Heinke (Seibe-Heinke, Seibe Gorman) Regulators.

[vance]

Ranier, I looked at your image, too, and it don't matter what size it is!

Yikes! I'm sure there are some on the board who can follow that (like Graeme) but it's too much for me.

I am expert at fractions and the metric system, but that must be something even more complicated, like long division!

[nikeajax]

John gets that mumbo-jumbo too

Math hates me--everything about me!

JB

EDIT: I have dyscalculia so it's not about I just need to learn math:

en.wikipedia.org/wiki/Dyscalculia

[SeaRat]

Dec 5, 2023 15:54:32 GMT -8 vance said:

Ranier, I looked at your image, too, and it don't matter what size it is!

Yikes! I'm sure there are some on the board who can follow that (like Graeme) but it's too much for me.

I am expert at fractions and the metric system, but that must be something even more complicated, like long division!

I have not had time to review this calculation. So until I do, I’ll withhold my judgement on the drawing.

What he is trying to do is to show how using the diameter to determine the surface area of the tube, the calculation shows the weight required close the tube and just balance the force of the air against the opening which would cause a free-flow of air. It’s an interesting concept. The angles he’s using show that the weight will close the tube in various positions equally well. But in a few weeks I’ll go through the calculation and see whether it makes sense to me.

John

[cg43]

Hello John 
Perfect ! I agree with everything you posted . Thank's .

Hello

I draw a line between the center of the main diaphragm and exhaust valve .

The angel betweenthis line and horizontal is  alpha , the distance the lenght d .

hw is the hight of the exhaust valve over the center main diaphragm : alpha = 0°   produces hw =  0
w = index for water  or water colume                                                 alpha = 90° produces hw =  d
c = index for compensation weigth                    for every alpha applies :  hw = d * sin alpha
v = index for exhaust valve : Av = area exhaust valve
The math. sin funktion is :      sin =   adjacent / the hypotenuse       (in an right angeled triangel) .

Now we must translate Hw in the force that try's to open the exhaust valve . Sorry that's a littel bit physik's .
But in the end we can cancel out most and the formular will be very simpel and easy to calculate .

First the calculation for an water zylinder with hight hw and area 1 cm^2
volumn = hw*1cm^2
density water = rho w :     mass = volumen*density  : Mw = hw*1cm^2 * rho w
To translate mass in mass force we need the physik. constant g= 9,81m/sec^2 
Force =  mass * g  : Fw = hw * 1cm^2 * rho w * 9,81m/sec^2
For  hw = d * sin alpha   and an exhaust valve area Av we finaly come to :  Fw = Av * d * sin alpha * rho w * g

That's the force from the water pressure we had to compensate .
Continuation in part two .

[cg43]

Weight compensated exhaust valve part 2

In part 1 we get : Fw = Av * d * sin alpha * rho w + g

Fa = force achsial = force that try's to close the exhaust valve .
We use the same physics  as for Fw .
But we must consider that the compensation weight has buoyancy in the water .
We can simply substract the density of the water from the density of the compendstion weight material .
In the drawing we see that we can split the force Gc in Fa and Fn .
The drawing shows :
                                  Fa = Gc * sin alpha
Fn is the force that goes to the outer case , and Fa is the compensation force .
Compensation is achieved if Fw and Fa have the same amount and opposite direction .
With Vc = volume compensation weigth .

Fa = Vc * (rho c - rho w) * g * sin alpha

Fa = Fw  :                                                 Vc * ( rho c - rho w) * g * sin alpha = Av * d * sin alpha * rho w * g

After canceling g and sin alpha we get :     Vc = Av * d * rho w /(rho c - rho w)

With  compensation weight mass  Mc = Vc * rho c

we finaly get :                             Mc = Av * d * (rho c  * rho w)/(rho c - rho w)

For a compensation weight material with rho about 8 times of water (steel or brass) the compensation weigth is
only a littel bit heavier than a water colume with the lenght d and the area of the exhaust valve .
What an effort for this result , but now we have the proof  and calculation for perfect compesation .

The heinke design is a littel bit diverent from my calculation .
The compensation weight's are not fixed to the exhaust valve plate's . They can only create a compensation force
if the exhaust valve is above the center main diaphragm . Otherwise they useless rest on the outer case .
To open such an valve now we need the pressure of the ambient water , and that's below the center main diaphragm .
The Heinke avoid this disadvantage by using two horns .


Greetings   Rainer