Say I have a container with some mass $m$ of particles each with the same size, $x$. New particles with size $x_1$ are entering the container at a mass flow rate of $\dot m_1$. I need an expression for the rate of change of the average particle size in the container, $\frac{d\bar x}{dt}$.

I started by writing the equation for an average of the two sizes:

$$ \bar x = \frac{mx+m_1(t)\cdot x_1}{m+m_1(t)} $$

then taking the derivative to get

$$ \frac{d\bar x}{dt} = \frac{\dot m_1 \cdot m \cdot (x_1-x)}{(m+m_1(t))^2} $$

however the derivative still has an $m_1(t)$ term in it, for which I don't have an expression or a value. I'm not even sure what the physical significance of $m_1$ would be in the context of an instantaneous rate of change ($\dot m_1$). I'd appreciate any suggestions or other approaches to this problem.

  • $\begingroup$ Is $\dot m_1$ constant? If so, $m_1(t)=\dot m_1t$. By the way, I'm a little unclear about your distinctions between “mass,” “size,” and “number.” Your solution computes “average particle size” as a mass-weighted average of the two sizes, as opposed to a count-weighted average, I think. $\endgroup$
    – Steve Kass
    Dec 3, 2019 at 18:40
  • $\begingroup$ $\dot m_1$ is not constant, nor do I have the value of the time or time-step to do that integration. To clarify the problem, instead of particles with size, this problem could be restated as two liquids with different temperatures, if that helps. $\endgroup$
    – Carlton
    Dec 3, 2019 at 18:44
  • $\begingroup$ Actually, I could use particle count instead of mass, but I'd have the same problem which is that I'd only know the incoming rate of particles, not the total number of particles. $\endgroup$
    – Carlton
    Dec 3, 2019 at 18:49
  • $\begingroup$ I think the solution is simpler than I first thought. The $m_1(t)$ term in the denominator is not just small, it's exactly zero, because it represents the number of new particles in the system before any new particles are added. This is the part I was struggling with; how to interpret $m_1(t)$. $\endgroup$
    – Carlton
    Dec 4, 2019 at 14:09
  • $\begingroup$ $m_1(0)=0$, but $m_1(t)$ is not, if $t>0$. $m_1(t)=\int_0^t \dot m_1(x)\,dx$. $\endgroup$
    – Steve Kass
    Dec 4, 2019 at 18:25

1 Answer 1


Okay so right now you have $m$ mass of particles with the same size of $x$. Let's say you have $\displaystyle \frac{m}{x}$ particles.

Now, you have $m_1$ mass flow rate with particles of size $x_1$, so $\displaystyle \frac{m_1}{x_1}$ particles per second.

Your average mass size is $\displaystyle \overline{x}(t)=\frac{m+m_1t}{\frac{m}{x}+\frac{m_1t}{x_1}}$.

Now, the rate of change is $\displaystyle \frac{d\overline{x}(t)}{dt}$. $m,m_1x,x_1$ are all constants.

If $m_1$ is a time variable, then we denote it as $m(t)$, and we have $\displaystyle \frac{m_1(t)}{x_1}$ particles per second.

Then, the average mass size is $\displaystyle \overline{x}(t)=\frac{m+\int_0^{t}m_1(a)\,da}{\frac{m}{x}+\frac{\int_0^{t}m_1(a)\,da}{x_1}}$.

Similarly, the rate of change is $\displaystyle \frac{d\overline{x}(t)}{dt}$. Little tougher, but doable using the Second Fundamental Theorem of Calculus.

  • $\begingroup$ I don't have the value of $t$ though, so I can't perform these integrations. The solution needs to be in terms of only $m$, $x$, $\dot m_1$, and $x_1$. $\endgroup$
    – Carlton
    Dec 4, 2019 at 14:12

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