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

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.