Dear colleagues,
in many publications the head loss coefficient K for a conduit component is determined as
K=Delta ptot/(rho u2/2)
with ptot i= rho um i2/2 + pi as the so called "total pressure", i.e. the sum of an approximate kinetic energy u2 /2 (or rho u2 /2 as its dynamic pressure) and the flow work within a cross section i of size A.
Taking into account that the velocity is non-uniformly distributed, the actual kinetic energy however is
Ekin= int u2/2 d dot_m = int_A u2/2 rho u dA
and
ekin=Ekin/dot_m= alpha u2/2
with alpha deviating from 1 for any viscous flow with no-slip flow at the walls.
Following the RANS-Approach, a part of the kinetic energy is not resolved. In nearly all two-equations models, however, it is the crucial quantity of a transport equation, the k-equation. Should therefore the kinetic energy be expressed as
Ekin= int (u2/2 + k) d dot_m = int_A (u2/2 + k) rho u dA
resulting in ekin= alphauku2/2 ? This increased alpha-correction factor results in very different results for the head loss due to nozzles and diffusers.
I am very interested in your opinion.
Best regards
Bastian
in my opinion It is hardly depend on turbulent intensity of you flow which is varying 0 up to 20 percent and will cause a pressure fluctuation.for low intensities may not be important.
regards
Dear Schmandt,
Clearly, turbulent kinetic energy should be counted. If there exist a non-turbulent, unsteady (sinusoidal) wavy motion, it also should be considered. -Ji-
Thank you for the replies,
in fact, the turbulent kinetic energy has a large impact on the results for a nozzle as demonstrated in our attached paper . Thus the "0 up to 20 percent" can make a significant change.
I currently wonder if these findings have been reported elsewhere since I did not find anything in the open literature. Maybe there also is a literature source for the "unsteady (sinusoidal) wavy motion" reported by Ji?
Best regards
Bastian
Article Diffuser and Nozzle Design Optimization by Entropy Generatio...
Just to add--another way of thinking is: how much does the (parameterized) unresolved turbulent kinetic energy contribute to the fluctuating pressure? If you have a Poisson solver, how much does it "see" in p' via the unresolved TKE? [In some LES codes they use e.g. a 'pseudopressure' which includes the unresolved (subgrid) normal stresses...]
This is an interesting point added by Mark Kelly.
In RANS, however, no pressure fluctuation is computed. Would your point mean, that the pressure computed in RANS is wrong?
From my current understanding i would suppose, that the pressure fluctuation vanishes due to time averaging, since the pressure gradient appears as a linear term in the momentum equation. Thus the effect of a fluctuating pressure (flow work done by pressure fluctuations) should cancel out in the mean.
Best regards
Bastian
If by u you mean the instantaneous Velocity Vector field, then the averaged Kinetic energy per unit mass would be
R(u.u) = R((U+u').(U+u')) = R(U.U+U.u'+u'.U+u'.u') **= R(U.U + u'.u')
= R(U.U) + R(u'.u') = 2(Kmean + k) ,
where R( ) is a linear Reynolds Operator, k=1/2R(u'.u') is the turbulent kinetic Energy per unit Mass and K=1/2R(U.U) is the kinetic Energy per unit Mass of the MEAN Flow.
Thank you Fernando,
this is the confirmation i was looking for. Are you aware of a publication in which k in addition to U^2/" is used to determine the head loss using the mechanical enrgy equation?
Best regards
Bastian
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