Commercial Vacuum System Design

GUIDE TO CENTRAL CLEANING SYSTEM DESIGN
A central cleaning system is generally designed to pick up and convey dry and free flowing material that can enter and pass through the vacuum cleaning tool and hose. Also, the system
should be designed to allow for a certain number of users. In order not to over-design or under-design the central cleaning system, a number of things should be taken into consideration.

– Material to be handled
– Number of vacuum inlets to be used simultaneously
– Layout of system
– Pressure losses
– Air flow required for cleaning
– Selection of vacuum power units

LEARN MORE ABOUT DESIGNING A COMMERCIAL VACUUM SYSTEM

MATERIAL TO BE HANDLED

The central vacuum system is capable of handling all dry granular material or low viscosity fluids; however, care should be taken in handling toxic, hygroscopic, or sticky materials.

The area to be cleaned or the amount of material to be moved will determine the size of the hose, wand, or tools necessary for the particular installation. For cleaning small areas or moving small amounts of material, a 1-1/4″ hose and tools will suffice, but for larger volume or areas, it may be necessary to use 1-1/2″ hose and tools. For special applications 2″ diameter hose and tools may be required.

The size of hose and tools, in turn, help determine the size of the vacuum power unit necessary for the particular job.

NUMBER OF VACUUM INLETS TO BE USED SIMULTANEOUSLY

The number of inlets in a given vacuum system has little to do with determining the size of power unit required for the installation. The number of inlets that will be in use at one time, however, does govern the design of the system.

If the vacuum system is primarily used for cleaning, the first thing to establish is the maximum number of operators available to use the system at one time. The next thing to consider would be the minimum number of operators required to do the necessary cleaning in the allotted time. The designer should keep in mind that cleaning with a central system is less time-consuming than other cleaning methods such as the use of brooms, mops, and portable cleaning units (see page 13 by the Dutch Government study on portables vs. central vacuums). In one hour, one operator should be able to clean approximately 3000 square feet of smooth floor when using a central system if he does not encounter too many obstructions to be moved or swept around. Rougher floors, or ones with many obstructions, may require an hour for approximately 2000 square feet of floor space. The size of the particular cleaning system and how the operators will be using it will determine how many vacuum power units will be required and where they will be located.

It is good to remember that if your estimate of operators using the system is to low, then the system you design will probably be under-powered and give unsatisfactory or poor service; however, if you overestimate the number of operators using the system, then the system’s initial cost will be unnecessarily high.

LAYOUT OF THE CENTRAL VACUUM SYSTEM

The vacuum inlets should be located in accessible places so that they are convenient to the area to be cleaned. Actually, the more inlets available, the more convenient the system will be to use; however, the cost of the system will be unnecessarily increased if the inlets are installed where they are not absolutely necessary. The inlet should be spaced to allow for the 20% overlap in cleaning areas. If a 20 foot hose is to be used, the inlets should be spaced approximately 30 feet apart; if a 25 foot hose is used, inlets should be spaced approximately 30 feet apart; and if a 30 foot hose is used, the inlets should be spaced approximately 46 feet apart. In some areas, the figures may not hold due to obstructions such as furniture, walls, machinery, or storage cabinets.

It is always desirable to use the minimum length of hose possible, since it is easier to work with and store. It also will have less pressure loss in the hose and thus requires less vacuum to operate. A few extra wall inlets may make it possible to reduce the number or size of power units required for the installation. The cost of the extra piping and wall valves will have to be weighed against the cost of larger or additional power units.

The power units should be located in an area where they can be serviced conveniently and the piping should be laid out to require the minimum amount of pipe and fittings, thus minimizing the friction losses in the pipe and the cost of the installation. From the standpoint of friction losses in the pipe, it is better to use several feet of pipe rather than a 90 degree or 45 degree turn. It is also better from the friction standpoint to use a 45 degree bend rather than a 90 degree bend.

The size of the pipe to be used in the installation will depend on the volume of air flowing, so it is dependent upon the number of inlets that will be in use at any one time. It is good to maintain a velocity of air through the pipes greater than the minimum values given in Table II. In general, air flow through the pipes should be in the range of 2000 to 6000 feet per minute.

As a general rule, it is permissible to operate two 1-1/4″ diameter hoses but only one 1-1/2″ or 2″ hose on a 2 inch diameter pipe line and a maximum of three 1-1/2″ hose or five 1-1/4″ hoses on a 3 inch diameter pipe line.

The piping should be laid out to minimize the possibility of filling the pipes with material. In large installations, provisions should be taken to see that there is a way of cleaning out the pipes if they do become clogged with materials that should not have been picked up.

NOTE: 3-1/2 inch diameter pipe is referred to as 3 inch (inside) diameter pipe. The 2 inch diameter pipe has an inside diameter of approximately 1.875 inches so because of the small variations in outside/inside diameters and to make it easier to read the graphs we will use the 2 inch diameter.

In some installations, it will be necessary to install an automatic pressure release valve in the line so that the cleaning tools will not stick to the carpet or floors when only one inlet is being used or when the system is being used in a dental operation or barber shop. This valve must be installed at the end of the pipe run that is the greatest distance from the unit so that air velocity through the pipes will be great enough to carry the material being picked up.

PRESSURE LOSSES

To select the proper vacuum power unit for the system, the total system loss (pressure drop
or resistance) must be determined and is measured in water lift (” H2O) or inches of mercury (” Hg)
vacuum. The system losses consist of the following:

A. Loss through the hose and tools
B. Loss through the piping (straight runs and bends)
C. Loss through the separator (s)

HOSE AND TOOL FRICTION LOSS
1. Hose and Tool                                                                           Friction Loss

Hose size with tool                                                                     (” H2O) or (” Hg)
1-1/4″ (25 feet @ 40 CFM)                                                        32″                 2.34″
1-1/4″ (30 feet @ 80 CFM)                                                        34″                 4.65″
1-1/2″ (25 feet @ 40 CFM)                                                         24″                3.27″
1-1/2″ (30 feet @ 80 CFM)                                                         26″                 3.56″

NOTE: CFM is cubic feet per minute

2. Friction loss in the line (tubing and piping) can be obtained from Chart I “Friction Loss Chart”.

3. The friction loss through 45 degree and 90 degree elbows are higher than the loss
through the same diameter of straight piping or tubing. The average friction loss for 2″
tubing through 2″ 45 degree and 90 degree elbows are equivalent to 5 feet and 10 feet
of straight tubing, respectively. The average friction loss for 3″ piping through 3″ 45
degree and 90 degree elbows are equivalent to 7 feet and 12 feet of straight pipe respectively.

AIR FLOW REQUIRED FOR CLEANING


The hose diameter, the particle size and the quantity of material to be conveyed determines
the air volume. The air required for each 1-1/4″ cleaning tool is approximately 40 CFM; for each
1-1/2″ cleaning tool approximately 80 CFM. However, for light cleaning, such as an office with
carpets or wooden floors, the air flow can be reduced slightly. For heavier material, the air flow
can be increased to suit the system requirements.

SELECTION OF VACUUM POWER UNITS


The size or number of power units required for the installation will be determined by the amount
of water lift and air flow required to make the system operate efficiently.

A) CR-1400 Series
The CR-1400 series power units are designed for one user only. In some cases when there
is very light dust, two users may be used, but in this case please verify with the factory for proper
use.

The CR-1500 series power unit is designed for multiple users. Up to five operators
at one time may use 1-1/4″ cleaning hoses depending on the length of hose, type of tool used and
type of floor to be cleaned (tile or concrete vs. high or low pile carpet) in a balanced system. Up
to three operators at one time may use 1-1/2″ cleaning hoses depending on the items listed above.
It is important to contact the factory for assistance in designing when the maximum number of users
may be used as outlined above.

B) Power Unit Reference
LMI                               Main Line Size      ” H2O            CFM          Sq. Ft.              Length of Run2            # of Valves
CR-1400 Series1                      2″                    112             210        max 12,000           250 Ft.                             10-15
CR-1500 Series                    ***3″                 112             410        max 80,000          450 Ft.                         Up to 100

*** Internal pipe diameter

1Must use 3 to 5 feet of metal tubing in the exhaust line coming out of the unit. It is recommended
(CR-1400 Series only) to use the PF-200 muffler on this unit to reduce exhaust noise.

2Length of run is for equivalent feet of tubing. Please refer to page 3 , “Hose and Tool Friction
Loss”, for the calculation of equivalent feet of tubing.

NOTE: The above figures are approximate and is for reference only and not intended as a strict guide.

C. Number of Units
In some cases when two users are required it may be more economical to use two CR-1400’s
(separate system) than one CR-1500. Also, if one unit goes down for repair you could hook
the working unit into the other system and still be able to clean for both areas.
The same applies for the CR-1500. If a job requires a larger unit you may be able to
use two or more CR-1500 units by placing them in different cleaning zones and arrange
them in such a way that you could connect one of the units into another zone and still clean both
areas if one of the units goes down for repair.

DESIGNING A SYSTEM


(example – See Figure 1)
Piping and tubing layout:
Legend: [  ]Inlet valves
[X]Inlet valves selected for calculation (A,B,and C)
Three Inlet valve locations have been selected for design purpose, to establish the friction
losses as well as line sizes for the system. Inlet valves (A) and (B) are the farthest from the vacuum
pump. Inlet valve (C) is located in a branch about equal distance to the vacuum pump. This
selection of active Inlet valve locations will aid in designing the system with proper line sizes to
insure the system capability. After looking over the requirements for this plant we have decided
to use a 1-1/2″ x 25 foot hose size and tools.
Conversion from inches of mercury (Hg) to inches of water (H2O) as follows:
1″ Hg = 13.62″ H2O
or 1″ H2O = .0734″ Hg
LOSSES
TOOL ENTRANCE AND HOSE AT POINT (A)
From paragraph 4, “Hose and Tool Friction Loss” when handling 80 CFM using a 1-1/2″ x 25 foot
hose the hoses and tool total friction loss is ——-1.76″ Hg (24″ H2O)
Looking at Figure I at point (A), 80 CFM enters the system and flows to point

(D). The total equivalent piping length from point (A) to point (D) is:
2″ line                          28 Ft. in length
2 – 90 Deg. Elbows 20 Ft. in equivalent length
TOTAL                         48 Ft. in equivalent length

From chart I “Friction Loss Chart”, at 80 CFM and a 2″ diameter line we read:
0.75″ Hg loss/100 Ft. line
or 10.24″ H2O loss/100 Ft. line
48 x 10.24 = 4.92″ H2O (0.36″ Hg)
100

At point (D) to (F) we have:
3″ line                           198 Ft. in length
0 – 90 deg. elbows  0 Ft. in equivalent length
TOTAL                        198 Ft. in equivalent length

From Chart I “Friction Loss Chart” at 80 CFM and 3″ diameter line, we read:
0.18″ Hg loss/100 Ft. line
or 1.29″ H2O loss/100 Ft. line
198 x 1.29 = 2.55″ H2O (0.19″ Hg)
100

From Figure I
At point (B) 80 CFM enters the system. The total equivalent piping from (B) to (E) is as follows:
2″ line                          23 Ft. in length
2 – 90 deg. elbows  20 Ft. in equivalent length
TOTAL                        43 Ft. in equivalent length

From Chart I “Friction Loss Chart”, at 80 CFM and a 2″ diameter line we read:
0.75″ Hg loss/100 Ft. line
or 10.24″ H2O loss/100 Ft. line
43 x 10.24 = 4.40″ H2O (0.32″ Hg)
100
At point (E) to (F) we have:
3″ line                       63 Ft. in length
1 – 90 deg. elbow 12 Ft. in equivalent length
TOTAL                      75 Ft. in equivalent length

From Chart I “Friction and Loss Chart”, at 80 CFM and 3″ diameter line we read:
0.18″ Hg loss/100 Ft. line
or 1.29″ H2O loss/100 Ft. line
75 x 1.29 = 0.97″ H2O (0.07″ Hg)
100

From Figure I at point (C) 80 more CFM enters the system. The total equivalent piping from (C)
to (G) is as follows:

2″ line                           28 Ft. in length
2 – 90 deg. elbows  20 Ft. in equivalent length
TOTAL                          48 Ft. in equivalent length

From Chart I “Friction Loss Chart” at 80 CFM and a 2″ diameter line we read:
0.75″ Hg loss/100 Ft. line
or 10.24″ H2O loss/100 Ft. line
48 x 10.24 = 4.92″ H2O (0.36″ Hg)
100

From Figure I at point (F) 80 more CFM enters the system. This combines with the flow from point
(B) (80 CFM + 80 CFM) for a total of 160 CFM at point (F). The equivalent piping length from (F) to (G) is:
3″ line                          20 Ft. in length
0 – 90 deg elbows   0 Ft. in equivalent length
TOTAL                        20 Ft. in equivalent length

From Chart I “Friction Loss Chart” at 160 CFM and a 3″ diameter line we read:
0.38″ Hg loss/100 Ft. line
or 5.17″ H2O loss/100 Ft. line
20 x 5.17 = 1.03″ H2O (0.08″ Hg)
100

From Figure I at point (G) 80 more CFM enters the system. This combines with a flow from points
(A) and (B) (80 + 160) for a total of 240 CFM at point (G). The equivalent piping length from (G)
to the separator inlet is:
3″ line                           94 Ft. in length
3 – 90 deg. elbows   36 Ft. in equivalent length
TOTAL                          130 Ft. in equivalent length

From Chart I “Friction Loss Chart” at 240 CFM and a 3″ diameter line we read:
0.90″ Hg loss/100 Ft. line
or 12.26″ H2O loss/100 Ft. line
130 x 12.26 = 15.94″ H2O (1.17″ Hg)
100

Separator loss is added to the system loss. The loss is 3.40″ H2O (0.25″ Hg). In general, the line
between the separator and the vacuum pump is insignificant for all practical purposes.

Therefore, the total friction loss for the system is:
1. Hose and tool loss(page 3)…………………………………..24.00″ H2O
2. Line loss from (A) to (D)…………………………………………4.92″ H2O
3. Line loss from (D) to (F)…………………………………………2.55″ H2O
4. Line loss from (B) to (E)…………………………………………4.40″ H2O
5. Line loss from (E) to (F)………………………………………….0.97″ H2O
6. Line loss from (C) to (G)…………………………………………4.92″ H2O
7. Line loss from (F) to (G)…………………………………………1.03″ H2O
8. Line loss from (G) to separator………………………………15.94″ H2O
9. Separator loss……………………………………………………….3.40″ H2O
TOTAL SYSTEM LOSS                                                                      62.13″ H2O (4.56″ Hg)

The total system air volume is determined by 80 CFM/1-1/2″ diameter hose x 3 operators = 240 CFM.

The figure of 240 CFM (at standard conditions is 29.92″ Hg and 68 deg F) is then multiplied by the
ratio of the standard barometric pressure divided by the standard barometric pressure minus the
vacuum pump inlet vacuum in inches of H2O to obtain the volume under inlet vacuum conditions.
240 x (29.92/(29.92 – 4.56)) = 283.2 ICFM

Therefore we select the CR-1500 Series pump for the application:
CR-1500 Unit
360 to 380 CFM Airflow
100″ H2O Water Lift @ sealed conditions
Check: 360 CFM > 283.2 ICFM
100″ H2O > 62.13″ H2O

The CR-1500 Series should be ok for this application.

PIPING/TUBING INSTALLATIONS


It is very important that care be taken in connecting all piping/tubing joints to insure an airtight
system as air leaks will decrease the vacuum system efficiency.

ELEVATION INSTALLATIONS


When a central vacuum system is to be installed at an elevation above sea level, corrections
to the vacuum pump performance becomes necessary since the vacuum pump performance
graphs and charts are based on standard sea level conditions.

Example – 7,000 Ft. elevation

Referring to our previous example, it was based on total air volume of 240 CFM/283.2 ICFM.
For the adjusted barometric pressure of 23.1″ Hg (7,000 Ft. elevation), see Chart II

Air volume correction: 240 CFM x (29.92/(23.1 – 4.56)) = 387.3 ICFM

Therefore the CR-1500 Series may not work in this application since 387.3 ICFM >360 CFM. You could in this case use two CR-1500 Series units by splitting the system in
two separate systems with no more than two operator per unit or use the smaller 1-1/4″ hose which would allow three users and only one CR-1500 Series unit.

DIAGRAM OF EXAMPLE SYSTEM (FIGURE 1)

see pic


FRICTION LOSS (CHART I)

chart

CONVEYING VELOCITIES (TABLE I)

Conveying Velocities in Dust Exhaust System

Material Velocity Feet/Minute
Grain, Dust 2,000
Sawdust, Dry 2,000
Sawdust, Damp 3,000
Wood Shaving, Chips 3,000
Wood Hogging Material 4,000
Cotton 2,000
Lint 1,500
Metal Grindings 2,500 & Up
Lead Dust 5,000
Fine Brass Turnings 4,000
Pulverized Coal 4,000
Sand Blasting 3,500 & Up

DUTCH GOVERNMENT STUDY

STUDY BY DUTCH GOVERNMENT On Vacuuming Rates
Comparison of vacuuming rates for portable and central vacuums in square
feet of office area cleaned per hour. Savings in annual cleaning costs for offices can quickly justify the cost of a central system.
Portable Central
Thorough Vacuuming 2,500 – 4,000 Sq. Ft. 3,150- 4,800 Sq. Ft.
Quick Vacuuming 3,600 – 4,500 Sq. Ft. 4,800 – 6,000 Sq. Ft.
Note: We plan systems based on a conservative vacuuming rate of 2000 square
feet per hour.

ELEVATION (CHART II)

CHART II
Average Absolute Atmospheric Pressure
Altitude in Feet                        Inches of Mercury
reference to sea level                          (” Hg)
sea level 0                                                29.92
+ 500                                                         29.39
+ 1,000                                                     28.87
+ 1,500                                                      28.33
+ 2,000                                                     27.82
+ 3,000                                                     26.81
+ 4,000                                                     25.85
+ 5,000                                                     24.90
+ 6,000                                                     23.98
+ 7,000                                                     23.10
+ 8,000                                                     22.22
+ 9,000                                                     21.39
+ 10,000                                                  20.58

TERMS AND DEFINITIONS


C0 = 5/9 (F0 – 32) = Degrees Centigrade
F0 = (C0 x 9/5) + 320 = Degrees Fahrenheit
CFM = Cubic feet per minute (airflow)
FPM = Feet per minute (velocity)
Water Lift = H2O
Water Lift = Static pressure
= Suction
Hg = Mercury
1″ H2O = .073431″ Hercury (Hg)
1″ Hg = 13.6230″ H2O @ 680 F and barometric pressure of 29.92″ Hg
I.D. = Inside Diameter
O.D. = Outside Diameter
P.S.I. = Pounds per square inch (pressure)
1 P.S.I. = 26.680″ H2O (vacuum pressure)
Micron = The millionth part of a meter
1 Micron = .00003935 inches
mm = millimeter            cm = cubic meter
1″ = 25.4 mm                       1″ = 2.54 cm
m = meter                              L = liter
1m = 39.37″                         1L = 2.113 pints
g = gram                             1 oz. = 28.375 g.
l lb. = 454 g.
ohms = Volts/amperes                   amperes = Volts/ohms
Volts = Amperes x ohms                             Watt = Volts x amperes
Horsepower = (volts x amperes x efficiency)/746
Torque (1b-ft) = (horsepower x 5250)/RPM
Horsepower = (torque (1lb-ft) x RPM)/5250

 

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