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concentration

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JP123

Mechanical
Aug 25, 2004
28
hi,

i have 1000 ppm of iron (density = 181 lb/ft3) mixed with water.

i'm trying to find the solid concentration in %.

thanks


 
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Assuming no other contamination:

Let:
X = Volume Water
(1ft3-X) = Volume Iron
Density water = 62.4 lb/ft^3
Density Iron = 480 lb/ft^3
Total Density = 181 lb/ft^3
Total Volume = 1 ft^3
Therefore:

(Density Water)(Volume Water) + (Density Iron)(Volume Iron) = (Total Volume)(Total Density)

(62.4)(X)+(480)(1-X)=(1)(181)

X=.71599 ft^3

Since this is a Factor of 1 ft^3

The percentages are:
71.599% Water
28.401% Iron

The density of iron may be different then what you have. This should give you a starting point though.
 

If 1000 ppm is on a mass basis then it equals 0.1% w/w.
 
ok thanks.

but the density iron should be 181 lb/ft^3 and it's not the total density.

where 480 lb/ft^3 come from?
 
480 lb/ft^3 is the standard density of Iron.

Where did 181 lb/ft^3 come from?

Also did you have a total weight of the ft^3 of the solution?
 
ok..181 lb/ft^3 is the density of iron ore crusehd (dust)..sorry i did'nt specify before.

no i don't have it...i only have flow rate of slurry 70 gpm
 
Webster's New Millennium™ Dictionary of English
Main Entry: part per million

Definition: a unit of measure of the amount of dissolved solids in a solution in terms of a ratio between the number of parts of solids to a million parts of total volume; abbr. Ppm


Therefore

1,000 parts iron for 1,000,000 parts water.

means you have .1% Iron

I guess the next step is to find out if your PPM is based on mass, weight, or volume.
 
JP123:

You must specify whether the ppm is parts per million by volume (ppmv) or parts per million by weight (ppmw). You must also specify whether you want to know the percent by volume (%vol) or the percent by weight (%wt). And if you want to be extremely accurate, you should also tell us the temperature of your water because that affects the water density.

Until you specify that information, there are many possible answers to your question.

The density of the iron in your water does not change simply because the iron is crushed.

Milton Beychok
(Visit me at www.air-dispersion.com)
.

 
I agree. Crushing iron ore doesn't change the density. Even with that said, you are using the correct density value for iron ore.
 
Right. He is using Iron ore with a density of 181 lb/ft^3.
 
I don't know the water temperature but i suppose it's around 15 deg C.

what can be the results between the percent by volume and the percent by weight?
 
yes!

and by weight...the concentration should be 0.1% of solid in water??
 
If it is 1000 PPM by weight it is .1% concentration by weight. Since the temperature is very close to ambient, it will not cause much of a variance.

It will be much lower % by volume.

For every 1,000 lb Iron ore you have 1,000,000 lb water
or .001 lb Iron Ore per 1 lb water

That means you have .0624 lb Iron ore for Every 62.4 lb

Therefore for every 1ft^3 water you have .0003448 ft^3 Iron Ore.

(.0003448 ft^3) / (.0003448 ft^3 + 1 ft^3)

= 0.0345% Iron Ore concentration Based on Volume.
 
JP123:

I don't mean to preach, but there is a lesson to be learned here. Always state whether your ppm is by weight or by volume ... and whether your % is by weight or by volume.

Darken would have been able to help you correctly much sooner had you taken just a few minutes to be more explicit.

Regards,

Milton Beychok
(Visit me at www.air-dispersion.com)
.

 
ok! I'll remember.

Thanks!

JP.
 
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