AMIT 135: Lesson 2 Circuit Mass Balancing


Introductory narrative or video


Upon completion of this lesson students should be able to:

  • Differentiate between different type metallurgical accounting methods and benefit of metallurgical accounting.
  • Recall the definitions of grade, recovery, yield and other characteristics used in mass balancing.
  • Recognize the application of two product formula for balancing.
  • Provide a brief idea about multi component balancing for complex ores.


Mass and Volume Balancing

Mass and Volume Balancing
Outotec Concentrator
The most important rule governing the accounting of a processing plant or unit operation is that mass can be neither created nor destroyed.

Mass In = Mass Out

Volume In = Volume Out

Mass and Component Balance

Mass Component Balance
Outotec Concentrator
The component balance is true for any assay type such as any element content, % finer than, water-to-solids ratio etc.

Ff = Cc + Rr

Mass Yield or Mass Recovery

Mass Yield or Mass Recovery
Outotec Concentrator

\LARGE \mathbf{Mass\; Yield, Y = \frac{tons/hr\; Plant\; Concentrate}{tons/hr\; Plant\; Feed} = \frac{C}{F}}

Two-Product Formula

Given the mass and component balance equations :

F = C + R

Ff = Cc + Rr

Mass yield (Y) can be solved as a function of the stream, assays which yields the two-product  formula:

R = F- C

Ff = Cc + r (F – C)

Mass Yield, Y = \frac{C}{F} = \frac{(f-r)}{c-r}



Mass Yield Formula Example

Mass Yield, Y = \frac{C}{F} = \frac{(f-r)}{c-r}

Mass Yield = \frac{(0.50 - 0.14)}{(24.5 - 0.14)}\ast 100

Mass Yield = 1.5%

Mass Yield Formula Diagram

Component Recovery

In many mineral processing applications, the focus is to maintain maximum recovery of the valued mineral or element while also maximizing upgrade or concentrate assay.

Component Recovery = \frac{Cc}{Ff} = \frac{c}{f} \ast \frac{(f-r)}{c-r} = \frac{c}{f}\ast Y

Mass Yield or Mass Recovery
Outotec Concentrator


Component Recovery Example

Cu Recovery, R = Y\ast \frac{c}{f}

Cu Recovery, R = 1.5\ast \frac{(24.5)}{(0.50)}\ast 100

Cu Recovery, R = 73.5%



Concentration Ratio, CR

Concentration Ratio, CR = \frac{c}{f}

Mass Yield Formula Diagram



Complex Sulfide Ore Metallurgical Balance

Brunswick No.1 Concentrator

Process Stream Weight (%) Assays
Pb% Zn% Cu% Ag%
Mill Feed 100.00 3.21 7.93 0.33 2.26
Copper Concentrate 0.74 4.69 4.65 22.64 50.39
Lead Concentrate 4.80 42.20 9.62 0.33 15.48
Zinc Concentrate 10.52 1.12 57.76 0.18 1.75
Tailings 83.94 1.22 1.61 0.15 1.14
Metal Recovery (%) 63.1 76.6 50.9



Section Process Stream Mass Flow (tph) Copper Assay (%) Cu Recovery (%)
Complete Plant Mill Feed 709 0.53 100
Mill Concentrate 10 33.00 88.3
Mill Tailings 699 0.06 11.7
Rougher Feed 716 0.55
Concentrate 28 12.00 83.6
Tailings 688 0.09
Scavenger Feed 253 0.13
Concentrate 6 2.00 2.4
Tailings 247 0.09
Cleaner Feed 29 12.00
Concentrate 15 22.00
Tailings 14 1.30
Re-Cleaner Feed 18 20.00
Concentrate 11 31.00
Tailings 7 3.00
Final Cleaner Feed 11 31.00
Concentrate 10 33.00 88.3
Tailings 1 5.00

Mass/Volume Flow Determination

  • The relationship between the mass (M) and the volume flow (Q rates in a given stream is defined as:M=\frac{Q\rho_{p} X}{100}
    \rho_{p} = the pulp density (solids and water) in lbs/ft3
    X = solids concentration in % by weight
  • Volumetric flow rate can be measured by flow meters, P-Q curve relationships for pumps or directly measured.
  • Pulp density can be measured using a Marcy Density of nuclear density gauge.
  • The solids concentration can be estimated using a common expression or directly measured.

Pulp Density

  • Density (ρ) is the ratio of the mass weight (M) of a substance and the total volume (V):
    \rho = \frac{Mass}{Volume} = \frac{M}{V}
  • Water has a density of:
    • 1.0gm/ml or 1.0gm/cm3
    • 1000kg/m3 – 1tonne/m3
    • 62.4lbs/ft3
  • Specific gravity is the ratio of the material density over the density of water:\frac{\rho }{\rho _{w}} = \frac{\rho }{62.4lb/ft^{3}}

Marcy Scale

  • The Marcy scale is a common tool used in preparation plants to measure pulp density.
  • The scale uses a container that allows a volume of exactly 1000 ml.
  •  After collecting  the sample,  the container  is hung on a weight  scale which measures in grams.
  • The printed scale reads the pulp density directly in grams/ 1ml.
  • Since water density is 1.0 gm/ml, the scale also indicates the specific gravity of the pulp.\frac{\rho _{p}}{1.0gm/ml}
Marcy Scale
A Marcy scale



Mass Flow Determination Example

A classifying cyclone is being fed slurry at a volumetric flow rate of 1000 gpm. The specific gravity of the slurry is 1.1 and the solids concentration was determined to be 15.0°/o by weight. Determine the mass feed flow rate in tons per hour:

M=\frac{Q\rho _{p}X}{100}

M=\frac{(Q gal/min)(0.13368ft^{3}/gal)(^{3})(X)(60min/hr)}{(2000lbs/ton)(100)}

M=\frac{(Q\rho _{p}X)(}{(4)(100)}


M (tph) = \frac{(100 gal/min)(1.1)(15.0)}{(4)(100)}=41.3 tph

Volumetric Balancing Using Pulp Density

  • Pulp density measurements around unit operations such as dense medium separators and classifying, cyclones can be used to assess volume yield.
  • Consider the balance around a classifying cyclone:Q_{f}=Q_{u}+Q_{o}
    Q_{f}\rho _{f}=Q_{u}\rho _{u}+Q_{o}\rho _{o}
    since (Qvol/time)\ast (\rho\: mass/volume)= Pulp\: mass/time
  • Thus, the volumetric yield to the underflow stream can be obtained from the following expression:Volume\: Yield\: to\: Underflow=\frac{Q_{u}}{Q_{f}}=\frac{(\rho _{f}-\rho _{o})}{(\rho_{u}-\rho _{o})}\times 100
A diagram depicting formulas for figuring the balance around a classifying cyclone
Formulas for figuring the balance around a classifying cyclone
[diagram 135-2-1]

Slurry Solids Concentration by Weight

A classifying cyclone is treating 150 gallons/min of slurry. A plant technician has measured the pulp densities of the process streams using a Marcy density gauge and reported the following: ρf = 1.08 gm/ml, ρo =  1.03 gm/ml and ρu = 1.20 gm/ml. Determine the volumetric flow rates to each stream.


Y_{u}=\frac{(\rho_{f}-\rho _{o})}{(\rho _{u}-\rho _{o})}\times 100=  _________%

Q_{u}=Q_{f}Y_{u}=150 \times  _________ = _________ gallons/min

Q_{o}=150- _________ = 150 X (1 – _________) = __________ gallons/min

  • The solids concentration X of a slurry by weight can be measured directly by:
    • Collecting a sample,
    • Measuring the weight of the total slurry or pulp Mp
    • Drying the sample and
    • Measuring the dry weight of solids Ms

      X=\frac{M_{s}}{M_{p}}\times100= \frac{M_{s}}{M_{s}+M_{w}}\times 100Mw = water weight
  • This equation is often used to determine the amount of water in a process stream1 knowing the solids concentration by weight and the tons/hr of solids in a process stream.


Solids Concentration Example

The underflow stream from the classifying cyclone bank contains 100 tons/hour of ore which represents 45% of the total slurry mass.

The downstream concentrators require a feed solids concentration of 30% by weight. How much water is required to be added to dilute the stream to the required solids concentration?


Water in Cyclone Underflow = M_{w}=\frac{100tph}{0.45} - tph=  ________ tph

Water\: in\: Spiral\: Feed = M_{w} = \frac{100 tph}{0.30} - 100 tph =  ________ tph

Dilution\: Water\: Required = Q = \frac{(400)(m)}{(} = \frac{(400)(\; \; \; \; \; \; \; )}{(1)(100)} =  ________ tph

= _______ gpm


% Solids & Pulp Density Relationship

  • It often occurs that the knowledge of the solids content is needed within a time frame less than the sample preparation and analysis time.
  • When this situation arises, the solids content by weight can be estimated knowing the pulp density (ρp =) as measured with the Marcy scale and using the following equation:
    X% = \frac{\rho _{s}(\rho _{p}-\rho _{w})}{\rho _{p}(\rho _{s}-\rho _{w})}\times 100ρs = solids density or specific gravity
    ≈ 2.65 for most host rock minerals.ρw = water density (62.4 lbs/ft3 or 1gm/ml) or specific gravity (=1).

Estimation of Solid Density

  • The density of  solid can be estimated if you know the composition by weight or volume of each component in the solid and the respective solid densities.
  • For example, assume that raw ore is a two component system comprised of a valued mineral and host rock having relative densities of 6.30 and 2.65, respectively. The amount of valued mineral is 30% of the total ore.Total Solid Mass, M: Mineral + Host Rock – 30 + 70 = 100
    Total Volume, Vs:  {\color{Red} \frac{M_{M}}{\rho_{M}}+\frac{M_{HR}}{\rho _{HR}} = \frac{30}{6.30}+\frac{70}{2.65}= 31.18}

    Total Solid Relative Density {\color{Red} =\frac{M_{s}}{V_{s}} - \frac{100}{31.18} = 3.21}


Circuit Balancing

  • The data generated from process units and circuits are used to make important decisions.
    • Plant design considerations
    • Operational  efficiencies
    • Potential upgrades
  • As such, reliable data is very important for an operating plant.
  • All data must be checked for proper balancing based on the ‘laws of conservation’.
  • ‘What goes in’ = ‘What goes out’


Material Balancing

Balance Summary
Overall:  F = C+R

Pb: Ff1 = Cc1 + Rr1
Zn: Ff2 = Cc2 + Rr2
L/S Ratio: Ff3 = Cc3 + Rr3


  • Is this a good set of data?
  • Which values are not reliable?
A diagram of a material balance equation
[image 135-2-2]

Material Balance

Overall Balance

F = C+R

100 = 90 + 10

100 = 100

Image of a thumbs up

A diagram of a material balance equation
[image 135-2-3]

Lead Balance

Ff = Cc+ Rr

100(5.20) = 90(0.20) + 10 (50.20)

520.00=18.00 + 502.00

520.00= 520.00

Image of a thumbs up

A diagram of a material balance equation
[image 135-2-4]

Zinc Balance

Ff = Cc+ Rr

100(2.30) = 90(0.30) + 10 (22.30)

230.00= 27.00+223.00

230.00 ≠250.00

Image of thumbs down

A diagram of a material balance equation
[image 135-2-5]

Slurry Balance

Mass balance equations must be satisfied!

  • Must use 1/Solids%
Product StreamMass (tph)Solids (%)Inverse SolidsSlurry (tph)

Water Balance

L/S Ratio Balance

Ff = Cc+ Rr

100(3.0) = 90(4.0) + 10(1.0)

300= 360+10

300 ≠370

Example: 3.0 = 3 parts Water to 1 part Solids = 25%

Image of thumbs down

A diagram of a material balance equation
[image 135-2-6]

Two-Product Formula


F = C + R

Ff = Cc + Rr


Multiply by r:

Ff = Cc + Rr

-(Ff = Cc + Rr)


Ff- Ff = Cr – Cc


F (r-f) = C (c-r)




Yield = C/F = (r-f) / (r-c)


* Only assays needed to get product-to- feed ratio

Two-Product Formula (Pb assays)


C/F = 10/100

= 10%




= 10%

A diagram of a material balance equation
[image 135-2-7]

Two-Product Formula Characteristics

  • Formula only applies under steady-state conditions
  • Formula very sensitive to variations in the reject assay, “r”
  • Formula inaccurate when component is not separated
  • Formula may calculate different yields for each assay (due to experimental errors)

Two-Product Formula (various assays)


C/F = (5.2-0.2)/ (50.2-0.2)

= 10%




= 9%



C/F = (1/25-1/20) / ( 1/50-1/20)

= 33%

A diagram of a material balance equation
[image 135-2-8]

What to Do with Unbalanced Data

Eliminate  Data

  • avoid collecting or ignore data that conflicts
  • most common approach in industry
  • arbitrary and highly subject to user biases
  • not getting full value out of your data

Adjust  Data

  • create consistent balances by adjusting data
  • use method of “weighted least squares”
  • adjustments should (i) satisfy all mass balances equations and (ii) be as small as possible