chaptet 6 (1)

SOLUTION THERMODYNAMICS

CHAPTER 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:

{

∂ H ∂ S

}

P

=_{T Proses Isobar dan memiliki slope bernilai} positif

Membedakan persamaan di atas:

{

∂2_H

∂ S2

}

P

=

{

∂ T ∂ S

}

P

Jika di substitusikan dengan contoh 6.17

{

∂2_H ∂ S2

}

P = T

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 ( ∂ 2V/ ∂ T2)P

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 :

(

∂Cp ∂ P

)

T =

[

∂

{

V −T (∂V ∂ T )P

}

∂ T

]

P

(

∂Cp ∂ P

)

T =

(

∂V ∂T

)

P −T

(

∂ 2 V ∂ T2

)

P −

(

∂V ∂ T

)

P Sehingga ,

(

∂Cp_{∂ P}

)

_T=−T

(

∂ 2 V ∂ T2

)

P

Untuk gas ideal :

(

∂V_∂T

)

_P=R_Pdan

(

∂ 2 V ∂T2

)

P =0

(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

(

∂V ∂ T

)

=

(

∂ P ∂T

)

V dV +

(

∂ V ∂T

)

P dP

Dalam kondisi P konstan, menjadi : Cp=Cv+T

(

∂ P

∂ T

)

V

(

∂ V ∂ T

)

P

Dengan contoh persamaan 3.2 dan 6.34 :

(

∂V_∂T

)

_P=βV dan

(

∂ P_{∂ T}

)

_P=_Kβ

Substitusikan dengan persamaan pada P konstan : Cp−Cv=βTV

(

β

K

)

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 : (∂ U /∂ T )P ₌ Cp−P

(

∂ V ∂ T

)

P =Cp−βPV = Cv+

[

T

(

∂ P_{∂ T}

)

v−P

]

(

∂V_∂T

)

p=Cv+β_k (βT −kP )V

To what do these equations reduce for an ideal gas? For an incompressible liquid?

SOLUTION :

Untuk definisi H,U=H=PV, turunannya:

(

∂U ∂ T

)

P =

(

∂ H ∂ T

)

P −P

(

∂ V ∂ T

)

P or

(

∂U ∂ T

)

P = Cp−P

(

∂ V ∂ T

)

P

Substitusi turunan akhir dari persamaan 3-2 definisi dari β (∂ V /∂ T )P=Cp−βPV

Pembagian persamaan 6.32 dengan dT dan terbatas untuk P konstan hasilnya adalah

(

∂U ∂ T

)

p=Cv+

[

T

(

∂ P ∂T

)

v −P

]

(

∂ V ∂T

)

p

Pemecahan untuk turunan kedua dari persamaan 6.34 dan 3.2 substitusinya adalah :

(

∂U

∂ T

)

p=Cv+ β

k(β T −kP) V

6.4 The PVT behavior of a certain gas is described by the equation of state: P(V-b) = RT

Where b is constant. 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, P(V-b)ɣ _{= const.}

SOLUTION : a Berdasarkan persamaan 6.32: dU = CvdT +

[

T ( ∂ P ∂ T)V −P

]

oleh persamaan keadaan: P = _{(V −b) dimana (}RT

∂ P ∂T ¿ ¿_V= R V −b= P T

dengan mensubstitusikan persamaan keadaan ke persamaan 6.32, maka diperoleh: dU = CvdT +

[

T (_{V −b}R ) V − RT V −b

]

⟹ dU = CvdT +

[

(_{V −b}RT ) V − RT V −b

]

persamaan akhir menjadi: dU = CvdT , (terbukti bahwa U hanya merupakan

fungsi T)

b dH = dU + d(PV) …… (1)

kombinasikan persamaan (1) dan (2) dengan persamaan akhir dari jawaban (a), diperoleh: dH = C (¿¿v dT ) ¿ + (R d T + b d P) ⟹ C ¿ ¿ ¿ + R) dT + b d P …… (3) integralkan persamaan (3), gunakan persamaan ∂ H_{∂ T} =CV+R

hasil akhir menjadi: Cp = Cv + R, karena Cv konstan, maka: γ ≡CP

CV = konstan

c Untuk persamaan mekanik proses adiabatic reversibel, dU = dW , sesuai dengan persamaan keadaan: dU = CvdT , maka persamaan dU = dW berubah menjadi:

C_vdT =−P dV =−RT V −b dV =−RT d(V −b) V −b Disederhanakan menjadi: dTT = −R C_V d ln (V −b ) dari jawaban (b), γ ≡ C_P

CV = konstan, diubah menjadi R CV =CP−CV CV =_{γ−1 ,} maka: d lnT = - (γ−1) d ln(V-b) ⟹ d lnT + d ln (V −b)γ−1 = 0

hubungkan dengan persamaan: T (V −b)γ−1 = konstan, substitukan dengan persamaan keadaan:

P (V −b)(V −b)γ−1

R =konstan , maka: P (V−b) γ

= const (TERBUKTI)

6.5 A pure fluids is described by the canonical equation of state : G = Γ (T) + RT ln P, where Γ (T) is a substance-specific function of temperature. Determine for such a fluid expressions for V, S, H, U, Cp, and Cv. These results are consistent with those for an important model of gas-phase behavior. What is the model ?

SOLUTION :

V = ( ∂G_{∂ P}¿ T and S = - ( ∂G ∂ T ¿ P Differentation of the give equation of state yields:

V= RT_P and S = - d Γ (T )_dT - R ln P Once V and S ( as well as G ) are known, 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. Cp = ( ∂ H_{∂ T} ¿ P and Cv = ( ∂U ∂ T ¿ T Because Γ is a function of temperature only, these become:

Cp = -T d 2 T dT2 and Cv = -T d2T dT2 - R = Cp – R

The equation for V gives the ideal-gas value. The equations for H and U show these properties to be functions of T only, which conforms to ideal-gas behavior. The equation for S shows its relation to P to be that of an ideal gas. The equations for CP and CV show these properties to be functions of T only, which conforms to ideal-gas behavior, as does the result, CP = CV + R. We conclude that the given equation of state is consistent with the model of ideal-gas behavior.

6.6 A pure fluid, 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. Determine for such a fluid expressions for V, S, H, U, Cp, and Cv. These results are consistent with those for an important model of liquid phase behavior. What is the model ?

SOLUTION :

G = F (T) + K P

Diferensiasi dari persamaan yang diberikan :

 Dalam V V =

(

∂G ∂ T

)

T ; V = K  Dalam S S=−

(

∂ G ∂ T

)

P ; S=−dF (T ) dT

Maka V dan S telah diketahui dalam persamaan :

 H = G + T S dan U = H – PV

Maka persamaan menjadi,

 H=F (T )+KP−T dF (T )_dT dan U= H – PK Sehingga, U=F (T )+KP−TdF (T ) dT −PK U=F (T )−T dF(T ) dT

Sedangkan dari persamaan (2.16) dan (2.20) : Cp=

(

∂ H ∂ T

)

P ; Cv=

(

∂ U ∂ T

)

V Cp=−T d 2 F d T2 ; Cv=−T d2F d T2 F(T) dianggap konstan Maka kesimpulannya adalah Cp = Cv

Persamaan untuk V menunjukkan hal itu akan konstan dari kedua T dan P. Ini adalah definisi dari cairan termampatkan H dipandang sebagai fungsi dari kedua T dan P

sedangkan U, S, Cp, dan Cv adalah fungsi T saja. Kami juga memiliki hasil yang Cp = Cv. Semua ini konsisten dengan model, cairan mampat.

6.7 Estimate the changes in enthalpy and entropy when liquid ammonia at 270K is compressed from its saturation pressure of 381 kPa. For saturated liquid ammonia at 270K, Vt _{= 1.551 x 10}-3 _m3 _kg-1_{, and} β=2.095 x 10−3K−1 _.

SOLUTION :

At constant temperature Eqs. (6.25) and (6.26) can be written : dS = −β .V dP and dH = 1−(β . T . V . Dp)

For an estimate, assume properties independent of pressure T = 270 K P1 = 381 kPa P2 = 1200 kPa V = 1. 551 x 10-3 _m3 _kg-1 β=2.095 x 10−3 K−1 ∆S = - β.V (P2 – P1) ∆H = 1 - β.T.V (P2 – P1) = -2.661 J / KG. K = 551.7 J / kg

6.8 Liquid isobutane is throttled through a valve from an initial state of 360 K and 4000 kPa to a final pressure of 2000 kPa. Estimate the temperature change and the entropy change of the isobutane. The specific heat of liquid isobutane at 360 K is 2.78 J g-1○_c-1_.

Estimates of V and β may be found from Eq. (3.63).

SOLUTION : Isobutane: Tc = 408.1K Zc = 0.282 Cp = 2.78 J / gm K P1 = 4000 kPa P2 = 2000 kPa mol (wt) = 58.123 gm / mol

Vc = 262.7 cm3 _{/ mol}

Eq. (3.63) 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.

T = ( 359 ; 360 ; 361 ) K

Tr = T / Tc Tr = ( 0.88 ; 0.882 ; 0.885 )

(The elements are denoted by subscripts 1, 2, & 3) V = Vc . Zc ( 1 – Tr )0.2857

V = (131.662 ; 132.113 ; 132.767 ) cm3 _{/ mol}

V = 262,7 cm3_{/mol x (0,282)}0,5457 _{= 131,662 cm}3_/mol V = 262,7 cm3_{/mol x (0,282)}0,5430 _{= 132,113 cm}3_/mol V = 262,7 cm3_{/mol x (0,282)}0,5391 _{= 132,767 cm}3_/mol

Assume that changes in T and V are negligible during throtling. Then Eq.(6.8) is integrated to yield:

H = TS VP but H = 0 Then at 360 K, S = V1P2P1T1 =

(

−1,31 x 662 x 10 −4

₎

m3. mol−1.

(

−2 x 106

)

Pa 359 k = 263,32 x 4₃₅₉ Pa m3_{/mol K} S = 0.733 J / mol K.

We use the additional values of T and V to estimate the volume expansivity: V = 1.105 cm3 _{/ mol T} 1 → T2 K = (132,767 – 131,662) cm3_{/mol = (361 – 359) = 2K} = 1,105 cm3_/mol (V1 ) (V / T ) → 4.098755 x 103 K-1 = ( 1 131,662 cm3mol−1 ) . ( 1,105 cm3mol−1 2 K )

= _{131,662 .2 K}1,105

= _{263,324 K}1,105

= 4, 196351 x 10-3 _K-1

Assuming properties independent of pressure, Eq. (6.29) may be integrated to give S = Cp . ( ∂ T / T ) – ( ∂ V/ ∂ P) = 2,78 j/gmk . 5,55x10-3 _{– 4,196351x10}-3 _K-1 _{. 1,32181x10}-4 _m3_{/mol . -2x10}6 _Pa = (0,05429 + 1,11) J/ mol K = 1,125429 J/mol K Whence T = (V / Cp ) . [ ( ∂ S  ∂ V1 ∂ P) / mol wt ] = (359 K/2,78 J/gm K) . 1,125429 j molK +4,196351 x 10 −3 K−1 . 1,31662 x 10−4m3mol−1.−1 x 106Pa 58,123 gm .mol−1 = (129,14) . ( - _{58,123 ) K}1,105

6.9 One kilogram of water ( V1=1003 cm3kg-1 ) in a piston / cylinder device at 250C and

1 bar is compressed in a mechanically reversible, isothermal process to 1500 bar. Determine Q, W, ΔU, ΔH, and ΔS given that β = 250 x 10-6_K-1 _{and k = 45x10}-6 _bar -1_.

SOLUTION :

V2 = V1.EXP[-k(P2-P1)] = 1003 c m

3

= 1003 c m 3

kg . EXP[(-45x10-6 bar-1)(1499 bar)]

= 1003 c m 3 kg . EXP( -0,067455) = 1003 c m 3 kg . 0,9348 = 937,604 c m 3 kg Vave = V 1+V 2 2 = 1003 c m 3 kg + 937,604 c m3 kg 2 = 970,302 c m 3 kg ΔH = Vave . (1 – β.T) . (P2 – P1) = 970,302 c m 3 kg ( 1 – (250 x 10-6K-1.298,15K)).(1500 bar – 1 bar) = 970,302 c m 3 kg (1 – 0,0745375).(1499 bar) = 1454482,698 c m 3_. ¯¿ kg ¿ . (0.9254625) = 1346069,194 c m 3 . ¯¿ kg ¿ = 134606,9194 _kgJ

= 134,6069194 kJ_kg = 134,607 kJ_kg ΔU = ΔH – (P2.V2 – P1.P1) = 134,607 kJ_{kg – [(1500 bar . 937,604} c m 3 kg ) – (1 bar . 1003 c m3 kg )] = 134,607 kJ_{kg – (1406406} c m 3_. ¯¿ kg ¿ – 1003 c m3 kg ) = 134,607 kJ_{kg – (140,6406} kJ_{kg – 0,1003} kJ_{kg )} = 134,607 kJ_{kg – 140,5403} kJ_kg = -5,9333 kJ_kg ΔS = -β . Vave (P2 – P1) = (-250 x 10-6_K-1_)(970,302 c m 3 kg )(1500 bar – 1 bar) = (-0,2425755 c m 3 kg . K )(1499 bar) = -363,6207 c m 3_. ¯¿ kg . K ¿ = -0,03636207 _{kg . K}kJ = -0,0364 _{kg . K}kJ Q = T. ΔS

= 298,15K.( -0,0364 _{kg . K )}kJ

= 4,91936 kJ_kg

W = ΔU – Q

= (-5,9333 kJ_{kg ) – (-10,85266} kJ_{kg )}

= 4,91936 kJ_kg

6.10 Liquid water at 250_{C and 1 bar fills a rigid vessel. If heat is added to the water}

until its temperature reaches 500_{C. What pressure is developed? The avarege}

value of β between 25 and 500_{C is 36.2 x 10}-5 _K-1_{. The value of} ĸ _{at 1 bar}

and 500_{C is 4.42 x 10}-5 _bar-1 _{and may be assumed is independent of P. The}

specific volume of liquid water at 250_{C is 1.0030 cm}3_g-1_.

SOLUTION :

Berdasarkan persamaan 3.5 (perubahan volume konstan) β . ∆ T – ĸ .∆ P = 0 β _(T 2 – T1) – ĸ (P2 – P1) = 0 β _(T 2 – T1) = ĸ (P2 – P1) P2 = β ĸ (T2 – T1) + P1 = 4.42 X 10−5¿¯−1 36.2 x 10−5_K−1 ¿ (323.15 K – 298.15 K) + 1 bar = 205.75 bar

6.11 Determine expressions for GR_,HR_{,and S}R _{implied by the three-term virial} equation in volume,Eq (3.39) SOLUTION : From Eq. 3.39: Z = PV_{RT = 1+} VB+ C V2 Vig _{= gas ideal,dimana : P.V = n.R.T} Vig ₌ RT P Z = PV_{RT  V =} ZRT_P Sehingga, Vr _{= V- V}ig = V - RT_P = ZRT_P −RT_P Vr ₌ RT P ( Z - 1) Volum Residu : VR ₌ RT P (Z-1) … (1) Dimana : dG = VdP -

∫

dT … (4) T konstan sehingga : dG = V.Dp

Integrasi dari nol menuju P > GR ₌

∫

0 P

Dibagi RT GR RT =

∫

₀ P VR RTdP … (5) Maka, GR RT =

∫

0 P VR RT dP =

_∫

0 P RT P (Z−1) RT dP =

∫

0 P 1 P(Z −1)dP Finally, the answer is

GR RT=

∫

₀

P

(Z−1)dP P

Substitusi persamaan (3) ke persamaan (5) GR

RT =

∫

₀ P

(Z−1)dP

P … (6)

Substitusi persamaan (2) ke persamaan (6) GR RT =

∫

₀ P (Z−1)dP P + dZ Z (Z-1) Diferensiasi kondisi 2 menjadi :

GR RT = ¿

∫

0 P (Z−1)dP P +¿ Z-1) – ln Z … (7) Persamaan (4) dapat ditulis

d _{RT =} G _{RT dG -} 1 _{RT 2 Dt}G

dimana dG disubstitusi dengan persamaan (4) dan G dengan G = H – TS menjadi (dalam keadaan T konstan)

d GR RT ¿ ) = VR RT dP - HR RT 2 dT HR RT 2dT = (Z-1) dP P – d ( GR RT) … (8)

Persamaan (8) dibagi dT dan dalam P konstan, maka : HR RT 2 = Z−1 P ( ∂ P ∂T )P – [ ∂(GR RT) ∂ T ]P

Diferensiasi persamaan (1) menyediakan hasil di kanan dan diferensiasi persamaan (7) menyediakan hasil kedua. Substitusinya menghasilkan :

HR RT = -T ∂ Z ∂T ¿

∫

0 P ¿ )P dP P + Z-1 … (9)

Dan dari persamaan 3.39 didapatkan = V = 1/ρ Z-1 = Bρ + Cρ2

Pensubstitusian persamaaan (7) dan persamaan (9) akan menghasilkan : GR RT = 2Bρ + 3 2 Cρ2 – ln Z … 1 HR RT = T [ ( B T - dB dT )]ρ + ( C T - 1 2 dC dT )ρ2 … 2

Persamaan Gibbs umum : G = H – TS

Untuk residual GR _{= H}R _{- TS}R

Dan entropinya adalah SR R = HR RT - GR RT … 3

6.12 Nilai parameter dari persamaan van der walls telah ditunjukkan pada baris pertama tabel 3.1, halaman 99. Pada bagian bawah halaman 214 ditunjukkan pada persamaan I = β/Z. Maka persamaan menjadi :

Gr

RT = Z − 1 − ln(Z − β) – qβ/Z

Telah diberikan T, V, dan P pada fase vapor dan pada fase liquid denganσ = € = 0. Maka persamaan van der wall menjadi

β = _8TrPr dan q = _8Tr27

maka dapat disubstitusikan pada persamaan tersebut menjadi Hr RT = Z − 1 − qβ Z dan Sr RT = ln(Z − β)

6.13. Determine expressions for GR_{, H}R_{, and S}R _{implied by the Dieterici equation:}

Here, parameters a and b are functions of composition only.

SOLUTION : According to eq6.56 P = ZρRT I. eq. 6.63 GR R . T=(z−1−ln ( z−β ). ql) GR ₌ (z−1−ln(z −β). ql) RT

ZρRT = _{V −G}RT Rexp ⁡( −GR VRT) ZρRT = −(z−1−ln (z −β) . ql) RT VRT RT v−(z−1−ln( z− β) . ql) RTexp ⁡¿ ZρRT = _{v−(z−1−ln( z− β) . ql)}1 exp ⁡( −

(

z−1−ln( z− β) . ql

)

V ) Z = _{(v −}

₍

_{z−1−ln (z −β ). ql}1

₎

_)ρRT exp ⁡( −

(

z−1−ln (z −β ). ql

)

V ) II. Eq. 6.65 SR R=z−1+

(

d ln α(Tr) d ln Tr −1

)

ql SR_=z−1+

(

d lnα (Tr ) d ln Tr −1

)

qlR Substitusi eq. 6.56 and eq. 6.65 to eq 6.13

ZρRT = −SR VRT RT v−SRexp ⁡¿ ) ZρRT = −z−1+

(

d ln α (Tr ) d ln Tr −1

)

qlR VRT RT v−z−1+

(

d ln α (Tr ) d ln Tr −1

)

qlR exp ⁡¿

ZρRT = −z−1+

(

d ln α (Tr ) d ln Tr −1

)

ql VT T v−z−1+

(

d ln α (Tr ) d ln Tr −1

)

ql exp ⁡¿ Z= −z−1+

(

d ln α (Tr ) d ln Tr −1

)

ql VT T (v −z−1+

(

d ln α (Tr ) d ln Tr −1

)

ql) ρRT exp ⁡¿ Z= −z−1+

(

d ln α (Tr ) d ln Tr −1

)

ql VT 1

(

v−z−1+

(

d ln α (Tr ) d ln Tr −1

)

ql

)

ρR exp ⁡¿ III. Eq. 6.64 HR RT=Z−1+

(

d ln α (Tr ) d ln Tr −1

)

ql HR ₌ Z −1+

(

d ln α(Tr) d lnTr −1

)

qlRT

Substitusi eq. 6.56 and eq. 6.64 to eq 6.13

ZρRT = −HR VRT RT v−HRexp ⁡¿ )

ZρRT = −Z−1+

(

d ln α (Tr ) d ln Tr −1

)

qlRT VRT RT v−Z−1+

(

d ln α (Tr ) d ln Tr −1

)

qlRT exp ⁡¿ ) Z = 1 v−Z−1+

(

d ln α(Tr) d ln Tr −1

)

qlRTρ exp ⁡( −Z−1+

(

d ln α(Tr) d ln Tr −1

)

ql VRTρ )

14.Calculate Z,H and R by the Redlich/Kwong equation for one of the following and compare result with values found from suitable generalized correlations:

a.Acetylene at 300K and 40 bar b.Argon at 175 K and 75 bar c.Benzene at 575 K and 30 bar d.n-Butane at 500 K and 30 bar e.Carbon dioxide at 325 K and 60 bar f.Carbon monoxide at 175 K and 60 bar g.Carbon tetrachloride at 575 K and 35 bar h.Cyclohexane at 650 K and 50 bar

i.Ethylene at 300 K and 35 bar

j.Hydrogen sulfide at 400 K and 70 bar k.Nitrogen at 150 K and 50 bar

l.n-Octane at 575 K and 15 bar m.Propane at 375 K and 25 bar n.Propylene at 475 K and 75 bar

SOLUTION :

Redlich/kwong equation : Eq(3.53)

Guess Z 1

Z q

I 114

HRi RTi Ziqi 1 1.5qili Eq (6.67) The derivative in these SRi Rln Ziqi I 0.5qili Eq (6.68) Equations equal-0.5 Z i q I 0.695 Sri HRi 0.605 -5.461 -2.302·103 0.772 -4.026 -2.068·103 0.685 -6.542 -3.319·103 0.729 -5.024 -4.503·103 0.75 -5.648 -2.3·103 0.709 -5.346 -1.362·103 0.706 -5.978 -4.316·103 0.771 -4.12 -5.381·103 0.744 -4.698 -1.764·103 0.663 -7.257 -2.659·103 0.766 -4.115 -1.488·103 0.775 -3.939 -3.39·103 0.75 -5.523 -2.122·103 -8.767 -3.623·103

6.15 Calculate ZR_,HR_{, and S}R _{by the Soave/Redlich/Kwong equation for the substance}

and conditions given by one of the parts of Pb.6.14 and compare results with values found from suitable generalized correlations.

SOLUTION :

Soave/Redlich/Kwong equation:

 = 0,08664

 = 0,42748

α = [ 1+c.1- Tr0,5_]2 β = (. Pr_{Tr ), maka β} = ( Pr_Tr¿ Eq.(3.50) q = α(Tr )_Tr E.q.(3.51) Guess: z=1 Given: Z = 1+ β – q. β _{z .(z +β) E.q.(3.49)}z−β The derivative in the following equations equals; -ci. (Tri_{∝ i}) 0,5 dimana, i=1,...,14 Ii = ln βi, qi+βi ¿ βi , qi Z (¿) Z¿ ¿ ¿

HRi= R.Ti [ Z ( βi, qi)−1−¿ [ci. Tri_{∝ i )}0,5 _+1].qi,Ii]

Sri = R [ ln(Z)(βi ,qi)−βi [ci. Tri_{∝ i )}0,5_.qi,Ii] So,

Z ( βi, qi) = HRi=J/mol SRi = J/mol.K -2,595.103 -2,090.103 -3,751.103 -4,821.103 -2,585.103 -1,406.103 -4,816.103 -5,806.103 -1,857.103 -2,807.103 -1,527.103 -4,244.103 -2,323.103 -3,776.103 -6,412 -8,947 -4,795 -7,408 -5,974 -6,02 -6,246 -6,849 -4,451 -5,098 -7,581 -5,618 -4,482 -6,103 0,691 0,606 0,774 0,722 0,741 0,768 0,715 0,741 0,774 0,749 0,673 0,769 0,776 0,787

6.17 Estimate the change in enthalpy and entropy when liquid ammonia at 270 K is compressed from its saturaration pressure of 381 kPa to 1,200 kPa. For saturated liquid ammonia at 270 K, Vt _{= 1.55 x 10} -3_m 3 _kg-1_{, and β = 2.095 x 10}-3 _K-1_.

SOLUTION :

T = 323.15 K t = _KT 273.15 t = 50

The pressure is the vapor pressure given by the Antoine equation: P(t) = exp

(

13.8858−_t+220.792788.51

)

P(50) = 36.166

d

dt P(t) 1.375 P = 36.166 kPa dPdt =1.375 kPa

K a. The entropy change of vaporization is equal to the latent heat divided by

the temperatur.

For the Clapeyron equation, Eq. (6.69), we need the volume change of vaporization. For this we estimate the liquid volume Eq.(3.63) and the vapor volume by the generalized virial correlation.

For benzene: ω = 0.210 Tc = 562.2 K Pc = 48.98 bar Zc = 0.271 Vc = 295 cm3_{/mol Tr = T / Tc Tr = 0.575 Pr = P/Pc Pr = 0.007} By Eqs. (3.65), (3.66), (3.61), & (3.63) B0 = 0.083-0.422/Tr 1.6 B0 = 0.139-0.172/Tr 4.2 = -0.941 = -1.621 V vap = R .T / P (1 + (B0 +ω. B1) Pr/Tr) = 7.306 x 104 cm3/mol By Eq. (3.72), Vliq = Vc . Zc (1 – Tr 2/7) = 93.151 cm3/mol

Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change of vaporization:

∆S = dPdt. (Vvap – V liq) = 100.34 J / mol. K

(a) Here for the entropy change of vaporization: ∆S = R. T / P x dPdt

= 102.14 J / mol.

6.18 Let P1sat and P2sat be values of the saturation vapor pressure of a pure liquid at

absolute temperature T1 and T2. Justify the following interpolation formula for

estimation of the vapor pressure Psat _{at intermediate temperature T :}

ln Psat _{= ln P} 1sat + T₂(T −T ₁) T (T₂−T ₁) ln P 2sat P 1sat SOLUTION : Ln P2sat _{= A –} B T 2 ………A Ln Psat _{= A –} B T ………..B Ln P1sat _{= A –} B T 1 ………..C Eliminasi C dari A Ln P2sat _{– ln P1}sat _{= A – A – (} B T 2− B T 1 ) Ln P 2 sat P 1sat = 0 + B T 1− B T 2 Ln P 2 sat P 1sat = B( 1 T 1− 1 T 2 ) Ln P 2 sat P 1sat = B( T 2−T 1 T 1. T 2 )………..pers.1 Eleminasi C dari B

Ln Psat _{– ln P1}sat _{= A – A – (} B T− B T 1 ) Ln P sat P 1sat = 0 + B T 1− B T Ln P sat P 1sat = B( 1 T 1− 1 T ) Ln P sat P 1sat = B( T −T 1 T 1. T )………pers.2

Bandingkan pers.1 dan pers.2 Ln P sat P 1sat : Ln P 2sat P 1sat = B( T −T 1 T 1. T ) : B( T 2−T 1 T 1. T 2 ) Ln P sat P 1sat : Ln P 2sat P 1sat = T₂(T −T ₁) T (T₂−T ₁) Ln P sat P 1sat = T₂(T −T ₁) T (T₂−T ₁) . Ln P 2sat P 1sat Ln Psat - ln P1sat = T_{T (T}₂(T −T ₁)_{₂−T ₁) . Ln} P 2 sat P 1sat ln Psat _{= ln P} 1sat + T₂(T −T ₁) T (T₂−T ₁) ln P 2sat P 1sat ………(terbukti)

6.19 Assumsing the validity of Eq. (6.70), derive Edmister’”s formula for estemation

of the acentric factor:

Where Ѳ ≡ Tn/Tc, Tn is the normal boiling point, and Pc is an (atm)

SOLUTION :

Tuliskan persamaan (6.70) dalam log10 sehinggga menjadi : log P sat = A – B/T...(A) Masukan titik kritis : log Pc = A – B/Tc...(B)

masukan perbedaan T, log P sa = B(i/Tc – 1/T) = B ((Tr – 1)/T)...(C) jika P sat dalam (atm), lau aplikasikan (A) pada area titik didih normal:

log 1 = A – B/Tn or A = B/Tn

dengan θ ≡ Tn/Tc, Eq. (B) sekarang dapat ditulis menjadi ::

Dimana: Persamaan menjadi: Masukan Tr = 0.7, maka: Dari persamaan (3.54) ω = −1.0 − log(Psat r )Tr=0.7 Jadi,

6.21 The state of 1 (lbm) of steam is changed from saturated vapor at 20 (psia) to

superheated vapor at 50 (psia) and 1,000 (0_{F). 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.4 : H1 = 1156.3 BTU lbm H2 = 1533.4 BTU lbm

S1 = 1.7320 BTU lbm . rankine S2 = 1.9977 BTU lbm . rankine Sehingga, ∆ H=¿ _H 2 – H1 = 377.1 BTU lbm ∆ S=¿ _S 2 – S1 =0.266 _{lbm . rankine}BTU

Berdasarkan persamaan 4.9 dan 5.18 (uap sebagai gas ideal) T0 = (227.96 + 459.67)rankine T1 = (1000+459.57)rankine P1 = 20 psi P2 = 50 psi T0 = 382.017 K T1 = 810.928 K τ =T1 T0 = 810.928 K 382.017 K = 2.123 (Cp)H R = A + B 2T0(τ +1)+ C 3 T0 2

₍

_τ2_{+τ +1}

₎

+ D τ T0 2 (Cp)H R =3.470 + 1.450 . 10−3 2 382.017 (2.123+1)+ 0.121 . 105 2.123(382.017)2 = 3.51391 ( ¿ ¿Cp_H = 3.51391 x R = 3.51392 x 8.314 J/molK _{= 29.215 J/molK}

Cp ¿ ¿ ∆ H=¿ _{= 29.215 J/molK(428.911 K)} =12530.635 J (Cp ig )s R = A +

[

B T0+

(

C T0 2 + D τ2T0 2

)

(

τ +1 2

)

]

(

τ−1 ln τ

)

= 3.470 +

[

1.450 . 10−3.382.017+

(

0.121 .10 5 2.1232(382.017)2

)

(

2.123+1 2

)

]

(

2.123−1 0.753

)

= 4.338 ∆ S R = (Cigp)s R ln T1 T₀−ln P1 P₀ = 4.338(0.753) – 0.916 = 2.350514 ∆ S=2.350514 X 8.314 = 19.5422 J/K

6.22. A two-phase system of liquid water and water vapor in equilibrium at 8,000 kPa consist of equal volumes of liquid and vapor. If the total volume V’=0.15m³, what is the total enthalpy H’ and what is total entropy s’?

SOLUTION :

Data, table F.2 pada 8.000 kPa

Vliquiid = 1.384. cm³_{gm Hliquid = 1317.1. gm Sliquid = 3.2076.}J

J gm . K

Vvapor = 23,525. cm³_{gm Hvapor = 2759.9. . gm Svapor = 5.7471}J J gm . K 1. M liquid = -0.15 . 106 2 cm ³ Vliquid = 0.15 . 106 2 cm ³ 1,384cm ³ gr = _{2,768 gr} 150000 = 54191 gr = 54,191 kg 2. Mvapor = - 0.15 . 106 2 cm ³ Vvapor = 0.15 . 106 2 c m 3 23,525c m 3 gr = _{47,05 gr}150000 = 3188 gr = 3,188 kg

3. Htotal = Htotal = - m liquid. Hliquid + m vapor. Hvap = 54,191 kg. 1317,1 KJ_{kg + 3,188 kg. 2759} KJ_kg = 71374,96 KJ + 8798,56 KJ = 80173,5 KJ 4. Stotal = 54,191 kg. 3,2076 _{kg/ K + 3,188 kg. 5,7471} KJ _{kg/ K}KJ = 173,82 KJ/K + 18,321 KJ/K = 192,145 KJ/K

vapor occupies 70% of the volume of the vessel, determine H dan S for the 1kg of H2O

SOLUTION :

Vliq = 1.127 cm3_/gm Hliq = 762.605 J/gm Huap =14.29 cm3_/gm Hvap = 2776.2 J/gm

Mencari x = fraksi masa dari vapor x .Vvap (1 – x).Vliq= 70 30 x .194.29 cm3 /gm (1 – x).1.1 27 cm 3/ gm= 70 30 X = 0.013  H = (1 – x) . Hliq + x. Hvap = (1 – 0.013) 762.605 J/gm + 0.013 . 2776.2 J/gm = 789,495 J/gm  S = (1 – x) . Sliq + x.Svap = (1 – 0.013) 2.1382 J/gm K + 0.013 . 6.5828J/gmK = 2.198 J/gm K

6.24 Reaktor bertekanan mengandung liquid air dan uap air di keadaan setimbang pada suhu 350 °F total massa dari liquid dan uap adalah 3 lbm. Jika volum dari uap adalah 50 kali volum liquid, berapakah total entalpi dari reaktor?

SOLUTION :

Data dari tabel F.3 pada 350 °F diketahui: Vliq = 0, 01799 ft lbm Vvap = 3, 342 ft lbm Hliq = 321,76 BTU lbm Hvap = 1192,3 BTU lbm

mliq + mvap = 3 lbm mvap ×Vvap = 50. mliq. Vliq mliq + 50. mliq .Vliq Vvap + = 3 lbm mliq = 3× lbm 1+50 Vliq Vuap = 3 lbm 3 lbm 1+50 0,01799 ft lbm 3,342 ft lmb = 2, 364 lb

mvap = 3 lbm mvap - 2, 364 lb mvap = 0, 636 lb Htotal = mliq×Hliq + mvap ×Hvap

= (2, 364 lb x 321, 76 BTU_{lbm ) + (0,636 lb x 1192,3} BTU_{lbm )} Htotal = 1519,1 BTU

6.25 Wet steam at 2300_{C has a density of 0,025 g cm}-3_{. Determine x, H, and S.}

SOLUTION : ρ=m V ρ=1 V V = _{0,025 gr}1 cm3

 Data pada Tabel F.1 pada suhu 230o_C V liquid =1,209 cm3/gr

H liquid =990,3 J / gr H vapor=2802,0 J /gr K S liquid=2,61202 J /gr K S vapor =6,2107 J /gr K  Mencari x V =(1−x) V liquid+ x uap x= V −V liquid V uap−V liquid x= 1 cm3 0,025 gr−1,209 cm3 gr 71,45cm 3 gr −1,209 cm3 gr x= 1−0,030225 cm3 0,025 gr 70,241cm 3 gr x=38,791 70,241 x=0,552  Mencari H H=(1−x ) H liquid+ x H uap H=(1−0,552)990,3 J gr+0,552 . 2802,0 J gr H=443,6554 J gr+1546,704 J gr H=1990,3584 J gr  Mencari S S=(1−x) S liquid + x S uap

S=(1−0,552)2,61202 J gr+0,552 .6,2107 J gr S=1,17 J gr+3,4283 J gr S=4,5983 J gr

6.26 Sebuah vessel dengan volume 0.15m3 _{mengandung uap saturated pada 150} o_C

didinginkan menjadi 30 o_{C. Tentukan volume akhir dan massa air liquid pada}

vessel.

SOLUTION :

Berdasarkan persamaan 6.73b, Vtot = mtot.Vliq + mvap.Vlv Dari tabel F.1 didapat:

 Pada suhu 150o_{C, Vvap}=392.4cm 3

gr =0.3924 m3 kg

 Pada suhu 30o_{C, Vliq}=1.004cm 3 gr =1.004 . 10 −3m3 kg ∆ Vlv=32930 cm3 gr =32.93 m3 kg m_tot=Vtot Vvap = 0.15 m 3 0.3942m3/kg = 0.382 kg m_vap=Vtot−mtot.Vliq

∆ Vlv 0.382 kg .1.004 .10−3m 3 kg ¿ 0.15 m3−¿ ¿ ¿ ¿4.543 ×10−3 kg

m_liq=m_tot−m_vap

¿0.382 kg−4.543 ×10−3_kg = 0.37772 kg

= 377.72 gram V_{tot liquid}=m_liq.V_liq

¿377.72 gram.1 .004cm 3 gr = 379.23 cm3

6.27 Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.325 kPa, where its temperature is 378.15 K (105°C). What is the quality of the steam in its initial state?

SOLUTION :

According from table F.2, 1100 kPa : Hliq = 781.124 J gm Hvap = 2779.7 J gm H2 = 2686.1 J gm Interpolate pada 101.325 kPa & 105 degC:

Const. –H throttling : H2 = Hliq + x . (Hvap – Hliq)

x =

H2−Hliq Hvap−Hliq

x = _{2779.7−781.124} 2686.1−781.124

6.28 Steam at 2,100 kPa and 260o_{C expands at constant enthalpy (as in a throttling}

process) to 125 kPa. 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, Table F.2 at 2100 kPa and 260 degC, by interpolation :

H1 = 2923.5 J S1 = 6.5640 J mol wt=18.015 gm

gm gm.K mol

H2 = 2923.5 J gm

Final state is at this enthalpy and a pressure of 125 KPa

By interpolation at these conditions, the final temperature is 224.80 degC and

S2 = 7.8316 J ∆S = S2 - S1 ∆S= 1.268 J

gm.K gm. K

For steam as an ideal gas, there would be no temperature change and the entropy change would be given by :

P1 = 2100 KPa P2 = 125 KPa

∆S = -R ln P2 ∆S=1.302 J

molwt P1 gm. K

6.29 Steam at 300(psia) and 500(o_{f) expands at constant enthalpy (as in a throttling}

process) to 20(psia). 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 :

P1 = 300 psia T = 500oF H1 = 1257.7 BTU lbm H2= 1257.7 BTU lbm S1 = 1.5703 BTU lbm rankin

Namun entalpi yang diperlukan pada tekanan akhir yaitu 20 psia Untuk itu diperlukan interpolasi dan diperoleh hasil interpolasi yaitu : P2 = 20 psia T = 438.87 S2 = 1.8606 BTU lbm rankin ∆S = S2 – S1 =1.8606 – 1.5703 BTU lbm rankin = 0.2903 BTU lbm rankin

Untuk steam pada gas ideal, tidak ada perubahan temperature maka

∆S =

−RT ln(P 2 P 1)

mol _{molwt = 18 lbmol}

= 8.314 x 500 x ln( 20 300) ¿ −¿ ¿ = 0.2903

6.30 Superheated steam at 500 kPa and 3000_{C expands insentropically to 50 kPa.}

What is its final enthalpy?

SOLUTION :

S2 = S1 = Sliq + x. ( Svap – Sliq)

x = _{Svap−Sliq =} S 1−Sliq (7.4614−1.0912) J gm. K (7.5947−1 .0912) J gm . K = 0.98

H = Hliq + x. ( Hvap – Hliq )

= 340.564 _{gm + 0.98. ( 2646.9} J _{gm - x340.564} J _{gm )}J

= 2599.6 _gmJ

6.31 What is the mole fraction of water vapor in air that is saturated with water at 25ºC and 101,33 kPa ? at 50ºC and 101,33 kPa ?

SOLUTION : At 25ºC Psat = 3.166 kPa P = 101.33 kPa Xwater = Psat P = 3.166 kPa = 0.031 101.33 kPa At 50ºC Psat = 12.34 kPa Xwater = Psat P = 12.34 kPa = 0.122

101.33 kPa

6.32 A rigid vessel contains 0,014 m3 _{of saturated-vapor in equilibrium with 0,021 m}3

of saturated-liquid water at 1000_{C. Heat is transferred to the vessel until one}

phase just disappears, and single phase remains.Which phase (liquid or vapor) remains, and what are its temperature and pressure? How much heat is transferred in the process?

SOLUTION :

Berdasarkan tabel F.1 pada suhu 1000_C,maka: Volume vapor = 0,014 m3 Volume liquid = 0,021 m3 Vliq = 1,044 cm3_/gm Uliq = 41,9 J/gm Vvapor = 1673,0 cm3_/gm Uvap = 2506,5 J/gm Vtotal = Vliq + Vvap

= (0,014 + 0,021) m3 = 0,035 m3

Mass = mliq + mvap = 0,021 m 3 1,044 cm3/gm

+

0,014 m3 1673,0 cm3/gm = 0,02011 gm + 8,3682 x 10-6 _gm = 0,02012 gm X = m. vap_mass

=

8,3682 x 10 −6 gm 0,02012 gm

= 4,159 x 10

-4 V2 = Vtotal mass

=

0,02012 cm3/gm0,035 = 1,739 cm3_/gm

Mencapai saturated liquid pada suhu 349,83 K ( fase liquid) P = 16,5001 kPa

U2 = 1641,7 J/gm

U1 = Uliq + X(Uvap – Uliq)

= 41,9 J/gm + 4,159 x 10-4 _{( 2506,5 – 419,0) J/gm} = 419,0 J/gm + 0,8682 J/gm = 419,8682 J/gm Q = U2 – U1 = 1641,7 J/gm - 419,8682 J/gm Q = 1221,832 J/gm

6.33 A vessel of 0,25 m3 _{capacity is filled with saturated steam at 1500 kPa. If the}

vessel is cooled until 25% f the steam has condensed, how much heat is transferred and what is the final pressure?

SOLUTION :

Of this total mass, 25% condenses making the quality 0,75 since the total volume and mass don’t change. We have for the final state:

Sementara itu: X=

V 1−V (l) V ( v )−V (l)

Find P for which (A) yields the value X=0,75 for wet steam

Since the liquid volume is much smaller than the vapor volume,we make a preliminary calculation to estimate:

Vvap = V 1_X = 131,66 cm3/_{gm/0,75 = 175,547 cm}3/_gm

This Value Occurs at a pressure a bit above 1100 kPa. Evaluate X at 1100 and 1150 kPa by (A). Interpolate on X to find P = 1114,5 kPa and

Vliq = 782,41 J/gm Vvap = 2584,9 J/gm V2 = Vliq + X (Vvap-Vliq) = 782,41 J/gm + 0,75 (2584,9 J/gm) -782,41 J/gm) = 782,41 J/gm + 0,75 (1802,49 J/gm = 782,41 J/gm + 1351,9 J/gm =2134,31 J/gm Q = mass (V2-V1) = 1898,9 gm (2134,3 J/gm -2592,4 J/gm) = 1898,9 gm ( -458,1 J/gm ) = -869886,09 J = -869,9 KJ

6.34 A vessel of 0,25 m3 _{capacity is filled with saturated steam at 1500 kPa. If the}

vassel is cooled until 25 % of the steam has condensed,how much heat is transferred and what is the final pressure?

SOLUTION : Vliq = 1.044 cm3_/gr Vvap = 1673.0 cm3_/gr Uliq = 418.959 J/gr

Uvap = 2506.5 J/gr

m liq = = = 19.157 x 103 _gr

m vap = = = 1.184 x 103 _gr

m total = m liq + m vap = 20.341 x 103 _gr

x = = = 0.058

V1 = Vliq + x (Vvap - Vliq)

= 1.044 cm3_{/gr + 0.059 (1673.0 cm}3_{/gr - 1.044 cm}3_/gr) = 1.044 cm3 _{/gr+ 0.059 x 1671.956 cm}3_/gr

= 1.044 cm3 _{/gr+ 98.645 cm}3 _/gr = 99.689 cm3 _/gr

U1 = Uliq x Uvap x Uliq

= 418.959 J / gr + 0.059 (2506.5 J / gr - 418.959 J / gr ) = 418.959 J / gr + 0.059 x 2087.541 J / gr = 418.959 J / gr + 123.165 J / gr = 542.124 J / gr V1 = V2 = 99.689 cm3 /gr

Berdasarkan tabel F.1 maka didapat T = 212 o_C U liq = 904.5 J/gr

U vap = 2598.0 J/gr

= 904.5 J/gr + 0.058 (2598.0 J/gr - 904.5 J/gr ) = 904.5 J/gr + 0.058 x 1693.5 J/gr = 904.5 J/gr + 98.233 J/gr = 1002.723 J/gr Q = m total ( U2 – U1 ) = 20.341 x 103 _{gr (1002.723 J/gr - 542.124 J / gr)} = 20.341 x 103 _{gr x 460.599 J / gr} = 9369.044 x 103 _J = 9369.044 kJ

6.35. A rigid vessel of 0,4 m³ volume is filled with steam at 800 kPa and 350 c. How much heat must be transferred from the steam to bring its temperature to 200c.

SOLUTION : Appendiks F2 : P = 800 kPa T = 350c V1 = 354,34 Cm ³ gr U1 = 2878 J gr Vtotal = 0,4 m³ T = 2000_c U2 = 2638,7 _gr J Q = Vtotal_{V 1} (U₂-U₁) = 0,4.106 354,34cm ³ gr cm ³

6.36 1 Kg of steam is contained in a piston / cylinder device at 800 kPa dan 200o_C.

a. If it undergoes a mechanically reversible, isothermal expansion to 150kPa. How much heat does it absorb?

b. If it undergoes reversible, adiabatic expansion to 150kPa. What is its final temperature and how much work is done ?

SOLUTION :

Dari Tabel F.2 (Superheted Steam) pada 800 kPa dan 200 o_CU

1 = 2629.9 J gm S1 = 6.8148 J gm . K

a. Keadaan Isotermal, Pada Saat 150 kPa dan 200 o_C Pada Tabel F.2 U2 = 2656.3 J gm S2 = 7.6439 J gm . K T = 473 o_K Q=m .T . ∆ S Q=1 kg .473 K .(7.6439−6.8148) J gm. K Q=473 kg . K (0.8291) J gm . K Q = 392.1643 KJ W =(m . ∆ U)−Q W=

{

1 kg . (2656.3−2629.9) J gm

}

−392.1643 KJ W=26.4 J gmkg−392.1643 KJ W=−365.7643 K B. Entropy konstant pada saat 150 kPa

 Sliq = 1.4336 J gm . K Uliq = 466.968 J gm  Svap = 7.2234 J gm . K

Uvap = 2519.5 J gm S_vap−¿Sliq X =S1−Sliq ¿ X = (6.8148−1.4336) J gm. K (7.2234−1.4336) J gm. K X =0.929

U₂=U_liq+X

(

U_vap−U_liq

)

U2=2.367 x 10 3 J gm W=m.

(

U₂−U₁

)

W=1 Kg

(

2.367 x 103 −2629.9

)

J gm W=−262.527 KJ

6.37 Steam at 2000 kPa containing 6% moisture is heated at constant pressure to

848.15 K. (575°C). How much heat is required per kj ?

SOLUTION :

Data, Table F.2 at 2000 kPa: H_{vap = 2797,2 x} J

gm Hliq = 908,589 x J gm

For superheated vapor at 2000 kPa and 575 degC, by interpolation: H2 = 3633,4 x

J gm Q = mass H2 H1

= 949.52 kJ.

6.38 Steam at 2.700 kPa and with a quality of 0,90 undergoes a reversible.adiabatic expansion in a nonflow process to 400 kPa.It is than heated at constant volume until it is saturated vapor. Determine Q and W for the process?

SOLUTION :

Kondisi pertama adiabatik ekspansi P1=2700kPa

X1=0,90 P2=400 kPa

Kondisi ke dua pemanasan dengan volume konstan 1 Q12 = 0

U12= Q12 + W12,,karena Q12 = 0 . 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 , 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

Uliq=977,968 _gmJ

Uvap= 2601,8 _gmJ

Sliq=2,5924 _{gm . K}J

U1 Uliq x1Uvap Uliq

U1 977,968 _gm J 0,92601,8 _gm J 977,968 _gmJ 

U1 977,968 _gm J  gmJ 

U1= 2,439 103 J gm S1 Sliq x1Svap Sliq

S1 2,5924 _{gm . K} J 0,96,2244 _{gm . K} J 2,5924 _{gm . K}J 

S1 2,5924 _{gm . K} J  gm . KJ S1= 5,861 _{gm . K}J

Untuk tekanan P2=400 kPa Dari tabel F2 halaman 700 Sliq = 1,7764 _{gm . K}J Svap= 6,8943 _{gm . K}J Uliq =604,237 _gmJ Uvap = 2552,7 _gmJ Vliq = 1,084 cm 3 gm Vvap = 462,22 cm 3 gm

Since step 1 is isentropic,

x2= _Svap−SliqS 1−Sliq x2= 5,861 J gm . K−1,7764 J gm. K 6,8943 J gm. K−1,7764 J gm. K x2= 4,0846 J gm. K 5,1179 J gm . K x2 0.798

U2 Uliq x2Uvap Uliq

U2 604,237 _gmJ 0.7982552,7 _gmJ 604,237 _gmJ 

U2 604,237 _gmJ  gmJ U2= 2,159 103 J

gm

V2 Vliq x2Vvap Vliq

V2 1,084 cm 3 gm 0.798462,22 cm3 gm 1,084 cm3 gm  V2 1,084 cm 3 gm  cm3 gm V2= 369,070528 cm 3 gm

V2=V3 (karena pemanasan dengan volume tetap/konstan)

The final state is sat. vapor with this specific volume.Interpolate to find that this V occurs at T = 509,23 0_C

Dari tabel F1 halaman 691

Volume spesifik U spesifik

373,2(X1) 2560,3 (Y1)

369,070528 (X) / V3 ???? (Y)/ U3

355,1 (X2) 2562,1 (Y2)

Y=Y1 + ( _{X 2−X 1 ) (Y2-Y1)}X −X 1 Y= 2560,3+ ( 369,070528−373,2_{355,1−373,2} ) (2562,1 -2560,3) Y= 2560,7 _{gm Y= U3, maka} J U3 =2560,7 _{gm = 2,5607} J 103 J gm Q U3 U2 Q 2,5607 103 J gm 2,159 103 J gm Q=0,4017 103 J gm Work U2 U1 W= 2,159 103 J gm - 2,439 103 J gm W= - 0,28 103 J gm W= - 280 _gm J

Tanda minus menunjukkan bahwa sistem melakukan kerja.

6.39 Four kilogram of steam in a piston/cylinder device at 400 kPa and 1750_C

undergoes a mechanically reversible, isothermal compression to a final pressure such that stem it just saturated. Determine Q and W for the process.

SOLUTION :

Untuk P = 400 kPa & T = 1750_{C diperolehlah nilai ( Dari tabel F.2 ),yaitu :} U1 = 2605,8 J/gm

S1 = 7,0548 J/gm.K

Untuk saturated steam pada suhu 1750_{C diperoleh lah nilai ( Dari tabel F.1 ), yaitu :} U2 = 2578,8 J/gm

Jadi, hasilnya ialah : a. Q = m x T x ( S2 - S1 ) = 4 kg x 448,15 K x (6,6221 J/gm.K - 7,0548 J/gm.K ) = - 775,66 kJ b. W = m x (U2 - U1 ) – Q = 4 kg x ( 2578,8 J/gm - 2605,8 J/gm ) – (- 775,66 kJ ) = 667,67 kJ

6.40. Steam undergoes a change from an initial state of 450o_{C and 3,000kPa to a final}

state of 140o_{C and 235 kPa. Determine ∆H and ∆S:}

(a) From steam-table data (b) By equations for an ideal gas

(c) By appropriate generalized correlations (a) Table F.2, 3000 kPa and 450 o_C

SOLUTION : H1=3344.6 J gm S1=7.0854 J gm. K Table F.2, interpolate 235 kPa and 140 o_C

H₂=2744.5 J gm S2=7.2003 J gm . K ∆ H=H₂−H_{1 ∆ H=−600.1} J gm ∆ S=S₂−S_{1 ∆ S=0.115} J gm . K (b) T1 = (450 + 273.15) K T2 = (140 + 273.15) K T1 = 723.15 K T2 = 413.15 K P1 = 3000 kPa P2 = 235 kPa ICPH(723.15, 413.15, 3.470, 1.450×10-3 _{, 0.0,0.121×10}5_{) = -1343.638} ICPS(723.15,413.15, 3.470,1.450×10-3 _{, 0.0, 0.121×10}5_{) = -2.415901} ICPH = -1343.638 K ICPS = -2.415901

Eqs. (6.86) & (6.87) for an ideal gas: molwt=18_molgm ∆ H_ig=R . ICPH molwt _{∆ S} ig= R .(ICPS−ln

(

P2 P₁

)

)⁡ molwt ∆ H_ig=−620.6 J gm ∆ Sig=0.0605 J gm . K (c) Tc = 647.1 K Pc = 220.55 bar w = 0.345 T_r1=1.11752 P_r1=0.13602 T_r2=0.63846 P_r2=0.01066

The generalized virial-coefficient correlation is suitable here

HRB(1.11752, 0.13602, 0.345) = -0.13341 HRB1 = -0.13341 SRB(1.11752, 0.13602,0.345) = -0.08779 SRB1 = -0.08779 HRB(0.63846, 0.01066, 0.345) = -0.04422 HRB2 = -0.04422 SRB(0.63846, 0.01066,0.345) = -0.05048 SRB2 = -0.05048 ∆ H=∆ H_ig+R .Tc.(HR B2−HR B1) molwt ∆ H=−593,95 J gm ∆ S=∆ S_ig+R .(HR B2−HR B1) molwt ∆ S=0.078 J gm. K

6.41 A piston / cylinder device operating in a cycle with steam as the working fluid executes the following steps:

 Steam at 550 kPa and 473.15 K (200°C) is heated at constant volume to a pressureof 800 kPa.

 It then expands, reversibly and adiabatically, to the initial temperature of 473.15 K(200°C).

 Finally, the steam is compressed in a mechanically reversible, isothermal processto the initial pressure of 550 kPa.

SOLUTION :

Data table F.2 Superheat Steam pada 550 kPa dan 200O_C: V1 = 385,19 cm3/gm

U1 = 2640,6 J/gm S1 = 7,0108 J/gm.K

Pemanasan Volume konstan pada tekanan 800 kPa, pada volume spesifik awal P, interpolasi yang didapat t = 401,74 o_{C. V = Konstan}

U2 = 2963,1 J/gm S2 = 7,5782 J/gm.K Sehingga, Q12 = U2- U1 = ( 2963,1 - 2640,6 ) J/gm = 322,4 J/gm

Ekspansi ke T awal, maka: Q23 = 0

S3 = S2

S3 = 7,5782 J/gm.K

Temperatur konstat dikompresi ke P awal: T = 473,15 K

Q31 = T ( S1 – S3 ) = 473,15 (7,0108 - 7,5782 ) = -268,465 J/gm Untuk Siklus perubahan energy dalah adalah = 0

Wsiklus = -Qsiklus = - Q12 – Q31 n = - Wsikuls / Q12

n = 1+Q 31_{Q 12} = 1+−268,465_322,4 = 0,1672

6.42 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, reversibly and adiabatically, to the initial temperature of 417,35ºF

 Finally, the steam is compressed in a mechanically reversible,isothermal proses to the initial state

What is the thermal efficiency of the cycle ?

SOLUTION :

Table, F.4, sat. Vapor, 300(psi) :

T1 = (417.35 + 459.67) rankine H1 = 1202.9 BTU Lbm. Rankine T1 = 877.02 rankine S1 = 1.5105 BTU

Lbm. Rankine Superheated steam at 300(psi) and 900ºF

H2 = 1473.6 BTU S2 = 1.7591 BTU

lbm Lbm. Rankine

Q12 = H2-H1 Q31 = T1. (S1-S3) Q31 = -218.027 BTU

lbm

6.43 Uap masuk ke dalam turbin pada tekanan 400 kPa dan suhu 400 °C terekspansi secara reversibel dan adiabatis.

a. Berapakah tekanan yang dikeluarkan dalam bentuk aliran uap jenuh? b. Berapakan tekanan yang dikeluarkan dalam bentuk aliran uap basah dengan nilai 0,95?

SOLUTION :

Dari data tabel F.2 superheated steam pada 4000 kPa dan 400 °C: S1= 6,7733

J gmK

Dari kedua masalah didapatkan: S2 S 1

a. Pertama kita mencari tekanan dimana tekanan uap jenuhnya memiliki entropi.

Hal ini terdapat pada tekanan di bawah 575 kPa, jadi kita mencarinya dengan cara interpolasi.

b. Untuk uap basa, entropinya di dapatkan: x = 0,95 S2 = Sliq xSvap Sliq

Jadi kita harus mendapatkan tekanan dimana persamaannya diketahui. Tekanan ini terdapat jika dibawah tekanan 250 kPa. Pada tekanan 250 kPa nilainya:

Sliq = 1,6071 J

gmK Svap = 7,0520 J gmK S2= Sliq x Svap Sliq

S2 = 1,6071 J gmK + 0,95 ( 7,0520 J gmK - 1,6071 J gmK ¿ = 6,7798 _{gmK atau > 6,7733}J

Didapatkan dari interpolasi P2 = 250,16 kPa

6.46 Table F.2 for superheated vapor at the initial conditions, 1300 kPa and 400 degC, and for the final condition of 40 kPa and 100 degC:

SOLUTION :

H1 = 3259.7 kJ/kg S1 = 7.3404 kJ/kgK H2 = 2683.8 kJ/kg If the turbine were to operate isentropically, the final entropy would be

S2 S1

Table F.2 for sat. liquid and vapor at 40 kPa:

Sliq = 1.0261kJ/kgK Svap = 7.6709 kJ/kgK Hliq = 317.16 kJ/kg Hvap = 2636.9 kJ/kg x2= (S2 Sliq) / (Svap Sliq)

x2= (7.3404 kJ/kgK - 1.0261kJ/kgK) / (7.6709 kJ/kgK - 1.0261kJ/kgK) x2= 0.95

H' Hliq x2Hvap Hliq

H'= 317.16 kJ/kg + 0.95(2636.9 kJ/kg - 317.16 kJ/kg) H' =2.522 103 kJ/kg

H2 H1) / (H' H1)

0.78

6.47 From steam table data estimate values for the residual properties VR _{, H}R _{, S}R _for

steam at 225 o_{C and 1600 kPa and compare with values found by a suitable}

generalized correlation? SOLUTION : P = 1600 kPa = 1.6 x 10 5 _{Pa = 1.6 atm} V = 132.85 cm3 _{/ gr} H = 2856.3 J / gr S = 6.5503 J / gr K Hig = 2928.7 J / gr Sig = 10.0681 J / gr K T = 225 o_{C = 498.15 K} Molwt = 18 gr / mol VR _{= V –} R molwt T P VR _{= 132.85 cm}3 _{/ gr –} 0.08206 atm dm 3 mol K 18 gr mol 498.15 K_{1.6 atm} VR _{= 132.85 cm}3 _{/ gr – 3633 cm}3 _{/ gr} VR _{= - 3500.15 cm}3 _{/ gr} HR _{=H - H} ig HR _{=2856.3 J / gm - 2928.7 J / gm} HR _{= - 72.4 J / gm} Δ Sig = _{molwt ln} −R _PoP

= −o .08206 atm dm3 mol K 18 gr mol ln 1600 kPa_{1 kPa} = - 0.0336 atm dm3_{gr K} = - 3.36 _{gr K}J SR _{= S – ( Sig + Δ Sig)} = 6.5503 _{gr K - ( 10.0681} J _{gr K - 3.36} J _{gr K )}J = - 0.2298 _{gr K} J Reduced condition ω = 0.345 Tc = 647.1 K Pc = 220.55 bar Tr = _{Tc =} T 498.15 K_{647.1 K} = 0.76982 Pr = _{Pr =} P 220.55 ¯¿ 1.6¯¿_¿ ¿ = 0.00725 Bo = 0.083 – 0.422_Tr1.6 = 0.083 – 0.422 0.769821.6 = - 0.558 Bi = 0.139 – 0.172 Tr4.2 = 0.139 – 0.172 0.769824.2 = - 0.377

Z = 1 + ( Bo + ω Bi ) Pr_Tr Z = 1 + ( -0.558 + 0.345 x -0.377 ) 0.00725_0.76982 Z = 1 – 0.00648 Z = 0.99352 VR ₌ R T P molwt ( Z – 1 ) VR ₌ 0.08206 atm dm3 mol K 498.15 K 1.6 atm18 gr mol ( 0.99352 – 1 ) VR _{=- 9.19755 cm}3 _{/ gr} HRB = - 0.17858 SRB = - 0.167101 HR ₌ R Tc molwt HRB HR ₌ 0.08206 atm dm 3 molK 647.1 K 18 gr mol -0.17858 HR _{=-52.643 J / gr} SR ₌ R molwt SRB SR ₌ 0.08206 atm dm 3 molK 18 gr mol -0.167101 SR _{= - 0.0076 J / gr K}

6.48 From data in the steam tables:

(a) Determine values for Gl _{and G}v _{for saturated liquid and vapor at 1000 kPa.}

Should these be the same?

(b) Determine values for ∆Hlv_{/ T and ∆S}lv _{at 1000 kPa. Should these be the}

same?

(c) Find values for VR_{, H}R _{, and S}R _{for saturated vapor at 1000 kPa.}

(d) Estimate a value for d P sat_{/dT at 1000 kPa and apply the Clapeyron}

equation to evaluate ∆Slv _{at 1000 kPa. Does this result agree with the}

steam-table value?

Apply appropriate generalized correlations for evaluation of VR_{, H}R _{, and S}R _for

saturated vapor at 1000 kPa. Do these results agree with the values found in (c)?

SOLUTION : So, ∆Vlv _{= V}v _{- V}l _{= 193.8 cm}3 _gr -1 _{- 1.128 cm}3 _gr -1 _{= 192.672 cm}3 _gr -1 ∆Hlv _{= H}v _{- H}l _{= 2776.3 J gr} -1 _{- 763.1 J gr} -1 _{= 2.0132 J gr} -1 ∆Slv _{= S}v _{- S}l _{= 6.5819 J gr}-1 _k-1 _{- 2.1393 J gr}-1 _k-1 _{= 4.4426 J gr}-1 _k-1 a. Gl _{= H}l _{– T. S}l _{= 763.1 J gr} -1 _{– 453.15 K x 2.1393 J gr}-1 _k-1 _{= -206.32 J gr}-1 Gv _{= H}v _{– T.S}v _{= 2776.3 J gr} -1 _{- 453.15 K x 6.5819 J gr}-1 _k-1 _{= -206.29 J gr}-1 b. ∆Hlv _{= 2.0132 J gr} -1 _{; T = 453.15 K} r = ∆Hlv _{/ T = 2.0132 J gr} -1 _{/ 453.15 K = 4.4427 J gr}-1 _k-1 ∆Slv _{= S}v _{- S}l _{= 6.5819 J gr}-1 _k-1 _{- 2.1393 J gr}-1 _k-1 _{= 4.4426 J gr}-1 _k-1 c. VR _{= V}v _{– ( R. T / molwt. P)} = 193.8 cm3 _gr -1 _{– ( 8.314 J mol}-1 _k-1 _{x 453.15 K / 18.015 gr mol}-1 _{x 1000} = -14.875 cm3 _gr-1

For enthalpy and entropy, assume that steam at 179.88 degC and 1 kPa is an ideal gas. By interpolation in table F.2 at 1 kPa

The enthalpy of an ideal gas is independent of pressure ; the entropy does depend on P : HR _{= H}v _{– H}

ig = 2776.3 J gr -1 - 2831.2 J gr-1 = -36.9 J gr-1

∆Sig = (-R / molwt ) x ln (P/P0) = ( -8.314 / 18.015) ln (1000/1) = -3.188 J mol-1 k-1 SR _{= Sv – Sig + ∆Sig = 6.5819 J gr}-1 _k-1 _{- 8.7994 J gr}-1 _k-1 _{+ 3.188 J mol}-1 _k-1

= -0.1126 J gr-1 _k-1

d. Assume ln P vs 1/T linear and fit three data pts @ 975, 1000, 1050 kPa

Data : pp: ( 975, 1000, 1050 ) kPa ; t: ( 178.79, 179.88, 182.02 ) (degC) i = 1..3 Xi = 1/ ti + 273.15 ; Yi = ln (ppi / kPa) ; slope = slope (x,y) slope = -4717 dPdT = (-P / T2_{) ln slope K = 22.984 kPa k}-1 ∆Slv _{= ∆V}lv _{. dPdT = 192.672 cm}3 _gr -1 _{x 22.984 kPa k}-1 _{= 4.4 J gr}-1 _k-1 Reduced conditions : ώ = 0.345 ; Tc = 647.1 K ; Pc = 220.55 bar Tr = T / Tc = 453.15 / 647.1 = 0.7001 K Pr = P / Pc = 10 / 220.55 = 0.0453 bar

The generalized virial-coefficient correlation is suitable here B0 = 0.083 – ( 0.422 / Tr 1.6 ) = -0.664

B1 = 0.139 – ( 0.172 / Tr 4.2 ) = -0.63

By Eqs (3.61) + (3.62) and (3.63) along with Eq. (6.40) Z = 1 + B0 + ώ . B1 . Pr / Tr = 1 + -0.664 + 0.345 . -0.63 . 0.0453 / 0.7001 = 0.943 VR _{= ( R.T / P. molwt ) Z-1} = ( 8.314 . 453.15 / 10. 18.015) 0.943 – 1 = -11.93 cm3 _gr-1 HR _{= ( R.Tc / molwt) (HRB. Tr)}

= -43.18 J gr-1

SR _{= R / molwt ( SRB. Tr)} = -0.069 J gr-1 _k-1

6.49 From data in the steam tables :

a) Determine values for Gl _{and G}v _{for saturated liquid and vapor at 150(psia).}

Should these be the same?

b) Determine values for ΔHlv _{/T and ΔS}lv _{at 150(psia). Should these be the same?}

c) Find values for VR_{, H}R_{, and S}R _{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). Does this result agree with the steam-table value?}

Apply appropriate generalized correlations for evaluation of VR_{, H}R_{, and S}R _for

saturated vapor at 150(psia). Do these results agree with the values found in (c)? T= (358,43+459,67) Rankine = 818,1 Rankine

SOLUTION :

Dari tabel F.4 didapat :

Vl = 0,0181 ft3/lbm Vv = 3,014 ft3/lbm Hl = 330,65 btu/lbm Hv = 1194,1 btu/lbm

Sl = 0,5141 btu/lbm rankine Sv = 1,5695 btu/lbm rankine

Vlv = Vv - Vl

= (3,014 – 0,0181) ft3_/lbm

Hlv = Hv - Hl

= (1194,1 – 330,65) btu/lbm = 863,45 btu/lbm

a. Gl=Hl - TSl

= 330.65 btu/lbm - (818,1 R . 0,5141 btu/lbm Rankine) = -89,935 btu/lbm

Gv =Hv

- TSv

= (1194,1 btu/lbm) – (818,1 R . 1,5695 btu/lbm Rankine) = -89,91 btu/lbm

= (1,5695 – 0,5141) btu/lbm R = 1,0554 btu/lbm R . ∆ H_lv T = 863,45 btu/lbm 818,1 R = 1,0554 btu/lbm R c. VR=VV− RT BM . P ¿3.014 ft3_/lbm−10.73 ft3psia/lbmol R . 818.1 R 18.015lbm /lbmol .150 psia ¿−0.2345 ft 3 lbm

Untuk entalpi dan entropi asumsikan sistem pada 358.43oF dan 1 psi adalah gas ideal. Dengan interpolasi pada tabel F.4 pada 1 psi, didapat:

T −T_b Ta−Tb =Hig−Hb Ha−Hb T −T_b Ta−Tb =Sig−Sb Sa−Sb 358.43−350 400−350 = H_ig−1218.7 1241.8−1218.7 358.43−350 400−350 = S_ig−2.1445 2.1722−2.1445 H_ig=1222.6Btu lbm Sig=2.1492 Btu lbm R Po = 1 psi

Entalpi gas tidak bergantung pada tekanan, tetapi entropi selalu tergantung pada tekanan, sehingga, HR =Hv −Hig lbm−1222.6 btu/¿lbm ¿1194.1btu/¿ = -28.5 btu/lbm

∆ Sig= −R BM ln

(

P P_o

)

¿−1.987 btu/lbmol R 18.015 lbm/lbmol ln 150 1 = - 0.552 btu/lbmol R S_R=S_v−

₍

S_ig−∆ S_ig

)

¿_{1.5695 btu/lbmol R−(2.1492−0.552)btu/lbmol R} = - 0,0277 btu/lbmol R

d. Asumsikan ln P vs _{T linear untuk data (145, 150, dan 155 psia).}1

Data :

[

145 150 155

]

psi t =

[

355,77 358,43 361,02

]

0_F i=1, 2,3 . Xi= 1 ti+459,67 . X1= 1 355,77+459,67=1,23 .10 −3 . X2= 1 358,43+459,67=1,22. 10 −3 . X3= 1 361,02+459,67=1,218 .10 −3

. y=ln

(

Ppi_Psi

)

. y1=ln

(

145 150

)

= -0,0339