Page 310 - Intro to Tensor Calculus

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EXAMPLE 2.5-1. (One-dimensional fluid flow)
Construct an x-axis running along the center line of a long cylinder with cross sectional area A. Consider
the motion of a gas driven by a piston and moving with velocity v 1 = u in the x-direction. From an Eulerian
point of view we imagine a control volume fixed within the cylinder and assume zero body forces. We require
the following equations be satisfied.
∂% ∂% ∂
~
Conservation of mass + div(%V ) = 0 which in one-dimension reduces to + (%u)= 0.
∂t ∂t ∂x
∂ ∂ 2 ∂p
Conservation of momentum, equation (2.5.28) reduces to (%u)+ %u + =0.
∂t ∂x ∂x
Conservation of energy, equation (2.5.48) in the absence of heat flow and internal heat production,

∂e ∂e ∂u
becomes in one dimension % + u + p =0. Using the conservation of mass relation this
∂t ∂x ∂x
∂ ∂ ∂u
equation can be written in the form (%e)+ (%eu)+ p =0.
∂t ∂x ∂x
In contrast, from a Lagrangian point of view we let the control volume move with the flow and consider
advection terms. This gives the following three equations which can then be compared with the above
Eulerian equations of motion.
d D% ∂u
Conservation of mass (%J) = 0 which in one-dimension is equivalent to + % =0.
dt Dt ∂x
Du ∂p
Conservation of momentum, equation (2.5.25) in one-dimension % + =0.
Dt ∂x
De ∂u
Conservation of energy, equation (2.5.48) in one-dimension % + p =0. In the above equations
Dt ∂x
D() ∂ ∂
Dt = ∂t () + u ∂x (). The Lagrangian viewpoint gives three equations in the three unknowns ρ, u, e.
In both the Eulerian and Lagrangian equations the pressure p represents the total pressure p = p g + p v
where p g is the gas pressure and p v is the viscous pressure which causes loss of kinetic energy. The gas pressure
is a function of %, e and is determined from the ideal gas law p g = %RT = %(c p − c v )T = %( c p − 1)c v T or
c v
p g = %(γ − 1)e. Some kind of assumption is usually made to represent the viscous pressure p v as a function
of e, u. The above equations are then subjected to boundary and initial conditions and are usually solved
numerically.

Entropy inequality

Energy transfer is not always reversible. Many energy transfer processes are irreversible. The second
law of thermodynamics allows energy transfer to be reversible only in special circumstances. In general,
the second law of thermodynamics can be written as an entropy inequality, known as the Clausius-Duhem
inequality. This inequality states that the time rate of change of the total entropy is greater than or equal to
the total entropy change occurring across the surface and within the body of a control volume. The Clausius-
Duhem inequality places restrictions on the constitutive equations. This inequality can be expressed in the
form
Z Z Z n
D i X
%s dτ ≥ s n i dS + ρb dτ + B (α)
Dt
V S V α=1
| {z }
| {z }
Rate of entropy increase
Entropy input rate into control volume
i
where s is the specific entropy density, s is an entropy flux, b is an entropy source and B (α) are isolated
entropy sources. Irreversible processes are characterized by the use of the inequality sign while for reversible