L14 & 15 - Distillation & Assisted distillation Flashcards
Describe simple distillation (Rayleigh distillation)
-In stillpot, initially filled with mixture that is heated producing a vapour
-Vapour is condensed overhead and collected in reciever; no reflux, plates or packing
- Concentration of more volatile component (mvc) is higher in vapour than liquid - over time liquid mvc decreases and amount of mvc in overhead increases
-Quality of vapour decreases over time
Describe VLE curve at constant pressure - 2 component
Temp v mole fraction of mvc
Mvc has lower BP at mole fraction = 1; less volatile has higher BP at mole fraction = 0
Dew point (vapour) higher T; bubble point (liquid) lower T
Mass balance for simple distillation & Rayleigh eq. proof
No. moles of mvc in stillpot, Lm = x*L
Small amount of liquid dL evaporates- change in no. of moles of mvc is: y*dL = dLm
Differentiating (chain rule): dLm = xdL + Ldx
-> xdL + Ldx = ydL
-> ydL - xdL = Ldx
-> (y-x)dL = L*dx
Therefore, dL/L = dx/(y-x)
Integrating from Li to Lf & xi to xf:
ln[Li/Lf] = integral xf to xi: dx/y-x
Mean composition of what has been distilled over, xD
Lixi - Lfxf = xD*(Li-Lf)
-> xD = (Lixi - Lfxf)/(Li-Lf)
Relative volatility
y = ax/[1+(a-1)*x]
Rayleigh in terms of relative volatility
ln[Li/Lf] = [1/(a-1)]{ln[xi/xf] - aln[(1-xi)/(1-xf)]}
Overall mass balance and rectification operating line
Assume constant molal overflow:
V = L + D
Vyn+1 = Lxn + DxD
yn+1 = (L/V)xn + (D/V)xD
yn+1 = (R/R+1)xn + (1/R+1)*xD
Constant reflux ratio
Distillation proceeds, slope of operating line constant
As stillpot mole fraction, Xs, decreases, the distillate mole fraction, XD, decreases
In reality, performance of distillation better than equation prediction, why?
Vapour condenses on lid of vessels/pipes so refluxes back into boiler - rectification
(Vapour entering condenser richer in mvc)
Finding Li or Lf at constant reflux ratio
Graphically:
1) plot 1/(xD-xS) against xS
2) Find area under curve between xiS and xiF = ln[Li/Lf]
Minimum reflux ratio
Rmin/(Rmin+1) = [xD-y*(xS)]/[xD-xS]
Variable reflux ratio
xD remains constant -> vetter quality of distillate
Reflux ratio must increase as distillation proceeds as more difficult to get distillate of desired composition from stillpot residue of smaller and smaller mvc mole fraction
As distillation proceeds, amount of material in stillpot falls: heat transfer area falls as level falls so more time & fewer xD and vaporisation rate slower
Rayleigh eq. at variable reflux ratio
ln[Li/Lf] = integral xf to xi: dx/y-x
Now: ln[Li/Lf] = ln[(xD-xf)/(xD-xi)]
-> [Li/Lf] = [(xD-xf)/(xD-xi)]
Running costs of batch distillation
Depends on amount of material vaporised:
Constant reflux ratio:
Total amount vaporised over whole batch, Vb:
Vb = (R+1)*D, where Db = distillate amount collected
Variable reflux ratio:
Amount of vaporisation required to collect same distillate quantity will increase as distillation proceeds
Developing mass balance for Bogart Eq.
Mass balance of vaporisation from stillpot to rate of depletion of content of stillpot:
dV/dt = (R+1)[-dL/dt]
Variable reflux Rayleigh: [Li/L] = [(xD-x)/(xD-xi)]
-> L = Li[(xD-x)/(xD-xi)]
dL = (dL/dx)(dx/dt) = [Li(xD-xi)]/[(xD-x)^2]*(dx/dt)
xD against Lsi/Lf graph: fractions of distillate
1) xD meets specificataions - product distllate
2) Meds added to feed of next batch- ensure all mvc recovered
3) Residue discarded - too little mvc for worthwhile further processing
Steam distillation & when is it used?
Carrier (Steam) sparged into column base in heated stillpot
Used for low volatility, high sensitive products (thermal degradation) and with high BP
What effect does steam distillation have on organics?
The partial pressure of organics are reduced by dilution, reducing the temperature
Cases for steam distillation
1) Steam is superheated to provide enough heat to vaporise the material without condensing itself (no liquid water)
2) Some steam condenses, forming liquid water phase in stillpot
How many components are assumed in steam distillation?
Assumed vapour consists only of water and required component (initially assumed only one organic component take part in vle, though other present that are not part of vle)
Partial pressure balance of component and carrier
(ma/Ma)/(ms/Ms) = pA/ps = yA/ys = pA / (P-pA)
Relative volatility, aSA
aSA = ps* / pA*
No water phase formed: phase rule & vapour pressure allowed
F = C - P + 2
If no water phase in stillpot, F = 2 -2 + 2 = 2; can fix both T & P independently
ps = P - pA so must not exceed vapour pressure of water at the operating temp or water phase will form
Water phase formed: phase rule & vapour pressure allowed
F = C - P +2
No. of phases (P) increase by one, so F = 1 and only choose T or P
Both water and other component will exert a partial pressure equal to vapour pressures at T of mixture.
Distillation temp always less than BP of water at total pressure, so at atmospheric pressure a high boiling organic steam distilled below 100deg C.
Phase diagram for water with an immiscible organic material
Hetero-azeotrope: for fixed pressure, minimum temp in which vapour and 2 liquid phases co-exist in equilibrium
Temp v Composition graph:
Bottom is where both liquid exist
Top is vapour exists
Varying vapour & either liquid and mainly one liquid with drop of other component
When water phase present in stillpot, how to drive off water?
If overall liquid composition on organic side, the vapour boiling off contains more water than liquid
If organic phase returned by reflux, water steadily driven off and solvent dried
When water phase present in stillpot, how to have water left?
If overall liquid composition is on water side, the vapour contains more organic than the liquid.
If water returned by reflux, only water left.
