A "Rather Small" Fluid Flow Question...

Flow rate from distal end = flow rate into proximal end.

@Doorman:

"Flow rate from distal end = flow rate into proximal end."

Should I rephrase the question for you?

What's this--a hydraulic whistle?? It's got "boundary layer" and "supersonic" written all over it, so I wouldn't trust any equations. Just build it, and they will come! (Bucket and stopwatch project.)

I'm going to have to side with Tornado here. You have some pretty extreme conditions and the typical first semester fluids equations your looking at will likely lead you in the wrong direction.

Let me convey a short story though... An operator at a power plant noticed a strange hovering gray/black cloud, naively he reached to touch it only to have his fingers neatly severed. That was a pin hole leak from a hydraulic system running at 2500psi.

What you're contemplating will cut (more than fingers)... if you do build it, be VERY cautious.

If you want pointed in the right direction, I would start with the Navier-Stokes equation.

@Tornado & ChaoticIntellect:

Thanks so much for the "heads up." Thought there might be a tuffy lurking in here; but I was of the persuasion that we'd be looking at more of a dribble effect than a water laser...

Indeed, as a minor digression, and while I'm not interested in putting anyone's fingers in harm's way (just a thought problem right now), shouldn't the viscosity value/channel length which we're looking at here militate against the emergence of a coherent stream? In effect, wouldn't this type of scenario more likely act as a practical impasse for the working fluid (only a few drops/sec)???

(I always save the best ones for CR4 )

Thanks again for riding along --

I don't have much experience with "choked flow" calculations, nor am I sure that the formulas used for relief valve system calcs would include this unusual geometry.

If you were to graph flow vs pressure, the result might even be chaotic rather than continuous, such as almost zero flow at various system resonances/harmonics.

Standing waves? Organ pipes? Handel's Oil Music?

I'm second guessing myself now... It has been a while, but there are some relatively simple differentials I have used to solve for estimating leakage across a typical piston seal. I'll dig it up tonight and look what the assumptions are... you might be right in that its a dribble. I just saw the 4000psi drop in a one inch length and just assumed....

Thanks again, folks

@ChaoticIntellect:

Thanks for the interest. I thought the leakage path length would have some play here. Looking forward to your further thoughts!

@Tornado:

Perhaps Handel's Messiah would be a better bet, as it might turn out that only God will know: The rest of us, as you've already said, will need to just grab a bucket and see

Cheers!

A good look at the basic formulae will give a pretty good start to the answer - is it to be a dribble or a "water jet" or somewhere in between.

Look up Reynolds number in Wikipedia and start from there. The careful use of related items like the Moody diagram and simplified formulae for pipes, including Darcy-Weisbach and Hagen-Poisuille (N-S already mentioned) will get you close to the right answer. You might need to do a "guess and calc approach" as a start.

Your problem is best initially tackled as "two parallel plates", since the duct is much wider than it is deep - and a quick and dirty start is to take the friction drop for that as half that for a square duct of the smallest dimension eg. 0.002 inch. The head drop will be the same for both cases.

The complexity of this problem is that, like many fluids problems, it is a simultaneous equation deal, and where there are end effects and possibly non linearities as well.

If your solution velocities are greater than the velocity of sound in the medium, then you are looking at choked flow, and the rules change a bit !!

At the entrance for example flow my well start laminar and then go turbulent as it proceeds along the duct. That "characteristic distance", in Reynolds, is distance from the start as to where turbulence develops, and distance between the walls in the case of fully developed flow (i.e. in a long pipe). All very confusing.

But if you start with some assumptions as to (say) velocity and do a head loss calculation (due to vel change) and a friction loss calculation (due to fluid friction) you will start to see convergence - the total of these two losses will ultimately have to equal the applied inlet pressure. The only easy thing about this problem is that you won't have to worry about "potential loss" due to height change.

I would tend to equate this with a crack in something thick, which, at that pressure, are quite dangerous. I have never actually measured a crack, but almost invisible ones let out a lot of oil, especially if hot. I doubt it will be a 'trickle'. I don't know how you would measure the 'exit pressure' either, but I think you need to get well below .002 to see large "line drop".

If you are thinking of testing this, do make the 'duct' ridiculously sturdy and out of hard material, or the .002" won't be .002" due to distortion and/or erosion.

Start with Hagen-Poiseuille equation:

V* = p*B*H³/(12*eta*L) fluid flow if non-turbulent in m³/s

B = L = 1" = 0.0254 m H = 0.002" = 51 µm = 51*10^-6m

eta = dynamic viscosity (N*s/m²) = kinematic viscosity (m²/s)*density(Kg/m³)

1cSt=10^-6m²/s

9.4cSt will give 9.4*10-6m²/s*900Kg/m³ = 8.5*10-3Ns/m² =eta

p = pressure 4kpsi = 2800N/cm² = 28*10⁶ N/m²

V*=28*10⁶*(51*10^-6)³/(12*8.5*10^-3) = 28*10⁶*1.3 * 10^-12 m³/s = 36*10^-6m³/s

or in cm3/s better adapted to our feeling: V* = 36 cm³/s

Channel cross section is 51µm*25.4mm = 1.2 mm²

Fluid velocity v = 36000/1.2 mm/s = 30m/s

Reynolds number Re = density*gap-width*velocity/viscosity

Re = (900Kg/m³*25*10-³M*30m/s)/(8.5*10^-3Ns/m²) = approx 75,000

So this is highly turbulent (as naturally expected).

Now assume an apparent viscosity/Reynolds-number function or look in the tables there are some data. As a rough first estimate I would assume from Re=75 to Re = 75000 or a factor of 100 in Re giving a factor of 30 in apparent viscosity.

So now recalculate with eta=function of Re and you will get a sound estimate of the flow rate.

Caveat: this was done in early morning without the first coffee and purely by memorising the different relations! So please check everything if really ok.

Have a Happy New Year

RHABE

Rhabe... You beat me to the function (this was the one I was thinking of). One must be careful in it's use though. The assumptions that are required to use that formula are ... 1) Steady flow 2) incompressible 3) fully developed and 4) laminar flow.

As with any calc using that equation the Reynolds number must be checked to ensure the assumptions were valid. Assuming your calcs are correct, the Reynolds number does not show laminar flow therefore invalidating assumption 4 making the equation invalid.

There are 2 ways to look at a problem :

-think how complex it is and do nothing and

- try a way to estimate how to solve it.

The second requires an organised Intellect.

The numbers are correct, before reading what RHABE wrote I went an other way computing the pressure limit drop for still a laminar flow. It is about 0.8...1 N/mm².

Which leads to same result direction : turbulent flow so that the usual formula for turbulent flow can be used. The hydraulic diameter of the section can be computed as dh= 4*A/P where A is the area of section and P its wetted perimeter. With this the flofriction can be estimated and because it is a short channel the entry and outlet flow coefficients have to be considered. If I am not (wrong it is an evening computation after a full day) then the flow is about 0.2 dm^3/s.

As long as we're using assumptions that are invalid, let's go ahead and make one more just for a bit of fun...

Let's assume... 5) inviscid flow

Now, with assumption 5, I can simply apply Bernoulli's equation and here's what I get...

an exit velocity of 817 ft/s yielding a flow rate of 5 gallons per minute out of that wee little crack.... IMPRESSIVE!!!

I say we go with that!

I say there are 2 solutions to a problem...

- a wrong one

- a right one

The second requires organized intellect.

The Re number far exceeds laminar flow - no doubt.

But beyond laminar flow apparent viscosity is rising roughly by square-root of RE.

Apparent viscosity is used as if laminar flow would be existing.

One problem: the transition between laminar (Re=1..?), first vortices (re=40) up to fully turbulent Re=1000...2000 is not really well known and a function of many other parameters. (Roughness, edges upstream, obstacles - also very small ones ...)

So I assumed this transition to be near Re = 75 (to be calculated as a corner frequency in Bode plots).

Your point 3 should be fulfilled as length to gap is 500.

Incompressible flow: the pressure here is "only" 280 bar. Compressed density at entrance may be 20% higher than without pressure.

RHABE

Oh my goodness! Where to start

Well, it looks as though God just might be holding the winning hand here; at least as far as a precise answer goes...

In that spirit, I'd love to dig in & start with TrevorM's roadmap; but I'm afraid, with the unseasoned guessing and assumptions which I'd likely need to make along the way, I would bumble into a result which looks more like Picasso than Michelangelo Indeed, Leonardo would not be pleased...

Nonetheless, thank you much for the fine thoughts and gold nuggets. I've gathered it all into the bag here--

@34point5: Thanks for the tips.

One back: Two steel blocks with two feeler gages clamped therebetween for side walls. That'd make the "duct;" but we'd still have to supply the monster with juice (haven't figured out how that would happen yet ).

A thought problem worthy of the forklift modification thread?

(just kidding)

@RHABE & ChaoticIntellect:

So, what I might glean is that, due to the particular assumptions here, if RHABE's calculations were brought into play we'd get a result which represents a flow rate that may be HIGHER than actual modelling would produce?

Forgive me if I have this skewed in the wrong direction, but that is what I seem to understand. Please let me know if my "trick knee" is wrong...

Thanks again --

BTW: Happy New Year to you all!

So, what I might glean is that, due to the particular assumptions here, if RHABE's calculations were brought into play we'd get a result which represents a flow rate that may be HIGHER than actual modelling would produce?

Great question. It is one I have never asked. The method I was taught was to try a different method if the assumptions turned out to be false. We never tried to interpret the results of equation with invalid assumptions. I'll have to ponder that.

"Two steel blocks with two feeler gages clamped therebetween for side walls. That'd make the "duct;" but we'd still have to supply the monster with juice (haven't figured out how that would happen yet)"

There should be no flow problem tapping in at 90⁰, provided the port diameter is greater than 1" circumference by the 0.002 height. So say a 1/2" drill, or counter sink or counter bore

You could mill locally to produce a tangent step, a few thou deep to give you a finite 1" to the block edge.

Structurally though, rather than vandalizing feeler gauges, buy shim. That will enable you to cut out the 1" slot of x depth, surrounding the feed port and "duct" with a "C".

I would be bolting through a full face of shim to assure the blocs are not distorted and have 'heel' support.

The combined area of duct and port means potentially around 6000 lbs of separating force. Not much in steel terms, but given you want an inelastic duct, I'd think two 4" squares of 1½" plate, and, as you really want a ground finish and flatness for this 'sandwich', you'd probably end up with 1¼ finished t.

This 'full shim' approach also means you can change the shim thickness and slot width, and maintain repeatability for a given shim/setup.

Yokey dokey.

Well, in my vast ignorance and present confusion, and, running out of fingers and toes a couple of minutes into things after cranking up the precision beyond belief, I came up with

V* = (about) 10.04 cm³/s

based upon an apparent viscosity of 30cP (0.03 N*s/m²) for eta...

I did this as just an exercise: Am I hip to what we're trying to do here so far?

Thanks again for all the input

Assuming laminar flow gives lower volume flow, greater pressure loss than actual (turbulent) flow

@34point5:

Great idea!

Don't see a future market in this for us now; but, who knows: A little press, some time on the Discovery Channel: Hey, the sky's the limit

Next, I propose to push the stick into the hive and see what comes a' shakin.

Starting with the assumption that RHABE's usage of Hagen-Poiseuille is what we need to get the job done (yes?), and, considering that I don't have a life, I decided to reduce the clearance to .001 just for fun:

V* = p*B*H³/(12*eta*L)

V* = flow in m³/s

p = pressure in N/m²

B = width in m

H = height in m

eta = dynamic viscosity (N*s/m²) = kinematic viscosity (m²/s) * density(Kg/m³)

L = length in m

...

p = 27579029.172713

B = 0.0254

H = 0.0000254

eta = 0.0000094 * ((1000000 * 0.89) / 1000) = 0.008366

L = 0.0254

...

V* = (27579029.172713 * 0.0254 * 0.000000000000016387064) / (12 * 0.008366 * 0.0254)

V* = 0.0000000114792586292223206096528 / 0.0025499568

V* = 4.5017463155541774706351103673599e-6 m³/s

V* = 4.50 cm³/s

...

Barring the compulsively annoying level of precision which I posted and any assorted math errors along the way, the above should be only really wrong; as the apparent viscosity is not factored in.

If this is all that's off, could someone give me some rough guidance as to how one might go about estimating the apparent viscosity for eta here?

As a footnote, here's what I got for Re through a .001" passage:

Re = (890Kg/m³*25.4*10-³M*6.98m/s)/(8.366*10^-3Ns/m²) = approx 18,900 (n)

However, still a wee bit lost at this juncture:

"Now assume an apparent viscosity/Reynolds-number function or look in the tables there are some data. As a rough first estimate I would assume from Re=75 to Re = n or a factor of x in Re giving a factor of y in apparent viscosity."

Lovin' every minute of it

Thanks bunches --

Since I was not any more sure about the numbers I made once more the computations, in fact I had an error:

p= 2.758 E7 N/m² b=l= 25.4 E-3 m h= 5.08 E-5 m ν = 9.4 E-6 m²/s ρ = 890 kg/m^3

η = 8.37 E-3 Ns/m² Q= p*b*h^3/(12*η*l) = 3.601 E-5 m^3/s = 2.16 dm^3/min

A=b*h= 1.29 E-6 m² w=Q/A= 27.9 m/s dh= 4*A/P= 1.01 E-4 m Re= w*dh/ν =301

The tube can be considered as a round tube with a d=dh and the friction factor can be read at Re= 300. As I mentioned the inlet and outlet losses have to be considered since the tube is short.

I would like to add some remarks. The viscosity value gave a rough indication with respect to the fluid (oil SAE 20 or some similar), the bulk modulus for oil at 80 °C and an air content of 1% which usual in the industry is E oil=730 N/mm², without air as solution in the oil the bulk modulus goes at same temperature higher (E oil=1123 N/mm²). The "compressibility" of such a fluid at the indicated pressure is thus in % ΔV/V= 100*p/E oil → 100*p/E oil= 2.46...3.78. It is not so important to consider it for such a computation.

It is not always necessary to make the problem more complex than it is.

My immediate reaction to your post, without checking your maths, is that:

(1)treating the wide slit as a "round" pipe will give a conservative result (i.e. lower flow) - essentially because the "side drag area" in a pipe is a bigger proportion of the cross section. BUT it will be close enough for a rough first approximation - as I already posted.

(2)the end effects are probably minimal (and less than that referred to above) because the length of the duct is quite large compared to the smallest cross section dimension, i.e. by times 50-100.

I'm struggling with this "round pipe will give a conservative result (i.e. lower flow)"

My might be wrong guesstimate was, this duct area is equivalent to a 0.025 dia pipe (even at an inch long, attached to 4000 psi, I would call that 'leak' a premium finger cutter)

But Shirley the boundary layer is Pi d of 0.025" = 0.078" - compared to 2" and a bit?

Even if it was a molecule thick, due to the shear, 2" has got to be more drag area than round.

Or am I reading this all wrong?

I see that you are not familiar with the "hydraulic diameter " notion. It is defined as

dh= 4*A/P for a round section it is dh= 4*π*d²/4/(π*d) = d.

For a square section it is dh= 4*a²/(4*a) = a and for a rectangle dh= 4*b*h/(2*(b+h)), if b/h>>1 then the value is near to dh= 2*h

"dh" is used to approximate sections which are not round with a round section since most if not all experimental results in flow coefficients are related to round sections. It used in computing the Reynolds number for any kind of geometries (Re= w*dh/ν).

It is very often used in Europe, I do not know if and how it is used in other parts of the world.

The "round pipe" approximation is ONLY for estimation of the friction coefficients in the different graphs or with the different equations.

I agree with the relative length correction, I looked at absolute values and neglected the "relative". Good remark.

Have a Happy new Year!

I am familiar with hydraulic radius, but am certainly not an expert.

Does not your example actually prove my point. i.e. that the "effective dia" of a round pipe is "d" and for a slit is (essentially) 2 * d. This means that in respect of (for example) drag force/s (and so too pressure) these will be higher in the round pipe for a given flow through a unit area of cross section.

The visual model to explaina this would be to consider the drag force on a small piece of fluid in the middle of a square pipe. It will experience shear generated drag from the top, bottom and side "walls" of the pipe. If the pipe is very very wide (i.e. a slit), it will only experience significant drag from the top and bottom "walls".

Nice. GA

I'll be in touch in 2011. I have a very related situation and need a bit of a hand out (brains out). My situation is immensely more complicated and will have to be resolved in a true hard ware situation in the end. I think the same goes for the OP, you can only do so much on paper.

I have been following this thread for a while and thought there would be some clues. There were but all in the common knowledge range, department.

I'll be in touch once I get to that bridge, maybe even here in CR4. Some of the old timers here are not with us anymore which is a shame. PM's will have to be involved so not to drift off too far. I will also have to phrase my question properly and just that will take more time than I can afford at the moment.

See you and all the best, Ky.

There are some remarks to be done:

- there are 2 frictions : at the wall and between layers. Friction is τ=η*dw/dy where "y" is measured normal to flow direction, η being the dynamic viscosity. At the wall a boundary layer will appear since velocity at wall is nil.

- "dh" indicates the influences of the wall friction (thus the P)

- if you look at the velocity graphs in the comment following yours you will notice that for a turbulent flow the inter-layer friction is very little in the middle of flow. An indication should have been given in the graph: the velocity profiles are "relative" to the maximal velocity. The maximal inter-layer friction occurs in a laminar flow. If you consider the velocity profiles and above equation which applies to boundary layers as well it is clear that the higher the velocity the higher the "dw/dy" value and thus the higher the friction losses at the wall level. In fact there are 2 aspects one is the velocity growth and the other is the due to a boundary layer thinner when velocity increases.

- in a pipe the boundary layer appears all around. The friction (and correlated pressure DROP per pipe length unit) will be less in a larger pipe than in a narrower one since the ratio P/A decreases when d increases. This means there is less wall friction for the flow which means that pressure drop will decrease if specific friction at wall is the same.

@nick_name:

Thanks for the thoughtful work in this. But, 2160 cm³ per second?! I must be missing something really huge (it has happened once before) --

So, in my limited faculty, here's what I can understand so far:

The guy from Skunkworks got his math mixed up.

Germany says, "Watch it leak."

France says, "Run for cover."

America says, "Build a rig."

God says, well, you know, He's got to be smiling anyway . . .

I say,

(sigh)...

And, in any event, have a great New Year, everyone.

@all:

Fabulous thread!

Been passing out goodies for the "cut-&-try" input; but haven't yet gotten my bearings on the winning pathway for the theory/math end of things. I have a feeling that, despite the widely varied talents/experience levels in play here, I'm not alone (although, for what little it's worth, my instinct tends toward the flow rates yielded by the Hagen-Poiseuille example thus far)

So, to be fair, I've popped open my lawnchair to watch and learn. I believe that might be my most valuable "contribution" for a bit. Truly, I had no idea we'd wind up with such a widely-varied spread when I put together the OP!

Thanks once again to all of you --

OK, so I asked myself what DP is required to maintain the assumptions valid for the equation that RHABE introduced. I was surprised to find my calc above your stated parameters.

I used MATHCAD to double check and sure enough, the flow in your situation IS laminar and therefore the solution using that equation is valid (all 4 assumptions made are satisfied).

My results...

Q= 0.56 gpm (35.6 cc/s)

v= 90.5 ft/s (27.6 ft/s)

Re= 149

All of which agree with RHABE original calc, with the exception of the Reynolds number. I opted to use the 0.002in value for the "D" in the calc, which accounts for the difference. I believe this is valid since the original derivation for the governing equation to find the flow is in terms of a "per unit width" of the plate, making the "D" value for the Reynolds number the distance between the plates.

@ChaoticIntellect:

Awesome. In consideration of the fact that we're hearing crickets in the room after your latest post, it seems as though the Germans did it again

Viva la MathCad!

Now, if you could, would you please expand on how you arrived at Re in this case? I'm sure that, once again, I'm missing something really obvious; but it just isn't ticking over. In that spirit, could you provide a worked-through example of the formula which you used?

Also, if you have a chance, could you briefly comment on RHABE's contention that we can take recourse to apparent viscosity when flow is turbulent through the duct?

Of note, the Engineering Toolbox is giving the following out as flow conditions for pipe or duct with respect to Reynolds:

• laminar when Re < 2300
• transient when 2300 < Re < 4000
• turbulent when Re > 4000

Are these parameters applicable in the present context?

Kudos to you and RHABE for the GAs!

Thanks again, everyone

Very good presentation less chaotic than usual. progress has been made, positive effect of CR4 participation. You deserve a GA

There is a mention still to be done the Reynolds thresholds established for round pipes are NOT always valid for other geometries.

Thanks (and sorry about the cynicism earlier...)

You're right about the Reynolds number thresholds. In fact, I think for the parallel plate example, the transition is something less (1300 or 1400?).

For very narrow slits the threshold is less 200.

"In fact, I think for the parallel plate example, the transition is something less (1300 or 1400?)."

This transition is a complicated behaviour: above the stated Re number it is fully developed turbulence equivalent to many small vortices everywhere and many within the smallest dimension (the gap width).

Transition starts at roughly 1% of this Re of fully developed turbulence.

One example for low Re vorticity is the well known Karman street of vortices.

That is regular line of individual vortices that are shed by a small or large obstacle and attached to this obstacle and attached (mostly but sometimes not) to across the gap endpoints. One vortex is cw the next ccw. Vortex frequency is a function of Re number, if vortex frequency matches natural frequency this may be disastrous, see Tacoma bridge accident. At higher Re these vortices may detach from the point of start and reattach at the same point or somewhere else- this is in parallel with hysteresis in drag.

The more approaching the fully established turbulence the less adherence to a starting point of vorticity can be seen - many small fast changing vortices.

Up to Re near 40 (in nearly parallel gaps) the viscous flow is a very good approximation but beware of roughness. And viscosity is a function of temperature and maybe pressure for fluids and of temperature only in gases. In gases the lower limit of viscous behaviour is Knudsen flow - the transition region to molecular flow.

The switching behaviour from one vortex frequency to the next (and changing attachment) at small Re numbers will show up as an unpredictable and abrupt changing hysteretic drag behaviour - but all apparent viscosities are above real viscosity (that is seen at laminar conditions).

There is known to me one condition where first vortices may form slightly above Re = 1, that is for concentric cylinders with the inner cylinder rotating. This is giving more instability as the centrifugal force will try to transport inner regions of fluid outwards.

I had to learn this (many years ago) by a rotor that turns inside a gas filled housing at these Re numbers. This was a dry tuned gyro, to be optimised by designing the gaps and the gas and the pressure and outgassing and ball-bearing lubrication without outgassing and many more error contributions.

Any vortex between rotor and housing will introduce severe instability in the gyro's drift performance.

RHABE

I have to add a note.

When a fluid enters a duct a boundary layer will form. Its thickness grows along the wall proportional to a power (less 1) of the distance to the entry border. The boundary layer is laminar due to the high friction between fluid layers. The core is still NOT laminar. The boundary layer thickness is also dependent on the fluid velocity so that this plays also a role.

The full flow can become laminar is the duct is long enough so that the boundary layer will reach the center (or the middle).

This explain why in narrow and short ducts even at low Reynolds the flow is not any more laminar.

This is the reason why orifices have a sharp edge: to avoid a laminar layer from very low flows up.

The problem of the laminar layer is that if it becomes too thick it looses stability so that is a critical length over which a laminar layer becomes partly turbulent. This can be explained by the reduction of the derivative dw/dy and by the following reduction of the friction between layers which leads to the flow "order". If this friction is small then other effects can disturb the flow thus turbulences!

It is interesting to compare the comments to the title: "A "Rather Small" fluid flow question. If there is a fields where questions are never "small" then it is the fluid technology!

Fascinating. Isn't nature wonderful?

I hope not to derail the more in-depth discussion which is proceeding at this juncture, but, after a good amount of effort grappling with the collective arithmetic, I seem to have run into a problem.

So, moving from the sublime to the banal, I'd like to post the following for any input which might be forthcoming.

Here's the "sum" of my grasp from the proceedings thus far:

A thumbsketch review (for anyone not with us from the beginning),

"...we have a rigid duct which is an inch long, an inch wide, and .002 inches in height. We'll be feeding hydraulic fluid (9.4 cSt @ 100°C) through it from a steadily pressurized source of 4kpsi."

"Start with Hagen-Poiseuille equation"

V* = p*B*H³/(12*eta*L)

V* = flow in m³/s
p = pressure in N/m²
B = width in m
H = height in m
eta = dynamic viscosity (N*s/m²) = kinematic viscosity (m²/s) * density(Kg/m³)
L = length in m

p = 27579029.172713
B = 0.0254
H = 0.0000508
eta = 0.0000094 * ((1000000 * 0.89) / 1000) = 0.008366
L = 0.0254

V* = (27579029.172713 * 0.0254 * 0.000000000000131096512) / (12 * 0.008366 * 0.0254)
V* = 0.0000000918340690337785648772224 / 0.0025499568
V* = 3.601397052443341976508088293888e-5 m³/s (or 36.013970524433 cm³/s)

Area = H * B

Area = 0.0000508 * 0.0254

Area = 0.00000129032 m²

Velocity (m/s) = V* / Area

Velocity (m/s) = 3.601397052443341976508088293888e-5 / 0.00000129032

Velocity (m/s) = 27.910882978201856721651127579887

"Plug that velocity into the Reynolds number equation to obtain a value for Re."


"The definition of the Reynolds number is the ratio of the inertial force to the viscous force. In math terms, Re = velocity * length / kinematic viscosity."

Re = (Velocity * L) / kinematic viscosity (m²/s)

Re = (27.910882978201856721651127579887 * .0254) / 0.0000094
Re = 0.70893642764632716072993864052913 / 0.0000094
Re = 75418.768898545442630844536226504

Far from Re = 149, I wind up with what RHABE worked out in his original post as "approx 75,000."

Forgive my "viscosity," but what simple twist might I be missing here?

Thanks once again, folks --

The error is in the characteristic length. It isn't 1 inch (or 0.0254m). The characteristic length is the distance between the plates (0.002in or 5.08*10^-5m)

Using your numbers and the gap distance...

V = 27.9 m/s

L = 0.0000508 m

kv = 0.0000094 m^2/s

Re = 27.9 * 0.0000508 / 0.0000094

Re = 151

Thanks to RHABE and Nick for showing me I still have much to learn...

For your attitude you deserve a GA from me!

@ChaoticIntellect:

I see. Thanks for clarifying how you arrived at Re for this arrangement. Glad to know that at least my arithmetic was OK...

So, I guess we're at a stopping place with this particular thread for now. If nobody has anything else to add, it looks like this one's parked here for the time being

Great run!

Thanks to you all for your thoughtful comments and input.

Have a terrific week, everyone --