Mechanical power flow density

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I am trying to determine the flow of mechanical power through a structure like a drive shaft or even a simple lever. The components of the Cauchy stress tensor, ##\sigma_{ij}##, have units of force/area, so if I multiply by the velocity, then ##\sigma_{ij} v_j## would have units of power/area. And since it is a vector I would interpret it as the mechanical power density crossing a surface normal to the vector.

Is this correct? If anyone has taken some mechanical engineering courses that discussed this quantity or another quantity which can describe the flow of mechanical power within a structure, I would greatly appreciate any insight.
 

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  • #2
jrmichler
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Mechanical engineers typically get stress tensors in grad school. I did about 1993, and promptly forgot it all, because it was not useful in real world systems.

We are a simple minded bunch. Torque is torque, RPM is RPM, and power is what you get when you multiply them (in consistent units). Most of the time we use formulas with unit conversions embedded in the formula. As an example: Power = TN / 5250, where:
Power = horsepower
T = torque in ft-lbs
N = RPM
5250 makes the units work out

For a complex machine, we typically start at the point where the machine is acting on something, then work our way back to the drive motor, adding friction as needed.
 
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  • #3
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Mechanical engineers typically get stress tensors in grad school.
Oh, I didn’t realize this would be a grad level topic. I thought it would be undergrad mechanical Engineering.

I think that I may have had the sign convention backwards above. I worked out two examples.

If you are pulling a rope in the -x direction then power flows in the +x direction.
51244DC4-E4D5-4E60-8306-D8B80077964C.jpeg

The only nonzero component of the stress tensor is ##\sigma_{xx}## and the only nonzero component of the velocity is ##-v_x## so ##-\mathbf{\sigma} \cdot \mathbf v## gives power flow in the +x direction.

Similarly, if you are turning a shaft in the direction shown then power flows in the +x direction

03E76862-263C-4F7B-AD50-C965EBD9E8C8.jpeg

The only nonzero components of the stress tensor are ##\tau_{xy}=\tau_{yx}## and the only nonzero component of the velocity is ##-v_y## so again ##-\mathbf{\sigma} \cdot \mathbf v## gives power flow in the +x direction.
 
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Mechanical engineers typically get stress tensors in grad school. I did about 1993, and promptly forgot it all, because it was not useful in real world systems.
My experience with real world systems as an engineer in industry for 35 years was very different from yours. I worked in both chemical engineering and mechanical engineering, and used the stress tensor extensively in both areas. It was essential in analyzing and designing complex systems involving fluid mechanics and solid deformation mechanics.
 
  • #5
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Dale, if the contraction you described between the stress tensor and the velocity vector is then contracted subsequently with a unit normal vector to a surface (either internal or external), the quantity obtained is the rate of doing work per unit area of the surface. This quantity is included in the differential energy balance on. parcels of material within the body or the fluid.
 
  • #6
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if the contraction you described between the stress tensor and the velocity vector is then contracted subsequently with a unit normal vector to a surface (either internal or external), the quantity obtained is the rate of doing work per unit area of the surface
OK, that sounds like it is what I wanted, although I didn’t think of that subsequent contraction. Do you have a reference for this that I could read more from.

It seemed to work for a rope and for a driveshaft, but at first glance it looks like it will not work for a lever since the power flux should go to zero at the fulcrum where v goes to zero. I have not yet worked it out to be sure, so hopefully I am wrong.
 
  • #7
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OK, that sounds like it is what I wanted, although I didn’t think of that subsequent contraction. Do you have a reference for this that I could read more from.

It seemed to work for a rope and for a driveshaft, but at first glance it looks like it will not work for a lever since the power flux should go to zero at the fulcrum where v goes to zero. I have not yet worked it out to be sure, so hopefully I am wrong.
I don’t know much about your power flux interpretation, but the use of the approach I described for the rate of doing work at a boundary can be found in all books on continuum mechanics. See also Transport Phenomena by Bird, Stewart, and Lightfoit.
 
  • #8
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See also Transport Phenomena by Bird, Stewart, and Lightfoit.
Thanks.

I don’t know much about your power flux interpretation, but the use of the approach I described for the rate of doing work at a boundary can be found in all books on continuum mechanics.
I think it is approximately the same thing, although I don’t know the terminology. The rate of doing work is mechanical power, and I am interested in the mechanical power that crosses an arbitrary boundary inside a simple machine like a rope or a shaft or a lever. I think that if you take the vector field that I described as the power flux and then integrate it across an arbitrary boundary that you get the power at that boundary.

The idea is that if I am doing work at one end of such a device and the device is doing work at the other end then the energy should not just magically disappear at my end and appear at the other end, but there should be a clear path within the device where energy moves from place to place internally in a systematic fashion.
 
  • #9
vanhees71
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To get the interpretation of the local quantities occurring in the partial-differential equation of fluid or elastic-body dynamics usually one brings them in integral form using Gauss's and Stokes's integral theorems (the former for balance equations you are after the latter for circulation, vortices and the like). A very good discussion is found (from the physicists' point of view) in Landau and Lifshitz vol. VI. For the theoretical-engineering point of view I can recommend

K. Hutter, Y. Wang, Fluid and Thermodynamics, 2 vols., Springer Verlag
 
  • #10
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See also publications of the well-known continuum mechanics group that was headed up by Clifford Truesdell at Johns Hopkins back in the day.
 
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  • #11
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Do you know if there is any specific name for either of these quantities? The rate of work done at a surface (##\sigma \cdot v \cdot dA##) or what I was calling the power flow density (##\sigma \cdot v##).
 
  • #12
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I don’t think there is any specific name for the former other than the rate of doing work over the differential area. I don’t recall ever seeing the latter one being used for anything or having physical significance. In momentum balances, Bird et al refer to the contraction of the stress tensor with the unit normal as a momentum flux.
 
  • #13
vanhees71
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Take the equation of motion of some continuous mechanical system. Then if ##\vec{g}(t,\vec{x})## is the momentum density (momentum per unit volume). Then the equation of motion reads
$$\partial_t g_j = f_j -\partial_{i} \Pi_{ij},$$
using the usual Ricci calculus including Einstein's summation convention (sum over repeated indices from 1 to 3). Here ##f_j## is the force per unit volume (like the force of an external electric field or gravity of the Earth acting on the medium) and ##\Pi_{ij}##. We'll see that ##\Pi_{ij}## is the momentum-current density.

Now take an arbitrary volume ##V## with boundary ##\partial V## at rest (a "control volume") and integrate this equation of motion, which gives
$$\dot{p}_j=\mathrm{d}_t \int_V \mathrm{d}^3 x g_j = \int_V \mathrm{d}^3 x f_j - \int_{\partial V} \mathrm{d}^2 f_i \sigma_{ij}.$$
Thus the total force acting on the fluid momentarilying being in the control volume is the total force from the external field (volume force) and due to the momentum lost by streaming out of it through the surface.

Maybe it's more intuitive when taking as a simple example a perfect fluid with mass density ##\rho## and pressure ##P##. Then you have the continuity equation (mass conservation)
$$\partial_t \rho + \partial_j (\rho v_j)=0,$$
where ##\rho \vec{v}## is the mass-current density (mass streaming through a surface with normal vector ##\vec{n}## per unit time is ##\rho \vec{v} \cdot \vec{n}##) and Euler's equation of motion
$$\rho \mathrm{D}_t v_j=f_j - \partial_j P,$$
which you can rewrite in the above form using
$$\mathrm{D}_t v_j = \partial_t v_j + v_k \partial_k v_j:$$
You get using the continuity equation and Euler's equation
$$\partial_t g_j = \partial_t (\rho v_j) =v_j \partial_t \rho + \rho \partial_t v_j = -v_j \partial_k (\rho v_k) + f_j-\partial_j P -\rho v_k \partial_k v_j = f_j - \partial_{k} (\rho v_j v_k + P \delta_{jk}).$$
Thus we have for the momentum-current density in this case
$$\Pi_{kj}=\rho v_k v_j + P \delta_{jk},$$
of which the first term in the balance equation takes into account the momentum transported out of the volume per unit time through the fluid streaming through the surface of the control volume and the second term is the momentum change due to the stress at the control volume, which in this case of an ideal fluid is just due to the isotropic pressure acting at the surface due to the surrounding fluid.
 
  • #14
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Take the equation of motion of some continuous mechanical system. Then if ##\vec{g}(t,\vec{x})## is the momentum density (momentum per unit volume). Then the equation of motion reads
$$\partial_t g_j = f_j -\partial_{i} \Pi_{ij},$$
using the usual Ricci calculus including Einstein's summation convention (sum over repeated indices from 1 to 3). Here ##f_j## is the force per unit volume (like the force of an external electric field or gravity of the Earth acting on the medium) and ##\Pi_{ij}##. We'll see that ##\Pi_{ij}## is the momentum-current density.

Now take an arbitrary volume ##V## with boundary ##\partial V## at rest (a "control volume") and integrate this equation of motion, which gives
$$\dot{p}_j=\mathrm{d}_t \int_V \mathrm{d}^3 x g_j = \int_V \mathrm{d}^3 x f_j - \int_{\partial V} \mathrm{d}^2 f_i \sigma_{ij}.$$
Thus the total force acting on the fluid momentarilying being in the control volume is the total force from the external field (volume force) and due to the momentum lost by streaming out of it through the surface.

Maybe it's more intuitive when taking as a simple example a perfect fluid with mass density ##\rho## and pressure ##P##. Then you have the continuity equation (mass conservation)
$$\partial_t \rho + \partial_j (\rho v_j)=0,$$
where ##\rho \vec{v}## is the mass-current density (mass streaming through a surface with normal vector ##\vec{n}## per unit time is ##\rho \vec{v} \cdot \vec{n}##) and Euler's equation of motion
$$\rho \mathrm{D}_t v_j=f_j - \partial_j P,$$
which you can rewrite in the above form using
$$\mathrm{D}_t v_j = \partial_t v_j + v_k \partial_k v_j:$$
You get using the continuity equation and Euler's equation
$$\partial_t g_j = \partial_t (\rho v_j) =v_j \partial_t \rho + \rho \partial_t v_j = -v_j \partial_k (\rho v_k) + f_j-\partial_j P -\rho v_k \partial_k v_j = f_j - \partial_{k} (\rho v_j v_k + P \delta_{jk}).$$
Thus we have for the momentum-current density in this case
$$\Pi_{kj}=\rho v_k v_j + P \delta_{jk},$$
of which the first term in the balance equation takes into account the momentum transported out of the volume per unit time through the fluid streaming through the surface of the control volume and the second term is the momentum change due to the stress at the control volume, which in this case of an ideal fluid is just due to the isotropic pressure acting at the surface due to the surrounding fluid.
I think it would have been preferable if you had shown the comparable development for a Newtonian viscous fluid, rather than an inviscid fluid.
 
  • #15
vanhees71
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Ok, then you have to add the next terms of the gradient expansion in ##\vec{v}## to get at the next (linear) order the Navier Stokes equations. Assuming a homogeneous fluid this introduces two scalars, the shear and bulk viscosity since only the symmetric part of the velocity gradient can lead to friction (the antisymmetric part means simply a rigid rotation of the fluid element). Thus the corresponding viscous correction to the momentum-current density is
$$\sigma_{jk}=\eta \left (\partial_j v_k + \partial_k v_j -\frac{2}{3} \delta_{jk} \vec{\nabla} \cdot \vec{v} \right) + \zeta \delta_{jk} \vec{\nabla} \cdot \vec{v}.$$
The decomposition is due to the irreducible parts of the symmetric 2nd rank tensor into a traceless part (quadrupole) and the trace (monopole) part corresponding to the ##l=2## and ##l=0## contribution of the irreducible representations of the rotation group.
 
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  • #16
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at first glance it looks like it will not work for a lever since the power flux should go to zero at the fulcrum where v goes to zero
I worked this out and it seems like it works for a lever also. The key was that a real lever has thickness. This means that there is only one point with v=0.

For example, when the lever is horizontal the top surface has a large vertical velocity at the end which decreases linearly to 0 at the fulcrum. This is combined with a constant shear stress to produce a power flow in the horizontal direction which decreases linearly to zero at the center. However, the lever also has a small horizontal velocity which is constant across the surface. This combined with the linearly increasing longitudinal stress from the bending moment produces a linearly increasing power flow in the horizontal direction. These two flows together combine to make a constant power flow in the horizontal direction, even around the fulcrum.
 

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