INERGEN® Pressure Relief Venting Guide

 

INERGEN PRESSURE RELIEF VENTING GUIDE 2-14-97

Page 1

Revised 1-5-98

Introduction Venting 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 overpressurization 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 system. In both cases a gas is moving through an opening which restricts the flow. The design must deal not only with the pressure 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 calculation of the venting system. In addition, if the venting is to be accomplished through multiple enclosures, pressure build-up created 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/ft 2. Any pressure greater than that listed for the enclosure or equipment will likely do some damage to that component. Early venting recommendations recommendations were based on enclosure wall strengths as listed in NFPA NFPA 12. It appears that these figures were based on old construction methods employed in the 1950’s and 60’s. Today’s Today’s construction methods, using materials such as metal studs, movable partitions, and concrete blocks, create walls with lateral strengths much weaker than those encountered in walls built when the NFPA NFPA 12 wall strengths were developed. The 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. 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 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 Technical Services Department at Ansul. The following Flowsystem. Chart gives a general overview of the steps required to complete the design and calculation of an INERGEN venting

 

INERG EN VEN VENTIN TING G GU GUIDE IDE INERGEN 1-5-98

Page 2

Rev. 1

Venting Guide Flow Chart CALCULATE INERGEN SYSTEM PEAK FLOW RATE

DETERMINE ALLOWABLE PSF

CALCULATE ALLOWABLE PSF PER ARE AREA A

YES

MULTIPLE AREAS? NO CALCULATE FREE CALCULATE VENT VEN T ARE AREA(S) A(S)

DEDUCT DEDU CT EL ELA A FR FROM OM DOOR FAN TEST

DUCTS?

YES

1 CALCULATE PRESSURE LOSS

NO

ADDITIONAL RESTRICTIONS? NO

SUM ALL PRESSURE LOSSES

CALCULATE PRESSURE LOSS

RESIZE FREE VENT VEN T ARE AREA A

NO

TOTAL PRESSU TOTAL PRESSURE RE LOSS ACCEPTABLE? YES

COMPONENT SIZE CHANGE? NO

NO

FINAL FIN AL FRE FREE E VENT AREA ARE A AND COMPONENT COMPO NENT SIZES

YES

DUCT SIZ DUCT SIZE E CHANGE?

YES

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

YES 2 CALCULATE PRESSURE LOSS FOR EACH DEVICE

RESIZE COMPONENT

RESIZE DUCT

 

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Page 3

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. 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 peak flow rate) and knowledge of what the maximum allowable internal pressure is that the enclosure can withstand. 1. The flow rate out of the enclosure enclosure can be determined determined by calculating calculating the INERGEN INERGEN system peak flow rate (Qp). This flow rate is determined from the formula: 2.7 IG  _____  _  Qp = ____ T Where: Qp = INERGEN system peak flow rate IG = INERGEN Agent Agent Quantity Supplied (ft 3) T = 90% 90% di disc scha harg rge e tim time e (mi (min. n.)) The flow rate calculated is the peak flow which results results within the first few seconds of the INERGEN system 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 architect to determine building wall strength. If this information information is not available, Ansul recommends the use 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 venting for an INERGEN system is to calculate the minimum size size of the Free Vent Area. The formula used to determine the size of the Free Vent is: 0.0855 Qp  _________  _____  X =  ____ P Where: X = The Free Vent Area in square inches. Qp = The INERGEN agent agent peak flow rate calculated from the formula formula listed above. P = The maximum maximum allow allowable able wall stren strength gth in lbs./f lbs./ftt2 of the weakest surface (walls, ceiling, floor, doors, windows, etc.) of the enclosure. Free Vent Calculation

 L E  E X  XA A M P

12 FT.

30 FT.

P  R   O  T   S  P   E  C  T   A C   E  D   E  

INERGEN Agent Quantity = (4) 439 ft3 cylinders 90% Discharge Time = 45 Seconds Maximum Allowable Allowable Enclosure Pressure = 5 lbs./ft2

10 FT.

Calculate Agent Peak Flow Rate: 2.7 (1756) 4741.2 Qp = ____  ________ _____ _ = ____ ______  __  = .75 .75

FIGURE 1

6322 CFM

000917

Calculate Free Vent Area: 0.0855 (6322) X =  ____  ________ ________ ____ = 5

0.0855 (6322) ____ ______  __   ________  ____ ____ = 2.236

540.5 ____  = 2.236

241.7 in2

= 1.68 ft2

The Free Vent Area calculated is the minimum area required to prevent over-pressurization of the enclosure when the opening is to atmosphere. Atmosphere Atmosphere is described as open air on the outside of a building.

 

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Rev. 2

Free Vent Calculation (Continued) An opening into another enclosure regardless of the size of that enclosure will cause a pressure build up inside both enclosures. 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 the formula: 995 V__  1  ______  P = ____ V2 Where Wh ere:: P = Final Final pre pressu ssure re in in both both encl enclosu osures res (lb (lb/sq /sq.. ft.) ft.) V1 = Protected enclosure enclosure volume volume (ft3) V2 = Total volume of both enclosures (ft3) As indicated by calculations, this pressure is prohibitively 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 movement may have more of an effect on leakage than the location of the opening. It will be advantageous to prevent any air movement 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 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 duct system to vent the gases to a safe location 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.

 

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1-5-98

Page 5

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, 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 restriction to the system which will decrease the effectiveness of the Free Ven Ventt Area previously calculated. Therefore, 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 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 protected 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 total 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 allowable pressure for the unprotected space.

FIGURE 2 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) Pa1 Pd1 =  ____  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)

000834

 

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Rev. 1

Venting Venting Through Through an Unprotected Space (Continued) For example, if the allowable wall pressure in a protected space, Pa1 = 6 lb/sq ft, both vents, X 1 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 sized with flow rates in the formulas formulas being the same and equal to the flow of the INERGEN system, Qp. During the discharge, discharge, the pressure buildup buildup in the unprotected space 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 decreased in size by using the design pressure of 4 lb/sq ft, or Pd1 = Pa1 – Pd2. The system system flow rate is Qp in both formulas. In no case can the sum sum of the design pressures used to calculate the vents in the two spaces exceed the allowable pressure, P a1, in the protected protected space. space. In other words, Pd1 + Pd2  ! Pa1. Venting Through an Unprotected Space 30 FT.

 E  L  P  E XA M

60 FT.

10 FT. 12 FT.

  D   T  E   E C   T   O   E   P  R  A C   U  N   S  P

P  R   O  T   S  P   E  C  T   A C   E  D   E  

Qp (INERGEN Agent Flow Rate) = 6322 CFM Pa1 (Maximum Allowable Enclosure Enclosure Pressure Pressure for Room 1) = 5 lbs./ft2 Pa2 (Maximum Allowable Enclosure Enclosure Pressure Pressure for Room 2) = 5 lbs./ft2

FIGURE 3 000832

Calculate Design Pressures: 5 = 2.5 lbs. Pd1 =  ___  lbs./ft /ft2 2 Pd2 = 5 – 2. 2.5 5 = 2. 2.5 5 lb lbs. s./f /ftt2 To find the Vent Area (X): 0.0855 (6322)  ____  _________ _______ __ X=

2.5

0.0855 (6322) ____ ________ ________ ____ =

1.581

540.5 _____  ____ _  =

1.581 =

2

341.9 in

2

=

2.4 ft for each vent

 

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Page 7

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 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 flow rates 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 will utilize Pd2 and the summation of flow rates, Q p1 + Qp2 = Qpd2, which is the design flow rate to be used to determine the area of the second vent in the series.

FIGURE 4 000835

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:

Pa1 Pd2 = ______________ .666

1 +

Q p1 ____ Q pd2

( )

This 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, pressures, the free vent areas can be calculated from the standard venting formula. formula. Remember 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 exceed the allowable pressure in hazard number one.

 

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Rev. 1

Venting Through a Protected Space (Continued) Venting Through a Protected Space 30 FT.

60 FT.

 E X  XA A M  M P  P L  L E  E

10 FT. 12 FT.

  2   O  M   O   R

R  O   O  M   1 

Qp1 (INERGEN Agent Flow Flow Rate for Room 1) = Qp2 (INERGEN Agent Flow Flow Rate for Room 2) = Qpd2 = 31608 + 6322 = 37930 Pa1 (Maximum Allowable Enclosure Enclosure Pressure Pressure for Pa2 (Maximum Allowable Enclosure Enclosure Pressure Pressure for

6322 CFM 31,608 CFM CFM

FIGURE 5 000832

Room 1) = 5 lbs./ft2 Room 2) = 5 lbs./ft2

To obtain the smallest total free vent area, complete the following calculations: Calculate the Design Pressure for Room 2: 5 5 ___  ________ _____ _ ________ _______ Pd2 =  ____ = ____ .666 1 + 6322 1 + 0.1667.666 37930

(

)

5 __  ____ = ______  = 1.3029

3.84 lb lbs./ft2

Calculate the Design Pressure for Room 1: Pd1 = 5 – 3.84 = 1.16 lbs./ft2 Calculate the Free Vent Area for Room 1: 0.0855 (6322) 0.0855 (6322)  ________ _______ ___ = ____ ________ _______ ___ = 540.5 ____  = 502 in2 = 3.49 ft2 X1 (vent area) =  ____ 1.16 1.077 1.077 Calculate the Free Vent Area for Room 2: 0.0855 (31608 +______ 6322) (37930)  ____ _________ _________ __ = 0.0855 _________ _____ _______ ___ = 3243 ____  = 1655 in2 = 11.49 ft2 X2 (vent area) =  ________ 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.

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 allowable 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 multiple iterations can be done much more easily and will yield acceptable results.

 

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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 individual 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 complexity 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. parameters. For additional additional information, refer refer to HVAC HVAC SYSTEMS DUCT DESIGN published published by SHEET SHEET MET METAL AL AND AIR CONDITIONING CONTRACTORS NATIONAL ASSOCIATION, INC (SMACNA), 4201 Lafayette Center Dr., Chantilly, VA 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 couplings on 48 inch inc h centers. These values may be used without correction for temperatures t emperatures between 50 °F to 140 °F °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 crosssectional 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 SHARP EXIT, NO LOUVERS 90° SHARP MITERED BEND 90° SMOOTH BEND

L/D 78 86 27

Gradual reductions ! 60°have 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 assign one 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: Wher e: d = diameter diameter,, in inches, inches, of free vent area area X = Free Vent Area

 

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Venting Through Ducts (Continued) Choose Choo se a duct size with with diameter diameter approxim approximately ately 15% 15% larger than than calculate calculated d above, above, (d x 1.15) 1.15).. Proceed to the duct friction loss chart and locate the INERGEN system system flow rate on the bottom of the chart Find the intersection 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 pressure drop versus velocity, 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 of water 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 beginning of the design. This total pressure drop must not exceed the allowable room pressure in pounds per square foot as determined from the building construction. The duct design pressure must not exceed the pressure rating of the duct. Venting Through Ducts Variables d da PSF PS F #P #P/L V HW L

= = = = = = = =

Equ quiv iva ale len nt Rou Round nd Duc uctt dia diam met ete er (in (inch ches es)) Assumed Assu med Equivale Equivalent nt Round Round Duct diameter diameter (inche (inches) s) POUN PO UNDS DS PE PER R SQ SQUA UARE RE FO FOOT OT Pressure Drop (PSF) Pressu Pre ssure re Drop Drop per per Foot Foot of Lengt Length h (PSF) (PSF) Velocity (ft./minute) Inches Inch es of Wa Water ter Column Column (in. W.G.) W.G.) Length (feet)

LEQ L/D CFM A

= = = =

Equivalen Equiva lentt Length Length (fe (feet) et) Equiva Equ ivalen lentt Length Length of a Resi Resista stance nce to to Flow Flow,, in duct duct diame diameter ters s CUBIC FEET PER MINUTE Area (ft.2)

 E  L  P  M  E XA

(Continued)

 

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Venting Through Ducts (Continued) Venting Through Ducts

FIGURE 6 000844

FREE VENT AREA IG = T = P = 1/2 P =

1756 ft.3 45 seconds = .75 minutes 5 PSF 2.5 PSF

Estimate Agent Flow Rate: 2.7 IG_ 2 .7 (1____ 756)  ______  _____ ________ ____ Qp = = = T .75

 E  L  P  M  E XA

4 741.__  2 ______  ____ = 6322 CFM .75

Calculate Free Vent Area Required: 0.0855____ Qp 0 .085_______ 5 (63___ 22) 5 41  _________ ________ ____ ___  X =  _____ = = P 2.5 1.58

342.4 in.2

Estimate Duct Diameter: d

=

 ____ 4X = "

4 _______ ____ (34___ 2.4) "

=

____  1370 = "

436

= 20.9 in.

Increase Duct Diameter: da = 20.9 (1.15) = 24 in. Calculate Actual Duct Area: 2   " 576   " da2   ____ " 24_ 810 _____ _ = 1____  = _____ = ____ = 453 in.2 X =  _____ 4 4 4 4 Calculate Actual Pressure Drop: 0.0855 Qp 2 0.0855 (6322)  ________  ____ ____ ________ ____ _______ ___ #P = = X 453

(

) (

)

2

=

541 ___  453

( )

2

=

1.192

=

1.42 PSF (Continued)

 

INERG INE RGEN EN VEN VENTIN TING G GU GUIDE IDE 2-14-97

Page 12

Venting Through Ducts (Continued) Venting Through Ducts

DUCT 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. Circular Equivalents of Rectangular Ducts for Equal Friction and Capacity (U.S. Units) (2) Dimensions in Inches PSF/ft. #P/L = 0.007 PS (from Duct Friction Loss Chart, Page 21) Calculate Friction Loss through Straight Duct: #P = L (#P/L) = 70 (0.007) = 0.49 PSF Calculate 90°Sharp Bend Equivalent Length: LEQ =  ____ L/D (1 2) = 8 6 (1__ 2) = 1 032 =  ______ __ ______ ____ ____  d 26.2 26.2 39.4 ft. Calculate 90°Smooth Bend Equivalent Length: LEQ =  ____ L/D (1 2) = ____ 27 (1__ 2) = ____  324  ______ __ ______ = 12.4 ft. d 26.2 26.2 Calculate Total Equivalent Length: LEQ = L + 2 (LEQ Sh Shar arp p Be Bend nd)) + LEQ Sm Smoo ooth th Be Bend nd = 70 + 2 (3 (39. 9.4) 4) + 12.4 = 161 Calculate Total Pressure Loss: #P = LEQ (#P/ P/L) L) = 161 161.2 .2 (0 (0.0 .007 07)) = 1.13

 XA A M P  P L  L E  E  E X

000836

PSF

Duct Friction Loss Chart (U.S. Units) (2) Air Quantity, cfm at 0.075 lb/ft3 .007

6322 CFM 000840

(Continued)

 

INERGE INE RGEN N VEN VENTIN TING G GUI GUIDE DE 2-14-97

Page 13

Venting Through Ducts (Continued) Venting Through Ducts

INLET GRILLE

Steel S590 & Aluminum S590 – Gymnasium Grilles

36 in. x 24 in. size, NC 30-40 application Qp = 63 6322 22 CF CFM M Hw = .21 in. W.G. (from (from chart) chart) Convert in. W.G. to PSF: #P

= Hw (5.204) = 0.21 (5 (5.204) =

1.09 PSF

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

Damper Size 24” x 24”

DAMPER 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 =  ___ A

6322 = 1581 ft./min. = ___  4

Hw = 0.10 in. W.G. W.G. (Damper (Figure (Figure 5.3 Style) Pressure Pressure Drop Chart) Chart) Convert in. W.G. to PSF: #P

= Hw (5.204) = 0.10 (5 (5.204) =

   )  .    G  .    W    N    I    (    S    S    O    L    E    R    U    S    S    E    R    P

0.52 PSF

 E  L  P  M  E XA 1581

000841

(Continued)

 

INERG INE RGEN EN VEN VENTIN TING G GU GUIDE IDE 2-14-97

Page 14

Venting Through Ducts (Continued) Venting Through Ducts

VENT SUBTOT SU BTOTAL AL #P

=   #P (Fr Fre ee Ve Vent nt)) +   #P (Duct) +   #P (I (Inl nlet et Gr Gril ille le)) +   #P (Da (Dampe mper) r)

#P

= 1.42 + 1.13 + 1.09 + 0.52 = 4.16 PSF

 P L  L E  E  E XA M P

5 PSF PSF – 4.1 .16 6 PSF = 0.8 .84 4 PS PSF F fo forr Lo Louv uver er

LOUVER PSF 0.84 _  _____  ____ Hw =  ____ = = .16 in. W.G. 5.204 5.204 V

= 97 975 5 ft./ ft./mi min. n. (fr (from om Ai Airf rflo low w Resi Resist stan ance ce Cha Chart rt))

X

Qp =  ___ V

=

6322 ____  6.49 ft.2 = 975

Louv Lo uver er Si Size ze = 48 in in.. W x 42 in in.. H A

= 7.19 ft.2 (Free Area Chart)

V

Qp =  ___ A

#P

= Hw (5.204) = 0.13 (5 (5.204) =

=

.13

6322 ____  879.3 ft./min. = 7.19

0.68 PSF

 E  L  P  M  E XA

880 000842

000843

VENT TOTAL

 E  L  P  M  E XA

#P

=   #P (F (Fre ree e Ven Ventt Ar Area ea)) +   #P (Duct) +   #P (Grille) +   #P (Damper) +   #P (Lo (Louve uver) r)

#P

= 1.42 + 1.13 + 1.09 + 0.52 + 0.68 = 4.84 PS PSF

(Continued)

 

INERGE INE RGEN N VEN VENTIN TING G GUI GUIDE DE Rev. 1

1-5-98

Page 15

Venting Through Ducts (Continued) Venting Through Ducts 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.

 P L  L E  E  E XA M P

Allowance for Room Pressurization Test Results The ELA (Equivalent Leakage Area) Area) determined from a Room Pressurization Test Test may be deducted from the amount of Free Ventt Area calculated for an enclosure. If the venting and leakage is into another room, calculations must be performed as outVen lined in this guide. However 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 NFPA modified. The most accurate way to determine if the integrity has been changed is to perform a Room Pressurization Test. Test. This must be done if the ELA is deducted from the calculated Free Vent Vent Area. If an annual test indicates that the ELA has changed, calculations to verify that proper venting is supplied must be performed.

 

INERG INE RGEN EN VEN VENTIN TING G GU GUIDE IDE 2-14-97

Page 16

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 expensive option. ELECTRICALL ELECTRICAL LY OPERA OPERATED TED 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 complexity of design. • Additional installation installation cost. cost. • Do not use manual mechanical actuators on cylinder. cylinder. PNEUMATICALLY PNEUMATICALL Y 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 electrically operated dampers. dampers. PRESSURE OPERATED Pressure operated vent which is calibrated to open under a specific pressure. • Available in styles to provide for various applications concerning concerning security. security. • Must be tested for calibration prior prior to discharge. NATURAL NA TURAL VENTING Venting which is accomplished by existing openings in the protected enclosure i.e., ceiling openings, door openings, exhaust Venting 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 motorized or pneumatic operated dampers) shall not be used.

 

INERGE INE RGEN N VEN VENTIN TING G GUI GUIDE DE 2-14-97

000836

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

Page 17

 

INERG INE RGEN EN VEN VENTIN TING G GU GUIDE IDE 2-14-97

Page 18

000837

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

 

INERGE INE RGEN N VEN VENTIN TING G GUI GUIDE DE 2-14-97

Page 19

000838

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

 

INERG INE RGEN EN VEN VENTIN TING G GU GUIDE IDE 2-14-97

Page 20

000839

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

 

INERGE N VEN VENTIN TING G GUI GUIDE DE INERGEN 2-14-97

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