ME 501 Advanced Thermodynamics
Solution for Assignment#4
Prob#1
The following is the composition of acid which is vaporized and burnt in a hazardous
waste plant: H2SO4 : 92 %, Hydrocarbons : 4 %, H2O : 4%(mass). Lump hydrocarbons
with water. The vapor pressure relations are as follows: ln (p)= A - B/(T(K)+C), p in bar
>the A, B and C are as follows: Water: 11.9559, 3984.849, -39.4856; 8.346772,
4240.275, -119.155 for H2SO4. Determine the vapor phase mole fraction of each
component at a pressure of 1 bar if T = 100 C. Assume ideal solution, b) Will the vapor
phase composition in mole % change if N2 is present in the vapor phase at same P of 1
bar and T= 100 C. If so, estimate them. Assume ideal solution, c) If pH2O and pH2SO4 at
270 C for a strong acid are given as 0.335 bar and 0.0525 bar, estimate the activity
coefficients for H2O and H2SO4. Strong acid  95 % of H2SO4 by mass
Solution
Ideal solution model
YH2SO4 () = 92 %, YH2O= 0.08; MH2SO4 = 98.1MH2O = 18.02; XH2SO4 =
92/98.1/(92/98.1+8/18.02)=0.938/(0.938+0.444)= 0.679,XH2O = 0.321,Mmix =
72.36 kg/kmol
Let (1)beH2SO4, (2) be H2O
P1ideal= x1liq  p1 sat (373) =0.679 exp ( 8.346772 - 4240.275/(373 -119.155))=
0.00237bar,
Other results are given below.
X2(liq)
Pideal Pnonid
p1ideal
p2ideal
p1nonid
3.21E-01
3.26E-01 8.71E-04 1.61E-04
3.26E-01
1.19E-04
p2nonid
7.52E-04
X2id
1.00E+00
X2nonid
8.64E-01
Non-ideal case
From Text
Recall that
(X1() X2() R T/ g E )= B´ + C´ (X1() - X2())
Rewrite (A) as
( g E / X1() X2() R T)= AB /(
(A)
AX1() + B X2())
N g = G = R T ( N1() N2() )/ /( AN1() + B N2())
Differentiate
( GE /N1)(1/RT) = ln 1 = N2() /( AN1() + B N2()) - A ( N1() N2() )/ /( AN1() +
B N2()) 2
= X2() /( AX1() + B X2()) - A ( X1() X2() )/ /( AX1() + B X2()) 2
(B)
E
( G /N2) (1/RT) = ln 2 = N1() /( AN1() + B N2()) - B ( N1() N2() )/ /( AN1() +
B N2()) 2
= X1() /( AX1() + B X2()) - B ( X1() X2() )/ /( AX1() + B X2()) 2
(C)
E
E
Solve for A and B from Eqs. (B) and (C)
ln 1 = A X2()2/((A/B) X1() + X2())2, ln 2 = B X1()2/((X1()
+ (B/A) X2())2
At 95 % H2SO4, XH2SO4(liq) =
(D)
0.777288413, XH2O(liq) = 0.222711587
P1 non-deal= 1 x1liq  p1 sat (503)= 1 p1ideal (543, Y=xH2SO4= 0.78) where p1ideal =
0.777 exp ( 8.346772 - 4240.275/(543 -119.155))
P1ideal
= 0.148683064, p2ideal=
12.70896563 bar
0.0525/0.149 = 1 ( 544,0.,78)
gamma1 = 0.353100069, gamma2= 0.026359344. Using these gammas,
A = ln 1() (1+ (X2() ln 2())/(X1() ln 1()))2,
(1)
B= ln 2() (1+ (X1() ln 1())/(X2() ln 2()))2.
(2)
One can obtain A and B as -14.53288172, -4.167124513
Using Eqs. (51a,b) in
ln 1 = A(T) X2()2/((A(T)/B(T)) X1() + X2())2, ln 2 = B(T) X1()2/((X1()
+ (B(T)/A(T)) X2())2,
Treating A as constant and with X1liq = 0.679, X2liq = 0.321, 1 and 2 are
obtained.
p1nonid = 1  p1ideal ( 100 C, XH2SO4 =0.679) = 1.19E-04
Results below:
X2(liq)
Pideal
Pnonid
p1ideal
p2ideal
1
3.21E-01
3.26E-01
1.30E-02
1.61E-04
3.95E-02
8.13E-01
2
p1nonid
p2nonid
X2id
X2nonid
3.95E-02
1.31E-04
1.29E-02
1.00E+00
9.90E-01
Prob#2
Suppose we have 7 g mole of methanol (1) ( liquid + vapor) and 3 g mole of water (2)
(liquid + vapor) in a piston cylinder assembly at 60 C, 433 kPa. At T= 60 C, p 1sat = 625
mm of Hg, p2sat = 144 mm of Hg. Determine
a) x1, x2, Y1, Y2
b) vapor fraction or quality ¯,w
c) moles of vapor of species 1 and 2
Solution
a) P = p1 + p2 = x1 p1 sat + x2 p2 sat = x1 p1 sat + (1-x1 ) p2 sat
Solving for x1 = 0.6, x2 = 0.4
Y1 = p1 /P =x1 p1 sat /P = 0.83, Y2 = 0.13
b)
¯,w= Ng /N ?
N1 = N1 + N1g
Dividing by N
Z1 = N /N + N1g/N
= (N /N ) (N /N) + (N1g /Ng) (Ng/N)
0.7
= x1 (1- ¯,w) + Y1 ¯,w = 0.4 (1-¯,w) + 0.83 ¯,w,
Solving for ¯,w = 0.698
c)
Ng = 0.698  10 = 6.98 g moles, liquid moles will be 10 -6.98 =
3.02
N = 0.6  3.02 =1.81, Similarly N = 1.21,
N1g = 0.83  6.98 = 5.79, N2g = 1.19.
Prob#3
Determine the maximum temperatures to which liquid water can be superheated and
vapor can be subcooled at 133 bar.
Solution
a)
P = RT/(v-b) - a /(Tnv2)
At points M and G,
∂P/∂v = - RT/(v-b)2 + 2a/(Tn v3) = 0
RT(n+1) v3 = 2a (v-b)2
Solve for v at given T. Obtain the two solutions vM, vG. Knowing vM and vG we
can obtain the spinodal pressures PM and PG from Eq.(A).
We can also get the spinodal curve P vs. v by eliminating T between equations
(B) and (A).
Thus for the case n = 0 (VW equation of state),
P = a(v-2b)/v3
(Note :If v < 2b, then P < 0 from Eq. (B). The negative pressure implies tensile
stress on the fluid. Recall that b = vc /3 for VW; thus for v > 2vc/3, P > 0 ;
otherwise tensile stress may be existing).
b) In reduced form for VW
TR = 54/64 ( vR' - 0.125)2 / vR' 3
and
PR = ((27/64)/vR' 2 ) (1 - (1/(4 vR')))
See Fig. for the spinodal curve based on VW equation of state.
c) ln PR sat
- 1/TR);
TR = 7/(7- ln PR sat) = 7/(7- ln (133/220.9)) = 0.932
T = TR Tc = 0.932647.3 = 603.6 K
P = 133 bar, vA = 0.069, vB = 0.16 m3 /kmol, TA = 600 K (?), TB = 545 K, Tsat =
604 K according VW equation of state. Tsuperheat =, Tsubcool = -55 K
(A)
(B)