# Confusion abt expt / non-dim simulation result comparison

 Register Blogs Members List Search Today's Posts Mark Forums Read

 October 27, 2015, 11:15 Confusion abt expt / non-dim simulation result comparison #1 Senior Member   TWB Join Date: Mar 2009 Posts: 125 Rep Power: 9 Hi, I have got an expt result e.g. a wing in a wind tunnel with inflow. Supposed the inflow (U) is 2m/s and the chord of the wing (c) is 0.1m. Density (rho) is 1.18kg/m3, viscosity = 0.002 and wing planform area = 0.02m2. So ref velocity and length are U and c. I can get the thrust, lift forces and coefficient from the expt. Now, the NS solver has been non-dimensionalized, using the chord and inflow as the reference length and velocity, I got Re = 118. pressure (p) and time (t) are non-dimensionalized using rho*U^2 and c/U. From the simulation, I can get the non-dim force (nd_F) as output. So how can I compare with the actual expt results? How do I convert non-dim force to dimensional force (F)? I know F = p*area and F = nd_F*(ref_force) Ref_force = (rho*U^2)*ref_area. But what is the ref_area? That's where I am stuck. Is ref_area = area / c^2? Thanks

 October 27, 2015, 12:34 #2 Senior Member   Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,484 Rep Power: 22 Are you asking what the reference area should be defined as or are you asking what the reference area in your black box code is? The reference area needs to also be explicitly stated. I don't know if this matters but experimentally, the reference area is commonly taken as the planform area for the lift coefficient and cross-sectional area for the drag coefficient. But this definition also varies as sometimes the same area is used for both coefficients simply for convenience. It sounds like you have a 2D non-dim N-S solver. Unfortunately I don't know how your solver treats this reference area. The reference area would have come into play to calculate the non-dimensional force. Different codes treat the missing 3rd dimension differently (some assume 1m depth, other codes accept an explicit reference area). Basically, you need to look into the details of your solver to see how it computes the non-dimensional force. Or do you have access to the solver and are trying to specify the reference area?

 October 27, 2015, 23:01 #3 Senior Member   TWB Join Date: Mar 2009 Posts: 125 Rep Power: 9 Well, I have a non-dim NS 3D solver. The forces are computed by taking the surface pressure/stress and multiplying by its area. All of these quantities are non-dim. In the code, chord of the wing (c) is 0.1m and that is the reference length. So in that case, can I use non_dim_area = area / c^2? In that case, the ref area = c^2. Or it is up to me? say can I choose ref area = planform of wing? However, in my non-dim NS eqns, there is no ref area req. I only need ref length and velocity.

 October 27, 2015, 23:30 #4 Senior Member   Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,484 Rep Power: 22 No I was confused about the solver. If x and y coordinates are both non-dimensionalized by the chord length then the reference area is indeed c^2. x_nd = x/c y_nd = y/c dA=dxdy dA_nd = dx_nd*dy_nd = dx/c*dy/c = dA/c^2 dF = PdA dF_nd = P_nd*dA_nd Then dF = dF_nd*rho*U^2*c^2 If your N-S is non-dimensionalized by Lx and Ly then the reference area is Lx*Ly.

 October 28, 2015, 10:49 #5 Senior Member   TWB Join Date: Mar 2009 Posts: 125 Rep Power: 9 ok that clarifies the confusion. then I have another problem involving my 2D NS solver. Supposed I do an expt about flow past a cylinder which is infinitely long, so this is essentially a 2D problem. The inflow U = 2m/s, diameter d = 5m, viscosity v = 0.1 So Re = 100. From expt, I get cd = K (~1.59). If I repeat my expt with U = 4, d = 2.5, v = 0.1, should I also get cd = K, since Re = 100? Or any other combination of values, so long as Re = 100, cd = K, right? Then I run a 2D simulation of the flow past circle (2D cylinder) using a non-dim NS solver ref U = 2 m/s, ref length = diameter = 5m, pressure p = rho*U^2 Re = Ud/v. I simulated a circle of nd_d = 1, Re = 100 and I got using the above rules, cd = nd_F*2 and I got cd = K. However, I tried to use circle of nd_d = 0.5 but Re = 100. Now my cd = J, not K(exact values J = 0.87, K = 1.59). But I thought I fixed my Re = 100. So is there something wrong with my NS solver? Thanks

 October 28, 2015, 12:46 #6 Senior Member   Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,484 Rep Power: 22 So do you have a 2-D or 3-D solver? I just want to point out that in a 2D solver, there is no natural characteristic length for the third dimension. This is normally not important except when you try to compute forces. Hence for a 2D solver the reference area is arbitrary. If you have a 3D non-dim N-S solver it's not a problem because there is a reference length but it's very important for a 2D solver. Anyway, assuming you are able to get a non-dimensional and dimensional force correctly: What did you change for your "nd_d=0.5" simulation? Is it d = 2.5 mm and U = 2m/s? Remember that you always scale by the reference velocity and length of the new problem. nd_u = u / uref nd_d = d / Lref And uref and Lref are always the velocity and diameter of that problem (not any other problem). In other words, you cannot run a nd_d = 0.5 case, you can only run a nd_1 = 1 case. That's what makes a non-dimensional solver non-dimensional is it takes the physical dimensions (u & d) and puts them always into the same non-dimensional variables (nd_u=1 & nd_d=1). What happens is Re is different depending since u & d will be different from before. Last edited by LuckyTran; October 30, 2015 at 21:36.

 October 29, 2015, 11:25 #7 Senior Member   TWB Join Date: Mar 2009 Posts: 125 Rep Power: 9 Ok I got it, although I'm still trying to sort it out. Btw, can I choose the ref length as radius (2.5m) instead of diameter, the others remaining the same? But then Re becomes 50 and nd length = 1 now. Then it is not correct? Or can I choose ref d = 2.5m and ref U = 4? Then Re = 100 and nd_d = 2 and nd_u = 0.5. Is this also wrong? If that's the case, does it mean choosing the wrong ref value will give the wrong ans?

 October 30, 2015, 22:02 #8 Senior Member   Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,484 Rep Power: 22 The reference length is arbitrary. The difference is that the Reynolds number will be different. Just know that Re_d = 2*Re_r

 November 1, 2015, 09:03 #9 Senior Member   TWB Join Date: Mar 2009 Posts: 125 Rep Power: 9 Thanks for the clarification LuckyTran!

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Kushagra CFX 2 July 8, 2008 21:14 amine CFX 3 June 14, 2008 05:01 sham81 CFX 3 March 22, 2004 17:41 Mason CFX 0 February 22, 2004 08:54 mounir Main CFD Forum 4 March 3, 1999 13:07

All times are GMT -4. The time now is 11:16.