
#1
Dec413, 04:05 AM

P: 2

Hi folks,
I'm currently in an internship at Cassidian, and my supervisor has asked me to calculate the velocity of a fluid. I didn't spend too long on the subject, maybe an hour or two, but I wasn't particularly busy that day. I concluded that it's impossible because not enough parameters were supplied. Parameters available are: *Total, static and dynamic pressure *Mach number *Reynolds number That's it. No temperature information, no density, kinematic viscosity, mass flow rate or area. To make it worse, this was a closed section test. So I can't even estimate ambient conditions. Also, I wasn't present at the test itself. It's been a bit quiet, so I've had some time to ponder and research this problem, and I can't find a way to do it. Is my initial conclusion that this is not possible to solve, correct? Confirmation or an explanation (if it is possible) would be greatly appreciated. Cheers, Matthew 



#2
Dec713, 05:31 PM

P: 136

I thought Cassidian is only dealing with software, IT and all these network/information security stuff...
Did your supervisor tell you which fluid you're using? If he did, or if you know somehow, then you DO know the density. I doubt it if you're gonna need that. It seems to me that you have plenty of data. For example if the test was conducted in a test facility somewhere in the UK (I don't know, just saying...) then you do have some information for the altitude and assuming ISA conditions you do know the temperature... You don't even need that... You do know the total pressure so, again assuming ISA conditions, you can find temperature and the speed of sound for these conditions. Now, since you know the Mach number, you can just solve for the fluid's velocity. 



#3
Dec913, 08:35 AM

P: 2

Aero_UoP,
Surprisingly they actually do a lot more than just IT related stuff, they just aren't very well known (hence the planned name change to Airbus Defence and Space). ISA was one of the first assumptions I made, however the test was conducted using a closed circuit wind tunnel, so my supervisor didn't think it was correct. I am assuming the fluid is air. Thanks for your reply, Matthew 



#4
Dec913, 11:47 AM

P: 136

Determine velocity of a fluid  not enough info given?
Yeap, I know about the whole rebranding and renaming thing ;)
EADS held an event here a few weeks ago... Cassidian focused only on the info security stuff. Can you please give us all the available parameters (specific values) and tell us how the problem is exactly stated? 



#5
Dec1013, 09:13 PM

P: 362

You can write the Mach number in terms of static pressure and density by using the equation of state for a perfect gas assuming the gas is a perfect gas. Then you have equations for Mach number and dynamic pressure where your unknowns are density and velocity. 2 equations, 2 unknowns.




#6
Dec1113, 10:45 AM

P: 136





#7
Dec1113, 11:17 AM

P: 362

Solve the equation of state for temperature and substitute this into the equation for Mach number thereby eliminating temperature.




#8
Dec1113, 02:12 PM

P: 136

He doesn't have the density :p




#9
Dec1113, 03:23 PM

P: 362

Read my original post. He doesn't need density since he has static pressure and dynamic pressure.




#10
Dec1113, 04:00 PM

P: 136

I would, and I'm sure Matthew would too, appreciate it if you could right down the solution process with all the equations...




#11
Dec1113, 04:34 PM

P: 136

Meanwhile Matthew, I think I found the solution, provided you know the characteristic length of the tested body.
Re=[itex]\frac{ρuc}{μ}[/itex] [1], where c the characteristic length. M= [itex]\frac{u}{α}[/itex] hence u=M*α [2] [itex]\frac{1}{2}[/itex]ρu^{2}=K , where K is the known value of the dynamic pressure. Hence, ρ=[itex]\frac{2K}{u^{2}}[/itex]. In this relation replace u^{2} by [2]. You get: ρ=2K/(M*α)^{2}. Put this in [1] and replace u by [2]. What you get is: Re=[itex]\frac{2Kc}{μMα}[/itex]. Replace α=[itex]\sqrt{γ\frac{p_{s}}{ρ}}[/itex], where p_{s} is the static pressure and μ from Sutherland's Law: μ=μ[itex]_{ref}[/itex]([itex]\frac{T}{T_{ref}}[/itex])[itex]^{\frac{3}{2}}[/itex][itex]\frac{T+110}{T_{ref}+110}[/itex]. Now in Sutherland's Law replace T by the equation of state T=[itex]\frac{p_{s}}{ρR}[/itex]. You'll see that the only unknown in this function is the density (provided as I told you that you know the characteristic length). Solve for density. Use the equation of state to find the temperature, put it in the formula for the speed of sound and then, finally, you get the velocity from the Mach number equation. 



#12
Dec1113, 05:49 PM

P: 1,438

Consider that he knows the following:
Mach number: [tex]M = \dfrac{u}{a}[/tex] Dynamic pressure: [tex]q = \dfrac{1}{2}\rho u^2[/tex] Reynolds number: [tex]Re = \dfrac{\rho u L}{\mu}[/tex] plus static pressure, [itex]p[/itex], and total pressure, [itex]p_0[/itex]. He also knows these standard relations for a perfect gas: [tex]p = \rho R T[/tex] [tex]a^2 = \gamma R T[/tex] You can easily use those two standard relations to find Mach number in terms of density and velocity (two unknowns in this case): [tex]M = \dfrac{u}{a} = \dfrac{u}{\sqrt{\gamma R T}} = \dfrac{u}{\sqrt{\frac{\gamma p}{\rho}}}[/tex] and then change that over to [tex]\boxed{u = M\sqrt{\dfrac{\gamma p}{\rho}} = M\sqrt{\gamma R T}}[/tex] so there are two unknown variables, velocity and either density or temperature, though I'd argue it is much easier to measure temperature than density, so you are better off leaving it in terms of temperature. I know of no way to eliminate that variable. If you try to relate density to the Mach number and static pressure using the definition of dynamic pressure, you come up with nothing. Consider that relationship: [tex]q = \dfrac{1}{2}\rho u^2 = \dfrac{1}{2}\gamma p M^2.[/tex] You can try a number of ways to use this to eliminate density from the boxed equation but you always end up just eliminating both it and velocity. That's because the equality [itex]\rho u^2 = \gamma p M^2[/itex] reduces directly to the boxed equation anyway. You can reduce it to 2 equations for 2 unknown variables but those two equations are not independent of one another. You could do what Aero_UoP said and use Reynolds number and Sutherland's law with the boxed relation to either explicitly solve for density and then velocity or implicitly solve for velocity directly, but in practice, just put a thermocouple in your test section next time. Might I also ask how Reynolds number is being "measured" anyway without knowing density and viscosity already? That seems like nonsense to me. You can't measure Reynolds number directly; you have to calculate it from other quantities, which typically means knowing the temperature since you can't measure density directly either. To me, this implies that somebody knows the temperature already and your boss just doesn't know that. 



#13
Dec1113, 05:55 PM

P: 136

Indeed, this is kind of odd especially in a wind tunnel facility where you want the temperature approximately constant.




#14
Dec1113, 05:58 PM

P: 1,438

That depends on the situation. Most situations you want to hold the Reynolds number constant, but you can't typically do that without knowing the temperature. The only time you also want to control temper is if your phenomenon being investigated is temperature sensitive, in which case the whole process becomes a royal pain in the butt.




#15
Dec1113, 06:11 PM

P: 136

Yeah, I had in mind the ARA's transonic wind tunnel in Bedford, which is cooled for that reason ;)
http://www.ara.co.uk/services/experi...cwindtunnel/ 


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