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Pumping a power law slurry

This example was adapted from the textbook chemical Engineering Fluid Mechanics 2nd ed. by Ronald Darby. Problem 6.33.

The situation

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A coal slurry that is characterized as a power law flluid has a flow index of 0.4 and an apparent viscosity of $200\,cP$ at a shear rate of $1\,s^{1}$. The coal has a specific gravity of 2.5 and the slurry is $50%$ coal by weight in water. What pump horsepower will be required to transport 25 million US ton of coal per year (365 days) through a $36\,in$ ID , $1000\,mi$ long pipeline? Assume that the entrance and exit of the pipeline are at the same pressure and elevation and that the pumps are $60%$ efficient.

A sketch of the situation


Solution approach

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First, some interesting things are noticed from the situation statement: both pressure and elevation are the same at entrance and exit of the pipeline. In other words,

$P_1=P_2$    and    $z_1=z_2$        Eqs. (01)

and since the pipeline does not change in diameter: $V_1=V_2$. Now, if we attemp a mechanical energy balance with the general energy equation,

$\dfrac{P_1}{\gamma}+z_1-h_L+h_A+\dfrac{V_1^2}{2g}=\dfrac{P_2}{\gamma}+z_2+\dfrac{V_2^2}{2g}$        Eq. (02)

the following is obtained,

$h_L=h_A$        Eq. (03)

Therefore, using the Darcy equation and the definition of $h_A$ for a pump it follows,

$4\,f_{FF}\dfrac{L}{D}\dfrac{V_1^2}{2g}=\dfrac{\mathbf{e_M}PO}{\gamma\, Q}$        Eq. (04)

from which the power of the pump can be isolated,

$PO=4\,f_{FF}\dfrac{\gamma\, Q}{\mathbf{e_M}}\dfrac{L}{D}\dfrac{V_1^2}{2g}$        Eq. (05)

Please, read post: Mechanical energy added to a fluid - A pump, for more details on the rhs of Eq. (04).

All we need to do is to find the data required in Eq. (05). For the pipeline length $L$ we have,


$L=1000\,mi$

$L=5.28\times 10^6\, ft$

For the internal pipe diameter we have,


$D=36\,in$

$D=3\,ft$

For the density of the slurry, being this fluid a sollution made of water and coal, the following formula applies,

$\rho=\dfrac{2\rho_{coal}\, \rho_{water}}{\rho_{coal}+\rho_{water}}$        Eq. (06)

were the density of the coal can be found using its specific gravity and the density of the water can be searched in tables (at room temperature, for example).

Please, read the post: The density of a mixture, for more details on Eq. (06) an a sample calculation for the present situation.

Thus,

$\rho=2.77\,slug/ft^3$

For volumetric flow rate $Q$, the mass flow rate $Q_m$ needs to be determined first. This is as follows,


$Q_m=\dfrac{\left(25\times 10^6\,US\,ton\right)}{365\,day}\dfrac{\left(907.19\,kg \right)}{1\,US\,ton}\dfrac{1\,slug}{14.59\,kg}\dfrac{24\,hr}{1\,day}\dfrac{3600\,s}{1\,hr}$

$Q_m=49.28\,slug/s$

In this way, the volumetric flow rate $Q$ can be estimated using the fluid density too,

$Q=17.79\,ft^3/s$

The specific weight $\gamma$ is easily estimated to be,


$\gamma=\left( 2.77\, slug/ft^3 \right)\left( 32.17\,ft/s^2 \right)$

$\gamma=89.11\,lb_f/ft^3$

The cross section area of the pipe is calculated as follows,


$A=\dfrac{\pi\left( 3\,ft \right)^2}{4}$

$A=7.07\,ft^2$

The fluid velocity is then,


$V=\dfrac{17.79\,ft^3/s}{7.07\,ft^2}$

$V=2.52\,ft/s$

The fluid parameter $m$ is calculated as follows,


$m=\dfrac{\eta}{\dot{\gamma}^{n-1}}$

$m=\dfrac{200\,cP}{1^{0.4-1}}\dfrac{2.09\times\,10^{-5}\,lb_F\cdot s/ft^2}{1\,cP}$

$m=4.18\times 10^{-3}\,lb_F\cdot s^{0.6}/ft^2$

Using the above data, the Reynolds number for the power law fluid can be estimated to be,

$N_{Re}=429.43$

Please, read the post: Hydraulic equations for non newtonian fluids, for more details on the calculation of the Reynolds number above given.

Therefore, the friction factor for the power law fluid, under laminar flow conditions, at hand can be readily estimated,

$f_{FF-L}=\dfrac{16}{429.43}$

$f_{FF-L}=3.73\times 10^{-2}$

Substitution of all known data into Eq. (05) produces the power required by the pump (or pumps),

$PO=4\left( 3.73\times 10^{-2} \right)\, \dfrac{\left( 89.11\, lb_F/ft^3 \right)\left( 17.79\,ft^3/s \right)}{0.6}\dfrac{5.28\times 10^6\,ft}{3\,ft}\dfrac{\left( 2.52\,ft/s \right)^2}{2\left(32.17\,ft/s^2\right)}$

$PO=68,230,593.85\,lb_F\cdot ft/s$

$PO=92,508.26\,kW$

All calculations were performed in a spreadsheet so tha if you are using a hand calculator, discrepancies may arise.

Finally, please be carefully while watching at this solution since involuntary errors may have introduced.

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

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