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I'm looking for a good intuition for understanding the mass flow rate and specifically why it's a product.

The mass flow rate is:

$$density \text{(of fluid)} \times area \text{(of a face)} \times velocity\space component \text{(normal to the face)}$$

Which I find intuitive physically (fluid of some density flows at some area and some velocity), but not mathematically.

Why is it a product? (what's the mathematical intuition)
Why does it work? (how does mathematical intuition $\implies$ physical intuition)

mavavilj
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4 Answers4

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Consider your fluid to be a solid for a moment - and for simplicity, let's use a cylindrical pipe of constant cross section.

Now at t=0, draw a line across the solid to indicate "start here". Wait for a time $t$, and draw a second line on the solid that is passing the point where you are observing.

What you have now is a cylinder with a "start" and "stop" mark on it - it contains all the mass that passed a point in time $t$.

The length of that cylinder is $v\cdot t$; the volume is therefore $A\cdot v\cdot t$. Multiply the volume by the density, and you have the mass that passed in time $t$. The mass that passes per unit time is then

$$M = \frac{\rho A\cdot v\cdot t}{t} = \rho A\cdot v$$

Which I believe is a nice way to tie your intuition to the mathematics.

Floris
  • 119,981
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In my opinion the other answers make it unnecessarily difficult:

A fluid parcel travels a distance: $$\Delta x=v\Delta t$$ in time $\Delta t$ at constant velocity $v$.

The volume taken $\Delta V$ by such a parcel in time $\Delta t$ is: $$\Delta V=A\Delta x=Av\Delta t$$

The volumetric flowrate $\phi_v$ is the rate of change in volume taken: $$\phi_v=\frac{\Delta V}{\Delta t}=Av$$

Converting to a mass flowrate $\phi_m$ requires: $$\phi_m=\rho\phi_v=\rho Av$$

nluigi
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Mathematically, you can understand this from dimensional analysis. Just considering the units of all of the quantities that you have there, you can multiply them together like the following: $$\frac{\text{kg}}{\text{m}^3}\cdot \text{m}^2\cdot \frac{\text{m}}{\text{s}}$$ Each quantity representing the density $\left(\text{kg}/\text{m}^3\right)$, area $\left(\text{m}^2\right)$, and velocity $\left(\text{m}/\text{s}\right)$ respectively. The units of $\text{m}^3$ cancel out and you are left with the units of $\frac{\text{kg}}{\text{s}}$, which is a flow rate of mass.

tmwilson26
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The mass of a fluid parcel is a scalar property of a flow field.

We can derive the desired expression which is part of Reynolds Transport Theorem which applies to the time evolution of any physical quantity $P$ associated with a flow field.

$$\frac{d}{dt}\iiint_{F_p} P \, dV = \iiint_{F_p} \bigg(\frac{DP}{Dt} + P\nabla \cdot \mathbf{u}\bigg)\,dV$$

Applying the divergence theorem to the right hand side we have

$$\frac{d}{dt}\iiint_{F_p} P \,dV = \iiint_{F_p} \frac{\partial P}{\partial t}\,dV + \iint_{F_p} P\mathbf{u} \cdot \mathbf{n} \,dS$$ Which is the time evolution of a physical quanity P associated with the flow field. Specifically for mass, the last equation can be written as $$\frac{d}{dt}\iiint_{F_p} \rho \, dV = \iiint_{F_p} \frac{\partial \rho}{\partial t}\,dV + \iint_{F_p} \rho\mathbf{u} \cdot \mathbf{n}\, dS$$ where the second term on the right hand side is the convective transport of mass from a control volume.