For a mixture of 15 mol % ethanol and 85 mol % water at 50 degree C, calculate t

Question

For a mixture of 15 mol % ethanol and 85 mol % water at 50 degree C, calculate the bubble point pressure and the dew point pressure. If you want to flash this mixture at 50 degree C to get a vapour with a high ethanol concentration, would you operate the flash near the dew point pressure or the bubble point pressure?, explain. Antoine constants for water: A = 8.10765, B = 1750.29, C = 235; P^degree = 10^(A-B/(T+C)) P^degree in mmHg, T in Celsius. Antoine constants for ethanol: A = 8.1122, B = 1592.864, C = 226.184; P^degree = 10^(A-B/(T+C)) P^degree in mmHg, T in Celsius.

Explanation / Answer

Solution:

We know, bubble point pressure can be given by the following equation:

i xipsati = p ------------ (1)

Since T is given, we can simply calculate p from (1). We start by computing the vapor pressures for the three components at T = 273+50 = 323K. Using the Antoine data, we get:

psat1 (323K) = 104.97 = 93325.4 mmHg

psat2 (323K) = 105.21 = 162181.0 mmHg

At the bubble point, the liquid phase composition is given, so the partial pressure of each component can be calculated as follows:

p1 = x1psat1 = 0.85x93325.4 = 79326.59 mmHg

p2 = x2psat2 = 0.15x162181.0 = 24327.15 mmHg

Thus, from (1) the bubble pressure is p = p1 + p2 = 103653.74 mmHg

In this case, p and xi are given, so we can find the vapor composition as and

y1 = p1/p = 0.765

y2 = p2/p = 0.234

We know, dew point pressure can be given by the following equation:

i yi/psati = 1/p ------------ (2)

Since T is given, we can simply calculate 1/p from (2). With the data from the above we get:

1/p = 0.765/93325.4 + 0.234/162181.0

1/p = 8.2x10-6 + 1.44x10-6

1/p = 9.64x10-6

P = 1x105 mmHg

I would operate near the dew point pressure since at the dew point pressure the pressure exerted by the ethanol is of the order of 106 (higher concentration), which is higher than that at the bubble point pressure.

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