User:Eml5526.s11.team7.flater/EGM6611.HW3

=Homework 3=

Description
Derive the conservation of mass equation.

Solution

 * Adapted from Malvern, Section 5.2.

Consider an arbitrary volume V bounded by the surface S. If a continuous medium of density $$\rho$$ fills the volume at time t, the total mass in V is


 * {| style="width:100%" border="0"

$$ \displaystyle M=\int_V \rho dV $$. (1.1)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

Since density is a function of position and time, the rate of increase of the total mass in the volume is the partial derivative


 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial M}{\partial t}=\int_V \frac{\partial \rho}{\partial t}dV $$.     (1.2)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

From the conservation of mass we assume that the rate of increase in V is equal to the rate of inflow of mass through the surface S. The mass flux inflow per unit area (dS) is defined as


 * {| style="width:100%" border="0"

$$\displaystyle \text{Flux} = -\int_S \rho \mathbf{v}\cdot \mathbf{\hat{n}}dS $$.     (1.3)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

Recall that the divergence theorem states that the integral of the outer normal component of a vector over a closed surface is equal to the integral of the divergence (designated $$\displaystyle \nabla $$ or "nabla") of the vector over the volume bounded by the closed surface.


 * {| style="width:100%" border="0"

$$\displaystyle \int_S \mathbf{v}\cdot \mathbf{\hat{n}}dS = \int_V \mathbf{\nabla} \cdot \mathbf{v}dV $$.     (1.4)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

Therefore equations 1.3 and 1.4 can be combined and equated to 1.2.


 * {| style="width:100%" border="0"

$$\displaystyle \int_V \left[ \frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot (\rho \mathbf{v}) \right]dV=0 $$.     (1.5)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

For any arbitrary volume V the integral becomes


 * $$\displaystyle\frac{\partial \rho}{\partial t} (x,t) + \nabla (\rho \mathbf{v}) =0$$


 * or


 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial \rho}{\partial t} + \rho \nabla \mathbf{v} =0 $$     (1.6)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

Thus we are left with the Eulerian description of the mass conservation.

Description
Derive the balance of linear momentum equations.

Solution

 * Adapted from Malvern, Section 5.3.

Let us consider the same control volume from problem 1, but in this case the mass is acted upon by external surface and body forces - t being applied per unit area ($$\displaystyle dS$$) and b applied per unit mass ($$\displaystyle \rho dV$$). The momentum principle as defined above can then be written as


 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dt}\int_V \rho \mathbf{v}dV = \int_S \mathbf{t}dS + \int_V \rho \mathbf{b}dV $$     (2.1)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }


 * where the left-hand side of the equation represents the rate of change of the total momentum. In rectangular coordinates, equation 2.1 takes the form


 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dt}\int_V \rho \ v_i \ dV = \int_S t_i \ dS + \int_V \rho \ b_i \ dV $$ (2.2)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

First, consider the definition of a traction vector in indicial notation: $$\displaystyle t_i = T_{ji}n_j$$. Second, recall the divergence theorem from equation 1.4. Now equation 2.2 can be expressed as


 * {| style="width:100%" border="0"

$$\displaystyle \int_V \rho \frac{dv_i}{dt} dV = \int_V \left( \frac{\partial T_{ji}}{\partial x_j} + \rho b_i \right) dV \qquad \Rightarrow \qquad \int_V \left( \frac{\partial T_{ji}}{\partial x_j} + \rho b_i - \rho \frac{dv_i}{dt}\right) dV =0$$. (2.3)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

As in problem 1, for any arbitrary volume V the integral becomes


 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial T_{ji}}{\partial x_j} + \rho b_i = \rho \frac{dv_i}{dt} $$     (2.4)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * 
 * }


 * or


 * {| style="width:100%" border="0"

$$\displaystyle \mathbf{\nabla \cdot T} + \rho \mathbf{b} = \rho \frac{d\mathbf{v}}{dt} $$     (2.5)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

Description
Derive the balance of moment of momentum equations.

Solution

 * Adapted from Malvern, Section 5.3.

Similar to the balance of linear momentum, the moment of momentum principle states that the time rate of change of the total moment of momentum is equal to the vector sum of the moments of the external forces acting on the system. In other words


 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dt}\int_V (\mathbf{r} \times \rho \mathbf{v}) dV = \int_S (\mathbf{r} \times \mathbf{t}) dS + \int_V (\mathbf{r} \times \rho \mathbf{b}) dV $$. (3.1)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

The cross product of two vectors in a right-handed coordinate system is defined as $$\mathbf{c} = \mathbf{a} \times \mathbf{b} \quad \rightarrow \quad c_p = e_{pqr}a_q b_r = e_{qrp} a_q b_r $$. So 3.1 can be rewritten in indicial notation as


 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dt}\int_V (e_{rmn} x_m \rho v_n) dV = \int_S (e_{rmn} x_m t_n) dS + \int_V (e_{rmn} x_m \rho b_n) dV $$. (3.2)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

Following the same procedure in problem 2 for equations 2.2 and 2.3, the surface integral can be transformed to a volume integral.


 * {| style="width:100%" border="0"

$$\displaystyle \int_V e_{rmn} \frac{d}{dt} (x_m v_n) \rho \ dV = \int_V e_{rmn} \left[ \frac{\partial (x_m T_{jn})}{\partial x_j} + x_m \rho b_n \right] dV $$ (3.3)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

Since $$\displaystyle \frac{dx_m}{dt}=v_m$$, equation 3.3 becomes


 * {| style="width:100%" border="0"

$$\displaystyle \int_V e_{rmn} \left( v_m v_n + x_m \frac{dv_n}{dt} \right) \rho \ dV = \int_V e_{rmn} \left[ x_m \left( \frac{\partial T_{jn}}{\partial x_j} + \rho b_n \right) + \delta_{mj}T_{jn}\right] dV $$. (3.4)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

Note the following:
 * Since $$\displaystyle v_m v_n$$ is symmetric and $$\displaystyle e_{rmn}$$ is antisymmstric, $$\displaystyle e_{rmn} v_m v_n=0$$.
 * From the conservation of linear momentum (equation 2.4 above) the last term of the left-hand side cancels with the first term of the right-hand side.
 * $$\displaystyle \delta_{mj} T_{jn}=T_{mn}$$.

Therefore the only remaining terms of equation 3.4 are


 * {| style="width:100%" border="0"

$$\displaystyle \int_V e_{rmn} T_{mn} dV=0 $$.     (3.5)
 * style="width:10%; |
 * style="width:10%; |
 * 
 * }

For any arbitrary volume V the integral becomes


 * {| style="width:100%" border="0"

$$\displaystyle e_{rmn} T_{mn}=0 $$     (3.6)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }


 * proving that the stress matrix, $$\displaystyle T_{mn}$$, is symmetric.


 * {| style="width:100%" border="0"

$$\displaystyle T_{mn}=T_{nm} $$     (3.7)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Description
Derive the conservation of energy equation.

Solution

 * Adapted from Malvern, Section 5.4.

The conservation of energy equations in continuum mechanics is derived from the first law of thermodynamics. In other words the total energy of the system ,E, is equal to the rate of heat, Q, and mechanical work, W, input to the system. The total energy is considered as sum of two parts, kinetic energy, K, and internal energy, U.


 * {| style="width:100%" border="0"

$$\displaystyle K + U = W + Q $$ (4.1)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * where,


 * {| style="width:100%" border="0"

$$\displaystyle K = \frac{d}{dt} \int_V \left( \frac{1}{2} \rho v_k v_k \right) dV $$, (4.2)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\displaystyle U = \frac{d}{dt} \int_V \rho u \ dV $$, (4.3)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\displaystyle W = \int_S \mathbf{t} \cdot \mathbf{v} \ dS + \int_V \rho \mathbf{f} \cdot \mathbf{v} \ dV $$, (4.4)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\displaystyle Q = - \int_S \mathbf{q} \cdot \mathbf{\hat{n}} \ dS + \int_V \rho h \ dV $$. (4.5)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

The heat input rate Q consists of conduction through the surface S, and a distributed internal heat source of strength h per unit mass. q is the heat flux vector. The power input (i.e. rate of work) is the rate at which the external surface tractions t per unit area and body forces b per unit mass are doing work on the system. Considering equation 4.4 in indicial notation and recalling the stress vector/tensor relationship (see equation 2.3), the surface integral is transformed to a volume integral. Equation 4.4 becomes


 * {| style="width:100%" border="0"

$$\displaystyle W = \int_V \left[ v_i \left( \frac{\partial T_{ji}}{\partial x_j} + \rho b_i \right ) + T_{ji} \frac{\partial v_i}{\partial x_j}\right ]dV $$     (4.6)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * and can be further reduced recognizing that the term inside the parentheses is equal to the balance of linear momentum equation (2.4).


 * {| style="width:100%" border="0"

$$\displaystyle W = \int_V \left[ v_i \left( \rho \frac{dv_i}{dt} \right ) + T_{ji} \ v_{i,j}\right ]dV \qquad = \qquad \frac{d}{dt} \int_V  \left( \frac{1}{2} \rho v_i v_i \right)dV + \int_V T_{ij}D_{ij}\ dV $$ (4.7)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * where $$\displaystyle D_{ij}$$ is the rate of deformation tensor.

Now equations 4.2, 4.3, 4.5, and 4.7 can be substituted into 4.1.


 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dt} \int_V \left( \frac{1}{2} \rho v_k v_k \right) dV + \frac{d}{dt} \int_V \rho u \ dV = \frac{d}{dt} \int_V \left( \frac{1}{2} \rho v_i v_i \right)dV + \int_V T_{ij}D_{ij}\ dV - \int_S \mathbf{q} \cdot \mathbf{\hat{n}} \ dS + \int_V \rho h \ dV $$ (4.8)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

This reduces to


 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dt} \int_V \rho u \ dV = \int_V T_{ij}D_{ij}\ dV - \int_S \mathbf{q} \cdot \mathbf{\hat{n}} \ dS + \int_V \rho h \ dV $$ (4.9)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

From the divergence theorem (equation 1.4) the surface integral can be expressed as a volume integral. The remaining volume integrals can be expressed for an arbitrary volume V within the continuous medium. Then the conservation of energy equation takes the form below.


 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = T_{ij}D_{ij} - \frac{\partial q_j}{\partial x_j}  +  \rho h $$ (4.10)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Description
When the displacements and displacement gradients are small, the second Piola-Kirchoff stress tensor $$\displaystyle \mathbf{\tilde{T}}$$ is approximately equal to the Cauchy stress tensor $$\displaystyle \mathbf{T}$$. Thus for small vibrations of an elastic system the balance of linear momentum equations, in terms of $$\displaystyle \mathbf{\tilde{T}}$$, reduces to


 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial T_{ji}}{\partial X_j} + \rho_0 b_{0i} = \rho_0 \frac{\partial^2 u_i}{\partial t^2} $$     (5.1)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Consider Hooke's Law $$\displaystyle \left( T_{ij} = \lambda e \delta_{ij} + 2G \epsilon_{ij} \right)$$ where $$\displaystyle e=\epsilon_{kk}$$ and $$\displaystyle \lambda$$ and $$\displaystyle G $$ are the Lame elastic moduli.


 * Show that the first equation of motion then leads to the following displacement equation of motion: $$\displaystyle (\lambda + G) \frac{\partial e}{\partial X} + G \nabla^2 u_x + \rho_0 b_{0x}=\rho_0 \frac{\partial^2 u_x}{\partial t^2}$$.
 * Write the vector equation of which is the first component. ← (I'm not sure what this question is asking.)

Solution
First substitute Hooke's Law for the stress tensor into equation 5.1 to get


 * {| style="width:100%" border="0"

$$\displaystyle \lambda \frac{\partial}{\partial X_j} \left( e \delta_{ij} \right ) + 2G \frac{\partial \epsilon_{ij}}{\partial X_j} + \rho_0 b_{0i}=\rho_0 \frac{\partial^2 u_i}{\partial t^2} $$.     (5.2)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

The strain tensor is defined as


 * {| style="width:100%" border="0"

$$\displaystyle \epsilon_{ij} = \frac{1}{2} \left( \frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i} \right) $$.     (5.3)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Substituting this into equation 5.2 yields


 * {| style="width:100%" border="0"

$$\displaystyle \lambda \frac{\partial}{\partial X_j} \left( e \delta_{ij} \right ) + G \frac{\partial }{\partial X_j} \left(\frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i} \right ) + \rho_0 b_{0i}=\rho_0 \frac{\partial^2 u_i}{\partial t^2} $$.     (5.4)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

The Laplacian of a vector $$\displaystyle u_x$$ is defined as


 * {| style="width:100%" border="0"

$$\displaystyle \nabla^2 u_x = \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} +\frac{\partial^2 u_x}{\partial z^2} $$.     (5.5)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Substitution of this into 5.4 yields


 * {| style="width:100%" border="0"

$$\displaystyle \lambda \frac{\partial e}{\partial X} + G \left( \nabla^2 u_x + \frac{\partial e}{\partial X}\right ) + \rho_0 b_{0x}=\rho_0 \frac{\partial^2 u_x}{\partial t^2} $$.     (5.6)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Simplifying results in the displacement equation of motion.


 * {| style="width:100%" border="0"

$$\displaystyle (\lambda + G) \frac{\partial e}{\partial X} + G \nabla^2 u_x + \rho_0 b_{0x}=\rho_0 \frac{\partial^2 u_x}{\partial t^2} $$     (5.7)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Description
In an ideal frictionless fluid there are no shear tractions on any plane.
 * 1) Show that his implies that $$\displaystyle T_{ij}=-p \delta_{ij}$$.
 * 2) Show that for this case the stress power, $$\displaystyle T_{ij}D_{ij}$$, is given by


 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=\frac{p}{\rho} \frac{d \rho}{dt} $$     (6.1)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * Express this result in terms of the specific volume $$\displaystyle \nu = \frac{1}{\rho}$$.

Solution
The stress vector/tensor relationship is defined by $$\displaystyle t_i^{(n)} = T_{ji}n_j$$ where the superscript n corresponds to the number of equations or rows of the stress tensor. Since $$\displaystyle T_{ji}$$ is a second order tensor with 9 components, $$\displaystyle n=1,2,3$$.


 * No shear tractions implies that $$\displaystyle T_{ji}=0 $$ when $$\displaystyle i \neq j$$.
 * For hydrostatic stress, the static pressure is defined as $$\displaystyle -p= \frac{1}{3} T_{ii}$$. Therefore, $$\displaystyle T_{ji}=-p $$ when $$\displaystyle i=j$$. This can also be written using the Kronecker delta as


 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}=-p \delta_{ij} $$     (6.2)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

The definition of the rate of deformation tensor is $$\displaystyle D_{ij} = \frac{1}{2} \left( v_{i,j} + v_{j,i} \right)$$. Since $$\displaystyle \mathbf{T}$$ is symmetric, $$\displaystyle D_{ij} = v_{i,j} $$. So from equation 6.2 it can be noted that


 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=-p \delta_{ij} \ v_{i,j} \qquad \Rightarrow \qquad T_{ij}D_{ij}=-p v_{i,i} \qquad \text{when} \quad i=j $$.     (6.3)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Recognizing that $$\displaystyle v_{i,i}$$ is the time derivative of the displacement vector,


 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=-p \frac{d}{dt} \left(u_{i,i} \right) $$     (6.4)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }


 * which can be expressed alternatively below by recalling that $$\displaystyle v_{i,i} = \frac{\dot{\rho}}{\rho}$$
 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=-p \left( \frac{1}{\rho} \frac{d \rho}{dt} \right) \quad = \quad \frac{-p}{\rho} \frac{d \rho}{dt} $$     (6.5)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Description
Show that in the frictionless fluid of Problem 6, if there is no internal heat source and if the heat conduction is governed by Fourier's Law $$\displaystyle \mathbf{q}=-\kappa \nabla \theta $$ where $$\displaystyle \theta $$ is the temperature, then the energy becomes


 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = \nabla \cdot \left( \kappa \nabla \theta \right) - p \nabla \cdot \mathbf{v} $$     (7.1)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Solution
Substituting equation 6.3 into 4.10 when $$\displaystyle h=0 $$ returns


 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = - p v_{i,i} - \frac{\partial q_j}{\partial x_j} $$.     (7.2)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Inputting the expression for Fourier's Law and recognizing that the first expression on the right-hand side is the divergence of the velocity vector


 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = - p \nabla \cdot \mathbf{v} + \frac{\partial}{\partial x_j} \left( \kappa \nabla \theta \right) $$.     (7.3)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Now the second expression on the right-hand side is the divergence, so equation 7.3 can be rewritten and rearranged to obtain the conservation of energy expression in a frictionless fluid.


 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = \nabla \cdot \left( \kappa \nabla \theta \right) - p \nabla \cdot \mathbf{v} $$     (7.4)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Description
By using the energy equation show that the Clausius-Duhem inequality takes the form


 * {| style="width:100%" border="0"

$$\displaystyle \rho \left( \theta \dot{s} - \dot{u} \right) + \mathbf{T}:\mathbf{D} - \frac{1}{\theta} \mathbf{q} \cdot \nabla \theta \ge 0 $$.     (8.1)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Solution
From the equation of energy (4.10),


 * {| style="width:100%" border="0"

$$\displaystyle \rho \dot{u} = \mathbf{T}:\mathbf{D} - \nabla \cdot \mathbf{q} +  \rho h $$ (8.2)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Based on the second law of thermodynamics, the Clausius-Duhem inequality states that the rate of entropy increase is greater than or equal to the entropy input rate. In other words


 * {| style="width:100%" border="0"

$$\displaystyle \dot{s} - \frac{h}{\theta} + \frac{1}{\rho \theta} \nabla \cdot \mathbf{q} - \frac{\mathbf{q}}{\rho \theta^2} \cdot \nabla \theta \ge 0 $$.     (8.3)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Multiplying through by $$\displaystyle \rho \theta$$ yields


 * {| style="width:100%" border="0"

$$\displaystyle \rho \theta \dot{s} - \rho h + \nabla \cdot \mathbf{q} - \frac{\mathbf{q}}{\theta} \cdot \nabla \theta \ge 0 $$.     (8.4)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Solving equation 8.2 for $$\displaystyle \rho h $$ and substituting into 8.4


 * {| style="width:100%" border="0"

$$\displaystyle \rho \theta \dot{s} - \left( \rho \dot{u} - \mathbf{T}:\mathbf{D} + \nabla \cdot \mathbf{q} \right) + \nabla \cdot \mathbf{q} - \frac{\mathbf{q}}{\theta} \cdot \nabla \theta \ge 0 $$.     (8.5)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

The two divergence terms cancel out, and the remaining terms can be collected to express the Clausius-Duhem inequality as requested.


 * {| style="width:100%" border="0"

$$\displaystyle \rho \left( \theta \dot{s} - \dot{u} \right) + \mathbf{T}:\mathbf{D} - \frac{1}{\theta} \mathbf{q} \cdot \nabla \theta \ge 0 $$     (8.6)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Description
As an example of how the Clausius-Duhem inequality palces restrictions on possible constitutive equations, consider a class of viscous fluid assumed defined by four constitutive equations of the forms below in terms of entropy $$\displaystyle s$$, density $$\displaystyle \rho$$, and rate of deformation $$\displaystyle \mathbf{D}$$.


 * Stress $$\displaystyle \Rightarrow \quad \mathbf{T} = \mathbf{T} (s,\frac{1}{\rho},\mathbf{D})$$
 * Heat flux vector $$\displaystyle \Rightarrow \quad \mathbf{q} = \mathbf{q} (s,\frac{1}{\rho},\mathbf{D})$$
 * Internal energy $$\displaystyle \Rightarrow \quad u = u (s,\frac{1}{\rho},\mathbf{D})$$
 * Temperature $$\displaystyle \Rightarrow \quad \theta = \theta (s,\frac{1}{\rho},\mathbf{D})$$

Show that the result of Problem 8 takes the form


 * {| style="width:100%" border="0"

$$\displaystyle \rho \left( \theta -\frac{\partial u}{\partial s}\right) \dot{s} -\rho \frac{\partial u}{\partial D_{km}}\dot{D_{km}} + \frac{\partial u}{\partial(1/ \rho)} \frac{\dot{\rho}}{\rho}+ T_{km}D_{km} - \frac{1}{\theta} q_k \frac{\partial \theta}{\partial x_k} \ge 0 $$.     (9.1)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

If this is to hold for arbitrary independent prescriptions of $$\displaystyle \dot{s}$$ and $$\displaystyle \dot{D_{km}}$$ we must have $$\displaystyle \theta=\partial u / \partial s $$ and $$\displaystyle \partial u / \partial D_{km}=0$$. Hence $$\displaystyle u$$ is independent of $$\displaystyle \mathbf{D}$$. If we define the thermodynamic pressure by $$\displaystyle p=-\frac{\partial u}{\partial (1/\rho)}$$ and use the equation of continuity, show that the inequality then reduces to


 * {| style="width:100%" border="0"

$$\displaystyle (T_{km} + p \delta_{km})D_{km} - \frac{1}{\theta}q_k \frac{\partial \theta}{\partial x_k} \ge 0 $$.     (9.2)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Solution
Equation 8.6 can be rewritten in Cartesian coordinates as


 * {| style="width:100%" border="0"

$$\displaystyle \rho \theta \dot{s} - \rho \dot{u} + T_{km}D_{km} - \frac{1}{\theta} q_k \cdot \frac{\partial \theta}{\partial x_k} \ge 0 $$     (9.3)
 * style="width:10%; |
 * style="width:10%; |
 * <p style="text-align:right">
 * }

Description
If $$\displaystyle \sigma_{ij} = -p \delta_{ij}$$, determine an equation for the rate of change of specific entropy during a reversible thermodynamic process.