Oct 24, 2014
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?