Multistage thermal separation

Multistage Thermal Separation

Condensation and Vaporization Processes in Series

First phase process we are not able to reach high purity. A solution for that problem could be to condense the vapor product and let that steam enter in another flash tank, and so on (point will move along the diagonal, reaching increasingly higher values, as the vapor product is enriched with the most volatile component.)

So we use partial condensation stages in order to increase the purity of products

Disadvantages:

• It requires many condensers with a variable heat exchange along the chain
• It loses most of the useful product in the liquid stream
• The product flow rate is not an independent variable and decreases as its purity increases

A similar situation would occur when considering a series of multiple vaporizations of the liquid streams exiting each stage. In that case we would like to have phi 1, so close to horizontal line, in that way stages down (we move downward in the composition plot)

How can we avoid having low product flow rates at the last stage? How can we supply/withdraw the required heat duty in each stage? How can we obtain, from the same feed, both the pure light component and the heavy one?

Let's analyze how we can solve the problem introduced before for the previous condensation process in series (goal: obtain a vapor product richer in light component). For a vaporization process the goal is: obtain a liquid product richer in heavier component

Feed

The flash tank i works at lower temperature respect to the tank i-1, because the feed stream of i is richer of the more volatile component than the i-1 one (the opposite of vaporization process, T is going to increase)

Therefore, we can use the liquid product of i stage in a previous stage in order to exploit that stream as the cooling utility. Doing so, we don't discard the liquid product but we send it into a 2-phase stream (L1) and all the intermediate stages would be unbalanced (except the last one) => the condensation of Vi occurs because of the mixing with Li+1 which is cooler. For the last stage a reboiler (Qc) is needed

For a single stage we have: Vaporization process

xj and yj are in thermodynamic equilibrium with each other

Using a single feed we can obtain two product streams (a liquid and a vapor product) with an higher purity than those we would obtain with the flash process)

correnely the condensation process with the vaporization one. We call that system multistage process with recycles or distillation

We recover a distillate D and a residue R

(vapor product) (liquid product)

Two heat duties are needed to run the process, namely Qc and Qr, which allow to fractionate the feed F into D and R

In order to obtain the product we need a certain number of stages along the chain and we can realize it through a graphical methodology called McCabe-Thiele methodology (only for binary mixtures)

FIRST AND SECOND ASSUMPTIONS OF McCabe-Thiele method

Vaporization process: (proving that the molar flow rates are constant along the process)

The stages are assumed adiabatic. Starting from the energy balance around the j-th stage it is possible to prove that the molar flow rates of condensed vapor and vaporized liquid streams are independent of the stage

Energy Balance:

V_j+1H_j+1 + L_j-1h_j-1 = V_jH_j + L_jh_j

H - h = lambda → H = h + lambda

V_j+1h_j+1 + V_j+1lambda_j+1 + L_j-1h_j-1 = V_jh_j + V_jlambda_j + L_jh_j

h_j ~ h_j-1 ~ h_j+1 assumption

V_j+1lambda_j+1 + (V_j+1 + L_j-1 - L_j + V_j)h = V_jlambda_j

=0

MATERIAL BALANCE of j-th stage

V_j+1lambda_j+1 = V_jlambda_j

lambda_j+1 = lambda_j Assumption that lambda latent heat of vaporization is constant

V_j+1 = V_j = V = const

L_j-1 = L_j = L = const

only on a molar basis

The assumption of lambda respects thereby if the Trouton rule is valid and it states that:

λ_c / T_{boil, c} = const

mostly true for hydrocarbons with which T_b,1, T_b,2 T1 - T2 < 10°C

So the latent heat of two components is constant along the stages, so lambda is not a function of composition

Also he stage's the composition changes and BC of lambda is the same for the species

THIRD ASSUMPTION OF McCabe-Thiele method (ideal stage assumption)

If the mixture is well mixed within the stage and the vapor and liquid phases have enough time for effective interaction, it can assumed that the compositions of the streams coming out from the same stage are in equilibrium

That assumption requires an perfect mass transfer within the stage => in reality the mass transfer is limited.

y _j = k ₁ x _j

1 - y _j = k ₂ (1 - x _j)

OVERALL MATERIAL BALANCES

F = D + R

F z _F = D y _D + R x _R

Fixed some specifications, we can solve the system => 4 unknowns, 2 fixed, so 2 calculated

ex. y _D, x _R fixed

R and D calculated

y _D, D fixed ...

pretty

we have at least 2 DoF

Mass balance around the condenser: ENRICHMENT SECTION

From 1-th plate to condenser: control volume

V = L + D

V y _j+1 = L x _j + D y _D

From these equations, the upper or enrichment operating straight line can be obtained:

y = L/V x + D/V y _D

The origin of that curve is in the point D of coordinates (y _D, y _D). If we set x = y _D then it turns out that y = y _D. So the origin will be at the bisector of the coordinates.

For the slope we have to consider the ratio L/V

There are two limiting cases:

1. L = 0, D = V, tb of enrichment section

The line is horizontal

1. L = V, D = 0, all vapor is condensed

The line coincides with the bisector

The slope is between these two cases

Mass balance Around the Reboiler : Stripping Section

From n+2th stage to reboiler: molar values

L' = V' + R

L x yⁿ⁺¹ = V' y' + R x_R

From these equations, the lower or stripping operating straight line can be obtained:

y = (L'/V') x + (R/V') x_R

The origin of that line is in the point R of coordinates (x_R, x_R), if we set x = x_R then it turns out that y = x_R. So the origin will be at the bisector of the coordinates plot

For the slope we have to consider the ratio L'/V'

There are two limiting cases:

1) R = L', V' = 0

The line is vertical

2) V' = L', R = 0

The line coincides with the bisector

The slope is between these two cases

The feed stage

It represents the transition between the enriching and the stripping section

(The feed stage is usually included in the stripping number of stage)

Material balance:

F + L + V' = V + L''

F z_F = (V - V') y + (L' - L) x

The feed stream cause a thermal effect on the process, which depends on the physical state and temperature of the feed. We take into account that characteristic with the fraction of the feed that is in vapor phase

q_F

phi_F = (V - V^')/F => V = V^' + phi_FF

L^' = L + (1 - phi_F)F

dividing for F the feed is:

z_F = phi_Fy + (1 - phi_F)x

y = -(1 - phi_F)/phi_F x + 1/phi_F z_F

feed straight line

it represents the straight line of flash process

it passes through the point F(z_F, z_F)

subtracting the two quantity lines of lower and upper section we will obtain:

Vy = Lx + D y_D -

V'y = L'x + Rx R

(V - V')y = (L - L')x + D y_D + Rx R

= F x F for the overall material balance

(V - V')/F y = (L - L')/F x + F/F z_F => z_F = phi_Fy + (1 - phi_F)x

linear combination of the two

So geometrically obtaining the straight line means that the three lines have a point in common (the intersection between them)

1 0 < phi < 1 partial vaporized feed biphasic feed

2 phi = 0 saturated liquid feed T = T_boil

3 phi < 0 subcooled liquid T < T_boil

4 phi = 1 saturated vapor feed T = T_boil

5 phi > 1 superheated vapor T > T_boil

If the feed is a subcooled liquid, at the meet stage its thermal condition is re-equilibrated, to secure a saturated liquid. To do this heat is required, that heat is supplied thanks to the condensation of a part of the vapor coming from the stripping section

Thermal balance at feed stage:

F cp,F (T_3,F - T_F) = lambda (V - V')

dividing to

phi_F = (V - V')/F = - Cp,F (T_3,F - T_F)/lambda < 0

remable heat needed to heat up F from Tf to T_3,F = latent heat given by the condensation of part of V'

If the feed is a superheated vapor, at the feed stage its thermal condition is re-equible, heated, to become a saturated vapor. Todo this heat is released, that heat is released thanks to the vaporization of part of the liquid coming from the enrichment section

Thermal balance at feed stage:

F c_p,F(T_F-T_d,F) = d(L-L')

removable heat needed to cool down F from T_F to T_d,F = latent heat taken from the vaporization of apart of L

phi_F = 1 - L'/F = 1 + c_p,F(T_F-T_d,F)/l > 1

OVERALL ENERGY BALANCES

F h_F + Q_R = D h_D + R h_R + Q_C

The feed column is decomposed in two parts: liquid and vapor parts

F phi_F h_F + F(1-phi_F) h_f + Q_R = D h_D + R h_R + Q_C    assumption: h_F = h_R It doesn't change much with the composition

F phi_F h_F + F(1-phi_F) h_F + F(1-phi_F) h_F + Q_R = D h_D + D h_D + R h_R + Q_C

F phi_F h_F + (F-D-R) h + Q_R = D h_D + Q_C

F phi_F h_F + Q_R = D h_D + Q_C

Q_C/F = Q_R/F + phi_F h_F - D/F h_D

In the general case Q_C and Q_R are not the same but are dependent from each other

In terms of utilities:

Q_R = l_s W_s => W_s = Q_R/l_s

STEAM    W_s = Q_R/l_s

Q_C = m w o c_{p,w o} k_{w o}(T_{k w o,out} - T_{k w o,in})

COOLING WATER

Reboiler:

We are assuming that x_R and x_e are in equilibrium so the reboiler is an ideal stage

Q_R = l V' / m

the flow rate of vapours

Condenser

X_D and y_D are in equilibrium, so the vessel is an ideal stage

Q_C = - l L

flow rate we are condensing

PARTIAL CONDENSER

Generally we would like to have the differential in liquid phase bc of troposes reasons, so we don't operate with that condenser

2) V, yF

L, xD

D, xD

Total condenser

XD = y1 & we are not interested in any separation in that stage so it is not an ideal stage

Qc = -lambda (L+D) = -lambda V

We derive the overall energy balance considering partial condenser. Now, we switch to total condenser

F hF + QR = D hD + R hR + Qc

F phiF hF + F(1-phiF) hF + QR = D hD + R hR + Qc

F phiF hF + F phiF lambda + F(1-phiF) hF + QR = D hF + R hR + Qc

F hF + F phiF lambda + QR = D hF + R hR + Qc

F-R-D = 0

F phiF lambda + QR = Qc

Qc/F = QR/F + phiF lambda

So QR = lambda V' = Qc - lambda phiF F

Qc and QR are connected depending on phiF

NUMBER OF IDEAL STAGES: McCabe-Thiele method

DESIGN PROBLEM: we have to find out how many stages there are

Data we need:

1. We know F, zF and phiF => we can draw the feed line
2. We can fix two specifications ex. xD, xR
3. Equilibrium line we fix P -> we are able to calculate the composition plot, we fix x and then we are able to find y and T -> approaches to calculate x
4. We may choose a value for L/V or L'/V' and the operating line are set

V₁y₃

V₂

D, y₁

L, x₁

L, x₂

Qc

We have y₁ and we can find x₁ BC they are in equilibrium with each other

So that point will be on the equilibrium curve (y₁, x₁)

The molar fraction x₁ is connected with y₂ through the material balance. So the point (x₁, y₂) will be on the enriching operating line

mole RLC balance

equilibrium line

At the feed stage we switch from enriching section to stripping section

located at intersection point (I)

We can go on following the enriching operating line, in that way we are moving the feed stage at the point where we decide to change section

We can do different values based on our decision about where we get the feed stage. Normally the feed stage is set where we can find the point I

doing so, the number of stages are minimised

The first stage of enriching section is the partial condenser

The last stage of stripping section is the reboiler, the first one is the feed

RATIING PROBLEM: We know the number of stages, where the feed is and we have to find the composition of products by alternative procedure

We fix and change L/V so L'/V' until we obtain the exact number of stages

the composition of products varies

THE CONVENTIONAL DISTILLATION COLUMN

The conventional distillation column

A cylindrical tower filled with either trays or packings

A reboiler, to produce the vapor stream V' at the bottom

A condenser, to produce the liquid stream L at the top

Two products: distillate D and residue R

One feed F, which cuts the column into two sections: the enrichment (upper) section and the stripping (lower) section

Due to gravity the vapor flows upwards and the liquid downwards

Description of a conventional distillation column