Power done by work on a continuous body isn't the derivative of work?

[Klaus3]

The definition of work and power done over a continuous body is:

$$ W = \int Tn \cdot u dA + \int b \cdot u dV $$

$$ P = \int Tn \cdot v dA + \int b \cdot v dV $$

$T$ is the stress tensor, $b$ is the body force, $u$ is the displacement vector, $v$ is the velocity, $n$ is the normal vector

I've been using these definitions for years, but it always looks like, by the definition of power $P = dW/dt$ that the second equation is not the derivative of the first. So i tried deriving it to settle my curiosity.

What i tried:

First, use the relation:

$$ \int Tn \cdot v dA + \int b \cdot v dV = \int \frac{1}{2} \rho \frac{d|v|^2}{dt} dV+ \int T:\nabla v dV$$

This is the theorem of expended power.

So i need to show that

$$ \frac{d}{dt} \left( \int Tn \cdot u dA + \int b \cdot u dV \right) = \int \frac{1}{2} \rho \frac{d|v|^2}{dt}dV + \int T:\nabla v dV $$


$$ \int Tn \cdot u dA + \int b \cdot u dV = \int\left( \nabla \cdot T \cdot u + T:\nabla u\right)dV + \int b \cdot u dV $$


$$ = \int\left( \left( -b + \rho \frac{dv}{dt} \right) \cdot u + T:\nabla u\right)dV + \int b \cdot u dV $$


$$ = \int\left( \left( \rho \frac{dv}{dt} \right) \cdot u + T:\nabla u\right)dV $$

$$ \frac{d}{dt} \int\left( \left( \rho \frac{dv}{dt} \right) \cdot u + T:\nabla u\right)dV = \int\left( \left( \rho \frac{d^2v}{dt^2} \right) \cdot u + \frac{1}{2} \rho \frac{d|v|^2}{dt} \right)dV +
\int\left( \frac{dT}{dt} : \nabla u + T:\nabla v + (T:\nabla u) \nabla \cdot v \right)dV $$


$$ = \int \frac{1}{2} \rho \frac{d|v|^2}{dt} dV +
\int T:\nabla v dV + \int \left( \rho \frac{d^2v}{dt^2} \cdot u + \frac{dT}{dt} : \nabla u + (T:\nabla u) \nabla \cdot v \right) dV $$

First and second terms is what i want to get to, but there is a third term. So, unless this term is zero, it looks like the relation $P = dW/dt$ doesn't hold here. I cannot continue further because i don't know any relations that involve $\frac{dT}{dt}$ so i can simplify further. Did i make any mistake? is that term really zero?

 

[Chestermiller]

In Transport Phenomena by Bird, Stewart, and Lightfoot, the differential Mechanical Energy Balance equation is obtained by dotting the Equation of Motion (momentum balance equation) with the velocity vector. See chapter 3.

 

[Klaus3]

I'm aware, but the equation on Bird chapter 3 only deals with power (it is essentially the third equation but with the splitting of the Cauchy stress tensor into deviatoric and pressure terms). My doubt is related with the definition of work itself, not with the balance of mechanical energy.

By the way work is defined (First equation) i can't seem to use the definition of Power = $dW/dt$ to obtain the second.

 

[Chestermiller]

I'm aware, but the equation on Bird chapter 3 only deals with power (it is essentially the third equation but with the splitting of the Cauchy stress tensor into deviatoric and pressure terms). My doubt is related with the definition of work itself, not with the balance of mechanical energy.

By the way work is defined (First equation) i can't seem to use the definition of Power = $dW/dt$ to obtain the second.

$$\int{\mathbf{F}\centerdot \mathbf{v}dt}=W$$

 

[Klaus3]

Using this definition:

$$ \int \int \left( Tn \cdot v dt \right)dA + \int \int \left( b \cdot v dt \right)dV $$
$$ = \int \int \left( Tn \cdot du \right)dA + \int \int \left( b \cdot du \right)dV $$

Thats still different, can you take $T$ and $b$ outside the inner integral so that $du$ becomes $u$?

 

[Chestermiller]

Using this definition:

$$ \int \int \left( Tn \cdot v dt \right)dA + \int \int \left( b \cdot v dt \right)dV $$
$$ = \int \int \left( Tn \cdot du \right)dA + \int \int \left( b \cdot du \right)dV $$

Thats still different, can you take $T$ and $b$ outside the inner integral so that $du$ becomes $u$?

Definitely not, and why are there two integrals in the 2nd equation?

 

[Klaus3]

I'm integrating the power equation(second equation of OP) with respect to time. There are two integrals because one is the volume/surface integral, the other is the time integral, transformed into displacement integral by $du/dt = v; du = vdt$

 

[Chestermiller]

I agree with your first equationn

 

[Chestermiller]

Is this basically a body immersed in another material (like, say, a fluid), and the body is capable of large deformations under the loads imposed by the other material?

 

[Klaus3]

There isn't anything specific, its just a continuous body subject to continuous contact and body forces, may or may not deform, but is deformable.

You can see on this entrance on the wiki that its just defined that way

https://en.wikipedia.org/wiki/Virtu..._of_virtual_work_and_the_equilibrium_equation

(i'm not talking specifically about virtual work here, its just one entry that i can easily find without having to screenshot books)

And books also do the same, what "bothers" me is that the work is defined based on the raw displacement vector $u$ instead of the classical differential of the displacement vector $du$ or more commonly, $dr$

As in:

$$W = \int F \cdot dr$$

But instead here its defined more like

$$W = \int dF \cdot \Delta r$$

where $u = \Delta r$

 

[Juanda]

@Klaus3 is there any news about it?
It looks like a pretty interesting topic even if I cannot fully follow it.

 

[Klaus3]

No, i never found an answer. It seems impossible to me to either derive the definition of work and find the definition of power, or integrate the definition of power to get the definition of work. There is always a $$du$$ that i can't get rid of.