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Assuming normal heptane and normal octane to form ideal solutions, do the following:
d) Calculate the dew pressure of the solution at 40 ◦C.
e) Calculate the bubble temperature at 1 bar.
f) Calculate the dew temperature at 1 bar.
g) Calculate the amount and composition of vapor and liquid when a solution with z1 = 0.45 is flashed to 40 ◦C, 0.065 bar. The saturation pressures of the two components are given by the Antoine equation:
ln Pi sat = Ai − Bi/(Ci + T)
where P sat is in mm-Hg, t is in Kelvin, and the parameters
Ai , Bi , Ci , are
nC8: A = 15.9426 B = 3120.29 C = –63.63
nC7: A = 15.8737 B = 2911.32 C= –56.51
Given sata
xf = 0.45
Mixture of n-heptane and n-octane
n-heptane is more volatile than n-octane
Antonie constants are given in question
For n-heptane
A: 15.8737
B: 2911.32
C: -56. 51
For n-octane
A: 15.9426
B: 3120.29
C: -63. 63
ln Pi= A-(B/(T+C) )
P is in mmHg and T is in K
D) dew pressure at 40°C
T = 40°C = T = 313 K
At dew point x = 1
P = 44.0541 mmHg
Dew pressure = 44.9541 mmHg = 0.0599 bar
E)
Bubble temperature at 1 bar
At bubble point
Subsituting P = 1 bar
1 bar = 750.062 mmHg
Solving we get
T = 383.6 K
Bubble temperature = 383.6 K = 110.45°C
F) dew temperature at 1 bar
At dew point
Solving we get
T = 388.3 K = 115.15°C
g)
P = 0.065 bar
T = 40 °C
Dew pressure at T = 40°C is 0.0599 bar
Bubble pressure at 40°C =
Bubble pressure = 58.433 mmHg = 0.07790 bar
Operating pressure = 0.065 bar
(V/L) =( 0.065-0.0599) /(0.07790-0.065)
V/L = 0.3953
Basis
F = 100 mol
Material balance
F = V+ L
V/L = 0.3953
V = 0.3953 L
100 = V+L
100 = 0.3953L+L
L = 71.669 mol
V = 28.330 mol
Applying Raoults law
yP = x(V.P)
For heptane
V. P for heptane at 40°C = 92.129 mmHg = 0.1228 bar
y(0.065) = x(0.1228)
y = 1.889x
Doing component balance
100(0.40) = V(y) +L(x)
100(0.40) = 28.330(1.889x) +71.669(x)
x = 0.3195
y = 1.889(0.3195) = 0.6035
Composition of liquid
L = 71.669 mol
Heptane- 31.95%
Octane -68. 05%
Composition of vapor
V = 28.330
Heptane - 60.35%
Octane- 39.65%
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