# Water Tunnel Testing (photos)

I'm starting a new topic so I can edit it later with more content.

I went with some friends to a different lab section that was also playing with the tunnel (my official turn is on Friday).

Things put in:

1. PVC tube -- to show detachment of a cylinder

2. Golf Ball -- to show how dimples create turbulence and keep flow attached longer

3. Tennis Ball -- to show how fuzz acts almost like golf ball dimples

4. Porsche 911

5. Atlantis Space Shuttle

6. Porsche 911-turbo (or possibly a 935 or 993 -- can't remember)

7. Chain Mail Cube -- wtf? why did someone make that?

8. Surf Board Fin

9. Optimus Prime

10. Unknown Porsche (I just can't remember what it was)

11. Batting Cage Ball

The pump was set to 3.55 inches per second. So for a 1/64 scale model (for the cars) - that's just under 13mph equivalence (in water).

My camera had a low battery -- so the shutter was slower than normal. I'll get fresh batteries and a tripod on Friday I'll have some better pictures (hopefully).

As for what I'm bringing in... Still don't know -- but I think I am going to do a comparison of apples and oranges :P

Videos of the 911, 911turbo, unknown porsche and of the overall tunnel in operation ;)

**Update November 3rd**

Items Tested:

1. Plastic Cone (with hole in top)

2. Apple

3. Eyeglass Case

4. Obligatory Altoids Tin

5. Plastic Water Bottle

6. Highlighter Cap

7. Reverse Gear from a manual gearbox (Nissan)

8. Pen Cap thing

9. Rubber Duck

10. Nerf Dart (velcro top)

11. Batman Shield (this was really cool)

12. Welded Plate

13. Golf Ball

14. Guitar Bar

15. Wing Foil

Again, it was set to 3.55 inches per second. I couldn't find an orange on campus (weird considering I'm in central Florida :P). So I'd say an orange is kinda like a golf ball :P

The Bat Shield video is very cool - if you watch just one of them, make it the bat shield ;)

Updated! We couldn't get the spring to work - we just didn't have enough drag force to get a measured result (and we couldn't find a spring with a lower spring constant). I'm also a little disappointed that I couldn't find a stupid orange, but we had some cool results nonetheless. If anyone ever does this for fun in the future. Put two model cars right behind each other so you can see how drafting looks. I didn't get to see it, but I'm told it was pretty interesting to watch.

Select as Best AnswerUndo Best Answer

Remember that you can convert to air velocities by equating Reynold's numbers. I'm getting an airflow of 2.23 mph for the scale model. Which results in an incredibly slow flow for of 0.035 mph over the cars at full scale.

Select as Best AnswerUndo Best Answer

I got 3.33mph air velocity @ scale --- for a full scale of .05mph... still painfully slow and for some reason does not seem completely intuitive... I would have thought that water (with a density around 1000 times that of air) would have had a greater effect on air velocity (the water seems to magnify the effects somewhere around 10X - 20X). Well, that just goes to show... don't guess :P

Select as Best AnswerUndo Best Answer

Well, remember that Re=V*L/nu=rho*V*L/mu, where V=velocity, L=characteristic length, nu=kinematic viscosity, mu=dynamic viscosity, rho=density, and nu=mu/rho. Nu_water~1.4*10

^{-6}m^{2}/s, nu_air~1.5*10^{-5}m^{2}/s for fluids near room temperature.So given a certain object and flow velocity, the Reynolds number for a water flow is about an order of magnitude greater than for air.

The other thing to notice is that for a 1/10th scale model, you need to have 10 times the flow velocity of the 1/1 scale object for flow similarity in the same working fluid.

Just means you modeled a parked car with a slight breeze :-P

Side note: does anyone know how to make Greek characters here?

Select as Best AnswerUndo Best Answer

Hrmmm.. help me find where my error is (I haven't gotten to this stuff in classes yet - but the subject has been briefly talked about)..

I have the same equation -- but my kinematic viscosity for water is different (8.91E-7m

^{2}/s). My kinematic viscosity for air came out to be 1.47E-5My density for water is 999kg/m

^{3}My dynamic viscosity for water is: 8.90E-4 (N*s)/m

^{2}-- or Pa*sI am using a V of 0.09017 meters per second.

I am setting Re_water = to Re_air and leaving V_air as the variable to solve for. L is constant (and therefore taken away) because the object does not change.

So, my Re_water turns out to be 1.01E+5*L - solving for V_air (L cancels itself out) I get (1.01E+5)*(1.47E+5) = 1.49mps which converts to 3.33mph.

EDIT: Okay, found first problem... I used viscosity -- not dynamic viscosity.. New Values (using a new table from my textbook @ 20C):

WaterDensity:997 kg/m

^{3}Dynamic Visc:1.002E-3 Pa*s

Kin. visc: 1.01E-6m

^{2}/sV=0.09017 meters per second

AirDensity:1.204kg/m

^{3}Dynamic Visc:1.82E-5

kin. visc:1.51E-5

Re*L for water = 8.97E+4*L

Solve for V_air = 1.36Emps == 3.04mph

3.06/64 =.048scale mph

I know I'm so close, I just want to get this right (figuring I'll be tested on this within a month or so :P).

Select as Best AnswerUndo Best Answer

Well, my values are slightly different than yours for one. (My kinematic viscosity for water @ 20ºC is 0.987*10

^{-6}m^{2}/s). Actually that's probably the main difference, now I see that I was going too fast and was using 1.4, which is not the correct value. When taking that into account, my result was off by a factor of 1.5, which is the difference between our answers. That's what I get for not paying attention to the columns in my tables.Another thing, I use nu_air=1.5*10

^{-5}for sea level conditions, because it's the value I can remember (and it's close enough for back-of-the-envelope calculations).Also, when working these problems it's often easier just to use kinematic viscosity instead of dynamic viscosity; it makes it one less term to screw up!

Still, I'd like to see the flow over the car in a more realistic regime. And it's hard for me to tell based on your pictures, but see if you can measure drag on it. You could compare your drag coefficient to the value for the car of 0.28-0.32 depending on model.

PS: C_d=Drag/(1/2*rho*V

^{2}*Area)Select as Best AnswerUndo Best Answer

That makes me feel better :P I wasn't what margin of error this type of calculation has to have two answers within a reasonable range :P I didn't see any force gauges :/

Select as Best AnswerUndo Best Answer

Put your object in connected to a compression spring. With your tripod, you should be able to roughly measure the drag force by visual inspection.

Select as Best AnswerUndo Best Answer

I figured that was going to be the case, oh well.

I'm actually more disapointed you can see the vortex shedding on that cylinder.

Select as Best AnswerUndo Best Answer