigp ) H ( T 2−T 1 ) +H 2R−H 1R ∆ S=−1431.3 J mol ∆ S=( C igp ) S ln T2 P −R ln 2 + S2R−S1R T1 P1 ∆ S=−25.287 J mol . K 6.52 A steam of propane gas is partially liquetied by throttling from 200 bar and 370K to 1 bar. What fraction of the gas is liquefied in this process? The vapor pressure of propane is given by Eq. (6.72) with parameters: A=-6,72219, B=1,33236, C=2,13868, D=-1,38551. SOLUTION : W=0.152 Tc=369,3 K Pc=42,48 bar Zc=0,276 Vc=200 cm3/mol Jika keadaan akhir adalah campuran dua fase, maka harus ada pada suhu jenuh pada 1 bar. suhu ini ditemukan dari tekanan uap. P= 1 bar A= -6,72219 B= 1,33236 C= -2,13868 D= -1,38551 Panas laten penguapan pada kondisi akhir akan diperlukan untuk energi keseimbangan. Hal ini ditemukan oleh persamaan Clapeyron. Kami melanjutkan persis seperti dalam Pb. 6,17. P=Pc exp [ A . τ (T ) + B .() τ ( T )1,5 + C .()τ ( T )3+ D.( )τ ( T )6 ] 1−τ (T ) T =230,703 K Panas laten penguapan pada kondisi akhir akan diperlukan untuk energi keseimbangan. Hal ini ditemukan oleh persamaan Clapeyron. Kami melanjutkan persis seperti dalam Pb. 6,17. T = 230,703 P( T )=Pc exp [ A . τ ( T ) + B .() τ ( T )1,5 +C .() τ (T )3 + D .() τ ( T )6 ] 1−τ (T ) d kPa P ( T )=4,428 dT K dPdT =4,428124 kPa K P = 1bar Pr= P Pc Pr = 0,024 Tr= T Tc Tr = 0,624 B ()=0,083− 0,422 1,6 Tr B() = -0,815 B 1=0,139− B1 = -1,109 0,172 Tr 4,2 Vvap= RT Pr [1+() B( )+ W . B 1. ] P Tr Vvap=1,847 x 10 4 cm 3 mol 2 7 Vliq=Vc . Zc[−Tr ] cm3 Vliq=75,546 mol Δ H 1 v=T . () Vvap−Vliq . dPdT Δ H 1 v=1,879 x 10 4 J mol Untuk Langkah (1), menggunakan korelasi umum dari Tabel E.7 & E.8, dan biarkan Jumlah dari perubahan entalpi untuk langkah-langkah ini ditetapkan sama dengan nol, dan persamaan yang dihasilkan diselesaikan untuk fraksi dari aliran yang cair. ENERGY BALANCE: Untuk proses throttling ada entalpi tidak berubah. Jalan kalkulasional dari keadaan awal ke akhir dibuat lanjut dari langkah-langkah berikut: (1) Merubah gas awal menjadi gas ideal di awal T & P. (2) Melakukan perubahan suhu dan tekanan untuk T & P pada akhir negara ideal gas. (3) Merubah gas ideal menjadi gas (4) Sebagian memadatkan gas pada akhir T & P. R H ¿1 R . Tc () HR r ( )= r 1=¿ R . Tc ( ) T1 = 370 K Tr= T1 P1 Pr= Tc Pc Tr = 1,001 Pr = 4,708 nyata pada akhir T & P. 6239 .023 . SOLUTION : Untuk 1.86).568 Δ H 1=−RTc ()r ()+ r 1.3-butadiene at 380 K is 1.07555 ∆ H 3=R .77 bar Zc = 0.0) = -1260.703 .248 x 104 J mol Untuk Langkah (3) perubahan entalpi diberikan oleh Persamaan.824.703 K Tr=0.78).4 cm3/mol Tn = 268.28 HRB = .190 Pc = 42.919.0235 Pc Pr =¿ HRB (0.The enthalpy and entropy are set equal to zero for the ideal-gas state at 101. (6.267 Tc = 425.7733 r1 = -3.enthalpy and entropy for 1. 230.07555 J mol ∆ H 4=−x .152 ) = . HRB ∆ H 3=−232.405 K ∆ H 2=R (−1260.3-butadiene : ω = 0. ∆ H lv ∆ H 1+ ∆ H 2+ ∆ H 3−x .2 K Vc = 220.405 K ) ∆ H 2=−1. 1. ω Δ H 1=1. 0. 0.3-butadiene as a saturated vapor and as a saturated liquid at 380 K. yang. Tr= 230.136 ∆ H lv 6.10-6 . -8.53 Estimate the molar volume.10-3 .6239 Tc ¯ 1 ¿ Pr =0. (6.7 K .785.0.r() = -3.The vapor pressure of 1. ∆ H lv =0 x= ∆ H 1+ ∆ H 2+ ∆ H 3 x=0. Tc . 28.27 x 10 4 J mol Untuk Langkah (2) perubahan entalpi diberikan oleh Persamaan.0. 0.4 kPa.33 Kpa and 0 derajat celcius. yang. ICPH (370.213 . The vapor pressure of n-Butane at 370 K is 1435 kPa.718 Vvap = Z.96bar = 37. Vliq = Vc.7442 Z1 = -0.15 K Po = 101.∆H/T Sliq = -38.96x 105 Pa=3796 kPa.689 .274 .Tr/ 1-Tn/Tc)0. Tc HRo = -2.33 Tr = T / Tc Tr = 0.383 J/mol. R. T Vvap = 1182.4 J/mol Sliq = S uap . SOLUTION : Untuk n-Butane B.Tn (1. K HR = HRo + ω HR1 SR = SRo + ω SR1 HR = -3.449 Zo = 0.035 x 103 J/mol SR = -5.4 KPa To = 273.540 R SR1 = -0.∆H Hliq = -7687.and entropy for n-Butane as a saturated vapor and as a saturated liquid at 370 K.892 R.9 J/mol Svap = -1.1 halaman 654.2 cm3 / mol KPa P HRo = -0. = 109.153 x 103 J/mol SRo = -0.892 J/mol.K SR1 = -7. R.89 cm3/mol ∆Hn = R.K Untuk saturated vapor.96 x105 )Pa =37.092 ((ln (Pc/bar) – 1.15 K .49 J/mol. enthalphy. 0.930 – Tn/Tc) ∆Hn = 22449 J/mol ∆Hn = ∆Hn (1.1K Pc 37. Zc 0.The enthalpy and entropy are set equal to zero for the ideal-gas state at 101. K 6.894 Pr = P / Pc Pr = 0.436 x 103 J/ mol HR1 = -3. Tc HR1 = -0.888 R SRo = -4.1366 Z = Zo + ω Z1 Z = 0.38 ∆H = 14003 J/mol Hliq = Hvap .624 J/mol.33 kPa and 273.200 Tc 425.013) / 0.475 J/mol. Zc (1-Tr 2/7) Vliq.54 Estimate the molar volume. K Hvap = 6315.T = 380 K P = 1919. 9015 (M21) 0.85 + (( .0661) 0.378(Pr) (x) ???? (M) 0.378 0.7800 (M22) X 2−X X −X 1 M = (( X 2−X 1 )M11 + ( X 2−X 1 ) M12) X −X 1 ( X 2−X 1 ) M22) M = (( Y 2−Y Y 2−Y 1 X 2−X + (( X 2−X 1 )M21 + Y −Y 1 Y 2−Y 1 0. but the following values are as good as any: Dari tabel E2 halaman 668 didapatkan: Untuk Z0 0. These must NOT be used for interpolation.87 Pr= P Pc = 0.2 0.4−0. extrapolations must be made from the vapor side.4−0. the values for a saturated vapor lie on the very edge of the vapor region.378−0.7800) M= 0.3 Vc = 255 cm mol Tn 272.2 0.7K T 370K P 1435kPa T0 273.90 (Y2) 0.378 Pr 0.87 0.15K P0 101.378 0.4−0.90−0.2 )0.378 Use Lee/Kesler correlation.33kPa Tr = T Tc = 370 K 425.4−0.4 (X2) 0.8810 (M11) 0.90−0.87−0.7692 0.4−0.87(Tr) (Y) 0.85 0.8810 + ( 0.0661 (M12) 0. and some adjacent numbers are for the liquid phase.85 0.2 )0.85 (Y1) 0.2 ) 0.9015 + ( 0.378−0.4−0.2 (X1) 0.1 K = 1435 kPa 3796 kPa =0. however.87 Tr 0. There may be some choice in how this is done.90−0.2 ) 0. Rather. K 425.7692 .378(Pr) (x) ???? (M) .30881 J mol = .74176=0.378−0.7692 + 0.2 0.1 cm3 mol cm3 V = 1590.0142 + ( 0.314 x 370 K Z x RxT mol K mol K V = = = = P 3 1435 kPa 1435 x 10 Pa 1590.607RTc 0.90 (Y2) 0.0268) + (( 0.87 0.7692 Dari tabel E2 halaman 669 didapatkan: Untuk Z1 0.742 x 8.2 ) -0.2 0.4−0.6078.742 j j 0.0142 (M21) -0.Z0 0.314 x 370 K 0.0268 (M12) Y 2−Y Y 2−Y 1 X 2−X + (( X 2−X 1 )M21 + Y −Y 1 Y 2−Y 1 0.378−0.2 ) -0.378 0.1118 (M22) X 2−X X −X 1 M = (( X 2−X 1 )M11 + ( X 2−X 1 ) M12) X −X 1 ( X 2−X 1 ) M22) M = (( 0.0.0.378 0.87(Tr) (Y) 0.742 x 8.314 2.87−0.145 10 3 J mol J mol .1K= 2145.02744= 0.1 mol HR0 0.0.2 ) 0.0715 + ( 0.90−0.742 Z 0.4−0.90−0.1372) = 0.2 (X1) -0.85 -0.2 ) -0.200 (0.4 (X2) .4−0.4−0.0715 (M11) 0.4−0.90−0.85 M0.4−0.1372 Z Z0 Z1 = 0.85 0.85 (Y1) 0.1118) 0.1372 Z1 0. 048 ICPH 1222.145 10 3 HR1 0.93536.314 J mol .421 J + 0.153701.93536.8318.200 x( 2.402106 0.048K ICPS273.942 J mol .032 mol .942 J mol .200 x (6. K J mol .402106 0.2.032 mol . K SR1 0. K J mol .835 x 8.942 J mol .915103 11. K ICPH273. K HR HR0 HR1 HR 2. K SR = 5.314 J mol .0= 3.0= 1222. K )= 5.987843 J mol = 2.80735 J mol . K 425.J mol HR0= 2.733 103 J mol SR SR0 SR1 J SR 4.80735 ICPS 3.937 103 mol ) = . K = 6.7324 103 + 0.485R 0.937 103 mol SR0 0.831RTc 0.937 J 103 mol J HR1= 2.15370 1.145 103 J mol HR = 2.915103 11. K J SR0= 4.835R 0. K = -4.4204 J mol .1K = .314 SR1= 6.032 J mol .485 x 8.2936. 013 ¿ ¿ ¿ x 272.33 kPa ))+ (5.63) & (4.12) 3 Vliq = 255 Vliq =123.197 1435 kPa (3.Hvap RICPH HR J Hvap 8.85 cm mol (1−0.1 P Svap = R (ICPS – ln( Po ))+ SR J Svap = 8.314 mol .314 mol .274 ¿ cm3 mol J Hn = 8. K ¯¿ ¯¿ 37.80735– ln( 101.7K( ) ln ¿ 1.048K 2. by Eqs.87) ¿ ¿ x ¿ 0. K For saturated vapor. (3.092¿ ¿ J mol . K ) .421 J mol .96 −1. K Svap = 4. K 1222.733 103 J mol ) J mol Hvap = 7427.314 mol . 9 ¿ ¿ = 183.0 cm3) = 585 cm3 n = 5 mol 0.55 Five moles of calcium carbide is combined with 10 mol of liquid water in a closed. 6298 . SOLUTION : Vgas = ( 750 cm3 – (5 mol x 33.187 Tc = 308.9 bar = 2.291 Pc = 61.15 K T 398.1948 Z Z0 Z1 = 0.6298 + 0. Start with reduced pressure from guess value above.187 x 0. and requires use of Lee/Kesler correlation. For a final temperature of 1250C.07 is: Z0 0.15 K 585 61.15 K Tr = Tc = 308.996 Pressure is clearly high. (b) The heat transferred At 125 0C.666 . 65 Z nRT P = V gas = Pr = P Pc 0.0 cm3 mol-1. The eventual result for a reduced pressure of 3.00364276 Z 0.3 K = 1.3 K T = 398.39 bar Z = 0.39 ¯¿ ¯¿ = 183.6298 + ( 0. because P is unknown but is required to find Z.14 x 398.1948) Z = 0. Solution is by trial.Hn = 22514 J mol 6. high-pressure vessel of 750-cm 3 capacity. determine : (a) The final pressure . Z1 0. the molar volume of Ca(OH) 2 is 33. rigid. and the reaction gas to completion. Acetylene gas is produced by the reaction : CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(s) Initial condition are 250C and 1 bar. Ignore the effect of any gas present in the vessel initially.65 x 5 mol x 83. 39 ¯¿ ¯¿ = 188. Q = nU = nH (PV) Q = nH VP= nH VgasP The enthalpy change is evaluated by a three-step process: (1) Reaction at 298.P= Pr = Z nRT V gas P Pc = 0. The heat transferred By the first law.5 bar b).3 K = 5.384 x 8.5 ¿ ¿ = 188.15398.340 x 8. states (3) Transformation to state of real gas H = H298 HP HR Step (1): From data of Table C.07 a).00.5 bar = 3.0= 11.71731 11.43510-3 0.666 x 5 mol x 83.15398.384RTc = 0. from Tables E.159.998 103 mol J HR1 0.5975. MCPH R(5.314 x 308.1321.299105= 5.5 J mol Step (3).340RTc = 2.15 K (2) Change in T for products in std.48920 For the products.271 mol .3 K = 984.48920) HP MCPH100K = 143.314 x 308.95210-30.2. for acetylene(g): MCPH298.4 J ∆ H=⌈ −986090+227480−(−59800 )−2(−285830)⌉ x mol ∆H J 298 = .14 x 398. The Final pressure is P = 188. Table C.7 & E.127150 mol Step (2).15 K 585 61.156.055 x 100 = 14305. for calcium hydroxide(s): MCPH298.8 at the reduced conditions of Part (a) for acetylene: J HR0 2.1 data.71731 Table C.01. may be split into two steps: (1) Transform into an ideal gas at the initial conditions.6K Tr 1.65bar Pr 0.140 Tc 365. interpolate.5 -1 ) bar Q = 606101 6.187x(984.19 105 J 3 mol ) – 585 cm ( 188.815 Step (1): Use the Lee/Kesler correlation.15K/ 365.998 10 3 J mol J + 0.19 105 J mol + (6182 J mol ) J mol Q nH Vgas(P 1bar) Q= 5 mol x ( 1. For the initial conditions: SOLUTION : Tr= T/Tc Tr 400.6K Pc 46.56 Propylene: 0. evaluating property changes from a generalized correlation. occurring at constant enthalpy. evaluating property changes by equations for an ideal gas.15K P 38bar P0 1bar The throttling process.095 Pr= P/Pc Pr38bar/ 46.271 mol )) =6182 J mol H H298 HP HR J H = 127150 mol + 14305. .HR HR0 HR1 = 5. (2) Change T and P in the ideal-gas state to the final conditions.5 H = 1. Property changes for the two steps sum to the property change for the process.65bar T 400. T(C/3 T3( Find 0.124J/molK SR S0 S1 SR = 5.496R S1 = 4. (ICPA – ln (p0/p)) Sig =22. SOLUTION : Dari Appendix B1 pada Tabel B. A 1. (4. dengan mengevaluasi perubahan properties dari korelasi umum. 1 (guess) Given HR= R ((A.623 103J/mol H1 0. Estimate the entropy change of the propane caused by this process.27K ICPS400. propane maybe assumed to be an ideal gas.898338 ICPS 0.271.898338 Sig = R . See Eq.15363.706103 6.915106 0. : Properties of Pure Species untuk propane : ωpropane = 0.0= 0.623 103J/mol HR H0 H1 HR = 2.908 T T T 363. Transformasi / perubahan bentuk menjadi gas ideal pada keadaan awal.57 Propane gas at 22 bar and 423 K is throttled in a steady state flow process to 1 bar.706103)/K C=(6.85 103J/mol S0 0.049J/molK 6.63722.48 bar Proses throttling terjadi ketika entalpi konstan.152 Tc = 369.637 B= (22.275J/molK Step (2): For the heat capacity of propylene.8 K Pc = 42.774J/molK S SR Sig S =28. .565R S0 =4.915106)/ K2 Solve energy balance for final T.7). In its final state.H0 0.1.697J/molK S1 0.534RTc H1 = 1. dapat terbagi menjadi dua proses : 1.863RTc H0 2. 144 Y1= 1. 1.6 -0.4 -0. 0 1 R ( HR) HR (H ) = +ω RTc RTc RTc Tentukan nilai HR dengan mencari nilai (HR)0 dan (HR)1.6 Y= 1.367 0. Nilai (HR)0 diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E. Nilai ω dapat dilihat melalui Appendiks B Tabel B.518 0.48 ¯¿ ¯ 22 ¿ Pr = = 0.15 X = 0.1.152.10 1. Tr/Pr 1.4 X2 = 0. Ubah T dan P dari keadaan gas ideal ke kondisi akhir.523 .518 ¿ ¿ Langkah pertama : Menggunakan rumusan korelasi umum virial.2. dengan mengevaluasi perubahan properties dengan menggunakan persamaan gas ideal. Nilai ω untuk propane adalah 0. Perubahan properties untuk dua tahap dijumlahkan ke perubahan properties untuk proses.144 42.334 -0. Dengan keadaan awal : Tr = T Tc Tr = 423 K 369.144 1.1 Y2= 1.8 K Pr = P Pc = 1.5.581 ? -0.518 X1 = 0.15 0. 1 + M 2.6 -0.4 -0.518 0.144 Y1= 1. 8.6−0.518 0.1 0.1 + M 1.2648064 (HR)1 = -0.314 J mol-1 K -1 .2 X 2−X 1 X 2−X 1 Y 2−Y 1 X 2−X 1 X 2 −X 1 Y 2−Y 1 M= [( 0.4 ) ( ) ) ] ( [( ) ( ] ) ) ] [( ) ( M =−0.6.251 0.518 0.15−1.2 2 + M 2.4 1.345 Jmol-1 2.15 M= [( X 2− X X− X 1 Y −Y X 2− X X−X 1 Y −Y 1 M 1.2648064 .6−0. Nilai (HR)1 diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E.15−1.518−0.6−0.144 0.6 Y= 1.518 0 (−0.581 ) + (−0.381 ? -0.4 0.8 K = -1387.4 ) ( ) ) ( ] [( ) ) ( ] ) [( M =−0.144 1.4 X2 = 0. .1 0.1 + M 1.199 )+ 0.296 X1 = 0.518 0.381 ) + (−0.M= [( X 2− X X− X 1 Y −Y X 2− X X−X 1 Y −Y 1 M 1.4 1.1 Y2= 1.6−0.1518315 Jmol-1 ] ) ( .518−0.15−1.6−0. 369.314 J mol-1 K -1 . 8.6−0.251 ) + (−0.2 X 2−X 1 X 2−X 1 Y 2−Y 1 X 2−X 1 X 2 −X 1 Y 2−Y 1 M= [( 0.45124 (HR)0 = -0.2 2 + M 2. 369.15 X = 0.1 + M 2.15−1.4 0.45124 .334 ) + 0.144 0.2648064 ( H R) 1 RTc =−0.8 K = -814.199 -0.4 1. Tr/Pr 1.518 0 (−0.367 )+ (−0.4 1.6−0.6−0.45124 ( H R) 0 RTc =−0.6−0.10 1.6−0. 4 1.518−0.8 8.229 0.6 Y= 1.152 8.2757284 4.371 ? -0.345 −814.6−0.6−0.8 HR = -1511.4 -0.144 0.314 .6−0. Pr/Tr 1.15 X = 0.371 ) +( (−0.201 )+ ( ([ 0. .183 -0.6−0.10 1.144 1.518 X1 = 0.144 0. 369.15 X = 0. Nilai (SR)0 / R diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E.2 X 2−X 1 X 2−X 1 Y 2−Y 1 X 2−X 1 X 2 −X 1 Y 2−Y 1 ) ( ) ) ( ) M =−0.10.23 ) + ( (−0.4 X2 = 0.6−0.369.15 M= [( ] [( ] M= 0.10 1.2757284 (SR)0 / R = -0.15−1.R 0 R 1 R (H ) (H ) H = +ω RTc RTc RTc HR −1387. (−0.1 Y2= 1.8 8.15−1.144 1. Nilai (SR)1 / R diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E.23 0.4 -0.518 0.144 Y1= 1.319 X1 = 0.35 ? -0.4 0.518 ) ) ) ] 1.518 0.1518315 = +0.4 0.096 J mol -1 3.9.1 [ 0.314 . .1 + M 2.1 + M 1.6 -0.518 0.518 0.314 . Pr/Tr 1.2 2 + M 2.6 -0.201 -0.275 . 369.4 X2 = 0.6 Y= 1.4 0 X 2− X X− X 1 Y −Y X 2− X X−X 1 Y −Y 1 M 1. 6−0.4 1.6998 J/mol K ΔSig – ΔS = -ΔSR 1 ) .518−0.518 ) ) ) ] 1.2757284+0.15−1.10-6 Cpig / R = A + BT + CT2 + DT-2 Cpig / R = 1.4 0.213 + 28.213 + 28.824.785.152(−0.028785 T −0.6−0.2448 (SR)1 / R = -0.2 2 + M 2.15 M= [( ] [( M= 0.785.518 0.2448) 8.1 Y2= 1.2 X 2−X 1 X 2−X 1 Y 2−Y 1 X 2−X 1 X 2 −X 1 Y 2−Y 1 ΔS = R/T P ∫ 1.Y1= 1. (−0.1 + M 1.000008824 T 2 dT −R ln P2 T1 ΔSig = 0 – 8. ln (1/22) = 25.6−0.213+0.314 J/mol K .183 ) +( ([ 0.824.144 0.1 [ 0.824.229 ) + ( (−0.213 B=28.601766 J mol-1 K-1 Untuk kapasitas panas propane dapat dilihat pada Appendix C Tabel C1 A= 1.10-6T2 ) R T2 ΔSig = ∫ Cpig T1 T2 ΔSig = ∫ Cpig T1 P2 dT −R ln T P1 P dT −R ln 2 T P1 T2 ig ] X 2− X X− X 1 Y −Y X 2− X X−X 1 Y −Y 1 M 1.4 0.785.10-6T2 Cpig = ( 1.10-3 C=-8.35 ) +( (−0.15−1.6−0.2448 R 0 R 1 (S ) SR ( S ) = +ω R R R SR =−0.1 + M 2.10-3T – 8.6−0.314 S R = -2.10-3T – 8.4 0 ) ( ) ) ( M =−0. 639 J mol J mol . Determine the tenperatur of the expanded gas and the work produced if the properties of ethylene are calculated by : a. HRB(1. Equation for an ideal gas.6 bar. Determine the temperature of the expand gas and the work produced if the properties of ethane are calculated by : a.00906.58. 0.15)K T = 373. SOLUTION : .152) = -0.b.260821 HRB = -0.6998 J/mol K + -2. ¿ SRB−ln ( ∆ S=−20.800 KPa expands isentropically in a turbine to 120 KPa.152 T = (100 + 273. Use virial correlation at final conditions.8 K Pc = 42.61 a stream of ethylene gas at 250 0C and 3. equations for an ideal gas b.009 P0 = 1 bar Pr= P Pc P = 10 bar Pr = 0. Estimate ∆H and ∆S SOLUTION : For propane: Tc = 369.23540.0.152) = -0. Appropriate generalized correlation.235 Assume ideal gas at initial conditions.48 bar = 0. Propane gas at 100oC is compresed isothermally from an initial pressure of 1 bar to a final pressure of 10 bar.601766 J/mol K = 23.179862 SRB = -0.179862 ∆ H=−801.9 ∆ H = R×T ×HRB c P ) P0 ∆ S=R .15 K Tr= T Tc Tr = 1.00906.260821 SRB(1. 0.62 A stream of ethane gas at 220 oC and 30 bar expands isentropically in a turbine to 2.098034 J/mol K 6. 0. appropriate generalized correlations SOLUTION : 6.ΔS = ΔSig + SR = 25.23540. K 6. 131+0.019225 T −5. 10-6 Cpig = (1.05337 Menggunakan korelasi koefisien virial : T = 362.6153 Pr0 = P0/Pc Pr0 = 30 bar / 48.314 .15 K P0 = 30 bar P = 2.019225T .131+0.20404 Pr0 = P0/Pc Pr0 = 2.59 K a.15 ΔHig = -8735 J/mol Ws = ΔHig = -8735 J/mol b.3K Pc = 48.T0 = 220OC = 493.61576 Untuk kondisi akhir : Tr = T/Tc Tr = 367.225 . Untuk kapasitas panas dari etana dapat dilihat pada Appendix C Tabel C1 A = 1.000005561T 2) dT 493. T ∫ 2.5.019225 T −0.15 K / 305.6 bar / 48.59 ig ΔH = R ∫ (1.59 K / 305. 10-3 C = -5. Dari Tabel Appendix B Tabel B1 maka diperoleh data untuk ethane adalah : ω =0.131 + 0.561 .561 .15 -ln 8.6 30 T = 367.1 Tc = 305.3 K = 1. 10-6T2 ) R T2 ΔHig = ∫ C Pig dT T1 367.131 B = 19.561.72 bar Untuk kondisi awal : Tr0 = T0/Tc Tr0 = 493.72 bar = 0.10−6 T 2 dT 0 = 493.3 K = 1.72 bar = 0.6 bar J Δ S = 0 mol K T ig C p dT ΔS / R ¿∫ -ln RT T 0 T P P0 1.73 K ΔHig = -9034 J/mol Ws = -8476 J/mol . 6.63 Estimate the final temperature the work required whan 1 mol of n-butane is comprassed isentropically in a steady-flow process from 1 bar and 50 0C to 7.8 bar. SOLUTION : A = 1.935 −3 B= 36.915 . 10 K −11.402 . 10−6 C= K2 ∆ S=0 J mol . K 323.15 K 425.1 K Tro = ¿ Tc = Pr0 = P0 Pc 37.96 ¯¿ ¯ 1 ¿¿ = = 0.02634 ¿ Pr = P Pc 37.96 ¯¿ ¯¿ = 7.8 ¿ = 0.205 ¿ HRB0 = -0,05679 τ = 1.18 T= τ . T0 T = 1.18 . 323.15 K = 381.43 K Tr = T Tc = = 0.89726 381.43 K 425.1 K = 0.76017 ∆ Hig = R. 3231,5 K. 381,43 K. 1,935. 36,915 . 10−3 , - 11,402. 10−6 3 J ∆ Hig = 6.551 x 10 mol Ws = ∆ Hig + (R. Tc. Tr . ω . Pr) – (Tr0. Pr0) 3 = 6.551 x = 5680 10 J mol + (425,1 K . 0,89726 . 0.200 . 0,205) – (0.76017 . 0.02634) J mol 6.65 liquid water at 325 K and 8000 KPa flows into boiler at the rate of 10 kg/s and is vaporized, producing saturated vapor at 8000 KPa. What is the maximum fraction of the heat added to the water in the boiler that can be converted into work in a process whose product is water at initial conditions, if T ð = 300 K ? what happen to the rest of the heat ? what is the rete of entropy change in the surroundings as a result of the work producing process? In the system ? total ? SOLUTION : T = 325 K P1 = 8000 K Pada table : Psat = 12.87 Kpa Dari table F1 : β = 460 x 10-6 /k Hliq = 217 kj/kg Sliq = 0.7274 kj/kg K 1.o13 cm3/ gr H1 = H liq + V liq x ( 1 –β. T) ( P1 – Psat ) H1 = 217 kj/kg + 1.013 ( 1 – 460 x 10-6 x 325 ) ( 8000 – 12.87 ) kj/ kg 1000 H1 = (217 + 6.8813) kj/ kg H1 = 223.88 kj/kg S1 = S liq –β. V liq ( P1- Psat ) S1 = 0.7274 kj/kg – (460 x 10-6 x 1.013 ) ( 8000 – 12.87 ) kj/ kg Vliq = 1000 S1 = (0.7274 – 3.722 . 10-3 ) Kj/Kg K S1 = 0.7236 Kj/ Kg K untuk uap saturated di 8000 KPa dari table F2 : H2 = 2759.9 kj/kg S2= 5.7471 Kj/Kg Tð = 300 K PENAMBAHAN PANAS Q = H2 –H1 Q = 2759.9 – 223.81 ) Kj/ kg Q = 2536.09 Kj/Kg Kerja maximum dari sistem W ideal = ( H1 – H2 ) - Tð ( S1 – S2 ) W ideal = ( 223.88 – 2759.9 ) – 300 ( 0.7236 – 5.7471 ) W ideal = ( 1507.05 – 2536.02 ) kj/ kg W ideal = - 1028.97 Kj/ kg Kerja sebagai fraksi sebagai penambahan panas F frac = I W ideal I / Q F frac = 1028.97 / 2536.09 F frac = 0.4057 Panas yang tidak converted untuk kerja akhir di surrounding S surr = Q + W ideal . 10 kg / sec Tð S surr = ( 2536.09 + (- 1028.97 ) / 300 ) 10 S surr = 50.237 kw/K S systm = ( S1 – S2 ) 10 Kg/ sec S systm = ( 0.7236 – 5.7471 ) x 10 S systm = 50.235 Kw/ K Total dari generasi entropi = 0, karena kerja ideal prosesnya reversible. 724 KJ/Kg.K .12.2 diperolah: H2= 2759 KJ/Kg S2= 5.87 KPa) = 0.T). dari table F.28 dH= Cp.881 KJ/Kg S1= Sliq + β. (8000 KPa . (P1 – Psat) = 0. Vliq. pakai rumus 6.7471 KJ/Kg.7274 KJ/Kg. V.dT+(1-βT).013 cm3/gr Psat = 12.K >>Untuk saturated vapour pada 8000 KPa. harus di cari dl data dari soal no.87 KPa) = 223.6. What is the total rate of entropy generation as a result of the heating process? What is Wlost? SOLUTION : *untuk menghitung SG dan Wlost .460.87 KPa P1 = 8000 KPa T = 325 K >>untuk compressed liquid pada 325 K dan 8000 KPa.7274 KJ/Kg. diperoleh: Hliq= 217.0 KJ/Kg + 1.12.10-6 K-1 x 325 K) .013 cm3/gr + (1.0 KJ/Kg Sliq = 0. 65 >>Dari table F.K + 460.10-6 K-1 x 1.10-6 K-1 H1= Hliq + Vliq + (1-β. dP β = 460.013 cm3/gr .66 Suppose the heat added to the water in the boiler in the preceding problem comes from a furnace at a temperature of 600°C.K Vliq = 1. (8000 KPa . (P1-Psat) = 217.1 untuk saturated liquid pada 325 K. 10 KW K Untuk menghitung SG pada soal no.234 SG system = (S1 – S2) .5.surrounding = Q+Wideal T 2536 = x 10 KJ KJ −1029 Kg Kg T x 10 Kg sec KW K = 50. dengan persamaan 5.T.K) = -1029 KJ/Kg Kerja pada saat panas di tambahkan: ¿ Wideal∨ ¿ Q Frac = ¿ ¿−1029∨ ¿ 2536 = ¿ = 0.50.K . (S2 – S1) = (2759 KJ/Kg – 223.881 KJ/ Kg) .4058 Panas yg tidk terkonversi menjadi kerja di akhir pada surrounding: SG. 10 Kg sec = (0.Panas yang ditambahkan ke boiler: Q = H2 – H1 = 2759 KJ/Kg – 223.724 .27 diperoleh: Widel = (H2 – H1) .0. 66: Kg sec .7471 KJ/Kg. K .5.234 Kg sec KJ Kg .881 KJ/ Kg = 2536 KJ/Kg Kerja maksimal dari steam.724 KJ/Kg.7471) = . 2963 kJ/kg.5 kg s⁻1 of flake ice at 0°C from water at 20°C (T ) in a continuous process.15)K = 873.15K .15)kJ/kgK S2= 1.221kJ/kgK T293.15 K SG = Q T + SG.86 kJ/ kg S1= 0.1: H0= 0.1: H1= 83.67 An ice plant produces 0.4 kJ kg⁻1 and if the thermodynamic efficiency of the process is 32%. what is the power requirement of the plant? SOLUTION : Untuk saturated liquid air pd 20˚ C.surrounding KJ sec 873.333.9 kW 6.Q = 2536 KJ Kg x 10 Kg sec T (600 + 273.34) untuk mencari harga Wlost T300K Wlost Tx SG 6356.234 KW K KW K *dari persamaan (5.(333.19 +50.4kJ/kg H2= 333.K Untuk saturated liquid air pada 0˚C. table F.0000 kJ/kgK Untuk es pada 0˚C H2= H0 .4/273.15 K −25360 = = 21. If the latent heat of fusion of water is 333.44 kJ/kg S2= S0. table F.04 kJ/kg S0= 0. 83.32 42.77 kW 6.26) dan (5.0.15K .S1)] kg/sec x [333.15 K (100°C) is the only source of energy.686 kW 0. Assuming that there is plenty of cooling water available at 273.44 kJ/kg . (1.15 K (O°C).32 Dengan persamaan (5.68 An inventor has developed a complicated process for making heat continuously available at an elevated temperature.m = 0.28): Wideal = m x [ H2 .86 kJ/ kg .686 kW W = Wideal ηt 13. Saturated steam at 373.2963 kJ/kg.5kg/sec t 0.K) 13.H1 -TS2.293. what is the maximum temperature level at which heat in the amount of 2000 kJ can be made available for each kilogram of steam flowing through the process? SOLUTION : .221kJ/kgK . reservoir = S2 – S1 - ) . K ) .reservoir kJ 0.reservoir = H2-H1-Q’ kJ = ( 0.3554 kg .3554 kg .0 kJ kg 2000. K kJ kg .0 kg 676.reservoir ΔH apparatus.0 kg = .3554 kg .ΔS apparatus.0 kJ kg ) kJ kg Q' T' = (0. K 2000. K kJ = (273.diketahui W=0 Wideal = ΔH apparatus.reservoir – Tσ.reservoir – Tσ.0 = (.( T' kJ =– (7.676.0 kJ kg ) – (-2000.0 ΔS apparatus.15 K)[(7.3554 kg . K ) – (7.15 K SOLUTION: Wideal = ΔH apparatus.0 kJ kJ kg kg .0 kJ kg H2= 0.0 kJ kg Q’= -2000.0 kJ −2000.3554 kg .0 Tσ = 273.0 kJ kJ kg ) – Tσ.0 )-( T' kJ kg T' )] kJ kg )] .ΔS apparatus. K S2= 0.0 )+ ( kJ kg T' ) ) Jadi .( 2676.[– (7.kJ kg H1= 2676.0 kJ S1= 7. K )+( 2000.676. rankine Svapour= 1.discharge equal amounts of steam into the same steam main.127 T’ 6. Steam from the first boiler is superheated at 420( 0F) and steam from the second is wet with a quality of 96%.5 ( 1.on the basis of 1 pound mass of steam after mixing.both operating at 200(psia).rankine Hliquid= 355.30) yields for the exit stream : Wet steam H= 0.165 x 103 BTU/lbm S2=1.3 BTU/lbm Sliquid= 0.6 BTU /lbm S1= 1.( 676.Hvapour-Hliquid S2= Sliquid + x.5438 H2=1.5737 BTU/lbm.Svapour-Sliquid H2 = 355.505 BTU/lbm.3)-355.5 ( 1222.5438 + 0.96 H2= Hliquid + x.rankine Neglecting kinetic and potential energy changes.5454 BTU/lbm.69 546300 K T' = 490.165 x103) H= 1193.0) =(2009.6) + 0.78 K Two boilers.4 at 200 (psia): At 4200 F H1 = 1222.Assuming adiabatic mixing and negligible changes in potential and kinetic energies.51 BTU/lbm Hvapour= 1198.51 S2= 0.51 + 0.96 (1.96 ( 1198.rankine x= 0.5438 BTU/lbm.What is the equilibirium condition after mixing and what is S G for each (lbm) of discharge steam ? SOLUTION : From Table F.6 BTU/lbm . (2.Eq.127) - 546300 K T' = 1333.5 H1 +0.5454)-0.5 H2 H= 0. x=_ H .505) = 2. determine the mass of steam vented. liq.54-0.22 ) on the basis of 1 pound mass of exit steam. Assuming no heat transfer to the contents of the tank. and vapor): V tank = 80 ft3 Mliq = 4180 lbm Vliq = 0. Since a bit more vapor space in the tank is wanted.5 (1.rankine 6.3496 ft3/lbm Uliq = 406.51 S= Sliquid + x.994 1198.Svapour-Sliquid S = 0.5438 + 0. SG = S-0.5454)-0.01909 ft3/lbm Vvap = 1. a valve at the top of the tank falls to 420 ( 0F). the small remaining volume being occupied by saturated-vapor steam.5 S2 = 1.3-355.70 A rigid tank 0f 80 (ft3) capacity contains 4180 ( lbm) of saturated liquid water at 430 (0F).5438 S = 1.51___ = 0.Hliquid____ Hvapour-Hliquid x = _1193.70 Btu/lbm Uvap = 1118.01909 ft3/lbm .5 S1-0.rankine By Eq. (5.0 Btu/lbm VOLliq = mliqx Vliq = 4180 lbm x 0.5(1.994 (1.895 x10-4 BTU/lbm. SOLUTION : From Table F. This amount of liquid almost completely fills the tank.6-355.5737)-0.3 at 430 degF (sat.54 BTU/lbm. 151 lbm 4180 lbm x 406.1(Btu/lb) as the temperature drops from 430 to 420 degF.3 we see that the enthalpy of saturated vapor changes from 1203.796ft3 VOLvap Vtank VOLliq = 80 ft3 .) Diintegralkan: m m2.U1+ ∫ Hdm 0 =0 From Table F. Uliq+ mvap. Then.70 Btu lbm By Eq.5 Btu/lbm m2(mass) m1 mass . This change is so small that use of an average value for H of 1203.151 lbm U1 mliq . m2U2 m1U1 Havem = 0 Have = 1203. dmtUtHdm = 0 (Subscript t denotes the contents of the tank.U2 – m1.796ft3 = 0.79.= 79.204 ft 3 1.3496 ft 3 /lbm = 0. H and m refer to the exit stream.29) multiplied through by dt. we can write.204ft3 Mvap = VOLvap Vvap = 0.9 to 1203.5(Btu/lb) is fully justified.151 lbm x 1118. Uvap mliq+ mvap Btu Btu +0. (2.0 lbm lbm 4180 lbm+0. 2 at this entropy and 3500 kPa : V2 = 78.71 A tank of 50 m3 capacity contains steam at 4.500 kPa and 400 0C.500 kPa.K By interpolation in Table F.01894 ft3/lbm Vvap = 1.Property values below Vliq = 0.721 cm3/gr S1= 6. liq.460C m1 = _Vtank__ V1 = _ 50 x 106 cm3___ 64.7093 J/gr.4997 ft3/lbm are for sat.81 Btu/lbm Uvap = 1117. from Table F.4 Btu/lbm V2 = Vtank m2(mass) X mass = V 2 ( mass )−Vliq Vvap−Vliq U2(mass) Uliq x(mass)Uvap Uliq U 1−U 2(mass) Mass = Have−U 2(mass) Mass = 55.2 at 4500 kPa and 4000C : V1 = 64.726 cm3/gr t2 = 362.estimate the final temperature of the steam in the tank and the mass of steam vented.721 cm3/gr m2= _Vtank___ V2 m2= _ 50 x106cm3__ 78. and vap.K Vtank= 50 m3 S2=S1 = 6.Steam is vented from the tank through a relief valve to the atmosphere until the pressure in the tank falls to 3. SOLUTION : The steam remaining in the tank is assumed to have expanded isentropically. at 420 degF Uliq = 395.7093 J/gr.36 lbm 6.If the venting process is adiabatic.726 cm3/gr Data . H liq – H-y.∆mt y.137 x 106 gr Δm = m1 – m2 = 0. A tank of 4 m3 capacity contains 1. The only two enthalpy changes within the tank result from : 1. ∆mt Here. The 1500 kg of liquid initially in the tank is unchanged during the process. vapor to sat.Vt = V liq. ∆mt – y. addition of 1000 kg of liquid water.H liq – H vap = -y. This contributes an enthalpy change of : Hliq. Ht = H liq. the symbols with subscript to refer to the contents of the tank. which fills the rest of the tank. This is contributes an enthalpy change of Q=∆mt. liq.772 x 106 gr = 0. A quantity of 1.1 are at 50 degC H = 209.772 x 106 gr – 0.79 cm3 gm y = V liq. H lv ∆mt.000 kg of water at 50oC is pumped into the tank. Ht – H. ∆mt – y. condensation of y kg of sat. ∆V lv = 0 Q= H liq. We illustrate here development of a simple expression for the first term on the right. the vapor initially in the tank that does not condense is unchanged. ∆mt – y. How much heat must be added during this process if the temperature in the tank is not to change? SOLUTION : Q = ∆mt.Ht – Ht. Similarly.∆H lv ∆mt. whereas H refers to the entering stream.641 kg ∆mt ∆V lv Q = ∆mt.8 kJ V liq = 1251 cm3 kg gm ∆H lv = 1714 kJ kg ∆V lv = 48.3 kJ kg at 250 degC H liq = 1085.∆mt 2.= 0.137x106 gr = 0.500 kg of liquid water at 250 oC in equilibrium with its vapor.∆H lv . y = 25.137 x106 gr = 137 kg 6.∆H lv – H ∆mt ∆mt = 1000kg Required data from table F.72. 003 cm3 / gm ΔV lv.(H – U2) = m1. Δ Ulv.115 cm3 / gm U liq..239 m3 / kg Δ Ulv.(H – U1 ) Also U2 = Uliq.74.2 V2 = V tank m2 Eliminating x2 from these equation gives V tank .115) cm / gm Δ Ulv.2 = (240.2 Which is later solved for m2 V tank = 50 m3 m1 = 16000 kg V1 = V tank V1 = 3.1 = 43400 cm3/ gm Uliq. What mass of steam is added ? Penyelesaian : m2.1 + x2.2 .3 .855 ´ 103 kJ / kg Data from Table F2 : 1500 kPa H = 2789.2 = 1. Saturated steam at 1500 Kpa is atmitted to the tank until the pressure reaches 800 Kpa. Δ U lv.2 = 0.1 + x1.720.1 U1 = U liq.1 X1 = 4.125 x 10 -3 m3 / kg m1 Data from table F1 : 25 0C V liq.889 x 10-5 U1 = 104.( H – U1) Δ lv. (H – U liq.086 x 104 kg . Δ U lv.( U2 –H ) – m1.2 m2 = m2 = 2.2 m2 m2.043) kJ / kg Δ Vlv. Δ V lv.26 . A well insulated tank of 50 m 3 volume initialy contains 16000 kg of water distributed between liquid and vapor phases at 25 o C.2 m1.2 + x2.2 ) = m1. ( H – U1 ) + V tank ( ) Δ U lv.(U1 – H) = Q = 0 when m2.8 kJ / kg Δ Ulv.V liq.Q = 832534 kJ 6.913 kJ / kg Data from Table F2 : 800 kPa V liq.1.1 = 2305 kJ / kg X1 = v1 – v liq.2 V2 = Vliq.1 = 104.2 = (2575.2 = 1.1 Δ V lv.1 = 1.9 kJ / kg Δ U lv.043 kJ / kg 3 Δ Vlv.1 = 720. 9−2884.Δ U lv.5 Y −350 = 2964. ( ) Δ U lv.75.2 .9 400−350 0.2 Y −350 = 2966.9−2889. prepare graphs showing the mass of steam in the tank and its temperature as a function of pressure in the tank. Penyelesaian : Interpolasi dari table F.2 pada buku hal.57 Pada P = 400 kPa x−x 1 y= y 1 + x2 −x1 ( y 2− y 1 ) 2943. Steam flows into the tank until the pressure in the tank reaches 400 kPa.75m3 volume is attached to a line containing steam at 400 kPa and 2400C.1−2889. 696 U = 2943.9 Y −350 = 2969.6818 = Y −350 50 Y = 384.2 H – U liq.9 kJ/mol Mencari temperature dengan cara interpolasi Pada P = 1 kPa x−x 1 y= y 1 + (y −y ) x2 −x1 2 1 2943.09 Pada P = 200 kPa x−x 1 y= y 1 + (y −y ) x2 −x1 2 1 2943.2 m steam = m2 – m1 m steam = 4.5 400−350 .7114 = Y −350 50 Y = 385.9−2887.2 400−350 0.2 + V liq.9−2887.6−2884. An insulated evacuated tank of 1.855 x 103 kg 6. Assuming no heat flow from the steam to the tank. 8 0.23 P ( kPa) 1 T(0C) 384.61 Pada P = 400 kPa x−x 1 y= y 1 + (y −y ) x2 −x1 2 1 2943.2 1549.8 116.61 400 387. 7415 = Y −713.5 772.8 757.1−2889.9−2887.5 Y −713.09 V ( cm3/gm) 303316 200 385.85 58.6−2884.85 0.9 310660−287580 0.9−2889.65 Y = 757.9 Y −287580 = 2969.85 = 2964.0.9−2884.9−2887.4 Y = 1515.50−713.2 Y −1432.23 .7114 = Y −1432. 8 Mencari volume dengan cara interpolasi Pada P = 1 kPa x−x 1 y= y 1 + (y −y ) x2 −x1 2 1 2943.2−1432. 7415 = Y −350 50 Y = 387.57 1515.8 = 2966.6818 = Y −287580 23080 Y = 303316 Pada P = 200 kPa x−x 1 y= y 1 + (y −y ) x2 −x1 2 1 2943. 23.216 x 10-3 kg . Of the total mass.5 1 0.2 at 3000 kPa (page 712) : x1 = 10% vapor 3 3 cm m Vliquid = 1.5 2 1.75 / 303316 . 10-3 = 5.216 gm = 1. A m3 tank contains a mixture of saturated-vapor steam and saturated-liquid water at 3. If during the process the temperature of contents of the tank is kept constant.61. Saturated-liquid water is bled from the tank through a valve until the total mass in the tank is 40% of the initial total mass.23. Data from table 7. 10-3 1.V TANK = 1.155 400 757.75 m3 MASSA = VTANK / V INTERPOLASI Maka : P ( kPa) 1 V ( m3) 303316 . 10-3 Massa ( kg ) 1.75 / 1515. 10-3 200 1515. .61.77 . 10-3 = 2.311 Sumbu Y : Massa ( kg ) 2. 10% is vapor.000 kPa. how much heat is transferred? Penyelesaian : Vtank = 2 m3 .5 0 1 200 400 Sumbu X : Tekanan ( kPa) 76.75 / 757. 10-3 = 1. 10-3 1. dan H adalah entalpi dari aliran yang mengalir keluar dari tangki.757x10 -3 m kg V tank m1= V 1 2 m3 3 = 7.626x10-3 kg Hliquid = 1008.41 gm ) ) cm3 cm3 +¿ =1.757 x 10−3 m kg =257.832kg Q= ∆ ( mt . H t ) +H.757 gm 3 =7.216 gm 6.216 gm +0.3 3 cm m Vvapor=66.626 gm =66.4 kJ kg kJ Hvapor=2802. ∆m tank V1=Vliquid+x1.626 −1. H t ) +H.1. Perubahan yang mempengaruhi entalpi dari isi tangki adalah: .541 gm 3 cm =7.216 gm + 0.3 kg Q= ∆ ( mt .216 gm gm 3 3 ( cm cm =1.(Vvapor-Vliquid) cm3 =1.1 x 65. ( cm3 cm 3 66. ∆m tank dimana t menunjukkan kondisi di dalam tangki. 832 kg).1. H t )= y ( H vapor −H liquid )−¿ 0.9 kJ kg ) )( ) + H . 1.6 m1 .6m1liquid dari tangki : H liquid 0.216 x 10−3 kg kg 0.V liquid . H liquid Volume tangki konstan.Penguapan(evaporation) y ( H vapor −H liquid ) : 2.V ( V vapor−V liquid ) liquid Maka: Q = ∆m 0.6m1.V . . 1793.6 m1 = tank Persamaannya menjadi : 0.(H ( V vapor−V liquid ) liquid vapor −H liquid ) −¿ 0.4 3 3 kg kg m m 66. maka: ∆ ( mt .1881 m 3 −3 65. ∆ ( mt .6 m .216 x 10−3 0.(100%–40%)untuk liquid 60%=0. dengan demikian.626 x 10−3 −1.3 −1008.6 m1 . H t )= y ( H vapor −H liquid )−¿ V =0 0.(257.6 m1 . H liquid 1 tank dengan H= H liquid ∆m dan 0. liquid Dimana: y= 0. ( H vapor −H liquid ) Q= V vapor−V liquid ( Q= ( Q= ( ) m3 kg kJ kJ .6m1.41 x 10 3 m kg ) ( .6 . 2802.6 keluaran : 0.6 m1 . at what rates are the steam and water fed to the mixer ? Penyelesaian : Data from Table F.100 kPa dan T1 = 500C P2 = 3.Q= 337.432 kJ ( 65.900kPa flow from the desuperheater at the rate of 15 kg/s assuming adiabatic operation. (2.000kPa and 3750C in an amount such that a single stream of saturated-vapor steam at 2. In a desuperheater.9 kJ / kg Data from Table F2 for sat.516 kg / sec 6.900kPa dan m3 = 15 kg/s a. following at the rate of 5 kg -1 is formed by mixing water at 24oC with saturated steam at 400 Kpa.484 kg / sec mdot1 = 4.30). liquid water at 3.900kPa didapat : H3 = 2802.K M3 = 15 kg/s .100 kPa and 50 0C is sprayed into a stream of superheated steam at 3.6 kJ / kg By Eq.77. Stream of water at 85oC.2 kJ/kg S3 = 6. Liq H1 = 100.1 for sat.2 saturated vapor untuk tekanan 2.H1mdot1 .H2mdot2 = 0 mdot1 = mdot3 . neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3mdot3 .6 kJ / kg H3 = 355.000kPa dan T2 = 3750C P3 = 2. Vapor : 400 kPa H2 = 2737.( H1 – H3 ) Mdot2 = H1 – H2 Mdot1 = mdot3 – mdot2 mdot2 = 0. Dari tabel F.730 kJ 6.41 x 10 ) −3 Q = 5158. what is the mass flowrate of the water? What is SG for the process ? What is the irreversible feature of the process? Penyelesaian : P1 = 3.78.1969 kJ/kg. Assuming adiabatic operation.mdot2 mdot3 = 5 kg / sec Whence mdot3. 1.K x 13. M2 – H1.2 saturated liquid dengan suhu 500C didapat : Vliq = 1.K – 4.941 x 10-4/K)T (3. M1 = 0 Juga M2 = M3 .1969 kJ/kg.000kPa dan suhu 3750C didapat : H2 = 3175.K x 1.H2.2 superheated vapor untuk tekanan 3.010 )cm3/gr ΔT = 10 K P = 3.941 x 10-4/K x 1.11kg/s Sg = 1.2 sat.89 kg/s M2 = 13.15 K Perubahan volum ekspansi antara suhu 450C dan 550C (pada tabel F. M3 – S1.89 kg/s Maka : M2 = 15 kg /s .015 – 1. M2 Sg = 6.6 kJ /kg) 211.012 cm3/gr (1-4. Pencampuran antara 2 aliran yang berbeda temperatur adalah irreversible 6.34kPa) S1 = 0. Vliq ( P-Psat ) S1 = 0.702 kJ/kg. M3 .8385 kJ/kg.2 kJ /kg−3175.79 Superheated steam at 700kPa dan 280oC flowing at the rate of 50 kg/s is mixed with liquid water at 40 oC to produce steam at 700 kPa and 200 oC.liquid ) ΔV = (1.K b.Dari tabel F.100kPa -12. M1 – S2. Assuming .3 kJ/kg Sliq = 0.8385 kJ/kg.M1 M 3 ( H 3−H 2) Dimana : M1 = H 1−H 2 M1 = 15 kg /s (2802.702 kJ/kg.926 kJ /kg−3175.012 cm3/gr (3.973 kJ/s.11 kg/s Untuk kondisi adibatis persamaa yang digunakan adalah : Sg = S3.100kPa -12.012 cm3/gr Hliq = 209.7035 kJ/kg.K Psat = 12.6 kJ /kg = 1.K Dari tabel F. Dengan persamaan pada bab 2 energi kinetik dan potensial diabaikan sehingga penjumlahan antara panas dan kerja sama dengan nol : H3.89 kg/s – 6.100kPa 1 ΔT ΔV = 5 x 10-3 cm3/gr β = V liq Δ V β = 1 10 K 1.K c.34kPa T = 323.K x 15 kg / s – 0.941 x 10-4/K Persamaan dengan temperatur konstan : H1 = Hliq + Vliq (1-β)T ( P-Psat ) H1 = 209.012 cm 3/gr 5 x 10−3 cm3 /gr = 4.34kPa ) H1 = 211.6 kJ/kg S2 = 6.7035 kJ/kg.3 kJ/kg + 1.926 kJ/kg S1 = Sliq – β. K m1 = 50 Kg/s Pada tabel F.K o P1 = 700 Kpa & T1 = 280 C.8859 KJ/KgK.58.m1 = 0 sehingga persamaan dapat menjadi m3 = m2 + m1 m1 ( H 1−H 3 ) m 2= ( H 3−H 2 ) 50 m 2= Kg KJ KJ .2 super heated vapor pada P3 = 700 kPa & T3 = 200oC.3.7.m2 SG = 6.2250 Kj/Kg. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2. bahwa : H3 = 2844.K Pada persamaan 2.5 times the mass flowrate. If the process were accomplished reversibly and adibatically.5721KJ/KgK.241Kg/s .m2 – H1.241Kg/s SG = 3.5 KJ/Kg Sliq = 0.7 −2844.8859 KJ/Kg.7 KJ/Kg S1 = 7. Bahwa : H1 = 3017. 3017.adiabatic operation.0.Liquid water T = 40oC (saturated) .m1 = 50 kg/s Pada tabel F.241 s ( ( ) ) Untuk kondisi adiabatik: SG = S3.30 dinyatakan bahwa pada keadaan steady state pengurangan energy kinetik dan energi potensial serta semua usaha dan panas yang bekerja pada sistem adalah nol. .P1 = 700 kPa & T1 = 280oC . Sehingga : H2 = Hliq dan S2 = Sliq H3.K 80. 50 Kg/s .2 KJ/Kg S3 = 6.5 Kg Kg Kg m2 =3.1 saturated T = 40oC Hliq = 167. at what rate is water supplied to the mixer? What S G for process? What is the irreversible feature of the process? Penyelesaian : Superheated Steam .P3 = 700 kPa & T3 = 200oc .7 −167.2 s Kg Kg KJ KJ 3017.m3 – S1.m1 – S2. what would be temperature and pressure of the resulting air stream? Assume air to be an ideal gas for which CP = (7/2) R.m3 – H2.2250 Kj/KgK.5721 KJ/Kg.508 Kj/s. 859 K [( ( n1 .5 mol at P and T T −T 1) T −T 2) n1 . K . K mol .859 K J J 542.1 ) K T= ( 55288.314 . ln ( PP )))+( n .099 =0 )( ( J 542. K 400 K ¿ ¿ ( ) ( ) .099 .846T = ( 55288. C P ( + n2.5 mol .099T – 26189. C P .314 29.5 mol .859 K .099 = 2 mol . ln ( TT )−R .846 ) K T = 542. K Basis : 1mol air at 12 bar and 900 K + 2. K mol .( C . ln ( PP )))] 2 1 P 1 2 2 =0 12 ¯¿ P ¿ 2 ¯¿ P ¿ ( 1 mol .5mol air at 2 bar and 400 K = 3. K 900 K mol .Penyelesaian : T1 =900 K T2 = 400 K 7 CP = 2 R CP = P1 = 12 bar n1= 1 mol P2 = 2 bar n2 = 2. ( T −900 K ) + 2. ln − mol . 29. 29. ln ( TT )−R . K =0 (29. ln −8. 29.747T – 29099) K = 0 101.5 mol 7 J J 8.099 J ( T −400 K ) mol .099 J mol .1) K + (72.1 101. CP ( =0 1 mol . 29. ln ( ¿ ) + 2. 248 BTU/lbm.319 bar 6. Nitrogen Keluar pada 325 F T3 = 1209. if the Nitrogen is cooled to 325 (oF) and if heat is lost to the surroundings at rate of 60 BTU for each lbm of steam generated. Saturated liquid air pada 212 F masuk 2. Hot Nitroge gas at 750(oF) and atmospheric pressure flows into a waste heat boiler at the rate of 40 (lbm)(S-1). and transfers heat to water boiling at 1 atm. Rankine Ms = steam rate in lbm/sec Mn = nitrogen rate in lbm/sec Mn = 1. The water feed to the boiler is saturated liquid at 1 atm and 300( oF). what is the steam generation rate? If the surroundings are at 70 (oF). Penyelesaian : Mr N2 = Cp = 7/2 R = 0.4.3.P = 4.67 Rankine Dari table F.8158 btu/lbm Rankine . H2 = S2 = 1. Uap steam keluar pada 1 atm dan 300 F 3.81. what is SG for the process? Assume Nitrogen to be an ideal gas for which Cp =(7/2)R. Nitrogen masuk pada 750 F 4. H1 = S1 = Dari Tabel F.67 Rankine T4 = 784. tanpa memperhatikan energy potensial dan energy kinetic dan kerja sama dengan nol sehingga transfer panas adalah Ms = (asumsi) Q= Maka persamaan 2.3 menjadi Persamaan 5.22 menjadi seperti berikut : .Pada persamaan 2.3. liq. Hot nitrogen gas at 400oC and aunospheric pressure flows into a waste heat boiler at the rate of 20 kg. and transters heat to water boiling at 101.33 kPa dan 150 C (3) = nitrogen dalam : 400 C (4) = nitrogen keluar pada 170 C .If the nitrogen is cooled to 325oF and if heat is lost to the surroundings at a rate of 8kJ for each kilogram of steam generated what is the steam generation rate? If the surroundings are at 25oC.33kPa.33 kPa masuk (2) = keluar uap pada 101.33 kPa.The water feed to the boiler is saturated liquid at 101.Q= . Maka nilai akhir dapat kita tentukan dengan memasukkan semua variable 6. What is SG for the process? Assume nitrogen to be an ideal gas for which Cp=(7/2)R. and it leaven the boiler as superheated steam at 1 atm and 300 oF.82.s-1 . Water :101. Penyelesaian : Ms = steam rate in kg/sec Mn= nitrogen rate in kg/sec (1) = masuk. 039 2 molwt gmK Mn=20 Kg .961 Kg sec Eq (5. mengabaikan energi kinetik dan potensial dan pengaturan istilah work ke nol dan dengan tingkat perpindahan panas yang diberikan oleh.15 K sec Tabel F.2 Kg Kg Tabel F. (2.064 Kj Kj H 2=2776.6075 Kg Kg Dengan Persamaan.194 Kj sec K Q Tσ Kj Ms Kg . T 3=673.014 Cp= gm mol 7 R J Cp=1.15 K Q=−80 Cp .Cp ( T 4−T 3 )=−80 Ms=1.30).22) Here becomes: sdotG=Ms ( S 2−S 1 )+ Mn ( S 4−S 3 )− S 4−S 3=Cp .molwt=28. ln Kj Ms Kg ( TT 43 )− TσQ SdotG=Ms ( S 2−S 1 ) + Mn ¿ SdotG=4.2 H 1=419. Ms sec Kg Ms ( H 2−H 1 ) + Mn .15 K .2 S 1=1. ln ( T4 ) T3 Tσ=298. T 4=443.3069 Kj Kj S 2=7. Ms=1 Kg Kj Q=−80 . quantitatively. Berdasarkan gambar 3.53) trough (6.11. HR. the temperature for which B = 0 corresponds to a reduced temperature of about Tr = 2.55) that the residual properties G R.7 for many gases. Dengan Cp > Cv .) . isokhorik lebih curam Sebuah persamaan kurva isobarik dari perbedaan persamaan diatas: 2 ( ) ∂T 2 ∂S P = T Cp 2 P 1 Cp = [ 1− ( ∂TTS ) T ∂ Cp p Cp ∂ T ( ) Dwngan Cp = a + bT.6.30 ( ∂TTS ) P T Cp = dan ( ∂T∂S ) V = T Cv Kedua slope bernilai positif. which is sheeper an isobar or an isochore? Why? Note that Cp>Cv SOLUTION : Slope isobar dan isokhorik pada diagram TS ditunjukkan pada persammaan 6.11 bahwa B adalah ( . 6.83 Shows that isobars and iscochorics have positive slopesin the single phase regions of a TS diagrams. qualitatively. 3.17 dan 6. and SR are negative for most gases at modest pressures and normal temperatures.85 The temperature dependence of the second virial coefficient B is shown for nitrogen on Fig. ( ∂TTS ) P - T ∂Cp 2 Cp ∂ S ( ) P = T Cp 2 - T Cp 2 ∂T ( ∂Cp ) ( ∂T ∂T ) P ] P= b dan 1 - T Cp ( ∂Cp ∂T ) P= 1 - bT a+bT = a a+bT Karena nilainya positif. (6. Suppose that Cp=a+bT where a and b positive constants.7 adalah jauh diatas temperatur gas normal. For specified T and S. jadi kurvanya adalah isobar.) dan dB / dT adalah ( + ) Selain itu d2B / dT2 adalah ( . Show that the curvature of an isobar is also positive. the shape of B(T) is the same for all gases. What can you say about the signs of VR and C R ? SOLUTION : Karena pada saat Tr = 2. Use this observations to show by Eqs. Dengan mensubstitusi GR .500 kPa and 900 C at the rate of 1. ∂HR dB ∂T P T d2B dT PT d2B dB dT 2 dT dT2 Sehingga. dan VR adalah ( − ).37 dan 6.) Dari definisi GR. If the velocity in the discharged line is not to exceed 30 m s-1.Contoh 6. Contoh 3. dan HR adalah ( − ).500 Kpa = 55 bar T = 900C = 363.55 GR = BP SR = −P(dB/dT) Dimana GR dan SR adalah ( .15 K .4 kg s-1. CR = ∂HR ∂T adalah ( + ) P 6. HR = GR + T SR. what is the minimum diameter of the discharged line? SOLUTION : Diketahui Data Sebagai Berikut: P = 5.40 VR = B.53 dan 6. SR dan HR HR = P B – T dB dT Dimana.86 An equimolar mixture of methane and propane is discharged froma compressor at 5. 006+0.2 K Ppc=Ymetana x Pc metana+Ypropana x Pc propana ¯ x 42.5 x 369.8010+0.296 Tpc 280.11 Z =0.8 K Tpc=95.48 ¯¿ Ppc=0.8010 Z1 = 0.243 ¯ 55 ¿¿ P Tpr = =¿ Tc Tpc=Ymetana x Tc metana+Ypropana x Tc propana Tpc=0.6 K +0.082 x 0.81 .5 x 190.3 K +184.152 w=0.012+0.00902 Z =0.9 K=280.99 +0.235 ¯ ¯¿ Ppc=22.2 44.5 x 45.3 And E.15 K = =1.995 +21.8010+0.5 x 0.5 ¯ + 44.082 Z =Zo+W x Z 1 Z =0.1100 w=Y x W .0160 w=0.235 ¯¿=1.5 x 0.Answer: Tpr= T 363.4 Zo = 0.24 By Interpolation In Table E.+Y 2W 2 w=0. 3-Butadiene at 500 K and 20 bar.964 cm ❑ π 3.79108 cm 2 = 2.15 K = =14.+Y 2 x 44. .81 x 0.5 x 16.+0.87.4 gm sec cm3 sec V ¿˙ 2.07 gm/mol m ¿˙ 1.097 ) gm /mo l V= mol wt =30.Massa Molar Campuran mol wt =( Y 1 x 16.901)cm2 4A 27.901 cm 2 m A=¿ D= √ √ √ 2 4 (6.07 x 104 30 m s cm 3 s 3 =690 cm =6.mol wt 55 bars x 30.788 cm3 P .07 gm/mol ZRT 0.4 kg sec U=30 m/ sec V ¿˙ V x m ¿˙ V ¿˙ 14.788 cm 3 kg x 1.082 x 363. HR.097 ) gm/mol mol wt =( 0. and SRfor one of the following by appropriate generalized correlations: (a) 1. 604 cm = = =√ 8. Estimate VR.07 x 104 ˙ V ¿ x U 2.043.043.5 x 44.14 6. 10 36.568 0.841 0.055 -0.252 0. b. d.012 0.029 0.297 0. -0.303 0.6 0.2 0.77 73.6 154.00 21.148 0.2 0.666 0.315 0.069 0.199 -0. j.065 1.331 1. g.333 0.510 = .217 x 10-3 9. (j) Tetrafluoroethane at 400 K and 15 bar. f.522 0.675 T r 2.728 0.703 0.190 0.971 1. h.730 0.312 0. SOLUTION : T P Tc Pc Tr= T Tc Pr = ω P Pc a.0 617.7 430.555 1.492 0.6 469. (c) Carbon disulfide at 450 K and 60 bar.084 -0.2 304.172 T r 4.005 1. (i) Sulfur dioxide at 450 K and 35 bar.8 374.052 0.468 2.957 0.056 6.60 1 B = 0.603 0.2 190.2 42.139- 0.443 0.43 33.948 0.370 -0.7 617.345 -0.224 0.083- 425.245 0. g.722 T r 5.815 0. c.022 0.091 0. f.397 0. (e) Ethylbenzene at 620 K and 20 bar. e.242 -0.306 1.576 0.6 a.(6) Carbon dioxide at 400 K and 200 bar.267 -0.845 0.175 0.574 0. i. j.639 DB0 0.311 0.512 -0.718 x 10-3 -4.70 78. (h) n-Pentane at 500 K and 10 bar.06 45. (d) n-Decane at 600 K and 20 bar.970 1.422 T r 1.045 1.444 0. d.084 -0. h.83 79.84 40.320 -0. (g) Oxygen at 150 K and 20 bar.99 50. 500 400 450 600 620 250 150 500 450 400 B0 = 20 200 60 20 20 90 20 10 35 15 0. b.369 -0. i.709 0.111 0. c. (f) Methane at 250 K and 90 bar. e.759 0.009 x 10-3 0.173 2.200 -0.2 552.308 -0.176 1.327 = DB1 0. Tc 0 1 Pc .53 -83. j.226 x 103 -1.DB1 ] -19.97 -503.44 -68.56 Bar.723 -314. g. i.907 -146.K .37 x 103 -2.52 -24.bar.mol-1K-1 R V = [ R.R = 83.358 x 103 -559.Pr.01 x 103 -23. B )] -208.ω. e.Tr.69 -17.40 -27.27 -87.593 -355.21 x 103 -23.501 -1.cm3/mol. d.Pr.(DB0 + ω.18 -61.DB0) + (B1–Tr.454 HR = [R.16 x 103 -4. f.45 -22.100 -232. a.55 -54.(B0.Tc. h.560 -94.99 -40.251 x 103 SR = [ -R.08 -89.DB1) -1.14 cm3.65 -963.746 x 103 -1. c.(B + ω. b.377 x 103 -5.