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Specific Gravities in Large Volumes

Date: 04/09/2002 at 18:05:33
From: Rob 
Subject: Specific gravities in large volumes

I am trying to create a chart that could be used to help determine 
what weight of methanol/water to use to either raise or lower the 
specific gravity of a volume of corrosion inhibitor in bulk mixing 

We mix 20,000 gallons of these fluids a day in each tank. We would 
like a simple equation that could be used that utilizes the measured 
specific gravity of the fluid in the mixing tank, the known current 
volume of fluid in the mixing tank, and the ideal specific gravity of 
the fluid in the mixing tank to determine what volume of a second 
fluid with a known specific gravity should be added to the bulk mixing 
tank to achieve the ideal specific gravity in the bulk mixing tank. 

Currently we are doing it by trial and error. Although I have a 
chemistry degree I am not sure where to go to get this equation. I 
figured since there is only one unknown in this mystical equation 
there must be one equation that will solve for it.


Rob Gustafson
BP Exploration Alaska

Date: 04/09/2002 at 19:22:47
From: Doctor Rick
Subject: Re: Specific gravities in large volumes

Hi, Rob.

Sounds good, let's try it! I'll start by defining variables:

t = target specific gravity of fluid
a = actual specific gravity of fluid in tank
b = specific gravity of fluid to be added
v = volume of fluid in tank
x = (unknown) volume of second fluid to be added to reach target s. g.

Specific gravity is the density of a fluid divided by the density of 
water. In SI units, 1 kg/l = 1 g/ml is the density of water, so 
specific gravity is essentially the density in kg/l.

What is the specific gravity of a mixture of two fluids? The mass of 
v liters of fluid a (specific gravity = a) is v*a; the mass of 
x liters of fluid b (specific gravity = b) is x*b. The total mass of 
the mixture is (va+xb); the total volume is (v+x). Therefore the 
specific gravity of the mixture is the mass divided by the volume, 

You want the specific gravity of the mixture to be t. This gives us an 

  (va+xb)/(v+x) = t

Solve this for x:

  va + xb = (v+x)t
  va + xb = vt + xt
  xb - xt = vt - va
  x(b-t) = v(t-a)
  x = v(t-a)/(b-t)

How's that for a formula? The desired volume of the second fluid is 
the volume in the tank times the ratio of the deficiency of the 
specific gravity in the tank to the excess of the specific gravity of 
the second fluid.

- Doctor Rick, The Math Forum 

Date: 04/10/2002 at 12:52:05
From: Rob 
Subject: Specific gravities in large volumes

Doctor Rick,

I sincerely appreciate the help you have provided and the formula for 
our needs. I will have the mixing guys try this out and see if it 
works for actual conditions. Looking at the equation it now seems so 
intuitively obvious. I guess I did not think to set it up that way 
and gave up. 

Thanks again,
Associated Topics:
College Physics
High School Physics/Chemistry

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