Numasizing
What Do I use?
What Is Numasizing?
Compressed air is an energy medium that
must be conserved, Numasizing has been
developed to use this resource more
effectively.
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 customer’s
targets.
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.
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
The Numasizing Process is...
A.
Establishing Customer Objectives
1.
Increase Productivity
2.
Optimum Energy Utilization
3.
Minimum Component Size
The Numasizing Process is...
B.
Delineating All Circuit Specifications
1.
Extend Load & Retract Load
2.
Available Pressure
3.
Desired Extend & Retract Times
4.
Conductor Length
5.
Required Cylinder Stroke, etc.
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.
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
Effecting Pneumatics
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.
CVV = 4.0
CVFC = 4.5
CVO = 1.12 CVP = 4.8
CVC = 6.0
CVF = 7.0
2.99
1 +
2
CVV
1 =
2
CVFC
1.05
1
1.03
CVS 2
1.01
1.000
1 +
2
4
CVV
CVFC
CVO
CVP
CVC
CVF
1 =
2
4.5
=
=
=
=
=
=
1
2.99 2
Valve CV
Flow Control CV
Orifice CV
Pipe CV
Cylinder Port CV
Fitting CV
CVV = 4.0
CVFC = 4.5
CVO = 4.37 CVP = 4.8
CVC = 6.0
2.99
2.47
2.19
2.06
2.000
CVF = 7.0
Relation Between Cvs & Actuator
Stroke Time (T)


Stroke time is inversely proportional to Cvs
Stroke time is directly proportional to system
exhaust volume
Cvs
1
2
4
8
T (sec)
4.0
2.0
1.0
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)
Cv of Fittings are Established by Calculating an
Equivalent Conductor Length Utilizing a K
Factor
Formula:
Le’ = n K d”
Equivalent Length = number of fittings x K
(feet)
factor x ID (inches)
Formula:
Le + L c = L t
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.
Equivalent Length + Conductor Length = Total Length
Cv of Fittings are Established by Calculating an
Equivalent Conductor Length Utilizing a K
Factor
Fitting K Values:
Device
Swing Check Valves, fully open
90 Degree Standard Elbow
45 Degree Standard Elbow
90 Degree Long Radius Elbow
Reducer (1 Size)
Enlarger (1 Size)
“Y” Fitting
Standard Tee
Flow Through Run
Flow Through Branch
K Factor
11
2.5
1.3
1.5
1.5
2
1
1.5
5
R=3 x I.D. Long Radius K =1.5
R=1.5 x I.D. Std. Elbow K = 2.5
R=0 x I.D. SHARP 90° K = 5.0
BRANCH TEE K = 5
ELBOWS
10 x I.D. minimum
K=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
K=1
Y Fitting
K=1
What is the cylinder’s 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.
3/8 NPT
The smallest orifice will determine
the effective Cv .
ID .493 - 3/8 NPT Pipe
SCH. 40
Formula:
2
Cv (port)= 23 x Ds
Ds = smallest I.D. in a port
1/4 NPT
Cv 5.6
Cv 3.0
C 3.0
ID .364 - 1/4 NPT Pipe, SCH .40
The Cv of a cylinder port is based on the
smallest I.D. entering that port
Ftg. 1/2 Tube to 3/8 NPT
ID = .406
3/8
NPT
.406” ID Controlling
Dimension
1/2 Steel Tube - .43”
ID
How shall we decide which ID conductor to use?
The Cv of a cylinder port is based on the
smallest I.D. entering that port
Ftg. 1/2 Tube to 3/8 NPT
ID = .406
3/8
NPT
.406” ID
1/2 Nylon Tube = .38” ID -- NOW
becomes the Controlling Dimension
for the CYLINDER PORT.
A fitting that accepts steel tubing will have a larger effective orifice than
a fitting for plastic or poly tubing because the wall thickness for plastic /
poly tubing is so much greater.
Flow
Cd = 0.98
Cv = 28.8d 2
Flow
Cd = 0.92
Cv = 27.0d 2
Flow
Cd = 0.82
Cv = 24.1d 2
Flow
Cd = 0.80
Cv = 23.5d 2
Flow
Cd = 0.72
Cv = 21.2d 2
Flow
Cd = 0.65
Cv = 19.1d 2
Flow
Cd = 0.61
Cv = 18.0d 2
Flow
Cd = 0.53
Cv = 15.6d 2
All openings = d” (diameter)
Various orifice discharge coefficients Cd and their related Cv
Optimized Drilling for Intersecting Holes
There is an optimum conductor ID for
each application.
3/16” I.D. Tubing
1/8” I.D. Tubing
Conductor
I.D.
1/16” I.D. Tubing
0
0
Time
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.
The Bends - a practical example:
Given: 3/8 NPT Sch.40 pipe has an I.D. of 0.493” or essentially 1/2”
Conductor Length
Cv
10’ (3/8 NPT)
3.0
20’ (3/8 NPT)
2.1
30’ (3/8 NPT)
1.75
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 (K=2.5) 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.
The most common way of adjusting a cylinder’s speed is with a
FLOW CONTROL.
TO CYLINDER
A flow control cannot
conserve
compressed air,
reduce force, or
speed up a circuit.
MODEL SELECTION AND FLOW CAPACITY (CV) CHART
Model Number
Controlled
Size
Free Flow
*
Flow
1/4 NPTF
2FC2
2.3
2.0
3/8 NPTF
3FC2
2.7
2.4
1/2 NPTF
4FC3
6.0
5.5
3/4 NPTF
5FC3
7.5
6.0
** For optional bottom port, add B to model number (e.g..., 4FC3B
FREE FLOW
CL
BOTTOM PORT
If utilize bottom port
option:
2.44
2.86
6.36
7.95
Quick Exhaust Valve
E
P
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.
ACTUATOR
ACTUATOR
EXHAUST
PRESSURE
APPLIED
EXHAUST
PRESSURE
BLOCKED
Quick exhaust allows the cylinder to exhaust at the cylinder port, not
back through the valve or through a flow control.
Loads on a Cylinder
F
Lx
Lf
Pe = 75
Le
Pr = 75
On any cylinder, there are three
LOADS resisting movement. The
cylinder must overcome each load
before it can extend or retract.
Let’s examine each load
separately.
Cylinders, typically, must have
some friction in sealing the piston
against the cylinder wall and also
the rod --- Load Lf.
Lx
Lf
Pe = 75
F
Pr = 75
Lr
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.
Extend Stroke
LOAD
60 # in .4 SEC
Return Stroke
Supply
Exhaust
LOAD
10 # in 0.45 SEC
75 PSIG
SAMPLE PROBLEM:
Move a load in the
specific times, given
75 psig supply.
75 PSIG
Te
=
=
=
Tm
Te or Tr
Ty
DP
=
=
=
=
Tr
Tm
D
Pdm
Pd
0
Pdm
Pd
Td
or
0
Exhaust
Supply
P
75psig
75psig
End of Stroke
Ty
Pressure PSIG
Cylinder Response
Time in Seconds
Td
Time - Seconds
36psi Pressure differential at inception of motion
Running pressure differential varies during cycle (see curve)
0.14 Time delay primarily due to exhaust preload, a minimum
delay due to static friction of cylinder seals and a negligibl
solenoid time delay (Ty).
0.26 Time cylinder piston is in motion
0.40 Time to extend or Time to retract
(Td+Tm)
0.008 Solenoid time delay
10psi Pressure drop from Supply Pressure
Pressures Fit The Load
F
Select the cylinder pressures
based on required force.
Lx
Lf
Pe = 49
Consider what the cylinder
must do in each direction
and select the required
pressure for each individual
action.
Lx
Lf
Pe = 49
Le
Pr = 21
F
Pr = 21
Lr
Extend Stroke
Supply
Exhaust
LOAD
60 # in 0.40 SEC
Now that the ideal
pressures have been
selected, what results
can be
observed?
Return Stroke
LOAD
10 # in 0.45 SEC
49 PSIG
21 PSIG
49 PSIG
21 PSIG
Te
Ty
Pressure PSIG
=
=
=
Tm
Te or Tr
Ty
DP
=
=
=
=
Tr
Tm
Exhaust
Supply
21psig
49psig
DP
Pd
Pdm
0
Pdm
Pd
Td
or
0
End of Stroke
Cylinder Response
Time in Seconds
Td
Time - Seconds
36psi Pressure differential at inception of motion
Running pressure differential varies during cycle (see curve)
0.14 Time delay primarily due to exhaust preload, a minimum
delay due to static friction of cylinder seals and a negligibl
solenoid time delay (Ty).
0.26 Time cylinder piston is in motion
0.40 Time to extend or Time to retract (Td+Tm)
0.008 Solenoid time delay
2 psi Pressure drop from Supply Pressure
=
=
=
Tm
Te or Tr
Ty
DP
=
=
=
=
Tr
Tm
D
Pd
Pdm
0
Exhaust
Supply
20psig
75psig
Initial:
Exhaust
Supply
75psig
75psig
Time saved from
initial pressure
conditions (0.11)
P
Time - Seconds
End of Stroke
or
End of Stroke
Ty
0
Pdm
Pd
Td
Te
Td
Pressure PSIG
Dual Pressure
Response Time
in Seconds
00
36psi Pressure differential at inception of motion
Running pressure differential varies during cycle (see curve)
0.09 Time delay primarily due to exhaust preload, a minimum
delay due to static friction of cylinder seals and a negligibl
solenoid time delay (Ty).
0.19 Time cylinder piston is in motion
0.28 Time to extend or Time to retract (Td+Tm)
0.008 Solenoid time delay
0.4 psi Pressure drop from initial Supply Pressure
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 Cv‘s 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 ($85+$25+$102+$20+$25 = $257)
O.E.M. would select #5 because of minimum capital investment
Survey #2B ($85+$25+$152+$102+$50 = $287)
End user would select #2B because of low cost/pc
Survey #3 ($58+$20+$102+$50 = $230)
Facilities would select #3 because of lowest pressure
Survey #4 ($45+$20+$102+$50 = $217)
demands or #5 because of minimum HP requirement
Survey #5 ($45+$20+$75+$50 = $190)
depending on the major objective of facilities engineer
2-1/2 Seconds
40 Minutes
300 Hours
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
cylinder’s ports to allow exhaust to escape; the Cv of the cylinder.
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
5.0
5.55*
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
MK 55
.6
2.0
2.85 4.0
1.5*
*there is no improvement in the valve’s 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
Critical Pressure Ratio: P1
Q (SCFM)
P2
Flow
meter
P
Needle Valve
P1 = P+ P2
P1
P P2
Q
80
“
“
“
“
“
“
“
“
“
0
28
38
43
45.5
46
“
“
“
“
0
10
20
30
40
45
50
60
70
80
80
70
60
50
40
35
30
20
10
0
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.
OBSERVE
Even with a larger DP --- FLOW DOES NOT
CONTINUE TO INCREASE.
Air has reached CRITICAL FLOW and FLOW
cannot increase.
Terminal velocity occurs at the CRITICAL PRESSURE RATIO
P2
P1 = .53
P
P1 = .47
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
P any faster. This
occurs at P1 = .47
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.
H
C
F
E
2
1
0
3 4 5
6
7
8
x1000r/min
If Energy Waste Was This Obvious,
You’d Put A Stop To It
Benefits of Dual Pressure:
Retract Pressure (PSIG)
90
Why NUMASIZE?
Extend Pressure (PSIG)
Percent Cost Savings
Attainable
Utilizing Numasizing in
Industrial Pneumatic
Systems
90
80
0% 5.0
70
60
50
40
30
9.2 14.8 19.1 24.0 28.9
80
10.0 14.1 19.8 24.0 28.9 33.9
70
18.4 24.0 28.2 33.9 38.8
60
29.6 33.9 38.8 43.7
50
38.1 43.0 47.9
40
47.8 52.9
30
57.8
Air Leaks:
Annual Cost of Compressed Air
Leakage
Through Orifices of Various
PSIG 1/32 3/64 1/16 3/32
1/8
Diameters
30 $ 78 $172 $310 $ 689 $1239
50
$112 $252 $448 $1009 $1784
70 $146 $330 $566 $1316 $2337
90 $181 $403 $722 $1611 $2891
110 $216 $486 $861 $1931 $3444
Compressed air is ENERGY
that must be CONSERVED.
24 hrs/day - 365 days/yr.
30¢/1000 SCFM
(6.75¢/kwh)
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
Cost to Generate Compressed Air ($1,000’s)
Resource Cost:
4000
3750
3500
3250
3000
2750
2500
2250
2000
1750
1500
1250
1000
750
500
250
1000 2000 3000 4000 5000 6000
Horsepower
Total Energy Production:
1990 FUEL SOURCES -Production of ENERGY
OIL
41.9%
16,120,000 BPD
COAL
23.3%
8,960,000 BPDOE
NAT. GAS
24.0%
9,230,000 BPDOE
HYDRO
3.5%
1,350,000 BPDOE
NUCLEAR
7.0%
2,690,000 PBDOE
OTHER
0.3%
38,470,000 BPDOE (Barrels Per
Day Oil Equivalent)
120,000 PBDOE
(81.44 Quadrillion BTU's or Quads)
Transportation
10,400,000 PBDOE
27%
INDUSTRIAL
COMMERCIAL
RESIDENTIAL
14,200,000 BPDOE
6,170,000 BPDOE
7,700,000 BPDOE
37%
16%
20%
CONSUMPTION
Electrical Segment of Industry
4,860,000 BPDOE (10.26 Quads) 34%
Electricity Production:
OIL
4.2%
COAL
56.8%
597,000 BPD
8,081,000 BPDOE
NAT. GAS
9.8%
1,392,000 BPDOE
HYDRO
9.5%
1,350,000 BPDOE
NUCLEAR
18.9%
2,690,000 PBDOE
OTHER
0.8%
1990 FUEL SOURCES -Production of ELECTRICITY
120,000 PBDOE
14,230,000 BPDOE (Barrels Per
Day Oil Equivalent)
(30.1 Quadrillion BTU's or Quads 37% of US. Energy System of
81.26 Quads)
ELECTRICITY
14,230,000 BPDOE
(30.1 Quads)
DISTRIBUTION
10% Savings 107,000 BPDOE (.22 Quads)
20% Savings 214,000 BPDOE (.45 Quads)
OTHER
compared to
390,000 PBDOE
(0.88 Quads)
Prudhoe Bay Oil Fields 1,800,000 BPD
If Numasized, a
minimum savings of
15% would be realized.
3%
INDUSTRIAL
COMMERCIAL
RESIDENTIAL
34%
30%
33%
4,860,000 BPDOE
(10.26 Quads)
4,290,000 BPDOE
(9.0 Quads)
Compressed Air Segment
1,070,000 BPDOE (2.26 Quads) 22%
4,690,000 BPDOE
(9.96 Quads)
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 BTU’s (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
BTU’s of heat (1 Kwhrs = 3413 BTU’s).
Formulae:
Those familiar with
electricity will recognize
Ohm’s 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.
ELECTRICITY
OHM’S LAW
E = IR
PNEUMATICS
BERNOULLI’S LAW
DP x P2
Q = 22.48 CV
T1 G
I =
E
R
Q = 22.48 CV
.47 P1 x .53P1
528 x 1
(AT CRITICAL FLOW)
I = E Cd
Q = 0.488 P1CV
Q = P CV
Q = (K) P CV
Same General Formula
•Flow of electrons is very similar to flow of air molecules
•Think of CONDUCTANCE as the reciprocal of RESISTANCE
•Whose formula came first?
Your Most Important Design Tool