Hydraulic equations for non newtonian fluids

 Here the dydrualic equations for two knwon non newtonian fluids are presented. These are,

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

These equations are to be 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,

$Q=\dfrac{\pi\,R^3}{(1/n)+3}\left(\dfrac{\left(P_0-P_L\right)R}{2mL}\right)^{1/n}$        Eq. (05)

where $R$ is the pipe internal radius and subscripts $0$ and $L$ indicate the end and final of the pipe. Formula in Eq. (05) was taken from the familiar book Dynamics of Polymeric Liquids by Bird R. B. et al.

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 $g_c=32.17\,lb_m/lb_f\,\cdot \,ft/s^2$ is a units correction factor used in the Britsh 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:

$f_{FF-L}=\dfrac{16}{N_{Re}}\left[ 1+\dfrac{1}{6}\dfrac{N_{He}}{N_{Re}}-\dfrac{1}{3}\dfrac{N_{He}^4}{f_F^3N_{Re}^7} \right]$        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)].

On the vapor pressure data for different NaCl dilutions at different temperatures

 The data presented in this post were extracted from the International Critical Tables.

As is usual in evaporation operations the boiling temperature elevation (BPE) is a key data for engineering calculations. Then, the vapor pressure for different combinations of solute concentration and temperatures are to be combined with the Duhring approximation (Duhring's lines).

The data was just rewritten from the source previously mentioned.

	Vapor pressure, mm Hg
Wt %	0.0	2.5	5.0	7.5	10.0	12.5	15.0	17.5	20.0	22.5	25.0	27.5
t °C
0	4.579	4.5	4.4	4.4	4.3	4.2	4.1	4.0	3.8	3.7	3.5
10	9.21	9.1	8.9	8.8	8.6	8.4	8.2	8.0	7.7	7.4	7.1
20	17.54	17.3	17.0	16.7	16.4	16.1	15.7	15.3	14.8	14.2	13.6
30	31.83	3.4	30.9	30.4	29.8	29.2	28.5	27.7	26.8	25.8	24.7
40	55..34	54.5	53.6	52.7	51.7	50.7	49.5	48.1	46.6	44.9	43.0
50	92.54	91.2	89.7	88.1	86.4	84.7	82.8	80.5	78.1	75.3	72.2
60	149.46	147.2	144.8	142.3	139.7	136.8	133.7	130.0	126.0	121.7	116.8
70	233.79	230.2	226.4	222.4	218.3	213.9	208.9	203.5	197.5	190.7	183.1
80	355.47	350	344	338	332	325	318	309.5	300.5	290.2	278.9	266
90	526	517	509	500	491	481	470	458	445.1	430	414	395
100	760	748	736	723	710	695	680	665	643	622	599	572
110	1075.4	1057	1040	1022	1003	983	961	936	911	881	849	810
B. P., °C	100	100.44	100.9	101.4	101.93	102.51	103.16	103.89	104.72	105.68	106.78	108.12

On the heat capacity for NaCl solutions

 This post is based on the results published by James C. S. Chou and Allen M. Rowe Jr. in their paper Desalination, 6(1969) 105-115.

For enthalply calculations the heat capacity is key. However, this last parameter depends on temperature and pressure, and in the case of dilutions on the concentration of solute.

From the thermodynamical point of view the formal relationship between enthalpy and heat capacity is expressed as,

$h=h_0+\int_{T_0}^Tc_P\,dT+\int_{P_0}^P\left[ v-T\left( \dfrac{\partial v}{\partial T} \right)_P \right]dT$    Eq. (01)

where the subscript $0$ indicates a reference data or condition that must be known in order to estimate the data at another set of conditions (without subscripts). $c_P$ and $v$ are the heat capacity at constant pressure and the specific volume, respectively. Chou and Rowe provide math expressions for these two parameters (fortunately):

$c_P=1.3041791 - 8.1519942x + 16.203997x^2 -\left( 0.19159475\times 10^{-2}\right.$

$\left. -0.029952864x+0.0037589577x^2\right)T+\left( 0.29944976\times 10^{-5} \right.$

$\left. -0.498581\times 10^{-4}x-0.89329066\times 10^{-6}x^2 \right)T^2$        Eq. (02)

where $x$ is the mole fraction and the temperature is given in $K$. The units of $c_P$ are $cal/g\,C$. The specific volume is:

$v=A(T)-P\, B(T)-P^2\, C(T)+w\, D(T)+w^2\, E(T)-wP\,F(T)$

$-w^2P\, G(T)-\dfrac{1}{2}wP^2\, H(T)$        Eq. (03)

where $w$ is the salt weight fraction in the solution and the remperature $T$ must be given in $K$. The $A$ through $H$ temperature functions are defined as,

$A(T)=5.916365-0.010357941T+0.92700482\times 10^{-5}T^2$

$-\dfrac{1127.5221}{T}+\dfrac{100674.1}{T^2}$

$B(T)=0.52049144\times 10^{-2}-0.10482101\times 10^{-4}T+0.83285321\times 10^{-8}T^2$

$-\dfrac{1.1702939}{T}+\dfrac{102.27831}{T^2}$

$C(T)=0.11854697\times 10^{-7}-0.65991434\times 10^{-10}T$

$D(T)=2.5166005+0.011176552T-0.17055209\times 10^{-4}T^2$

$E(T)=2.8485101-0.015430471T+0.22398153\times 10^{-4}T^2$

$F(T)=-0.0013949422+0.77922822\times 10^{-5}T-0.17736045\times 10^{-7}T^2$

$G(T)=0.0024223209-0.13698670\times 10^{-4}T+0.20303356\times 10^{-7}T^2$

$H(T)=0.55541298\times 10^{-6}-0.36241535\times 10^{-8}T+0.60444040\times 10^{-11}T^2$

Finally, for the purpose of a reference situation we may take the data of enthalpy at $25\,C$ and pressure of $1\,atm$. Again, Chou and Rowe provide an expression for it in the range of salt weight fraction $w$: $28.8524\%$ to $0.0006\%$. This is,

$h_0=24.953(1-w)+30.805561w^{1.5}-161.50632w^2$

$+79.059598w^{2.5}+114.83149w^3$        Eq. (04)

where the enthalpy $h_0$ is given in $cal/g\,solution$.

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.

Fitting data to a power law equation

 This is a common task for experimentalists looking for relationship among data sets and groups of these.

The common feature of data behaving according to a power law equation,

$y=\alpha\,x^\beta$        Eq. (01) 

where $\alpha$ and $\beta$ are free parameters or parameters for adjustment is that once $\ln y$ vs $\ln x$ is plotted, a straight line is obtained.

Schematic representation of data behaving as power law Eq. (01) (left) and data producing a straight line in log log plot (right).

For short, rather than having to use logarithms to get data from a log-log plot, $\alpha$ and $\beta$ can be determined so that all data can be easily represented by Eq. (01).

How to determine the free parameters

Vey easy. Since in the log-log plot the curve becomes a straight line, it should have an equation like,

$\ln y=m\, \ln x + b$        Eq. (02)

where $m$ is the slope and $b$ is the ordinate at which the line cuts the axis. On the other hand, if operate with $\ln$ on Eq. (01) we obtain,

$\ln y=\ln\left| \alpha \, x^\beta \right|$

$\ln y = \ln \alpha + \beta\ln x$        Eq. (03)

Next, comparison of Eqs. (02-03) leads to,

$m=\beta$ and $b=\alpha$        Eq. (04)

Well, this is not all because the real question is not being answered, how can we estimate $\alpha$ and $\beta$? Equation (4) is just a partial answered. A reliable way of finding $m$ and $b$ is by a linear regression using the minimum squares method so that the reader is then referred to,

Minimum squares method

I hope you can find this useful.