3/8 NPT. 1/2 Nylon Tube = .38' ID -- NOW becomes th

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Numasizing
?
What Do I use?
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What Is Numasizing?
Compressed air is an energy medium that must be
conserved, Numasizing has been developed to use
this resource more effectively.
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What Is Numasizing?
Numasizing is a technique which takes into
account all of the physical specifications of a
circuit and results in a tailor made
circuit designed to the customers
targets.
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What Is Numasizing?
Numasizing is not based on theoretical approach
or mathematical model. A data base of over
250,000 test firings of cylinders allows Numatics
to predict components and pressures with
confidence.
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The Numasizing Process is...
A B C D A. Establishing Customer
Objectives B. Delineating All Circuit
Specifications C. Selecting All Circuit
Components D. Results in Obtaining Customer Goals
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The Numasizing Process is...
A. Establishing Customer Objectives 1. Increase
Productivity 2. Optimum Energy
Utilization 3. Minimum Component Size
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The Numasizing Process is...
B. Delineating All Circuit Specifications 1. Ext
end Load Retract Load 2. Available
Pressure 3. Desired Extend Retract
Times 4. Conductor Length 5. Required Cylinder
Stroke, etc.
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INTO.
The Numasizing Process is...
Numasizing Computer Bank
Generating
C. Selecting All Circuit Components All
components in the pneumatic system starting at
the valve through the exhaust port such as the
valve, fittings, conductor, regulator and
actuator.
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The Numasizing Process...
D. Results in Obtaining Customer
Goals 1. Actuator Response Time 2. Compressed
Air Cost 3. Air Consumption 4. Pressure
Optimization 5. Compatible Sizes
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Effecting Pneumatics
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Knowledge of a pneumatic circuit design begins
with an understanding of the term Cv is
essentially a dimensionless number used to
express the CONDUCTANCE VALUE of a pneumatic
device. All fixed orifice devices in a pneumatic
system have a conductance value and therefore a
certain capability to flow.
Cv
The larger the Cv --- the greater the flow.
Typically, the greater the Cv of the entire
circuit, the faster the devices in the circuit
will respond.
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CVO 1.12
CVV 4.0
CVFC 4.5
CVP 4.8
CVC 6.0
CVF 7.0
2.99
1.05
1.03
1.01
1.000
CVV Valve CV CVFC Flow Control CV CVO
Orifice CV CVP Pipe CV CVC Cylinder Port
CV CVF Fitting CV
CVO 4.37
CVV 4.0
CVFC 4.5
CVP 4.8
CVC 6.0
CVF 7.0
2.99
2.47
2.19
2.06
2.000
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Relation Between Cvs Actuator Stroke Time (T)

• Stroke time is inversely proportional to Cvs
• Stroke time is directly proportional to system
  exhaust volume

Cvs T (sec) 1 4.0 2 2.0 4 1.0
8 0.5
1,092 ft/sec (vel of air - unconfined) 400 ft/sec
(vel of air - in a straight conductor) 80 in/sec
(vel of actuator piston rod)
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Cv of Fittings are Established by Calculating an
Equivalent Conductor Length Utilizing a K Factor
Once the ID and the line length has been
determined --- plus all the fittings equivalent
length has been added, we can calculate the Cv of
the CONDUCTOR.
Formula Le n K d Equivalent
Length number of fittings x K (feet)
factor x ID (inches)
Formula Le Lc Lt Equivalent
Length Conductor Length Total Length
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Cv of Fittings are Established by Calculating an
Equivalent Conductor Length Utilizing a K Factor
Fitting K Values Device
K Factor Swing Check Valves, fully
open 11 90 Degree Standard Elbow
2.5 45 Degree Standard Elbow 1.3 90
Degree Long Radius Elbow 1.5 Reducer (1
Size) 1.5 Enlarger (1 Size)
2 Y Fitting 1 Standard Tee Flow
Through Run 1.5 Flow Through Branch
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R3 x I.D. Long Radius K 1.5 R1.5 x I.D. Std.
Elbow K 2.5 R0 x I.D. SHARP 90Â K 5.0
BRANCH TEE K 5
ELBOWS
10 x I.D. minimumK 0
RUN TEE
K 1.5
x
BENDS
The more significant the change in flow
direction, the greater the restriction and
therefore the greater the K factor.
K 0
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What is the cylinders Cv ?
To calculate Cv we need to determine the smallest
I.D. (DS) of whatever component is in the port of
the cylinder.
Reducing the port size with a bushing reduces the
flow capability of the port. The smallest
orifice will determine the effective Cv .
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The Cv of a cylinder port is based on the
smallest I.D. entering that port
How shall we decide which ID conductor to use?
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(No Transcript)
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All openings d (diameter)
Various orifice discharge coefficients Cd and
their related Cv
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Optimized Drilling for Intersecting Holes
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There is an optimum conductor ID for each
application.
Choose the best commercially available size. Too
small a conductor ID and there is RESTRICTION
too large and it becomes a VOLUME CHAMBER. The
LENGTH of the conductor now must be considered.
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The Bends - a practical example
The direct distance between the valve and the
cylinder is 10. We choose to pipe the circuit
with an additional 10 (because it looks
better). If we used (8) 90 degree standard elbows
(K2.5) between the valve and cylinder, it
would have an equivalent length of 10 Le
nKd Le 8 x 2.5 x 1/2 10 We have
penalized the cylinders speed by reducing the
efficiency of the circuit. By identifying the
weak link, we can improve our circuit.
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The most common way of adjusting a cylinders
speed is with a FLOW CONTROL.
A flow control cannot conserve compressed air,
reduce force, or speed up a circuit.
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Quick Exhaust Valve
The most common device selected for improving
cylinder speed is a Quick Exhaust Valve.
This device has a limited life -- the disc is
slammed (full line pressure) twice every cycle.
No breaking or control of the cylinder will occur
either.
Quick exhaust allows the cylinder to exhaust at
the cylinder port, not back through the valve or
through a flow control.
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Loads on a Cylinder
On any cylinder, there are three LOADS resisting
movement. The cylinder must overcome each load
before it can extend or retract. Lets examine
each load separately. Cylinders, typically, must
have some friction in sealing the piston against
the cylinder wall and also the rod --- Load
Lf. Work that the cylinder performs is expressed
as Le for extend and Lr for retract directions
of movement. The crucial load, however, is Lx
--- the EXHAUST BACK PRESSURE LOAD.
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SAMPLE PROBLEM Move a load in the specific
times, given 75 psig supply.
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Cylinder Response Time in Seconds
Pdm 36psi Pressure differential at inception
of motion Pd Running pressure differential
varies during cycle (see curve) Td 0.14
Time delay primarily due to exhaust preload, a
minimum delay due to static
friction of cylinder seals and a negligible
solenoid time delay (Ty). Tm 0.26
Time cylinder piston is in motion Te or Tr
0.40 Time to extend or Time to
retract (TdTm) Ty 0.008 Solenoid time
delay DP 10psi Pressure drop from Supply
Pressure
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Pressures Fit The Load
Select the cylinder pressures based on required
force.
Consider what the cylinder must do in each
direction and select the required pressure for
each individual action.
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Now that the ideal pressures have been selected,
what results can be observed?
LOAD 10 in 0.45 SEC
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Cylinder Response Time in Seconds
Pdm 36psi Pressure differential at inception
of motion Pd Running pressure differential
varies during cycle (see curve) Td 0.14
Time delay primarily due to exhaust preload, a
minimum delay due to static
friction of cylinder seals and a negligible
solenoid time delay (Ty). Tm 0.26
Time cylinder piston is in motion Te or Tr
0.40 Time to extend or Time to retract
(TdTm) Ty 0.008 Solenoid time delay DP 2
psi Pressure drop from Supply Pressure
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Dual Pressure Response Time in Seconds
Pdm 36psi Pressure differential at inception
of motion Pd Running pressure differential
varies during cycle (see curve) Td 0.09
Time delay primarily due to exhaust preload, a
minimum delay due to static
friction of cylinder seals and a negligible
solenoid time delay (Ty). Tm 0.19
Time cylinder piston is in motion Te or Tr
0.28 Time to extend or Time to retract
(TdTm) Ty 0.008 Solenoid time delay DP 0.4
psi Pressure drop from initial Supply Pressure
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2 75/75 1.50 .40/.45 70
yes 476 257 4.75
1.34 2B 75/75 2.00 .23/.14 163
no 1733 287 2.08
4.89 3 49/21 1.50 .40/.45
70 no 229 230 4.75
0.65 4 61/35 1.50 .40/.45
70 no 289 217 4.75
0.81 5 70/31 1.25 .40/.45
70 no 219 190 4.75
0.62
Compressor pressure for all surveys are kept at
100 PSIG Valve Cvs and costs for survey s 2,
2B, 3, 4 and 5 are respectively, 2.05 - 85, 2.05
- 85, 0.41 - 58, .25 - 45, 0.25 - 45 3/8 NPT
Pipe conductor cost 25 and 1/8 NPT costs
20 Cylinder costs for 2.0, 1.5 1.25 bores
are respectively 152, 102 and 75 A pair of
flow controls cost 20 and a pair of regulators
cost 50 Survey 2 (85251022025 257)
O.E.M. would select 5 because of
minimum capital investment Survey 2B
(852515210250 287) End
user would select 2B because of low
cost/pc Survey 3 (582010250 230)
Facilities would select 3
because of lowest pressure Survey 4
(452010250 217)
demands or 5 because of minimum HP
requirement Survey 5 (45207550 190)
depending on the major
objective of facilities engineer
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Dynamic Seals With Lubrication
Flow of O-Ring into Metallic Surfaces
The theory has been proposed and generally
accepted that the increase of friction on
standing is caused by the rubber O-ring flowing
into the microfine grooves or surface
irregularities of the mating part. As a general
rule for a 70º durometer rubber against an 8
micro-inch surface, the maximum break-out
friction which will develop in a system is three
times the running friction.
Friction is always a factor with dynamic seals.
Obviously, the more dynamic seals --- the greater
the friction. We can assume the cylinder has
been sized properly to overcome the Le and Lr
loads. Lx --- the exhaust back pressure load
--- is determined by the ability of the
cylinders ports to allow exhaust to escape the
Cv of the cylinder.
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Effective Orifice --- Valves
As with cylinders, VALVE Cv is determined not by
the port of the valve, but by the smaller orifice
of the fitting in the port.
Port Size (NPT) 10/32 1/8 1/4 3/8 1/2 3/4 1
1 1/4
Valve Size MK 3 .18 .35 MK 7 .2 .4 MK
8 .3 .8 1.0 MK 15 .4 1.1 1.4 1.5 MK
55 .6 2.0 2.85 4.0 5.0 5.55 there is no
improvement in the valves Cv beyond this point
Consider System Cv equals the combination
of Cylinder Cv Conductor and Fittings Cv
Valve Cv and any additional devices in the
circuit
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Critical Pressure Ratio
At start, there is 80 psig available and the
needle valve is closed. As the needle valve is
slowly opened, in DP increments of 5 and 10 psi,
the flow is noted. OBSERVEEven with a larger DP
--- FLOW DOES NOT CONTINUE TO INCREASE. Air
has reached CRITICAL FLOW and FLOW cannot
increase.
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Terminal Velocity
A jet engine takes air in, compresses the air
with fuel, and then ignites the mixture. At some
point, air cannot enter the engine any faster. No
matter how much more fuel is added --- the plane
cannot go any faster. The limiting factor is AIR
FLOW, supply to the engine. In a pneumatic
circuit, the cylinder reaches TERMINAL
VELOCITY when the air cannot enter and/or exit
the circuit any faster. This occurs at
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Terminal Velocity for a Cylinder
Therefore HOW FAST CAN AN AIR CYLINDER CYCLE?

• IF a properly sized cylinder had
• air supplying and
• exhausting at the
• CRITICAL PRESSURE DROP RATIO THE PNEUMATIC
  CYLINDER WOULD BE AT
• TERMINAL VELOCITY for that particular circuit.
• To improve cycle time we must improve the
  smallest Cv of the system and maintain the
  critical pressure ratio.

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If Energy Waste Was This Obvious, Youd Put A
Stop To It
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Benefits of Dual Pressure
Why NUMASIZE?
Percent Cost Savings Attainable Utilizing
Numasizing inIndustrial Pneumatic Systems
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Air Leaks
Annual Cost of Compressed Air LeakageThrough
Orifices of Various Diameters
Compressed air is ENERGY that must be CONSERVED.
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Resource Cost
Compressed Air Cost
Average Compressor H.P. In Use based upon 4
s.c.f.m./H.P. - 30/1000 s.c.f. - 6.75/kwh
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Total Energy Production
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CONSUMPTION
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Electricity Production
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Conceptual View Point
The preceding two slides/charts are based on
Energy Information Administration (EIA) surveys.
There were some slight differences between the
1990 and the updated figures for 1990 as
published in the October 1991 report. Some
variations also showed up when comparing the
figures of National Business, Electric World,
Pipe and Gas Journal, etc. with those of EIA.
Since the deviations were minor (mostly due to
independent rounding), we have reconciled all of
them as to render uniform results. From a
conceptual viewpoint and from the ultimate
potential savings and increased productivity
aspect, they have no bearing as the differences
are miniscule. Also, we have incorporated in
these figures the fact that 1,033,100 BTUs (EIA
from utilities) is required to generate 100 Kwhrs
of electricity (due to losses in conversion and
transmission), while 300 Kwhrs are needed to
produce 1,024,000 BTUs of heat (1 Kwhrs 3413
BTUs).
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Formulae
Those familiar with electricity will recognize
Ohms Law. In pneumatics, FLOW is represented by
Q, T is temperature in Rankin, G is the specific
gravity (assume 1 for air), P1 and P2 are
pressure expressed in PSIA.

• Flow of electrons is very similar to flow of air
  molecules
• Think of CONDUCTANCE as the reciprocal of
  RESISTANCE
• Whose formula came first?

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Your Most Important Design Tool