Given information:
It is given that ethane expands isentropically from 30 bar and 493.15 Kto 2.6 bar .
Process is isentropically, so ΔS=0
From equation (2),
ΔS=∫T1T2CPigdTT−Rln P 2 P 1+S2R−S1RR∫T1T2CP igRdTT−RlnP2P1+S2R−S1R=0R∫T1T2CPigRdTT=RlnP2P1−S2R+S1R
Initial statE
For pure species ethane, the properties can be written down using Appendix B, Table B.1
ω=0.1, Tc=305.3 K, Pc=48.72 bar, ZC=0.279
Tr=T1TcTr=493.15 K305.3 K=1.615
Pr=P1PcPr=30bar48.72 bar=0.61576
So, at above values of Tr and Pr, The values of ( H R)RTC0 and ( H R)RTC1 can be written from Appendix D
Tr=1.615 lies between reduced temperatures Tr=1.6and Tr=1.7 and Pr=0.61576 lies in between reduced pressures Pr=0.6 and Pr=0.8 .
At Tr=1.6 and Pr=0.6
( H R)RTC0=−0.261, ( S R)R0=−0.12
At Tr=1.6 and Pr=0.8
( H R)RTC0=−0.35, ( S R)R0=−0.162
At Tr=1.7 and Pr=0.6
( H R)RTC0=−0.231, ( S R)R0=−0.102
At Tr=1.7 and Pr=0.8
( H R)RTC0=−0.309, ( S R)R0=−0.137
And
At Tr=1.6 and Pr=0.6
( H R)RTC1=−0.013, ( S R)R1=−0.057
At Tr=1.6 and Pr=0.8
( H R)RTC1=−0.011, ( S R)R1=−0.073
At Tr=1.7 and Pr=0.6
( H R)RTC1=0.009, ( S R)R1=−0.044
At Tr=1.7 and Pr=0.8
( H R)RTC1=0.017, ( S R)R1=−0.056
Applying linear interpolation of two independent variables, From linear double interpolation, if M is the function of two independent variable X and Y then the value of quantity M at two independent variables X and Y adjacent to the given values are represented as follows:
X1 X X2Y1 M1,1 M1,2Y M=? Y2 M2,1 M2,2
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( HR )RTC0=[( 0.8−0.61576 0.8−0.6)×−0.261+( 0.61576−0.6 0.8−0.6)×−0.35] ×1.7−1.6151.7−1.6 +[( 0.8−0.61576 0.8−0.6)×−0.231+( 0.61576−0.6 0.8−0.6)×−0.309] ×1.615−1.61.7−1.6( HR )RTC0=−0.263
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( SR )R0=[( 0.8−0.61576 0.8−0.6)×−0.12+( 0.61576−0.6 0.8−0.6)×−0.162] ×1.7−1.6151.7−1.6 +[( 0.8−0.61576 0.8−0.6)×−0.102+( 0.61576−0.6 0.8−0.6)×−0.137] ×1.615−1.61.7−1.6( SR )R0=−0.1205
And
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( HR )RTC1=[( 0.8−0.61576 0.8−0.6)×−0.013+( 0.61576−0.6 0.8−0.6)×−0.011] ×1.7−1.6151.7−1.6 +[( 0.8−0.61576 0.8−0.6)×0.009+( 0.61576−0.6 0.8−0.6)×0.017] ×1.615−1.61.7−1.6( HR )RTC1=−0.0095
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( SR )R1=[( 0.8−0.61576 0.8−0.6)×−0.057+( 0.61576−0.6 0.8−0.6)×−0.073] ×1.7−1.6151.7−1.6 +[( 0.8−0.61576 0.8−0.6)×−0.044+( 0.61576−0.6 0.8−0.6)×−0.056] ×1.615−1.61.7−1.6( SR )R1=−0.056
Now, from equation,
H1R=(H1R)0+ω(H1R)1
Or
H1RRTC=( H 1 R)0RTC+ω( H 1 R)1RTCH1RRTC=−0.263+0.1×−0.0095H1RRTC=−0.26395H1R=−0.26395×8.314 Jmol K×305.3 KH1R=−669.97 Jmol
And
S1R=(S1R)0+ω(S1R)1
Or
S1RR=( S 1 R)0R+ω( S 1 R)1RS1RR=−0.1205+0.1×−0.056S1RR=−0.1261S1R=−0.1261×8.314 Jmol KS1R=−1.048 Jmol K
Final state
Final state temperature is calculated in part (a), T=367.4 K
i A 103B 106 C 10−5DEthane 1.131 19.225 −5.561 0
Tr=T2TcTr=367.4 K305.3 K=1.204
Pr=P2PcPr=2.6 bar48.72 bar=0.0534
So, at above values of Tr and Pr, The values of ( H R)RTC0 and ( H R)RTC1 can be written from Appendix D
Tr=1.204 lies between reduced temperatures Tr=1.2and Tr=1.3 and Pr=0.0534 lies in between reduced pressures Pr=0.05 and Pr=0.1 .
At Tr=1.2 and Pr=0.05
( H R)RTC0=−0.036, ( S R)R0=−0.021
At Tr=1.2 and Pr=0.1
( H R)RTC0=−0.073, ( S R)R0=−0.042
At Tr=1.3 and Pr=0.05
( H R)RTC0=−0.031, ( S R)R0=−0.017
At Tr=1.3 and Pr=0.1
( H R)RTC0=−0.063, ( S R)R0=−0.033
And
At Tr=1.2 and Pr=0.05
( H R)RTC1=−0.02, ( S R)R1=−0.019
At Tr=1.2 and Pr=0.1
( H R)RTC1=−0.04, ( S R)R1=−0.037
At Tr=1.3 and Pr=0.05
( H R)RTC1=−0.013, ( S R)R1=−0.013
At Tr=1.3 and Pr=0.1
( H R)RTC1=−0.026, ( S R)R1=−0.026
Applying linear interpolation of two independent variables, From linear double interpolation, if M is the function of two independent variable X and Y then the value of quantity M at two independent variables X and Y adjacent to the given values are represented as follows:
X1 X X2Y1 M1,1 M1,2Y M=? Y2 M2,1 M2,2
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( HR )RTC0=[( 0.1−0.0534 0.1−0.05)×−0.036+( 0.0534−0.05 0.1−0.05)×−0.073] ×1.3−1.2041.3−1.2 +[( 0.1−0.0534 0.1−0.05)×−0.031+( 0.0534−0.05 0.1−0.05)×−0.063]×1.204−1.21.3−1.2( HR )RTC0=−0.0383
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( SR )R0=[( 0.1−0.0534 0.1−0.05)×−0.021+( 0.0534−0.05 0.1−0.05)×−0.042] ×1.3−1.2041.3−1.2 +[( 0.1−0.0534 0.1−0.05)×−0.017+( 0.0534−0.05 0.1−0.05)×−0.033]×1.204−1.21.3−1.2( SR )R0=−0.022
And
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( HR )RTC1=[( 0.1−0.0534 0.1−0.05)×−0.02+( 0.0534−0.05 0.1−0.05)×−0.04] ×1.3−1.2041.3−1.2 +[( 0.1−0.0534 0.1−0.05)×−0.013+( 0.0534−0.05 0.1−0.05)×−0.026]×1.204−1.21.3−1.2( HR )RTC1=−0.02106
M=[( X2 −X X2 −X1 )M1,1+( X−X1 X2 −X1 )M1,2]Y2−YY2−Y1 +[( X2 −X X2 −X1 )M2,1+( X−X1 X2 −X1 )M2,2]Y−Y1Y2−Y1( SR )R1=[( 0.1−0.0534 0.1−0.05)×−0.019+( 0.0534−0.05 0.1−0.05)×−0.037] ×1.3−1.2041.3−1.2 +[( 0.1−0.0534 0.1−0.05)×−0.013+( 0.0534−0.05 0.1−0.05)×−0.026]×1.204−1.21.3−1.2( SR )R1=−0.02
Now, from equation,
H2R=(H2R)0+ω(H2R)1
Or
H2RRTC=( H 2 R)0RTC+ω( H 2 R)1RTCH2RRTC=−0.0383+0.1×−0.02106H2RRTC=−0.0404H2R=−0.0404×8.314 Jmol K×305.3 KH2R=−102.561 Jmol
And
S2R=(S2R)0+ω(S2R)1
Or
S2RR=( S 2 R)0R+ω( S 2 R)1RS2RR=−0.022+0.1×−0.02S2RR=−0.024S2R=−0.024×8.314 Jmol KS2R=−0.1995 Jmol K
Now,
∫T1T2 C P igRdTT=ln P 2 P 1−S2RR+S1RR∫T1T2CPigdTT=ln2.630−(−0.1995 Jmol K)8.314 Jmol K+(−1.048 Jmol K)8.314 Jmol K∫T1T2CPigdTT=−2.5477
Hence,
∫T1T2CPigRdTT=[A+{BT0+(CT02+Dτ2T02)(τ+12)}(τ−1lnτ)]×lnτ
τ=TT0
∫T1T2 C P igRdTT=[A+{BT0+(CT02+D τ2 T0 2 )( τ+12)}(τ−1lnτ)]×lnτ∫T1T2CP igRdTT= [1.131+{19.225×10−3×493.15+(−5.561×10−6×493.152+0 τ 2× 493.15 2)(τ+12)}(τ−1lnτ)]×lnτ−2.5477= [1.131+{19.225×10−3×493.15+(−5.561×10−6×493.152)(τ+12)}(τ−1lnτ)]×lnττ=0.7352
τ=TT00.7352=T493.15KT=362.56 K
Now, for work produced
W=ΔH
And enthalpy change by generalized correlations is given by,
ΔH=∫T1T2CPigdT+H2R−H1R
∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)
Where τ=TT0
∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫T0TΔC∘PRdT =1.131×493.15×(0.735−1)+19.225×10−32×493.152×(0.7352−1)+−5.561×10−63×493.153(0.7353−1)∫T0TΔC∘PRdT=−1088.593 K
Hence,
W=ΔH=R∫T1T2 C P igRdT+H2R−H1RW=8.314 Jmol K×−1088.593 K+(−102.561 Jmol)−(−669.97 Jmol)W=−8453.15 Jmol
Hence,
W=ΔH=R∫T1T2 C P igRdTW=8.314 Jmol K×−1425.78 KW=11853.93 Jmol