# Mass Balances

# Mass Balance

If we add up all the mass entering a CV and subtract the mass leaving a CV we get a **mass balance**. If the sum is positive then the amount of mass in the CV is increasing. If negative, then the mass is decreasing. A value of 0 means that the amount of mass remains constant. Mathematically, the mass balance for an open system is

$$\frac{dm}{dt} = \sum_{i} \dot{m}_{i,in} - \sum_{i} \dot{m}_{i,out} + \sum_{i} \dot{m}_{i,gen} - \sum_{i} \dot{m}_{i,con}$$

where \(\dot{m}_{i,in}\) is the mass flow rate (mass/time) into the CV and \(\dot{m}_{i,out}\) is the mass flow rate out of the CV. The variables \(m_{i,gen}\) and \(m_{i,con}\) are the rate of mass generation and consumption by chemical reactions, respectively. The derivative \(\frac{dm}{dt}\) is the change in mass with respect to time. In other words, how much the mass is increasing or decreasing over time.

## Constant flow rates

Assume we want to perform a mass balance on the coffee being poured into the cup below.

The mass balance for the system is

$$\frac{dm}{dt} = \sum_{i} \dot{m}_{i,in} - \sum_{i} \dot{m}_{i,out} + \sum_{i} \dot{m}_{i,gen} - \sum_{i} \dot{m}_{i,con}$$

We assume:

- No chemical reactions
- No mass leaves the system
- Coffee is being poured into the cup a constant rate of 10 g/sec
- There is 50 g of coffee already in the cup

Then our mass balance reduces to

$$\frac{dm}{dt} = \dot{m}_{in} - 0 + 0 - 0$$

or

$$\frac{dm}{dt} = \dot{m}_{in}$$

No we want to see how the mass in the system changes with time. We know it changes because the \(\frac{dm}{dt}\) term is still in the equation, but by how much? To find out, we need to integrate the equation. Rearranging we have

$$\int dm = \int \dot{m}_{in}\,dt$$

Integrating yields

$$m(t)=\dot{m}_{in}t + C$$

where \(C\) is the constant of integration. At \(t = 0\) sec there was already 50 g of coffee in the cup, so

$$m(0)=\dot{m}_{in}0 + C=50$$

or

$$C=50$$

Remembering that \(\dot{m}_{in}=100\) g/sec we can write the final equation as

$$m(t)=100t + 50$$

## Variable flow rates

Another situation we may encounter is mass flow into or out of a system that is not constant. This situation arises when using tracers. A tracer is a chemical that is used to measure a process or event within the body. Often a tracer is injected in a single large bolus and it is slowly excreted. The rate at which the tracer is eliminated depends on how much remains in they body.

Assume we inject a tracer into a patient’s body. The mass balance is

$$\frac{dm}{dt} = \sum_{i} \dot{m}_{i,in} - \sum_{i} \dot{m}_{i,out} + \sum_{i} \dot{m}_{i,gen} - \sum_{i} \dot{m}_{i,con}$$

We assume:

- No chemical reactions (the tracer does not react with anything in the body)
- No mass enters the system (our injection counts as the start)

Then our equation reduces to

$$\frac{dm}{dt} = 0 - \dot{m}_{out} + 0 - 0$$

As mentioned previously, the change in tracer concentration depends on the amount still in the body, so we re-write the mass balance as

$$\frac{dm}{dt} = 0 - km + 0 - 0$$

where \(k\) is a variable called the *first-order rate constant* that describes how fast the tracer is eliminated from the body. In engineering problems we get this rate in one of three ways: (1) look it up, (2) it is given in the problem statement, or (3) calculate it. In the real world (i.e., in a lab) you would have to measure it if it is unknown. The variable \(m\) on the right side of the equation is the mass of tracer present in the body. **Note that \(m\) is no longer \(\dot{m}\).** In the previous problem \(\dot{m}\) was equal to a number (i.e., a constant). Here, we are saying that \(\dot{m}\) is equal to the varable \(m\). This is an important distinction from the first example. Simplifying the equation we get

$$\frac{dm}{dt} = - km$$

Rearranging and preparing to integrate we get

$$\int \frac{dm}{m} = \int -k\,dt$$

Integrating yields

$$\ln(m) = -kt + C$$

Taking both sides to the \(e\) power gives

$$m(t) = e^{-kt+C}$$

Simplifying slightly gives

$$m(t) = e^{-kt}e^{C}$$

The second exponential is a constant. We can find its value if we know what the concentration of mass is at \(t=0\). Here’s how

$$m(0) = e^{-k0}e^{C}$$

And since \(e^0=1\) this reduces to

$$m(0) = e^{C}= M_0$$

The value \(M_0\) means “the mass at time zero”. So the final equation is

$$m(t) = M_0e^{-kt}$$

This type of equation is known as an *exponential decay*. It starts at the value \(M_0\) and decreases as time increases. The speed of decrease depends on the value of \(k\).