Important equations in an absorption tower

 Well, these equations correspond to a general overview. These relevant equations are given for each stream. In this case, only one component is being separated from the mixture.

Simple sketch of an absorption equipment showing its streams.

Notice that in the figure above, $x$ is used for all mol fractions in the streams for the sake of clarity. However, $x$ for the liquid and $y$ for the vapor phase are to be used. Also, $L_S$ and $G_S$ represent the solvent streams.

For the gas mixture inlet at the bottom

For this stream, the mol ratio is defined as,

$Y_1=\dfrac{y_1}{1-y_1}=\dfrac{\bar{p_1}}{p_t-\bar{p_1}}$        Eq. (01)

The stream of gas mixture entering the tower is composed of,

$G_1=\text{component of interest} + \text{solvent}$        Eq. (02)

The solvent part of $G_1$ is represented as $G_S$. The flow rate of solvent $G_S$ in $G_1$ or $G_2$ is defined as,

$G_S=G_1y_{1S}=G_1\left( 1-y_1 \right)=\dfrac{G_1}{1+Y_1}$        Eq. (03)

where $y_{1S}$ is the mol fraction corresponding to the solvent in the gas mixture at the inlet. Since $G_1$ is a gas stream, it is natural to think that $Y_1$ can be expressed in terms of the total $p_t$ and partial $\bar{p}$ pressures too.

For the gas mixture outlet at the top

The gas stream at the top is $G_2$. Since no solvent in $G_2$ is passed to the liquid phase, it is straightforward that $G_S$ is the same at the bottom and at the top of the tower. Also, $y_{1S}=y_{2S}$, too.

For this stream, the mol ratio is defined as,

$Y_2=\dfrac{y_2}{1-y_2}=\dfrac{\bar{p_2}}{p_t-\bar{p_2}}$        Eq. (04)

The solvent stream can also be estimated using the top data, with,

$G_S=G_2y_{2S}=G_2\left( 1-y_2 \right)=\dfrac{G_2}{1+Y_2}$        Eq. (05)

For the liquid mixture, or not, inlet at the top

For the liquid phase, the story is very similar to the phase stream. The mol ratio is defined as,

$X_2=\dfrac{x_2}{1-x_2}$        Eq. (06)

while the liquid solvent stream is defined as,

$L_S=L_2x_{2S}=L\left( 1-x_2 \right)=\dfrac{L_2}{1+X_2}$        Eq. (07)

where again $x_{2S}$ is the mol fraction corresponding to the solvent in the liquid mixture at the inlet $L_2$. Since no part of the liquid solvent is passed to the gas stream,

$x_{2S}=x_{1S}$        Eq. (08)

too.

For the liquid mixture outlet at the bottom

For the bottom of the tower, the equations are barely the same as Eqs. (06-07).

Ildebrando.

What are the units of $m$ in the power-law model?

 In this post, I present a little algebraic calculation to estimate the units so that the Reynolds number $N_{Re}$ remains dimensionless. You should recall that incorrect units on $m$ would lead you to numerical errors.

I shall then start with the formulas already presented for non-Newtonian fluids in the post: Hydraulic equations for non-Newtonian fluids. Then, the Reynolds number is defined as,

$N_{Re}=\dfrac{(4n)^{n}\,D^n\,V^{2-n}\rho}{g_c\,m\,(3n+1)^n8^{n-1}}$        Eq. (01)

 where the constant $g_c$ is defined as $32.174\,lb_m\,\cdot \,ft/lb_f\, \cdot s^2$. Also, the diameter $D$ is used in $ft$, the fluid velocity is used in $ft/s$, and the fluid density must be used in $lb_m/ft^3$. The flow index $n$ is dimensionless.

Thus, we may envisage the units of $m$ by considering, from the Reynolds number in Eq. (01), solely,

$\dfrac{D^n\,V^{2-n}\rho}{g_c\,m}$        Eq. (02)

Next, if we substitute the units of all variables listed above, we obtain,

$\dfrac{(ft)^n\cdot \left(\dfrac{ft}{s}\right)^{2-n}\cdot \dfrac{lb_m}{ft^3}}{\dfrac{lb_m \cdot ft}{lb_f \cdot s^2}\cdot m}$        Eq. (03)

A simplification process leads to,

$\dfrac{\dfrac{lb_m}{ft \cdot s^{2-n}}}{\dfrac{lb_m \cdot ft\cdot m}{lb_f \cdot s^2}}$        Eq. (04)

$\Rightarrow \dfrac{lb_m \cdot lb_f \cdot s^2}{ft \cdot s^{2-n} \cdot lb_m \cdot ft\cdot m}$        Eq. (05)

Further simplification leads to,

$\dfrac{lb_f }{ft^2 \cdot s^{-n}\cdot m}$        Eq. (06)

From Eq.  (06), it is easily seen that to get all units cancelled, the parameter $m$ must have units:

$m=\left[\dfrac{lb_f}{ft^2 \cdot s^{-n}}\right]$        Eq. (07)

or using slugs, since $1\, lb_f\cdot s/ft^2= 1\, slug/ft \cdot s$,

$m=\left[\dfrac{slug}{ft \cdot s^{2-n}}\right]$        Eq. (08)

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

Ildebrando.

Key Excel features into Python add-in

 In this post, two very common features of Excel are tested within an add-in for Python from Anaconda. These features are:

• cell fixing with $ and
• repeated calculation by the drag and drop function of Excel.

For short, you should relax since YES, these two features are actually available in the Python add-in from Anaconda. In other words, you may use the $ to fix cells so that some data remains constant within a formula. See for reference the image below,

Also, if you inserted your code in a given cell, say B9, for example, you can drag and drop from that cell so that other cells fill in with the same formula while some cells remain fixed, just like B4, B5 and B8.

I hope you find this post helpful.

A case of conversion - From $lb_m$ to $gpm$

 Imperial units prove to be somtimes hard to use. Here the case of conversion from mass to volumetric flowrate is presented.

As one may imagine a physical property of the fluid is required: the density $\rho$. This can be written as,

$\dot{Q}= Q\, \rho$        Eq. (01)

where $\dot{Q}$ is the mass flow rate, in $lb_m/hr$ for example, and $Q$ is the volumetric flow rate, in $gpm$, for example.

Let us now consider the case of $40,000.00\, lb_m/hr$ flowing liquid water at $190\,^\circ \, C$. In this case, its density would be $\rho=54.70\,lb_m/ft^3$.

Using Eq. (01), it follows,

$Q=\dfrac{\dot{Q}}{\rho}=\dfrac{40,000.00\, lb_m/hr}{54.70\, lb_m/ft^3}$

$Q=924.56\,ft^3/hr=115.27\,gpm$

which is the desired conversion. If you were working with steam, working pressure must be considered to get the proper fluid density.

Commercial steel pipe Schedule 80 data

 You should be familiar with the data provided in the Appendix B of Flow of fluids through valves, fittings and pipes, for commercial steel pipes with schedule 40 for the flow of water and air.

Several questions always arise: what is the data for schedule 80 commercial steel pipe? what are the data for other pipes (PVC, copper, etc.)? This  data is available on the internet and you should search for it. However, this post is devoted to provide the data for commercial steel pipe schedule 80 and to serve as proof that this information is provided by the manufacturers.

Here, two files are provided:

• Data for flow rate and friction loss for commercial steel pipe schedule 80 #01
• Data for flow rate and friction loss for commercial steel pipe schedule 80 #02

Make a comparison with the data provided in the reference mentioned early. These files data are for water too.