The good old topic of organ air supply surfaced again in the MMD Pipes Forum in 1999. At that time I learnt a lot from our experienced community about what devices you employ to get a stable air supply. Here is a brief review of these components and some of their properties described in acoustical engineering terms. Special attention is with the problem of regulation of transient loads, how to keep pressure constant across sudden changes in air demand. It is no cookbook on how to make a complete supply system - a prominent reason is the difficulty to specify what is required - but hopefully you may find some dimensioning guidelines from examples given. Basic formulas and diagrams are in SI units , but translations into traditional units are mostly shown in parallel.
In an organ pipe the tone onset is very important and characteristic
and there are several factors to determine this onset. One is the pipe
itself as set up by the voicer. To keep a predictable pipe tone you must
also have control of how rapidly the playing valve opens and how well the
wind supply stands up to the increase in air demand when the pipe goes
on. Also any subsequent pressure changes will modulate the pipe tone, notably
when it sounds while other pipes go on or off, in particular in wide chords.
Small as well as big street organs are usually powered from bellows,
agitated from a hand operated crankshaft. Modern modifications of these
most often include accumulator driven electric motors or gas engines to
drive them over a belt. Here are the outlines of a classical street organ
The bottom two wedges constitute a double acting feeder with valves for the air into the top, spring loaded reservoir. (The second crank connecting rod drives one more such feeder, hidden in the drawing). Cranking at constant speed the feeders basically deliver constant airflow, no matter how much you need and at what pressure. The first regulation here comes with the feeder output valves, they are joined with adjustable strings to the reservoir lid. When the reservoir is full they are held open such that the surplus air returns back into the feeders. This is a very essential feature: when air consumption is low, then the mechanical power to drive the feeder is also reduced - the different feeder bellows then exchange the same air between them and you only have to supply power to overcome friction. Note removable panels and valves for easy maintenance and a string operated safety pallet at top of the reservoir.
Without this tying up the feeder valves, the feeder would blow up the reservoir, were it not for the safety pallet. This one actually implements another principle of regulation, it is a spill regulator. Any surplus air is blown off, but in the same time you lose the energy you spent to pressurize it. Using spill regulation alone you have to provide full power to the crankshaft all the time. Apart from being wasteful this also makes noise, so spill regulation is mostly used for very small systems where simplicity and low cost may be important.
Fancy drive alternatives from over the times are now displaced by ubiquitous electric centrifugal fans. These have radically different properties from a bellows, you can show them with a load characteristic graph. Fig. 2 is an example of such, how the output pressure can vary as function of the flow delivered. The red dashed curve, where VC stands for a typical vacuum cleaner fan, shows a very high pressure at low flow, dropping considerably as load increases. This common fan type is about the least suitable there is for an organ. But fans may be radically different from each other. By proper engineering, including swept rotor blades, you can get a much superior characteristic having almost constant pressure, as exemplified with the black solid curve. The figures nearby show typical rotational speed (for a 3 phase, 50 Hz motor) and diameter for a fan that can deliver pressures in the 1 kPa (4 inWC) range. A fixed fan design delivers pressure in proportion to radius and to the square of speed. For higher pressures you would normally select a bigger fan (or possibly more than one in cascade) rather than belt drive to a faster and noisier one.
There are a few makers of specialized organ fans . Beyond their load characteristic these are distinguished by silent operation due to careful balancing, solid housings, often cast, and a preference for sleeve bearings rather than ball bearings. Such fans have a noise level about 50 to 60 dBA and cost typically three times more than equivalent industrial fans which are 20 to 30 dB more noisy.
The pneumatic power you supply to an organ is pressure times flow, W=PU. It is convenient to use SI units for P and U such that the power comes out in Watts without conversion factors. For instance, in the fan diagram P=1.1 kPa and U=200 m3/h=0.056 m3/s gives W=1100*0.056=61 Watts. Installed drive motor nominal power is normally several times bigger than this pneumatic power. And finally delivered sound power to the order of 1% of it.
The reservoir is conventionally a single wedge bellows or a double parallel moving bellows (church organs). Another variant is the sack reservoir where the lid is joined to the support frame with a rubber cloth collar that rolls inside and out as the lid moves, fig. 3 . This picture also shows a 'curtain regulator' that shuts off the supply coming in from lower right by way of a chain to the reservoir lid.
The lid force may in stationary cases be exerted by weights (iron bars, stones, concrete blocks), now more often by springs, always so in portable applications. An easy way to estimate it is to start from the required pressure, expressed in terms of water column height. Multiply this height by the lid area - then you get a volume of water having the required weight.
|Pressure 2.5 kPa = 0.25 meters water column, lid area 0.6*0.6=0.36 m2. Water volume would be 0.25*0.36=0.09 m3, weighing about 90 kg. Lid force has to be 90 kp, about 900 Newtons. 900/0.36=2500 Pa.|
|Pressure 10 inch WC, lid area 2*2=4 sq. ft. Water volume would be (10/12)*4=3.33 cubic ft, at 62.4 lb./cubic ft weighing about 208 lb. Lid force has to be 208 lb, about the same 900 Newtons.|
The obvious part of the lid force is that from the weight or spring. A weight will give a constant force while a spring generally delivers a greater force the more the reservoir is filled. But there is also another force acting on the lid, namely from the bellows folds. With a sack reservoir this is essentially small and independent of lid position. But with a gusseted ribs reservoir this force will vary depending on filling degree.
Fig. 4  shows one corner of a double reservoir where I have inserted red arrows to show the forces from internal pressure acting on the folding ribs. The magnitude of these forces is pressure times rib area. They can further be split up into vertical and horizontal components. These will vary in size, depending on the filling state, indirectly defined by the rib angle a.
The lower bellows is inward folded, between frames A and B, and here the vertical component on the ribs try to pull the frames together, they counteract the pressure on the lid which balances the weight or spring. This effect is most pronounced when the bellows are almost collapsed since the vertical component decreases with growing a as the lid lifts. The horizontal component works the opposite way, it grows with a, and it makes the ribs try to wedge the frames apart, it cooperates with the pressure on the lid. The total result is that when you fill a single, weight loaded inward folded reservoir the pressure decreases as you fill it.
Contrarily, in the upper outward folded bellows between frames B and C, the rib forces change sign. So when you fill a single, weight loaded outward folded reservoir the pressure increases as you fill it. The specific feature of the illustrated double bellows design is to have these opposing actions to neutralize each other. But to make this work you must force frame B to be mid-way between A and C, effected by the pantograph mechanism D. This double bellows then keeps constant pressure, independent of filling degree, when compressed by a weight.
If you make a square 'camera folded' bellows as in fig. 5 - cylindrical cloth, two folds out, two in - this is also essentially neutral to the rib forces.
Accounting for those extra forces from the folding ribs, the relation between the external bellows compression force F and pressure P can be formulated as a bellows equation:
F/P = A ± B*0.5*(sin(a)*tan(a)-cos(a)) [m2]where A is the lid area, B is folding rib area (one layer only, as seen in a top projection drawing of the bellows), and a is the opening angle of the folds. Plus sign for inward folds, minus for outward. An important special case is at a=45 degrees, ribs at right angles to each other. Then the complicated coefficient for B becomes zero and pressure equals the nominal P=F/A.
With a spring loaded reservoir the spring force increases in proportion to the lid lift. Then the best is to use an inward folded reservoir. Ideally, for constant pressure, the force should then grow with a the same way as the complicated right hand expression. But we can reasonably well match this with a straight line. For given areas A and B, and pressure P, we may use for a rule of thumb that the spring force at full lift should be F(100%) = P*A, and with the bellows collapsed, F(0%) = P*(A - 0.6*B). That magic coefficient 0.6 is derived from the criterion that pressure around 20% lift should be the same as at 100%. Deviation from the ideal increases with B/A. Using B/A=0.5 as drawn in fig. 6, then pressure will be some 6% too high around 60% lift. For better precision you may reduce B, but that implies that the bellows volume capacity goes down in proportion.
|In example 1 above, where A = 0.36 m2, P = 2500 Pa, we already found F(100%) = 900 N, about 90 kp. Now we also prescribe that 100 % lift is to be 0.1 m. This settles the rib width to 0.1/sqrt(2) = 0.0707 m. The free window area inside the ribs is then A-B = (0.6-2*0.0707)2 = 0.21 such that B = 0.36-0.21 = 0.15 m2. The rule of thumb renders F(0%) = 2500*(0.36-0.6*0.15) = 675 N, about 68 kp. This may be reformulated into a required spring constant to be K = (900-675)/0.1 = 2250 N/m, or 230 kp/m, or 0.23 kp/mm. At full lift the spring(s) should be elongated from rest length a distance F/K = 900/2250 = 0.4 m, going down to 0.3 m when bellows is collapsed.|
|In example 1 above A = 22 ft2 = 242 in2 = 576 in2, P = 10 inWC. For dimensional consistency we convert that into P = 0.361 PSI. This in an alternative way renders, as we already found, that F(100%) = A*P = 576*0.361 = 208 lb. Now we also prescribe that 100 % lift is to be 4 inches. This settles the rib width to 4/sqrt(2) = 2.83 in. The free window area inside the ribs is then A-B = (24-2*2.83)2 = 336 such that B = 576-336 = 240 in2. The rule of thumb renders F(0%) = 0.361*(576-0.6*240) = 156 lb. This may be reformulated into a required spring constant to be K = (208-156)/4 = 13 lb/in. At full lift the spring(s) should be elongated from rest length a distance F/K = 208/13 = 16 in, going down to 12 in when bellows is collapsed.|
|The viscous pressure drop per unit length in a cylindrical trunk for
air at about atmospheric pressure and room temperature.
Step 1: Locate on a bottom horizontal scale the flow rate you are going to transport.
Step 2: From this point move vertically until you intersect the (right upward sloping) line for the diameter of your trunk. Mark this intersection point.
Step 3: Go horizontally to a left scale and find the pressure drop per unit length.
Step 4: Multiply with the length of your trunk to obtain the total pressure drop due to viscous friction.
Step 5: At the intersection point of step 2, find from the right downward sloping lines what is the linear speed of the air inside the trunk.
Step 6: Follow the direction of these downward sloping lines to the right scale and read the corresponding dynamic pressure.
Pkin is the sum of additional pressure drops that come about when you change the shape of the trunk, which implies that the air is accelerated some way. For instance at the trunk inlet air has to be accelerated from essentially standstill to a speed V=U/A where A is the trunk area. The pressure required to do this acceleration is close to what is often called the dynamic pressure
Pdyn = 0.5*r*V2 [Pa=N/m2] The Bernoulli equation for dynamic pressureAlso other area changes like in valves, or any bends in the trunk, will add to Pkin. In handbooks and catalogs these contributions are often said to be a certain fraction K of the dynamic pressure, for instance K=0.3 for a 90 degree large radius bend. Another classical rule of thumb says that for a sharp elbow the pressure drop is the same as for 32 diameters length of tube. In other words, when you lay out the proposed trunk, count the elbows, and add 32 times the diameter of the proposed trunk diameter for every elbow, to the total length.
|Air is conducted from a reservoir holding a pressure of 2.5 kPa, (25
cmWC, 10 inWC) to a chest, using a 50 mm (2 in) diameter tube of total
length 5 m (16.4 ft). The tube has two 90 degree elbows. The flow rate
is 10 liters/second (21 CFM). How much lower is the chest pressure
than that of the reservoir?
The diagram tells the viscous loss to be 1 mmWC/m, totally 5 mmWC for the 5 meter length. The air speed is 5 m/s and the dynamic pressure 1.5 mmWC, being lost when air accelerates into the tube. With two slow bends you might lose additionally 2*0.3*1.5=0.9 mmWC. The elbow thumb rule says 2*32 diameters is 3.2 meters, hence drop 3.2 mmWC, considerably more. Total pressure drop is thus estimated into 8-10 mmWC. Also you would regain some of the dynamic pressure should the air speed slow down entering the chest volume. The drop is to the order of 4% of the reservoir pressure. This is measurable with a water manometer and you might believe this is an acceptable loss, that you could allow such a long, narrow trunk.
It is important to note that these pressure drops are for the maximum desired flow rate chosen for the calculation. Pressure loss is proportional to the square of the flow, so at lower flows, the loss will be much lower. In any typical organ, the airflow for any chord in the top octave (1/2') will consume only about 4.5% of the wind required for a similar chord played in the bottom (8') octave. At the lower flow of the treble chord, the pressure drop will be essentially zero. So the calculated loss will be equivalent to the expected range of deviation in the chest pressure, depending on the character of the music being played.
Luckily, the effect of increasing the trunk size just a little has an equally dramatic effect in the opposite direction. For example, going from a 3" trunk to a 4" trunk cuts the loss in half. The practical range of available pipe sizes (in America, 1/2", 3/4", 1", 1-1/2", 2", 3" 4", 6", 8", 12", 16") is designed to do approximately that. So if the calculated range of chest pressure variations seems a little high, upping the trunk one size will cure that.
The fan load characteristic above also shows a curve for dynamic pressure. This is found with the Pdyn formula and the linear air speed V as computed from the flow and the fan outlet diameter. When the fan is run free, without anything connected to it (giving 400 m3/h with that specific example fan), then there will be zero static (over)pressure at both ports. Then all work done by the fan is used to develop a dynamic pressure, manifest only in the speed of the output air.
To handle this we must leave statics and DC flow aerodynamics and switch into acoustics. The big issue is that now we are talking about higher frequencies and transients, not the continuous DC flow of the 'plumber's diagram' above. Now, looking at a trunk, we can represent it mathematically two different ways, either as an acoustical mass, or as an acoustical transmission line.
The simpler way is to regard the air inside the trunk as an acoustical mass, a simple discrete element. This mass has an inertia and it will take time and pressure exerted on it to make it accelerate. 'Discrete element' presumes the enclosed air moves as a single unit, all parts of the air column move in synchrony. Whether this is reasonable or not depends on what frequency range or what time scale we are interested in. A convenient way to formulate this is that the trunk must be 'acoustically small', meaning small compared to wavelength.
The acoustical mass of a tube is simple to derive from Newton's law in mechanics: Force=mass*acceleration. In acoustical terms the same will read: Pressure=(acoust. mass)*(time derivative of flow). From this you can eventually derive
Mtube = r L/A [Ns2/m5 = kg/m4] Acoustical mass of a tubewhere L is the length and A is the area of the tube, r is density of air.
|With the 5 meter by 5 cm diameter tube Mtube = 1.2*5/(p*0.0252) = 3056 Ns2/m5. Let us suddenly by means of a valve put the 2.5 kPa (2500 N/m2, 10 inWC) pressure to the input end of it. Then flow will accelerate, dU/dt=P/Mtube=2500/3056=0.818 m3/s2. Queer unit? In each second, flow will increase by 818 liters/second (this will not go on for a very long time, soon flow will be limited by viscosity as per section 3.1). But it means that for the flow to reach 10 liters/sec will take some 10/818=12 milliseconds. But in that time sound will propagate a distance approximately 0.012*c=4.1 m which is less than the tube length. This tells there is something wrong, flow cannot possibly propagate faster than c. The trunk is too long to regard as a discrete element on this time scale, however you come somewhere near a plausible answer when you add the 5 meter propagation time of 15 milliseconds to those 12 milliseconds rise time.|
The more complex and accurate way to treat a trunk should be used when the dimensions, particularly the length, are not small. Then we could regard it an acoustical transmission line which can be tackled using two different key properties, the speed of sound c, and its acoustical impedance Z.
In a wide trunk c is essentially the same as in free air, but in a narrow channel (like in the tubes for pneumatic control in organs or player pianos) it is up to 15% lower because of isothermal conditions rather than adiabatic.
Acoustical impedance by definition is the ratio between pressure and flow, Z = P/U. Pressure is measured in Pascals = Newtons/meter2, flow in meter3/second, so impedance is consequently in Newton*seconds/meter5. Alas nobody has any intuitive feeling for that unit.
Now hopefully knowing what impedance is, what is its value for a trunk
or tube? Derivation of this is a bit complex, so I just tell you it is
Ztube = rc/A [Ns/m5 ] Acoustical impedance of a tube
Air density times speed of sound, divided by tube cross-sectional area A. This impedance giving the ratio between P and U now makes it possible to estimate the pressure at the loaded end when you give it a sudden, transient load. To do this in detail requires computers and programming, but using sheer reasoning we can still look at the following:
Horror example 5:
|With our 5 meter by 5 cm diameter tube the impedance in SI units is
= 1.2*343/(p*0.0252) = 2.1*105
Ns/m5. This trunk comes from a remote feeding reservoir with
2500 Pa (25 cmWC, 10 inWC) pressure. Suppose we now suddenly start drawing
liters/second from the tube. By way of the impedance we can then find how
much the pressure drops at the tube end: P = ZU = 2.1*105
*0.010 = 2100 Pa. Essentially a short circuit, almost none of the original
pressure remains at the receiving end!. What happens next is a wavefront
rushes upstream the tube at the speed of sound, sending backwards the required
flow. The substance of this flow comes from the decompression of the air
in the tube. The wavefront will reach the reservoir after 5/c= 0.015 seconds.
Not until now the reservoir 'knows' what happened at the far end of the
tube and gets its chance to feed the tube with a flow. But the wave that
traveled up the tube hits a low impedance in the reservoir, it is almost
totally reflected back toward the loaded end with opposite pressure polarity
and the reservoir starts feeding with twice the required flow. After
another 15 ms this surge reaches the receiving end, and if we still insist
on extracting 10 l/s then pressure will jump up to the reservoir 2500 Pa
plus the 2100 Pa surge (minus some because of losses), considerably more
than the nominal pressure!
The surge wave will continue for some time bouncing back and forth in the trunk which actually behaves just like an organ pipe resonator. This idealized fig. 8 shows pressure vs. time at the receiving end of the trunk.
In a practical situation things are even more complex, for instance the flow drawn will not naturally stay constant when the available pressure fluctuates this way, rather it will inherit some of the modulation from the pressure. This very fact together with viscous damping is the reason the air bouncing dies out in a few periods. The dotted line suggests the average pressure at the far end. It is a little bit lower than the reservoir pressure, namely that tiny drop we calculated from the 'plumber's diagram' in section 3.1, example 3.
With a shorter trunk of same diameter the variations will be just as large. The trunk area and the load change alone define the magnitude of the pressure variations. But with a shorter trunk the oscillations are faster, less sound propagation time from end to end. So with a short trunk the transients may be over in just a few tens of milliseconds.
|Same 10 liter/second load change, but we will allow no more than 4% drop in the 2500 Pa pressure, that is 100 Pa. The impedance will then have to be at most Z=P/U=100/0.01=10000 Ns/m5, tube area A=rc/Z=1.2*343/10000=0.041m2, at least 23 cm (9 in) diameter!|
Here are two illustrating examples taken from the real world. Recently Carlsson  showed the pressures at the receiving end of the trunk and in the note channel (on the downstream side of the playing valve), recorded from a church organ, fig. 9. This trunk is 13.5 m (44.4 ft) long with a square cross section of 15 cm (6 in) and the transient load change was 8.6 l/s. The initial transient pressure loss is about 20%. This is believed to be representative, it exhibits the same behavior as in example 5, but not to such extreme degree. The pressure oscillates at about 4 Hz, somewhat lower than ideally expected for this length of the trunk. The difference is ascribed to a lowered sound speed and to additional effects from reservoir (wedge) shape, wall elasticity and adjacent channels.
Similarly the wind chest itself contributes variations due to internal waves in its channels that are generated downstream when a note valve opens. Fig. 10 is an example, adapted from Finch and Nolle . At bottom it shows note channel pressure near the foot of the pipe after the note valve has begun opening at time 0. At top the tone from the pipe, clearly modulated by the pressure variations. The filled circle marks supply pressure and the solid bar the somewhat lower pressure entering the note channel, after the playing valve.
Mementos of this section are: With transient loads the pressure variations at the end of a trunk may be much bigger than you see with an ordinary slow manometer. They may severely affect the pipe sound. To study them you will need a fast pressure transducer. To reduce them you can use a wider trunk, i.e. having a lower impedance. But to reach a sufficiently low impedance the trunk area probably would have to be inordinately large.
To efficiently combat the pressure variations due to a varying load to a trunk you must introduce additional components.
The prototype for an acoustical capacitance is a closed stiff container of volume V. When you compress the air inside it by pumping in that DV, then pressure will rise the amount DP. By use of the gas law you can eventually derive the formula
C = V/(rc2) [m5/N] Acoustical capacitance of a closed volume VExample 7, recalling the horror example 5:
|During the first 30 ms interval the rarefaction wave runs to the reservoir and then comes back, apparently with fresh air. During that time interval the 10 liter/second flow means we would consume DV=0.3 liter=0.0003 m3. Suppose we connect a capacitance container at the receiving end hoping to let this one supply the necessary air while we are waiting. Also suppose we will accept a pressure drop DP=250 Pa (2.5 cmWC, 1 inWC). The container must then have a capacitance C=DV/DP=0.0003/250 and with the above formula we find its necessary volume to be V=C*rc2=0.0000012*1.2*3432=0.169 m3 (6 ft3). Probably we would not accept such a big device for the purpose.|
But there is a marvel called a bellows. Ideally a bellows can render an infinite capacitance, meaning the pressure is constant (DP =0) when you push in or take out a limited volume DV. - A concussion bellows is the standard decoupling device used to subdue pressure variations such as those we discuss here. There are two common forms, 'winkers' and 'schwimmers'.
A winker is a small hinged bellows, often mounted in the wall of a trunk. It is pushed on by a compass spring to match the average pressure. You can fine adjust the point where the spring force is applied to a suitable distance from the hinge of the moving board.
In a schwimmer the moving wall is a flat bellows cloth supporting a light panel and this is often mounted in the wall of the pipe chest itself, the ultimate location to avoid any ducting between compensator and point of use.
A difficulty with concussion bellows is that you have to match the spring force quite precisely to the actual pressure. It is a passive device that adapts to the pressure, whatever this happens to be, it does not regulate it. If the spring force is too high or too low, then the bellows will stay in one of its end positions and will not be able to supply any volume displacements. With a schwimmer adjustable pantograph like in fig.11 you can adjust both the spring constant and the spring force operating on the plate. You can then come very close to infinite capacitance, i.e. constant pressure independent of plate position, but for this to be practical pressure must be closely regulated.
Many organ chest designs, such as Aeolian, Pitman or WurliTzer, have the bottom of the chest occupied with various machinery, and there is no room for a schwimmer. These makers solved the problem differently, by placing another regulator close-coupled to the chest with a large opening. This essentially isolated the chest from any dynamic or pressure loss effects of the trunking. See Section 7 for an analysis of the entire wind supply configuration, including this feature.
Ma = Mm/A2 [kg/m4]= [Ns2/m5] Acoustical mass of a pistonwhere Mm is the mechanical mass in kg and A is the lid area in m2.
Examples 8: Table relating mehanical mass and
area to acoustical mass. The first two items represent air in trunks. The
last three represent the lids for a small bellows and a big reservoir.
One might believe that 12 tiny grams of air in the 5 meter, 5 cm dia. tube should not do too much to slow down the air supply. But the resulting Ma says it indeed does so quite a lot! It is 25 times more difficult to accelerate the air in this trunk than it is to accelerate the small spring loaded bellows lid. So such a small bellows would do a lot of good as a compensator if you have that trunk. On the other hand, if you want it to maintain a sizable pressure using a weight instead of a spring, this weight would slow down the response of the bellows to make it good for nothing in the present circumstances.
The last example underlines the significance of area. Despite an enormous amount of iron to accelerate it is very fast responding. Mostly valid: bigger is better.
A regulator may be characterized by its 'droop', telling how much pressure drops, DP, when you increase flow DU. The ratio between those two is then measured in [N/m2]/[m3/s] = [Ns/m5]. This is nothing else but acoustical impedance once again. Low droop and low (output) impedance is the same thing. Two factors decide this droop, or impedance. One is the sensor sensitivity. With our simple mechanically operating regulators the sensor is a spring and bellows. The spring compliance together with the bellows characteristic tell how far the bellows lid is displaced by any pressure change DP. The other factor is how big flow change DU is induced by the effecting valve when displaced this much.
Next interesting property is the speed of regulation. This is where the preceding section 5 about bellows speed comes in, the acoustical mass (not its mechanical in kg) of the sensor bellows puts a limit to its reaction speed.
An additional important feature with a regulator is its stability. Theoretically, the simple regulators considered here are mass-spring systems, so called first order, which should always be stable. But even within this narrow scope practical solutions are not always so, there may be additional elastic and inertial elements introducing extra delays and a more complicted higher order behavior. In particular the load can make for problems, it may for instance be another trunk where waves can propagate back and forth like in example 5 and present a complicated situation for the regulator. Also the wedging action of bellows ribs may cause a negative acoustical capacitance (meaning pressure goes down rather than up when you inject an air volume) which in itself is enough to trigger instability.
Fig. 12 shows four regulator variants a - d. In all cases the unregulated input enters at the left side, and all have the same sensor element at the output, namely a bellows at top where the lid area and the spring force define what the regulated pressure is going to be. An essential feature for a good regulator, and for all these, is that the input unregulated pressure does not put any direct force on the sensor, neither any force to operate the valve - the output pressure should be independent of the input..
a has a sliding cylinder valve, the principle is the same as a knife valve regulator, common in player pianos. It may be difficult to make the slider with enough precision. If there is some play between it and the outer cylinder to ensure smooth running, then it will not be airtight enough when there is no load and the output pressure may rise beyond control. This can be somewhat repaired by making another hole in the output tube, bleeding it to atmosphere when the plunger goes up above the closed position.
b is a variation using a "stove damper", "butterfly", or rotating disc as the inlet valve, the usual arrangement on the Spencer Orgoblo blower system so common on old American organs. In practice, the damper is weighted closed, and opened by a long chain through pulleys, from the top of the reservoir to the damper crank on the fan discharge. Crude, but effective if you accept the low speed caused by the weight.
c with a rolling rubber cloth curtain against a grid is a quite good workaround to ease leakage problems. The soft curtain will rest tightly against its supporting grid and its actuating rod is easy to lead through a leather packing or just a small hole. This item is the same device as the sack reservoir in section 2. In all cases a-c the regulating valves have a relatively long mechanical travel between their fully open and closed positions. This impedes regulation accuracy but improves stability.
d is a classical design, inspired from the Aeolian organ with Zeb Vance. Here the regulating valve is just a disk covering a hole, easy to make it airtight with a leather gasket. Its very essential distinctive feature is the small relief bellows below the disk. This one insures independence from the input pressure with the important proviso that its bellows cross sectional area must be about the same as the valve hole area. The valve disk requires a quite small displacement between open and closed positions. Speed of regulation accords with the acoustical mass of the lid and there are no critical friction elements. Indeed it may easily become too fast, meaning unstable. This is very conveniently damped by the viscous gas friction in the hole connecting the interior of the relief bellows to the ambient - in case of instability you can slow regulation down using a smaller hole.
The area of the regulator valve opening should be perhaps twice the areas of the connecting trunks. Because the flow path is rather irregular in the valve region a small opening would give an unnecessary kinetic pressure drop in it when fully open.
All the illustrated regulators are shown with a comparatively big bellows lid sensing the output regulated pressure. This is very deliberate. First reason is this enables the lid to exert a considerable force on the valve to overcome friction and other disturbances, this insures precision. Second, it automatically serves as a secondary reservoir/compensator.
The left basic supply, a feeder and reservoir, is supposed to deliver a constant pressure Pr whatever you load it with. To that end it is normally a complete regulated system in itself, it has basically the same elements and works the same way as the following regulation system. With fan blowers it is common that the first regulating valve (controlled by the reservoir lid) alternatively is put at the fan intake rather than at its outlet. Then follows a trunk, the problem child, leading to the wind chest where pressure is Pc. The trunk causes temporary pressure drops and surges at changes in the load. To close or open the switch in the electrical diagram corresponds to start or stop playing pipes. From the circuit diagram we may describe the middle supply regulator system, between Pr and Pc, to have three branches, there are three different ways for flow to arrive at the note valve switch:
#1. A 'distant' branch comprising the trunk, slowly reacting because of the inertia of the enclosed air mass. From the definition of acoustical mass we can estimate a characteristic time constant like in example 4 Dt=(Md*DU)/(Pr-Pc). A high acoustical mass Md for the trunk and a large flow increase DU will make it takes a long time Dt before supply stabilizes. You can shorten that time by widening the trunk (decreasing Md) or increasing the regulator margin Pr-Pc, the pressure difference between reservoir and chest.
#2. A 'near' branch where the movement of the regulator bellows lid compliance Cr delivers the temporarily required extra puffs of air at transient loads, until the 'heavy' air in the trunk has had its time to accelerate. Its speed to do this is limited by the acoustical mass Mr of the lid, examples 8. To be efficient the lid acoustical mass must be appreciably smaller than that of the trunk. To be sufficient the lid area and displacement must be big enough to supply the flow during the transient time Dt. This can be translated into a rough estimate for necessary bellows displacement volume to be Vb=DU*Dt.
#3. A 'fast' branch with the acoustic capacitance Cv of the local regulator bellows volume and other nearby volumes. This one delivers air immediately by virtue of its decompression, essentially without inertia, but has little capacity (!). V/(r*c2) is small with reasonable volumes, this amount of air is enough only for very short times. It is conventional good practice to have a large volume here, but this will hardly be sufficient anyway, recall example 7.
When you stop the loading airflow, then the trunk flow carries on by
its inertia and rapidly fills up the regulator bellows such that its valve
closes abruptly. The trunk flow then suddenly 'hits the wall' and there
will be an appreciable pressure surge at the regulator input which can
be computed from Ztube and U, compare horror example
5. This underlines it is necessary the regulator valve be insensitive
to input pressure, otherwise it might not be capable of preventing this
surge to proceed into the regulated chest area.
|Quantity||From traditional units to SI||From SI to traditional units||Note|
|Length||1 in = 0.0254 m
1 ft = 0.3048 m
|1 m = 39.37 in
1 m = 3.281 ft
|Area||1 in2 = 6.4516 cm2
1 ft2 = 0.0929 m2
|1 m2 = 1550 in2
1 m2 = 10.764 ft2
|Volume||1 in3 = 16.387 cm3
1 ft3 = 28.317 l
|1 l = 61.023 in3
1 m3 = 35.31 ft3
|Flow||1 m3/min = 16.67 l/s
1 m3/h = 0.2778 l/s
1 CFM = 0.472 l/s
|1 l/s = 0.06 m3/min
1 l/s = 3.600 m3/h
1 l/s = 2.119 CFM
cubic ft per min
|1 oz = 28.35 g
1 lb = 0.4536 kg
|1 kg = 35.27 oz
1 kg = 2.205 lb
|Force||1 kp = 9.810 N
1 lb = 4.450 N
|1 N = 0.102 kp
1 N = 0.225 lb
|Pressure||1 cmWC = 98.1 Pa
1 inWC = 249 Pa
1 PSI = 6.897 kPa
|1 kPa = 10.2 cmWC
1 kPa = 4.01 inWC
1 kPa = 0.145 PSI
|cm of water column
inches of water column
pounds per sq inch