NUMERICAL STUDY OF THERMOVISCOUS EFFECTS IN . . . PHYSICAL REVIEW E 90, 043016 (2014)
TABLE I. IAPWS parameter values for pure water at ambient
temperature 25◦C and pressure 0.1013 MPa. For references, see
Appendix A.
Parameter Symbol Value Unit
Thermodynamic parameters:
Mass density ρ 9.970 × 102 kg m−3
Heat capacity cp 4.181 × 103 J kg−1K−1
Speed of sound cs 1.497 × 103 m s−1
Compressibility κT 4.525 × 10−10 Pa−1
Thermal expansion αp 2.573 × 10−4 K−1
Heat capacity ratio γ 1.011 × 100
Transport parameters:
Shear viscosity η 8.900 × 10−4 Pa s
Bulk viscosity ηb 2.485 × 10−3 Pa s
Thermal conductivity kth 6.065 × 10−1 Wm−1 K−1
Thermodynamic derivatives:
1
η
∂η
∂T
−2.278 × 10−2 K−1
1
η
∂η
∂ρ
−3.472 × 10−4 kg−1 m3
1
ηb
∂ηb
∂T
−2.584 × 10−2 K−1
1
kth
∂kth
∂T
2.697 × 10−3 K−1
1
kth
∂kth
∂ρ
2.074 × 10−3 kg−1m3
data. To first order in the acoustic perturbation, we thus write
the dynamic shear viscosity η, the bulk (second) viscosity ηb,
and the thermal conductivity kth as
η(T,ρ) = η0(T0,ρ0) + η1(T0,T1,ρ0,ρ1), (4a)
η1
=
∂η
∂T
T=T0
T1
+
∂η
∂ρ
ρ=ρ0
ρ1, (4b)
ηb(T,ρ) = ηb
0(T0,ρ0) + ηb
1(T0,T1,ρ0,ρ1), (4c)
ηb
1
=
∂ηb
∂T
T=T0
T1
+
∂ηb
∂ρ
ρ=ρ0
ρ1, (4d)
kth(T,ρ) = kth
0 (T0,ρ0) + kth
1 (T0,T1,ρ0,ρ1), (4e)
kth
1
=
∂kth
∂T
T=T0
T1
+
∂kth
∂ρ
ρ=ρ0
ρ1. (4f)
For the acoustic amplitudes used in this model, the
maximum relative perturbations, such as |η1
|/η0, due to the
temperature dependence of the transport coefficients, are
0.33%, 0.53%, and 0.034% for η, ηb, and kth, respectively,
and the perturbations due to the density dependence are 0.37%
and 0.82% for η and kth, respectively. We could not find any
literature on the density dependence of ηb of water.
C. Governing equations
Besides the above thermodynamic relations, the governing
equations of thermoviscous acoustics requires the introduction
of the velocity field v of the fluid as well as the stress tensor
σ, which is given as 29
σ = −p1 + τ , (5a)
τ = η∇v + (∇v)T +
ηb − 23
η
(∇ · v)1. (5b)
Here, 1 is the unit tensor, and the superscript “T” indicates
tensor transposition.
Mass conservation implies that the rate of change ∂tρ of
the density in a test volume with surface normal vector n is
given by the influx (direction −n) of the mass current density
ρv. In differential form by Gauss’s theorem, it is
∂tρ = ∇ · −ρv. (6a)
Similarly, momentum conservation implies that the rate
of change ∂t (ρv) of the momentum density in the same
test volume is given by the stress forces σ acting on the
surface (with normal n), and the influx (direction −n) of
the momentum current density ρvv. In differential form,
neglecting body forces f , this becomes
∂t (ρv) = ∇ · τ − p1 − ρvv. (6b)
12
Finally, energy conservation implies that the rate of change
∂(ρε + ρv2) of the energy density (internal plus kinetic)
t is given by the power of the stress forces v · σ on the
surface (direction n), and the influx (direction −n) of both
heat conduction power −kth∇T and energy current density
(ρε + 12
ρv2)v. In differential form, neglecting heat sources in
the volume, this becomes
∂t
ρε + 12
ρv2
= ∇ ·
v · τ
−pv + kth∇T − ρ
ε + 12
v2
v
. (6c)
D. First-order equations of thermoviscous acoustics
The homogeneous, isotropic quiescent state (thermal equilibrium)
is taken to be the zeroth-order state in the acoustic
perturbation expansion. Following standard first-order perturbation
theory, all fields g are written in the form g = g0
+ g1,
for which g0 is the value of the zeroth-order state and g1 is the
acoustic perturbation, which must be much smaller than g0. In
our work, |g1
|/g0 10−3 20. We assume that the acoustic
perturbations g1 are oscillating harmonically with the angular
frequency ω of the acoustic actuation,
g1(r,t ) = g1(r)e
−iωt, ∂tg1
= −iωg1. (7)
For the velocity, the value of the zeroth-order state is v0
= 0,
and thus v = v1. The zeroth-order terms solve the governing
equations for the zeroth-order state and thus drop out of the
equations. Keeping only first-order terms, we obtain the firstorder
equations.
The continuity Eq. (6a) becomes
∂tρ1
= −ρ0
∇ · v1, (8a)
which, by using Eq. (3b) in the form
ρ1
= ρ0κT p1
− αp T1, (8b)
is rewritten as
αp ∂tT1
− κT ∂tp1
= ∇ · v1. (8c)
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