# Expression of mass flow rate for steady flow

According to Wikipedia the mass flow rate is given by:

$${\dot {m}}=\iint _{CS}\rho \mathbf {v} \cdot {\rm {d}}\mathbf {A}$$

And using the Reynolds Transport Theorem (if there is no sources):

$$\dfrac{dm_{sys}}{dt} =\iint_{CV}\dfrac{\partial \rho}{\partial t}\mathrm dV\llap{--} + \iint _{CS}\rho \mathbf {v} \cdot {\rm {d}}\mathbf {A} = 0$$

If the flow is steady then:

$$\dot m = 0$$

But as far as I know the mass flow rate for steady flow is also given by $$\dot m = \rho\cdot V \cdot A$$ And it will never be zero (if $$\rho$$, $$V$$, and $$A$$ are not zero)

What I am missing?

## 1 Answer

Working through the math:

$$\dot{m} = \frac{dm}{dt} = \iiint_{CV} \frac{\partial \rho}{\partial t} dV + \iint_{CS} \rho \vec{V}\cdot dA$$

If the flow is steady, all time derivatives are zero and you are left with:

$$0 = \iint_{CS} \rho \vec{V} \cdot dA$$

If we consider a 1D flow just as an example, this will give you:

$$0 = -u_{\text{left}} \rho_{\text{left}} A_{\text{left}} + u_{\text{right}} \rho_{\text{right}} A_{\text{right}}$$

or

$$u_{\text{left}} \rho_{\text{left}} A_{\text{left}} = u_{\text{right}} \rho_{\text{right}} A_{\text{right}}$$

which is an expression that is probably familiar. So, that all shows that $$\dot{m} = 0$$.

Now to your other equation, $$\dot{m}_{\text{system}} = \rho V A$$. That doesn't hold for steady flow. So the part you are missing is that your second statement isn't true, and there's no way to get this expression from the equation we started with if we assume the flow is steady. This expression will sometimes show up when specifying boundary conditions, as in $$\dot{m}_{\text{in}} = \rho V A$$ -- but if it is steady, that means that $$\dot{m}_{\text{out}} = \rho V A$$ also and the total $$\dot{m}$$ of the system is zero.

In other words, if you see something that says $$\dot{m} = \rho V A$$ and it also says the flow is steady, it means they are specifying the $$\dot{m}$$ across one of the control surfaces. Not all of the control surfaces.

• Sorry you are explaining assuming that I am an expert. I am totally a beginner please use simple explanation. I cannot understand it sorry. – IamNotaMathematician Mar 3 at 17:05
• @IamNotaMathematician Which part don't you understand? Are you familiar with derivatives and integrals, dot products? I can try to clarify specific parts you are unsure about depending on what the confusion is. – tpg2114 Mar 3 at 17:08
• Now to your other equation, $m_{system}=ρVA$. That doesn't hold for steady flow. – IamNotaMathematician Mar 3 at 17:10
• I cannot understand what is the difference – IamNotaMathematician Mar 3 at 17:11
• Right -- so your question said "But as far as I know... for steady flow... $\dot{m} = \rho V A$." I'm telling you that isn't true. It does not hold for steady flow for the system mass flow. The only true expression for the system is $\dot{m} = 0$. – tpg2114 Mar 3 at 17:12