cpp - bubble and dew points for ideal systems (1)

• Plot y against x for more volatile component at fixed Ptotal

Vapour-Liquid EquilibriumK-values and Relative Volatility

xA

yA

0 1

1 equilibrium

curve

y = x

x1

y1

AAB

AABA x1-α1

xαy

Higher α gives more easy separation by distiller

• Also, no such thing as “boiling point” any more

Vapour-Liquid EquilibriumK-values and Relative Volatility

Temperature (ºC)

0.0 1.0Mole fraction of more volatile component

Cool a vapour sample of 40% benzene, 60% toluene

Boiling point of pure toluene

Boiling point of pure benzene

Starts to condense here

Condensation now complete

Continues to cool as liquid

Dew point

Bubble point

• Liquid has N components and total pressure Ptotal

• Defining bubble point mathematically:

Vapour-Liquid EquilibriumBubble point calculations

1xKN

1iii

K-value of component i at this temperature/pressure/composition

mole fraction of component i in liquid phase

• For ideal liquid phase, Ki = f (T, P, composition) only

• Calculation of bubble point is stepwise procedureGuess a

temperatureCalculate K1,

K2, K3 etcDoes (K1x1)+(K2x2)+

…=1?

Pick a new temperature SuccessNO

YES

Vapour-Liquid EquilibriumBubble Point Example

• Find bubble point to nearest 0.1ºC

• Vapour of 80 mol% benzene and 20 mol% toluene

• Pressure is 1000 mmHg

• From now on call benzene (1) and toluene (2)

• SOLUTION assuming this binary combination is ideal in both vapour and liquid phases

• First we need to guess our first temperature

• Rather than a random guess, make an educated estimate…


• Find boiling points of pure components at 1000 mmHg

• Use Antoine equation

• A1 = 6.90565 B1 = 1211.033 C1 = 220.790A2 = 6.95464 B2 = 1344.800 C2 = 219.482

(1) log10 (1000) = 6.90565 – 1211.033 (t + 220.79) t = 89.38°C

(2) log10 (1000) = 6.95464 – 1344.8 (t + 219.482) t = 120.6°C

• Make first guess at temperature to be weighted by mole%:

i.e.0.8(89.3) + 0.2(120.6) = 95.6°C


• Calculate both values for Pi0 at this temperature:

log10 (P10)= 6.90565 – 1211.033 (95.6 + 220.79)

P10 = 1197 mmHg

log10 (P20)= 6.95464 – 1344.8 (95.6 + 219.482)

P20 = 486 mmHg

• Both liquids assumed ideal so:

total

0i

i P

PK 197.1

0010

1197K1 486.0

0001

486K2

0548.10.4860.21.1970.8xK2

1iii


• Answer gave > 1, i.e. too high

• Need lower K-values, ie. a lower temperature

• Not below 89.3°C (pure benzene boils); try t =93°C


log10 (P10)= 6.90565 – 1211.033 (93 + 220.79)

P10 = 1112 mmHg

4477.00001

447.7K2

9795.04477.02.0112.18.0xK2

1iii

112.10010

1112K1

log10 (P20)= 6.95464 – 1344.8 (93 + 219.482)

P20 = 447.7 mmHg


• Answer gave < 1, i.e. too lowNeed higher K-values, ie. a higher temperature

• Pick random temperature between 93°C and 95.6°C ?

• Better to use a method to seek next guessLinear interpolation suggests the following:

Temperature (ºC)

Σ(Kixi)

93 95.6

1.0548

0.9795

1.0000

• Model curve as straight line between the two known points

• Straight line gradient is:

39t

0.9795-1.0000

396.59

0.9795-1.0548

t



log10 (P10)= 6.90565 – 1211.033 (93.7 + 220.79)

P10 = 1135 mmHg

log10 (P20)= 6.95464 – 1344.8 (93.7 + 219.482)

P20 = 457.8 mmHg

C7.930.9795-1.0548

0.9795-1.0000396.5993t

4578.00001

457.8K2

9993.04578.02.0135.18.0xK2

1iii

135.10010

1135K1

• Effectively converged on 1 – bubble point = 93.7ºC


• Also interesting to know vapour composition at equilibrium with this liquid, ie. that of the first few bubbles

i

ii x

yK yi = Kixi

y1 = 0.8 × 1.135 = 0.908 y2 = 0.2 × 0.4578 = 0.092 y1 + y2 = 0.908 + 0.092 = 1.000

• Vapour at equilibrium with liquid at 1000 mmHg and 93.7°C is 91 mol% benzene and 9 mol% toluene

Vapour-Liquid EquilibriumDew Point Example

• Now find dew point to nearest 0.1ºC for same mixture at 1000 mmHg

• SOLUTION assuming this binary combination is ideal in both vapour and liquid phases

• Again, first guess at temperature is weighted average of boiling points:

i.e. 0.8(89.3) + 0.2(120.6) = 95.6°C

• This time, we use dew point criterion, which is:

1K

y2

1i i

i

K-value of component i at this

temperature/pressure/composition

mole fraction of component i in vapour phase



log10 (P10)= 6.90565 – 1211.033 (95.6 + 220.79)

P10 = 1197 mmHg 197.1

0010

1197K1

486.00001

486K2

08.10.486

0.2

1.197

0.8

K

y2

1i i

i

log10 (P20)= 6.95464 – 1344.8 (95.6 + 219.482)

P20 = 486 mmHg

• Both liquids assumed ideal so:

> 1, i.e. too high. K-values not high enough


• A higher temperature will increase K values.

• Cannot be >120.6°C (pure toluene boils); try t =105°C


log10 (P10)= 6.90565 – 1211.033 (105 + 220.79)

P10 = 1543 mmHg

log10 (P20)= 6.95464 – 1344.8 (105 + 219.482)

P20 = 645.9 mmHg 6459.0

0001

645.9K2

828.00.646

0.2

1.543

0.8

K

y2

1i i

i

543.10010

1543K1


• Answer gave < 1, i.e. too lowNeed lower K-values, ie. a lower temperature

• Use linear interpolation like before:

Temperature (ºC)95.6 105

1.08

0.828

1.0000

• Model curve as straight line between the two known points

• Straight line gradient is:

t-95.6

1.0000-1.08

1056.59

0.828-1.08

i

i

K

y

t

C6.980.828-1.08

1.0000-1.081056.59-95.6t



log10 (P10)= 6.90565 – 1211.033 (98.6 + 220.79)

P10 = 1300 mmHg

99.00.533

0.2

1.300

0.8

K

y2

1i i

i

log10 (P20)= 6.95464 – 1344.8 (98.6 + 219.482)

P20 = 533 mmHg 533.0

0001

533K2

300.10010

1300K1

•Try linear interpolation, as = 0.99 < 1: too lowIf 95.6°C gives = 1.08 and 98.6°C gives = 0.99

C98.398.6-95.60.991.08

0.99-198.6try t



log10 (P10)= 6.90565 – 1211.033 (98.3 + 220.79)

P10 = 1289 mmHg

999.00.5282

0.2

1.289

0.8

K

y2

1i i

i

5282.00001

528.2K2

289.10010

1289K1

log10 (P20)= 6.95464 – 1344.8 (98.3 + 219.482)

P20 = 528.2 mmHg

• Dew point at 1000 mmHg for vapour mixture of 80 mol% benzene and 20 mol% toluene is 98.3°C


• Also interesting to know liquid composition at equilibrium with this vapour, ie. that of the first few drops

i

ii x

yK

x1 = 0.8 1.289 = 0.6206 x2 = 0.2 0.5282 = 0.3786 x1 + x2 = 0.6206 + 0.3786 = 0.9992 1

• Liquid at equilibrium with vapour at 1000 mmHg and 98.3°C is 62 mol% benzene and 38 mol% toluene

i

ii K

yx

Vapour-Liquid Equilibriumde Priester charts for K-values

• Common hydrocarbon reference

• All in psia and °F!

• Use chart instead of Antoine equation to find K-value

• eg. find dew point at 1.36 atm. of 10% n-pentane, 40% n-hexane and 50% n-heptane


• Conversion:• 14.7 psia = 1 atm.

• 1.36 × 14.7 = 20.0 psia

• Make first guess 100°F

K5 = 0.73

K6 = 0.25

K7 = 0.086

137.00.73

0.1

K

y

5

5

6.10.25

0.4

K

y

6

6

8.50.086

0.5

K

y

7

7


• Σ = 0.137 + 1.6 + 5.8 = 7.55

• Reduce Σ by increase in K

• Higher temperature needed

• Try 200°F

K5 = 3.2

K6 = 1.4

K7 = 0.60

0313.03.2

0.1

K

y

5

5

2857.01.4

0.4

K

y

6

6

8333.00.60

0.5

K

y

7

7


• Σ = 0.031 + 0.285 + 0.833 = 1.149

• Much better

• Try 210°F

K5 = 3.4

K6 = 1.5

K7 = 0.70

0294.03.4

0.1

K

y

5

5

2667.01.5

0.4

K

y

6

6

7143.00.70

0.5

K

y

7

7


• Σ = 0.029 + 0.267 + 0.714 = 1.010

• Best yet

• Is 212°F any better?

K5 = 3.5

K6 = 1.6

K7 = 0.72

0286.03.5

0.1

K

y

5

5

2500.01.6

0.4

K

y

6

6

6944.00.72

0.5

K

y

7

7


• Σ = 0.029 + 0.250 + 0.694 = 0.973

• Overshot target

• Answer for 210°F best

• Conversion to °CSubtract 32Then multiply by 0.556

• Tdew = 0.556(210 – 32)

• Tdew = 98.9°C ie 99°C

Gas-Liquid EquilibriumHenry’s Law

• Consider a 2-phase system (e.g. water & a gas)• gas is mixture of A & B (e.g. NH3 & air)• A is soluble in liquid, B is not (ideally)• constant pressure, Ptotal

• temperature fixed for each run, T

• Apply HENRY'S LAW: PA = HCA (dilute solution only)

PA

CA

T2 (high)

T1 (low)

solubility of A lower with high T


• e.g. For O2 in water @ 20°C, H = 4.01×104 atm/mol.

• What is saturated O2 concentration in water for dry air @ 1 atm/20°C?

• HENRY'S LAW: PA = H.xO

(air is 21% O2 & 79% N2)

• DALTON'S LAW: PO2 = 0.21×1 atm = 0.21 atm

• xO2 = PO

2÷H = 0.21÷(4.01×104) = 5.24×10-6 mol

fraction


• Take 1 kg solution

• 1 kg water occupies one litre volume at 20°C

• xA = moles of oxygen/moles of solution ≈ NO2÷NH

2O

• NH2

O = 1000÷18 = 55.56 mol

• NO2

= 55.56×(5.24×10-6) = 2.91×10-4 mol/kg water

• Molecular weight of oxygen is 32 g/mol

• CO2 = 9.32×10-3 g/kg water = 9.32×10-3 g/litre

cpp - bubble and dew points for ideal systems (1)

Documents

n liquid

liquid phases

calculation of bubble

vapour sample

higher temperature

liquid mole fraction

inphase i

mmhg2 i