# Giuseppe Mensitieri » 5.State and equilibrium. Part 2

### Constitutive equation for Newtonian fluids

Generalized Newtonian Fluid

On the basis of the ‘frame indifference theorem‘ and of the application of the ‘theorem of tensor representation‘ it can be demonstrated that the most general constitutive expression for $\underline{\underline T}$ in a fluid for which the state is identified by T, V and $\underline{\underline D}$, is of the following type (at constant T):

$\underline{\underline T}=G(\underline{\underline D})=k_0\underline{\underline I}+k_1\underline{\underline D}+k_2\underline{\underline D}\cdot \underline{\underline D}\;\;\;\;\text{eq. (1)}$

Where

$k_k=k_k(I_D, II_D, III_D)$

with $\underline{\underline D}$ the rate of deformation tensor, and the $I_D,II_D,III_D$ 1st, 2nd and 3rd invariants of the rate of deformation tensor, defined as:

$I_D=tr\underline{\underline D}=div\;\underline v$

$II_D=\frac 1 2 \left[(I_D)^2-tr(\underline{\underline D}\cdot \underline{\underline D})\right]$

$III_D=det\;\underline {\underline D}$

From eq. (1), it follows that the most general linear relation between $\underline{\underline T}$ and $\underline{\underline D}$ that is consistent with the principle of material frame indifference is:

$\underline{\underline T}=-(\alpha+\mu_1div\;\underline v)\underline{\underline I}+2\mu \underline{\underline D}\;\;\;\;\text{eq. (2a)}$

### Constitutive equation for Newtonian fluids

By applying the 2nd law of thermodynamics, it can be demonstrated that α = p* (where p*, again, is the thermodynamic pressure) and we obtain from eq. (2a) the constitutive equation for the generalized Newtonian fluid:

$\underline{\underline T} =-(p^*+\mu_1div\;\underline v)\underline{\underline I}+2\mu\underline{\underline D}\;\;\;\;\text{eq. (2b)}$

We remind here that:

$\underline\nabla\; \underline v=\frac 1 2 (\underline \nabla\; \underline v+\underline \nabla\;\underline v^T)+\frac 1 2 (\underline \nabla\; \underline v -\underline \nabla\; \underline v^T)$

$\underline \nabla\;\underline v =\;\;\;\;\;\;\;\;\;\;\;\;\underline {\underline D}\;\;\;\;\;\;\;\;\;\;+\;\;\;\;\;\;\;\;\;\;\underline{\underline W}$

and that the bulk viscosity, k, is defined as follows:

$k=-\mu_1+\frac 2 3 \mu$

We now specify the difference between mechanical pressure, p, and thermodynamic pressure, p*. The mechanical pressure is the average normal stress one would measure in a point in the material along three mutually orthogonal directions and is consequently defined as:

$p=-\frac 1 3 tr\;\underline{\underline T}$

Consequently, the isotropic part of the stress tensor is defined as: $\text{Isotropic part of}\;\; \underline{\underline T}=-p\underline{\underline I}$

### Constitutive equation for Newtonian fluids

The thermodynamic pressure is the value of p one would measure in the fluid if it was at equilibrium, once the site has been fixed at the value it has.
Let us determine p – p* for the case of a) pure compression or expansion and b) in a more general case.

For a pure compression or expansion we have that:

$\underline{\underline D}=\frac 1 3 (div\;\underline v)\underline {\underline I}\Rightarrow\underline{\underline T}=\left[-p^*+\left(-\mu_1+\frac 2 3 \mu\right)div\:\underline v\right]\underline{\underline I}$

Based on the definition of p, we conclude that:

$p=-\left[-p^*+\left(-\mu_1+\frac 2 3 \mu\right)div\;\underline v\right]\Rightarrow p-p^*=-\left(-\mu_1\frac 2 3 \mu\right)div\;\underline v=-k(div\;\underline v)=-k\frac{\dot V}V\Rightarrow B=\frac{-3\mu_1+2\mu}{3V}$

From the previous expression it is evident that p = p* only if one or both of the following events happen:

• fluid is uncompressible (i.e. $div\;\underline v=0$);
• the Stokes’ hypothesis holds true (i.e. $k=0\Rightarrow \mu_1=\frac 2 3 \mu$).

In the first case actually no compression or expansion is possible. In the second case, when Stokes’ hypothesis holds true, the fluid is not able to dissipate energy neither in compression nor in expansion.

### Constitutive equation for Newtonian fluids

b) The result obtained above for p – p* is identically valid for more general cases and not only in the case of pure compression or expansion processes. In fact:

$\underline{\underline T}=-(p^*+\mu_1div\;\underline v)\underline {\underline I}+2\mu\underline{\underline D}}\Rightarrow tr\;\underline{\underline T}=-3p^*-3\mu_1div\;\underline v+2\mu(tr\;\underline{\underline D})=-3p^*-3\mu_1div\;\underline v+2\mu(div\;\underline v)\Rightarrow$

$\Rightarrow -p^*=\frac 1 3 tr\;\underline {\underline T}+\mu_1 div\;\underline v-\frac 2 3 \mu(div\;\underline v)\Rightarrow p-p^*=-k(div\;\underline v)$

It is now a useful exercise to identify the various parts of stress tensor in the case of a Newtonian fluid. Every tensor can be expressed as the sum of its isotropic part (i.e. $1\big / 3 ~~tr\;\underline{\underline T}$) and of its deviatoric part (which, for the case of stress tensor we will indicate with the symbol $\underline{\underline T'}$):

$\underline{\underline T}=\underline {\underline T'}+\text{isotropic part}=\underline{\underline T'}+\left(\frac 1 3 tr\;\underline{\underline T}\right)\underline{\underline I}=\underline{\underline T'}-p\underline {\underline I}$

This is, obviously, also true for the tensor rate of deformation, $\underline{\underline D}$:

$\underline{\underline D}=\underline{\underline D'}+\left(\frac 1 3 tr \underline {\underline D}\right)\underline {\underline I}$

Equation (2b) can hence be recast in the following form:

$\underline{\underline T'}=(p-p^*)\underline{\underline I}+\left(-\mu_1+\frac 2 3 \mu\right)(tr\;\underline{\underline D})\underline{\underline I}+2\mu\underline{\underline D'}$

### Constitutive equation for Newtonian fluids

By definition we have that:

$tr\underline{\underline T'}=0 \;and\; tr\underline{\underline D'}=0$

Consequently it follows that:

$(p-p^*)\underline{\underline I}+\left(-\mu_1+\frac 2 3 \mu\right)tr\underline{\underline D}=0\Rightarrow \underline{\underline T'}=2\mu\underline{\underline D'}$

It then follows:

$\underline{\underline T}=-p^*\underline{\underline I}+\left(-\mu_1+\frac 2 3 \mu\right)(tr\;\underline{\underline D})\underline{\underline I}+2\mu\underline {\underline D'}\;\;\;\;\text{eq. (3)}}$

The tensor $\underline{\underline T}$, as already mentioned above, can be split in different contributions in several ways:

$\underline{\underline T}=p^*\underline{\underline I}+\underline{\underline S}~~~\Rightarrow~~~~~~~~~~~~\underline{\underline S}=\left(-\mu_1+\frac 2 3 \mu\right)(tr\underline{\underline D})\underline{\underline I}+2\mu\underline{\underline D'}$

$\underline{\underline T}=p\underline{\underline I}+\underline{\underline T'}~~~~~~~\Rightarrow~~~~~~~~p\underline{\underline I}=\left(\frac 1 3 tr\underline {\underlineT}\right)\underline{\underline I}=p^*\underline {\underline I}+\left(-\mu_1+\frac 2 3 \mu\right)(tr\underline{\underline D})\underline{\underline I};\;\;\;\underline{\underline T'}=2\mu\underline{\underline D'}$

$\underline{\underline T}=p^*I+p^V\underline {\underline I}+\underline{\underline S'}~~~~~\Rightarrow~~~~~p^V\underline{\underline I}=\left(-\mu_1+\frac 2 3 \mu\right)(tr\underline{\underline D})\underline{\underline I};\;\;\;\;\underline{\underline S'}=2\mu\underline{\underline D'}$

### Constitutive equation for Newtonian fluids

As we will see, the 2nd law imposes two constraints to viscosity:

1) B ≥ 0 (B non negative): pure compression or expansion phenomena are irreversible phenomena if B>0.

2) μ ≥ 0 (μ non negative): Deformations at constant density are irreversible if μ > 0.

In the case of uncompressible fluids we then have:

$div\;\underline v=tr\underline{\underline D}=0\Rightarrow \underline{\underline S}=\underline{\underline T'}=2\mu\underline{\underline D'}\;\;\;and\;\;\; \underline{\underline D}=\underline{\underline D'}$

In the cases in which the bulk viscosity k $(k=-\mu_1+\frac 2 3 \mu)$ is zero we have:

$\underline{\underline T}=-p^*\underline{\underline I}+\underline{\underline S}=-p^*\underline{\underline I}+2\mu\underline{\underline D'}=-p^*\underline{\underline I}-\frac 2 3 \mu(tr\underline {\underline D})\underline{\underline I}+2\mu\underline{\underline D}\;\;\;and \;\;\; \underline{\underline T'}=\underline{\underline S}=2\mu\underline{\underline D'}$

In this case $\underline{\underline D}\neq \underline{\underline D'}$

### Constitutive equation for Newtonian fluids

On the basis of the previous discussion on the stress tensor, we have that the rate of increase of internal energy per unit volume, $\underline{\underline T}:\underline\nabla\;\underline v$, due to the work of contact forces, can be subdivided in several contributions:

$\underline{\underline T}:\underline\nabla\;\underline v=$

$p^*(\underline\nabla\;\cdot\underline v)\;\;\;\text{which is the reversible rate of work related to volume change} +$

$p^V(\underline\nabla\;\cdot\underline v)\;\;\;\text{which is the irreversible rate of work related to volume change} +$

$\underline{\underline S'}:\underline \nabla\;\underline v\;\;\;\text{which is contributed by the reversible and irreversible rate of work realated to shape change}$

### On the balances of mechanical and internal energies

As we have seen in the course on Transport Phenomena (see for example the book by Bird, Stewart and Lightfoot) the expression for the linear momentum balance reads:

$\frac{\partial(\rho\underline v)}{\partial v}=-\underline \nabla(\rho \underline v\;\underline v)-\underline\nabla p^*-\underline \nabla\cdot\underline{\underline S}+\rho \underline g\;\;\;\;\text{eq. (4a)}$

Which can be recast in the following form:

$\rho\frac{d_{(m)}\underline v}{dt}=-\underline \nabla p^*- \underline \nabla \cdot\underline {\underline S}+\rho \underline g \;\;\;\;\text{eq. (4b)}$

Multiplying scalarly eq.(4b) by the velocity vector, we obtain after some elaboration the local forms of mechanical energy balance:

$\rho\frac{d_{(m)}\underline{\left(\frac 1 2 v^2}\right)}{dt}=-\underline \nabla\cdot (p^*\cdot \underline v)+p^*\underline\nabla \cdot\underline v+\rho(\underline v \cdot\underline g)-\underline \nabla \cdot(\underline S\cdot\underline v)+\underline{\underline S}:\underline \nabla\;\underline v\;\;\;\;\text{eq. (5a)}$

$\frac{\partial \left(\frac 1 2 \rho v^2\right)}{\partial t}=-\underline \nabla \cdot\left(\frac 1 2 \rho v^2 \underline v\right)-\underline \nabla \cdot (p^*\underline v)+p^*\underline \nabla\cdot\underline v+\rho(\underline v\cdot g)-\underline \nabla(\underline{\underline S}\cdot\underline v)+\underline{\underline S}:\underline{\underline \nabla}\;\underline v\;\;\;\text{eq. (5b)}$

### On the balances of mechanical and internal energies

We recall now that the total energy balance reads:

$\rho\frac{d_{(m)}\left(U+\frac 1 2 v^2\right)}{dt}=-\underline \nabla \cdot \underline q +\rho (\underline v \cdot \underline g)-\underline \nabla\cdot (p^*\underline v)-\underline \nabla\cdot (\underline{\underline S}\cdot\underline v)\;\;\;\;\text{eq. (6)}$

By subtraction of eq (5a) from eq. (6) we finally obtain the thermal energy balance:

$\rho\frac{\partial U}{\partial t}=-\underline \nabla\cdot \underline q-p^*(\underline \nabla\;\underline v)-\underline{\underline S}:\underline\nabla\;\underline v\;\;\;\;\text{eq. (7)}$

Terms appearing both in equation of variation of mechanical energy and of thermal energy are exchange terms between the two forms of energy (i.e. kinetic and internal). These terms are different from zero only for non rigid motions.

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