6. balance laws jan 2013

Balance Laws
Conservation of Mass, Momentum & Energy

Conservation Laws
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

Department of Systems Engineering, University of Lagos 2 oafak@unilag.edu.ng 12/30/2012

Balance of Mass
 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.


Open & Closed System
 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.


Leibniz-Reynolds Transport Theorem

 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 𝜕Φ
Φ(𝐱, 𝑡)𝑑𝑣 = Φ𝐯 ⋅ 𝐧 𝑑𝑠 + 𝑑𝑣
𝐷𝑡 Ω 𝜕Ω Ω 𝜕𝑡


Proof
 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 = Ω Φ 𝐱, 𝑡 𝑑𝑣 then

𝐷 𝑑𝑣 𝐷
𝐼 𝑡 = Φ 𝐱, 𝑡 𝑑𝑉 = Φ 𝐱, 𝑡 𝐽𝑑𝑉
𝐷𝑡 Ω 𝑑𝑉 𝐷𝑡 Ω0
𝐷
= Φ 𝐱, 𝑡 𝐽 𝑑𝑉 = Φ 𝐱, 𝑡 𝐽 + 𝐽Φ 𝐱, 𝑡 𝑑𝑉
Ω0 𝐷𝑡 Ω0
𝐽
= Φ 𝐱, 𝑡 + Φ 𝐱, 𝑡 𝐽𝑑𝑉
Ω0 𝐽
= Φ 𝐱, 𝑡 + div 𝐯 Φ 𝐱, 𝑡 𝑑𝑣
Ω


We now use of the fact that, for any spatial function,𝜙(𝐱, 𝑡), the
𝐷𝜙 𝐱,𝑡 𝜕𝜙 𝐱,𝑡 𝜕𝜙 𝐱,𝑡
material time derivative, = + 𝐯⋅ . Consequently,
𝐷𝑡 𝜕𝑡 𝜕𝐱
𝐷
 𝐼 𝑡 = Ω0 𝐷𝑡
Φ 𝐱, 𝑡 𝐽 𝑑𝑉 = Ω0
Φ 𝐱, 𝑡 𝐽 + 𝐽Φ 𝐱, 𝑡 𝑑𝑉
𝐽
= Φ 𝐱, 𝑡 + Φ 𝐱, 𝑡 𝐽𝑑𝑉
Ω0 𝐽
= Φ 𝐱, 𝑡 + div 𝐯 Φ 𝐱, 𝑡 𝑑𝑣
Ω
𝜕Φ 𝐱, 𝑡
= + 𝐯 ⋅ grad Φ 𝐱, 𝑡 + div 𝐯 Φ 𝐱, 𝑡 𝑑𝑣 =
Ω 𝜕𝑡
𝜕Φ 𝐱, 𝑡
= + div 𝐯Φ 𝑑𝑣
Ω 𝜕𝑡
which after applying the divergence theorem of Gauss, we find to be,
𝜕Φ 𝐱, 𝑡 𝜕Φ
𝐼 𝑡 ≡ + div 𝐯Φ 𝑑𝑣 = Φ𝐯 ⋅ 𝐧 𝑑𝑠 + 𝑑𝑣
Ω 𝜕𝑡 𝜕Ω Ω 𝜕𝑡
as required.


Conservation of Mass
Based upon the above theorem, we can express the balance
of mass compactly, considering the fact that,
ϱ 𝐱, 𝑡 𝑑𝑣 = ϱ0 𝐗 𝑑𝑉
Ω Ω0
The right hand of the above equation is independent of time
t. Hence a time derivative,
𝐷 𝐷
ϱ 𝐱, 𝑡 𝑑𝑣 = ϱ 𝐗 𝑑𝑉 = 0
𝐷𝑡 Ω 𝐷𝑡 Ω0
Invoking the Leibniz-Reynolds theorem, we conclude that,
𝜕ϱ 𝜕ϱ
+ div 𝐯ϱ 𝑑𝑣 = ϱ𝐯 ⋅ 𝐧 𝑑𝑠 + 𝑑𝑣 = 0
Ω 𝜕𝑡 𝜕Ω Ω 𝜕𝑡


Conservation of Mass
The mass generation inside the system plus the net mass
transport across the boundary sum up to zero.
𝜕𝜚
𝑑𝑣 + 𝜚𝐯 ⋅ 𝒏 𝑑𝑠 = 0
𝛺 𝜕𝑡 𝜕𝛺
Equivalently, in differential form, The time rate of change
of spatial density 𝜚 plus the divergence of mass flow rate
equals zero.
𝜕𝜚
+ div 𝐯𝜚 = 0
𝜕𝑡
Or equivalently,
𝐷𝜚
+ 𝜚 div 𝐯 = 0
𝐷𝑡

Implications for Scalar Fields
For a scalar field 𝜙(𝐱, 𝑡), using the continuity equation, we
can write,
𝜕 𝜚𝜙 𝜕𝜙 𝜕𝜚 𝜕𝜙
= 𝜚 + 𝜙 = 𝜚 − 𝜙div 𝜚𝐯
𝜕𝑡 𝜕𝑡 𝜕𝑡 𝜕𝑡
𝜕𝜙
= 𝜚 − div 𝐯𝜚𝜙 + 𝜚𝐯 ⋅ grad𝜙
𝜕𝑡
𝜕𝜙 𝜕𝜙
= 𝜚 − div 𝐯𝜚𝜙 + 𝜚𝐯 ⋅
𝜕𝑡 𝜕𝐱
𝐷𝜙
= 𝜚 − div 𝐯𝜚𝜙
𝐷𝑡
from which we can conclude that,
𝐷𝜙 𝜕 𝜚𝜙
𝜚 = + div 𝐯𝜚𝜙 .
𝐷𝑡 𝜕𝑡

Mass Measure
Furthermore, in spatial coordinates, using Leibniz
theorem,
𝐷 𝐷 𝜚𝜙
𝜚𝜙 𝑑𝑣 = + div 𝐯 𝜚𝜙 𝑑𝑣
𝐷𝑡 Ω Ω 𝐷𝑡
𝐷𝜙 𝐷𝜚
= 𝜚 + 𝜙 + 𝜚 div 𝐯) 𝑑𝑣
Ω 𝐷𝑡 𝐷𝑡
𝐷𝜙
= 𝜚 𝑑𝑣
Ω 𝐷𝑡
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].

Vector Fields

Such a process remains valid for any spatial vector or tensor
𝝍 𝐱, 𝑡 , the above expression remains valid, for,
𝐷 𝐷 𝜚𝝍 𝐱, 𝑡
𝜚𝝍 𝐱, 𝑡 𝑑𝑣 = + div 𝐯 𝜚𝝍 𝐱, 𝑡 𝑑𝑣
𝐷𝝍 𝐱, 𝑡 𝐷𝜚
= 𝜚 + 𝝍 𝐱, 𝑡 + 𝜚 div 𝐯 𝑑𝑣
Ω 𝐷𝑡 𝐷𝑡
𝐷𝝍 𝐱, 𝑡
= 𝜚 𝑑𝑣
Ω 𝐷𝑡
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 & Tensor Fields

For a vector field 𝝍 𝐱, 𝑡 , we can also write,
𝜕 𝜚𝝍 𝐱, 𝑡 𝜕𝝍 𝜕𝜚 𝜕𝝍
= 𝜚 + 𝝍 = 𝜚 − 𝝍 div 𝐯𝜚
𝜕𝑡 𝜕𝑡 𝜕𝑡 𝜕𝑡
𝜕𝝍 𝜕𝝍
= 𝜚 − div 𝜚𝐯 ⊗ 𝝍 + 𝜚𝐯 ⋅ [See Ex 3.1 28 ]
𝜕𝑡 𝜕𝐱
𝑑𝝍
= 𝜚 − div 𝜚𝐯 ⊗ 𝝍
𝑑𝑡
from which we can conclude that,
𝑑𝝍 𝜕 𝜚𝝍
𝜚 = + div 𝜚𝐯 ⊗ 𝝍
𝑑𝑡 𝜕𝑡


Vector & Tensor Fields

Furthermore for a tensor field, the arguments are
exactly the same and the result is in objects of one
degree higher:
𝐷𝚵 𝜕 𝜚𝚵
𝜚 = + div 𝜚𝐯 ⊗ 𝚵
𝐷𝑡 𝜕𝑡
In particular, for the velocity field, we can see clearly
that,
𝑑𝐯 𝜕 𝜚𝐯
𝜚 = + div 𝜚𝐯 ⊗ 𝐯
𝑑𝑡 𝜕𝑡

Mass Balance on Control Volume
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 𝜕
𝜚𝜙 𝑑𝑣 = 𝜚𝜙 𝑑𝑣
𝑑𝑡 𝑅 𝑅 𝜕𝑡
𝜕𝜙 𝜕𝜚 𝜕𝜙 𝜕𝜚
= 𝜚 + 𝜙 𝑑𝑣 = 𝜚 𝑑𝑣 + 𝜙 𝑑𝑣
𝑅 𝜕𝑡 𝜕𝑡 𝑅 𝜕𝑡 𝑅 𝜕𝑡
𝜕𝜙 𝜕𝜚
= 𝜚 𝑑𝑣 + 𝜙 𝑑𝑣
𝑅 𝜕𝑡 𝑅 𝜕𝑡
𝜕𝜙
= 𝜚 𝑑𝑣 − 𝜙𝜚𝐯 ⋅ 𝒏 𝑑𝑠
𝑅 𝜕𝑡 𝜕𝑅


Mass Balance on Control Volume
For a vector field 𝝍 𝐱, 𝑡 , applying the conservation of mass
as before, we can write,
𝑑 𝜕 𝜕𝝍 𝜕𝜚
𝜚𝝍 𝐱, 𝑡 𝑑𝑣 = 𝜚𝝍 𝑑𝑣 = 𝜚 + 𝝍 𝑑𝑣
𝑑𝑡 𝑅 𝑅 𝜕𝑡 𝑅 𝜕𝑡 𝜕𝑡
𝜕𝝍 𝜕𝜚
= 𝜚 𝑑𝑣 + 𝝍 𝑑𝑣
𝑅 𝜕𝑡 𝑅 𝜕𝑡
𝜕𝝍
= 𝜚 𝑑𝑣 − 𝝍𝜚𝐯 ⋅ 𝐧 𝑑𝑠
𝜕𝝍
= 𝜚 𝑑𝑣 − 𝜚 𝝍 ⊗ 𝐯 𝐧 𝑑𝑠
so that,
𝜕𝝍 𝑑
𝜚 𝑑𝑣 = 𝜚𝝍 𝐱, 𝑡 𝑑𝑣 + 𝜚 𝝍 ⊗ 𝐯 𝒏 𝑑𝑠
𝑅 𝜕𝑡 𝑑𝑡 𝑅 𝜕𝑅


Cauchy’s Laws of Motion

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 𝑑𝑣 = 𝜚0 𝐗 𝐕 𝐗, 𝑡 𝑑𝑉
Ω Ω0
where Ω is the spatial configuration volume and Ω0 the
reference configuration.


Linear Momentum
The balance of linear momentum, according to the
second law of Newton is that,
𝐷𝑷(𝑡)
= 𝑭(𝑡)
𝐷𝑡
where 𝑭(𝑡) is the resultant force on the system. Hence
by the conservation of linear momentum, we may write,


Cauchy’ Law
𝐷
𝑷 𝑡 = 𝜚 𝐱, 𝑡 𝐯 𝐱, 𝑡 𝑑𝑣
𝐷𝑡 Ω
𝐷
= 𝜚0 𝐗 𝐕 𝐗, 𝑡 𝑑𝑉 = 𝑭(𝑡)
𝐷𝑡 Ω0
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 + 𝐛 𝐱, 𝑡 𝑑𝑣
𝜕Ω Ω


Cauchy law of Motion

𝒏
By Cauchy’s stress law, 𝑻 = 𝝈 ⋅ 𝒏. Consequently, we
may write,
𝐷 𝐧
𝜚𝐯𝑑𝑣 = 𝐓 𝐱, 𝑡 𝑑𝑠 + 𝐛 𝐱, 𝑡 𝑑𝑣
𝐷𝑡 Ω 𝜕Ω Ω

= 𝝈𝐧𝑑𝑠 + 𝐛𝑑𝑣
𝜕Ω Ω

= (grad𝝈 + 𝒃)𝑑𝑣
Ω


Cauchy’s Law
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 𝐷
𝜚𝐯𝑑𝑣 = 𝜚𝐯 𝑑𝑣
𝑑𝐯 𝐱, 𝑡
= 𝜚 𝑑𝑣
Ω 𝑑𝑡
= (grad 𝝈 + 𝒃)𝑑𝑣
Ω
We can write the above equation in differential form as,
𝑑𝐯 𝐱, 𝑡
grad 𝝈 + 𝒃 = 𝜚
𝑑𝑡
when we remember the definition of the derivative with respect to
the position vector.


Formal Statement

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.
D𝐯
grad 𝝈 + 𝐛 = 𝜚
D𝑡


Cauchy Second Law of Motion

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
𝝈 𝑇 (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

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.


Work and Energy Balance

We are now in a position to compute such quantities as
work rate and energy. Beginning with Euler-Cauchy first
law of motion,
𝐷
div𝝈 + 𝐛 = 𝜚𝐯 = 𝜚𝐯
𝐷𝑡
if we assume density is constant. A scalar product of
this equation with velocity gives,
div𝝈 + 𝐛 ⋅ 𝐯 = 𝜚𝐯 ⋅ 𝐯


Integrating over the spatial volume,
div 𝝈 + 𝒃 ⋅ 𝐯𝑑𝑣 = div 𝝈 ⋅ 𝐯𝑑𝑣 + 𝒃 ⋅ 𝐯𝑑𝑣 =
Ω Ω Ω

𝜚𝐯 ⋅ 𝐯 𝑑𝑣 = div 𝝈 ⋅ 𝐯 𝑑𝑣 − 𝝈 ∶ 𝑳 𝑑𝑣 + 𝒃 ⋅ 𝐯𝑑𝑣
Ω Ω Ω Ω

= div 𝝈 ⋅ 𝐯 𝑑𝑣 − tr 𝝈 ⋅ 𝑳 𝑑𝑣 + 𝒃 ⋅ 𝐯𝑑𝑣
Ω Ω Ω
where 𝑳 is the velocity gradient and it is obvious that
𝜎 𝑖𝑗 𝑣 𝑗 , 𝑖 = 𝜎 𝑖𝑗 , 𝑖 𝑣 𝑗 + 𝜎 𝑖𝑗 𝑣 𝑖 , 𝑗
or
div 𝝈 ⋅ 𝐯 = div 𝝈 ⋅ 𝐯 + 𝝈 ∶ 𝑳


Applying the divergence theorem, we have
𝜚𝐯 ⋅ 𝐯 𝑑𝑣 + tr 𝝈𝑳 𝑑𝑣 = 𝝈𝐯 ⋅ 𝐧 𝑑𝑠 + 𝐛 ⋅ 𝐯𝑑𝑣
Ω Ω 𝜕Ω Ω
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 𝑑
𝜚𝐯 ⋅ 𝐯 𝑑𝑣 + 𝝈 ∶ 𝑫 𝑑𝑣 = 𝝈𝐯 ⋅ 𝐧 𝑑𝑠 + 𝒃 ⋅ 𝐯𝑑𝑣
2 𝑑𝑡 Ω Ω 𝜕Ω Ω


Power Balance

The quantity, 𝜌𝐯 ⋅ 𝐯 is the kinetic energy density of the
1 𝑑
body, and 𝜌𝐯 ⋅ 𝐯 𝑑𝑣 is the rate of change of the
2 𝑑𝑡 Ω
kinetic energy. Ω 𝝈 ∶ 𝑫 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.


Conjugate Pairs
The stress power expression can be written in terms of
other conjugate pairs as follows:
𝝈 ∶ 𝑫 𝑑𝑣 = 𝝉 ∶ 𝑫 𝑑𝑉 = 𝒔 ∶ 𝑭 𝑑𝑉 = 𝚵 ∶ 𝑬 𝑑𝑉
Ω Ω0 Ω0 Ω0
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.

Stress Power Law

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.


Kirchhoff Stress

Kirchhoff Stress. It is straightforward to show that,
𝑑𝑣
𝝈 ∶ 𝑫 𝑑𝑣 = 𝝈∶ 𝑫 𝑑𝑉
Ω Ω0 𝑑𝑉
= 𝐽𝝈 ∶ 𝑫 𝑑𝑉 = 𝝉 ∶ 𝑫 𝑑𝑉
Ω0 Ω0


First Piola-Kichhoff Tensor
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 𝒔𝑭 𝑻 𝑳 𝑻 = 𝐽 tr 𝝈𝑳
1
It therefore follows that 𝝈 ∶ 𝑫 = 𝝈 ∶ 𝑳 = 𝒔 ∶ 𝑭 in the
𝐽
reference configuration. Using this in the stress power
integral provides the necessary Jacobian in the volume ratio 𝐽
so that,
𝑭
𝝈 ∶ 𝑫 𝑑𝑣 = 𝒔 ∶ 𝑑𝑣 = 𝒔 ∶ 𝑭 𝑑𝑉
Ω Ω 𝐽 Ω0


Second Piola-Kirchhoff Tensor
1
The Lagrange strain tensor is given by 𝐄 = 𝐅T 𝐅 − 𝟏
2
1
𝐄= 𝐅T 𝐅 + 𝐅T 𝐅
2
1 𝐓 𝐓 1
= 𝐅 𝐋 𝐅 + 𝐅 T 𝐋𝐅 = 𝐅 𝐓 𝐋 𝐓 + 𝐋 𝐅
2 2
𝐓
= 𝐅 𝐃𝐅
𝝈 ∶ 𝑫 𝑑𝑣 = 𝝈 ∶ 𝐅 −𝐓 𝐄𝐅 −𝟏 𝑑𝑣
Ω Ω

= tr 𝐅 −𝟏 𝝈𝐅 −𝐓 𝐄 𝑑𝑣 = 𝐽𝐅 −𝟏 𝝈𝐅 −𝐓 : 𝐄 𝑑𝑉
Ω Ωo

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


Thermodynamical Balances

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.


First Law of Thermodynamics

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
𝐸 𝑡 + 𝐾 𝑡 = 𝑄 𝑡 + 𝐿(𝑡)
𝑑𝑡


First Law

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 Ω 𝜚𝑟𝑑𝑣 .
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.


Energy Balance

With these, we can write the energy balance equation as,
𝑑 1
𝜚 𝜀+ 𝐯 2 𝑑𝑣 =
𝑑𝑡 Ω 2
= 𝝈𝐯 ⋅ 𝐧 𝑑𝑠 + 𝒃 ⋅ 𝐯𝑑𝑣 − 𝒒. 𝒏 𝑑𝑠 + 𝜚𝑟𝑑𝑣
𝜕Ω Ω 𝜕Ω Ω
1 𝑑
= 𝜌𝐯 ⋅ 𝐯 𝑑𝑣 + 𝝈 ∶ 𝑫 𝑑𝑣 − 𝒒. 𝒏 𝑑𝑠 + 𝜚𝑟𝑑𝑣
1 𝑑
since 𝐿 𝑡 = 𝜌𝐯 ⋅ 𝐯 𝑑𝑣 + 𝝈 ∶ 𝑫 𝑑𝑣 =
2 𝑑𝑡 Ω Ω
1 𝑑
𝜕Ω
𝝈 ⋅ 𝐯 ⋅ 𝐧 𝑑𝑠 + Ω
𝒃 ⋅ 𝐯𝑑𝑣 and 𝐾 𝑡 = 𝜌𝐯 ⋅ 𝐯 𝑑𝑣
2 𝑑𝑡 Ω


Energy Balance
We can write the above in terms of the specific internal energy as,
𝑑
𝜚𝜀𝑑𝑣 = − 𝒒. 𝒏 𝑑𝑠 + 𝜚𝑟𝑑𝑣 + 𝝈 ∶ 𝑫 𝑑𝑣
𝑑𝑡 Ω 𝜕Ω Ω Ω
and as a result of the continuity of mass, the constancy of the mass
measure ( Ω 𝜚𝑑𝑣 )allows us to bring the derivative under the
integral sign and apply it only to the internal energy so that,
𝜚𝜀 𝑑𝑣 = − grad𝒒 𝑑𝑣 + 𝑟𝜚𝑑𝑣 + 𝝈 ∶ 𝑫 𝑑𝑣
Ω Ω Ω Ω
after application of the divergence theorem. This is the same as,
𝜚𝜀 + grad𝒒 − 𝜚𝑟 − 𝝈 ∶ 𝑫 𝑑𝑣 = 0
Ω
which gives a local energy balance,
𝜚𝜀 + grad𝒒 − 𝜚𝑟 − 𝝈 ∶ 𝑫 = 0.


Spatial Energy Balance
Recall that for any scalar field 𝜙, as a consequence of
𝑑𝜙 𝜕 𝜚𝜙
the balance of mass, 𝜚 = + 𝛻 𝑦 𝐯𝜚𝜙 .
𝑑𝑡 𝜕𝑡
Applying this to specific internal energy,
𝑑𝜀 𝜕 𝜚𝜀
𝜚 = + grad 𝐯𝜚𝜀
𝑑𝑡 𝜕𝑡
Using this, the local energy equation now becomes,
𝜕 𝜚𝜀
= 𝜚𝑟 + 𝝈 ∶ 𝑫 − grad 𝒒 + 𝐯𝜚𝜀
𝜕𝑡
where the derivative with respect to time here is the
spatial time derivative and heat flux term now has the
convective addition 𝐯𝜚𝜀 which is a result of heat
transfer due to motion.

Material 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
𝑑𝑣
𝜚𝑟𝑑𝑣 = 𝜚𝑟 𝑑𝑉 = 𝐽𝜚𝑟𝑑𝑉
Ω Ω0 𝑑𝑉 Ω0

= 𝜚0 𝑟𝑑𝑉
Ω0
And for the heat flux through the boundary, a Piola Transformation yields
− 𝒒. 𝒏 𝑑𝑠 = − 𝐽𝑭−1 𝒒 ⋅ 𝑑𝑨 = − 𝒒0 ⋅ 𝒏𝑑𝐴 = − Div𝒒0 𝑑𝑉
𝜕Ω 𝜕Ω0 𝜕Ω0 Ω0
Hence, the energy balance in terms of referential (material) frame, becomes,
𝜚0 𝜀 𝑑𝑉 = − 𝛻𝒒0 𝑑𝑉 + 𝑟𝜚0 𝑑𝑉 + 𝒔 ∶ 𝑭 𝑑𝑉
Ω0 Ω0 Ω0 Ω0
from which we can now obtain the local material energy balance,
𝜚0 𝜀 = 𝑟𝜚0 − 𝛻 ⋅ 𝒒0 + 𝒔 ∶ 𝑭


Second Law
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.


Importance of the Second Law

"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):


Entropy . . .
𝜂 is defined as the measure of entropy per unit mass so that total
entropy
𝑆 𝑡 = 𝜚𝜂𝑑𝑣
Ω
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 𝒒 𝜚𝒓
𝜚𝜂𝑑𝑣 ≥ − ⋅ 𝒏 𝑑𝑠 + 𝑑𝑣
𝑑𝑡 Ω 𝜕Ω 𝑇 Ω 𝑇
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 𝒒 𝜚𝑟
𝜚𝜂𝑑𝑣 ≥ − div 𝑑𝑣 + 𝑑𝑣
We apply the law of conservation of mass, observe the constancy of the mass
measure in the flow, so that
𝒒 𝑟
𝜚𝜂 + div − 𝜚 𝑑𝑣 ≥ 0
Ω 𝑇 𝑇
or
𝒒 𝜚𝑟
𝜚𝜂 ≥ −div + .
𝑇 𝑇
Applying the law of continuity to the material derivative of specific entropy, we
have,
𝑑𝜂 𝜕 𝜚𝜂
𝜚 = + div 𝐯𝜚𝜂
𝑑𝑡 𝜕𝑡
which we now substitute to obtain,
𝜕 𝒒 + 𝐯𝑇𝜚𝜂 𝜚𝑟
𝜚𝜂 ≥ −div +
𝜕𝑡 𝑇 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,
𝒒 1 𝒒
−div = − div 𝒒 + 2 ⋅ 𝛻 𝑦 𝑇
𝑇 𝑇 𝑇
Using this, we can write,
𝒒 𝜚𝑟 1 𝒒 𝜚𝑟
𝜚𝜂 ≥ −div + = − div 𝒒 + 2 ⋅ grad 𝑇 +
𝑇 𝑇 𝑇 𝑇 𝑇
1 𝒒
= −div 𝒒 + 𝜚𝑟 + ⋅ grad 𝑇
𝑇 𝑇
1 𝒒
= 𝜚𝜀 − 𝝈 ∶ 𝑫 + ⋅ grad 𝑇
𝑇 𝑇
using the local spatial expression of the first law. Bringing everything
to the RHS, we have,
𝒒
𝜚 𝜀 − 𝑇𝜂 − 𝝈 ∶ 𝑫 + ⋅ grad 𝑇 ≥ 0.
𝑇


Free Energy
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 = 𝜀 − 𝑇𝜂. Consequently, then we can write,
𝒒
𝜚 𝜓 + 𝑇 𝜂 − 𝝈 ∶ 𝑫 + ⋅ 𝛻 𝑦 𝑇 = −𝑇Γ ≥ 0
𝑇


Free Energy

Recalling the conventional energy expression that
1 𝑑
𝜚𝐯 ⋅ 𝐯 𝑑𝑣 + 𝝈 ∶ 𝑫 𝑑𝑣 = 𝝈𝐧 ⋅ 𝐯 𝑑𝑠 + 𝒃 ⋅ 𝐯𝑑𝑣
we can integrate the above inequality and obtain,
𝑑 1
𝑇Γ 𝑑𝑣 = 𝝈𝐧 ⋅ 𝐯 𝑑𝑠 + 𝒃 ⋅ 𝐯𝑑𝑣 − 𝜚 𝜓+ 𝐯 𝟐 𝑑𝑣
Ω 𝜕Ω Ω 𝑑𝑡 Ω 2
𝒒
− 𝜚𝑇 𝜂 + ⋅ grad𝑇 𝑑𝑣 ≥ 0
Ω 𝑇
That is,
𝐸𝑛𝑒𝑟𝑔𝑦 𝐷𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑜𝑛
= 𝐶𝑜𝑛𝑣𝑒𝑛𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 – 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐹𝑟𝑒𝑒 𝑎𝑛𝑑 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑖𝑒𝑠
– 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸𝑛𝑒𝑟𝑔𝑦

Energy Dissipation
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 𝜀0 , 𝜂0 such that 𝜀0 = 0, 𝜂0 = 0,
𝑑 𝑑
𝜚𝜂𝑑𝑣 = 𝜚 𝜂 + 𝜂0 𝑑𝑣
𝑑𝑡 Ω 𝑑𝑡 Ω
and
𝑑 𝑑
𝜚𝜀𝑑𝑣 = 𝜚 𝜀 + 𝜀0 𝑑𝑣 .
𝑑𝑡 Ω 𝑑𝑡 Ω
This implies invariance with respect to translations
𝜀 → 𝜀 + 𝜀0 , 𝜂 → 𝜂 + 𝜂0
by these values. It can also be shown (Ex. 22) that for a
vector 𝝎 a the heat flux Ω 𝒒 ⋅ 𝒏 𝑑𝑠 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.


Entropy Rate, Isolated Body
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 1
𝜚 𝜀+ 𝐯 2 𝑑𝑣 = 𝒃 ⋅ 𝐯𝑑𝑣 + 𝜚𝑟𝑑𝑣
𝑑𝑡 Ω 2 Ω Ω
If in addition to these, there are no body forces and the radiative or other bulk
heat generation vanishes, then,
𝑑 1
𝜚 𝜀+ 𝐯 2 𝑑𝑣 = 0
𝑑𝑡 Ω 2
And the second law,
𝑑 𝒒 𝑟
𝜚𝜂𝑑𝑣 ≥ − div 𝑑𝑣 + 𝑑𝑣
becomes,
𝑑
𝜚𝜂𝑑𝑣 ≥ 0
𝑑𝑡 Ω
which shows that the rate of change of entropy in such a situation is always
positive.


Second Law of Thermodynamics in
Material Frame
Similar to the internal energy and internal heat source, we
can express the net entropy in terms of material coordinates,
𝑑𝑣
𝜚𝜂𝑑𝑣 = 𝜚𝜂 𝑑𝑉 = 𝐽𝜚𝜂𝑑𝑉
Ω Ω0 𝑑𝑉 Ω0

= 𝜚0 𝜂𝑑𝑉
Ω0
And proceed to write,
𝑑 𝒒0 𝑟
𝜚0 𝜂𝑑𝑉 ≥ − Div 𝑑𝑉 + 𝜚0 𝑑𝑣
𝑑𝑡 Ω0 𝜕Ω0 𝑇 Ω0 𝑇
where, as before, 𝒒0 ≡ 𝐽𝑭−1 𝒒 the Piola transformation of
the heat flux.


Examples
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 −1 𝐮. Using the Piola identity
that 𝛻 ⋅ 𝐽𝐅 −1 = 0, Show that 𝛻 ⋅ 𝐔 = 𝐽𝛻 𝑦 ⋅ 𝐮

4. The components of Cauchy stress in Cartesian coordinates are
2
𝑥1 𝑥2 𝑥1 −𝑥2
2
𝑥1 0 0
2 2
−𝑥2 0 𝑥1 + 𝑥2
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 0
0 𝑥2 + 𝛼𝑥3 Φ 𝑥 2 , 𝑥 3 (a) Find Φ 𝑥 2 , 𝑥 3 so that the equilibrium
0 Φ 𝑥 2, 𝑥 3 𝑥2 + 𝛽𝑥3
equations are satisfied assuming body forces are zero.
(b) Use the value of Φ 𝑥 2 , 𝑥 3 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 −1 𝐮. Use
the Piola identity to show that 𝛻 ⋅ 𝐔 = 𝐽𝛻 𝑦 ⋅ 𝐮
7. Given the Cauchy stress components,
𝟐
𝒚𝟏 𝒚𝟐 𝒚𝟏 −𝒚 𝟐
𝟐
𝒚𝟏 𝟎 𝟎
𝟐 𝟐
−𝒚 𝟐 𝟎 𝒚𝟏 + 𝒚𝟐
Find the body forced that keeps the body in equilibrium.

8. Given the Cauchy stress components,
𝜶 𝟎 𝟎
𝟎 𝒚 𝟐 + 𝜶𝒚 𝟑 𝚽(𝒚 𝟐 , 𝒚 𝟑 )
𝟎 𝚽(𝒚 𝟐 , 𝒚 𝟑 ) 𝒚 𝟐 + 𝜷𝒚 𝟑
(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 = 𝒚 𝟏 + 𝒚 𝟐 + 𝒚 𝟑 .
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 𝑤 𝑖𝑛𝑡 = 𝐽𝛔: 𝐝 = −𝐽𝑝𝐈: 𝐝 =
𝐽𝑝 𝑑𝜌
− 𝐽𝑝 𝑇𝑟 𝒅 = −𝐽𝑝𝛁 ⋅ 𝐯 = 𝜌 𝑑𝑡
11. A rigid body is rotating about a fixed point 𝑂 with angular velocity 𝝎 show
1
that the kinetic energy may be expressed as 2 𝛚 ⋅ 𝐃𝛚, where
𝐷 = Ω ϱ 𝐈 𝐫 ⋅ 𝐫 − 𝐫 ⊗ 𝐫 𝑑𝑣, 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 𝑪 𝟏 Prove that
𝝏𝑽
𝒖 ⊗ 𝝈 ⋅ 𝒏𝒅𝑆 = 𝑽 𝒖 ⊗ 𝛻 ⋅ 𝝈 + 𝛻 ⊗ 𝒖 ⋅ 𝝈 𝒅𝑉

13. Let 𝑉 be a region of 𝐸3 bounded by 𝜕𝑉 and outward normal 𝒏. Let 𝒖 be a
vector field of class 𝑪 𝟏 . Show that 𝑽 𝛻 ⊗ 𝒖 𝑑𝑉 = 𝝏𝑽 𝒏 ⊗ 𝒖 𝒅𝑆
14. By evaluating the divergence operation on a tensor show that the
equilibrium equations can be expressed in Cylindrical Polar coordinates as,
𝜕𝜎 𝑟𝑟 1 𝜕𝜎 𝜃𝑟 𝜎 𝑟𝑟 −𝜎 𝜃𝜃 𝜕𝜎
+ + 𝜕𝑧𝑧𝑟 + 𝑏 𝑟 = 0
+
𝜕𝑟 𝑟 𝜕𝜃 𝑟
𝜕𝜎 𝑟𝜃
2𝜎 𝑟𝜃 1 𝜕𝜎 𝜃𝜃 𝜕𝜎 𝑧𝜃
+ + + + 𝑏𝜃 = 0
𝜕𝑟 𝑟 𝑟 𝜕𝜃 𝜕𝑧
𝜕𝜎 𝑟𝑧 𝜎 𝑟𝑧 1 𝜕𝜎 𝜃𝑧 𝜕𝜎 𝑧𝑧
+ + + + 𝑏𝑧 = 0
𝜕𝑟 𝑟 𝑟 𝜕𝜃 𝜕𝑧
15. By evaluating the divergence operation on a tensor show that the
equilibrium equations can be expressed in Spherical Polar coordinates as,
1 𝜕𝜎 𝜌𝜃 1 𝜕𝜎 𝜌𝜙 𝜕𝜎 𝜌𝜌
𝜎 cot 𝜃 + + + 𝜌 − 𝜎 𝜃𝜃 + 2𝜎 𝜌𝜌 − 𝜎 𝜙𝜙 + 𝑏 𝜌 = 0
𝜌 𝜌𝜃 𝜕𝜃 sin 𝜃 𝜕𝜙 𝜕𝜌
1 𝜕𝜎 𝜌𝜃 1 𝜕𝜎 𝜃𝜙 𝜕𝜎 𝜃𝜃
𝜎 𝜃𝜃 − 𝜎 𝜙𝜙 cot 𝜃 + 𝜌 + 3𝜎 𝜌𝜃 + + +𝑏 𝜃 = 0
𝜌 𝜕𝜌 sin 𝜃 𝜕𝜙 𝜕𝜃
1 1 𝜕𝜎 𝜙𝜙 𝜕𝜎 𝜌𝜙 𝜕𝜎 𝜃𝜙
+ 𝜌 + 3𝜎 𝜌𝜙 + 2𝜎 𝜃𝜙 cot 𝜃 + + 𝑏𝜙 =0
𝜌 sin 𝜃 𝜕𝜙 𝜕𝜌 𝜕𝜃


16. The law of conservation of mass is expressed in the vector form, 𝛻 ⋅ 𝑽𝜚 +
𝜕𝜚
= 0. Express this law in tensor components and find the equivalent in
𝜕𝑡
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 2 𝑰 − 𝒓 ⊗ 𝒓 𝑑𝑣 = 𝐼 𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 𝑑𝑣
Ω Ω
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 𝐼 𝑖𝑗 = 𝜚 𝑥 𝛼 𝑥 𝛽 𝑔 𝛼𝛽 𝑔 𝑖𝑗 − 𝑥 𝑖 𝑥 𝑗 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, 𝜚0 𝜀 = −𝛻 ⋅ 𝒒0 + 𝑟𝜚0 + 𝒔 ∶ 𝑭
𝒒 1
20. For a vector field 𝒒 = 𝑞 𝑖 𝐠 𝑖 and a scalar field 𝑇, show that 𝑑𝑖𝑣 = 𝑑𝑖𝑣 𝒒 −
𝑇 𝑇
𝒒
𝑇2
⋅ 𝑔𝑟𝑎𝑑 𝑇


21. For an arbitrary vector 𝝎 and scalar 𝑇, such that 𝑔𝑟𝑎𝑑𝝎 = 0 show that the
integral Ω 𝒒 ⋅ 𝒏 𝑑𝑠 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
𝒒
integral Ω 𝑇 ⋅ 𝒏 𝑑𝑠 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
𝑑 1 2
the first law of thermodynamics becomes, 𝜚 𝜀+2 𝐯 𝑑𝑣 =
𝑑𝑡 Ω
𝑑 𝒒
𝜕Ω
𝝈 ⋅ 𝐯 ⋅ 𝐧 𝑑𝑠 − 𝜕Ω 𝒒. 𝒏 𝑑𝑠, and the second law 𝑑𝑡 Ω 𝜚𝜂𝑑𝑣 ≥ − 𝜕Ω 𝑇 ⋅ 𝒏 𝑑𝑠
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
𝑑 1 2 1 2
becomes, 𝑑𝑡 R
𝜚 𝜀+2 𝐯 𝑑𝑣 + 𝜕R
𝜚 𝜀+2 𝐯 𝐯 ⋅ 𝒏 𝑑𝑠 =
𝜕R
𝝈 ⋅ 𝐯 ⋅ 𝐧 𝑑𝑠 + R
𝒃 ⋅ 𝐯𝑑𝑣 − 𝜕R
𝒒. 𝒏 𝑑𝑠 + R
𝜚𝑟𝑑𝑣 and the second law,
𝑑 𝒒 𝒓
becomes, 𝜚𝜂𝑑𝑣 + 𝜚𝜂𝐯 ⋅ 𝒏𝑑𝑠 ≥ − ⋅ 𝒏 𝑑𝑠 + 𝑑𝑣
𝑑𝑡 R 𝜕R 𝜕R 𝑇 R 𝑇
26. Obtain the local dissipation inequality, 𝜓 − 𝝈 ∶ 𝑫 ≡ −𝛿 ≥ 0


6. balance laws jan 2013

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