Darcy friction factor $f_F$ solver. Based on the Colebrook equation.

Darcy Friction Factor ($f_F$) Calculator

Colebrook Equation Numerical Solver

Fluid Velocity, $v$ (m/s):

Kinematic Viscosity, $\nu$ ($\text{m}^2/\text{s}$):

Pipe Internal Diameter, $ID$ (m):

Pipe Roughness, $\epsilon$ (m):

Reynolds Number ($N_{Re}$): -

Flow Regime: -

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Darcy Friction Factor ($f_F$): -

Calculadora de la densidad del agua

Calculadora de la densidad del agua

La densidad del agua varía en función de la temperatura. Para obtener un cálculo preciso en un amplio rango, se utiliza la siguiente ecuación polinomial empírica:

\[\rho(T) = \frac{999.84 + 16.95 T - 7.99\times 10^{-3} T^2 - 46.17\times 10^{-6} T^3 + 105.56\times 10^{-9} T^4 - 280.54\times 10^{-12} T^5}{1 + 16.88\times 10^{-3} T}\]

Donde:

• \(\rho\) es la densidad del agua en kg/m³.
• \(T\) es la temperatura en °C.

Ingresa la temperatura (°C):

Densidad Calculada:

--- kg/m³

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.