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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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  • If not a product, what else would it be? – Kyle Kanos Dec 30 '15 at 21:16
  • @KyleKanos Well since summing vectors does produce "flowing" (moving) of the vector, then it could be some kind of sum. – mavavilj Dec 30 '15 at 21:21
  • Explain further how you can get a flow from the tip to tail method? – Kyle Kanos Dec 30 '15 at 21:23
  • I think more specifically: without multiplication, how can you get a mass flow rate (kg/s) since the units themselves require a product? – Kyle Kanos Dec 30 '15 at 21:28
  • @KyleKanos Which I think is part of the question. I'm studying maths myself and I don't understand the physics unit system well. That is, that the units system is sort of a "different" system in addition to maths, e.g. vector maths. – mavavilj Dec 30 '15 at 21:29
  • You probably want to start with Newtonian mechanics to become familiarized with units before diving into fluid dynamics. You probably would be interested in reading over this question, this one, and this one. – Kyle Kanos Dec 30 '15 at 21:35
  • Surely the question cannot be unclear now that 4 answers have already be given? – mavavilj Jan 01 '16 at 11:33
  • You have 4 very different answers to the question, surely that suggests to you that the question isn't clear what you're asking? If it were clear, the answers would be fairly similar, no? – Kyle Kanos Jan 01 '16 at 11:58

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
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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.