INERGEN® Pressure Relief Venting Guide

8/12/2019 INERGEN Pressure Relief Venting Guide

1/22

Revised 1-5-98

INERGEN PRESSURE RELIEFVENTING GUIDE

2-14-97 Page 1

IntroductionVenting of enclosures protected by INERGEN systems can take many forms. Examples include: Free Venting directly to the

exterior of the building, venting through ducts, venting through unprotected spaces, and venting through protected spaces. It

is the intent of this guide to address the most common methods of venting and offer design parameters to prevent over-

pressurization of enclosures protected by Ansul INERGEN systems.

Pressure venting for an INERGEN system is similar to the design and calculation of the piping for the agent distribution sys-

tem. In both cases a gas is moving through an opening which restricts the flow. The design must deal not only with the pres-

sure loss caused by the device the gas flows through, but also with the friction loss developed by the walls of the piping or

ducts. Friction loss increases as the flow is restricted through orifices regardless of whether these orifices are manifold orifices

and nozzles, or inlet grilles, dampers, louvers, bird screens, etc. All devices must be accounted for in the design and calcula-

tion of the venting system. In addition, if the venting is to be accomplished through multiple enclosures, pressure build-up cre-

ated by restrictions in each of the rooms must also be accounted for. This pressure build-up is cumulative and will definitely

have an effect on the venting for the room being protected.The main difference between pressure venting systems and distribution piping systems is that the venting system deals with

enclosures and equipment typically rated at pressures less than 50 lbs/ft2. Any pressure greater than that listed for the enclo-

sure or equipment will likely do some damage to that component.

Early venting recommendations were based on enclosure wall strengths as listed in NFPA 12. It appears that these figures

were based on old construction methods employed in the 1950s and 60s. Todays construction methods, using materials such

as metal studs, movable partitions, and concrete blocks, create walls with lateral strengths much weaker than those encoun-

tered in walls built when the NFPA 12 wall strengths were developed. The only sure way to determine the wall strength of an

enclosure is to obtain that information from the owner of the building or the architect responsible for the buildings design. If this

information is not available, the wall strength will have to be estimated using an appropriate safety factor. This estimate should

be made by the INERGEN system designer after consultation with the owner and/or architect. If the system designer, along

with owner and/or architect, are not able to estimate a maximum wall strength, Ansul recommends the use of 5 lbs./ft2. This is

a conservative estimate which will generally calculate a vent larger than actually required, but will not require more agent to

reach or hold concentrations.

This guide will provide the INERGEN system designer with the information necessary to design and calculate venting systems

for most spaces protected by INERGEN systems. Complex duct systems and specialized venting system components may

require support by the Distributor Technical Services Department at Ansul.

The following Flow Chart gives a general overview of the steps required to complete the design and calculation of an

INERGEN venting system.


2/22

INERGEN VENTING GUIDE1-5-98 Page 2 Rev. 1

Venting Guide Flow Chart

DETERMINE

ALLOWABLE PSF

DEDUCT ELA FROM

DOOR FAN TEST

MULTIPLE AREAS?

CALCULATE INERGEN

SYSTEM PEAK FLOW RATE

RESIZE

COMPONENT

RESIZE

DUCT

CALCULATE

PRESSURE LOSS

COMPONENT

SIZE CHANGE?

DUCT SIZE

CHANGE?

RESIZE FREE

VENT AREA

CALCULATE

PRESSURE LOSS

CALCULATE

PRESSURE LOSS

FOR EACH DEVICE

CALCULATE ALLOWABLE

PSF PER AREA

CALCULATE FREE

VENT AREA(S)

DUCTS?

ADDITIONAL

RESTRICTIONS?

SUM ALL

PRESSURE LOSSES

TOTAL PRESSURE

LOSS ACCEPTABLE?

FINAL FREE VENT

AREA AND

COMPONENT SIZES

YES

YES

YES

YES

YES

YES

NO

NO

NO

NO

NO

NO

NOTES: 1 Use tables and charts attached to this guide.2 Obtain data from component manufacturer or architect.

1

2


3/22

Free Vent Calculation

Ansul has always indicated that venting is an integral part of any INERGEN system and the requirement for pressure relief

venting is mandatory. Per the Ansul Design Manual, a calculation is required to determine the amount of free venting area, in

square inches, to prevent an excessive pressure buildup within the protected volume. The information needed to complete this

calculation includes a calculation to determine the flow rate of air out of the enclosure (equal to the INERGEN system peakflow rate) and knowledge of what the maximum allowable internal pressure is that the enclosure can withstand.

1. The flow rate out of the enclosure can be determined by calculating the INERGEN system peak flow rate (Qp). This flow

rate is determined from the formula:

Qp =2.7 IG_____

T

Where: Qp = INERGEN system peak flow rate

IG = INERGEN Agent Quantity Supplied (ft3)

T = 90% discharge time (min.)

The flow rate calculated is the peak flow which results within the first few seconds of the INERGEN system discharge. This

flow rate is required for the calculations which will be used later in this guide when using ducts in the venting system.

2. Consult owner or architect to determine building wall strength. If this information is not available, Ansul recommends theuse of 5 lbs./ft2 The use of a low wall strength (5 lbs./ft2) will require larger venter sizes but will not require more agent to

reach or hold concentration.

3. The next step in designing pressure venting for an INERGEN system is to calculate the minimum size of the Free Ven

Area. The formula used to determine the size of the Free Vent is:

X =0.0855 Qp_________

P

Where: X = The Free Vent Area in square inches.

Qp = The INERGEN agent peak flow rate calculated from the formula listed above.

P = The maximum allowable wall strength in lbs./ft2 of the weakest surface (walls, ceiling, floor, doors, windows

etc.) of the enclosure.

The Free Vent Area calculated is the minimum area required to prevent over-pressurization of the enclosure when the open-

ing is to atmosphere. Atmosphere is described as open air on the outside of a building.

INERGEN VENTING GUIDE2-14-97 Page 3

Free Vent Calculation

INERGEN Agent Quantity = (4) 439 ft3 cylinders

90% Discharge Time = 45 Seconds

Maximum Allowable Enclosure Pressure = 5 lbs./ft2

Calculate Agent Peak Flow Rate:

Qp = 2.7 (1756) = 4741.2 = 6322 CFM_________ ______

.75 .75

Calculate Free Vent Area:

X = 0.0855 (6322) = 0.0855 (6322) = 540.5 = 241.7 in2 = 1.68 ft2____________ ______________ ____

5 2.236 2.236

10 FT.

30 FT.

12FT.

PROTECTEDSPACE

FIGURE 1000917

EXAMPLE


4/22

Free Vent Calculation (Continued)

An opening into another enclosure regardless of the size of that enclosure will cause a pressure build up inside both enclo-

sures. The pressure generated in the enclosures will be directly dependent upon the size of the secondary enclosure in

relation to the amount of agent supplied. The larger the secondary enclosure, the lower the pressure generated.

The potential pressure resulting within both enclosures, if completely sealed and not vented, can be approximated with theformula:

P =995 V1______

V2

Where: P = Final pressure in both enclosures (lb/sq. ft.)

V1 = Protected enclosure volume (ft3)

V2 = Total volume of both enclosures (ft3)

As indicated by calculations, this pressure is prohibitively high unless the second enclosure is extremely large. A second

volume 100 times the protected volume will still retain a pressure of approximately 10 lb/sq.ft. within both enclosures.

Location of Vents and Effects on Concentration

Since the density of INERGEN is very close to that of air, the agent is not affected by gravity to the same extent as Halon 1301

and other clean agents. Gravity will have an effect on leakage of agent from a room; however, that leakage will be slower than

if the agent were heavier as is the case for Halon 1301 and the other clean agents. Therefore, openings will have less of an

effect on leakage if they are located higher on the walls than if they are at the floor level.

Hold time of an INERGEN system will also be affected by air movement through the enclosure. In many cases, this air move-

ment may have more of an effect on leakage than the location of the opening. It will be advantageous to prevent any air move-

ment through the enclosure. Openings at different heights will result in faster leakage than if all openings were at the same

height.

Rooms with normal openings and without any air movement through the enclosure should hold concentration for a minimum

of 10 minutes. Discharge tests have indicated hold times in excess of 30 minutes where the air flow through the enclosure is

negligible. The hold time predictions calculated using the European ISO procedures have been found, through tests, to be

more accurate than the procedures in NFPA 2001, 1996.

Vents should not be located on a wall in the immediate vicinity of a nozzle, where the discharge of agent from the nozzle could

be directed through the opening. Care should be taken to locate the vents away from the flow of agent so as to vent air from

the room instead of INERGEN agent or agent/air mixture.

If multiple vents are installed, all vents should be at the same height on the walls. This will minimize the rate at which the agent

is lost.

Gravity or spring operated dampers will normally close after operation. Pneumatically or electrically operated dampers should be

closed after discharge of the system. If concentration must be held after discharge, dampers must be closed to eliminate airflow

through the enclosure.

When designing a venting system through an unprotected space it is important to consider the affect that the vented products will

have on the unprotected space. Depending on the type of fuel being suppressed, it is feasible that smoke, sparks, or flame may

travel into the unprotected space. Products of combustion and/or heat from the protected space could potentially spread the fire

or put detectors into alarm. A venting system designer should consider the use of a duct system to vent the gases to a safe loca-tion instead of venting through an unprotected space.

If venting must be through protected spaces, it is recommended that agent be discharged simultaneously into all of the spaces

along the venting path.

Recommended venting design for a selector valve system is to vent each space separately to a safe location rather than through

a protected space.



5/22

Venting System Components

An INERGEN pressure vent can be thought of as a venting system. In most cases the vent will not consist of an open hole

through a wall, but will include components such as inlet grilles, dampers, and exhaust louvers. A venting system may also

include ducts and special components such as weather hoods, and bird screens. Each of these components adds a restric-

tion to the system which will decrease the effectiveness of the Free Vent Area previously calculated. Therefore, the friction loss

through these components must be allowed for in the pressure venting calculations. Many of the pressure drops associated

with these components relate to edge effects as discussed in the previous section.

The location of a device, like a damper in the duct system, can also have an effect on the pressure drop across that device. If

the device has a similar size duct leading into and out of it, the friction loss will be less than if the size of the duct changes

immediately before or after the device. If the device is at the end of the duct, an increase in pressure drop will occur due to the

effect of the sharp exit from the damper. Pressure drop also increases if the duct reduces in size drastically immediately before

or after the damper. This is due to the sudden changes of area which creates edge effect or sharp exit pressure losses.

Pressure drop calculations for all of these components must be completed for every system to assure that the total system

provides enough venting to prevent over-pressurization and damage to the enclosure being protected.

Pressure loss data for these components are not standard. This information must be obtained from the manufacturer of the

component, the architect, or the supplier of the components. Without this information, the Pressure Venting calculations

cannot be completed for the system.

Venting Through an Unprotected Space

When an INERGEN system is protecting a hazard which must be vented into an unprotected space, the second space will

require venting if the resulting pressure from the formula P = 995 V1 is greater than the allowable pressure from the room._______V2

The second space must be at least 50% of the volume of the protected space, otherwise another venting option must be used

One of the steps for a vent calculation is to determine the pressure the walls can withstand. The allowable pressures must be

determined for all spaces that will be involved in the venting process, protected, as well as unprotected spaces. The protect-

ed space, which is the first in a venting sequence, will always encounter the highest pressure. The pressures encountered

through the venting sequence are additive and the sum of these pressures cannot exceed the maximum allowable pressure

in the protected space. Pressure effects on the walls of the protected space, the unprotected space, and the wall between the

two spaces, must be examined and care taken not to exceed the allowable pressure on any wall.

The most cost effective way to size the vents is to cause the vent pressure drops to be equal, or use the allowable pressure

in the protected space and divide it by two to obtain the design pressure to be used in the vent formulas. This will yield the tota

minimum vent area required by both spaces. The unprotected space must be able to withstand the pressure used in sizing

the vent in that space. Once these pressures are determined, the maximum total pressure drop across vents added together

cannot exceed that of the protected space, while any pressure utilized within the unprotected space cannot exceed the allow-

able pressure for the unprotected space.

FIGURE 2000834

Qp = INERGEN Peak Flow rate to be used in vent design formula (CFM)

Pa1 and Pa2 = Allowable pressures in respective spaces, wall strength (lb/sq ft)

Pd1 =Pa1____

2

Pd2 = Pa1 Pd1

Where: Pd1 and Pd2 = Design pressures to be used in venting formula (lb/sq ft)

X1 and X2 = Vent areas for respective spaces (in2)

INERGEN VENTING GUIDERev. 1 1-5-98 Page 5


6/22

Venting Through an Unprotected Space (Continued)

For example, if the allowable wall pressure in a protected space, Pa1 = 6 lb/sq ft, both vents, X1 and X2 should be sized with

a design pressure, Pd, of 6 divided by 2, or 3 lb/sq ft. Both vents should be sized with flow rates in the formulas being the

same and equal to the flow of the INERGEN system, Qp. During the discharge, the pressure buildup in the unprotected space

will be 3 lb/sq ft. and the pressure in the protected space will be 3 + 3 = 6 lb/sq ft.If, in the example above, the unprotected space can only withstand an allowable Pa2 pressure of 2 lb/sq ft, the vent in that

space will require sizing at Pd2 ! Pa2 = 2 lb/sq ft. The protected space vent can then be decreased in size by using the design

pressure of 4 lb/sq ft, or Pd1 = Pa1 Pd2. The system flow rate is Qp in both formulas. In no case can the sum of the design

pressures used to calculate the vents in the two spaces exceed the allowable pressure, Pa1, in the protected space. In other

words, Pd1 + Pd2 ! Pa1.


Venting Through an Unprotected Space

Qp (INERGEN Agent Flow Rate) = 6322 CFM

Pa1 (Maximum Allowable Enclosure Pressure for Room 1) = 5 lbs./ft2


Calculate Design Pressures:

Pd1 = 5 = 2.5 lbs./ft2___2

Pd2 = 5 2.5 = 2.5 lbs./ft2

To find the Vent Area (X):

X =0.0855 (6322)

=0.0855 (6322)

=540.5

= 341.9 in2 = 2.4 ft2 for each vent___________ ____________ _____

2.5 1.581 1.581

10 FT.

30 FT.

60 FT.

12FT.

PROTECTEDSPACE

UNPR

OTEC

TED

SPACE

FIGURE 3000832

EXAMPL

E


7/22

Venting Through a Protected Space

When two adjacent spaces are being simultaneously protected and one space is vented through the second, the guidelines

for determining design pressures are the same as in venting through an unprotected space. Whenever possible, the smaller

space should be vented through the larger space. If not possible, the smaller space must be at least 50% of the volume of

the larger space; otherwise, another venting option must be used. When both spaces are protected, the vent design flowrates are additive and must be summed as well as the design pressures. The design of the first free vent area in the series

will utilize system flow rate Qp1 and the appropriate Pd1 as determined above. The design of the second vent in the series wil

utilize Pd2 and the summation of flow rates, Qp1 + Qp2 = Qpd2, which is the design flow rate to be used to determine the area

of the second vent in the series.

FIGURE 4000835

Qp1 = INERGEN Flow rate in first hazard (CFM)

Qp2 = INERGEN Flow rate in 2nd hazard (CFM)

Qpd2 = Qp1 + Qp2 = Design flow rate for 2nd hazard

Pa1 and Pa2 = Allowable pressures in respective spaces, wall strength (lb/sq ft)

Pd1 and Pd2 = Design pressures to be used in venting formula (lb/sq ft)

X1 and X2 = Vent areas for respective spaces (in2) determined by venting formula

Any combination of pressures that do not exceed the space allowable can be utilized. However, the most effective vent sizing

will be experienced when the pressure drop across the second vent is larger than the drop across the first vent. The specific

values of these design pressures can be determined by: utilizing the flow rates, the allowable pressure in the first hazard, and

the formula:

Pd2 =Pa1______________

.666

1 + ( Q p1 )____Q pd2This predicts the design pressure for the second hazard. The design pressure for the first hazard then becomes:

Pd1 = Pa1 Pd2

Knowing flow rates and design pressures, the free vent areas can be calculated from the standard venting formula. Remembe

to keep in mind the fact that the design pressures are additive and the highest actual pressure will be experienced in the first

hazard in the venting series. Check to be sure the sum of the pressures does not exceed the allowable pressure in hazard

number one.



8/22

Venting Through a Protected Space (Continued)

Venting Through Three or More Enclosures

It is possible to calculate a venting series through multiple spaces, both protected and unprotected. The pressures used for

design will remain additive with the highest pressure to be experienced in the first hazard in the venting series. The design

pressures of all hazards downstream must be added together and checked to be sure the pressure does not exceed the allow-

able for each specific hazard. Likewise, flow rates from each succeeding hazard will require the flow from the preceding

hazards to be added to it for the design flow rate for each specific vent.

The most cost effective venting area combinations can be determined through the use of complicated calculus equations

or multiple iterations. A series of vent area calculations using multiple iterations can be done much more easily and will yield

acceptable results.


Venting Through a Protected Space

Qp1 (INERGEN Agent Flow Rate for Room 1) = 6322 CFMQp2 (INERGEN Agent Flow Rate for Room 2) = 31,608 CFM

Qpd2 = 31608 + 6322 = 37930



To obtain the smallest total free vent area, complete the following calculations:

Calculate the Design Pressure for Room 2:

Pd2 =5

=5

=5

= 3.84 lbs./ft2_________ ___________ ______

1 + ( 6322 1 + 0.1667.666 1.302937930).666

Calculate the Design Pressure for Room 1:

Pd1 = 5 3.84 = 1.16 lbs./ft2

Calculate the Free Vent Area for Room 1:

X1 (vent area) =0.0855 (6322)

=0.0855 (6322)

=540.5

= 502 in2 = 3.49 ft2___________ ___________ ____

1.16 1.077 1.077

Calculate the Free Vent Area for Room 2:

X2 (vent area) =0.0855 (31608 + 6322)

=0.0855 (37930)

=3243

= 1655 in2 = 11.49 ft2___________________ ____________ ____

3.84 1.960 1.960

If the pressures of 2.5 lb/ft2 for each enclosure had been picked, which would be acceptable, the free vent areas would have

been 2.37 ft2 and 14.24 ft2 for a total vent area of 16.61 ft2. This is larger than the 14.98 ft2 using the optimum vent sizes

calculated above of 3.49 and 11.49 ft2.

10 FT.

30 FT.

60 FT.

12FT.

ROOM1

ROOM

2

FIGURE 5000832

EXAMPLE


9/22

Venting Through Ducts

Venting through ducts can be treated somewhat the same as venting through unprotected enclosures. Pressure drops are

additive and optimum areas will be obtained when the drop in the room and the duct system are equal. The pressure loss

developed in the duct and all associated components will be additive and the sum of the total duct pressure drop and the free

vent area design pressure will be the total pressure developed in the room during discharge.Determining the pressure drop through the duct can become very complex. Each duct system has a combined set of pressure

drops dependent upon the individual duct system components. The amount of total pressure drop due to flow for the individ-

ual duct system components can be obtained from the component manufacturer for dampers and louvers. The friction loss for

straight sections of duct can be obtained from the included duct friction loss charts based on equivalent diameters (See charts

on Pages 17-20). Equivalent duct diameters can be obtained from the tables for circular equivalents of rectangular ducts (See

charts on Pages 17-20) The equivalent length factors of a few common duct fittings, the L/D factor, is also included. The ratio

L/D is the equivalent length, in diameters of straight ducts, that will cause the same pressure drop as the obstruction under the

same flow conditions.

As mentioned, duct design can become very complicated. Details discussed here are intended to show a designer the com-

plexity of this type of design. This section can only be considered a basic guideline and cannot be relied upon as a method for

designing any and all possible combinations or configurations of ductwork design. This guideline only covers simple basic

design parameters. For additional information, refer to HVAC SYSTEMS DUCT DESIGN published by SHEET METAL ANDAIR CONDITIONING CONTRACTORS NATIONAL ASSOCIATION, INC (SMACNA), 4201 Lafayette Center Dr., Chantilly, VA

22021-1209, telephone 703-803-2980, fax 703-803-3732, and also various handbooks and publications available from

American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (ASHRAE), 1791 Tullie Circle NE, Atlanta, GA

30329-2305, telephone 1-800-527-4723 or 404-636-8400, fax 404-321-5478.

Pressure drop in a straight duct section is caused by surface friction and varies with the velocity, the duct size and length, as

well as interior surface roughness. Friction loss can be easily determined from the duct friction loss charts which are based on

standard air with a density of 0.075 lb/cu. ft. flowing through average clean round galvanized metal ducts with beaded slip cou-

plings on 48 inch centers. These values may be used without correction for temperatures between 50 F to 140 F and up to

2000 foot altitude. Correction factors are available for other specific conditions, but are not included here. The density of air rather

than INERGEN is used because the medium flowing at the point of maximum pressure and flow is air. The INERGEN discharge

has just begun and with less than 10% out of the cylinder, has not yet had a chance to mix with the air in the enclosure.

HVAC duct systems are usually sized first as round ducts. Then, if rectangular ducts are desired, duct sizes are selected to

provide flow rates equivalent to those of the round ducts. Rectangular ducts should not be sized directly from actual duct cross-

sectional areas. Instead, the circular equivalent tables must be used, or the resulting duct sizes will be smaller, creating greater

duct velocities and higher pressure drops.

Changes of direction and changes in duct cross section cause increased friction loss or pressure drop during high flows. These

losses are termed dynamic losses and can be determined by various methods. The following chart gives equivalent length

(L/D) values for three common fittings likely to occur in simple ducted systems.

FITTING L/D

SHARP EXIT, NO LOUVERS 78

90SHARP MITERED BEND 86

90SMOOTH BEND 27

Gradual reductions ! 60have no additional length beyond the actual dimension and are recommended.

The first step in designing a ducted venting system is to determine the allowable room pressure. One starting point is to assignone half the allowable pressure to the room and one half to the duct, depending on the complexity of the duct. If the ductwork

has many components, a higher pressure drop may be required in the duct. The duct design pressure must not exceed the

pressure rating of the duct.

Utilizing a layout of the duct configuration, the total straight length of the duct must be known. Utilizing the INERGEN system

flow rate and the assigned design pressure, determine free vent area.

Then estimate the diameter (d), in inches, which gives the free vent area from the formula:

d = 4X___"

Where: d = diameter, in inches, of free vent area

X = Free Vent Area



10/22

Venting Through Ducts (Continued)

Choose a duct size with diameter approximately 15% larger than calculated above, (d x 1.15).

Proceed to the duct friction loss chart and locate the INERGEN system flow rate on the bottom of the chart Find the inter-

section of that flow rate with the estimated duct diameter and read the pressure drop on the left hand side of the chart in LB/sq.

ft per foot of duct length.

Multiply pressure drop per foot by length of duct in feet to obtain total straight length of duct loss.

Compare this number with the allowable pressure drop. If it is equal to or less than the allowable, the duct may be sized large

enough. At this point the duct size may be increased or decreased and re-calculated to obtain closer results. Keep in mind that

other duct components and fittings, such as dampers and louvers will also be adding to this pressure drop.

With an actual duct diameter, the added length due to elbows and sharp exits can be determined from the equivalent length

values.

To calculate the equivalent length of a duct fitting, the formula is:

L/D x (12)________

Fitting dia. in inches

L/D = Equivalent length of a resistance to flow in duct diameter (See values listed on Page 9).

From the table on Page 9, a sharp mitered bend has an L/D value of 86. A sharp mitered bend of 20 in. diameter duct would

be:86 x 12

= 51.6 feet of equivalent straight 20 inch duct______20

The length of this and any other fittings must then be added to the straight duct length and the new value then multiplied by

the pressure drop per foot to obtain the total drop across duct and fittings.

If the pressure drop across the duct and fittings is still low enough to be acceptable, the circular equivalent chart can be used

to determine an appropriate rectangular duct dimension. Once this dimension is determined, the drops across the dampers,

louvers and any other duct components can then be determined.

These drops can be read directly from charts supplied by the manufacturer of these components. If a chart is based on pres-

sure drop versus velocity, the velocity can be calculated by using the flow rate in CFM and dividing it by the area of flow of the

component, in square feet. If the chart is based on flow or velocity versus pressure drop or friction loss as in. w.g. (inches ofwater gage), the pressure drop in pounds per square foot can be determined by multiplying in. w.g. by 5.204. This pressure

drop must then be added to the duct loss pressure drop for a total duct system pressure drop. Be sure a pressure drop is

included for every component in the duct system that provides any resistance to flow.

The duct system pressure drop must then be added to the room pressure drop used to calculate the free area at the begin-

ning of the design. This total pressure drop must not exceed the allowable room pressure in pounds per square foot as deter-

mined from the building construction. The duct design pressure must not exceed the pressure rating of the duct.



Variables

d = Equivalent Round Duct diameter (inches)

da = Assumed Equivalent Round Duct diameter (inches)PSF = POUNDS PER SQUARE FOOT

#P = Pressure Drop (PSF)

#P/L = Pressure Drop per Foot of Length (PSF)

V = Velocity (ft./minute)

HW

= Inches of Water Column (in. W.G.)

L = Length (feet)

LEQ

= Equivalent Length (feet)

L/D = Equivalent Length of a Resistance to Flow, in duct diameters

CFM = CUBIC FEET PER MINUTE

A = Area (ft.2)

EXAMPLE

(Continued)


11/22




IG = 1756 ft.3

T = 45 seconds = .75 minutes

P = 5 PSF

1/2 P = 2.5 PSF

Estimate Agent Flow Rate:

Qp =2.7 IG 2.7 (1756) 4741.2

6322 CFM______ ________ ______

T=

.75=

.75=

Calculate Free Vent Area Required:X =

0.0855 Qp 0.0855 (6322) 541 342.4 in.2_________ ___________ ___

P=

2.5=

1.58

Estimate Duct Diameter:

d =4X 4 (342.4) 1370 436

20.9 in.____ _______ ____"

="

="

= =

Increase Duct Diameter:

da = 20.9 (1.15) = 24 in.

Calculate Actual Duct Area:

X =

" da2 " 242 " 576 1810

453 in.2_____ _____ _____ ____

4 = 4 = 4 = 4 =

Calculate Actual Pressure Drop:

#P =0.0855 Qp 2 0.0855 (6322) 2 541 2 1.192 1.42 PSF________ ___________ ___( X ) = ( 453 ) = (453) = =

FREE VENT AREA

(Continued)

EXAMPLE

FIGURE 6000844


12/22




CFM = 6322

24 in. x 24 in. duct (selected from the Circular

Equivalents of Rectangular Ducts Chart, Page 17)

Note: After calculating a 24 in. diameter duct in the previous step, a 24 in. x 24 in. square duct was selected from the chart

because the round equivalent diameter is larger than 24 in. and most devices are available in a 24 in. x 24 in. size.

d = 26.2 in.

L = 70 ft.

#P/L = 0.007 PSF/ft.

(from Duct Friction Loss Chart, Page 21)

Calculate Friction Loss through Straight Duct:

#P = L (#P/L) = 70 (0.007) = 0.49 PSF

Calculate 90Sharp Bend Equivalent Length:

LEQ = L/D (12) 86 (12) 1032 39.4 ft.______ ______ ____d

=26.2

=26.2

=

Calculate 90Smooth Bend Equivalent Length:

LEQ

= L/D (12) 27 (12) 324 12.4 ft.______ ______ ____d

=26.2

=26.2

=

Calculate Total Equivalent Length:

LEQ

= L + 2 (LEQ

Sharp Bend) +

LEQ

Smooth Bend = 70 + 2 (39.4)

+ 12.4 = 161Calculate Total Pressure Loss:

#P = LEQ

(#P/L) = 161.2 (0.007) = 1.13 PSF

DUCT

(Continued)

Duct Friction Loss Chart (U.S. Units) (2)Air Quantity, cfm at 0.075 lb/ft3

.007

6322 CFM

EXAMPLE

Circular Equivalents of Rectangular Ducts for Equal Friction andCapacity (U.S. Units) (2) Dimensions in Inches

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36 in. x 24 in. size, NC 30-40 application

Qp = 6322 CFM

Hw = .21 in. W.G. (from chart)

Convert in. W.G. to PSF:

#P = Hw (5.204) = 0.21 (5.204) = 1.09 PSF

24 in. x 24 in., Figure 5.3

Calculate Area of Damper:

A =24 x 24 576

4 ft.2______ ___

144=

144=

Calculate Velocity of Air Flow:

V =Qp 6322 1581 ft./min.___ ___A

=4

=

Hw = 0.10 in. W.G. (Damper (Figure 5.3 Style) Pressure Drop Chart)

Convert in. W.G. to PSF:

#P = Hw (5.204) = 0.10 (5.204) = 0.52 PSF

DAMPER

INLET GRILLE

1581

PRESSURELOSS(INW.G.)

(Continued)

EXAMPLE

EXAMPLE

Damper Size 24 x 24

Steel S590 & Aluminum S590 Gymnasium Grilles

*

*Because the cfm of 6322 was slightly higher than 6180, the in. of W.G. was interpolated at.21 in. of W.G. 000890

000841


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#P = #P (Free Vent) + #P (Duct) + #P (Inlet Grille) + #P (Damper)

#P = 1.42 + 1.13 + 1.09 + 0.52 = 4.16 PSF

5 PSF 4.16 PSF = 0.84 PSF for Louver

Hw =PSF 0.84

.16 in. W.G.____ _____5.204

=5.204

=

V = 975 ft./min. (from Airflow Resistance Chart)

X =Qp 6322 6.49 ft.2___ ____

V=

975=

Louver Size = 48 in. W x 42 in. H

A = 7.19 ft.2 (Free Area Chart)

V =Qp 6322 879.3 ft./min.___ ____A

=7.19

=

#P = Hw (5.204) = 0.13 (5.204) = 0.68 PSF

#P = #P (Free Vent Area) + #P (Duct) + #P (Grille) + #P (Damper) + #P (Louver)

#P = 1.42 + 1.13 + 1.09 + 0.52 + 0.68 = 4.84 PSF

VENT TOTAL

LOUVER

VENT SUBTOTAL

880

.13

EXAMPLE

EXAMPLE

EXAMPLE

(Continued)

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Allowance for Room Pressurization Test Results

The ELA (Equivalent Leakage Area) determined from a Room Pressurization Test may be deducted from the amount of Free

Vent Area calculated for an enclosure. If the venting and leakage is into another room, calculations must be performed as out-

lined in this guide. However, the size of the vent from the protected enclosure can be reduced by the ELA.

NFPA 2001 requires that at least every 12 months the enclosure be thoroughly inspected to determine if the integrity has been

modified. The most accurate way to determine if the integrity has been changed is to perform a Room Pressurization Test. This

must be done if the ELA is deducted from the calculated Free Vent Area. If an annual test indicates that the ELA has changed

calculations to verify that proper venting is supplied must be performed.



Performing the calculations from the example using a wall strength of 10 PSF results in a reduction in size for most of the

components as indicated in the following table.

DEVICE 5 PSF 10 PSF

Free Vent Area 453 in2 340 in2

Inlet Grill 36 in. x 24 in. 36 in. x 24 in.

Duct 24 in. x 24 in. 19 in. x 19 in.

Exhaust Louver 42 in. x 36 in. 36 in. x 36 in.

EXAMPLE


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Types and Sources of Vents

GRAVITY/STATIC

Damper installed in the wall of the enclosure which will open when minimal pressure is created in the protected space.

Does not maintain fire rating of wall.

Least expensive option.

ELECTRICALLY OPERATED

Damper installed in the wall of the enclosure which is opened when power is applied to the damper or power is removed if

powered closed.

Design must insure damper is open prior to discharge or damper will not be effective.

Adds to complexity of design.

Additional installation cost.

Do not use manual mechanical actuators on cylinder.

PNEUMATICALLY OPERATED

Damper installed in the wall of the enclosure which is opened when air pressure is applied to a pneumatic device on the

damper.

Same concerns as electrically operated dampers.

PRESSURE OPERATED

Pressure operated vent which is calibrated to open under a specific pressure.

Available in styles to provide for various applications concerning security.

Must be tested for calibration prior to discharge.

NATURAL VENTING

Venting which is accomplished by existing openings in the protected enclosure i.e., ceiling openings, door openings, exhaust

duct openings, or wall openings.

Time Delays and Actuation Methods Relating to Vent Types

Dampers must be fully opened prior to agent release. An agent release time delay is typically required to assure that those

dampers are fully open before the pressure in the room starts to increase. If the pressure in the room starts to increase before

the dampers are fully open, the dampers may be forced closed. It is very important to make sure the system is designed and

installed to prevent agent discharge prior to the dampers being fully opened. The manufacturer of the damper can provide the

time delay required to fully open the damper. If concentration must be held after discharge, dampers must be closed to eliminate

airflow through the enclosure.

System operation by electric manual pull stations must also incorporate a time delay prior to operating the release circuit to

allow time for the dampers to open. Consult the appropriate control system manual for details on how to program a time delay

with an electric manual release.


CAUTION!

Manual mechanical actuators located directly on the cylinder valve shall not be used

since operation would result in immediate valve opening and dampers would not have

sufficient time to open. If the system requires mechanical actuation or electric manual pull

stations with immediate release, dampers requiring a time delay to open (such as motor-

ized or pneumatic operated dampers) shall not be used.


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000836

* The chart above was used with permission of the copyright holder, the Sheet Metal and Air Conditioning Contractors National Association,Inc. (SMACNA)



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000837


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000838




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*Thechartabovewasusedwithpermissionofthecopyrighth

older,theSheetMetalandAirConditioningContractorsNationalAssociation,Inc.(SMACNA)

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No.

422793-02

1998AnsulIncorporated

LithoinU.S.A.

ANSUL INCORPORATED

ONE STANTON STREET

MARINETTE, WI 54143

INERGEN® Pressure Relief Venting Guide

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Transcript of INERGEN® Pressure Relief Venting Guide