gas properties and kinetics
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Gas Properties and Kinetics. Gas and Vapor. Gas and vapors have some properties in common. Both are composed of widely separated freely moving molecules. Both will expand to fill a larger differently sahped container. - PowerPoint PPT PresentationTRANSCRIPT
Gas Properties and Kinetics
Gas and Vapor
• Gas and vapors have some properties in common. – Both are composed of widely separated freely
moving molecules. – Both will expand to fill a larger differently
sahped container. – Both exert pressure in all directions. The
major difference lies in the internal eneryg of the molecules.
Gas and Vapor
• Gas if it is far removed from the liquid state. (Usually it means that the temperature of the substance is above its critical point)
• Vapor a substance in the gaseous state that is not far from being a liquid. Therefore a vapor can usally be adsorbed onto surfaces or condensed into a liquid relatively easily.
Vapor Pressure
• The pressure exerted by a pure component vapor in equilibrium with a flat surface of its pure component liquid at a certain temperature
• is a measure of the escaping tendency or volatility of the liquid (liquid’s vapor pressure)
Vapor Pressure• Vapor pressure increases rapidly with an
increase in the temperature
• The pressure/temperature relatin can be given by Antoine equation:
TC
BAP
i
iivi log Pv in mmHg and T in oC (if Table 9.2 is used)
Equilibrium• When a pure liquid is placed in contact
with air in a closed space, some of the liquid will volatilize until vapor-liquid equilibrium is established
ivii PPyP
Pi partial pressure of component i, atm
yi mole fraction of component i in the gasP= total pressure, atm
Example 10.1
• An airstream is presently at 1 atm and 160 F., and contains 40,000 ppm toluene. To what temperature must the air be cooled to remove two-thirds of the toluene vapor?
Solution• 40,000 ppm = 4% = 0.04 mole fraction
40 ppmy=0.04
Pi=0.04 atm
13.33 ppmy=0.013
Pi =0.013 atm
Sıcaklık
160 F
T = ?
ivii PPyP At equilibrium:
0.013 atm = 0.196 psi
Solution, contd.
40 F
Solution, contd.
• Pvi =0.013 atm = 10.13 mmHgTC
BAP
i
iivi log Pv in mmHg and T in oC (if Table 9.2 is used)
377.219)13.10log(95334.6
943.343,1
log
i
vii
i CPA
BT
T = 6.5 C
Comments on the Example• In the design of a heat exchanger to cool
the air in the previous example, provide enough surface area to not only cool the air (remove sensible heat) bu also to condense the vapors (remove latent heat)
• Usually chilled water or a refigerant might be sufficiently cool to remove enough vapor from the air to meet the pollution control objective
• Even cooled to 6.5 C, the air stream still contains 13.333 ppm toluene.
Diffusivities• The matter will diffuse spontenously from a region of high
concentration to one of low concentration• Fick’s first law describes diffusivity: the proportionality
constant between the rate of flux of matter and the concentration gradient
dx
dCD
A
M
M=mass transfer rate, mol/s
A= area normal to the direction of diffusion, cm2
D=diffusivity, cm2/s
dC/dx=concentration gradient, mol/cm4
Since the concentration decreases in the direction of diffusion a minus sign is put
Diffusivities
• The diffusivity of a particular substance depends on the substance itself and the medium through which it is moving
• In all the gas and vapor control processes in which the gas flows past (and contacts) a solid or liquid, a laminar film layer is formed where the pollutant passes from one phase to the other is via molecular diffusion
• Diffusion is ofther the rate-limitin step in control processes• More detailed discussion on diffusion will be seen in
absorption process
Gas Liquid and Gas-Solid Equilibria
• In absorption process, once a gas molecule has diffused thorughthe stagnnt gas film it must be absorbed into the liqud.
• Once absorbed, the pollutant then diffuse through a stagnant ilquid film into the bulk liquid
• Even thoud the rate of absoprtion is very fast and thus not a limitation, the extent of absorption (solubility) is crucial to the overall objective of mass transfer
Gas Liquid and Gas-Solid Equilibria• Most air pollution control equipment works at a or
near atmospheric pressure and with relatively dilute solutions
• For dilute solutions, the concentrations of the pollutant in the gas and the liquid are often linearly related (Henry’s Law)
• Ci: concentration of pollutant i in the aqeuous phase• http://www.mpch-mainz.mpg.de/~sander/res/henry.html
i
ii p
CH
][Henry’s Law constant =
(M/atm)
Henry’s Law Constants
• Hi vary with T
• Hi has units depending on the Ci and Pi unit.
• The equation eventually becomes nonlinear at high Ci value and no longer valid.
Henry’s Law ConstantsTür HA (298K),mol/Latm -∆H/R, K
SO2 1.23 3120
H2O2 7.45E4 6620
HNO3 2.1E5
HNO2 49 4780
O3 1.13E-2 2300
O2 1.3E-3
NO2 1.0E-2 2500
NO 1.9E-3 1480
CO2 3.4E-2 2420
NH3 75 3400
CH3C(O)O2NO2 2.9 5910
HCl 727 2020
HCHO 6.3E3 6460
NO3 2.1E5 8700
OH. 25 5280
HO2. 2.0E3 6640
Adsorption
• Involves a gas-solid equilibrium that is quite similar in principle to the solubility equilibrium in liquid-gas systems
• Amount adsorbed on the solid depends on– Type of vapor– The partial pressure of the vapor– The type of solid (adsorbent)– The amount of sufrace area available for adsorption– Temperature
A Typical Adsorption Isotherm
Adsorption Types
• 1) PHYSICAL– Exothermic– Reversible (to separate VOCs from air)
• 2) CHEMICAL– İnvolves the breaking and re-forming of bonds– Much more energetic than physical adsorption
Chemical Reactions
• Once the pollutant is captured a chemical reaction takes place– In catalytic incineration of VOCs to from CO2
and H2O
– SO2 scrubbing to form CaSO4
Chemical Reactions-Kinetics
• Reaction rate: the rate of “disapperance” of reactants or the rate of “apperance” of products.
• Reaction rate =rP =-rR
• rP = rate of generation of product P• rR = rate of generation of ractant R• Is the rate constant?• r is a function of
» Reactants concentration» T» Catalyst or inhibitor existence» Light
Chemical Reactions-Kinetics
• Consider R + S P + Q• rP = kCx
RCyS
• k = rate constant f(T)• CR and CS= concentrations of reactants
(mol/L)• x,y = exponents
Chemical Reactions-Kinetics
• Consider R + S P + Q• rP = kCx
RCyS
• k can be expressed by an Arrhenius equation such as:
K=Ae-E/RT
A: frequency factorE: activation energyR: universal gas constant, in energy unitsT: absolute temperature
Example 10.3
Reactor Models
• Two ideal reactor models are commonly used to descirbe real reactors where reactions take place
• 1) CSTR (Continous stirred tank reactor
• 2) Plug flow
CSTR
• Continuous flow with inlet concentration (Cin) through a tank in which the contents are rapidly mixed
• Homogenous distribution of the all species• Concentration in the tank equals outlet
concentration (Cout)• At steady state: • dC/dt = QinCin-QoutCout+riV=0V: tank volume (L)Q:volumetric flow rate (L/s)
Plug Flow Reactor • Can be described as one dimensional flow through
a long tube. • Velocity is constant at all radial position in the tube• Axial dispersion is negligible• The stady-state material balance for component i:
• dC/dt = QVCi,V-QV+VCi,V+V+riV=0
If the flow rate is constant and ri in not a function of position then:
dCi/ri=(1/Q)dV
Example 10.4
• The reaction RP in Example 10.3 is to occur isothermally at 640 K in a) CSTR b)PFR. Calculate the required volume of each reactor to give 99% convesion of R to P when the volumetric flow rate is constant at 100 L/s.
Example 10.4, Solution
• From Example 10.3 A=5.28E7 s-1 , E=21.53 cal/mol. k at 640 K:
For a CSTR
• dC/dt = QinCin-QoutCout+riV=0 then
LC
CV
kC
CCQCV
r
CCQV
inR
inR
outR
inRoutRinR
R
outRinR
4230)01.0(34.2
)01.01(100
)/1(
)(
,
,
,
,,,
,,
34.2)640*987.1/(53.217)10(28.5 ek
Example 10.4, Solution
For a PFR if Q is constant:
dCR/rR=dCR/-kCR=(1/Q)dV
LV
V
k
Q
V
C
C
k
dVQkC
dC
inR
outR
VC
CR
RoutR
inR
197)01.0ln(34.2
100100
)01.0ln(1
ln1
1
,
,
0
,
,
Thermodynamics
• Reactions can be exothermic (release heat) and endothermic ( heat input)
• These heat exchange changes the temperature of the reacting mix and thus affects the reaction rates
• T also affects the volumetric flow rates and concentrations in mol/L unit
Thermodynamics• Many chemical reactions do not go to
100% completion
• An equilibrium depending on the temperature is establisehde between reactants and products
• SO2 + 1/2O2SO3
• For this gas phase reaction equilibrium constant Kp = PSO3/(PSO2*(PO2)1/2
• Kp=1.53E-5exp(11750/T) (T in Kelvin)
Thermodynamics• As T increases Kp decreases rapidly
at 298 Kp = 2(10)12 at 750 Kp = 97• Note that equilibrium concentrations are not often
observed in industrial reactors. • Generally the equilibrium constants for exothermic
reactions decrease with increasing T while kinetic constants always increase with temperature
• Therefore it is possible for an optium T to exist for a given reaction. It should be low enough to allow the equilibrium to achieve the desired degree of conversion yet high enough to allow the reactions to proceed rapidly.
Gas Properties and Kinetics
• Gas kinetics is a branch of statistical mechanics applied to gases
• Applying statistical mechanics to gain a microscopic understanding of gas properties such as pressure,temperature,viscosity, and diffusivity.
• Let’s examine gas transportation and interaction between gases and particulates
Ideal Gas Law• Ideal gas law is the result of two
experiments and is stated as:• 1)For a given mass of gas held at a
constant T, P is inversely proportional to the volume (Boyle’s law)
• 2) For a given mass of gas held at a constant pressure, the V is directly proportional to the T (Law of Charles and Gay-Lussac)
Given in an Equation:PV = nRT (for a fixed mass of gas)
Ideal Gas Law
PV = nRT
P in atm, V in L, T is in K, n is the number of moles of gas in the gas sample, and R is universal gas constant
R = 0.082 (Latm/molK)
Useful Links
• http://www.aim.env.uea.ac.uk/aim/aim.php