Lesson 10

Setting up an audit

A design calculation asks what the compressor must do. A plant already has a compressor, set to some pressure years ago, and the question is what that setting really delivers — and what it costs. This lesson turns the network of lesson 7 into a plant: the compressor is set too high, one consumer needs less than it gets, the pipes leak, and the audit puts a figure on each of these.

Design and audit: two questions

The scheme is the same in both modes; the audit only adds rows of its own to the cards. So an audit usually starts from a design scheme of the plant, as this one does.

1. Switch the scheme to audit mode

   Open lesson8.gf9 and save it at once as lesson10.gf9. Check that Diameter of P1 is 25 mm. Then switch the mode with the Design | Audit | Dynamic switch on the toolbar, or with View → Audit mode. On a narrow window the switch shows only the current mode — the single Design button on the screenshots of lessons 7–9 — click it and the mode moves on to the next one, Design → Audit; the screenshots of this lesson were taken in a wider window, where all three parts are shown.

   Audit mode is marked in teal wherever you look:

   • the Audit part of the switch is highlighted;
   • a teal frame surrounds the sheet, with an AUDIT MODE banner along its top;
   • the window title shows [Audit] before the file name;
   • the Calculate button reads Audit.

   Switching the mode deletes the results of the other mode and marks the file as changed. The mode is saved in the file: lesson10.gf9 will open in audit mode.

2. Tell the audit how the compressor is set

   Click COMP. Below its usual rows the Item tab now has rows whose names start with Audit; in design mode they are hidden. They are added to every card the moment the mode is switched, and kept in the file; back in design mode they are only hidden. Most are marked ○ — optional, they may stay empty. One is required:

   Row	Value
   Audit pressure	6.5 bar	the unload (upper) setting of the machine, as found on site

   Lesson 7 found that this network needs about 6.17 bar. The plant runs at 6.5: someone once added a margin, as plants usually do. The audit will show what the margin costs.

   The other rows of the compressor wait for later steps and lessons:

   Row	For
   Audit capacity	What the machine can deliver. Empty — not limited, unless a machine is accepted in Compressor selection. Step 9.
   Audit leak …	The leak test. Step 9.
   Audit power, Audit hours	The measured mean electrical power and the hours of this source a year — energy by measurement rather than by model. Lesson 11.
   Audit measurement data	Opens the import of logger files. Lesson 11.

   The consumers have audit rows too — Audit utilization, Audit required pressure, Audit fixed flow — and every pipe has Audit roughness, for an old or corroded line. All of them may stay empty for now.

3. Run the audit

   Press Audit. The audit first runs an ordinary design calculation of the same scheme: it gives each consumer its design point and each element its design pressure drop, and fits the element characteristics to them. Then it calculates the other way round — from the compressor at 6.5 bar down to the consumers.

   That is why an audit needs the same data as a design and may ask for a missing value the same way. With Audit pressure empty, the source gets the alarm Audit pressure is not set.

   About the screenshots of this lesson. As in lessons 7–9, they were taken on a file in which the items kept the IDs the program gave them, with the names as plain captions on the sheet. Where your windows show names, the screenshots show ID7 for COMP, ID13 for P1, ID3 for T1, ID5 and ID6 for P2 and P3, ID1 and ID2 for B1 and B2.

4. Read the consumers

   Click B1, then B2. A consumer already has an input row called Pin — its design point — so the pressure it really gets is called Pin actual. Under it come m and Qn, now the flow at that pressure, and Design pressure repeats what it needs. Above them sit the three optional audit rows of a consumer, ○ Audit utilization, ○ Audit fixed flow and ○ Audit required pressure, all empty for now.

   Row	B1	B2	Unit
   Pin (design point)	6	6	bar
   Pin actual	6.3275	≈ 6.36	bar
   Qn	1097.29	≈ 620	NL/min

   Both get more than their 6 bar, and both blow more than in lesson 7. In an audit a consumer is an orifice fitted to its design point: unless it is told otherwise, its flow follows the pressure it gets.

   Check by hand: the flow follows the pressure

   An open pipe at these pressures is choked (lesson 1), and the mass flow of a choked orifice is proportional to the absolute pressure in front of it. B1 blew 1048.25 NL/min at 6 bar gauge, 7 bar absolute. At 6.3275 bar gauge it blows 1048.25 × 7.3275 / 7 = 1097.29 NL/min — exactly the Qn on the screen. A third of a bar over the design costs B1 almost 5 % more air.

   Audit fixed flow. Some consumers take the same air at any pressure — a machine with its own regulator, a device with a flow controller. For them type 1 into Audit fixed flow: they then take their design flow whatever the network gives. Try it on B1 — its Qn drops back to 1048.25 — and clear the row again. An open pipe follows the pressure by default.

5. Read the network and the compressor

   The pipes have the same rows as in design — m, Pin, Pout, delta P, v — so Pipe display, the pressure profile and the Problems panel work in audit mode as they did in lessons 7 and 8. P1 now carries more air than in the design and loses a little more: about 0.11 bar.

   Row, COMP	Design (lesson 7)	Audit	Unit
   Pout	≈ 6.17	6.5	bar
   Qn	1637.89	1717.19	NL/min

   The same two consumers take about 5 % more air than the design asked for, and nothing on the sheet has changed — only the setting of the compressor.

   The source also gets rows of the audit. What it can deliver is not known yet: no machine is accepted in Compressor selection and Audit capacity is empty, so the source holds its setting at any demand and says so in one row, Audit: capacity not limited. Once the capacity is known (step 9) that row gives way to Qn capacity and Load at maximum. An accepted machine that runs load/unload adds Pressure band: it works in a band below its setting — at maximum load the network is taken at the bottom of the band, at average load at its middle.

   The cost rows — Cs, Electricity consumption, CO2 — are there too, under Audit: cost estimated. Without a machine of its own the audit prices the air at average load with Compressed air cost of Input parameters, as the design does, and turns that money into kWh with Electricity cost. Until the next step the average load is simply the maximum one.

6. Say how long each consumer runs

   So far both open pipes blow non-stop. A blow-off station rarely does. Fill Audit utilization — the share of the time the consumer runs:

   Item	Audit utilization
   B1	50 %
   B2	30 %

   Press Audit. The audit now works at two load points:

   Point	Demand	Where the results go
   Maximum	All consumers at their design demand.	The rows of all elements — the worst case. They do not change.
   Average	The flow of each consumer times its utilization.	Qn average, Pout average and the energy on the source.

   Row, COMP	Expected	Unit
   Qn	1717.19	NL/min	maximum load, unchanged
   Qn average	747.32	NL/min	what the year is paid for

   Check by hand

   Half of B1 and 30 % of B2 at maximum load would be 0.5 × 1097 + 0.3 × 620 = 735 NL/min. The audit gives a little more: at average load less air goes through P1, it loses less, and the open pipes get a little more pressure — 6.46 bar instead of 6.33 — so they blow a little more.

   With a cyclogram (lesson 14) the audit takes the running share of each consumer from the cycle instead, and checks the pressure at every moment of it.

7. Find the lowest setting that still works

   Click COMP and read the recommendation rows of the audit:

   Row	Expected	Unit
   Min audit pressure	6.1678	bar	the lowest setting with no consumer short of pressure
   Set point reduction	0.3322	bar
   Qn saving	33.13	NL/min	at average load

   Check: the audit meets the design

   Min audit pressure is the setting at which B1, the consumer hardest to reach, gets exactly its 6 bar. That is the question the design calculation of lesson 7 answered, so the two figures agree: about 6.17 bar. And the saving follows from step 4: 0.33 bar less at about 7.46 bar absolute is 4.4 % less air through the open pipes — 4.4 % of 747 NL/min is 33.

   Power saving, Energy saving and Cost saving on the compressor need the mean electrical power of an accepted machine (Compressor selection, lesson 9); without one the card shows only Qn saving. The money is not lost: the Problems panel prices the same saving with the compressed air cost (step 7). The minimum is searched only when you press Audit; after an automatic recalculation the last value stays, marked not updated - press Audit to search again.

   The row Bottlenecks: P2, P3, P1 names the three pipes with the highest loss per metre at maximum load — the thin branches first, then the 25 mm main. Each of them carries its own row Bottleneck #1…#3 with that loss in mbar/m.

8. Tell the audit what a consumer really needs

   B2 was given 6 bar in the design only because the open pipe of lesson 1 had it. Say the blow-off at B2 does its job at 4 bar. Type 4 into its Audit required pressure and press Audit.

   The flows do not change — the network still gives B2 about 6.36 bar and it still blows about 620 NL/min. What changes is the judgement: B2 is now checked against 4 bar, and everything it takes above what 4 bar would give is artificial demand — air bought only because the pressure is higher than needed. Min audit pressure stays where it was: B1 still decides it.

   The compressor counts that air in rows of its own:

   Row, COMP	Expected	Unit
   Qn artificial demand	62.45	NL/min	at average load
   Artificial demand share	8.36	%	of Qn average
   Artificial demand energy	1.4	MWh	a year
   Artificial demand cost	181	€	a year

   Check by hand: the artificial demand

   At average load B2 gets 6.47 bar and, running 30 % of the time, takes 188.80 NL/min. At 4 bar it would take 589.64 × 5 / 7 = 421.17 NL/min while running — the figure of lesson 8 — and 30 % of that is 126.35. The difference, 62.45 NL/min, is the artificial demand.

   Open the Problems panel. Besides the errors and warnings it now lists the measures the audit found, each with its effect — every figure calculated by running the same network again with the change. The largest effect comes first:

   Problem	On	Value	What it says
   The consumer gets more pressure than it needs and uses more air	B2	181 €/yr	6.471 bar instead of 4 bar; 188.797 NL/min instead of 126.352 NL/min at average load. A pressure regulator set to 4 bar before it saves these 62.4 NL/min.
   The same	B1	100 €/yr	6.459 bar instead of 6 bar; 558.521 NL/min instead of 524.126 NL/min. B1 too gets more than its design point, and a regulator set to 6 bar would save 34.4 NL/min.
   The compressor setting can be lowered	COMP	161 €/yr	6.5 bar → 6.168 bar without a pressure deficit at maximum load — the 33 NL/min of step 5. Lower it in steps of about 0.1 bar, watching B1.
   Undersized pipe	T1	0.015 bar	9.34 m/s and 397 Pa/m at maximum load against the limits of 9 m/s and 200 Pa/m: the 23 mm bore of the tee carries all the air of the 25 mm main. The value is the pressure a larger size would gain.

   The two open pipes share one problem, so they come as one group, [-] 2 × TUBE_BLOWING, with the sum of their effects, 281 €/yr (groups: lesson 8). The money is an estimate from the compressed air cost, because no machine is accepted; with one the figures come from its power.

   P2 and P3 lose more per metre than T1 — they head the Bottlenecks row — yet are not listed. A pipe enters the list when its velocity or its loss per metre is over the limit (2 mbar/m) and the audit can name a larger size that helps. T1 is over the 9 m/s of a main, so the size the velocity limit asks for is larger than its bore. P2 and P3 stay under the 15 m/s of a branch, and the audit found no larger size to offer for them.

   The effects cannot simply be added: the regulator before B1 and the lower setting save the same air. All measures together is a separate calculation with all of them at once; it is in the summary at the bottom left of the panel — in a narrow panel the summary shrinks to coloured squares, and the figure is in their tooltip. Do not press Apply yet — lesson 12 applies the measures in a scenario of their own and counts what they pay back.

9. Measure the leaks

   Every plant leaks, and the leaks run day and night, whatever the consumers do. They are measured with the consumers off. The quickest test needs nothing but a watch: the compressor alone keeps the network up, and you time how long it runs loaded and unloaded.

   The load/unload test needs the capacity of the machine. Fill on COMP:

   Row	Value
   Audit capacity	2500 NL/min	150 Nm³/h, the machine of lesson 9; typed in the unit of Qn
   Audit leak t load	6 s	loaded
   Audit leak t unload	54 s	unloaded

   Press Audit. The leakage is the capacity times the loaded share of the time: 2500 × 6 / (6 + 54) = 250 NL/min.

   Row, COMP	Expected	Unit
   Qn leak at test	250	NL/min	at the test pressure
   Qn leak total	247.39	NL/min	at the pressures of the network, a little below 6.5 bar
   Qn leak unidentified	247.39	NL/min	all of it: no leak is drawn on the sheet
   Leak share	25.04	%	five times the 5 % target of the leak measure
   Leak energy	5.6	MWh	a year
   Leak cost	723	€	a year, estimated from the compressed air cost
   Load at maximum	78.41	%	1960.35 NL/min of the 2500

   Audit leak test pressure was empty, so the first audit filled it in with the pressure of the test — the audit pressure, 6.5 bar here. It stays there: a lower setting later gives less leakage, as it should, because the test keeps the pressure it was really made at.

   The leaks drawn on the scheme — leak orifices, drains — are subtracted from the measured total; the rest, the unidentified leaks, are spread over the nodes of the network in proportion to the length of the pipes at each node. Here most of them land on the two ends of the 50 m main. They load the pipes at both load points, so Min audit pressure moves up a little — 6.1832 bar instead of 6.1678 — and Qn average grows by the leaks, to 995.68 NL/min. The lower setting is now worth more, 42.1 NL/min, since the leaks too lose less at a lower pressure. The Problems panel gets a measure, Leakage above the target: repair down to the 5 % advisory threshold. It is a recommendation, not a warning: the warnings of the leak share count only leaks drawn on the sheet as leak orifices, and none is drawn here. Now that the capacity is known, the compressor measures appear too: the average load is only 995.68 / 2500 ≈ 40 % of what the machine can deliver, and Low average load suggests comparing a variable speed drive or a smaller machine in Compressor selection.

   The other method: pressure decay. Switch the compressor off and note how fast the pressure falls: Audit leak P start, Audit leak P end, Audit leak t decay. The leakage is then V × (P start − P end) / (R × T × t), with V the volume of the pipes plus the receivers that have a Volume row; Audit leak volume overrides it. In this network the pipes hold only about 27 litres, so it would empty in seconds — a decay test needs a real plant with its receivers.

10. Save it

    Save lesson10.gf9. It opens in audit mode. Lesson 11 replaces the guesses with measurements — gauges on the network and logger files — and lesson 12 applies the measures and counts their payback.

When it does not work

Try this next

• Set Audit pressure to 6.0. B1 now gets less than 6 bar and shows a Pressure deficit; Min audit pressure is reported as an alarm because the current setting is too low. Set it back to 6.5.
• Type 0.5 into Audit roughness of P1 — a badly corroded main — and watch Min audit pressure climb.

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