Laboratory Testing & General Mineral Processing Engineering

Laboratory Testing & General Mineral Processing Engineering 2017-03-23T09:50:31+00:00
  • To participate in the 911Metallurgist Forums, be sure to JOINLOGIN
  • Use Add New Topic to ask a New Question/Discussion about Mineral Processing or Laboratory Work.
  • OR Select a Topic that Interests you.
  • Use Add Reply = to Reply/Participate in a Topic/Discussion (most frequent).
    Using Add Reply allows you to Attach Images or PDF files and provide a more complete input.
  • Use Add Comment = to comment on someone else’s Reply in an already active Topic/Discussion.

Slurry density (4 replies and 3 comments)

8 months ago
Leo5 8 months ago

how can one calculate or establish the slurry density comprising of two different solids at different solid concentrations and densities?

8 months ago
jpearcy 8 months ago
1 like by David

The derivation should be fairly straight forward. All you need to consider is:

1.- The definition of density for a pure solid or liquid (mass/volume)

2.- Volume and mass are extensive properties, i.e. they add-up to give you total mass or total volume respectively.


M (slurry) = Ms1+Ms2+Mw, and

V (slurry) = Ms1/SGs1+Ms2/SGs2+Mw/SGw

SG (slurry) = 1 /(xs1/SGs1+xs2/SGs2+xw/SGw),

where xs1 = weight fraction of solid 1, xs2 = weight fraction of solid 2, and xw = 1 - xs1-xs2 (weight fraction of water).

Note that SG (slurry) is a function of the densities of the two solids and water, and also of the concentration of each solid in the slurry. The slurry density equation above should simplify to the known formula for a single solid slurry when only one solid is present.


8 months ago

thanks very much for the insight. greatly appreciate.

6 months ago
geologist79 6 months ago

I am trying to prepare a ferrosilicon solution (powder density is 6.67g/cc). Target slurry density/SG is about 3.5. Can I use the above mentioned formula for this purpose? where,

M (slurry) = Ms1+Mw, and

V (slurry) = Ms1/SGs1+Mw/SGw

SG (slurry) = 1 /(xs1/SGs1+xw/SGw)

Basically I want to know amount of powder needed to make a soln.of 500 ml with target SG of 3.5.

6 months ago
Charlie 6 months ago

About 1475 grams of 6.67 SG powder with 279 grams water = about 500 ml of 3.5 SG solution.

6 months ago

Thanks Charlie.
Yes, before we even saw this post, we were trying to make the solution by trial and error method and after certain time the solution turned too colloidal to stir (at a geological exploration site).
I think this has to be made under some typical pressure/apparatus (as centrifuge) etc?
Basically we were doing a small lab scale test to prepare a small amount of slurry but failed.

Alan Carter
6 months ago
Alan Carter 6 months ago

Good day geo79

w = Weight (grams) of dry ore in 1 liter of pulp

w = K(W – 1000)

W = Weight (grams) of 1 liter of pulp

K = S / ( S – 1)

= 6.67/(6.67-1) x (3500-1000) = 2941

divide by 2 for 500ml = 1470.46 g of that ferrosilicon

see page 229+ of

Pretty sure you can prepare this without special pressure/apparatus/centrifuge.

6 months ago

Thanks Alan for your valuable input.
Now the problem we faced while preparing at the site (with basic beakers, stirrer etc) is the solution turned into a thick colloid. We started with a modest quantity of water. just 50 ml. In that we gradually added 160 gm of FeSi when it became too thick to stir (sp.gravity of solution was way below 2, whereas our minimum target is 2.6).
The powder has 90% Fe, 10% Si, atomized 325 mesh size. May be this fineness causes the hindrance.
Our objective is to prepare a small amount of solution, put some gravels to get an quantitative idea on float:sink before we can move to a pilot scale study.

Please join and login to participate and leave a comment.