Lesson 9

Ring mains and compressors

Plants lay their mains as a ring so that air reaches every take-off from both sides. This lesson closes a ring, lets the calculation find how the air splits, and then turns to the other end of the network: choosing the machine, a compressor room, and several compressors on one network.

Part 1 — the ring

1. Draw the ring

   Start a new file from lesson7.gf9 (Save as lesson9.gf9), keep the compressor and delete the rest. Build the scheme above:

   Item	What	Diameter	Length
   P0	feed pipe, pipe	25 mm	10 m
   T0, T1, T2	tees, d-1 in_Equal Tee	—	—
   RA1, RA2, RB1, RB2	ring pipes, pipe	25 mm	30 m
   G1, G2	open pipes on the free outlets of T1 and T2, Pin 6 bar	4 mm	0.1 m

   The lengths are in metres because the Length row keeps the unit of its own you gave it in lesson 7. The ring is closed by the LOOP element: type loop in the filter and place loop from # Generic elements # — its card is called Loop; rename it LOOP. It has two inlets: RA2 goes into inlet 1, RB2 into inlet 2. Every other line goes from a red outlet to a green inlet as usual, so both halves are drawn away from T0.

   A ring cannot be closed by joining two pipes head to head: an outlet holds one line, and two outlets cannot meet. The LOOP element is where the two halves meet, and it is also where the calculation reports what goes on there.

2. Calculate the symmetric ring

   Press Calculate. In design mode the ring is solved as a network: EasyCAS finds how the flow splits between the halves and the lowest feed pressure at which every branch still gets what it needs. Each element of the ring gets its own flow, Pin and Pout.

   Row	Expected	Unit
   Qn, COMP	2096.5	NL/min	two open pipes of lesson 1
   m, RA1 and RB1	0.0208	kg/s	each carries its own open pipe
   m, RA2 and RB2	0	kg/s	nothing to carry
   v, P0	≈ 10.1	m/s	over the 9 m/s of a main

   Check by symmetry

   The two halves are mirror images, so each feeds its own open pipe and nothing crosses the closure: RA2 and RB2 stand still. Each of them says so in a row Flow regime — LAMINAR — very low flow through this pipe. In a symmetric ring that is expected, not a fault. LOOP itself has no figures; what happens at the closure is read on the pipes next to it.

   P0 carries the air of both open pipes, so it runs at about 10 m/s and gets a warning with the row Min. diameter by velocity, about 26.5 mm. The next step lowers the demand and the warning goes.

   RA1 and RB1 carry a wrench although nothing is wrong with them. For a ring pipe the recommendation is the smallest catalog pipe that still passes every check of the whole network, and its rows say what that costs: Source pressure increase and Annual extra cost. Weigh that figure against the metal saved. RA2 and RB2 show Substitutions rejected instead: no thinner pipe would get air moving in them either.

3. Break the symmetry

   Set Diameter of G2 to 3 mm and calculate. G1 now needs more air than G2, so the B half has spare capacity, and part of its air goes on round the ring — through LOOP and back along RA2 — to help G1.

   Row	Expected
   m, RA2 and RB2	≈ 0.004 kg/s each	the air that goes round through LOOP
   RA2	Reverse flow	air runs against the drawing
   m, RA1 + RB1	0.0324 kg/s	always the total: 0.0208 + 0.0117
   Qn, COMP	1637.89 NL/min	the same demand as in lesson 7
   Pout, COMP	6.0626 bar	6 bar at G1 plus the losses on its way

   Where air runs against the direction an element was drawn in, the element gets the row Reverse flow, with the hint Air flows against the drawing direction - Pin and Pout are given along the flow. In a ring drawn all the way round in one direction, part of it always carries air backwards — that is normal, not a mistake. View → Pipe display → Flow animation shows the real direction in every part of the ring.

4. Know when the ring is simplified

   The network model takes pipes, fittings, tees, receivers, resistances, filters, dryers, drains, sensors and the LOOP itself. With anything else on the ring — a regulator, a consumer right on the ring — or with the ring fed from more than one point of the same main, EasyCAS falls back to the simplified model: every ring element carries the whole flow at one common pressure.

   LOOP then shows the error Isobaric ring model, and the row Reason under it says why. A value missing on a card (the Air condition of a filter, the catalog point of a dryer…) is asked for as usual, and the next calculation solves the ring as a network again. The flow animation does not run inside a simplified ring: there is no single direction to draw.

   Leave G2 at 3 mm: the demand is now that of lesson 7, 1637.89 NL/min, and part 2 sizes the compressor for it.

Part 2 — the compressor

5. Open Compressor selection

   After the calculation the compressor has a green row Compressor selection. Double-click it. The window is not a catalog search: you describe one candidate machine, and EasyCAS shows what each way of regulating it would cost on this network. The head of the window repeats the demand the calculation found.

   No cyclogram, worst case. Without a cyclogram (lesson 14) every consumer is taken at its maximum all the time, and the window says so in bold above the fields. That is right for the open pipes of this lesson, which blow non-stop; for real machines draw the cycle first, or the machine will be sized for a plant that never runs.

6. Describe the candidate

   Pick the Technology at the top left — SCREW for this lesson. The fields under it describe the machine (Rating standard, U, Candidate Qnom, Pmeas, EtaMot, EpsInt), the receiver and the pressure band (Vrec, Vaux, DpBand) and the control parameters of the class (Kidle, SigMin, Fvsd, Ki, TauStart). A field with a greyed value takes the value typical for the class when left empty. Fill:

   Field	Value
   Candidate Qnom	2500 NL/min	a size above the demand of 1637.89 NL/min
   Vrec	0.5 m³	the receiver; without it LOAD_UNLOAD cannot be compared

   The field is in the unit of the Air consumption list, as its label says — with the units of lesson 1 that is NL/min, so a 2.5 meant as cubic metres a minute would describe a machine a thousand times too small. The comparison is computed as soon as the technology is picked; Calculate at the bottom left recomputes it after you change a field. On the right, Machine requirements shows what the demand asks of a machine — the nominal flow needed (Q_nom required = Qn / U) and your Candidate Qnom; once the candidate can carry the demand, also its electrical power (P_nom_el at eta), its Specific power and the physical band of specific power at its pressure ratio — each with its source: input, ~ estimate or ~ calculated. Under it, Capacity control compares the four methods:

   Method	What it is
   LOAD_UNLOAD	Runs loaded or unloaded, with blow-down losses; needs a receiver and a pressure band on top of the network pressure.
   VSD	Variable speed drive — follows the demand.
   VAR_DISP	Variable displacement.
   MODULATION	Inlet throttling.

   For each: Pnom req — the nameplate pressure this method needs: the network pressure, 6.06 bar, for the methods that follow the demand, and the network pressure plus the pressure band DpBand (0.7 bar typical for a screw) for LOAD_UNLOAD, which must climb above the network to unload; ne — power per unit of air relative to the same machine at full load (VSD shows about 1.03: its drive loses 3 %); P mean — the mean power in kW; E in MWh a year and its cost in your currency, over your operating hours. The pane under the table lists the accepted method with every figure behind it.

7. Accept a method

   Double-click the row you choose (or select it and press Accept selected) and press OK; until then the pane under the table reads No control accepted yet. That is the only field of the window you choose rather than type. The compressor card now shows the energy of the machine: Pnom required, Mean electrical power, Annual electricity, Annual CO2, and At design load saying which point they refer to.

   From now on the energy and the money come from the machine, not from Compressed air cost — lesson 6 showed what happens without one. Zones, scenarios and the audit all take their cost from these rows. PDF card… writes the machine card with every figure marked by where it came from: entered, accepted, calculated, typical, not set.

   What you typed is stored with the compressor; the computed figures are not — they are recomputed whenever needed, so they never go stale.

8. Several machines: the compressor room

   A compressor room is one source with several machines on one header. Place the compressor room item instead of a single compressor — a source card with Element = comproom; type room in the catalog filter to find it; in its Compressor selection window two things appear: the Editing machine list at the top and the Compressor room group.

   Field	Meaning
   Compressors	Machines on the header, up to 8.
   Dispatch	How the demand is shared: CASCADE / SEQ fill machines in order of priority; BASETRIM runs base machines full and lets one trim machine follow; SHARE spreads evenly; BESTFIT tries every combination and takes the least power.
   Trim machine	Which machine follows the demand, for BASETRIM.

   On the sheet each machine gets a satellite symbol joined to the room by a bus, showing its own figures: Qnom with control and role, load factor, hours on load, mean power, energy a year. The satellites are not part of the network. The room itself shows the room rows, among them Room saving reserve — what BESTFIT would save against the law you chose.

9. Several compressors on one network

   A plant often has two compressor houses feeding one network at different points. Two compressors are joined into one network through the same LOOP element that closes a ring. Try it on a copy of the network of lesson 8: open lesson8.gf9 and save it as lesson9-two.gf9.

   1. Release the connection between P1 and T1 and put a tee there, d-1 in_Equal Tee as before: P1 into its inlet, one outlet on to T1.
   2. Place a second compressor from # Generic elements # below the main and rename it Comp2.
   3. Place loop from # Generic elements # between them. Join the free outlet of the new tee to one inlet of LOOP, and the red outlet of Comp2 to its other inlet.

   A LOOP whose two inlets lead to two different compressors is not a ring but a source tie: the two compressors and everything between them are calculated as one network. Press Calculate. In design mode all the compressors of a network hold one common outlet pressure — the lowest at which no consumer is short — and the network itself decides how much each one delivers.

   Click Comp2: Pout 6.0622 bar and Qn 1637.89 NL/min — all the air of B1 and B2. 6.0622 is 6 bar at B1 plus the losses of P2 (0.0466) and T1 (0.0156): there is no P1 on the way from Comp2. At one common outlet pressure Comp would have to push its air through 50 m of P1 that Comp2 does not need, so Comp delivers nothing, and P1 gets the warning Flow regime Re LAMINAR — very low flow through this pipe. That is what really happens when two machines with the same setting feed one network: the one nearer the consumers does the work.

   With a machine accepted in Compressor selection (step 6) for each of them, their capacity becomes a limit: a compressor that reaches it shows Capacity limit reached and the other delivers the rest; one that delivers nothing because the other holds a higher pressure shows Source unloaded. A compressor never takes air back.

   The Source list on the toolbar now offers both compressors. Each has its own rows, its own Compressor selection and its own accepted machine, and the report is written compressor by compressor. Joined into one network, they share its consumers: the cyclogram of either shows B1 and B2.

   If the common network cannot be solved, the tie is switched off: LOOP gets the alarm Sources not joined with a Reason, and each compressor supplies only its own part of the scheme. A ring fed from two points of one main, with no second compressor, is not a source tie: it is calculated with the simplified model (step 4).

When it does not work

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