How important is the estimation of minor head loss in transport of fluids

 It depends on several factors. In some cases minor head losses may be too small to waste time in its calculations but in some other it may not.

A discussion and some guidlines can be given about this matter.

Two pipe design cases

Head loss $h_L$, head added $h_A$ and head removed $h_R$ are adjustments to the Bernoulli equation so that these only make sense when the piping components are such that need to be taken into account. For example,

• when control valves are present in a pipe or branch,
• when the piping design is a network,
• when pumps or turbines are part of the pipe,
• among other cases,

heads need to be calculated for a proper balance. Head losses could be of the order of two digits.

However, if the pipe is not too long or intricated then heads could be expected to be small. In this case, head loss could be of the order of one digit.

Factors incresing/decreasing the head loss $h_L$

Head losses may be increased by some factors. Here are some:

• large pipe sections,
• pipe length to diameter ratio larger than three digits,
• pipe accesories such as: elbows, high pressure drop valves, invasive flow meters (orifice plate, venturi meter, etc.),
• pipe elevations.

Warning

You should always estimate head losses to give a reliable result and reasonable design. However, rough estimations discarding some head losses could only be possible in early design stages to have a quick idea of the overall piping.

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• LE01 - AC and DC voltage measurement and continuity test
• LE 02 - Start and stop push button installation 24V DC
• LE 03 - Turn on/off an 24V DC pilot light with a push button
• Minor losses  - Formulas
• What is a process variable?
• What are the most important process variables?
• Time dependence of process variables
• A list of process variables

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Ildebrando.

Solving the Colebrook equation

 Finding the friction factor $f_F$ can be big trouble for some students with poor numerical methods background while some other will try to go around by using the diagram of Moody. There is nothing wrong with the Moody diagram but you cannot automate any calculations with that. Besides, we are in the XXI century!

First, the friction factor $f_F$ in the Colebrook equation cannot be isolated because this is a trascendental equation. Since $f_F$ appears in the argument of the function and in other terms only numerical solutions are possible. There are several approximations to the Colebrook equation, like that due to Swamee-Jain (shown below),

$f_F=\dfrac{0.25}{\left[ \log_{10}\left( \dfrac{\epsilon}{3.7D} + \dfrac{5.74}{N_{Re}^{0.9}} \right) \right]^2}$        Eq. (1)

which give $f_F$ explicitly. However, these approximations are valid for a range of parameters only. It is not easy to take care of these restrictions all the time.

As the Colebrook equation has a more general scope and very practical tools are already available, its numerical solutions should not be painful anymore. This equation is,

$\dfrac{1}{\sqrt{f_F}}=-2\log_{10}\left( \dfrac{\epsilon}{3.72D}+\dfrac{2.51}{N_{Re}\sqrt{f_F}} \right)$        Eq. (2)

where $\epsilon$ is the pipe roughness, $D$ is the pipe inside diameter, and $N_{Re}$ is the Reynolds number.

Data for some common pipe roughness $\epsilon$

Pipe material	Roughness $\epsilon$ (m)	Roughness $\epsilon$ (ft)
Glass	Smooth	Smooth
Plastic	$3.0 \times 10^{-7}$	$9.8 \times 10^{-7}$
Drawn tubing; copper, brass, steel	$1.5 \times 10^{-6}$	$4.9 \times 10^{-6}$
Steel, commercial or welded	$4.6 \times 10^{-5}$	$1.5 \times 10^{-4}$
Galvanized iron	$1.5 \times 10^{-4}$	$5.0 \times 10^{-4}$
Ductile iron - coated	$1.2 \times 10^{-4}$	$4.0 \times 10^{-}$
Ductile iron - uncoated	$2.4 \times 10^{-4}$	$8.0 \times 10^{-4}$
Concrete, well made	$1.2 \times 10^{-4}$	$4.0 \times 10^{-4}$
Riveted steel	$1.8 \times 10^{-4}$	$6.0 \times 10^{-3}$

Numerical estimation of $f_F$

First, you will need to fixed all parameters $N_{Re}$, $\epsilon$, $D$ but $f_F$ (just in case). All parameters must be in the same unit system!

The friction factor $f_F$ can be iteratively approached by rewritting Eq. (2) as follows,

$f_F=\dfrac{0.25}{\log_{10}\left( \dfrac{\epsilon}{3.72D}+\dfrac{2.51}{N_{Re}\sqrt{f_F}} \right)^2}$        Eq. (3)

The iterative process is as follows. In iteration #1, substitute a guess for $f_F$ on the right hand side of Eq. (3) as shown,

$f_F^{New}=\dfrac{0.25}{\log_{10}\left( \dfrac{\epsilon}{3.72D}+\dfrac{2.51}{N_{Re}\sqrt{f_F^{Guess}}} \right)^2}$        Eq. (4)

and check for the error on satisfying this equation with,

$\left|\dfrac{f_F^{Guess}-f_F^{New}}{f_F^{Guess}}\right| \times 100$        Eq. (5)

For iteration #2, use $f_F^{New}$, from iteration #1, as the $f_F^{Guess}$. Check again for the % error, which should have decreased. Continue the iterations until a reasonable 0.1% error has been achieved.

Estimating $f_F$ with Google Sheets

As you may think, this iterative procedure is a perfect candidate for implementation in Google Sheets. Follow the link below to access a sheet automated to estimate the friction factor:

Friction factor estimation

Any question? Write in the comments and I shall try to help.

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Ildebrando.