# Total Pressure: RANS vs EXPT

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 February 15, 2001, 06:49 Total Pressure: RANS vs EXPT #1 Dave Hunt Guest   Posts: n/a Total Pressure: RANS vs EXPT Dear All, I am looking at comparisons between total pressure from an experiment and CFD data from a RANS calculation (k-w although the model shouldn't be relevant to the question). How do I do this? My assumption is that for incompressible flow; p0=p + rho.Ui.Ui/2 Reynolds averaging <...> this gives; = + /2 + /2 or = + /2 + k based on Ui=+ui In an experiment, the total pressure observed will be and the RANS CFD solution will solve for , and k. 1) Firstly, is the above argument correct. 2) Is there a reference that indicates this? 3) Is there a reference showing how to obtain the time-averaged total pressure based on the Reynolds averaged (strictly Favre averaged) quantities available from a compressible RANS solution. Many thanks David Hunt

 February 15, 2001, 14:10 Re: Total Pressure: RANS vs EXPT #2 John C. Chien Guest   Posts: n/a (1). The total pressure is a derived parameter or variable. If you bring the current state " isentropically " to a zero velocity state , then the resultant pressure will be the total pressure. (2). Remember that this is not a "real, local" process. It is only a definition. If you define the "total" based on the mean variables, then it will be a mean variable also. In other words, it will be the total pressure derived from the mean velocity, the mean density, the mean static pressure.

 February 16, 2001, 06:29 Re: Total Pressure: RANS vs EXPT #3 sylvain Guest   Posts: n/a If the flow is incompressible (Ma << 1), then you have : (1) P_t = P_s + 1/2 * Rho * U_i * U_i This expression is equivalente to the following where the Reynolds decomposition is introduce ( G = [G] + g' ) : (2) [P_t] + p_t' = [P_s] + ps' + 1/2 * ([Rho] + rho') * ([U_i] + u_i') * ([U_i] + u_i') Then, making the assumption that rho'= 0 and taking the Reynolds average of equation (2) gives : [P_t] = [P_s] + 1/2 * [Rho] * [U_i] * [U_i] + 1/2 * [Rho] * [ u_i'* u_i'] since [ u_i'* u_i'] is equal to 2k, it comes : [P_t] = [P_s] + 1/2 * [Rho] * [U_i] * [U_i] + [Rho] * k CQFD Using the same recipe, one can obtain any equation involving Reynolds average quantities. THIS IS THE WAY USED TO DERIVE THE RANS EQUATIONS.

 February 16, 2001, 07:07 Re: Total Pressure: RANS vs EXPT #4 sylvain Guest   Posts: n/a If the flow is incompressible (Ma << 1), then you have : (1) P_t = P_s + 1/2 * Rho * U_i * U_i This expression is equivalente to the following where the Reynolds decomposition is introduce ( G = [G] + g' ) : (2) [P_t] + p_t' = [P_s] + ps' + 1/2 * ([Rho] + rho') * ([U_i] + u_i') * ([U_i] + u_i') Then, making the assumption that rho'= 0 and taking the Reynolds average of equation (2) gives : [P_t] = [P_s] + 1/2 * [Rho] * [U_i] * [U_i] + 1/2 * [Rho] * [ u_i'* u_i'] since [ u_i'* u_i'] is equal to 2k, it comes : [P_t] = [P_s] + 1/2 * [Rho] * [U_i] * [U_i] + [Rho] * k CQFD Using the same recipe, one can obtain any equation involving Reynolds average quantities. THIS IS THE WAY USED TO DERIVE THE RANS EQUATIONS.

 February 16, 2001, 12:31 Re: Total Pressure: RANS vs EXPT #5 sylvain Guest   Posts: n/a If the flow is incompressible (Ma << 1), then you have : (1) P_t = P_s + 1/2 * Rho * U_i * U_i This expression is equivalente to the following where the Reynolds decomposition is introduce ( G = [G] + g' ) : (2) [P_t] + p_t' = [P_s] + ps' + 1/2 * ([Rho] + rho') * ([U_i] + u_i') * ([U_i] + u_i') Then, making the assumption that rho'= 0 and taking the Reynolds average of equation (2) gives : [P_t] = [P_s] + 1/2 * [Rho] * [U_i] * [U_i] + 1/2 * [Rho] * [ u_i'* u_i'] since [ u_i'* u_i'] is equal to 2k, it comes : [P_t] = [P_s] + 1/2 * [Rho] * [U_i] * [U_i] + [Rho] * k CQFD Using the same recipe, one can obtain any equation involving Reynolds average quantities. THIS IS THE WAY USED TO DERIVE THE RANS EQUATIONS.

 February 16, 2001, 13:01 Re: Total Pressure: RANS vs EXPT #6 sylvain Guest   Posts: n/a If the flow is incompressible, then you have : (1) P_t = P_s + 1/2 * Rho * U_i * U_i Introduciing the Reynolds decomposition : (2) P_t = P_t_av + p_t (3) P_s = P_s_av + p_s (4) U_i = U_i_av + u_i and assuming that : (5) Rho = Rho_avg it comes : (6) P_t_av + p_t = P_s_av + p_s + 1/2 * Rho_av * ( U_i_av + u_i ) * ( U_i_av + u_i ) taking the Reynolds average of this relation gives : (7) P_t_av = P_s_av + 1/2 * Rho_av * U_i_av * U_i_av + 1/2 * Rho_av * (u_i*u_i)_av since k = (u_i*u_i)_av, one gets : (8) P_t_av = P_s_av + 1/2 * Rho_av * U_i_av * U_i_av + Rho_av * k You can apply the same recipe to any kind of equation to get the Reynolds Average expression of it.

 February 16, 2001, 13:22 PS : Re: Total Pressure: RANS vs EXPT #7 sylvain Guest   Posts: n/a Dave, my previous mail is not in contradiction with John's point of vue. It's just a matter of what you call average total pressure.

 February 16, 2001, 13:41 Re: PS : Re: Total Pressure: RANS vs EXPT #8 Dave Hunt Guest   Posts: n/a Sylvain, I notice that my original message has been corrupted. It should include a derivation identical to yours. What I would like is a reference to back up this form for total pressure. I would also like an expression for the time-averaged total pressure for a compressible flow. I've tried a few approaches but they all get a bit messy. If you have any thoughtsm then I'd appreciate them. Thanks Dave

 February 16, 2001, 14:35 Re: PS : Re: Total Pressure: RANS vs EXPT #9 John C. Chien Guest   Posts: n/a (1). One thing I don't know is, will the turbulence kinetic energy k turn into pressure as indicated in the equation? Will it turn into heat through dissipation?

 February 19, 2001, 06:05 Re: PS : Re: Total Pressure: RANS vs EXPT #10 sylvain Guest   Posts: n/a Physically, total enthalpy takes into account all the "energy" of the fluid. One gets, at any time : H_t = Cp * T_t = Cp * T + 1/2 * U_i * U_i The Cp * T has to do with the kinetic energy of the molecules. In term of Reynolds decomposition, it comes : Cp * T = Cp ([T] + teta) The 1/2 * U_i * U_i takes into account all the kinetic energy of the flow. At any moment, introducing the Reynolds decomposition, one gets : 1/2 * U_i * U_i = 1/2 * ( [U_i] + u_i ) * ( [U_i] + u_i ) = 1/2 * [U_i] * [U_i] + [U_i] * u_i + 1/2 * u_i * u_i So : H_t = Cp * [T] + 1/2 * [U_i] * [U_i] + 1/2 * u_i * u_i + Cp * teta + [U_i] * u_i Then, the mean total enthalpy is : [H_t] = Cp * [T] + 1/2 * [U_i] * [U_i] + 1/2 * [u_i * u_i] = Cp * [T] + 1/2 * [U_i] * [U_i] + k So the turbulent kinetic energy has to be taken into account in the mean total enthalpy. In the other hand, the dissipation rate is the amount of turbulent kinetic energy that vanish into heat. That's just mean that if you look at the homogeneous decay of a turbulent flow : At the the beginning (time = t_0) : [H_t](t_0) = Cp * [T](t_0) + k(t_0) at the end of the experiment (time = t_1) : [H_t](t_1) = Cp * [T](t_1) since all the kinetic energy has vanish. And since : [H_t](t_0) = [H_t](t_1) it comes : [T](t_1) > [T](t_0) the turbulent kinetic energy has turn into heat. Moreover, since : dk/dt = - epsilon it comes : -k(t_0) = - int(from t_0 to t_1) epsilon dt so : [T](t_1) = [T](t_0) + 1/Cp * int(from t_0 to t_1) epsilon dt Remarks 1 : I hope that my calculus are OK. Remarks 2 : I hope that my english is understandable. Remarks 3 : I don't have any references about this.

 February 19, 2001, 12:58 Re: PS : Re: Total Pressure: RANS vs EXPT #11 John C. Chien Guest   Posts: n/a (1). I was thinking about the turbulence kinetic energy term in the total pressure formula. (2). Will the turbulence kinetic energy term actually be converted into the pressure just like the dynamic pressure term (converted from the kinetic energy of the mean flow )? Higher turbulence fluctuation (higher k) means higher total pressure?

 February 20, 2001, 05:35 Re: PS : Re: Total Pressure: RANS vs EXPT #12 sylvain Guest   Posts: n/a I believe that what is true for T_t is also true for P_t since they are link throught the equation of state. Then, the turbulence kinetic energy is converted into pressure as the kinetic energy of the mean flow. For exemple, if you have a complete DNS result, you could compute at any time and at any place the total pressure from the usual relationship for low mach number: P_t = P_o + 1/2 * rho * U_i * U_i That pressure is the pressure measure with a Pito tube align with a streamline. It takes into account the exchange of kinetic energy of the gaz with the wall. This kinetic energy is split in two part : - kinetic energy of molecules in the gaz (i.e P_o); - kinetic energy of fluids particules at speed U. For a given place and a given time, you can make several realisations of the measurement. Since the flow is turbulent, you won't get the same value, but you can compute a mean value [P_t] and a fluctuation p_t: P_t = [P_t] + p_t if you want to express this in term of P_o, p_o, [U_i] and u_i, it comes : p_t = p_o + Rho * u_i * [U_i] and [P_t] = [P_o] + 1/2 * Rho * [U_i] * [U_i] + 1/2 * Rho * [u_i*u_i] This is just a way of writing thing. But this is what is done implicitely when you use the Reynolds Average Navier Stokes equation, in that case you just know average [ ] quantities. So, if you increase the velocity U but not the mean velocity [U], you indeed increase the fluctuating part of the velocity. In term of total pressure, you increase P_t, so [P_t] also grow. Since [U_i] doesn't move, this is done by the convertion of turbulent kinetic energy.

 February 20, 2001, 12:54 Re: PS : Re: Total Pressure: RANS vs EXPT #13 John C. Chien Guest   Posts: n/a (1). Thank you for the detailed answer. (2). I was thinking that for a uniform, steady , turbulent flow through a screen, based on your formula, the total pressure behind the screen will be higher, because the turbulence kinetic energy is higher behind the screen. (3). At a downstream station, the turbulence kinetic energy will be lower because of dissapation, therefore, the computed total pressure will be lower.

 February 21, 2001, 05:28 Total Pressure: RANS vs EXPT #14 Johan Larsson Guest   Posts: n/a When passing the screen, energy is taken from the mean flow and put into turbulence. This will increase k but decrease the static pressure, so that the total pressure behind the screen is <= the total pressure in front of the screen. When the turbulence is being dissipated, the total pressure will decrease, due to the transfer of energy into heat. I agree with Sylvain that the total enthalpy should be constant, but is it really correct to say that the total pressure has to be?

 February 21, 2001, 06:30 Re: Total Pressure: RANS vs EXPT #15 Lars Ola Liavåg Guest   Posts: n/a In laminar flow, the total pressure drops along a streamline as kinetic energy is dissipated to heat, and my perception is that this should also be the case in turbulent flow for the total pressure formulation Sylvain has presented. However, the streamline considered should then be a true streamline of the turbulent flow field. Unless you do DNS, in which case Reynolds averaging is not used anyway, you cannot resolve the direction of a fluid "parcel". If, on the other hand, p_tot is calculated for turbulent flows by inserting mean flow values into the "laminar" expression, the streamline can be thought of in terms of the mean flow, which is solved for in a RANS. Now, the calculated total pressure will drop even as turbulence kinetic energy is generated from mean stream kinetic energy. Which one is more correct? I think that's an academic question, but I can certainly see the practical value of having irreversible loss generating processes, turbulence generation included, being characterised by loss of total pressure. Lars Ola

 February 21, 2001, 08:51 Re: Total Pressure: RANS vs EXPT #16 sylvain Guest   Posts: n/a I didn't mean that P_t should be constant, but I believe that it is easier to understand energy transfers using T_t than P_t. In fact, this two quantities are link together throught the following relationship : P_t = Rho * (Cp-Cv) * T_t + Cv/Cp * 1/2 * Rho * U_i * U_i and in term of mean quantities : [P_t] = Rho * (Cp-Cv) * [T_t] + Cv/Cp * 1/2 * Rho ( [U_i] * [U_i] + 2 * k ) That's why [P_t] decrease after the screen when k is converted into heat.

 February 21, 2001, 09:25 Re: Total Pressure: RANS vs EXPT #17 sylvain Guest   Posts: n/a The fact is that total pressure loss is a good way to identified the location of the dissipative structures in flows. This is done experimentaly behind bodies using pressure probes. In that case, the measurement is averaged in space (due to the size of the hole of the probe) and in time (the length of the sample). For DNS, it is possible to exactly compute the total pressure at any time and at any place. Moreover if the flow is statistically steady, it is possible to compute the mean values (with respect to time) of total and static pressure and of velocity. It should be interresting to compare the different values obtain for various length of the sample in time. Who could do that ?

 February 21, 2001, 14:28 Re: Total Pressure: RANS vs EXPT #18 John C. Chien Guest   Posts: n/a (1). For flow behind the screen, if we define a total pressure state by a local isentropic process, the mean kinetic energy term can be converted into dynamic pressure as usual. (2). Is it possible to convert the turbulence kinetic energy into the pressure term, through isentropic process? The instantaneous state has only the static pressure and the kinetic energy term, (no turbulence kinetic enery term), so, it is possible to define an instantaneous total pressure by converting the static pressure and the kinetic energy isentropically. (3). In the total pressure formula with a turbulence kinetic enery term in it, you will have to assume that it can be isentropically converted into pressure. But in reality, it can only be dissipated. In other words, the turbulence kinetic energy is always irreversible and can not be converted back, while the mean kinetic energy term is possible. (4). The question is: using Pitot total pressure probe to measure the total pressures at the station behind the screen, and at a station further downstream of the screen, will those two measurements the same? Based on the formula derived, they are different. But, what about the measurement? (will the Pitot total pressure probe pick up the contribution from the turbulence kinetic energy term? as the turbulent flow slow down to stand still in front of the Pitot probe, what will happen to the turbulence kinetic energy term?)

 February 22, 2001, 03:50 Total Pressure: RANS vs EXPT #19 Johan Larsson Guest   Posts: n/a To me, the main conclusion to be drawn from this discussion seems to be that one has to be careful when using the total pressure as a comparison between either measurements and CFD, or two different CFD simulations. We have said that turbulent kinetic energy cannot be isentropically recovered into static energy (pressure), but that the kinetic energy of the mean motion can be. What about large scale structures that are not turbulence, but more unsteady features of the flow? Would a pitot pressure tube "capture" those? This seems like a grey zone, and like I said above, I think the main conclusion is that one has to be careful when comparing data!

 February 26, 2001, 05:52 Re: Total Pressure: RANS vs EXPT #20 Dave Hunt Guest   Posts: n/a The discussion seems to be getting bogged down in dealing with what happens to k, when a pitot tube is placed in the flow to measure P_tot. It is stated that mean kinetic energy can be recovered as dynamic pressure but k cannot. However, the concepts of mean and 'k' are the result of time-averaging. The pitot tube only sees the instantaneous flow, which has instantaneous k.e. I do not see any reason why the instantaneous k.e. cannot be converted to total pressure through isentropic change. I think this is the crux of the issue. If what I have said is true, then we get back to Sylvain's original formula for incompressible flow; Instantaneously; P_tot = P + 0.5*rho*U_i.U_i (1) which can be time-averaged <...> to give; = + 0.5* + k (2) Is there any reason why (1) is not true instantaneously? David

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