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October 25, 2000, 23:16 |
help:negative entropy appears in cfd result
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#1 |
Guest
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I am in trouble with my cfd results for a transonic case(Ma(freestream)=0.7,blunt body). The solver is a full N-S equation program. The computational strategy is :Set a homogeneous initial condition,then the body is assumed impulsively put in it and solve the field time accurately. But the results show that in some part of the flow field, Total Pressure Loss(P0initial-P0) is negative. I tried to calculate the entropy change using: S2-S1=Cp*ln(T2/T1)-R*ln(P2/P1); (where "1" denote initial condition) The result also shows that at few points the entropy is negative (about -0.05). My question is:1. Is it possible that the Total Pressure Loss is negative in some part of the flowfield ? 2. What is the possible reason to explain the negative entropy in cfd calculation? How to avoid it? 3. Basically, I want to ask you: how is the Second Law of thermodynamics applied in the Cfd codes to ensure the correct output?
Thanks in advance |
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October 26, 2000, 01:39 |
Re: help:negative entropy appears in cfd result
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#2 |
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(1). The total pressure is a derived variable, at a local point. It is just something you create based on the computed local flow field variables. You can calculate it at any time, and location. (2). When you say , the total pressure loss=(P0initial-P0), You must identify the time and the location for the P0initial and the P0. From your definition, it is difficult to know your definition of the total pressure loss. All you can say is the total pressure difference. What I am saying is your P0initial should be P0initial(x,y,j,t) or something like that. The same applies to P0(x,y,z,t). Notice that P0(x,y,z,t) is derived from the local flow field as a function of (x,y,z,t). (3). Before you can say "loss", you also need to define it clearly first. The loss is usually applied to the same collection of fluid, as it entering and leaving the domain. So, your definition can at best be described as the total pressure difference. (4). Whether the local total pressure will increase or decrease, it depends on the problem or the system you have. (5). My suggestion is: instead of focusing on the derived variable and the difference, make sure that your flow field solution is accurate and reasonable. If you have errors in the primary dependent variable, then you will get errors in the derived variable. (6). The other possibility is: your definition simply does not make any sense. Then you will get strange results. So, focus on your definition of the total pressure loss. (Does your definition make sense?)
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October 26, 2000, 04:59 |
Re: help:negative entropy appears in cfd result
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#3 |
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It looks to me like you are computing a steady state flow with transient time-stepping. Am I right? In that case, how do you define the ptot difference? Are you looking at differences over time or spatially along the flow? In a steady state flow without work or heat input, the total pressure should always drop along a streamline. However, I am not sure if it needs to drop between subsequent time steps, although it seems to make sense in your case.
Are you confident that your temporal and spatial resolution is adequate? What differencing schemes are you using? Are there any indications of numerical problems/inaccuracy, e.g. exessive number of PISO correctors or SIMPLE iterations per time steps? How is your linear solver convergence in the different variables? Also note that if you define your total pressure in terms of the mean velocity, the generation of turbulence implies a drop in total pressure, but if you define your entropy in terms of mean pressure and temperature, it will increase as turbulence is dissipated, not generated Regs. Lars Ola |
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October 26, 2000, 12:04 |
Re: help:negative entropy appears in cfd result
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#4 |
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Hi rb,
why do you say that entropy cannot become negative? The thing to check is whether the production of entropy is non-negative. i.e. write the whole control-volume balance of entropy and calculate all terms except the production term and see if their sum is of a consistent sign. regards, chidu... |
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October 26, 2000, 12:38 |
make it clear
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#5 |
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I want to make it clear about my definition of P0. P0 is calculated from the local primary variables P(x,y,z,t),rho(x,y,z,t) and v(x,y,z,t).It's an imagined stagnation value at each point in flow field. Since in my case,P0initial is equal to P0 at farfield point. So P0initial-P0 can be interpreted either as the difference at the same point and different time or difference following the path of a paticle.
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October 26, 2000, 21:02 |
Re: help:negative entropy appears in cfd result
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#6 |
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(1). You can either lower the Mach number (free stream) or change the geometry of the body with a less blunt or smaller nose, to see the effect. (2). You can actually run a flow over a flat plate case first to identify the location and the source of error. (3). The mesh distribution and the Reynolds number are also other important issues, not to mention the turbulent models (if applicable). Do some parametric studies in these areas first.
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October 27, 2000, 01:14 |
Re: help:negative entropy appears in cfd result
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#7 |
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Hi respondents,
I had posted a similar question sometime back regarding total pressure increase in transient flow development in a blow down wind tunnel. I did get one response at that time but it was not satisfactory. I now feel that an answer of sorts is known. Firstly, we do know for certain that total pressure is preserved only for steady inviscid smooth flow, along a streamline. We also know for sure that in unsteady inviscid flow, DH/Dt=del(p)/del(t), clearly implying that total enthalpy is not constant even for a particle in transient flow. Obviously total pressure cannot be expected to be so. Introduction of a cylinder at t=0 in an uniform flow amounts to bringing the flow impulsively to rest at t=0 which implies an energy input to the system. Obviously the total pressure of particles which are suddenly brought to rest by the introduction of the boundary will exceed that of the particles far upstream. I think this explains rb's observation in the transient regime. Alternatively other initial conditions may be chosen where this argument can be put to test. As for negativel entropy, I think Chidu's response is essentially correct. Ravi |
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