• BME 200
  • 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:

    Then our mass balance reduces to

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


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

    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

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



    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:

    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 \(k\).

    Last updated:
    August 23, 2018