che111p gas mixtures
DESCRIPTION
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:GAS MIXTURE CH7 C2H6 C3H8 COMPOSITION 87% 12% 1%REQUIRED: %comp. in wt. %comp. in vol. V(m3) if m=80 kg, T = 90C, P = 600kPa Density at STP (kg/m3)CONDENSABLE GAS Vapor, liquid at room temperatureNONCONDENSABLE GAS Gas, gasTRANSCRIPT
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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 +……
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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%
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CONDENSABLE GAS
Vapor, liquid at room temperature
NONCONDENSABLE GAS
Gas, gases at room temperature
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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
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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
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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.
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MOLAL SATURATION (Sm)
The ratio of the moles of vapor to the moles of vapor-free gas
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ABSOLUTE SATURATION (Sabs)
Weight of vapor per weight vapor-free gas
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PERCENTAGE SATURATION (%S)
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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
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Given:
RH = 85% PV @ 900F = 35.64mmHg
T = 90 0F
PT = 14.696 psia= 760 mmHg
Required:
a) Hm
b) Habs
c) Saturation temperature
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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
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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
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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
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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
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18.10 18.11
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E Gas or Gas vapor
Dried material
L Gas vapor
Wet material
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Gas mixture E
Leaving solution P
Gas mixture L
Absorbing medium(solvent/sol’n) F
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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?