CONVERSION CHART ACFM / SCFM

Transcription

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CONVERSION CHART
ACFM / SCFM
Vacuum
Level
(Gauge
Conversion
Factor
Vacuum
Level
(Gauge
Conversion
Factor
Vacuum
Level
(Gauge
Conversion
Factor
1"
1.03
11"
1.58
21"
3.33
2"
1.07
12"
1.67
22"
3.75
3"
1.11
13"
1.79
23"
4.28
4"
1.15
14"
1.88
24"
5.00
5"
1.20-
15"
2.00
25"
6.00
6"
1.25
16"
2.14
26"
7.63
7":
1.30
17"
2.31
27"
10.00
9"
1.36
18"
2.50
28"
15.00
9"
1.43
19"
2.73
29"
30.00
10"
1.50
20"
3.00
TO FIND ACFM - Multiply SCFM by conversion factor
EXAMPLE:
99 SCFM @ 19” Hg
99 x 2.73 = 270.27 ACFM
99 SCFM @ 25” Hg
99 x 6 = 594 ACFM
TO FIND SCFM - Divide ACFM by conversion factor
EXAMPLE:
270.27 ACFM @ 19” Hg
270.276 divide by 2.73 = 99
TM
www.ohiomedical.com
594 ACFM @ 25” Hg
594 divide by 6 = 99
1111 Lakeside Drive, Gurnee, IL 60031-4099
Phone: 800-448-0770 - Fax: 847-855-6300
P/N 255277 (Rev.2) 07/2006
www.squire-cogswell.com
An Ohio Medical Corporation Brand
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1) Start at the column with the
units that you wish to convert from.
2) Follow down the chart to where
you find the “1”.
3) Horizontally move from that column to the column with the units
that you wish to convert to.
C
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FLOW CONVERSION CHART
GPM
L/Hr
m 3/Hr
m 3/min
L/min
1
7.481
1.69907
0.0283
1699.07
28.318
0.134
1
0.227
0.00379
227.1
3.785
0.589
4.403
1
0.0167
1000
16.67
35.31
264.2
60
1
60000
1000
0.000589
0.0044
0.001
1.67E-05
1
0.0167
0.0353
0.264
0.06
0.001
60
1
CFM
TO USE THESE CHARTS
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VOLUME CONVERSION CHART
Gal
L
M3
Ft
In3
1
1728
7.48
0.0283
28.32
0.000579
1
0.00433 1.64E-05 0.0164
0.134
231
1
0.00378
3.786
35.34
61060
264.3
1
1000
0.0353
61.02
0.264
0.001
1
3
4) Multiply the number found in the
box by the number that you have
to get the converted value. or use
the appropriate formula contained
in the box.
TEMPERATURE CONVERSION CHART
K
R
C
1
K x 1.8
K - 273.15
R x 0.5556
1
(R - 491.67) x 0.5556
C + 237.15
(C x 1.8) + 491.67
1
(F + 459.67) x 0.5556
F + 459.67
(F - 32) x 0.5556
#/Hr to ACFM:
ACFM to #Hr:
F
(K - 255.37) x 1.8
R - 459.67
(C x 1.8) + 32
1
MISCELLANEOUS CONVERSIONS
#/Hr(1/60)(379/MW)(760/P)(460+T/250)
ACFM(60/1)(MW/379)(P/760)(520/460+T)
NOTE: Where MW is the molecular w eight, P is the operating pressure in TORR,
and T is the operating temperature in oF.
In
Hg
Torr
(mmHg)
mBar
In
H2O
1
0.0394
0.0295
0.0735
0.883
0.0029
2.9
2.04
0.000295
0.295
29
29.921
25.4
1
0.75
1.868
22.42
0.0736
73.55
51.71
0.0075
7.5
735.7
760
33.86
1.333
1
2.49
29.89
0.0981
98.06
68.95
0.01
10
980.9
1013.25
13.6
0.535
0.402
1
12
0.0394
39.38
27.69
0.00402
4.016
393.9
406.9
PRESSURE CONVERSION CHART
Ft
m
mm H2O
PSI
H2O
H2O
1.133
0.0446
0.0335
0.0833
1
0.00328
3.281
2.307
0.000335
0.335
32.82
33.9
345.3
13.6
10.2
25.39
304.8
1
1000
703.1
0.102
102
10003
10333
0.345
0.0136
0.0102
0.0254
0.305
0.001
1
0.703
0.000102
0.102
10
10.333
0.491
0.0193
0.0145
0.0361
0.434
0.00142
1.422
1
0.000145
0.145
14.23
14.696
Pa
kPa
3386
133.3
100
249
2989
9.806
9806
6895
1
1000
98088
101325
3.386
0.133
0.1
0.249
2.989
0.00981
9.806
6.895
0.001
1
98.09
101.325
Kg/cm2
0.0345
0.00136
0.00102
0.00254
0.0305
0.0001
0.1
0.0703
0.0000102
0.0102
1
1.033
ATM
0.0334
0.00132
0.000987
0.00246
0.0295
0.0000968
0.0968
0.068
0.0000099
0.00987
0.968
1
NOTE: All pressure terms are absolute, not gauge.
TM
www.ohiomedical.com
Phone: 800-448-0770 - Fax: 847-855-6300
P/N 255033 (Rev.3) 07/2006
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Receiver Sizing Formula
In vacuum thermoforming it is customary to generate deep vacuum in a short period of
time. A vacuum receiver is usually employed to produce this quick initial pulldown. If
the installation is new, the following formula may be used to determine the size of the
appropriate vacuum receiver.
V2 = V1(P3-P1)/P2-P3
Where:
Equ. 1.1
V2 = Applicable vacuum receiver size [Ft3]
V1 = Total mold volume (forming volume + box volume) [Ft3]
P1 = Initial pressure in the mold [PSIA]
P2 = Pressure in vacuum system [PSIA]
P3 = Pressure in combined system [PSIA]
Example:
A vacuum receiver needs to be sized for a new installation. The forming volume is 0.8 Ft3, and the box volume is 1.2 Ft3. The initial pressure in the mold
is 29.75” Hg (barometric pressure). The required pressure to form the sheet in
the mold is 22” Hg. The pressure produced by the vacuum pump will be considered to be a maximum of 28” Hg.
First one must determine what are the appropriate variables, and then covert them to the
appropriate units.
V2 – To be determined
V1 – 2 Ft3 (0.8 Ft3 + 1.2 Ft3)
P1 – 14.61 PSIA (29.76” Hg(A) converted to PSIA)
P2 – 0.98 PSIA (28” Hg(G) first converted to 2” Hg(A), then converted to PSIA)
V2 = 2 (3.93 – 14.61)/(0.98-3.93)
V2 = 7.24 Ft3 or times 7.48 gal/1 Ft3;
V2= 54 gal
In this case, a 60 gal receiver would be appropriate.
TM
www.ohiomedical.com
Phone: 800-448-0770 - Fax: 847-855-6300
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Another formula is useful where an existing tank is to be used. If the receiver size is known,
one will wish to calculate the pressure in the combined system.
P3 = (P1V1+P2V2)/(V1+V2)
Equ. 1.2
Where:
V2 = Existing vacuum receiver size [Ft3]
V1 = Total mold volume (forming volume + box volume) [Ft3]
P1 = Initial pressure in the mold [PSIA]
P2 = Pressure in vacuum system [PSIA]
P3 = Pressure in combined system [PSIA]
Example:
Say that a 60 gal receiver is available from stock, what will be the final pressure
in the combined system?
First one must determine what are the appropriate variables, and then convert them to the
appropriate units.
V2 – 8.02 Ft3 (60 gal * 1 Ft3/7.48 gal)
V1 – 2 Ft3 (0.8 Ft3 + 1.2 Ft3)
P1 – 14.61 PSIA (29.75” Hg(G) converted to PSIA)
P3 – to be determined
P3 = (14.61 * 2 + 0.98 * 8.02)/(2 + 8.02)
P3 = 3.7 PSIA or converted to “Hg(G) (29.92” Hg(A)-3.7 PSIA (29.92” Hg(A)/14.7 PSIA))
P3 = 22.4” Hg(G)
In most cases the system volume is equivalent to the tank volume. Generally if the pipework
associated with the system is greater than 10% of the tank volume, that the volume of the
pipework should be taken into account. These formulas also do not account for the collapse
of the plastic in the mold. In most cases, the forming volume is so much smaller than the system volume that it does not add much to the calculations. These formulas also do not
account for leaks in the system.
Ohio Medical Corporation - 1111 Lakeside Drive, Gurnee, IL 60031-4099 - Phone: 800-448-0770 - Fax: 847-855-6300
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Pump Sizing Formula
After the vacuum receiver has produced the initial quick pulldown, the vacuum pump
can be used to generate the final deep vacuum. The following formula is the general
pump down formula.
t = V * LN (PI/PF)/QAVG Equ. 1.3
Where:
t – Time to evacuate volume [min]
LN – Natural Log
V – Total volume to be evacuated [Ft3]
P1 – Initial pressure (PSIA)
PF – Final pressure (PSIA)
QAVG – Average volumetric flowrate of the pump (ACFM)
Note that this formula does not account for leaks in the system
Example:
The 60 gal receiver from the previous example has been installed. The process
requires that the vacuum pump should pull the combined system from 22.4” Hg
to 28” Hg in approximately 10 seconds to increase the detail on the molded
plastic. Will the S3 be able to accomplish this?
Once again the appropriate variables need to be determined.
t = To be determined
V – 9.22 Ft3 (1.2 Ft3 = 8.02 Ft3) {note that the forming volume is approximately 0}
P1 – 3.7 PSIA (22.4” Hg(G) determined by equation 1.2)
PF – 0.98 PSIA (28” Hg(G) first converted to 2” Hg(A), then converted to PSIA)
QAVG – 37 ACFM
The average flowrate is determined by adding the flows at different vacuum levels. The average capacity of the pump should be taken between the starting vacuum level and the ending
vacuum level. For this example, the values are for the S3.
5” Hg
19” Hg
22” Hg
25” Hg
28” Hg
29” Hg
55 ACFM
44 ACFM
43 ACFM
35.9 ACFM
31 ACFM
21 ACFM
43 + 38 + 31
3
= 37.33 or 37 Avg. ACFM
Solving for time:
t = 9.22 * LN (3.8/0.98)/37
t = 0.331 min or times 60 s/1 min;
t = 20 s
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Basically the S3 will take 10 seconds too long to accomplish this pulldown. Try the S5 using
the same parameters. All of the values remain the same, except for the average flowrate of
the pump For the S5 from 22” Hg to 28” Hg:
QAVG = 82 ACFM
Once again solving for time:
t = 9.22 * :N (3.8/0/98)/82
t = 0.149 min or times 60 s/1 min;
t=9s
In answer to the question that was posed in the example, the S3 will not be able to satisfy the
requirement, but the S5 does satisfy the requirement.
Another way to work this problem is to determine what flowrate is required given a specific
time period.
Example:
The 60 gal receiver from the previous example has been installed. The process
requires that the vacuum pump should pull the combined system from 22.4” Hg
to 28 Hg in approximately 10 seconds to increase the detail on the molded plastic. Will the S3 be able to accomplish this?
Change the formula to the following format:
QAVG = V * LN (PI/PF)/t
Equ. 1.4
From the values that been used in the previous pump down examples:
QAVG = 9.22 * LN (3.7/0.98)/0.1667
{t = 10 s * (1 min/60 s) = 0.1667 min}
QAVG = 73 ACFM
From what has been shown here, knowing the average volumetric flowrates, is that the SC10TR will be the best choice.
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Converting from SCFM to ACFM is as follows:
V2 = V1(P1/P2)(T2/T1)
Where:V2 – ACFM
V1 – SCFM
Equ. 1.5
P1 – Absolute Pressure @ STP
P2 – Absolute Pressure @ Vacuum Condition
T1 – Absolute Temperature @ STP
T2 – Absolute Temperature @ Vacuum Condition
MW – Average Molecular Weight
STP (Standard Temperature and Pressure)
is 520oR and 760 mmHg(A)
An easy way to remember is:
V that you want = V that you have x
(P that you have / P that you want) x
(T that you want / T that you have)
Converting from a mass flowrate to a volumetric flowrate is as follows:
V = m (1/60)(379/MW)(P1/P2)(T2/T1)
Where:V – ACFM
m - #/Hr
Equ. 1.6
P1 – Absolute Pressure @ STP
P2 – Absolute Pressure @ Vacuum Condition
T1 – Absolute Temperature @ STP
T2 – Absolute Temperature @ Vacuum Condition
MW – Average Molecular Weight
STP (Standard Temperature and Pressure)
is 520oR and 760 mmHg(A)
Example:
Convert 20#/Hr of air to volumetric flowrate 25” Hg and 100oF.
V = m (1/60)(379/MW)(P1/P2)(T2/T1)
m = 20#/Hr
P1 = 760 mmHg(A) = 29.92” Hg(A)
P2 = 25” Hg(G) = 125 mmHg(A), or 4.92” Hg(A)
T1 = 460 + 60oF = 520oR
T2 = 460 + 100oF = 560oR
V = 20 (1/60)(379/29)(760/125)(560/520), or V = 20 (1/60)(379/29)(29.92/4.92)(560/520)
V = 29 ACFM @ 25” Hg and 100oF (Volumetric flowrates should be described at a specific
pressure and temperature
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Converting from a volumetric flowrate to a mass flowrate is as follows:
M = V (60/1)(MW/379)(P2/P1)(T1/T2),
using the same symbols as above.
Equ. 1.7
Example:
Convert 300 ACFM of air @ 28.5 Hg and 60oF to a mass flowrate.
m = V (60/1)(MW/379)(P2/P1)(T1/T2)
V = 300 ACFM
P1 = 760 mm Hg = 29.92” Hg
P2 = 28.5” Hg = 36 mmHg, or 1.42” Hg(A)
T1 = 460 + 60oF = 520oR
T2 = 460 + 100oF = 520oF, note that since T1 = T2, the division is 1, and it may be neglected
m = 300 (60/1)(29/379)(36/760), or m = 300 (60/1)(29/379)(1.42/29.92)
m = 65#/Hr
Note that all calculations involving pressure and temperature should be performed at absolute
conditions.
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Affects of Altitude on Vacuum
Systems
Series of Technical White Papers from Ohio Medical Corporation
Ohio Medical Corporation. • 1111 Lakeside Drive • Gurnee, IL 60031
Phone: (800) 448-0770 • Fax: (847) 855-6304 • [email protected] • www.ohiomedical.com
560809 Rev. 1
TERMINOLOGY
ACFM – Actual Cubic Feet per Minute
CFM – Cubic Feet per Minute
MDCFD – Thousand Standard Cubic Feet per Day
psia – Pounds Per Square Inch in Absolute pressure
SCFM – Standard Cubic Feet per Minute
Ohio Medical Corporation
1
INTRODUCTION
This paper discusses the affect of atmospheric variation on a vacuum pump’s
performance. To simplify the understanding, vacuum pumps are compressors
operating in reverse, where inlet pressure is below the atmospheric pressure,
and compressed to the discharge at atmospheric pressure. The operating range
of a vacuum pump will be between atmospheric pressure down to absolute
zero (a perfect vacuum). Realistically, we can not achieve a perfect vacuum
(29.9 in-Hg), and the vacuum pumps used for medical and industrial
applications require approximately 95% (28.5 in-Hg) of the atmospheric
pressure to be evacuated in a tank. Cryogenic applications require nearly a
perfect vacuum, and will achieve more than 99.9% (greater than 29.8 in-Hg)
of the atmospheric pressure to be evacuated from a chamber.
This paper supplements the Squire-Cogswell white paper titled: ACFM vs.
SCFM vs. ICFM published in 2004 and explains the differences in compressor
performance with respect to the varying atmospheric conditions. The paper
also addresses appropriate “CFM” terminology that should be use in
comparing compressors (SCFM) and sizing them properly (ACFM) for the off
“Standard” conditions for the altitude and conditions for the area.
Due to the atmospheric variation in air pressure, temperature and density – the
fluid properties are constantly changing (i.e. - conditions are dependent on
location, time of the year, altitude, etc.) Thus, it is important to understand
that the conditions in Los Angeles vary significantly from the conditions in
Denver, and a vacuum pump’s performance (capacity and operation) will vary
significantly. The intent of this paper is to provide a better understanding of
how vacuum pump’s capacity varies with respect to altitude, so we can
properly select and size vacuum pumps for their specified and intended
applications.
2
DESCRIPTION
The term cubic feet per minute (CFM) describes the fluid flow rate, (measured in volume
- ft3) not the weight per minute on the inlet side of a compressor. The vacuum pump’s
performance capability is measured in how many one ft3 cubes of fluid are able to move
per minute through the system.
1 ft
1 ft
1 ft
Figure 1 – One Cubic Feet of Volume
Now consider the conditions in Los Angeles, where one cubic foot of air weighs 0.075
lbs., and in Denver, where one cubic foot of air weighs 0.062 lbs. Even though the
volume is the same, the weight (mass) of the air is different.
1 ft
1 ft
Denver
Los Angeles
1 ft
1 ft
W=0.075 lbs.
1 ft
W=0.062 lbs.
1 ft
Figure 2 – Constant Volume
Condition
Now consider a constant weight (mass) condition. A balloon filled with 31 actual cubic
feet of air in Los Angeles is then taken up to Denver. The balloon now contains 38
standard cubic feet of air.
Los Angeles
Denver
V=31 ft3
V=38 ft3
Figure 3 – Constant Mass
3
The two examples illustrate the confusion of measuring volume due to the fact air is
compressible. In this instance, the number of gas molecules occupying a particular
volume, depends primarily on the pressure and temperature conditions of that location.
At a microscopic level, the air molecules are closer together (greater air density) in Los
Angeles compared to the air molecules in Denver.
Denver
Los Angeles
Figure 4 – Variation in Air Molecule Density
A variation in air pressure results in a variation in air density, as show in Figure 4, and is
consistent with constant volume concept in Figure 2. Another way to look at this is to
analyze the number of air molecules in a 120-gallon receiver tank at atmospheric pressure
at Los Angels and at Denver, where the former (higher pressure) tank occupies a greater
number of molecules. The weight and density vary primarily because the atmospheric
pressure is significantly different between the two cities, as show in Table 1. Note the
terms for “actual” and “standard” for the volumes described above leads us to “SCFM”
and “ACFM”.
City
Los Angeles
Denver
Altitude
(ft)
0
5280
Atmospheric Pressure
(psia)
14.69
12.12
(in-Hg)
29.92
24.68
Table 1 – Variation in Atmospheric Pressure between the Two Cities
THE “GENERAL RULE”
To simplify the understanding of the affect of vacuum pumps with respect to variation in
altitude, the following illustration simplifies and points out the concept to understand
prior to proceeding with a more theoretical view point on the matter. For this paper, we
will assume only the attitude is varying, while keeping other conditions constant (like
temperature, humidity, etc,).
4
The first general rule is to understand that the maximum vacuum level that can be
achieved is with respect to the atmospheric conditions in the area. For example, assume
the maximum vacuum that can be achieved theoretically in Los Angeles (sea level) is
29.92 in-Hg, but in Denver (5280 feet above sea level) is 24.68 in-Hg.
City
Los Angeles
Denver
Altitude
(ft)
0
5280
(in-Hg)
29.92
24.68
(in-Hg using the rule)
29.92
24.64
Table 2 – “Rule of Thumb” for Atmospheric Pressure
As a general “Rule of Thumb”, for every 1000 feet above sea level, the maximum
possible vacuum is reduced by approximately one in-Hg (0.491 psi). By using this rule
one can quickly determine the maximum possible vacuum for the area. Note the
accuracy of this “Rule”, as there is only a 0.16% difference between the approximated
and the actual pressure (shown in Table 2).
PERFORMANCE BASED ON CAPABILITY
Next consider that a vacuum system’s performance is a percentage of the atmospheric
pressure that it can exhaust from a closed system. At sea level, Los Angeles has a
barometric pressure at 29.92 in-Hg. Thus, a vacuum pump with a maximum capability of
24.00 in-Hg will have are rating of 80.2%.
24.00 psi
= 0.802
29.92 psi
Then, the 80.2% rating can be assigned to the vacuum pump to determine its capability in
Denver.
0.802 × 24.68 = 19.79 in − Hg
The 80.2% rating applied to the maximum possible vacuum (24.68 in-Hg) results in a
maximum vacuum of 19.79 in-Hg for this pump in Denver. This is a very important
point to understand and consider for vacuum performance and sizing for your location. If
the user needs a vacuum that can achieve 22 in-Hg in Denver, a pump with at least a 89%
vacuum capability is needed, or a pump that will achieve at least 26.7 in-Hg capability in
Los Angeles.
5
PERFORMANCE BASED ON CAPACITY
% CAPABILITY
In a closed system vacuum pumps use kinetic energy to move air through a closed
system. At low vacuum levels large volumes of air can be evacuated through the system,
but at higher vacuum levels, the capacity decreases, due to increased leakage from a
larger pressure differential with the environment and there is additional resistance to
flow. This phenomena is illustrated in Figure 5.
0
5
10
15
20
25
VACUUM (in-Hg)
Figure 5 – Vacuum Capacity for Los Angels and Denver
6
SUMMARY
This paper summarized the affect of atmospheric variation on a vacuum pump’s
maximum performance. By comparing conditions in Los Angeles from the conditions in
Denver (capacity and operation) and using the general “Rule of Thumb” (for every 1000
feet above sea level, the maximum possible vacuum is reduced by approximately one inHg), we can quickly determine the maximum possible vacuum for the area.
Finally, a specifier can use the “Rule of Thumb” for sizing a vacuum system properly. In
addition, the reference pressure, temperature, and required vacuum must be specified, in
addition to the required capacity and capability. When specifying the vacuum
requirement, the worst case conditions should be used (i.e. - generally hot days – lower
air density). Other important factors to consider in vacuum system sizing are:
• Vacuum requirement or demand in a given day
• Normal operating conditions
• Other operating conditions (hot days are the worst)
• Single-stage or two-stage vacuum
• Electrical characteristics and power requirement
• Area classification (Elevation)
7
ACFM vs. SCFM vs. ICFM
Ohio Medical Corporation • 1111 Lakeside Drive • Gurnee, IL 60031
560808 Rev. 2
TERMINOLOGY
ICFM – Inlet Cubic Feet per Minute
MDCFD – Thousand Standard Cubic Feet per Day
psia – Pounds Per Square Inch in Absolute pressure
SCFM – Standard Cubic Feet per Minute
NOMENCLATURE
P1
Barometric pressure at the non-standard site in psia
T1
Ambient air temperature in °R
PF
Pressure after the inlet filter in psia
TF
Air temperature after the inlet filter in °R
Psat
Saturation Pressure
φ
Relative Humidity at the non-standard site
Tstd
Standard temperature in degrees in °R (60°F = 60 + 460 = 520 °R)
Pstd
Standard air pressure in psia (14.696 psia)
Ohio Medical Corporation.
1
INTRODUCTION
The term CFM is often confusing and difficult to define for one condition, and
one definition does not satisfy all conditions we encounter in our customer’s
applications throughout the world. Simply put, CFM is an acronym for Cubic
Feet per Minute, and defines the volumetric flow rate of a fluid displaced by a
pump (like a compressor, a blower, or a booster). The term CFM is generally
used to describe a pump’s capacity, and is used to determine the size of the
source system for medical, industrial and other applications. The common
terms used to specify a volumetric flow rate in different industries are SCFM,
ACFM, ICFM, MCFM, MSCFD, etc. Often times these terms are very vague,
and in turn, misunderstood.
The primary reason for all the difficulties described above is because air is a
compressible fluid, due to the atmospheric variation in air pressure,
temperature and density - the fluid properties are constantly changing. The
conditions are dependent on location, time of the year, altitude, etc. Thus, it is
important to understand that the conditions in Los Angeles vary significantly
from the conditions in Denver. The terms SCFM, ACFM and ICFM are often
used to define the different instances and conditions of a compressor’s
capacity and operation. If the CFM terms are used appropriately, they can be
useful in the direct and relative comparison to their operating conditions, and
to other source systems. The intent of this paper is to provide a better
understanding of SCFM, ACFM and ICFM and their meaning, so we can
properly select and size compressors for their specified and intended
applications.
2
DESCRIPTION
The term cubic feet per minute (CFM) describes the fluid flow rate, (measured in volume
- ft3) not the weight per minute on the inlet side of a compressor. The compressor’s
performance capability is measured in how many one ft3 cubes of fluid are able to move
per minute through the inlet.
1 ft
1 ft
1 ft
Figure 1 – One Cubic Feet of Volume
Now consider the conditions in Los Angeles, where one cubic feet of air weighs 0.075
lbs., and in Denver, where one cubic feet of air weighs 0.062 lbs. Even though the
volume is the same, the weight (mass) of the air is different.
1 ft
1 ft
Denver
Los Angeles
1 ft
1 ft
W=0.075 lbs.
1 ft
W=0.062 lbs.
1 ft
Figure 2 – Constant Volume
Condition
Now consider a constant weight
(mass) condition. A balloon filled with 31 actual cubic
feet of air in Los Angeles is then taken up to Denver. The balloon now contains 38
standard cubic feet of air.
Los Angeles
Denver
V=31 ft3
V=38 ft3
Figure 3 – Constant Mass
3
The two examples illustrate the confusion of measuring volume due to the fact air is
compressible. In this instance, the number of gas molecules occupying a particular
volume, depends on the pressure and temperature conditions of that location. At a
microscopic level, the air molecules are closer together (greater air density) in Los
Angeles compared to the air molecules in Denver.
Denver
Los Angeles
Figure 4 – Variation in Air Molecule Density
A variation in air pressure results in a variation in air density, as show in Figure 4, and is
consistent with constant volume concept in Figure 2. Another way to look at this is to
analyze the number of air molecules in a 120-gallon receiver tank at 80 psia and 100 psia,
where the higher pressure tank occupies a greater number of molecules. The weight and
density vary primarily because the atmospheric pressure is significantly different between
the two cities, as show in Table 1. Note the terms for “actual” and “standard” for the
volumes described above leads us to “SCFM” and “ACFM”.
City
Los Angeles
Denver
Altitude
(ft)
0
5280
(psia)
14.69
12.12
Table 1 – Variation in Atmospheric Pressure between the Two Cities
SCFM and ACFM
The term standard cubic feet per minute (SCFM) is usually used as a standard reference
condition for flow rate performance for atmospheric pressure at sea level, as opposed to
actual cubic feet per minute (ACFM) is typically used to rate flow rate performance of
compressor systems for actual pressure and temperature. SCFM is defined as air at
14.696 psia and 520°R (60 °F). Sometimes other conditions are used, such as 530°R
(70°F), 528°R (68°F), 0% and 36% Relative Humidity for describing the standard
conditions. It is important to remember SCFM is defined by a fixed set of conditions or
common reference point for comparing different compressors systems. Otherwise the
consequences are the improper sizing of the compressor system for its true application.
This point will be apparent in the two examples to follow.
4
The conditions at sea level are generally not experienced by our customers and for
practicality purposes ACFM is typically used for sizing compressors for these
applications (+100°F and lower pressures). The conversion for ACFM from SCFM is
shown by the formula below.
ACFM = SCFM ×
Pstd
T
× 1
P1 − Psat1 × φ1 Tstd
(
)
Equation 1
Note, absolute units must be used in the equation.
SCFM – ACFM EXAMPLE 1 – Normal Day
For this example we will reference a 30-hp compressor operating at 1020 rpm, and use it
in Los Angeles and Denver to demonstrate how SCFM and ACFM should be used.
Assume the requirement for compressed air is 125 psig discharge pressure, and 100
SCFM of demand, and the site ambient conditions are T1 = 80 °F (540 °R), P1 = 14.7 psia
and φ = 75%, and this results in the following:
ACFM = 106.6
If we assume that all ambient conditions remain the same with the exception of moving
the compressor to Denver, where the atmospheric pressure will drop to P1 = 12.12 psia,
the resulting volumetric flow requirement becomes:
ACFM = 130.0
In order to deliver the same amount of work (100 SCFM at 125 psig), the compressor in
Denver must ingest larger quantities of the lower-density air, due to the change in the
atmospheric pressure.
Location
Los Angeles, CA
Denver, CO
SCFM
100
100
ACFM
106.6
130.0
% Diff.
6.6 %
30.0 %
Table 1 – Normal Day between the Two Cities
To provide the required 100 SCFM of work in Denver, the compressor must be able to
process 130 ACFM.
SCFM – ACFM EXAMPLE 2 – Hot Humid Day
For this example we will reference the first example for Los Angeles and Denver and
change the conditions to a hot humid day. Assume the same requirements and conditions
except for the ambient temperature of T1 = 100 °F (560 °R) and relative humidity of φ =
100%, which results in the following:
5
Location
Los Angeles, CA
Denver, CO
SCFM
100
100
ACFM
115.1
141.6
% Diff.
15.1 %
41.6 %
Table 3 – Hot Humid Day between the Two Cities
The results at the higher temperature and higher humidity condition show an even greater
amount of air is needed (approximately 10% more air is needed compared to Example 1)
to meet capacity requirements. Thus, we recommend using hot humid days to calculate
the worst case conditions for sizing a system. Tables can be constructed to size systems
from Equation 1, but it is up to the specifier to determine the proper conditions. Table 4
in Appendix A shows the expansion ratio conversions for SCFM and ACFM for the
conditions described in this example.
ICFM
The term Inlet Cubic Feet per Minute (ICFM) is used by compressor vendors to establish
the conditions at the inlet of compressor – in front of the inlet filter, blower, or booster.
If the pressure and temperature condition at the inlet is the same as after the filter, blower,
or booster, then the ICFM and ACFM values will be the same. However, as the air
passes through these components, there will be always be a pressure drop or rise, and
Equation 2 is used to approximate ACFM.
ACFM = ICFM ×
P1 TF
×
PF T1
Equation 2
Then ICFM is used to measure inlet capacity, which will approximate ACFM for this
type of a system. Note, when a blower or booster is added, the inlet may experience
significantly higher pressure and temperature conditions than the actual ambient
conditions. Greater the difference in pressure and temperature, greater the difference in
ACFM and ICFM. Finally, there are losses (air seal, heat, etc.) associated with the use of
these components and coupled with the pressure and temperature differences, the use of
ICFM will result in a misleading outcome in determining a compressor’s capability.
Note at higher altitudes, the specifier must account for the decrease in air pressure when
estimating a compressor’s performance and sometimes blowers or boosters are used for
economic reasons, but this is not always the best solution. Thus, in certain markets, like
the Medical, where tighter controls are employed, ICFM should not be used to determine
a compressor’s capacity, instead ACFM should be used.
6
CONCLUSION
This paper defined, summarized and applied the terms SCFM, ACFM and ICFM, and the
differences between them. The term CFM, at a fundamental level, is defined such that a
compressor will pump a specific volume of air in a given amount of time when the
compressor speed and flow resistance matches the test conditions.
A specifier’s most difficult task is sizing a compressor properly, by specifying the
compressor’s required capacity. It is important to note that the proper understanding of
these terms will help a specifier in selecting a compressor. The specifier should use
SCFM to compare differences in compressor capacities, and ACFM for actual nonstandard site conditions and proper load applications. ICFM should be used only when a
filter, a booster or a blower is added to the system, and should not be used in determining
compressor selection, due to misleading results.
Finally, the reference pressure, temperature, and discharge pressure must be specified, in
addition to the required capacity. When specifying the compressed air requirement, the
worst case conditions should be used (i.e. - generally hot humid days, as shown in
Example 2). Otherwise, there will be confusion in the sizing process. Other important
factors to consider in compressor capacity and system sizing are:
• Air requirement or demand in a given day
• Normal operating conditions
• Other operating conditions (hot humid days are the worst)
• Single-stage or two-stage compressor (compression ratio)
• CFM reduction due to flow resistance
• Electrical characteristics and power requirement
• Area classification (Elevation)
• Compressors with a higher CFM rating will pump more air than
compressors with lower CFM
7
APPENDIX A
Volume Expansion Ratio
Pressure
(psia)
14.70
14.50
14.25
14.00
13.75
13.50
13.25
13.00
12.75
12.50
12.25
12.00
11.75
11.50
11.25
11.00
10.75
10.50
10.25
10.00
9.75
9.50
9.25
9.00
8.75
8.50
8.25
SCFM to ACFM
ACFM to SCFM
1.151
1.168
1.190
1.213
1.237
1.261
1.287
1.314
1.341
1.371
1.401
1.433
1.466
1.500
1.537
1.575
1.615
1.658
1.702
1.749
1.799
1.851
1.907
1.966
2.029
2.097
2.168
0.869
0.856
0.840
0.824
0.809
0.793
0.777
0.761
0.745
0.730
0.714
0.698
0.682
0.666
0.651
0.635
0.619
0.603
0.588
0.572
0.556
0.540
0.524
0.509
0.493
0.477
0.461
Table 4 – Volume Expansion Ratio Conversion Chart
8
Pressure Drop in Air Piping Systems
Ohio Medical Corporation • 1111 Lakeside Drive • Gurnee, IL 60031
560807 Rev 2
1
TERMINOLOGY
GPM – Gallons per Minute
NOMENCLATURE
Constants:
g – Acceleration Due to Gravity (32.2 ft./sec2)
γ - Specific Weight (lb/ft3)
ρ - Density of Fluid (slugs/ft3)
µ - Dynamic Viscosity (lb*s/ft2)
ε - Equivalent Roughness (.0005 ft. for Galvanized Pipe)
KL – Loss Coefficient for Fittings (Found in Industrial Literature or College Text)
Variables:
zn – Height at Position “n” (ft.)
V - Velocity of Fluid (ft/sec.)
D - Diameter of Pipe (in. or ft.)
A – Cross Sectional Area of Pipe (in.2)
L – Pipe Length (ft.)
hL – Head Loss (ft.)
pn – Pressure at Node “n”
f – Friction Factor
2
INTRODUCTION
Ever since the development of piping systems throughout civilization there has been the
need to analyze pipe size for optimal flow. In 1738 Daniel Bernoulli had developed an
equation to represent all variables within a piping system, as shown below:
p1
2
2
V
p V
+ 1 + z1 = 2 + 2 + z 2 + ∑ hL
γ 2g
γ
2g
(1)
From Bernoulli’s equation, we can see how pressure, velocity and position relate to one
another, and many derivations can be created from this equation. Depending on the
assumptions made and under certain conditions, some of the variables become negligible
and can be removed to simplify the equation, and result in a simpler solution. The nature
of this paper is to solve for the pressure drop (or commonly referred to the back pressure)
which is essentially the pressure difference between two points in a piping system. The
pressure drop is critical when sizing pipe. A pump that is integrated within a piping
system is designed such that it will withstand certain forces at the inlet and exhaust. If
the pump is subjected to forces greater than the ones prescribed, there is a high potential
for damage to the internal pump components, and thus the designed flow will be affected.
This paper will focus on the flow of air, although it can be shown that other fluids can
also be modeled using the same methodology. Like other media that flows within a
piping system, air has its own characteristics which benefit and hinder the process.
There are numerous ways in which to solve for the pressure drop of piping system. That
is to say, one can incorporate the use of a computer with a plethora of software available.
However, the underlying equations used in these programs follow the same fundamental
laws of physics found in any collegiate Fluid Mechanics textbook. In addition, sound
engineering judgment should be used when sizing pipe. While a cost-effective solution
may look good on the bottom line, a safe and reliable system should have precedence in
any design.
For our discussion the following assumptions can be made:
Assumptions:
1. Air Flow will be turbulent.
2. The temperature for the ambient air will be 70oF
3. Air flow will be defined using ACFM.
With these assumptions we can model our system.
3
BACKGROUND:
As we all have either experienced, or heard about, the phenomenon referred to as
turbulence. Essentially turbulence is a random positioning of the flow of air. Whereas
in a laminar condition, the flow of air is uniform and follows a smooth, organized path.
For air piping, we assume that the air flow will be turbulent due to surface randomness in
the piping fabrication and/or power fluctuation in the air source equipment. A visual
depiction of laminar and turbulent flow is shown below:
Laminar Flow
Figure 1
Turbulent Flow
Figure 2
To see if the flow of air will be turbulent or laminar, we solve for a parameter referred to
as the Reynolds number. The Reynolds number is a dimensionless number which is
obtained from the following equation:
Re =
ρ ×V × D
µ
(2)
The conditions for if the flow is turbulent or laminar is as follows:
If Re < 2100 then the flow is Laminar
If Re > 4000 then the flow is Turbulent
If 2100 < Re < 4000 then the flow is classified as in Transition.
4
To determine if the flow rate is laminar or turbulent, the Reynolds number should be
calculated. The combination of the Reynolds number, equivalent roughness and pipe
diameter we can determine the friction factor from the Moody chart. (Moody charts can
be found in industrial literature and in college text books.) The friction factor is used in
the equations below.
OTHER EQUATIONS:
Some of the other equations used to determine the pressure drop are in this section. We
can solve for the velocity of the flow by dividing the ACFM by the cross sectional area:
Velocity ( ft / s) =
Volumetric Flow Rate( ft 3 / s)
Area( ft 2 )
(3)
Where the area is solved by:
A = π ⋅r2
(4)
Once the velocity is found, we can then check to see if the flow is turbulent, or laminar,
by utilizing the Reynolds equation (as mentioned previously).
The pressure drop in a piping system can be broken down into two (2) equation forms:
1. Major Pressure Losses (Pipe Losses)
2. Minor Pressure Losses (Losses through fittings, valves, etc.)
The Equation for Major Losses:
p1 − p2 = f
l 1
ρV 2
D2
(5)
The Equation for Minor Losses:
p1 − p2 = γ ∑ hL
Where hL is found by:
hL = K L
(6)
V2
2g
(7)
And the constant KL is found in tables of either college text, or industrial references.
Combining Equations (6) and (7) we can solve the pressure differential directly:
p1 − p 2 = γ ∑ K L
V2
2g
(8)
5
Once the pressure drop has been calculated for the pipe length and all of the
fittings/valves, the total pressure drop can be found by the summation of all components.
In Equation form:
(9)
pTotal = p Fittings + p Pipe + pValves
COMPRESSIBLE VERSUS INCOMPRESSIBLE:
The reader might be wondering that since air is a gas, shouldn’t the flow be characterized
as compressible? The answer to this question is dependent on many conditions.
Depending on the length of the pipe and the complexity of the arrangement of fittings and
valves, the pressure drop may, or may not, be small relative to the initial pressure. If the
pressure drop is small enough, then you can assume the fluid is incompressible.
Otherwise, the flow is Compressible, and complicates the analysis. An example to find
the pressure ratio is as follows:
We have a pipe length of 7’-0” and the pressure drop should be no greater than 1.0 psi
per 7’-0”. The pressure at the beginning is 14.7 psi.
( p1 − p2 )
(10)
p1
 1 psi 

(7 ft )
 7 ft 
= .068 = 6.8%
14.7 psi
This ratio is small enough to assume an incompressible flow. Sound judgment and
experience should be used when applying this equation. In different industries, different
values are used to make the difference.
6
EXAMPLE:
Given: A Squire-Cogswell S750TR-T2 system needs to have an exhaust line sized
properly. The customer needs to pipe the system to the outside the building. The
customer knows that there will be 100 ft. of pipe, three 90o elbows, and two 45o elbows.
All Piping will be galvanized (ε = 0.0005 ft.). Air temperature is 60oF and atmospheric
pressure is 14.7 psi.
Find: The proper exhaust pipe diameter for this system.
Known: The given flow rate is 163.8 ACFM per pump. The exhaust for the pump is 11/2”. (It is always advisable to check with the manufacturer of the pump for a back
pressure allowance. Depending on the manufacturer of the pump, the allowable back
pressure may vary.) The back pressure should not be greater than 1 psi.
Solution:
First we solve for the major lossesFind the velocity of the fluid from the flow rate using equation (3):
ft 3 1min
ft 3
V = 163.8
×
= 2.73
min 60 sec
sec
V = 2.73
ft 3
1
×
sec Cross Sectional Area of Pipe
Let’s assume a 1-1/2” (0.0625 ft.) Cross Sectional Diameter:
Therefore:
V = 2.73
ft 3
1
ft
×
= 222.6
2
sec (0.0625 ft ) × 3.14
sec
The next variable we look for is the Reynolds Number using equation (2):
Re =
ρVD
µ

slugs  
ft. 
 0.00238 3   222.6  (0.125 ft.)
s 
ft  
Re = 
= 176,471
− 7 lb ⋅ s
3.74 × 10
ft 2
7
Once the Reynolds number is found then the flow can be determined as either turbulent
or laminar. In this example, the flow is turbulent. The frictional factor can be found
from knowing the Reynolds number, relative roughness and diameter of the pipe.
From the Moody Chart f = 0.029
Using equation (5) to solve for the major losses:
 100 ft.  1 
slugs 
ft 
  0.00238
 222.6
p1 − p 2 = 0.029

3 
sec 
ft 
 0.125 ft.  2 
p1 − p 2 = 1368
p1 − p 2 = 1368
2
lb
ft.2
lb
1 ft 2
×
ft.2 144in 2
∆p = 9.49 psi
Second we solve for the minor losses (i.e. through fittings and valves) using equation (8):
p1 − p 2 = γ ∑ K L
2
 
ft 



 222.6 
lb  
s
p1 − p 2 = 0.0765 3 3 1.5 × 
ft 
ft  

2  32.2 2 
 
s 

 
∆p = 359
V2
2g
2


ft  



 222.6  


s  

 + 2  .4 × 
ft  


2  32.2 2  


s  



lb
ft 2
lb
1 ft 2
∆p = 359 2 ×
ft 144in 2
∆p = 2.50 psi
pTotal = pFittings + pPipe
8
pTotal = 9.49 psi + 2.50 psi
∆pTotal = 11.99 psi
As we can see from this solution, the pressure drop is much higher than what should be
observed in the pumps. Therefore, other iterations of the example are shown in Table 1:
Iteration
Pipe
Size
(in.)
Major Losses
Pressure
Drop in Pipe
(psi)
Minor Losses
Pressure Drop
in Fittings (psi)
Total
Pressure
Drop (psi)
1
1 1/2
9.49
2.20
11.99
2
3
0.24
0.14
0.38
3
2 1/2
0.64
0.28
0.92
Table 1
EXPERIMENT:
The above theory was tested in our lab to see if the data correlated with the solutions.
The experiment was set up as seen in the following pictures:
9
Pressure Measuring Device:
Merical DP2000I (Accuracy 0.05%R)
Pump:
45 lpm Volumetric Free Flow, 12 VDC, 4.0 Amp
Pipe Size:
1/8” (.269ID)
Fittings:
1/8” Tee’s (Qty. 2)
1/8” Pipe Nipples (as appropriate)
1/8” Pipe Couplings (as appropriate)
Conditions for the test:
Ambient conditions at Sea Level:
Temperature - 70oF
Pressure – 14.7 PSIA
Flow Rate:
45 lpm (Approximately 1.6 ACFM)
The pressure drop was measured at different pipe lengths. Refer to the data below.
Length
Major
Losses
Calculated
(psi)
6”
0.04
12”
0.09
24”
0.17
36”
0.26
48”
0.35
Minor Losses Calculated (psi)
Tee Branch Flow: 0.05(2) = 0.10
Elbow: 0.04(1) = 0.04
Elbow: 0.04(1) = 0.04
1 Coupling: 0.002(1) = 0.002
Elbow: 0.04(1) = 0.04
3 Couplings: 0.002(3) = 0.006
Elbow: 0.04(1) = 0.04
5 Couplings: 0.002(5) = 0.010
Elbow: 0.04(1) = 0.04
7 Couplings: 0.002(7) = 0.014
∆P
Calculated
(psi)
∆P
Measured
(psi)
0.18
0.19
0.23
0.23
0.32
0.30
0.41
0.37
0.49
0.42
Table 2
10
Pressure Drop Experiment
1
0.9
0.8
Pressure Drop (psi)
0.7
0.6
Calculated
0.5
Measured
0.4
0.3
0.2
0.1
0
6
12
24
36
48
Pipe Length (Inches)
Figure 3
As we can see from the data above, the calculated and measured data points correlate
with each other. The average difference is 0.03 psi between the calculated and measured
data points. The non-linear behavior of the measured values is attributed to the random
behavior of air. From Figure 3, one can see that the pressure data separates from the 12”
point. The separation of data points can be attributed to leaks or irregularities in the pipe
(i.e. burrs, surface irregularities from galvanizing etc.) The leaks in the piping system
will decrease the flow rate, this in turn decreases the pressure. It is also interesting to
note that as the pipe length increases, so does the pressure. This is both demonstrated in
the measured and calculated values.
11
CONCLUSION:
Optimization of piping is essential in today’s new construction of building systems. Due
to ever increasing costs of steel and copper, there is no other alternative but to take a
closer look at the piping system. In the past, a “rule of thumb” gave a large margin of
safety. However, the Engineer, Contractor and Architect must understand that the margin
of safety can be held while decreasing the excess size of the pipe. As long as the pressure
drop is within the pump manufacturer’s specifications, the performance will not be
affected.
The pressure drop in the piping system directly affects the performance of the pump.
Essentially the pressure drop produces additional forces that directly and indirectly affect
the internal parts. It should be noted that having an excess size diameter of pipe can be
detrimental as well (i.e. oversized). If the area is increased, the excess (stagnant) air will
act as an obstruction to the flow and therefore create a greater turbulence. Another
detrimental effect, besides damage to the internal components of the pump, is the
reduction of air capacity the pump removes. The back pressure acts as an obstruction to
the flow, therefore hindering capacity.
The example could be solved in different ways. In fact we could have different pipe sizes
in this line. That is to say, we could have calculated for a run 50’-0” of 1 ½” Dia. and
50’-0” of 3” Dia. The calculation above is good for approximating the pipe size for the
entire length of the pipe, and it is also good for solving for smaller sections of pipe. In
addition, this paper makes lot of assumptions regarding the condition of the air flow and
the ambient conditions. It can be argued that a more detailed analysis would be
appropriate, and in other cases a more general solving approach would be appropriate.
As shown in the example, there are many steps that need to be followed to give us an
optimized pipe size. There are many opportunities to overlook error by using these
equations. Therefore, it is recommended to use sound industry judgment when solving
for the pressure drop and interpreting the results.
12

CONVERSION CHART ACFM / SCFM

Transcription

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