6. balance laws jan 2013

Balance Laws Conservation of Mass, Momentum & Energy

Balance of Mass The principle of mass conservation

Balance of Momentum & Angular Momentum A reformulation of Newton’s second law of motion –

Emphasis on continuously distributed matter. Symmetry

Balance of Energy and the Work Principle. Conjugate Stress Analysis

Consistency in the scalar quantities Work and Energy.

Inbalance of Entropy – Statement of the second law of thermodynamics.

The principle of energy availability;

Implications on Processes & Efficiency

Conservation Laws

[email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 2

mailto:[email protected]

The mass of a continuously distributed body is defined in basic physics as the total amount of substance or material contained in the body. The basic idea behind the conservation law is that the mass of an identified quantity is not subject to change during motion. Here, we are obviously restricting ourselves to non-relativisitic mechanics. It is appropriate to reiterate certain basic definitions at this point:

System: A particular collection of matter in space that is of interest. The complement of this is the rest of matter – essentially, the rest of the universe. The boundary of the system is the surface that separates them. The kind of system depends on the nature of this surface – especially what are allowed to pass through.


Balance of Mass

A system is open if matter or mass can pass through the boundary. Otherwise the system is closed. In a closed system therefore, we are dealing with the same quantity of matter throughout the motion as no new mass comes in and old matter are trapped inside.

In addition to this, a system may also be closed to energy transfer. Such a system is said to be isolated. A thermally isolated system, closed to the transfer of heat energy across the boundary is said to be insulated. A system may also be only mechanically isolated. An intensive or bulk property is a scale-invariant physical property of a system. By contrast, an extensive property of a system is directly proportional to the system size or the amount of material in the system.


Open & Closed System

The rate of change of an extensive property Φ, for the system is equal to the time rate of change of Φ within the volume Ω and the net rate of flux of the property Φ through the surface 𝜕Ω, or

𝐷

𝐷𝑡 Φ(𝐱, 𝑡)𝑑𝑣Ω

= Φ𝐯 ⋅ 𝐧 𝑑𝑠𝜕Ω

+ 𝜕Φ

𝜕𝑡𝑑𝑣

Ω


Leibniz-Reynolds Transport Theorem

The fact that the volume is variable with time that is, Ω = 𝛺(𝑡) means that the derivative does not commute with the integral in spatial coordinates. A transformation to material coordinates simplifies the situation. Use the fact that in material coordinates, a derivative under the integral sign is the same as the derivative of the integral itself. If 𝐼 𝑡 = Φ 𝐱, 𝑡 𝑑𝑣

Ω then

𝐼 𝑡 =𝐷

𝐷𝑡 Φ 𝐱, 𝑡

𝑑𝑣

𝑑𝑉𝑑𝑉

Ω

=𝐷

𝐷𝑡 Φ 𝐱, 𝑡 𝐽𝑑𝑉Ω0

= 𝐷

𝐷𝑡Φ 𝐱, 𝑡 𝐽 𝑑𝑉

Ω0

= Φ 𝐱, 𝑡 𝐽 + 𝐽 Φ 𝐱, 𝑡 𝑑𝑉Ω0

= Φ 𝐱, 𝑡 +𝐽

𝐽Φ 𝐱, 𝑡 𝐽𝑑𝑉

Ω0

= Φ 𝐱, 𝑡 + div 𝐯 Φ 𝐱, 𝑡 𝑑𝑣Ω


Proof

We now use of the fact that, for any spatial function,𝜙(𝐱, 𝑡), the material time derivative,

𝐷𝜙 𝐱,𝑡

𝐷𝑡=

𝜕𝜙 𝐱,𝑡

𝜕𝑡+ 𝐯 ⋅

𝜕𝜙 𝐱,𝑡

𝜕𝐱. Consequently,

𝐼 𝑡 = 𝐷

𝐷𝑡Φ 𝐱, 𝑡 𝐽 𝑑𝑉

Ω0= Φ 𝐱, 𝑡 𝐽 + 𝐽 Φ 𝐱, 𝑡 𝑑𝑉

Ω0

= Φ 𝐱, 𝑡 +𝐽

𝐽Φ 𝐱, 𝑡 𝐽𝑑𝑉

Ω0

= Φ 𝐱, 𝑡 + div 𝐯 Φ 𝐱, 𝑡 𝑑𝑣Ω

= 𝜕Φ 𝐱, 𝑡

𝜕𝑡+ 𝐯 ⋅ grad Φ 𝐱, 𝑡 + div 𝐯 Φ 𝐱, 𝑡 𝑑𝑣

Ω

=

= 𝜕Φ 𝐱, 𝑡

𝜕𝑡+ div 𝐯Φ 𝑑𝑣

Ω

which after applying the divergence theorem of Gauss, we find to be,

𝐼 𝑡 ≡ 𝜕Φ 𝐱, 𝑡

𝜕𝑡+ div 𝐯Φ 𝑑𝑣

Ω

= Φ𝐯 ⋅ 𝐧 𝑑𝑠𝜕Ω

+ 𝜕Φ

𝜕𝑡𝑑𝑣

Ω

as required. [email protected] 12/30/2012 Department of Systems Engineering, University of Lagos 7

Based upon the above theorem, we can express the balance of mass compactly, considering the fact that,

ϱ 𝐱, 𝑡 𝑑𝑣Ω

= ϱ0 𝐗 𝑑𝑉Ω0

The right hand of the above equation is independent of time t. Hence a time derivative,

𝐷

𝐷𝑡 ϱ 𝐱, 𝑡 𝑑𝑣Ω

=𝐷

𝐷𝑡 ϱ 𝐗 𝑑𝑉Ω0

= 0

Invoking the Leibniz-Reynolds theorem, we conclude that,

𝜕ϱ

𝜕𝑡+ div 𝐯ϱ 𝑑𝑣

Ω

= ϱ𝐯 ⋅ 𝐧 𝑑𝑠𝜕Ω

+ 𝜕ϱ

𝜕𝑡𝑑𝑣

Ω

= 0


Conservation of Mass

The mass generation inside the system plus the net mass transport across the boundary sum up to zero.

𝜕𝜚

𝜕𝑡𝑑𝑣

𝛺

+ 𝜚𝐯 ⋅ 𝒏 𝑑𝑠𝜕𝛺

= 0

Equivalently, in differential form, The time rate of change of spatial density 𝜚 plus the divergence of mass flow rate equals zero.

𝜕𝜚

𝜕𝑡+ div 𝐯𝜚 = 0

Or equivalently, 𝐷𝜚

𝐷𝑡+ 𝜚 div 𝐯 = 0


Conservation of Mass

For a scalar field 𝜙(𝐱, 𝑡), using the continuity equation, we can write,

𝜕 𝜚𝜙

𝜕𝑡= 𝜚

𝜕𝜙

𝜕𝑡+ 𝜙

𝜕𝜚

𝜕𝑡= 𝜚

𝜕𝜙

𝜕𝑡− 𝜙div 𝜚𝐯

= 𝜚𝜕𝜙

𝜕𝑡− div 𝐯𝜚𝜙 + 𝜚𝐯 ⋅ grad𝜙

= 𝜚𝜕𝜙

𝜕𝑡− div 𝐯𝜚𝜙 + 𝜚𝐯 ⋅

𝜕𝜙

𝜕𝐱

= 𝜚𝐷𝜙

𝐷𝑡− div 𝐯𝜚𝜙

from which we can conclude that,

𝜚𝐷𝜙

𝐷𝑡=𝜕 𝜚𝜙

𝜕𝑡+ div 𝐯𝜚𝜙 .


Implications for Scalar Fields

Furthermore, in spatial coordinates, using Leibniz theorem,

𝐷

𝐷𝑡 𝜚𝜙 𝑑𝑣Ω

= 𝐷 𝜚𝜙

𝐷𝑡+ div 𝐯 𝜚𝜙 𝑑𝑣

Ω

= 𝜚𝐷𝜙

𝐷𝑡+ 𝜙

𝐷𝜚

𝐷𝑡+ 𝜚 div 𝐯) 𝑑𝑣

Ω

= 𝜚𝐷𝜙

𝐷𝑡𝑑𝑣

Ω

a relationship we can also arrive at by treating the “mass measure” 𝜚𝑑𝑣 as a constant under spatial volume integration [Gurtin et al. pg 130].


Mass Measure

Such a process remains valid for any spatial vector or tensor 𝝍 𝐱, 𝑡 , the above expression remains valid, for, 𝐷

𝐷𝑡 𝜚𝝍 𝐱, 𝑡 𝑑𝑣Ω

= 𝐷 𝜚𝝍 𝐱, 𝑡

𝐷𝑡+ div 𝐯 𝜚𝝍 𝐱, 𝑡 𝑑𝑣

Ω

= 𝜚𝐷𝝍 𝐱, 𝑡

𝐷𝑡+ 𝝍 𝐱, 𝑡

𝐷𝜚

𝐷𝑡+ 𝜚 div 𝐯 𝑑𝑣

Ω

= 𝜚𝐷𝝍 𝐱, 𝑡

𝐷𝑡𝑑𝑣

Ω

Again, bringing the derivative under the integral sign with the mass measure 𝜚𝑑𝑣 treated as a constant under the integral on account of continuity.


Vector Fields

For a vector field 𝝍 𝐱, 𝑡 , we can also write, 𝜕 𝜚𝝍 𝐱, 𝑡

𝜕𝑡= 𝜚

𝜕𝝍

𝜕𝑡+ 𝝍

𝜕𝜚

𝜕𝑡= 𝜚

𝜕𝝍

𝜕𝑡− 𝝍 div 𝐯𝜚

= 𝜚𝜕𝝍

𝜕𝑡− div 𝜚𝐯⊗𝝍 + 𝜚𝐯 ⋅

𝜕𝝍

𝜕𝐱 [See Ex 3.1 28 ]

= 𝜚𝑑𝝍

𝑑𝑡− div 𝜚𝐯⊗𝝍

from which we can conclude that,

𝜚𝑑𝝍

𝑑𝑡=𝜕 𝜚𝝍

𝜕𝑡+ div 𝜚𝐯⊗𝝍


Vector & Tensor Fields

Furthermore for a tensor field, the arguments are exactly the same and the result is in objects of one degree higher:

𝜚𝐷𝚵

𝐷𝑡=𝜕 𝜚𝚵

𝜕𝑡+ div 𝜚𝐯⊗ 𝚵

In particular, for the velocity field, we can see clearly that,

𝜚𝑑𝐯

𝑑𝑡=𝜕 𝜚𝐯

𝜕𝑡+ div 𝜚𝐯⊗ 𝐯


Vector & Tensor Fields

A control volume 𝑅 is a fixed region of space entirely encompassed by the deformed configuration. It is therefore permissible for material to pass through its boundary while the boundary itself is taken as being fixed in time. First consider the scalar function 𝜙(𝐱, 𝑡) which depends on both the spatial location and time.

𝑑

𝑑𝑡 𝜚𝜙𝑅

𝑑𝑣 = 𝜕

𝜕𝑡𝜚𝜙

𝑅

𝑑𝑣

= 𝜚𝜕𝜙

𝜕𝑡+ 𝜙

𝜕𝜚

𝜕𝑡𝑅

𝑑𝑣 = 𝜚𝜕𝜙

𝜕𝑡𝑅

𝑑𝑣 + 𝜙𝜕𝜚

𝜕𝑡𝑅

𝑑𝑣

= 𝜚𝜕𝜙

𝜕𝑡𝑅

𝑑𝑣 + 𝜙𝜕𝜚

𝜕𝑡𝑅

𝑑𝑣

= 𝜚𝜕𝜙

𝜕𝑡𝑅

𝑑𝑣 − 𝜙𝜚𝐯 ⋅ 𝒏 𝑑𝑠𝜕𝑅


Mass Balance on Control Volume

For a vector field 𝝍 𝐱, 𝑡 , applying the conservation of mass as before, we can write,

𝑑

𝑑𝑡 𝜚𝝍 𝐱, 𝑡𝑅

𝑑𝑣 = 𝜕

𝜕𝑡𝜚𝝍

𝑅

𝑑𝑣 = 𝜚𝜕𝝍

𝜕𝑡+ 𝝍

𝜕𝜚

𝜕𝑡𝑅

𝑑𝑣

= 𝜚𝜕𝝍

𝜕𝑡𝑅

𝑑𝑣 + 𝝍𝜕𝜚

𝜕𝑡𝑅

𝑑𝑣

= 𝜚𝜕𝝍

𝜕𝑡𝑅

𝑑𝑣 − 𝝍𝜚𝐯 ⋅ 𝐧 𝑑𝑠𝜕𝑅

= 𝜚𝜕𝝍

𝜕𝑡𝑅

𝑑𝑣 − 𝜚 𝝍⊗ 𝐯 𝐧 𝑑𝑠𝜕𝑅

so that,

𝜚𝜕𝝍

𝜕𝑡𝑅

𝑑𝑣 =𝑑

𝑑𝑡 𝜚𝝍 𝐱, 𝑡𝑅

𝑑𝑣 + 𝜚 𝝍⊗ 𝐯 𝒏 𝑑𝑠𝜕𝑅


Mass Balance on Control Volume

The momentum balance principles in this section are generalizations of Newton’s second law of motion in the context of a continuously distributed body instead of a particle. These principles lead to the Cauchy’s Laws of motion. We begin this section with the linear momentum balance. Continuing from the last section, we can express the linear momentum of a body in the spatial frame as,

𝑷(𝑡) = 𝜚 𝐱, 𝑡 𝐯 𝐱, 𝑡 𝑑𝑣Ω

= 𝜚0 𝐗 𝐕 𝐗, 𝑡 𝑑𝑉Ω0

where Ω is the spatial configuration volume and Ω0 the reference configuration.


Cauchy’s Laws of Motion

The balance of linear momentum, according to the second law of Newton is that,

𝐷𝑷(𝑡)

𝐷𝑡= 𝑭(𝑡)

where 𝑭(𝑡) is the resultant force on the system. Hence by the conservation of linear momentum, we may write,


Linear Momentum

http://en.wikipedia.org/wiki/File:Equilibrium_equation_body.svg

𝑷 𝑡 =𝐷

𝐷𝑡 𝜚 𝐱, 𝑡 𝐯 𝐱, 𝑡 𝑑𝑣Ω

=𝐷

𝐷𝑡 𝜚0 𝐗 𝐕 𝐗, 𝑡 𝑑𝑉Ω0

= 𝑭(𝑡)

We now look at the forces acting on the body from the categorization of surface and body forces. The surface forces

are measured by the tractions or force intensities 𝐓 𝐧 per unit area of the surface while the body forces are in terms of the specific body force 𝐛 per unit volume. Clearly,

𝑭 𝑡 = 𝑻 𝒏 𝐱, 𝑡 𝑑𝑠𝜕Ω

+ 𝐛 𝐱, 𝑡 𝑑𝑣Ω


Cauchy’ Law

By Cauchy’s stress law, 𝑻 𝒏 = 𝝈 ⋅ 𝒏. Consequently, we may write,

𝐷

𝐷𝑡 𝜚𝐯𝑑𝑣Ω

= 𝐓 𝐧 𝐱, 𝑡 𝑑𝑠𝜕Ω

+ 𝐛 𝐱, 𝑡 𝑑𝑣Ω

= 𝝈𝐧𝑑𝑠𝜕Ω

+ 𝐛𝑑𝑣Ω

= (grad𝝈 + 𝒃)𝑑𝑣Ω


Cauchy law of Motion

The law of conservation of mass allows us to take the above substantial derivative under the integral, we are allowed to treat the mass measure as a constant, hence we can write that,

𝐷

𝐷𝑡 𝜚𝐯𝑑𝑣Ω

= 𝐷

𝐷𝑡𝜚𝐯 𝑑𝑣

Ω

= 𝜚𝑑𝐯 𝐱, 𝑡

𝑑𝑡Ω

𝑑𝑣

= (grad 𝝈 + 𝒃)𝑑𝑣Ω

We can write the above equation in differential form as,

grad 𝝈 + 𝒃 = 𝜚𝑑𝐯 𝐱, 𝑡

𝑑𝑡

when we remember the definition of the derivative with respect to the position vector.


Cauchy’s Law

Euler-Cauchy First Law of Motion:

The divergence of the stress tensor 𝝈 plus the body force 𝐛 per unit volume equals material time rate of change of linear momentum.

grad 𝝈 + 𝐛 = 𝜚D𝐯

D𝑡


Formal Statement

The rate of change of angular momentum about a point, 𝑳 (𝑡) is equal to the sum of the external moments about that point. In a continuously distributed medium, the Cauchy Stress Tensor field is a symmetric tensor

𝝈𝑇(𝐱, 𝑡) = 𝝈(𝐱, 𝑡)

We do not offer a proof here of this important theorem except to say that it is a consequence of the balance of angular momentum. The full proof is in the accompanying notes to the slides.


Cauchy Second Law of Motion

In concluding this section, observe that the natural tensor characterizations of strain such as deformation gradient, displacement gradient are not symmetric.

Various strain tensors are defined. These were done essentially to separate the rigid body displacements from actual deformations.

The symmetry of Lagrangian and Eulerian strains are definitions. On the other hand, the Cauchy true stress is symmetric as a natural principle.


Cauchy Second Law

We are now in a position to compute such quantities as work rate and energy. Beginning with Euler-Cauchy first law of motion,

div𝝈 + 𝐛 =𝐷

𝐷𝑡𝜚𝐯 = 𝜚𝐯

if we assume density is constant. A scalar product of this equation with velocity gives,

div𝝈 + 𝐛 ⋅ 𝐯 = 𝜚𝐯 ⋅ 𝐯


Work and Energy Balance

Integrating over the spatial volume,

div 𝝈 + 𝒃Ω

⋅ 𝐯𝑑𝑣 = div 𝝈Ω

⋅ 𝐯𝑑𝑣 + 𝒃Ω

⋅ 𝐯𝑑𝑣 =

𝜚𝐯 ⋅ 𝐯Ω

𝑑𝑣 = div 𝝈 ⋅ 𝐯Ω

𝑑𝑣 − 𝝈 ∶ 𝑳Ω

𝑑𝑣 + 𝒃Ω

⋅ 𝐯𝑑𝑣

= div 𝝈 ⋅ 𝐯Ω

𝑑𝑣 − tr 𝝈 ⋅ 𝑳Ω

𝑑𝑣 + 𝒃Ω

⋅ 𝐯𝑑𝑣

where 𝑳 is the velocity gradient and it is obvious that 𝜎𝑖𝑗𝑣𝑗 ,𝑖 = 𝜎𝑖𝑗 ,𝑖 𝑣𝑗 + 𝜎𝑖𝑗𝑣𝑖 ,𝑗

or div 𝝈 ⋅ 𝐯 = div 𝝈 ⋅ 𝐯 + 𝝈 ∶ 𝑳



Applying the divergence theorem, we have

𝜚𝐯 ⋅ 𝐯Ω

𝑑𝑣 + tr 𝝈𝑳Ω

𝑑𝑣 = 𝝈𝐯 ⋅ 𝐧𝜕Ω

𝑑𝑠 + 𝐛Ω

⋅ 𝐯𝑑𝑣

and because the stress tensor is symmetric, we can write the trace in the above equation as

𝝈 ∶ 𝑳 = 𝝈 ∶ 𝑫 + 𝛀 = 𝝈 ∶ 𝑫

on account of the symmetry of 𝑫 and the antisymmetry of 𝛀 where 𝑫 is the deformation (or stretch) rate tensor and 𝛀 is the spin rate. Hence the mechanical energy balance of the body becomes 1

2

𝑑

𝑑𝑡 𝜚𝐯 ⋅ 𝐯Ω

𝑑𝑣 + 𝝈 ∶ 𝑫Ω

𝑑𝑣 = 𝝈𝐯 ⋅ 𝐧𝜕Ω

𝑑𝑠 + 𝒃Ω

⋅ 𝐯𝑑𝑣



The quantity, 𝜌𝐯 ⋅ 𝐯 is the kinetic energy density of the

body, and 1

2

𝑑

𝑑𝑡 𝜌𝐯 ⋅ 𝐯Ω

𝑑𝑣 is the rate of change of the

kinetic energy. 𝝈 ∶ 𝑫Ω

𝑑𝑣, the stress power is the rate of working of the stresses on the body. As the above derivation shows, the spin tensor does no work. In obtaining the stress power for the body, Cauchy stress tensor and the stretch rate appear in the integral. Because of this, Cauchy stress is said to be the stress measure conjugate to the stretch rate.


Power Balance

The stress power expression can be written in terms of other conjugate pairs as follows:

𝝈 ∶ 𝑫Ω

𝑑𝑣 = 𝝉 ∶ 𝑫Ω0

𝑑𝑉 = 𝒔 ∶ 𝑭 Ω0

𝑑𝑉 = 𝚵 ∶ 𝑬 Ω0

𝑑𝑉

which shows that the Kirchhoff, First Piola-Kirchhoff and the second Piola Kirchhoff stresses are conjugate to the stretch rate, rate of deformation gradient and the Lagrangian strain rates respectively. While the Cauchy stress expresses the stress power in terms of spatial coordinates, the other conjugate pairs are in material or reference coordinates.


Conjugate Pairs

The results above can be summarized in the

Power Balance Law:

In the absence of thermal effects, the conventional power expended by the body and surface forces on a body 𝛺 is balanced by the sum of the internal stress power and the rate of change of the kinetic energy.


Stress Power Law

Kirchhoff Stress. It is straightforward to show that,

𝝈 ∶ 𝑫Ω

𝑑𝑣 = 𝝈 ∶ 𝑫Ω0

𝑑𝑣

𝑑𝑉𝑑𝑉

= 𝐽𝝈 ∶ 𝑫Ω0

𝑑𝑉 = 𝝉 ∶ 𝑫Ω0

𝑑𝑉


Kirchhoff Stress

Recall the fact that 𝑭 = 𝑳𝑭. The First Piola Kirchhoff tensor is obtained by a Piola transformation of Cauchy stress

𝒔 = 𝐽𝝈𝑭−𝑇. Now,

tr 𝒔𝑭 𝐓 = tr 𝒔𝑭𝑻𝑳𝑻 = 𝐽 tr 𝝈𝑳

It therefore follows that 𝝈 ∶ 𝑫 = 𝝈 ∶ 𝑳 =1

𝐽𝒔 ∶ 𝑭 in the

reference configuration. Using this in the stress power integral provides the necessary Jacobian in the volume ratio 𝐽 so that,

𝝈 ∶ 𝑫Ω

𝑑𝑣 = 𝒔 ∶𝑭

𝐽

Ω

𝑑𝑣 = 𝒔 ∶ 𝑭 Ω0

𝑑𝑉


First Piola-Kichhoff Tensor

The Lagrange strain tensor is given by 𝐄 =1

2𝐅T𝐅 − 𝟏

𝐄 =1

2𝐅T 𝐅 + 𝐅T𝐅

=1

2𝐅𝐓𝐋𝐓𝐅 + 𝐅T𝐋𝐅 =

1

2𝐅𝐓 𝐋𝐓 + 𝐋 𝐅

= 𝐅𝐓𝐃𝐅

𝝈 ∶ 𝑫Ω

𝑑𝑣 = 𝝈 ∶ 𝐅−𝐓𝐄𝐅−𝟏

Ω

𝑑𝑣

= tr 𝐅−𝟏𝝈𝐅−𝐓 𝐄 Ω

𝑑𝑣 = 𝐽𝐅−𝟏𝝈𝐅−𝐓 : 𝐄 Ωo

𝑑𝑉

= 𝚵: 𝐄 Ωo

𝑑𝑉

where 𝚵 ≡ 𝐽𝐅−𝟏𝝈𝐅−𝐓 is the Second Piola-Kirchhoff stress tensor. This Work Conjugate of the Lagrange strain is symmetrical.


Second Piola-Kirchhoff Tensor

Thus far we have considered mechanical work and energy in the absence of thermal effects. In this section, the thermodynamic effects including heat transfer and entropy generation will be considered.


Thermodynamical Balances

The first law of thermodynamics is an expression of the principle of conservation of energy. It expresses the fact that energy can be transformed, i.e. changed from one form to another, but can neither be created nor destroyed. It is usually formulated by stating that the change in the internal energy of a system is equal to the amount of heat supplied to the system, minus the amount of work performed by the system on its surroundings.


First Law of Thermodynamics

We state here a version of the first law of thermodynamics due to HLF von Helmholtz:

The heat supply 𝑄(𝑡) and the power of external forces 𝐿(𝑡) lead to a change of the kinetic energy 𝐾(𝑡) in an inertial frame and of the internal energy 𝐸(𝑡) of the body.

In the deformed configuration, we can write, 𝑑

𝑑𝑡𝐸 𝑡 + 𝐾 𝑡 = 𝑄 𝑡 + 𝐿(𝑡)


First Law of Thermodynamics

Furthermore, the supply of heat into the body is assumed to come from two sources: Heat generation inside the body and heat energy crossing the boundary. If we assume that the rate of heat generation (eg by radiation) per unit volume is the scalar field 𝑟, then the heat generation rate is 𝜚𝑟𝑑𝑣

Ω.

The vector flux per unit area is denoted by 𝒒 so that the heat flux into the system along the boundary is

− 𝐪 ⋅ 𝐧 𝑑𝑠𝜕Ω

By the Fourier-Stokes Heat Flow theorem.


First Law

With these, we can write the energy balance equation as, 𝑑

𝑑𝑡 𝜚 휀 +

1

2𝐯 2

Ω

𝑑𝑣 =

= 𝝈𝐯 ⋅ 𝐧𝜕Ω

𝑑𝑠 + 𝒃Ω

⋅ 𝐯𝑑𝑣 − 𝒒. 𝒏 𝑑𝑠𝜕Ω

+ 𝜚𝑟𝑑𝑣Ω

=1

2

𝑑



𝑑𝑣 − 𝒒. 𝒏 𝑑𝑠𝜕Ω


since 𝐿 𝑡 = 1

2

𝑑



𝑑𝑣 =

𝝈 ⋅ 𝐯 ⋅ 𝐧𝜕Ω

𝑑𝑠 + 𝒃Ω

⋅ 𝐯𝑑𝑣 and 𝐾 𝑡 =1

2

𝑑


𝑑𝑣


Energy Balance

We can write the above in terms of the specific internal energy as, 𝑑

𝑑𝑡 𝜚휀𝑑𝑣Ω

= − 𝒒. 𝒏 𝑑𝑠𝜕Ω


+ 𝝈 ∶ 𝑫Ω

𝑑𝑣

and as a result of the continuity of mass, the constancy of the mass measure ( 𝜚𝑑𝑣

Ω)allows us to bring the derivative under the

integral sign and apply it only to the internal energy so that,

𝜚휀 𝑑𝑣Ω

= − grad𝒒 𝑑𝑣Ω

+ 𝑟𝜚𝑑𝑣Ω

+ 𝝈 ∶ 𝑫Ω

𝑑𝑣

after application of the divergence theorem. This is the same as,

𝜚휀 + grad𝒒 − 𝜚𝑟 − 𝝈 ∶ 𝑫 𝑑𝑣Ω

= 0

which gives a local energy balance, 𝜚휀 + grad𝒒 − 𝜚𝑟 − 𝝈 ∶ 𝑫 = 0.


Energy Balance

Recall that for any scalar field 𝜙, as a consequence of

the balance of mass, 𝜚𝑑𝜙

𝑑𝑡=

𝜕 𝜚𝜙

𝜕𝑡+ 𝛻𝑦 𝐯𝜚𝜙 .

Applying this to specific internal energy,

𝜚𝑑휀

𝑑𝑡=𝜕 𝜚휀

𝜕𝑡+ grad 𝐯𝜚휀

Using this, the local energy equation now becomes, 𝜕 𝜚휀

𝜕𝑡= 𝜚𝑟 + 𝝈 ∶ 𝑫 − grad 𝒒 + 𝐯𝜚휀

where the derivative with respect to time here is the spatial time derivative and heat flux term now has the convective addition 𝐯𝜚휀 which is a result of heat transfer due to motion.


Spatial Energy Balance

We can also express the first law of thermodynamics in the material description. We begin with heat flow into the system. The radiative and other mass heat generation

𝜚𝑟𝑑𝑣Ω

= 𝜚𝑟𝑑𝑣

𝑑𝑉𝑑𝑉

Ω0

= 𝐽𝜚𝑟𝑑𝑉Ω0

= 𝜚0𝑟𝑑𝑉Ω0

And for the heat flux through the boundary, a Piola Transformation yields

− 𝒒.𝒏 𝑑𝑠𝜕Ω

= − 𝐽𝑭−1𝒒 ⋅ 𝑑𝑨𝜕Ω0

= − 𝒒0 ⋅ 𝒏𝑑𝐴𝜕Ω0

= − Div𝒒0𝑑𝑉Ω0

Hence, the energy balance in terms of referential (material) frame, becomes,

𝜚0휀 𝑑𝑉Ω0

= − 𝛻𝒒0 𝑑𝑉Ω0

+ 𝑟𝜚0𝑑𝑉Ω0

+ 𝒔 ∶ 𝑭 Ω0

𝑑𝑉

from which we can now obtain the local material energy balance, 𝜚0휀 = 𝑟𝜚0 − 𝛻 ⋅ 𝒒0 + 𝒔 ∶ 𝑭


Material Energy Balance

An expression of the observed tendency that over time, differences in temperature, pressure, and chemical potential will equilibrate in an isolated physical system. The second law is an additional restriction on the first law that

forbids certain processes which on their own might have been compatible with the first law but are known by observation never to occur.

That natural processes are preferred choices out of many that are energy preserving.

It consequently defines the concept of thermodynamic entropy, dissipation and available or free energy (more accurately, Helmholtz and Gibbs functions). In this section, we shall also see that specific entropy is an invariant over linear transformations from a fixed origin.

The second law predicts that the entropy of an isolated body must never decrease.


Second Law

http://en.wikipedia.org/wiki/Entropy

"The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equations — then so much the worse for Maxwell's equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.” Arthur Stanley Eddington, The Nature of the Physical World (1927):


Importance of the Second Law

http://en.wikipedia.org/wiki/Arthur_Stanley_Eddington





𝜂 is defined as the measure of entropy per unit mass so that total entropy

𝑆 𝑡 = 𝜚𝜂𝑑𝑣Ω

A consequence of the second law is that, unlike energy, entropy may be produced (generated) in a system that is isolated. As a result, the net entropy can be greater than zero after accounting for the entropy crossing the system boundaries. We state here the second law in form of the Claussius-Duhem inequality

𝑑

𝑑𝑡 𝜚𝜂𝑑𝑣Ω

≥ − 𝒒

𝑇⋅ 𝒏

𝜕Ω

𝑑𝑠 + 𝜚𝒓

𝑇Ω

𝑑𝑣

meaning that the net entropy is at least as much as the entropy inflow through the boundary and the entropy generation by other sources into the system. 𝑇(𝐱, 𝑡) is the scalar temperature field.


Entropy . . .

𝑑


≥ − div𝒒

𝑇𝜕Ω

𝑑𝑣 + 𝜚𝑟

𝑇Ω

𝑑𝑣

We apply the law of conservation of mass, observe the constancy of the mass measure in the flow, so that

𝜚𝜂 + div𝒒

𝑇− 𝜚

𝑟

𝑇𝑑𝑣

Ω

≥ 0

or

𝜚𝜂 ≥ −div𝒒

𝑇+𝜚𝑟

𝑇.

Applying the law of continuity to the material derivative of specific entropy, we have,

𝜚𝑑𝜂

𝑑𝑡=𝜕 𝜚𝜂

𝜕𝑡+ div 𝐯𝜚𝜂

which we now substitute to obtain, 𝜕

𝜕𝑡𝜚𝜂 ≥ −div

𝒒 + 𝐯𝑇𝜚𝜂

𝑇+𝜚𝑟

𝑇

where the partial derivative denotes the spatial time derivative.


Entropy

Again with the convective term added to the heat flux vector resulting from the motion from point to point. Now, it is easily shown that,

−div𝒒

𝑇= −

1

𝑇div 𝒒 +

𝒒

𝑇2⋅ 𝛻𝑦𝑇

Using this, we can write,

𝜚𝜂 ≥ −div𝒒

𝑇+𝜚𝑟

𝑇= −

1

𝑇div 𝒒 +

𝒒

𝑇2⋅ grad 𝑇 +

𝜚𝑟

𝑇

=1

𝑇−div 𝒒 + 𝜚𝑟 +

𝒒

𝑇⋅ grad 𝑇

=1

𝑇𝜚휀 − 𝝈 ∶ 𝑫 +

𝒒

𝑇⋅ grad 𝑇

using the local spatial expression of the first law. Bringing everything to the RHS, we have,

𝜚 휀 − 𝑇𝜂 − 𝝈 ∶ 𝑫 +𝒒

𝑇⋅ grad 𝑇 ≥ 0.


Entropy

The Gibbs free energy, originally called available energy, was developed in the 1870s by the American mathematician Josiah Willard Gibbs. In 1873, Gibbs described this “available energy” as “the greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition.” [11] The initial state of the body, according to Gibbs, is supposed to be such that "the body can be made to pass from it to states of dissipated energy by reversible processes." The specific free energy (Gibbs Function) is defined as 𝜓 = 휀 − 𝑇𝜂. Consequently, then we can write,

𝜚 𝜓 + 𝑇 𝜂 − 𝝈 ∶ 𝑫 +𝒒

𝑇⋅ 𝛻𝑦𝑇 = −𝑇Γ ≥ 0


Free Energy

http://en.wikipedia.org/wiki/Josiah_Willard_Gibbs

http://en.wikipedia.org/wiki/Volume_(thermodynamics)

http://en.wikipedia.org/wiki/Heat

http://en.wikipedia.org/wiki/Dissipated_energy

http://en.wikipedia.org/wiki/Reversible_process_(thermodynamics)

Recalling the conventional energy expression that 1

2

𝑑

𝑑𝑡 𝜚𝐯 ⋅ 𝐯Ω


𝑑𝑣 = 𝝈𝐧 ⋅ 𝐯𝜕Ω

𝑑𝑠 + 𝒃Ω

⋅ 𝐯𝑑𝑣

we can integrate the above inequality and obtain,

𝑇ΓΩ

𝑑𝑣 = 𝝈𝐧 ⋅ 𝐯𝜕Ω

𝑑𝑠 + 𝒃Ω

⋅ 𝐯𝑑𝑣 −𝑑

𝑑𝑡 𝜚 𝜓 +

1

2𝐯 𝟐 𝑑𝑣

Ω

− 𝜚𝑇 𝜂 +𝒒

𝑇⋅ grad𝑇 𝑑𝑣

Ω

≥ 0

That is, 𝐸𝑛𝑒𝑟𝑔𝑦 𝐷𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑜𝑛= 𝐶𝑜𝑛𝑣𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 – 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐹𝑟𝑒𝑒 𝑎𝑛𝑑 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑖𝑒𝑠 – 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸𝑛𝑒𝑟𝑔𝑦


Free Energy

The second law therefore assures there is always a dissipation of energy. It is desirable to minimize such dissipation in order to

increase the amount of available energy. Once again, the constancy of the mass measure as a result of the law of conservation of mass is implicit in the above derivation.

This inequality must be satisfied for all admissible processes the material can undergo. They are therefore restrictions on material behavior. Constitutive equations must conform to these stipulations in order to be valid and admissible in physical processes.

Notice also that the quantities, specific internal energy, specific entropy and specific free energy appear in these inequalities via their time derivatives.


Energy Dissipation

Introducing base values 휀0, 𝜂0 such that 휀 0 = 0, 𝜂 0 = 0, 𝑑


=𝑑

𝑑𝑡 𝜚 𝜂 + 𝜂0 𝑑𝑣Ω

and 𝑑

𝑑𝑡 𝜚휀𝑑𝑣Ω

=𝑑

𝑑𝑡 𝜚 휀 + 휀0 𝑑𝑣Ω

.

This implies invariance with respect to translations 휀 → 휀 + 휀0, 𝜂 → 𝜂 + 𝜂0

by these values. It can also be shown (Ex. 22) that for a vector 𝝎 a the heat flux 𝒒 ⋅ 𝒏 𝑑𝑠

Ω is invariant under the

transformation 𝒒 → 𝒒 + 𝝎 × grad 𝑇 for any scalar 𝑇 provided grad 𝝎 = 0. In this particular instance, the scalar 𝑇 is the temperature field.


Energy Dissipation

A body can be considered isolated in the sense that boundary heat fluxes are nil (𝒒 = 𝒐), boundary surface tractions are either zero 𝝈𝐧 = 𝐨 or no work is done by any applied boundary forces 𝐯 = 𝐨 . The first law then becomes,

𝑑

𝑑𝑡 𝜚 휀 +

1

2𝐯 2

Ω

𝑑𝑣 = 𝒃Ω

⋅ 𝐯𝑑𝑣 + 𝜚𝑟𝑑𝑣Ω

If in addition to these, there are no body forces and the radiative or other bulk heat generation vanishes, then,

𝑑

𝑑𝑡 𝜚 휀 +

1

2𝐯 2

Ω

𝑑𝑣 = 0

And the second law, 𝑑


≥ − div𝒒

𝑇𝜕Ω

𝑑𝑣 + 𝑟

𝑇Ω

𝑑𝑣

becomes, 𝑑


≥ 0

which shows that the rate of change of entropy in such a situation is always positive.


Entropy Rate, Isolated Body

Similar to the internal energy and internal heat source, we can express the net entropy in terms of material coordinates,

𝜚𝜂𝑑𝑣Ω

= 𝜚𝜂𝑑𝑣

𝑑𝑉𝑑𝑉

Ω0

= 𝐽𝜚𝜂𝑑𝑉Ω0

= 𝜚0𝜂𝑑𝑉Ω0

And proceed to write, 𝑑

𝑑𝑡 𝜚0𝜂𝑑𝑉Ω0

≥ − Div𝒒0𝑇𝜕Ω0

𝑑𝑉 + 𝜚0Ω0

𝑟

𝑇𝑑𝑣

where, as before, 𝒒0 ≡ 𝐽𝑭−1𝒒 the Piola transformation of the heat flux.


Second Law of Thermodynamics in Material Frame

1. Given that 𝑇 𝑛 is the resultant stress vector on a surface whose outward unit normal is 𝑛, Find an expression for the normal stress and show that the shear stress on that surface is given by 𝐼 − 𝑛 ⊗ 𝑛 ⋅

𝑇 𝑛 . Express this shear in tensor component form.

2. Obtain an expression for the mass center of a region 𝛺 of a body B and (𝑏) obtain the first and second material time derivatives of the expression.

3. A material velocity field 𝐔 is given by its Piola Transformation 𝐔 = 𝐽𝐅−1𝐮. Using the Piola identity that 𝛻 ⋅ 𝐽𝐅−1 = 0, Show that 𝛻 ⋅ 𝐔 = 𝐽𝛻𝑦 ⋅ 𝐮


Examples


4. The components of Cauchy stress in Cartesian coordinates are

𝑥1𝑥2 𝑥12 −𝑥2

𝑥12 0 0

−𝑥2 0 𝑥12 + 𝑥2

2

Find the body forces in the system to keep the system in equilibrium. 5. The components of Cauchy stress in Cartesian coordinates are

𝛼 0 00 𝑥2 + 𝛼𝑥3 Φ 𝑥2, 𝑥3

0 Φ 𝑥2, 𝑥3 𝑥2 + 𝛽𝑥3

(a) Find Φ 𝑥2, 𝑥3 so that the equilibrium

equations are satisfied assuming body forces are zero. (b) Use the value of Φ 𝑥2, 𝑥3 found in (a) to compute the Cauchy Traction vector on the plane 𝜓 = 𝑥1 + 𝑥2 + 𝑥3.

6. A material vector field is given by its Piola transformation 𝐔 = 𝐽𝐅−1𝐮. Use the Piola identity to show that 𝛻 ⋅ 𝐔 = 𝐽𝛻𝑦 ⋅ 𝐮

7. Given the Cauchy stress components,

𝒚𝟏𝒚𝟐 𝒚𝟏𝟐

−𝒚𝟐

𝒚𝟏𝟐

𝟎 𝟎

−𝒚𝟐 𝟎 𝒚𝟏𝟐+ 𝒚𝟐

𝟐

Find the body forced that keeps the body in equilibrium.


8. Given the Cauchy stress components, 𝜶 𝟎 𝟎𝟎 𝒚𝟐 + 𝜶𝒚𝟑 𝚽(𝒚𝟐, 𝒚𝟑)

𝟎 𝚽(𝒚𝟐, 𝒚𝟑) 𝒚𝟐 + 𝜷𝒚𝟑

(𝒂) Find the values of 𝜶 and 𝜷 to satisfy the equations of equilibrium assuming zero body forces. (𝒃) Find the Cauchy traction vector on the plane 𝚿 = 𝒚𝟏 + 𝒚𝟐 + 𝒚𝟑 .

9. If only mechanical energy is considered show that the energy equation can be obtained directly from the Cauchy’s first law of motion.

10. A hydrostatic state of stress at a certain point is given by the Cauchy stress tensor in the form, 𝛔 = −𝑝𝐈, where 𝐈 is the identity tensor. Show that the stress power per unit referential volume is given by 𝑤𝑖𝑛𝑡 = 𝐽𝛔: 𝐝 = −𝐽𝑝𝐈: 𝐝 =

− 𝐽𝑝 𝑇𝑟 𝒅 = −𝐽𝑝𝛁 ⋅ 𝐯 =𝐽𝑝

𝜌

𝑑𝜌

𝑑𝑡

11. A rigid body is rotating about a fixed point 𝑂 with angular velocity 𝝎 show

that the kinetic energy may be expressed as 1

2𝛚 ⋅ 𝐃𝛚, where

𝐷 = ϱ 𝐈 𝐫 ⋅ 𝐫 − 𝐫⊗ 𝐫 𝑑𝑣Ω

, the moment of inertia tensor.

12. Let 𝑉 be a region of 𝐸3 bounded by 𝜕𝑉 and outward normal 𝒏. Let 𝝈 be the Cauchy stress field and 𝒖 a vector field both of class 𝑪𝟏 Prove that

𝒖⊗ 𝝈 ⋅ 𝒏𝒅𝑆𝝏𝑽

= 𝒖⊗ 𝛻 ⋅ 𝝈 + 𝛻 ⊗𝒖 ⋅ 𝝈𝑽

𝒅𝑉


13. Let 𝑉 be a region of 𝐸3 bounded by 𝜕𝑉 and outward normal 𝒏. Let 𝒖 be a vector field of class 𝑪𝟏. Show that 𝛻 ⊗ 𝒖 𝑑𝑉

𝑽= 𝒏⊗ 𝒖 𝒅𝑆

𝝏𝑽

14. By evaluating the divergence operation on a tensor show that the equilibrium equations can be expressed in Cylindrical Polar coordinates as, 𝜕𝜎𝑟𝑟

𝜕𝑟+

1

𝑟

𝜕𝜎𝜃𝑟

𝜕𝜃+

𝜎𝑟𝑟−𝜎𝜃𝜃

𝑟+

𝜕𝜎𝑧𝑟

𝜕𝑧+ 𝑏𝑟 = 0

𝜕𝜎𝑟𝜃𝜕𝑟

+2𝜎𝑟𝜃𝑟

+1

𝑟

𝜕𝜎𝜃𝜃𝜕𝜃

+𝜕𝜎𝑧𝜃𝜕𝑧

+ 𝑏𝜃 = 0

𝜕𝜎𝑟𝑧𝜕𝑟

+𝜎𝑟𝑧𝑟+1

𝑟

𝜕𝜎𝜃𝑧𝜕𝜃

+𝜕𝜎𝑧𝑧𝜕𝑧

+ 𝑏𝑧 = 0

15. By evaluating the divergence operation on a tensor show that the equilibrium equations can be expressed in Spherical Polar coordinates as, 1

𝜌𝜎𝜌𝜃 cot 𝜃 +

𝜕𝜎𝜌𝜃

𝜕𝜃+

1

sin 𝜃

𝜕𝜎𝜌𝜙

𝜕𝜙+ 𝜌

𝜕𝜎𝜌𝜌

𝜕𝜌− 𝜎𝜃𝜃 + 2𝜎𝜌𝜌 − 𝜎𝜙𝜙 + 𝑏𝜌 = 0

1

𝜌𝜎𝜃𝜃 − 𝜎𝜙𝜙 cot 𝜃 + 𝜌

𝜕𝜎𝜌𝜃

𝜕𝜌+ 3𝜎𝜌𝜃 +

1

sin 𝜃

𝜕𝜎𝜃𝜙

𝜕𝜙+𝜕𝜎𝜃𝜃𝜕𝜃

+𝑏𝜃= 0

1

𝜌

1

sin 𝜃

𝜕𝜎𝜙𝜙

𝜕𝜙+ 𝜌

𝜕𝜎𝜌𝜙

𝜕𝜌+ 3𝜎𝜌𝜙 + 2𝜎𝜃𝜙 cot 𝜃 +

𝜕𝜎𝜃𝜙

𝜕𝜃+ 𝑏𝜙 = 0


16. The law of conservation of mass is expressed in the vector form, 𝛻 ⋅ 𝑽𝜚 +𝜕𝜚

𝜕𝑡= 0. Express this law in tensor components and find the equivalent in

physical components for Cartesian, Cylindrical polar and Spherical polar coordinate systems.

17. Cylindrical Coordinates: 18. The moment of inertia of a continuum with volume Ω is given by,

𝜚 𝒓 2𝑰 − 𝒓⊗ 𝒓Ω

𝑑𝑣 = 𝐼𝑖𝑗𝐠𝑖 ⊗𝐠𝑗Ω

𝑑𝑣

where the scalar density 𝜚(𝒓) is a function of the position vector 𝒓, 𝑰 the identity tensor and 𝑣 the volume element. Show that the components of the

integrands 𝐼𝑖𝑗 = 𝜚 𝑥𝛼𝑥𝛽𝑔𝛼𝛽𝑔𝑖𝑗 − 𝑥𝑖𝑥𝑗 in Cartesian coordinates. Why is this

not correct in Spherical or Cylindrical polar coordinates? 19. Show that the material heat flux rate is the Piola Transformation of the

spatial heat flux rate and that the local material energy balance requirements are satisfied if, 𝜚0휀 = −𝛻 ⋅ 𝒒0 + 𝑟𝜚0 + 𝒔 ∶ 𝑭

20. For a vector field 𝒒 = 𝑞𝑖𝐠𝑖 and a scalar field 𝑇, show that 𝑑𝑖𝑣𝒒

𝑇=

1

𝑇𝑑𝑖𝑣 𝒒 −

𝒒

𝑇2⋅ 𝑔𝑟𝑎𝑑 𝑇


21. For an arbitrary vector 𝝎 and scalar 𝑇, such that 𝑔𝑟𝑎𝑑𝝎 = 0 show that the integral 𝒒 ⋅ 𝒏 𝑑𝑠

Ω is invariant under the transformation 𝒒 → 𝒒 + 𝝎 ×

𝑔𝑟𝑎𝑑 𝑇. 22. For an arbitrary vector 𝝎 and scalar 𝑇, such that 𝑔𝑟𝑎𝑑𝝎 = 0 show that the

integral 𝒒

𝑇⋅ 𝒏 𝑑𝑠

Ω is invariant under the transformation 𝒒 → 𝒒 + 𝝎 ×

𝑔𝑟𝑎𝑑 𝑇. 23. If the heat generation field 𝑟 = 0 and there are no body forces, Show that

the first law of thermodynamics becomes, 𝑑

𝑑𝑡 𝜚 휀 +

1

2𝐯 2

Ω𝑑𝑣 =

𝝈 ⋅ 𝐯 ⋅ 𝐧𝜕Ω

𝑑𝑠 − 𝒒. 𝒏 𝑑𝑠𝜕Ω

, and the second law 𝑑


≥ − 𝒒

𝑇⋅ 𝒏

𝜕Ω𝑑𝑠

24. Define terms “isolated body”. Show that the net energy in an isolated body does not change and that its entropy can never decrease.

25. Show that for a spatial control volume 𝑅, the first law of thermodynamics

becomes,𝑑

𝑑𝑡 𝜚 휀 +

1

2𝐯 2

R𝑑𝑣 + 𝜚 휀 +

1

2𝐯 2 𝐯 ⋅ 𝒏

𝜕R𝑑𝑠 =

𝝈 ⋅ 𝐯 ⋅ 𝐧𝜕R

𝑑𝑠 + 𝒃R

⋅ 𝐯𝑑𝑣 − 𝒒. 𝒏 𝑑𝑠𝜕R

+ 𝜚𝑟𝑑𝑣R

and the second law,

becomes, 𝑑

𝑑𝑡 𝜚𝜂𝑑𝑣R

+ 𝜚𝜂𝐯 ⋅ 𝒏𝑑𝑠𝜕R

≥ − 𝒒

𝑇⋅ 𝒏

𝜕R𝑑𝑠 +

𝒓

𝑇R𝑑𝑣

26. Obtain the local dissipation inequality, 𝜓 − 𝝈 ∶ 𝑫 ≡ −𝛿 ≥ 0

6. balance laws jan 2013

Education

university of lagos

new mass

mass measurefurthermore

balanceof mass

system size

thermallyisolated system

system mayalso

open closed system