DALTON’S LAW OF PARTIAL PRESSURE T and V are constant

Pressure fraction = mole fraction

PT = PA + PB + PC +…….

AMAGAT’S LAW OF PARTIAL VOLUME T and P are constant

Volume fraction = mole fraction

VT = VA + VB +VC +……

GIVEN:

REQUIRED: %comp. in wt.

%comp. in vol.

V(m3) if m=80 kg, T = 90C, P = 600kPa

Density at STP (kg/m3)

GAS MIXTURE COMPOSITION

CH7 87%

C2H6 12%

C3H8 1%

CONDENSABLE GAS

Vapor, liquid at room temperature

NONCONDENSABLE GAS

Gas, gases at room temperature

SATURATION (PP = PV)

Partial pressure of the vapor is equal to the vapor pressure at specified temperature.

UNSATURATION (PP < PV)

Partial pressure of the vapor is less than the vapor pressure at specified temperature

DEW POINT

Temperature at which the vapor starts to condense

Example: dew point = 300C [H2O]

PH2O = PVH2O at 300C

Vapor pressure calculation (Antoine Equation)

ln(p) = A – B/C + T

RELATIVE SATURATION (RS)

Defined as the partial pressure of the vapor divided by the vapor pressure of the vapor at the temperature of the gas.

MOLAL SATURATION (Sm)

The ratio of the moles of vapor to the moles of vapor-free gas

ABSOLUTE SATURATION (Sabs)

Weight of vapor per weight vapor-free gas

PERCENTAGE SATURATION (%S)

If a gas at 600C and 101.6 kPa, has a molal humidity of 0.030, determine:

the relative humidity

the dew point of the gas (in 0C)

PV @ 600C = 148.29 mmHg

Given:

RH = 85% PV @ 900F = 35.64mmHg

T = 90 0F

PT = 14.696 psia= 760 mmHg

Required:

a) Hm

b) Habs

c) Saturation temperature

Transformation of a liquid into a vapor in a non-condensable gas.

ENTERING, E LEAVING, L (dry gas, water vapor)

VAPOR, V

Dry gas or Dry gas, water vapor

Change of a vapor in a non-condensable gas to liquid.

ENTERING, E LEAVING, L (dry gas, water vapor)

CONDENSATE, C

Dry gas, water vapor saturated

CONDENSER E, AIR

V=30 m T= 1000C P=98.6kPa Dew pt. = 300C

C

T=140C P = 101.9 Kpa

P @ 300C = 31.38 mmHg = 4.18kPa P @ 140C = 11.7 mmHg = 1.56kPa Unknown = fraction H2O condensed

VAPORIZER

E, dry air T=200C P=100kPa

L T=200C P=100kPa Pv eth. Alc=5.76 kPa

V=6.0 kg eth. Alc Unknown = VE

18.10 18.11

E Gas or Gas vapor

Dried material

L Gas vapor

Wet material

Gas mixture E

Leaving solution P

Gas mixture L

Absorbing medium(solvent/sol’n) F

An absorber receives a mixture of air containing 12 percent carbon disulfide. The absorbing solution is benzene and the gas exits from the absorber with a CS2 content of 3 percent and a benzene content of 3 percent (because some of the benzene evaporates). What fraction of CS2 was recovered?