Question

For ethanol/water (70;30) determine;

A) the bubble point temperature if p=10 bar

B) the bubble point temperature if p=100 bar

Answer 1

A) the bubble point temperature if p=10 bar

B) the bubble point temperature if p=100 bar

The bubble point is the temperature at which a liquid mixture will start to boil. As with a dew point calculation, we can use Dalton’s and Raoult’s Laws to calculate the bubble point. By combining the two equations, we can calculate the vapour mole fractions for a given liquid composition, i.e.:

Yi = (Xi * Pi°) / P

Calculating the bubble point is iterative.The vapour pressure of each component can be estimated using the Antoine Equation.

log (P°) = a - (b / (T (°C) +c))

For ethanol:

a = 5.33675

b= 1648.220

c= 230.918

For water:

a= 5.11564

b= 1687.537

c= 230.170

We have the liquid molar fractions X but we need the molar fractions of the vapour ‘Y’. Also, we know that Y1 + Y2 = 1.

A ) We need a first guess to the temperature, so let’s look into the water vapour tables. At 10 bar, the temperature of saturated water vapour is around 180 °C but we also know that the boiling point of ethanol at atmospheric pressure is under 80 °C. I’ll try 160 °C:

Ethanol

log (P°) = 5.33675 - (1648.220 / (160 + 230.918)) = 1.12

P° = 13.18 bar

Water

log (P°) = 5.11564 - (1687.537 / (160 + 230.170)) = 0.79

P° = 6.17 bar

Yet = 0.70 * 13.18 bar / 10 bar = 0.923

Y wat = 0.3 * 6.17 / 10 bar = 0.184

Y = 0.923 + 0.184 = 1.12 > 1 We need to try a lower T.

Let’s se 155°C:

Ethanol

log (P°) = 5.33675 - (1648.220 / (155 + 230.918)) = 1.07

P° = 11.75 bar

Water

log (P°) = 5.11564 - (1687.537 / (155 + 230.170)) = 0.73

P° = 5.37 bar

Yet = 0.70 * 11.75 bar / 10 bar = 0.822

Y wat = 0.3 * 5.37 / 10 bar = 0.161

Y = 0.983 < 1

Use both data to interpolate the result:

T = 160 °C, Y = 1.12

T = 155 °C, Y = 0.983

T = 155.6 °C

B) Do exactly the same procedure and:

T = 273 °C

If you have a scientific calculator, you also can input all the equations and let a program solve it.