# I Reynolds transport theorem derivation - linear momentum

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1. May 10, 2017

### ussername

I've managed to derive the form of Reynolds transport theorem as a bilance of linear momentum of the system:
$$\left (\frac{\vec{\mathrm{d} p}}{\mathrm{d} \tau} \right )_{system}=\frac{\mathrm{d} }{\mathrm{d} x}(\int_{V}^{ }\vec{v}\cdot \rho dV)+\int_{A}^{ }\vec{a}dm+\int_{A}^{ }\vec{v}\cdot \rho \cdot (\vec{v}\cdot \vec{n_{0}})dA$$
where V is volume of the system and A is boundary area of system. The second integral in the equation is:
$$\int_{A}^{ }\vec{a}dm=\int_{A}^{ }\vec{dF}=\vec{F}_{boundaries}$$
which is the overal force applied on the boundary of the system. The integral was derived the following way: let's consider element dA of the system boundary, through which the linear momentum "flows": $$\vec{dp}(dA)=\vec{v}\cdot dm(dA)$$ But generally either $$\vec{v}$$ or $$dm(dA)$$ can be functions of time (the velocity and exchanged mass rate can both change with time in the system inlet/outlet) so the derivation of this equation with rescpect to time is:
$$\frac{\mathrm{d} }{\mathrm{d} \tau}(\vec{dp}(dA))=dm(dA)\cdot \vec{a}+ \vec{v} \cdot\frac{\mathrm{d} m(dA,d\tau)}{\mathrm{d} \tau }$$
which gives last two previous integrals after integration.

The derived bilance makes sense to me - first integral stands for the source of linear momentum inside the system, second integral expresses the change of linear momentum with forces acting on the system boundary and third integral means linear momentum flow with mass flow through system.

Nevertheless in literature I have not found the second integral in RTT for linear momentum. Can you tell me wheather this derivation is correct and why/why not?

2. May 10, 2017

That second integral shows up in most books that I've seen. They just usually make use of the fact that $dm = \rho\;dV$. It represents the unsteady motion of the control volume itself. It's equivalent to the motion of a frame of reference in dynamics.

3. May 11, 2017

### ussername

The second integral is nonzero when the mass incoming or outcoming from the control volume has nonzero accelaration. That is when the force is acting on the system through incoming mass / the system is acting with force to its surroundings through outcoming mass.

There are two types of changing of linear momentum of the system with mass exchange. The linear momentum of the system can simply change with changing mass being in the system + linear momentum of the system is changing when there is a force acting between system and surroundings through mass exchange.

Can you refer to some book that shows that integral?

In most book I've seen this integral was neglected, that is the mass flowing through control surface is not accelarating.

4. May 12, 2017

As far as I can tell, you have three integral terms there:
$$\dfrac{\partial}{\partial t}\int_{C\mathcal{V}}\rho \vec{v}\;d\mathcal{V}$$
This is the force related to the overall change in momentum contained in the control volume.
$$\int_{CS}\rho\vec{v}(\vec{v}\cdot\vec{n})dA$$
This is the force related to the momentum leaving the control volume.
$$\int_{CV}\vec{a}\;dm = \int_{CV}\rho\vec{a}\;d\mathcal{V}$$
This is the force term associated with the acceleration of the control volume itself.

A few sources that I know have all three integral terms are:
Fox & McDonald's Introduction to Fluid Mechanics by Pritchard and Mitchell
Fluid Mechanics by White

Munson, Young and Okiishi's Fundamentals of Fluid Mechanics by Gerhart, Gerhart, and Hochstein

5. May 12, 2017

### ussername

But you probably mean the force acting on the system on the other side of the equation:
$$\left (\frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} \right )_{system}=\vec{F}_{system}=\int_{V}^{ }\vec{a}\cdot dm(dV)$$

Actually I changed my mind. This derivation seems more meaningful to me:
The element of linear momentum flowing through surface dA in time Δτ is:
$$\vec{dp}(dA,\Delta \tau ) =\int_{\tau _{1}}^{\tau _{2}}\vec{v}(\tau)\cdot dm(dA,d\tau )$$
Thus the element of linear momentum flowing through surface dA in time dτ is:
$$\vec{dp}(dA,d \tau ) =\vec{v}(\tau)\cdot dm(dA,d\tau )$$
and the flow of linear momentum through surface dA is obtained simply by dividing this equation by dτ:
$$\frac{\vec{dp}(dA,d \tau ) }{\mathrm{d} \tau}=\vec{v}(\tau)\cdot \frac{dm(dA,d\tau ) }{\mathrm{d} \tau}$$

6. May 12, 2017

### Staff: Mentor

My understanding of this is as follows: If V is a control volume whose boundary is moving with the fluid velocity $\vec{v}(\vec{x},t)$, then

$$\frac{\partial}{\partial t}\left(\int{\vec {p}dV}\right)=\int{\frac{\partial \vec {p}}{\partial t}dV}+\int{\vec{p}(\vec{v}\centerdot \vec{n})dA}=\int{\vec{v}\centerdot (\vec{\sigma}\centerdot{n})dA}+\int{\rho \vec{b}dV}$$where $\vec{b}$ is the body force per unit mass, $\vec{\sigma}$ is the stress tensor, and $\vec{n}$ is an outwardly directed normal from the control volume. The first term on the left hand side represents the rate of change of momentum within the control volume if it were stationary. The second term on the left hand side represents the rate that momentum is being swept into the control volume by virtue of its boundary moving. The first term on the right hand side represents the rate at which work is being done on the surroundings at the surface of the control volume, and the second term on the right hand side represents the body force acting on the control volume. If we apply the divergence theorem to this equation, we obtain:$$\frac{\partial}{\partial t}\left(\int{\vec {p}dV}\right)=\int{\left(\frac{\partial \vec {p}}{\partial t}+\vec{\nabla }\centerdot (\vec{p}\vec{v})\right)dV=\int\left(\vec{\nabla} \centerdot (\vec{\sigma}\centerdot \vec{v})+\rho \vec{b}\right)dV}$$