A technique based on the fact that conservation of mass is a reasonable assumption for most processes in our macroscopic world. In its most basic form it looks something like this:

in - out + generation - consumption = accumulation

While this may seem like a painstakingly obvious observation, the equation sees broad application throughout many fields of engineering.

In order to properly apply this equation it is first necessary to specify a control volume over which the mass is to be balanced. This will typically be a mixing point, or some type of chemical reactor. In general one will want to perform a mass balance on each species of your process individually in order to solve the problem.

Common Variations of the Mass Balance:

Steady State:

Often when performing a mass balance, it is useful to assume that your process is operating at conditions that are independent of time. This is called the steady state assumption. The main characteristics of the steady state assumption are that your accumulation term is zero, and all flow rates and concentrations are constant.

Nonreactive Species:

Another assumption that will frequently apply to mixing processes is the assumption that the components involved in your process will not chemically react with each other. Under this assumption the generation and consumption terms of your mass balance will both reduce to zero.

An Example Mass Balance Problem:

A generic chemical process produces as a waste product significant amount of chemical X. However, large concentrations of X can be toxic to local fish, so before disposing of X it must be diluted with a stream of fresh water down to a safe concentration of 30 ppm. Given that water containing 1g/L of X is being produced at a rate of 10L/hr what must the flow rate be of freshwater stream F used to dilute the waste to safe levels?

Solution:

First, let's draw a diagram and define our control volume:

     
 10L/hr H2O (1g/lL X)  ___________
                                  \
                                   =======  Q L /hr H2O (30 ppm X)
          N L/hr H2O   ___________/

For this problem we will define our control volume as our mixing point. There are some assumptions we can make. In this case we will assume both that the system is operating at steady state (there is nowhere for anything to accumulate), and that the species are nonreactive (since this is a dilution, and one would think any reaction between X and water would be happening before it gets to this mixing point and relevant to mention in the problem statement). Next we will do is a mass balance on chemical X as it enters and leaves our control volume: (Recall that from our assumptions generation, consumption, and accumulation terms are all assumed to be zero.)


         IN               -                 OUT                 =   ACCUMULATION

1 g X          10 L H2O       30 g X     1000g H2O    Q L H2O
=======   X   ==========  -  ========= X ========= X =========  =        0
1 L H2O         1  hr        106 g H2O    1 L H2O     1 hr

Using some algebraic rearrangement and calculation, the units all nicely cancel to leave us with 333.3 = Q. The next step in our problem is to perform a mass balance on the water:

        IN           -      OUT        =  ACCUMULATION 
 
10 L H2O + N L H2O   -    333.3 L H2O  
========  ========       ============  =      0
  1 hr      1 hr              1 hr

It is trivial to find that N = 320.3, and voila! We have the solution to our problem: Our waste stream must be mixed with a freshwater stream flowing at 323.3 L H2O/hr in order to be diluted to acceptable levels. While in this particular example it may seem a bit cumbersome to go through the process of mass balances to find a solution, in more complex problems the mass balance can be a very effective tool and vital toward process design.

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