Blowoff Plate/Air Jet 1

[TangoCleveland]

I am trying to determine how high a jet of air will lift a guided steel plate used as a relief device. Pipe diameter is 24", air pressure is 7 psig. We want the plate to lift at 1 psig and got about a 1 7/8" thick plate (same as a blind flange) using a force balance. Once air starts flowing, we use F=m * v for fluid momentum, but how does the velocity and flow decay with distance from the opening?
Velocity of the air in the 24" pipe is about 824 ft/sec. How high will the air flow lift the plate? I'd appreciate any suggestions or approaches.


Larry

 

[quark]

Ok. I will try for it.

You have to provide enough pressure which causes the sufficient force to balance the weight of the plate.

If A is the area of the plate in square inches, d is density (in lb/cu.in, for steel it is 7843 Kg/Cu.M) then weight of the plate is

W = A x 1 7/8 x d, in lbs.

Area of 24" dia pipe is 3.142x24x24/4 = 452.448 sq.in

So force applied on the plate will be F = 6 x 452.448 = 2714.688 lbs,

If F is > W then your plate will lift up.

Kinetic Energy of flowing air is mv[sup2[/sup]/2, where m is mass of air and v is its velocity.

Potential energy required to lift the plate to a height of 'h' is mgh where mg is the weight of plate.

Equate these two and you will get 'h'.

Regards,

 

[TangoCleveland]

Thanks, but the air flow diverges and slows down as it exits from the 24" pipe. The 24" plate (blind flange) weighs 450lb, so the initial pressure force of 2715 lb will lift the plate. We need to determine at what point
we will have F=m * v is equal to 450 lb. If we hold mass flow constant at 100lbm/sec, this will occur at v=144 ft/sec. How does free jet velocity vary with distance from the orifice?


Larry

 

[quark]

Oops! I said density? But it is specific weight.

One more correction (a typo)

Kinetic Energy is 1/2 (mv[sup]2[/sup]) = 1/2(rho x a x v[sup]3[/sup])

Where rho = air density at the rated pressure

a = area of cross section of pipe

v = air velocity

I did a sample calculation with air density at atmospheric pressure and it seemed to me that you were test firing a missile. Where did I go wrong?

 

[TangoCleveland]

In my Fluids book, force on piping from flow is evaluated with fluid momentum, F = mass flow * velocity, not by kinetic energy. I'm guessing that the jet will dissipate roughly as the inverse square of distance, which would give me a lift of about 5.5 inches. Am I on the right track?


Larry

 

[hacksaw]

pallet type reliefs don't "pop". you need higher pressure and a "huddling chamber" to accomplish that.

we typically set the relief pressure according to weight(you can use a non-linear spring or a pilot actuated valve if you want full lift at setpoint). The capacity of a pallet calve (and its pressure accumulation) has to be inferred from tests.


Seem to recall that the civil guys have models for manhole covers, you might be able to establish the behavior for water and extend it to your application.

 

[quark]

Work done = Force times distance. This is what you have to calculate. It is not sufficient if you just calculate the Force. Energy is the ability to do the work.

My approach is correct but the final figure I got is misleading me. (That can launch me on Mount Everest )

Wait, I saw safety devices on Roots Blowers which lift up incase of excess pressure and are very much similar to your application. No. of plates are to be increased or decreased as per the requirement of pressure. Try any Roots Blower manufacturer.

Meanwhile I will try to rope in TD2K, the finest tipster.

 

[dydx]

Clarification request.
Can you confirm that:
a/ the arrangement is a vertically inclined pipe of Dia 24".
b/ the maximum static pressure is 7 psig when the top of the pipe is securely sealed; no air flow.
c/ the air flow is a maximum of 824 feet/second when the pipe opening is unobstructed.
d/ the air flow will leave radially from the pipe at the plate/pipe opening. If not radially, at what angle. This affects the momentum equation.

I can propose 2 possible approaches:
Approach 1/ Use Continuity, Energy & Momentum equations.
In the energy equation subtract out mgh term
[=(Mass_plate)*g*(lift of plate).]
In the momentum equation only vertical components of forces and vertical components of flowrates need be considered. See point d/ above.
Also Ideal gas equation. (Is it okay to assume air is dry and ~20degC???)
Assume that the pressure at the opening is atmospheric. [True for incompressible flow at the 'vena contracta' section; not true for compressible flow regimes. Okay for 1st estimate; +/-50%]
I tried this for radial flow, ended up with 3 equations with 3 unknowns. Solved a cubic equation numerically and got a "lift" of ~4.5 inches as one of the non-negative roots of the equation.
This sounds reasonable.

Approach 2/ Assume a converging nozzle being supplied by a supply reservoir at a pressure 21.5 psia and discharging into a receiver at 14.5 psia(atmosphere).
Apply appropriate "converging nozzle" equations.
It does seem that the flow is sub-sonic (Mach No. < 1) and therefore an iterative process may be required to find a solution.
I have not fully completed this calculation as of yet. Will inform you of the result.

Note: Regardless of method used and inaccuracies from assumptions, there is likely to be an oscillatory motion when the plate lifts if it is just being held up by the air flow regime. Is this acceptable?

 

[TangoCleveland]

d2dt,
Thanks for an excellent set of ideas. We also have a NASA researcher (I'm a NASA Contractor) looking at this as well.
Here are your answers:
a/ This is a 24&quot; Sch 10 pipe with a slipon flange with flow vertically upward.
b/ We are now working with a 1 psig static pressure with no flow. 1 psig against 454 sq inches will lift a 450lb. blind flange.
c/ Unobstructed air flow velocity-we want to relieve 100 lbm/sec of air at 1 psig (&quot;set pressure&quot; of the disk). 23 psig back pressure would be acceptable.
d/ When the disk is lifted, air is flowing out radially to atmosphere.

Also, the lifting disk is vertically guided by four rods. We are neglecting friction against the rods.



Larry

 

[dydx]

I have a few further queries and have thought of another possible approach.
a/ Acknowledged.
b/ Understood that 1 psig is initial lift pressure.
You mentioned a 7 psig pressure in your initial posing of the problem. What physical situation does the 7 psig pressure relate to?
c/ Your design criteria is that at:
0.9 psig there should be no flow.
1.0 psig there should be some initial flow.
1.1 psig there should be a flow >/= 100 lbm/sec of air.
Where does the 23 psig 'back pressure' come from?
23 psig versus 1 psig?
Is this the pressure drop from the base of the pipe to the top of the pipe?
What is the length of the pipe? It might need to be greater than ~ten pipe diameters ~10*24&quot;=240&quot; in order for a fully developed flow profile to become established. Otherwise it may not be possible to assume symmetry around the vertical centre-line.
d/ Radial flow acknowledged.
Where did the 824 ft/sec velcoity of air in pipe come from in your initial posing of the problem?
Are you able to make experimental measurements on a rig?

Approach 3/ Pressure profile on plate.
Conceptualise streamlines as the air flow progresses up the pipe and the flow diverges from the pipe centre-line towards an open annular exit. The central region of the plate will have a &quot;stagnation pressure&quot; pushing it up.
At the edge of the plate the static pressure will be much reduced; Bernoulli says that when velocity increases pressure drops and vice versa. Call this unknown plate edge pressure P[sub]edge[/sub].
There must therefore be a pressure profile accross the plate; this profile may be linear or parabolic or ....
Select a profile.
The integral of the pressure over the plate area (a radial summation) must be equal to the 450 lbm otherwise the plate would be accelerating.
By reversing this equality it may be possible to determine the P[sub]edge[/sub].
An annular exit velocity might be determined from this pressure.

If nothing else, the above concepualisation might lead one to believe that attaching a low mass profile form [such as an inverted cone or an inverted &quot;sugar loaf&quot;] to the underside of the plate, in the central region, might encourage radial flow.

 

[TangoCleveland]

We used data from a Protectoseal Rupture Pin Tank Vent, which flows 100 lb/sec with a 7.78 psi pressure drop. We worked up a three-part solution. Part 1 is with 1 psig pressure cracking the lift disk open. Part 2 uses Bernoulli to determine flows at partial disk openings. Part 3 is with the lift disk fully open against the stops, which is where we get the 100 lb/sec at 7.78 psi drop. I can supply an &quot;anonymous&quot; calculation if you're interested. Our Safety people are in agreement that the pressure relief is adequate.

Larry