Separation Unit Operations - Appunti (Notes)

V,MAX

For example, to have a maximum liquid entrapment of 5% one can rely on the K values

drum

​

recommended by the graph.

In design problems, the area must be greater than otherwise the u velocity will

V

exceed the reference velocity.

Then calculating the area that verifies , the diameter of the tank must be

=

4

≥ π → the minimum diameter

4 is usually identified and

= π

then the engineer finds the

industrial tank that has the nearer

value of diameter (in excess).

Usually the height of the tank is

chosen with a length between 3 and

5 times that of the diameter →

.

3< < 5

The inlet duct is usually located at

⅓ of the height of the tank.

Usually between the feed duct and

the liquid level there must be at

least 0.5 m to avoid the formation

of foam (foaming) due to the entry

of steam directly into the liquid phase → .

− > 0, 5

32

The demister reduces the entrainment of the liquid inside the gas phase by aggregating the

various drops of liquid which, becoming larger, fall downwards → in this way I can use the

graph for 5% entrainment.

EXERCISE

A binary mixture of n-esane (1) and n-octane (2) with a flowrate F=680 kmol/h and a

6

composition of z =0.4 is partially vaporized. The applied heat duty is Q=2.8⨯10 kcal/h and

F

the process condition are P=1 atm and T=378K.

Calculate the vapor and liquid flowrate and compositions, knowing that the relative volatility

=6.7

of the system is and the recovery.

Size the tank (D and H) of the flash.

Data:

●​ = 86

1

●​ = 114

2

●​ λ = 8070

●​ ρ = 696 3

Procedure: 6,7

(1−)

The relative volatility is → .

1

α = = = 6, 7 =

(1−) 1−+6,7

2

The energy balance allow to identify 6

2,8·10

.

ℎ

φ= = = 0, 51

λ 8070 ·680

ℎ

It is possible now to calculate the flow rates of vapor and liquid phase:

●​ = φ = 0, 51 · 680 = 346, 8

ℎ ℎ

●​ .

= (1 − φ) = 0, 49 · 680 = 333, 2

ℎ ℎ

Mass balances are:

= + −(1−φ) 0,4 1−0,51

→ →

= φ + (1 − φ) = = − = 0, 784 − 0, 961

0,51 0,51

φ

= 0, 784 − 0, 961

Equilibrium equations are:

=

1

(1 − ) = (1 − )

2

Using Excel, the system constituted by the mass balance and the relative volatility is solved,

starting from a random guess of x and y and minimizing the function

2

( )

2 6,7

( 0, 784 − 0, 961 − ) + −

1−+6,7 33

The solution is x =0,186 and y =0,605.

1 1 0,605·346,8

The recovery is calculated by .

= = = 0, 771 = 77, 1%

0,4·680

In order to size the flash drum we need to calculate F and K .

L,V drum

ρ ρ

The flux parameter is

= =

ρ ρ

,

To obtain, L , the ponderal flowrate of the liquid, we use:

P [ ]

= · = · + (1 − ) =

1 2

= 333, 2 · [

0, 186 · 86 + 0, 814 · 114

] = 36. 249 .

ℎ ℎ

With the same procedure: [ ]

= · = · + (1 − ) =

1 2

= 346, 8 · [

0, 605 · 86 + 0, 395 · 114

] = 33. 660 .

ℎ ℎ

The ponderal density of the vapor is given by

[ ]

ρ = · = · + (1 − ) = · [ 0, 605 · 86 + 0, 395 · 114

]

1 2

.

Assuming the vapor phase as a mixture of ideal gasses:

1 ·97

ρ = · 97 = · 97 = = 3, 1 = 3, 1 .

3

·

·378

0,082

·

The flux parameter is therefore:

3,1

ρ 36.249 3

ℎ

= = · = 0, 0718

.

ρ

, 33.660 696

3

ℎ

From the following diagram we can evaluate K to get a maximum 5% of entrainment:

drum 34

so .

= 0, 133

From the obtained value we can ρ −ρ 696−3,1

= = 0, 133 = 1, 99 .

ρ 3,1

,

To increase the caution, the max velocity is multiplied by a safety coefficient 0,85:

' = · 0, 85 = 1, 69

, ,

To obtain V , the volumetric flow rate of vapor: where c is the molar

=

V V

concentration of the vapor phase. 1

c has been used before: .

= = 0, 032

v 3

0,082·378

The volumetric flowrate is therefore: 3

= = 346, 8 /0, 032 = 10. 837, 5 .

3 ℎ

ℎ

Remembering that the minimum area of the drum is:

=

3 2

10.837,5

= = /1, 69 = 1, 78 .

3600

2 4

.

= = 1, 51

The minimum diameter is therefore: →

= π π

4 35

.

= 1, 6

Choosing from the industrial fabricated tank, we get

= 4 · 1, 60 = 6, 4

We choose →

= 4

ENTHALPY-COMPOSITION DIAGRAM)

(J-x DIAGRAM)

An adiabatic mixing (at constant enthalpy) of two compounds is considered.

The material balances are always valid:

●​ total = +

●​ for the components = +

and the energy balance = +

Combining the three equations we obtain: −

−

= =

− −

which is the equation of a straight line.

In a J-x diagram, an adiabatic mixing process is represented by a straight line.

The distance between M and A is proportional to B and the distance between M and B is

proportional to A → lever rule.

If heat is exchanged during mixing: = + +

3 schemes are shown which, from the point of view of heat balance, are equivalent → the

same point in the graph is reached: 36

Let's consider scheme b:

1.​ A is heated → the enthalpy of A increases but the composition does not change

2.​ mixing occurs in adiabatic conditions (A, B and M are aligned)

The same result occurs if:

1.​ B is heated first

2.​ adiabatic mixing with A occurs (alignment)

If we consider scheme c:

1.​ adiabatic mixing of A and B

2.​ heating of the mixture → the enthalpy of M increases 37

MAKING the J-x-y DIAGRAM

A binary mixture is considered.

In the case of pure components, the enthalpy is given by the integral .

= ∫

0

The temperature T indicates a reference in which the enthalpy is assumed to be zero.

0

In the J-x-y diagram the enthalpies of the pure compounds are those present at the edges of

the graph (in the same way as the T in the T-x-y diagrams or the p in the p-x-y diagrams).

sat sat

When two components are mixed at their reference temperature (both enthalpies are zero),

the resulting mixture always has J=0, but the temperature has increased → the isotherms in

the graph are curves.

A mixture of composition z, at a temperature T has an enthalpy given by

= + ∫

, ,

0 0

The mixing enthalpy at a temperature T is given by:

∆ = ∆ + ∫ ∆

, ,

0 0

[ ]

where is the specific heat at constant pressure.

∆ = + + (1 − )

, ,1 ,2

The bubble curve is given by the points where the liquid mixture is saturated: ∆ = ℎ

,

where T is the boiling temperature → for each value of z

B

.

ℎ = ∆ + ∫ ∆ ()

,

0 0 38

The dew curve is constructed by adding the latent heat of vaporization (which depends on the

composition) to the enthalpy of the saturated liquid → for each z

.

= ℎ + λ()

FLASH PROCESS in the J-x-y

DIAGRAM

The mass balance is a linear equation so it is represented by a line in the diagram → tie line.

1.​ increase of system pressure from p to p’ in a pump

2.​ heating of the feed

3.​ decrease of pressure from p’ to p through a throttling valve

4.​ partial adiabatic vaporization of the mixture and division into two streams.

It is possible to express the exchanged thermal energy in units of created vapor by

−

constructing the triangle LVV’ similar to LF’F → .

'

=

39

To obtain the tie lines, we start from the composition graph (x-y diagram) and draw a

horizontal line: it intersects with the curve of the diagram and with the diagonal → by

translating point A vertically in the J-x-y diagram we obtain point L, by translating B instead

we obtain point V. 40

AZEOTROPE J-x-y DIAGRAMS

If the reference compound is the most volatile ( ) the slope of the tie line is positive.

α > 1

At azeotrope → the slope is vertical because x=y.

α = 1

QUENCH PROCESS

The quen