Steady State Distillation Process

In summary: I don't understand why the heat balance equation includes a term for the enthalpy. (ii) Identify where any constitutive equations may be required for themodelling processes.I am not sure what you are asking.In summary,A steady state distillation process is shown diagrammatically as in FIGURE 1.There are two states that need balancing, and four mass and energy balance equations are involved. Any constitutive equations may be necessary for the modelling processes.
  • #1
topcat123
78
1
A steady state distillation process is shown diagrammatically as
FIGURE 1.

Assuming no heat losses to the atmosphere:
(i) Write four balanced equations for this system.
(ii) Identify where any constitutive equations may be required for the
modelling processes.There are two states that I can think of that need balancing.

mass and energy balances

mass in = mass out + accumulation
energy in = energy out + accumulation

I am not overly sure how this becomes 4 but
this is my idea is

Mass x in = mass x out + accumulation
Mass y in = mass y out + accumulation

energy in = energy out Top + accumulation
energy in = energy out Bottom + accumulation

Any help would be appreciated, and I am stuck with the constitutive equations.

Thanks
Ross
 

Attachments

  • FIG. 1.png
    FIG. 1.png
    6.5 KB · Views: 631
Physics news on Phys.org
  • #2
How do the words "steady state" and "accumulation" relate to one another?
Let's see the detailed mass balance equations you are considering?
What form of constitutive equation would you use to represent the enthalpy per unit mass (or mole) of a stream?
 
  • #3
Thank you for your response.
After reading my literature over and over, I believe I have answered the question, if you could have a look.

There are three mass balance equations.

Overall Mass balance

mass in = mass out + accumulation

Component Mass balance

mass of X in = mass of X out + accumulation

mass of Y in = mass of Y out + accumulation

The accumulations for both X and Y can be zero, positive or a negative value but in a steady state should be zero.

Energy balance.

energy in = energy out + accumulation

The accumulation in this equation is the condenser which will have a negative value.Constitutive equations

The specific heat capacity C will varies with temperature using the relationships of the type. The specific heat capacity will change the energy transfer rate.

CT=CO+aT+bT^2+aT^3+...

Where

CT is the specific heat capacity at temperature T

CO is the specific heat capacity at temperature 0 (K or °C)

T is temperature in K or °C

a, b, c … are constants for the given material.

The ideal gas equation may be used in the mass equation as the Density will change with temperature.

ρ=pMr/RT

Where

ρ = density (kg m–3)

p = absolute pressure (Pa)

Mr = relative molecular mass (kg kmol–1)

R is the ideal gas constant (J kmol–1 K–1)

T is the absolute temperature (K).
 
  • #4
I have problems with your assessment. There are too many issues to cover in detail. But, here are a few:

The term "steady state" means that there is no accumulation. So there is no accumulation in any of your mass and heat balances. I have no idea how you can say that there is negative accumulation in the condenser.

There are only two independent mass balances, not 3. Do you know why?

How in the world does the ideal gas law enter the picture? 1 liquid stream enters and two liquid streams exit.

The specific enthalpy of a stream should be expressed as a function h(x,T) of the mole fraction x and the temperature T. That's all you need.

You can't simply use heat capacity to get the enthalpy because you have a 2 component system, and the enthalpies of formation of the two species need to be included in expressing h(x,T) because their mole fractions change.

Let's see some actual mass balance and heat balance equations.
 
  • #5
Thank for your response.

topcat123 said:
(i) Write four balanced equations for this system.

I can see there are two mass balance equations and one enthalpy .

Component Mass balance
As we have a steady state

mass of X in = mass of X out

mass of Y in = mass of Y out

Energy balance.

Energy balance.

energy in = energy out + energy removed by the condenser

I don't see a 4th equation?

Constitutive equations
These are the constitutive equations covered in the info i have been provided with

• property relationships/equations of state
• transport flux relationships
• reaction rate expressions
• equilibrium expressions
• fluid flow relationships.

Think I am stuck a bit here.
 
  • #6
I didn't mean to write the equations in a general way. They give you symbols and mole fractions in the figure. Let's see the equations in terms of these. Also, for the heat balance, call h(X,T) the enthalpy per mole of a given stream, QB the rate of heat added at the boiler and QC the rate of heat removed at the condenser. Let's see the heat balance in terms of the flow rates and these.

Chet
 
  • #7
so
Mass Balance equation for component X
0.6qF = 0.99qT + 0.10qB

Mass Balance equation for component Y
0.4qF = 0.01qT + 0.90qB

I am struggling with the enthalpy.
The equation I have been given is:
Q = qm(To - Ti)

Chestermiller said:
h(X,T)
I don't recognise this.

Qh = qmbCTb + QC + qmtCTt - qmiCTi
 
  • #8
topcat123 said:
so
Mass Balance equation for component X
0.6qF = 0.99qT + 0.10qB

Mass Balance equation for component Y
0.4qF = 0.01qT + 0.90qB

The symbol q is usually used for volumetric flow rate. In the mass balances, you should be using mass flow rates F, T, and B.
I am struggling with the enthalpy.
The equation I have been given is:
Q = qm(To - Ti)I don't recognise this.

Qh = qmbCTb + QC + qmtCTt - qmiCTi
I don't recognize this. The heat balance I would write would be:

$$Fh(0.6,T_f)+Q_b-Q_t=Th(0.99,T_t)+Bh(0.1,T_b)$$
If x and y formed an ideal solution, I would represent h(X,T) by:
$$h(X,T)=Xh_x(T)+(1-X)h_y(T)$$where
$$h_x(T)=\int_{T_0}^T{C_x(T')dT')}$$and
$$h_y(T)=\int_{T_0}^T{C_y(T')dT')}$$ where T0 is a reference temperature, and T' is a dummy variable of integration.
 
  • #9
[tex]Fh(0.6,Tf) = \text{ energy of X in }[/tex]
Qb = energy out bottom
Qt = energy out top
Th(0.99,Tt) = energy out X top
Bh(0.1,Tb) = energy out bottom

Chestermiller said:
Fh(0.6,Tf)+Qb−Qt=Th(0.99,Tt)+Bh(0.1,Tb)
I don't see how this balances
do we not need something in for the condenser?

in the mass balance I should have used qm.
in the information i have been given qm = mass flow rate qv = volume flow rate

X = specific heat capacity where I use C

all the examples i have been given do not show any integration.
 

Attachments

  • excert.png
    excert.png
    6.7 KB · Views: 574
  • #10
topcat123 said:
[tex]Fh(0.6,Tf) = \text{ energy of X in }[/tex]

This is not the energy of x in. This is the rate of enthalpy entering the column in the feed stream.

Qb = energy out bottom
Qt = energy out top
Th(0.99,Tt) = energy out X top
This is the rate of enthalpy exiting in the product stream
Bh(0.1,Tb) = energy out bottom
This is the rate of enthalpy exiting in the bottoms.

I don't see how this balances
do we not need something in for the condenser?
My heat balance says that the rate of enthalpy exiting in the bottoms and product streams is equal to the rate of enthalpy entering in the feed stream + the rate of heat addition to the reboiler - the rate of heat removed at the condenser.
in the mass balance I should have used qm.
in the information i have been given qm = mass flow rate qv = volume flow rate

X = specific heat capacity where I use C

all the examples i have been given do not show any integration.
If the heat capacity is a function of temperature, you need to integrate with respect to temperature to get the enthalpy. Didn't they cover this in your thermo course?
 
  • #11
I am just looking at this question part (i)

Is it not just asking for 4 basic equations

Equation 1 Overall balance = Top Product + Bottom product
Equation 2 X Balance 0.6F = 0.99T + 0.1 B
Equation 3 Y Balance 0.4F = 0.01T + 0.9 B
Heat Balance = Heat in = Heat out
 
  • #12
Would the main balance equations be:

Overall Mass Balance
Component X Mass Balance
Component Y Mass Balance
Energy Balance

and then the Constitutive Equation would be the Heath Balance equation?
 
  • #13
I submitted my suggestions from post number 11 in this thread and was given a correct answer for it

Feed F = Top Product (T) + Bottom Product (B) (Equation 1)

X-balance 0.6qF=0.99qT + 0.10qB (Equation 2)

Y-balance 0.4qF=0.01qT + 0.90qB (Equation 3)

Heat Balance
Heat in = Heat Out
(Equation 4)
 
  • #14
David J said:
I submitted my suggestions from post number 11 in this thread and was given a correct answer for it

Feed F = Top Product (T) + Bottom Product (B) (Equation 1)

X-balance 0.6qF=0.99qT + 0.10qB (Equation 2)

Y-balance 0.4qF=0.01qT + 0.90qB (Equation 3)

Heat Balance
Heat in = Heat Out
(Equation 4)

Hi David

What did you but for the ii) part of the question?

I was thinking it would be:

Q=UA(Tf−Ts)
where
U = overall heat transfer coefficient for the system (Wm^−2∘C^−1)
A = heat transfer area (m2).
 
  • #15
Which question are you looking at? is it 2 (ii) The answer to this is disguised in one of the lessons but I basically answered as below

(ii) Identify where any constitutive equations may be required for the modeling process.

As per the lesson constitutive equations are used to better define individual terms within the balance equation, such things as specific heat capacity, density of gas, transport flux relationships, reaction rates, equilibrium expressions, fluid flow relationships
 
  • Like
Likes David Deal
  • #16
David Deal said:
Hi David

What did you but for the ii) part of the question?

I was thinking it would be:

Q=UA(Tf−Ts)
where
U = overall heat transfer coefficient for the system (Wm^−2∘C^−1)
A = heat transfer area (m2).
Hi David, did you find the answer for this?
 
  • #17
Tromso80 said:
Hi David, did you find the answer for this?
David J last visited Physics Forums on Jan 15 of this year.
 
  • #18
Good Morning, please remind me of the TMA number for this as its been some time since I completed the study.
 
  • #19
Write 4 balanced equations for this system

Feed F = Top Product (T) + Bottom Product (B)

(Equation 1)

X-balance 0.6qF=0.99qT + 0.10qB (Equation 2)

Y-balance 0.4qF=0.01qT + 0.90qB (Equation 3)

Heat Balance

Heat in = Heat Out

(Equation 4)

Identify where any constitutive equations may be required for the modeling process.

As per the lesson constitutive equations are used to better define individual terms within the balance equation, such things as specific heat capacity, density of gas, transport flux relationships, reaction rates, equilibrium expressions, fluid flow relationships
 

What is a steady state distillation process?

A steady state distillation process is a method used to separate and purify different components of a liquid mixture based on their boiling points. It involves continuously feeding the mixture into a distillation column and collecting the vaporized components at different heights in the column.

What are the main components of a steady state distillation process?

The main components of a steady state distillation process include a distillation column, a heat source, a condenser, and a collection system. The distillation column is where the separation of components takes place, the heat source provides the energy needed for vaporization, the condenser cools and condenses the vapor, and the collection system collects the separated components.

What are the factors that affect the efficiency of a steady state distillation process?

The efficiency of a steady state distillation process can be affected by several factors, including the composition and properties of the mixture, the design and operation of the distillation column, the temperature and pressure conditions, and the flow rate of the mixture.

What are the advantages of using a steady state distillation process?

There are several advantages of using a steady state distillation process, including its ability to handle large quantities of liquid mixtures, its energy efficiency, and its versatility in separating a wide range of components. It also produces high-purity products and can be easily controlled and automated.

What are the limitations of a steady state distillation process?

Despite its many advantages, a steady state distillation process also has some limitations. It may not be suitable for separating components with similar boiling points, and it can be costly to set up and maintain. It also requires a skilled operator to monitor and adjust the process, and it may not be suitable for separating heat-sensitive components.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
11
Views
1K
  • Thermodynamics
Replies
2
Views
856
  • Engineering and Comp Sci Homework Help
Replies
13
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
877
  • Engineering and Comp Sci Homework Help
Replies
2
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
847
Back
Top