Is a Heterogeneous Slurry Still Governed by the Darcy-Weisbach Equation?

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The discussion centers on the application of the Darcy-Weisbach equation to a hypothetical heterogeneous slurry pipeline, measuring 15 km in length with a diameter of 0.9 m. The user calculates a slurry density of approximately 1555.56 kg/m3 based on a solid-to-water ratio but struggles with determining the viscosity, initially using water's viscosity. This leads to an excessively high Reynolds number, resulting in an unrealistic friction factor and pressure head calculations. After correcting a miscalculation, the user concludes that the required pump power is around 600 kW, suggesting the need for multiple pumps along the pipeline. The conversation highlights the complexities of fluid dynamics in slurry transport and the importance of accurate parameter estimation.
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Hey folks,

I'm trying to do a bit of back-of-the-envelope stuff for a hypothetical slurry pipeline, but it's been a while since I've done any fluid mech./dyn., and I have no confidence in my numbers -- please jump in wherever you catch an error.

First of all, I'm making the assumption that a heterogeneous slurry still behaves according to the Darcy-Weisbach equation, and all of the associated paraphernalia. This might be shaky ground already, but bear with me..

The pipeline is to be in the order of 15 km long, with an internal diameter of, say, 0.9 m. I'm assuming a roughness of 0.025 mm (steel, right?), giving a relative roughness of 2.78E-5 (first sanity check: relative roughness has the 'units' of m/m, not mm/m, right?).

Continuing, I've taken the density of solid phase as 2000 kg/m3. With 1 part solid to 0.8 parts water, I get a slurry density of (1*2000 + 0.8*1000) / 1.8 = 1555.56 kg/m3.

Here's where things start to go really bad, because I've got no idea what the viscosity of the slurry is. As a sanity check I've started with that of water: 8.9E-4 Pa.s. All of this gives me a Reynolds number in excess of 5E10 -- way off the end of Wikipedia's Moody Diagram. Assuming those lines just go horizontal, drawing a line across to the left gives a friction factor of -- for the sake of nicety -- 0.01.

Finally, putting it all back into D-W, I get ... over 500 km of head. Well then.

Please help!
Rukt.
 
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(Sorry for the double-post; couldn't see an 'Edit' option -- must've expired)

Rukt said:
... over 500 km of head.

Strike that; I forgot the gravity term. 50 km or 500 km -- they're both excessive.

Additionally, given P = q.rho.g.h, I figure the pump has to supply 920 MW. That's got to be out by at least two orders of magnitude..

--

EDIT: I should really keep my mouth shut in the echo chamber. That wasn't 500 km of head, it was 500 kPa of pressure (requiring a pump power approaching 600 kW, given the volumetric flow rate of one-and-a-bit m3/s.
 
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so complicated! is this question so important to you?l try to find something that will help. one of my friends is busy now writing his diploma on the gravity maybe he'll find the answer)
 
I have no idea, either, whether your numbers are too huge or not, but if they are, maybe you should think about using more than one pump for the entire 15km span
 

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