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.

Hydraulic equations for non newtonian fluids

 Here, the hydraulic equations for two known non-Newtonian fluids are presented. These are,

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

These equations are to 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,

$Q=\dfrac{\pi\,R^3}{(1/n)+3}\left(\dfrac{\left(\mathbb{P}_0-\mathbb{P}_L\right)R}{2mL}\right)^{1/n}$        Eq. (05)

where $R$ is the pipe internal radius and subscripts $0$ and $L$ indicate the end and final of the pipe. The formula in Eq. (05) was taken from the familiar book Dynamics of Polymeric Liquids by Bird R. B. et al. Also, the velocity given as a function of the pipe radius is,

$v_z=\left( \dfrac{\tau_R}{m} \right)^{1/n}\dfrac{R}{(1/n)+1}\left[ 1-\left( \dfrac{r}{R} \right)^{(1/n)+1} \right]$        Eq. (05a)

where,

$\tau_{rz}=\dfrac{\left( \mathbb{P}_0-\mathbb{P}_L \right)r}{2L}$        Eq. (05b)

and 

$\tau_{R}=\tau_{rz}\vert_{r=R}=\dfrac{\left( \mathbb{P}_0-\mathbb{P}_L \right)R}{2L}$        Eq. (05c).

Recall that $\tau_{rz}$ is the shear stress and $\tau_{r}$ is the stress evaluated at the pipe wall. You should be aware that the pressure difference $\mathbb{P}_0-\mathbb{P}_L$ in Eqs. (05) includes the effect of gravity. 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 British 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!

Mechanical energy added to a fluid - A pump

 An important component of the general mechanical energy equation

$\dfrac{p_1}{\gamma}+z_1+h_A-h_R-h_L+\dfrac{v_1^2}{2g}=\dfrac{p_2}{\gamma}+z_2+\dfrac{v_2^2}{2g}$        Eq. (01)

is $h_A$ since it represents the mechanical energy added to the fluid so that it can continue its trip through the pipe. At this point, two cases are visualized,

• the energy $h_A$ is one the system requires in order to the pipe system to do its work, and
• the energy $h_A$ is brought from the technical features of a real life pump, for example, so that the conditions of the pipe system are changed in turn.

Of course, in most technical problems both cases presented above are part of the solution in an iterative procedure.

What is the relationsip between $h_A$ and a pump

Since the pump takes electrical energy and transforms it into mechanical energy to be later transferred into the fluid, these  two parameters should be related. This is done through the effeciency of the motor,

$\mathbf{e_M}=\dfrac{Power\, output\, from \, the\, motor}{Power\, delivered\, by\, fluid}=\dfrac{P_O}{P_R}$    Eq. (02)

where $P_O$ is the mechanical power the motor so that the impeller may turn at certain speed with certain force and $P_R$ is the power the fluid actually receives. Since part of the power $P_O$ is wasted as heat or lost due to wearing of mechanical parts, $P_O>P_R$ and as a consequence $\mathbf{e_M}<1$. If $\mathbf{e_M}=1$ you would have an impossible thermodynamical machine. $P_O$ is a parameter measured and supplied, in the name plate, by pump manufacturers so that this data is easy to get.

Notice that Eq. (02) is the same if instead of a pump the case were that of a turbine (actioned by a mechanical energy of the fluid). We would be talking about $h_R$ instead of $h_A$ too.

For a centrifugal pump the power transferred into the fluid and the energy added $h_A$ are related as follows,

$P_R=h_A\gamma Q$        Eq. (03)

where $Q$ is the volumetric flow rate and $\gamma$ a property of the fluid. Also, from Eq. (03) it is obvious that $P_R$ would be very hard to measured. Therefore, it is more common to speak of the efficiency of a pump which we know is smaller than 1 but with present technological advances could be in the range of 0.8 - 0.9 for new equipments. Then, the usual case would be that $P_R$ is unknown, so that,

$P_R=\mathbf{e_M}P_O$        Eq. (04)

and consequently, Eq. (01) becomes,

$\mathbf{e_M}P_O=h_A\gamma Q$        Eq (05)

From Eq. (05), $h_A$ is isolated to be,

$h_A=\dfrac{\mathbf{e_M}P_O}{\gamma Q}$        Eq. (06)

In this way $h_A$ can only be estimated if the efficiency of the pump $\mathbf{e_M}$ and power of the motor of the pump $P_O$ are known.

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

==========
Ildebrando.

Gross estimation of pipe size (no head calculations)

 Can pipe size be determined without engineering calculations involving the Bernoulli or the energy balance equations?

The short answer is yes; but some experience may be needed and the result may only be point of start or guess for formal calculations.

Here is how to do it, pros and cons.

Some restraints to this approach

Before using this approach you better know the limitations of it:

1. since no pressure gradient nor head losses are being considered the estimated pipe size may not be in agreement with the hydraulic load balance,
2. the pipe size may be larger or smaller than required so that experience is to be used to reduce the margin of error,
3. you will need data sheets for different pipe materials,
4. this approach works better for pipe sections without fittings,
5. most data sheets only have data for water as working fluid (air is sometimes used as well).

Some advantages of this approach

Despite the errors, possibly involved by this method, there are some good things about it:

1. only real pipe sizes are used so that the result is better adapted to already sized equipment characteristics (pumps, heat exchangers, etc.),
2. you can get a good approach to the pipe size in a shorter time than by performing large engineering calculations,
3. pipe material, ranges for flow rate and fluid velocity so that head loss may be balanced are only required.

Gross estimation of the pipe size (no calculations)

The key is in a table having the following data:

• pipe nominal size,
• flow rate,
• fluid velocity,
• working fluid,
• pressure drop (a head loss indication).

A good example for commercial steel pipe Sch. 40 is given in the book Flow of fluids through valves, fittings and pipe. Crane Co. Look in its engineering data appendix. We shall not reproduce the table here since you may find the bok over the internet.

An example of pipe size gross estimation

Consider the a situation in which about $2.5\times 10^{-2}$ m$^3$/s of water are to be transported across 200 m. The pipe to be installed will need a number of bendings and little elevations.

What would be an estimated pipe size for this case?

Solution.

Let us first take a flow rate margin of 5% for the flow rate. Then, we would be looking for pipes capable of transporting about:

$Q=1,500 \pm 75$ L/min of water

Now, if you take a look to the data table from the book Flow of fluids through valves, fittings and pipe. Crane Co several pipe candidates, capable of carrying this amount of fluid, appear!

In fact, pipes of nominal diameters: 3 1/2 inch, 4 inch, 5 inch, 6 inch and 8 inch fit our needs.

What is the best option?

Here is where your engineering experience is important and integration of other information (fittings, elevations, pumps, etc.) help to make a decision.

You may choose one of the pipe sizes by answering the following questions with approximate values:

• How fast will the fluid travel through the pipe?

You should recall that as the internal diameter increases the fluid velocity reduces and vice versa. You may compare different pipe sizes for the same flow rate and see how the fluid velocity dicreases with the size.

• Do you expect the fluid to be highly pressurizsed?

Two options may arise here: to increase the pipe size or to increase the wall thickness (schedule). One way of going around this issue is lby chosing a pipe size whose average capacity matches your needs.

• Economic factors

The larger the pipe size the larger the economic bill. If the extension of the pipe line is short then perhaps economicas may not be a trouble.

• Do you expect head losses to be important?

The parameter $h_L$ is becomes important with the number of fittings, pipe elevation and when pipe line is too long. If you forsee a large $h_L$ in the order of 10 m, for example, then perhaps a larger pipe size would work better. However, the fluid velocity will be reduced too.

Now, back to the question on the most suitable pipe size.

For this case the pipe of ND 4 inch has been selected. The main reasons are:

1. it is large enough to allow a maximum flow rate of 2,200 L/min which is 147 % more than required,
2. being only the second pipe in size it may give a reasonable economic advantage,
3. if fittings are installed this pipe may still give a good margin to keep $h_L$ at reasonable values.

Note: Remember that this is just a gross estimation before more detailed and formal calculations but it may be a good guess.

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

Other stuff of interest

• Estimation of internal pipe diameter
• Who was Clausius?
• Units conversion
• LE 04 - Latch contact with encapsulated relay for turning on/off an AC bulb light
• About PID controllers
• Ways to control a process
• About pilot lights
• Solving the Colebrook equation

==========
Ildebrando.

How to make sure the fluid goes in the direction it suppose to go?

 For short, the answer lies in a concept widely used in pipe engineering: hydraulic load or just load. Unfortunately, this concept is not well understood. 

The hydraulic load can be understood as the amount of energy available to drive a fluid through a pipe or conduit. And if you compare the load at two different points along a pipe you could easily check for the direction in which the fluid travels.

The hydraulic load can be, mathematically, written as:

$H_L=\dfrac{p}{\gamma}+z$        Eq. (01)

where:

$H_L$ stands for the  hydraulic pressure at a given point along th pipe,

$p$ for pressure at that point along the pipe,

$\gamma$ for specific weight of the fluid,

$z$ vertical height of that point along the pipe.

As you may  have already noticed: the hydraulic load only makes sense when talking about a given location along a certain pipe.

Fig. 01 Flow direction cases according to the hydraulic load $H_L$ at two different locations.

In case 1 in Fig. 01 the flow travels upward and the only way of making sure this is so is by checking that $H_{L1}$ is greater than $H_{L2}$. Otherwise, the fluid will travel backwards.

 This concept is a mechanical balance coming from the Bernoulli equation. However, estimations on how greater $H_{L1}$ is with respect to $H_{L2}$ depends on some pipe features and equipments taken into account in the general energy balance equation.

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

Other stuff of interest

• Some basic concepts on thermodynamics
• Who was Clausius?
• Units conversion
• LE 04 - Latch contact with encapsulated relay for turning on/off an AC bulb light
• About PID controllers
• Ways to control a process
• About pilot lights
• Solving the Colebrook equation

==========
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.

=========

Ildebrando.

Hardy - Cross method - Head balance method for flow loops

 The main idea behind the Hardy-Cross method is to force a head loss $h_L$ balance among all pipes which are part of a flow loop. Without the $h_L$ balance flow rates, which are usually the unknown, cannot be determined. There are other methods but this is the most familiar.

This methodology, as you may already know, is based on the general equation for mechanical energy balance, so that every loop in the above network must be decomposed in smaller parts, which are in fact the pipe sections.

Note: A loop is a circuit of flow. In other words, a loop is formed by interconnected pipe sections through wich the fluid may travel in a given direction. For example, pipe sections [1], [4], [6] and [3] form a loop.

Then, for any of the 12 pipe sections in the network above has a particular head loss defined as follows,

$h_L^{(n)}=f_F^{(n)}\frac{L_n}{D_n}\frac{v_n^2}{2g}$        Eq. (1)

or as,

$h_L^{(n)}=f_F^{(n)}\frac{L_n}{D_n}\frac{8Q_n^2}{\pi^2 D_n^4 g}$        Eq. (2)

where $n$  indicates any of the pipe sections in a loop ([1], [4], [6] or [3], for example). Also $n$ was written in parenthesis $(n)$ to avoid confusion with powers. Several textbooks authors like to write Eq. (2) as,

$h_L^{(n)}=K_nQ_n^2$        Eq. (3)

where $K$ is defined as,

$K_n=f_F^{(n)}\frac{L_n}{D_n}\frac{8}{\pi^2 D_n^4 g}$        Eq. (4)

Following, with Hardy-Cross technique a head losses balance forming a loop is made. In a very general form, this would look like,

$\sum_{n=1}^{n=m}h_L^{(n)}=0$        Eq. (5)

where $m$ is the total number of pipes forming the loop. For example, for each loop in the network above, $m=4$.

Now, as the $h_L$ balance is not zero per se the way to forcing it to be zero is by small adjustments on the flow rates. Substitution of Eq. (3) into Eq. (5) gives,

$\sum_{n=1}^{n=m}K_n Q_n^2=0$        Eq. (6)

and if you add a small flow rate correction $\Delta Q$ to each pipe section, Eq. (6) becomes,

$\sum_{n=1}^{n=m} K_n \left( Q_n+\Delta Q \right)^2=0$        Eq. (7)

 Expansion of Eq. (7) produces,

$\sum_{n=1}^{n=m} K_n \left[ Q_n^2+2Q_n\Delta Q+\left( \Delta Q \right)^2 \right] =0$        Eq. (8)

Since the flow corrections $\Delta Q$ are small (smaller than 1), then powers of it can be neglected. Thus, Eq. (8) can is reduced to,

$\sum_{n=1}^{n=m} K_n \left[ Q_n^2 + 2Q_n \Delta Q \right] =0$        Eq. (9)

and since $\Delta Q$ is not particular to any pipe section (do not depend of $n$) a further simplification is possible,

$\sum_{n=1}^{n=m} K_n Q_n^2 + 2\Delta Q \sum_{n=1}^{n=m} K_n Q_n=0$        Eq. (10)

Next, the flow rate correction $\Delta Q$ can be easily isolated form Eq. (10) to be,

$\Delta Q =- \frac{\sum_{n=1}^{n=m} K_n Q_n^2 }{2\sum_{n=1}^{n=m} K_n Q_n}$        Eq. (11)

Equation (11) can also be written in terms of $h_L$ to ease calculations,

$\Delta Q = - \frac{\sum_{n=1}^{n=m} h_L^{(n)}}{2\sum_{n=1}^{n=m} h_L^{(n)}/Q_n}$        Eq. (12)

Finally, the correction $\Delta Q$ given in Eq. (12) must be applied iteratively to each loop so that with each iteration $\Delta Q$ decreases assuring that the $h_L$ balance is approaching to zero.

At this point, the mass balance has not been mentioned but this is part of the weaknesses of the method.

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

=========

Ildebrando.

Some thoughts about pipe mechanical features

Perhaps what follows may be obvious for experienced professionals but this may  not be true for the engineering student. Hopefully, this will be useful.

1. Pressure and flow rate are related. This is, theoretically, stated by the energy and Bernoulli equation. Besides, this is also tru in practice. Increment and decrement of flow rate are related to changes in pressure.
2. Flow rate increases with diameter; but also with pressure. If you wish to increase the flow rate two options are available: a) you may choose a larger pipe diameter or b) you may increase pressure. If you choose to increase pressure you must know that this cannot be changed arbitrarily and without control.
3. The minimum flow rate is key. The reason is simple. When a pipe system is designed, it is so to operate under steady conditions at the lowest possible cost (without sacrifyce of safety, service life, etc.). Otherwise, you will be wasting money.
4. Different pipe sizes may tranport the same flow rate. This is that, commercial steel pipes with ND 1", 1.5" and 2" may transport the same volume of fluid per minute for certain conditions. In other words, you could, theoretically, estimate the internal pipe diameter but in practice different real pipes may be available to transport the very same flow rate at a lower cost.
5. Pipes may fail. This is that, in contrast with the classroom situations, the engineer should pay attention to the maximum pressure that the pipe may work at (before increasing the flow rate). Even steel pipes may burst or crack at overpressure conditions creating a real chaos.
6. Be aware of phase changes. Phase changes may induce vacuum in the pipe exerting enormous local mechanical stresses in the pipe and creating risk for the structure. This is not apparent; but accessories like safety valves could be very useful in some cases.

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

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Ildebrando

A general comment on pipes

The flow of a fluid through a pipe, in several bachelor courses, is studied merely as a theoretical topic. However, in several ocasions instructors tend to use a textbook where flowrate or pressure loss or pipe diameter are unknown and then estimated by means of the energy or Bernoulli equation.

Many textbooks show this. Take for example the following two popular sources:

A more realistic point of view

It comes out that estimation of flow rate or pressure loss or pipe diameter must be complemented with real pipe data that could possibly be installed. In my opinion, these kind of situations must be treated as a design problem. Otherwise, learning is reduced (I think).

In my opinion, it is important that whoever is taking a regular fluid mechanics course should know that there is pipe data indicating its mechanical features. For short, who performs hydraulic calculations should consider the following:

• nature of the working fluid, where densisty, viscosity, volumetric expansion coefficient, consensation temperature and material compatibility, for example;
• maximum and minimum flow rate, at which the pipe could work,
• maximum and minimum pressure expected to be used to move the fluid through the pipe, and
• maximum and minimum temperature that the fluid will have while being transported.

The previous statements are important because no mater the results of the theoretical calculations, in the end a real pipe satisfying the following factors:

• safety,
• life of service y
• available money, for example

must be selected.

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Ildebrando