MRF (aka frozen rotor) / 2d / fluent

everest20 · July 15, 2015, 08:00

Hi guys,

First of all, I know that there are some questions about MRF on the Forum, not only Fluent-related ones, but I've got some concerns I would like to clarify.

There are some papers in the literature regarding MRF method presented for machinery with rotating part which is then treated as non-inertial reference frame. These papers often cite work of Gosman et.al. as the first one that introduced MRF idea, I do not have access to this paper however and what Fluent documentation presents derived set of equations without refering to any sort of literature.

Talking about FLUENT, here for 2D case, e.g. 2D axial turbine cascade one can apply translational motion for the rotor. As far as I know this would introduce intertial frame of reference, thus no additional loads are introduced to the fluid.

I am familiar with the relevant set of equations used in FLUENT (e.g. eq. 2.9-11):
http://www.arc.vt.edu/ansys_help/flu...equations.html

Unfortunately I still cannot explain to myself what is the intuitive reasoning in these eqautions for 2D case without rotation. What are the assumption when one derives these equations?

I was trying to do some backward reasoning using momentum equation:

The RHS is the same as in stationary frame, and when one assumes no change in time and no rotational motion what lefts on the LHS is:

$\nabla \cdot (\rho \vec{v_r} \vec{v})$

Where (under assumption of translational motion): $\vec{v_r}=\vec{v}-\vec{u_t}, \vec{u_t}$ being the translational motion velocity vector.

Making quick transformation and assuming further on that density is constant one can get (unless I've made some mistake in the calcualtions, please confirm that):

$\vec{e}_k \left( v_{ri} \frac{\partial{v_k}}{\partial{x_i}} + v_{k} \frac{\partial{v_{ri}}}{\partial{x_i}} \right)$

Substituting $\vec{v_r}=\vec{v}-\vec{u_t}$ and noting that $\vec{u}_t$ is the same in every placewe obtain:

$\vec{e}_k \left( v_{i} \frac{\partial{v_k}}{\partial{x_i}} + v_{k} \frac{\partial{v_{i}}}{\partial{x_i}} \right) + \vec{e}_k \left( -u_{ti} \frac{\partial{v_k}}{\partial{x_i}} \right)$

First term relates to "normal" momentum equation in stationary frame and could be derived from the term:

$\nabla \cdot (\rho \vec{v} \vec{v})$

Therefore my conclusion is (since the RHS is the same) the only difference is on the LHS that when moved on the RHS can be treaded as forcing in stationary frame of reference:

$\vec{e}_k \left( u_{ti} \frac{\partial{v_k}}{\partial{x_i}} \right)$

Before I make any further commments could you please confirm that my derivation is correct, and maybe you would be so kind and comment on the origins of the MRF. I would really appreciate it.

Kind regards,
Lukas

July 15, 2015, 08:00	MRF (aka frozen rotor) / 2d / fluent	#1
everest20 New Member everest Join Date: Mar 2014 Posts: 3 Rep Power: 12	Hi guys, First of all, I know that there are some questions about MRF on the Forum, not only Fluent-related ones, but I've got some concerns I would like to clarify. There are some papers in the literature regarding MRF method presented for machinery with rotating part which is then treated as non-inertial reference frame. These papers often cite work of Gosman et.al. as the first one that introduced MRF idea, I do not have access to this paper however and what Fluent documentation presents derived set of equations without refering to any sort of literature. Talking about FLUENT, here for 2D case, e.g. 2D axial turbine cascade one can apply translational motion for the rotor. As far as I know this would introduce intertial frame of reference, thus no additional loads are introduced to the fluid. I am familiar with the relevant set of equations used in FLUENT (e.g. eq. 2.9-11): http://www.arc.vt.edu/ansys_help/flu...equations.html Unfortunately I still cannot explain to myself what is the intuitive reasoning in these eqautions for 2D case without rotation. What are the assumption when one derives these equations? I was trying to do some backward reasoning using momentum equation: The RHS is the same as in stationary frame, and when one assumes no change in time and no rotational motion what lefts on the LHS is: $\nabla \cdot (\rho \vec{v_r} \vec{v})$ Where (under assumption of translational motion): $\vec{v_r}=\vec{v}-\vec{u_t}, \vec{u_t}$ being the translational motion velocity vector. Making quick transformation and assuming further on that density is constant one can get (unless I've made some mistake in the calcualtions, please confirm that): $\vec{e}_k \left( v_{ri} \frac{\partial{v_k}}{\partial{x_i}} + v_{k} \frac{\partial{v_{ri}}}{\partial{x_i}} \right)$ Substituting $\vec{v_r}=\vec{v}-\vec{u_t}$ and noting that $\vec{u}_t$ is the same in every placewe obtain: $\vec{e}_k \left( v_{i} \frac{\partial{v_k}}{\partial{x_i}} + v_{k} \frac{\partial{v_{i}}}{\partial{x_i}} \right) + \vec{e}_k \left( -u_{ti} \frac{\partial{v_k}}{\partial{x_i}} \right)$ First term relates to "normal" momentum equation in stationary frame and could be derived from the term: $\nabla \cdot (\rho \vec{v} \vec{v})$ Therefore my conclusion is (since the RHS is the same) the only difference is on the LHS that when moved on the RHS can be treaded as forcing in stationary frame of reference: $\vec{e}_k \left( u_{ti} \frac{\partial{v_k}}{\partial{x_i}} \right)$ Before I make any further commments could you please confirm that my derivation is correct, and maybe you would be so kind and comment on the origins of the MRF. I would really appreciate it. Kind regards, Lukas

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