Mass balance on rectifying and stripping plates

 This post deals with the mass balance aplied to a distillation column. This has to do with something called fractional distillation. As you may know in this kind of operation, the column top section is called: rectifying section; while that of the bottom is called: stripping section; provided the feed is in the middle plate.

Mass balance on the rectifying section

Or better said, on the plates of the top section. This applies for a binary mixture. In Fig. 01 a general representation of this is presented. 

• All plates are numbered from the top
• $n$ is the last plate on the rectifying section. Of course, $n$ is also a counter going from 1 to the last plate.
• The referred component on the mol fractions $(x,y,z)$ is the most volatile

Fig. 01 Mass balance on the rectifying section.

In this way a global mass balance can be easily writen as,

$G_{n+1}=L_n+D$        Eq. (01)

On the other hand, a mass balance for the component of interest (say component $A$)  on each plate on this section can be easily written as,

$G_{n+1}\, y_{n+1}\, -\, L_n\, x_n \,=\,D\, z_D$        Eq. (02)

From Eq. (02), the mole fraction in the vapor entering the nth plate can be isolated to be,

$y_{n+1}=\dfrac{L_n}{G_{n+1}}x_n+\dfrac{D}{G_{n+1}}z_D$        Eq. (03) 

One important simplification on Eq. (03) can be made on the nature of the flow rates entering and leaving the plates: the flow rates of the liquid and vapor are constant always. This is a very strong assumption and is at the heart of the method of calculation called after Sorel and Lewis.

In this way, Eq. (03) becomes,

$y_{n+1}=\dfrac{L}{G}x_n+\dfrac{D}{G}z_D$        Eq. (04)

Mass balance on the stripping section

In this case, we just repeat the same process as before. Fig. 02 shows an schematic of this section.

Fig. 02 Mass balance on the stripping section.

In. Fig. 02 variables for vapor and liquid have been barred $(\bar{V},\bar{L})$ to avoid confusion with those of the rectifying section. Also, the counter $m$ has been introduced to number the plates of the stripping section. The global mass balance is,

$\bar{L}_m=\bar{G}_{m+1}+W$        Eq. (05)

In the same way as before, from the mass balance for the component $A$ the following equations arise.

$\bar{L}_m x_m=\bar{G}_{m+1}y_{m+1}+Wx_W$        Eq. (06)

and

$y_{m+1}=\dfrac{\bar{L}_m}{\bar{G}_{m+1}} x_m-\dfrac{W}{\bar{G}_{m+1}}x_W$        Eq. (07)

Again, using the idea of constant liquid and vapor flow rate art each plate, Eq. (07) reduces to,

$y_{m+1}=\dfrac{\bar{L}}{\bar{G}} x_m-\dfrac{W}{\bar{G}}x_W$        Eq. (08)

Recall that Eq. (08) comes from a rough approximation.

A simple mass and enthalpy balance on single stage distillation

 Here the case of distillation in a single stage is presented. This is a simple case corresponding to just boiling the mixture and then separating it in the column. Of course, neither plates nor packing is needed. Despite its simplicity it is worth to discuss the physics of the process.

Instantaneous evaporation

In sketch above a mixture feed $F$ is passed to a heat exchanger that boils the  liquid and then to a separator. In the separator, the vapor goes to the top while  the liquid goes to the bottom. A mass balance for this very schematic case is twofold: there is a global balance and a balance on the component of interest (which can be more than one). For the global mass balance we have,

$F=D+W$        Eq. (01)

and for the component of interest the balance is written as,

$Fz_F=Dy_D+Wx_W$        Eq. (02)

An energy balance based on the enthalpy of each stream can be written as,

$FH_D+Q=DH_D+WH_W$        Eq. (03)

where $Q$ is the heat supplied in the heat exchanger to boil the feed F. Equations (01-02) can be combined to give,

$\dfrac{W}{D}=\dfrac{y_D-z_F}{z_F-z_W}$        Eq. (04)

which can be a more useful relationship. Also, Eqs. (01,03) can also be combined. The result is,

$-\dfrac{W}{D}=\dfrac{H_D-\left( H_F+Q/F \right)}{H_W-\left( H_F+Q/F \right)}$        Eq. (05)

This is the end. Enjoy.

About slugs and pound units

 This post is intended to provide a common ground on the usage of these Imperial and US customary units.

The units $slugs$ and pound-mass ($lb_m$) are sometimes used as synonyms for mass units. However, only the $slug$ is a true unit of mass.

The $lb_m$ is something else. The equivalence between these two quantities is,

$1\, slug=1\, lb_f\cdot s^2/ ft= \dfrac{1\,lb_m}{32.17\,ft/s^2}$        Eq. (01)

where the unit $lb_f$, or pound-force, was introduced too. Many times, instead of writting $lb_f$, people just write $lb$. Also, $1.0\, lb_f$ has a mass of $1.0\, lb_m$ and the only way the above Eq. (01) really works is by introducing the constant $g_c$ as follows,

$g_c=\dfrac{32.17\, lb_m}{lb_f/\left( ft/s^2 \right)}=\dfrac{32.17\,lb_m\cdot ft/s^2}{lb_f}$        Eq. (02)

which is as you can see a adjusting parameter to fix the unit system.

One in which this $g_c$ occurs is in the well known formula,

$F=mg$        Eq. (03)

to calculate the force $F$ from the mass $m$ and the acceleration due to gravity $g$. Equation (03) works well in SI units but not if Imperial units are introduced. For these purposes, Eq. (03) must be changed to,

$F=m\dfrac{g}{g_c}$        Eq. (04)

Otherwise, physical inconsistencies would arise. However, another remedy is possible: make use of $slugs$ rather than $lb_m$, so that $g_c$ is no longer needed.

Working with Imperial units can be confusing but using the presented equivalencies you will succeed (hopefully).

The density of a mixture

- Two components of equal mass but different volume - 

Knowing the density of such a mixture is a common strange situation in engineering. This demonstration can be extended to more components provided the mass of each component is the same.

The situation

Consider that you have a solution, of a solid with certain density $\rho_S$ dilute in certain solvent of density $\rho_L$, in which both the solute and solvent are mixed in equal quantities. This is, the mass of the solute is equal to the mass of the solvent. What is the density of such a mixture?

An analytical approach

The solution to this case is not knew and can also be found elsewhere. However, in this post, some different flavor shall be given.

The density of the solute is,

$\rho_S=\dfrac{m_S}{V_S}$        Eq. (01)

while the density of the solvent is,

$\rho_L=\dfrac{m_L}{V_L}$        Eq. (02)

but since the mass of solute and solvent are equal, it follows,

$m=m_S=m_L$        Eq. (03)

Also, the density of the mixture should be,

$\rho=\dfrac{m_S+m_L}{V_S+V_L}=\dfrac{2m}{V_S+S_L}$       Eq. (04) 

The volumes, $V_S$ and $V_L$, in Eq .(04) can be determined from Eqs. (01-02) condering Eq. (03) as follows,

$V_S=\dfrac{m}{\rho_S}$        Eq. (05)

$V_L=\dfrac{m}{\rho_L}$        Eq. (06)

Substitution of Eqs. (05-06) into Eq. (04) produces,

$\rho=\dfrac{2m}{\dfrac{m}{\rho_S}+\dfrac{m}{\rho_L}}$

which can be simplified,

$\rho=\dfrac{2m}{\dfrac{\rho_L m+\rho_S m}{\rho_S \rho_L}}=\dfrac{2m}{\dfrac{m\left( \rho_L+\rho_S \right)}{\rho_S \rho_L}}=\dfrac{2\rho_S \rho_L}{\rho_L+\rho_S}$            Eq. (07)

Of course, Eq. (07) applies for the mixture of two liquids as well. Therefore, if you know the density of both substances, you can readily estimate the density of the mixture (provided the mass of the two components is the same).

A case to estimate the density of a mixture

Let us consider a slurry fluid made of the mixture of coal and water. The coal has specific gravity 2.5 while the slurry has composition 50% coal w/w. What is the density of the slurry?

Well, the slurry mixture is made of coal and water mixed in the same mass proportion. This is, the mass of the coal is the same as the mass of the water used to make the slurry. The volume is unknown: it can be the same but, who knows.

Starting with the specific gravity of the coal (solute), its density should be,

$\rho_S=\left(2.5\right) \left( 1000\,kg/m^3 \right)=2500\,kg/m^3$

For the water, let us consider that it is at room temperature (say, 25 °C). Similar temperatures around this one will give a very similar water density, so that you do not have to worry too much. Then,

$\rho_L=997.05\, kg/m^3$

Therefore, the density of such a slurry should be,

$\rho=1425.56\, kg/m3$

This is the end of the post. I hope you find it useful.

Ildebrando.

Hydraulic equations for non newtonian fluids

 Here the dydrualic equations for two knwon non newtonian fluids are presented. These are,

• the power law fluid and
• the Bingham plastic fluid.

These equations are to be focus on the flow rate $Q$, the Reynolds number $N_{Re}$ and the friction factor $f_F$, for the case of a fluid flowing through a pipe of circular section.

A comment on the friction factors

There are in fluid mechanics two different friction factors:

• the Darcy friction factor
• the Fanning friction factor

This is confusing if you are not aware of this fact. The Darcy friction factor (say $f_F$) and the Fanning's (say $f_{FF}$) are related as follows,

$f_F=4\,f_{FF}$

You should pay attention when using Darcy's equation for head loss since it is originally and commonly expressed in terms of $f_F$. This means that if you are using $f_{F}$ the head loss can be found from:

$h_L=2f_{FF}\dfrac{L}{D}\dfrac{V^2}{g}$

In this post, the formulas for the newtonian fluid use the Darcy friction factor $f_F$ while those for power law and plastic Bingham fluids use the Fanning friction factor $f_{FF}$.

Newtonian fluid

The flow rate $Q$ is given by,

$Q=\dfrac{\pi\,R^4}{8\mu}\dfrac{P_0-P_L}{L}$        Eq. (01)

where $R$ is the pipe internal radius and subscripts $0$ and $L$ indicate the end and final of the pipe.

The Reynolds number $N_{Re}$ is given by,

$N_{Re}=\dfrac{D\,V}{\nu}$        Eq. (02)

The friction factor $f_F$,for laminar conditions $N_{Re}<2000$, is given by,

$f_{F-L}=\dfrac{64}{N_{Re}}$        Eq. (03)

for turbulent flow regime $N_{Re}>4000$ by,

$\dfrac{1}{\sqrt{f_{F-T}}}=-2\log_{[10]}\left( \dfrac{\epsilon}{3.72D}+\dfrac{2.51}{N_{Re}\sqrt{f_{F-T}}} \right)$        Eq. (04)

Power law fluid

The flow rate $Q$ is given by,

$Q=\dfrac{\pi\,R^3}{(1/n)+3}\left(\dfrac{\left(P_0-P_L\right)R}{2mL}\right)^{1/n}$        Eq. (05)

where $R$ is the pipe internal radius and subscripts $0$ and $L$ indicate the end and final of the pipe. Formula in Eq. (05) was taken from the familiar book Dynamics of Polymeric Liquids by Bird R. B. et al.

The Reynolds number $N_{Re}$ is given by,

$N_{Re}=\dfrac{(4n)^{n}\,D^n\,V^{2-n}\rho}{g_c\,m\,(3n+1)^n8^{n-1}}$        Eq. (06)

where $g_c=32.174\,lb_m\,\cdot \,ft/lb_f\, \cdot s^2$ is a units correction factor used in the Britsh units system. The density of the fluid $\rho$ must be given in $lb_m/ft^3$. Equation (06) was taken from the article of Dodge and Metzner [AIChE J 1959 5 (2)].

The friction factor $f_{FF}$, for laminar conditions $N_{Re}<N_{Re-c}$, is given by,

$f_{FF-L}=\dfrac{16}{N_{Re}}$        Eq. (07)

and for turbulent flow $4000<N_{Re}<10^5$ regime by,

$f_{FF-T}=\dfrac{0.0682\,n^{-1/2}}{N_{Re}^{1/(1.87+2.39\,n)}}$        Eq. (08)

and for the transition region $N_{Re-c}<N_{Re}<4000$ by

$f_{FF-Tr}=1.79\times 10^{-4}\exp\left[ -5.24\,n \right]\,N_{Re}^{0.414+0.757\,n}$        Eq. (09)

Finally, the critical Reynolds number $N_{Re-c}$ is defined as,

$N_{Re-c}=2100+875(1-n)$        Eq. (10)

Equations (07-10) were taken from the paper of Darby et al. [Chem. Eng. 1992 99 (9)].

Plastic Bingham fluid

The flow rate $Q$ is given by,

$Q=\dfrac{\pi\,R^3\tau_w}{4\mu_\infty}\left[  1 - \dfrac{4}{3}\left( \dfrac{\tau_0}{\tau_w}\right)+ \dfrac{1}{3}\left( \dfrac{\tau_0}{\tau_w} \right)^4 \right]$        Eq. (11)

where $R$ is the pipe internal radius, $\mu_\infty$ is called the limiting viscosity and $\tau_0$ is the yield stress. Also, $\tau_w$

$\tau_w=\dfrac{\left(\mathbb{P}_0-\mathbb{P}_L\right)R}{2L}$        Eq. (12)

In Eq. (12) pressures $\mathbb{P}_0$ and $\mathbb{P}_L$ include the hydrostatic contribution in an inclined pipe. Formulas in Eqs. (11-12) were taken from the familiar book Dynamics of Polymeric Liquids by Bird R. B. et al. Subscripts $0$ and $L$ indicate the end and final of the pipe.

The Reynolds $N_{Re}$ and Hedstrom $N_{He}$ numbers are given by,

$N_{Re}=\dfrac{D\,V\, \rho}{\mu_\infty}$        Eq. (13)

$N_{He}=\dfrac{D^2\,\rho\,\tau_0 }{\mu_\infty^2}$        Eq. (14)

In this case the friction factor $f_{FF}$ is given for all flow regimes as,

$f_{FF}=\left( f_{FF-L}^m+f_{FF-T}^m \right)^{1/m}$        Eq. (15)

where,

$m=1.7+\dfrac{40000}{N_{Re}}$        Eq. (16)

and the friction factors for laminar $f_{FF-L}$ and turbulent $f_{FF-T}$ regimes are:

$f_{FF-L}=\dfrac{16}{N_{Re}}\left[ 1+\dfrac{1}{6}\dfrac{N_{He}}{N_{Re}}-\dfrac{1}{3}\dfrac{N_{He}^4}{f_{FF}^3N_{Re}^7} \right]$        Eq. (17)

$f_{FF-T}=\dfrac{10^a}{N_{Re}^{0.193}}$        Eq. (18)

$a=-1.47\left[ 1+0.146\exp\left( -2.9\times 10^{-5}N_{He} \right) \right]$        Eq. (19)

Notice that for plastic Bingham fluids there is no laminar-transition-turbulent regions reported, so that in order to find $f_F$ you must solve numerically Eq. (15). However, you may read the manuscript of Swamee and Aggarwal [J. Pet. Sci. Eng. 2011 76] if you would like to know about an effort to give a critical $N_{Re}$ for these fluids. Equations (13-19) were taken from the paper of Darby et al. [Chem. Eng. 1992 99 (9)].

Watch this video if you do not want to read. Enjoy!