FACTS AND RULES OF THUMB

**1. Compressed air is
the most inefficient utility in the plant **

to get 1 hp work from an air motor requires approximately 30 scfm inlet air at 90 psig

which requires 6-7 hp at the compressor shaft to produce this compressed air

assuming a 90% efficient motor, this translates into 7-8 hp of electrical power to deliver 1 hp of compressed air to the plant floor

Net efficiency is >12.5% of the input energy is available for useful work energy

But, on average half of the air is wasted

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**2. Operating costs reductions are worth more than sales and
productivity dollars **

Cost reduction vs. productivity / profitability

$100,000 in costs reduction = $2,000,000 sales at 5% net profit

Turning off 100 bhp of air compressor power @ $0.08 / kwh =

$56,000 year in operating costs at 8760 hrs = $1,120,000 sales

Savings potential in systems average

35% to 50% in smaller systems <300 bhp

20% to 30% in larger systems >1000 bhp

**3. RULE
OF THUMB Compressed air costs**

Total Compressed Air Costs @ $0.06 /kWh

A 100 scfm application costs $1.22 per hour to support or $10,655 per year at 8760 hrs

Or, $0.20 per 1000 scf

1 HP air motor costs - $3,516 per year (8760 hrs)

1 HP electric motor - $426 per year

Energy savings on the demand side of the system are at a factor of > 8X -10 X if electric vs. air powered

**4. Estimate energy
costs by :**

Compressors - ratio of actual amps to nameplate amps times the motor nameplate HP and convert to kW

or nameplate HP X .746 / .92 = kW

Refrig Dryers nameplate amps X 460v X .85 X 1.732 / 1000 = kW

or refrig dryer capacity / 200 cfm = kW

Cycling refrig dryers offer significant energy savings

Heated desiccant dryers capacity / 60 cfm = kW

Dewpoint based purge control offers significant energy savings

Cooling systems nameplate amps of pump motors and fans and convert to Kw

Or 3% of total compressor kW for cooling energy

Total kW x hours x $0.08 /kWh = total energy costs

Total energy costs / .70 = total operating costs $$$

(these calculations assume full load on each compressor but still provides an estimate of costs)

**5. Annual Electricity Cost (measurement formula)**

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__full load amps) x (voltage) x (1.732) x pf x hours x rate__

1,000

__Where:__

full load amps =
average of three phases

voltage =
line to line voltage

pf =
power factor

hours =
annual hours of operation

rate =
electricity cost in $/kWh

The full load amps and voltage are the measured values

Get power factor (pf) from motor manufacturer (use .85 as estimate)

__Example:__

__ __

__(230) x (460) x (1.732) x (0.85)
x 4,160 x $0.05__

1,000

= $ 32,398 per year

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**6. How to
Convert Acfm to Scfm **

SCFM = ACFM x
(Actual Inlet Pressure/14.5) X (520/(Actual Inlet Temperature + 460) X RH%
correction (.995 to .97)

Example: 500
ACFM compressor at 14.3 psia and 95F and summer conditions 95F and 60% RH

SCFM = 500 x
(14.3/14.5) x (520/(95+460) x .97

= 448 scfm

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Positive displacement compressor input power increases 1% for every 2 psi increase in discharge pressure

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**7. RECEIVER FORMULA**

l
Tank =__Tank
Height x (Tank Radius)² __(Gallons)

73.53

l
Tank =__Tank
Height X (Tank Radius)² __ (Cubic Ft)

l 550

l Tank =Pi(3.14) X (Tank Radius)²XHeight (Cubic Ft)

l Tank = 23.5 X (Tank Radius)² X Height (Gallons)

Gallons X 0.1337 = Cubic Feet

Cubic Feet X 7.48052= Gallons

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*8. Compressed Air Storage Primary Formula*

Capacitance calculations:

The size of the event in CF (rate of flow X duration)

Divided by the allowable pressure drop in PSI = the required capacitance in CF/PSI, then convert to gallons (X 14.5 psi per atmosphere X 7.48 gal/cf)

Or, divided by the capacitance in CF/PSI = the pressure drop in PSI

- (event CF) / (delta PSI) = capacitance CF/PSI , or
- (event CF) / (capacitance CF/PSI) = delta PSI

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**9. HEADER STORAGE**

*Compressed air moves at a limited velocity inside the pipe;
approximately 250 linear feet per second at 1.0 psid and 100 psig*

*Header Storage Problem *

How will the startup of a 600 scfm application located 1000 feet in header distance from the compressor room impact the header pressure? The piping consists of 600 ft of 4 and 1015 ft of 3. Also, assume that there is a flow controller in the system. What additional storage is required to control the pressure fluctuations to less than 2 psi?

Remember:

The size of the event in CF (rate of flow X duration)

Divided by the allowable pressure drop in PSI =

the required capacitance (convert to gallons)

Or, divided by the capacitance in CF/PSI = the pressure drop in PSI

*Header Storage Solution*

Size of the event

This application will remove air from the header for (1000 ft / 250 fps) = 4 seconds at a rate of 10 scf/sec (600 scfm / 60 sec) = 40 scf

Divided by the capacitance

with the piping capacitance at 7.25 scf/psi, the
pressure will drop

(40 scf / 7.25 scf/psi) = 5.5 psi.

Assuming we want to control the
pressure drop to less than 2 psi, the storage requirement would be:

(40 scf / 2 psid) X 14.5 psia = 290 scf X 7.48gal/cf = 2,169 gallons. You can subtract the existing volume contained in the header piping from this figure if it is significant (786 gallons in this example).

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