Single stage thermal separation

Single stage thermal separation

Relative volatility

The equilibm ratio can be expressed also as the ratio of the mole fractions in vapor and liquid phases for each component of the mixture

Ki = yi / xi

This parameter is an indicator of the component's tendency to move towards the vapor phase, called volatility

We are interested in defining the selectivity or relative volatility as:

where i and j are two component of the mixture

di,j = Ki / Kj

When alpha is equal to one the components have the same tendency to move towards the vapor phase, so the separation is not possible. we would like it to be different from 1.

For a binary system:

alpha = alpha_1,2 = y/x / (1-y)/(1-x)

alpha_2,1 = 1 / alpha_1,2

From the definition of binary system's volatility, the following equation, called Brown's equation, is obtained:

y = alpha x / (1 + (alpha - 1)x)

When the system is ideal, alpha is constant and the belly of the curve in composition plot is symmetrical

alpha is generally not constant and is a function of composition, in fact is equal to:

alpha = p₁^sat gamma₁(x,T) / p₂^sat gamma₂(x,T)

Let's consider the following plot with a constant alpha = 2.5

- what happens if alpha = 10? Using the Brown equation with a bigger value of alpha, we obtain bigger values for y for each x point, so the curve is larger

- alpha = 1? From Brown's equation, we obtain y = x so the curve overlaps the bisector. We are not able to separate the components by thermal separation

- alpha = 0.1? The curve goes below the bisector

The first and the third case are the same - alpha are equally far from 1. But in the first case the component 1 is more volatile, while in the third case the role volatile is the second one, which we find in vapor phase

To be possible the thermal separation process a limit must be respected: |x_i - 1| > 0.05

sufficiently distant from 1 (x > 1.05 or x < 0.95)

IF x -> 1 the separation becomes difficult and expensive

FLASH PROCESS: Partial Vaporization Process at Equilibrium

A separation problem is defined when, starting from an entire mixture (feed) of known amount (flowrate) F and composition z_F, some specifications are defined on the desired product -> specification related to flowrate, purity of the product, or to the recovery of a determined component

For a vaporization process, the specifications are:

• Amount (flowrate) V of the vapor phase and purity of the component in that phase y_i.
• Purity and recovery of one component in the vapor phase.

Recovery: R_i = V y_i / F z_F,i

for the heavy component

R_i = L (1 - x_i) / F (1 - z_i)

The flash process is a continuous process: heat must be supplied to partially vaporize the liquid mixture, which enters in a tank in order to separate it into two phases by gravity

V = phi F

phi is the degree of vaporization = V / F

fraction of feed vaporized

F

T_F

P_F

L = (1 - phi)F

1-2: Pure

Pressure increases and, as consequence, the temp is shifted to higher temperatures. The heat transfer coefficient for a liquid is higher then the gas one, so we need to increase the P before the heat exchanger because otherwise we would obtain bubble of vapor in the exchanger.

P1 transfer coefficient - we remain in the liquid phase

2-3: Heat Exchanger Heat is supplied, remaining in liquid phase due to high P

3-4: Lamination valve Drop of pressure and partially vaporization at atmospheric P

So the pump and the valve are used to avoid the formation of vapor in heat exchangers

Then there is the flash drum where the two phases are separated

Degrees of Freedom

The DoF are the operating variables that we can manipulate in order to effect the control

To predict the DoF, it is used Kwoak method:

Operating variables = independent variables - independent equation

OV = IV - IE

Intensive = NC-NP+2 Gibbs rule and extensive

F: NC+1+1 = NC+2

Extensive variable = flow rate

Since V and L are in equilibrium with each other, we can consider and study (V+L) > two-phase system

V+L: NC-2+2+2 = NC+2

Flow rates

IV = (NC+2) + (NC+2) + 1 = 2NC+5

F

V+L

Energy stream

LE -> number of mass balances for each component = NC + 1

Feels number of energy balances for each block

OV = 2NC + 5 - (NC + 1) = NC + 4

The feed is given, we know (NC + 2) of feed

So at the end we find OV = 2 -> not depending on number of components

The flash process has 2 degrees of freedom

usually P and Q (thermal duty)

(pressure of flash)

Te and composition are not easy variables to control so they are not set as operating variables

MATERIAL BALANCES AND EQUILIBRIUM CONDITIONS

The general material balance is: Accumulation = Inlet - outlet + Production

We are at steady-state conditions and there is not reaction, so Acc and Prod are = to 0

IN/LET = OUT/LET => Total material balance + (NC - c) component material balances

F = V + L

F, z_F,c = (V, y_c) + (L, x_c) for each c

phi = V/F => l = phi + L/F => L/F = 1 - phi

z_F,c = (phi, y_c) + [(1 - phi), x_c] for each c

Mathematical continui (material eq) that must be respected by components of the mixtures

The other constraint to be respected is the equilibrium condition: y_c = K_cx_c for each c

expressed based on the approach used

For a binary system:

y = K₁x

(1 - y) = K₂(1 - x)

ENERGY BALANCE

It is necessary in order to correlate the performance to the thermal duty we supply, therefore we want to obtain through the energy balance a relationship between Q and phi

IN - OUT = 0

[kcal/h], [kJ/h], kW... Units of Energy Flow rate

Enthalpy contribution of feed

Enthalpy of saturated liquid

Enthalpy of saturated vapor

H - h = LATENT HEAT OF VAPORIZATION ON MOLAR BASIS

Q/F + z_F = [phi H] + [(1-phi) h] = phi H + h - phi h = phi + h

If we supply more heat we will obtain more vapor

Q/F + z_F = phi h + h

phi = Q/(F . l) + (z_F-h)/l

Approximation: z_F - h => phi = Q/(F . l)

That approximation can't be done when the feed is far from its boiling temperature ex. subcooled feed (z_F-h)

If F is cold, the Q supplied is used to increase the temperature of F to its boiling value

(z_F-h)/l < 0 => phi ↓

We have to provide more heat to achieve the same value of phi obtained in the case where the approximation is valid (z_F = z_sat)

GRAPHIC REPRESENTATION

To represent the material balance and the equilibrium we use composition diagram

Benzene-Toluene Case:

Vapor-liquid equilibrium at P = 110 kPa

Vapor mole fraction

y of V

phi = 0

phi = 1

x of L

Liquid mole fraction Benzene

z_F = phi y + (1-phi) x

From the material balance we want to obtain how y changes with respect to x

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

It is a straight line

At x = z_F y = z_F that point is on the bisector

Always valid

The slope is depending on phi, obtained after fixed Q

The only point that belongs to both equilibrium and material balance is the point of intersection between the line, which represents the material balance, and the curve of equilibrium (I)

What is the maximum purity achievable for the light component in the vapor phase starting from a mixture with inlet composition z_F = 0.4 ? Let's consider that phi <= phi₁, when phi = 0 the slope is too vertical and we find the point

LIMITATION / limit

So with phi=0 we have an upper limit for the flash process, we can't go any further. At that condition we have a feed liquid with an infinitesimal amount of vapor characterized by the higher purity

V/F = 0 -> V=0 Recovery zero

At phi=1 we obtain the maximum purity achievable for the heavy component in the liquid phase the slope is equal to zero (horizontal line). We have at this condition F=V and an infinitesimal amount of liquid with the high purity

If we want an higher purity, the recovery decreases

fixed so the degree of vap.

Fixed Pf and Q(=phi): we are able to draw the equilibrium curve (isobaric curve) with phi-phi or y-x approach fixed phi we increase to draw the straight line -> we find the composition at the equilibrium Found x and y we can calculate the flash temperature as the boiling temperature of the liquid outlet

If we fix Pf and Tf: fixing the P creates fixing the composition plot and with vapor-liquid equilibrium equation, temp/Tf, we are able to find x and y Drawing the straight line that passes through F (we know it) and I (obtained), we can create the slope, so the value of phi. With energy balance we predict Q.

Not easy to control

UTILITY OF FLASH PROCESS

Discussion about the utility of a flash unit

• In a flash process it is always possible to satisfy a specification on the vapor flow rate, by adjusting Q.
• On the other hand, it is usually impossible to achieve the specification on product composition, because the equilibrium curve, for a given system, slightly depends on P_F.
• So, a flash process cannot simultaneously satisfy design specifications on both composition and recovery.
• Another drawback is the limitation in the range of concentration of the products (y_b and x_r, as stated above): unless the equilibrium curve is favorable, the fractionation is not significant. Therefore, the flash unit operation is effectively used only to recover, from a liquid feed, components whose volatilities are much higher than the others.
• It is useful when we want to remove highly volatile components (e.g.: crude oil refinery), as it allows to do it at relatively low energy cost.

The purity is constrained by the volatility which doesn't allow for high purity to be achieved => the phase compositions are determined by VLE which depends on pressure => R up purity down and viceversa

THE QUENCH PROCESS (opposite of flash)

It is a partial equilibium condensation process

The feed is a vapor stream and the thermal power is released by the system, not requiredThere is no need to use 2 different phases

l - q = q condensation degree

Generally the value of PE can be higher or equal to P atmif T is much lower than the ambient one

=> we don't cool a vapor following that scheme

The exchanger is more efficient if a liquid stream is processed

Example: recovery of light hydrocarbons

ENTHALPY-COMPOSITION DIAGRAM: we are able to draw the energy balance

Example: adiabatic mixing of two streams

Material balances

Energy balance

Lever rule

Material and Energy Balance are linear with respect to the composition

that relationship expresses the arithmetic condition of the points A, M and B

Exergi. Mixing with heat exchange

a fA + B fB + Q = r fR

(a) A fA' + B fB = r fR

fA' = fA + Q/A

b) A fA + B fB = r fR

fR = fR' + Q/r

By the graph we can say A is earger then B (Q/A < Q/B), with the same amount of Q

Representation of Enthalpy as a function of T and composition:

In that plot we want to represent isotherms + consider the equation to calculate the enthalpy for pure components

h1 = integral from T0 to T of C1 dT ; h2 = integral from T0 to T of C2 dT where T0 is the reference temperature

At T0 f(T0) = 0

h1,T = h2,T bc c1 < c2

f1,T - f1,T0 = integral from T0 to T of C2 dT - delta Hm,T0(z) = f2,T0(z) + mixing contribution

Due to mixing in adiabatic condition the T will increase or drop, if we want to come back to previous k we need to supply or remove an amount of heat equal to chem,T0

In that plot the isotherms are not straight lines but curves bc of continuous variation of mixing enthalpy depending on the composition

The bubble line can be constructed, considering the isobaric heating of the mixture of composition z1 up to the boiling temperature Tb(z)

h = delta Hm,T0 + integral from T0 to Tb Cp dT |z

corresponding h for all compositions - bubble curve

The dew curve is constructed considering that the heat required to go from saturated liquid to saturated vapor is equal to the latent heat of vaporization

h = h + lambda lv

P = const

To find x and y, which are in equilibrium, we use the tie lines (time connected)

The tie lines connect a liquid and vapor phase in equilibrium

Points L and V belong to the same isotherm which, in the biphasic domain, is represented by the LV segment

Both material and energy balances are energy equations

z_F = φy + (1 - φ)x

J_F = φl + (1 - φ)h

Points V(y, h), L(x, h) and F'(z_F, J_F) must be aligned following the lever rule

Representation of the flash process in a J-x diagram

P = cost

Saturated liquid @ P'

Bubble curve

Dew curve

P ↑ at point 1

Q/F

Saturated liquid @ P_F

x, y, z →

SUO- Prof. Elena Barbera

38

SIZING AND RATING OF A VERTICAL FLASH DRUM

We studied sizing a tank that minimize the presence of liquid droplets in the vapor flow rate, acting on the vapor flowrate velocity. With velocity we act on the area of the tank and on volumetric flow rate

u_v = V_v/A = m/s

For vapor we expect u_v = 1-5 m/s

Through the area we can obtain a relationship between the velocity and the tank diameter

A = πD²/4 => D = √(4A/π)

If ↑ D ↓ u_v and vice versa

We are going to size the diameter and the height of the tank (D and H)

RATING PROBLEM: We have to check if the given equipment is suitable, so check if the velocity of the vapor based on operating condition is adequate, changing parameters like degree of vaporization, heat duty...

The approach is based on hydrodynamics, based on the concept of ENTRAINMENT.

There is a maximum vapor velocity u_v,max to avoid excessive entrainment of liquid droplets in the outlet vapor stream. Accordingly, the drum cross-section A (and therefore the diameter D) must be large enough.

A > Amin = Vv / u_v,max

Two parameters must be evaluated:

1. Flux parameter F_e,v
2. Reference velocity K_drum

F_e,v = L_v/V_v √(p_L/p_v) = Φ_p/Φ_p √(p_v/p_L)

K_drum = u_v,max(p_v/(p_L-p_v))

related to droplet diameter d according to a force balance between gravity and friction

We calculate F_e,v and by the graph we can obtain K_drum value. Through the equation of K_drum we can predict u_v,max.

u_v = Vv/A ≤ u_v,max = K_drum√(p_L-p_v/p_v)

Sizing problem: With u_v,max we calculate Amin.

A > Vv / u_v,max

From that area we select D

D > √(4A/π)

And we take a large value of D

We use a safety factor of 0.8-0.85 on u_v,max.

We want to go too far below the value Vv-u_v,max & BC it means oversize the equipment (A↑)

Rating problem: Area is given, so after calculating u_v,max we check that the vapor velocity we have is below the maximum allowable value.

Vv ≤ u_v,max A

A = πD²/4

If Vv > Vvmax, then we reduce Vv (so reduce feed)

For the height H/D = 3-5 There has to be 0.5m between inlet pipe and liquid level