Given information:
It is given that n-butane gas compressed isentropically from 1 bar and 323.15 K to 7.8 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 n-butane, the properties can be written down using Appendix B, Table B.1
ω=0.2, Tc=425.1 K, Pc=37.96 bar, ZC=0.276
Tr=T1TcTr=323.15 K425.1 K=0.7602
Pr=P1PcPr=1 bar37.96 bar=0.0263
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=0.76 lies between reduced temperatures Tr=0.75 and Tr=0.8 and Pr=0.026 lies in between reduced pressures Pr=0.01 and Pr=0.05 .
At Tr=0.75 and Pr=0.01
( H R)RTC0=−0.017, ( S R)R0=−0.015
At Tr=0.75 and Pr=0.05
( H R)RTC0=−0.088, ( S R)R0=−0.078
At Tr=0.8 and Pr=0.01
( H R)RTC0=−0.015, ( S R)R0=−0.013
At Tr=0.8 and Pr=0.05
( H R)RTC0=−0.078, ( S R)R0=−0.064
And
At Tr=0.75 and Pr=0.01
( H R)RTC1=−0.027, ( S R)R1=−0.029
At Tr=0.75 and Pr=0.05
( H R)RTC1=−0.142, ( S R)R1=−0.156
At Tr=0.8 and Pr=0.01
( H R)RTC1=−0.021, ( S R)R1=−0.022
At Tr=0.8 and Pr=0.05
( H R)RTC1=−0.11, ( S R)R1=−0.116
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=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( H R )RTC0=[( 0.05−0.026 0.05−0.01)×−0.017+( 0.026−0.01 0.05−0.01)×−0.088] ×0.8−0.760.8−0.75 +[( 0.05−0.026 0.05−0.01)×−0.015+( 0.026−0.01 0.05−0.01)×−0.078] ×0.76−0.750.8−0.75( H R )RTC0=−0.04436
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( S R )R0=[( 0.05−0.026 0.05−0.01)×−0.015+( 0.026−0.01 0.05−0.01)×−0.078] ×0.8−0.760.8−0.75 +[( 0.05−0.026 0.05−0.01)×−0.013+( 0.026−0.01 0.05−0.01)×−0.064] ×0.76−0.750.8−0.75( S R )R0=−0.03884
And
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( H R )RTC1=[( 0.05−0.026 0.05−0.01)×−0.027+( 0.026−0.01 0.05−0.01)×−0.142] ×0.8−0.760.8−0.75 +[( 0.05−0.026 0.05−0.01)×−0.021+( 0.026−0.01 0.05−0.01)×−0.110] ×0.76−0.750.8−0.75( H R )RTC1=−0.06972
M=[( X 2 −X X 2 − X 1 )M1,1+( X− X 1 X 2 − X 1 )M1,2]Y2−YY2−Y1 +[( X 2 −X X 2 − X 1 )M2,1+( X− X 1 X 2 − X 1 )M2,2]Y−Y1Y2−Y1( S R )R1=[( 0.05−0.026 0.05−0.01)×−0.029+( 0.026−0.01 0.05−0.01)×−0.156] ×0.8−0.760.8−0.75 +[( 0.05−0.026 0.05−0.01)×−0.022+( 0.026−0.01 0.05−0.01)×−0.116] ×0.76−0.750.8−0.75( S R )R1=−0.07576
Now, from equation,
H1R=(H1R)0+ω(H1R)1
Or
H1RRTC=( H 1 R )0RTC+ω( H 1 R )1RTCH1RRTC=−0.04436+0.2×−0.06972H1RRTC=−0.058304H1R=−0.058304×8.314 Jmol K×425.1 KH1R=−206.063 Jmol
And
S1R=(S1R)0+ω(S1R)1
Or
S1RR=( S 1 R )0R+ω( S 1 R )1RS1RR=−0.03884+0.2×−0.07576S1RR=−0.053992S1R=−0.053992×8.314 Jmol KS1R=−0.4489 Jmol K
Considering at final state gas is an ideal gas.
Hence, H2R=0 and S2R=0
Now,
∫T1T2 C P igRdTT=ln P 2 P 1−S2RR+S1RR∫T1T2CP igRdTT=ln7.81−0+(−0.4489 Jmol K)8.314 Jmol K∫T1T2CPigRdTT=2
Hence,
∫T1T2CPigRdTT=[A+{BT0+(CT02+Dτ2T02)(τ+12)}(τ−1lnτ)]×lnτ
τ=TT0
Values of constants forn-butane in above equation are given in appendix C table C.1 and noted down below:
i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0
τ=TT0
∫T1T2 C P igRdTT=[A+{BT0+(CT02+D τ 2 T 0 2 )( τ+12)}(τ−1lnτ)]×lnτ∫T1T2CP igRdTT= [1.935+{36.915×10−3×323.15+(−11.402×10−6×323.152+0 τ 2× 323.15 2)(τ+12)}(τ−1lnτ)]×lnτ2=[1.935+{36.915×10−3×323.15+(−11.402×10−6×323.152)(τ+12)}(τ−1lnτ)]×lnττ=1.18
τ=TT01.18=T323.15KT=381.317 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 =1.935×323.15×(1.18−1)+36.915×10−32×323.152×(1.182−1)+−11.402×10−63×323.153(1.183−1)∫T0TΔC∘PRdT=−786.41 K
Hence,
W=ΔH=R∫T1T2 C P igRdT+H2R−H1RW=8.314 Jmol K×−786.41 K+0−(−206.063 Jmol)W=−6332.15 Jmol