This study evaluated two turbulence closure models (GI and GWJ models) for modeling rotating turbulence at different rotation rates. The models were tested on three cases with different rotation numbers and their constants were optimized using explained variance. After optimization, the GI model showed greater improvement than the GWJ model, especially for higher rotation rates. The A4 coefficients had the most effect on the GWJ model results.

1. Overview Equations
Case B
Case C
GWJ Model Improvement
GI Model Improvement
Models
Case A
Case Studied
Conclusion
Works Cited
Evaluation of Turbulence Closures for Rotating Turbulence
Cesar Galan, Adviser: Sedat Biringen
When there is a spanwise rotation in a turbulent flow, it is useful to un-
derstand how the flow behaves with the Reynolds stress. The flow is
divided in suction and pressure side, suction wall is y = 0 whereas pre-
ssure wall is y = 2. These Reynolds stresses can be accurately describe
by direct numerical simulation (DNS). However, the computational re-
sources for this process is extremely high. Therefore, an accurate
Reynolds Averaged Navier-Stokes (RANS) model with less computatio-
nal cost is desirable. The Reynolds number for all the models is
Reτ
= 200.
· These RANS models compute the Reynolds stress anisotropy (bij
),
equation (2) and equation (3) for GI model (Girimaji, 1996) and GWJ
model (Grundestam, Wallin, Johansson, 2005), respectively. Then
the Reynolds stress (u’i
u’j
) is computed with equation (1) where, δij
is
the kronecker delta, and K is the turbulent kinetic energy.
· The models contain constants derived from experiments, which were
optimized using explained variance. Explained variance is shown in
equation (4).
· After the constants were optimized, the model was tested for realizability.
Realizability is a physical test that check if the model follows the physi-
cal laws. All the models succesfully passed the realizability test.
· There are two models that are being optimized, GI model and GWJ model, for three different cases,
each with different rotation number (Rob
). Case A: Nonrotating case (Rob
= 0); Case B: Rotating case
(Rob
= 0.2); Case C: Rotating case (Rob
= 0.5).
· GI model: The GI model uses equation (2) to compute the Reynolds stress anisotropy, where Sij
is the
rate-of-strain tensor and Wij
is the normalized vorticity tensors. Normalized vorticity tensor contains the
rate-of-rotation tensor and the vorticity. G1
is a conditional variable dependent of η1
, L1
1
, D, and b. G2
and
G3
are dependent on G1
. For more information refer to Girimaji, 1996 .
· GWJ model: This model uses equation (3) to compute the Reynolds stress anisotropy where there are
ten different β coefficients dependent on invariants of S and Ω* and ten different T tensors that are
summed to obtain bij
. For more information refer to Grundestam, Wallin, Johansson, 2005.
· After the optimization the GI model demostrated a bigger improvement in all three
cases. Whereas, the GWJ model only showed a substantial improvement for case B.
· For the constants in the GI model, with a higher rotation rate, the optimal
coefficientsare much divergent from the original coefficient. The same conclusion
can be derived for the first three coefficients in the GWJ model.
· The A4
coefficients for the GWJ model had the most effect in the model.
· Girimaji, Sharath S. "Fully Explicit and Self-Consistent Algebraic Reynolds Stress Model." 1996. MS 387-402.
Institute for Computer Applictions in Science and Engineering, NASA Langley Center, Hampton, VA, USA, Hampton
· Grundestam, Olof, Stefan Wallin, and Arne Johansson. "An Explicit Algebraic Reynolds Stress Model Based on a
Nonlinear Pressure Strain Rate Model." 2005. MS 732-745. Department of Mechanics, Royal Institute of Technology, Stockholm.
· Schumann, U. "Realizability of Reynolds-stress Turbulence Models." 1997. MS. Institut Fur Reaktorentwicklung, Karlsruhe.
y
0 0.5 1 1.5 2
u’u’
+
0
1
2
3
4
5
6
7
8
(a)
y
0 0.5 1 1.5 2
u’v’
+
-1.5
-1
-0.5
0
0.5
1
1.5
(d)
y
0 0.5 1 1.5 2
w’w’
+
0
0.5
1
1.5
2
2.5
3
(c)
y
0 0.5 1 1.5 2
v’v’
+
0
0.5
1
1.5
(b)
y
0 0.5 1 1.5 2
u’u’
+
0
1
2
3
4
5
6
7
8
9
(a)
y
0 0.5 1 1.5 2
u’v’
+
-0.5
0
0.5
1
1.5
2
2.5
(d)
y
0 0.5 1 1.5 2
v’v’
+
0
0.5
1
1.5
2
2.5
(b)
y
0 0.5 1 1.5 2
w’w’
+
0
0.5
1
1.5
2
2.5
3
3.5
4
(c)
y
0 0.5 1 1.5 2
u’u’
+
0
1
2
3
4
5
6
7
(a)
y
0 0.5 1 1.5 2
v’v’
+
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
(b)
y
0 0.5 1 1.5 2
u’v’
+
-0.5
0
0.5
1
1.5
2
2.5
(d)
y
0 0.5 1 1.5 2
w’w’
+
0
0.5
1
1.5
2
2.5
3
3.5
(c)