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 $N_{Re}\leq N_{Re-c}$, 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 $N_{Re-c}<N_{Re}\leq 4000$ 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 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

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Ildebrando.

Composition variables for mixtures

 When composition of a mixture of several gases is to be considered it can be challenging to use all concepts to represent the right quantities. This is a brief explanation.

About mole $n$ and volume $V$

The number of moles for a pure substance is usually represented by $n$. However, for a mixture with several components the moles of each of these components are to be expressed as follows:

$n_1$, $n_2$, $n_3$,...

where the subscripts 1, 2, 3 indicate the component in the mixture. 

Important note: You should remember that moles are extensive variables which is not recommended for composition calculations purposes. You may go around this difficulty dividing $n$ by an intensive variable, which results in a new intensive variable.

On the other hand, you may also have volumetric concentrations [concentración volumétrica] $\bar{c}$ defined as:

$\bar{c}_i=\dfrac{n_i}{V}$

where $n_i$ stands for the mole of some component and $V$ for the volume of the mixture. When $\bar{c}$ is given in units such as mole/l or mole/dm$^3$ the volumetric concentration is also called molar concentration [molaridad].

Important note: volumetric concentration is recommended for liquid or solid mixtures since these change very little with temperature and pressure. However, the use of $\bar{c}_i$ is not advised for gas mixtures. 

Mole ratio $r_i$ and molal concentration $m_i$

This is another form for referring to composition in terms of moles of components in a mixture. Picking up the moles of component 1 as reference we may define the corresponding ratios $r_i$ for all others as:

$r_i=\dfrac{n_i}{n_1}$

On the hand, molal concentration $m_i$ is in fact a variation of the mass concentration (how it is expressed) of the single component  gas $m$. Remember that the mass $m$ can be defined as:

$m=nM$

where $M$ is the molar mass (molalidad) given in [mole/g]. However, the mass of a component in a gas mixture is defined as:

$m_i=\dfrac{n_i}{n_1M_1}=\dfrac{r_i}{M_1}$

In other words, the mass $m_i$ of a mixture component must be given in terms of the mass and moles of the other components.

Since mole and molality ratios are temperature and poressure independent, these are preferable for any physicochemical calcuation.

Mole fractions $x_i$

These are obtained dividing each of the number of moles ($n_1$, $n_2$,...), of each component, by the total number of moles $n_t$ (of the whole substance) which is defined as:

$n_t=n_1+n_2+n_3+...$

The mole fraction is then expressed as,

$x_i=\dfrac{n_i}{n_t}$

Also, the summation of the mole fractions is always equal to 1:

$x_1+x_2+x_3+...=1$

Important note: The composition of a mixture is determined when all mole fractions are given or can be determined. Since mole fractions are temperature and pressure independent, these are suitable, and possibly the most used, to describe the composition of any mixture.

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.

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 that are part of a flow loop. Without the $h_L$ balance flow rates, which are usually the unknown, they 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 into 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 which 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, a particular head loss is 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 parentheses $(n)$ to avoid confusion with powers. Several textbook 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 the Hardy-Cross technique, a head loss balance that forms a loop is created. 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, since the $h_L$ balance is not zero per se, you can force it to be zero by making minor adjustments to 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 minor 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 minor (smaller than 1), then powers of it can be neglected. Thus, Eq. (8) can be 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 (does not depend on $n$), 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 from 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 zero.

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

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

=========

Ildebrando.