in Article “Stationary UPS Sizing Calculations Part Four”, we explained Selection and sizing of UPS protective devices (CBs or Fuses).
Also, in Article “Stationary UPS Sizing Calculations – Part Five”, we explained the following:
 Selection and sizing of UPS Cables,
 Sizing a generator set for UPS system
Today, we
will explain the Battery Room Ventilation Calculations.
Battery Room Design Criteria 
There are many critical design issues that must be taken into consideration when planning, designing and constructing a safe and reliable battery room. Many of the model building codes and recognized standards such as IEEE, OSHA, NEC, and NFPA Life Safety Codes outline the requirements for the design and installation of battery rooms. They provide guidance on the performance
criteria for the various systems as well as requirements for related
equipment. These requirements are as follows:

1 General Requirements

2 Mechanical Requirements A Temperature Control
B Ventilation

3 Electrical Requirements A
General
B Lighting
C Monitoring and Instrumentation
D Cable Entry Facilities

4 Fire protection Requirements

Ventilation
Design Criteria The battery room ventilation design criteria include:
Fans and Motors
Note:

Battery
Room Ventilation Calculations 
Hydrogen
is produced during battery charging. If hydrogen gas is allowed to accumulate
in an enclosed area, it is readily ignitable and may result in an explosion. How
much hydrogen does a battery emit? As
a rule of thumb, when the battery is about fully charged, each charging
ampere supplied to the cell produces about 0.0158 cubic feet of hydrogen per
hour from each cell. This rate of production applies at sea level, when the
ambient temperature is about 77ºF, and when the electrolyte is "gassing
or bubbling." The
National Fire Protection Association lists the lower explosive level (LEL) of
hydrogen as 4% by volume. What
does this mean? LEL
is the point at which hydrogen can combust. For example the air in a box with
a volume of 100 cubic feet containing 4 cubic feet of hydrogen gas would be
expected to ignite when exposed to a spark or open flame. This can be
disastrous. Below is a picture depicting the extent of damage due to a
ventilation failure. To
ensure safety, most regulations such as the Uniform Fire Code and the
International Fire Code stipulate a maximum hydrogen concentration below the
level of 1% or 25% of the lower explosion limit LEL in a battery room. 
Ventilation calculation methods:

First Method: NFPA Method NFPA
1 (2006) Section 52.3.6 — Ventilation: Ventilation shall be provided for rooms and cabinets in accordance with the mechanical code adopted by the jurisdiction and one of the following:
The following steps shall be followed: Step 1: Calculate Hydrogen Release Amount of hydrogen release during normal float condition for a flooded battery is given as: H = N * I * k Where:
Step 2: Calculate Room Volume A room with a flat roof has a volume of: RV = w x l x h Where,
Step 3: Determine Critical Volume The critical volume is the maximum permissible hydrogen concentration to limit the value to below 1% and is given by: CV = RV * PC Where,
Step 4: Time to reach Critical Level of Hydrogen Concentration t= CV/H Where,
Step 5: Determining the Ventilation Rate Q = H/(60*PC%) Where,
Step 6: Fan Sizing Add a 25% safety margin in Step 5. This safety factor is to allow for hydrogen production variations with changes in temperature, charge controller failure, and reduction in net volume of battery room due to battery equipment and fixtures. It also allows for deterioration in ventilation systems. As such: QA = Q x FS Where,
Step 7: Determine Air changes per hour ACH = QA * 60/RV Where,
Step 8: calculate time required for one air change Time required for one
air change = 60 minute/ ACH Step 9: compare results from step 4 and step 8 If step 8 result = step 4 result, then the ventilation is
identical to the required and ventilation system is ok. If step 8 result < step 4 result, then the ventilation
is quicker than the required and ventilation system is ok. If step 8 result > step 4 result, then ventilation
system is not adequate. Note:

Example # 1 A 60cell leadacid battery, located in a room having a volume of 2000 cubic feet, is being charged at 50 amperes. The ventilation system is designed to provide three airchanges each hour. Determine the rate of hydrogen production and the adequacy of the air exchanges required for ventilation. Solution: Step 1: Calculate Hydrogen Release Hydrogen (H2) production in cubic meters per hour is: H = N * I
* k 50 amps * 60 cells * 0.0158 ft3 /cell/hour = 47.4 ft3 /hour Step 2: Calculate Room Volume Room having a volume of 2000 cubic feet Step 3: Determine Critical Volume Critical volume, based on 1 percent by volume is: 2000 ft3 * 0.01 = 20 ft3 Step 4: Time to reach Critical Level of Hydrogen Concentration
1% t= CV/H = 20 ÷ 47.4 = 0.42 hour (25.2 minutes) The ventilation system must clear the 2000 ft3 room within 0.42
hour (25.2 minutes) before the batteries can produce 20 cubic feet of
hydrogen. Step 5: Determining the Ventilation Rate Q = H/(60*PC%) = 47.4 / (60*0.01) = 79 ft3 Step 6: Fan Sizing QA
= Q x FS QA
= 79 x 1.25 QA = 98.75 ft3 Step 7: Determine Air changes per hour Room Volume = 2000 sq.ft. ACH = QA * 60/RV ACH = 98.75 * 60
/ 2000 = 2.96 say 3 air changes per hour. Step 8: calculate time required for one air change Time required for one air change = 60 minute/ ACH = 60/3 =20
minutes Step 9: compare results from step 4 and step 8 Step 4 time = 25.2 minutes Step 8 time = 20 minutes So, step 8 result < step 4 result, Then the ventilation is quicker than the required and ventilation system is ok. Critical hydrogen concentration will not be reached with
continuous operation of the ventilation system. Example# 2 Same condition as previously mentioned in Example 1, except that the battery is located in a smaller room size of 1000 ft3. Solution: Step 1: Calculate Hydrogen Release Hydrogen
(H2) production in cubic meters per hour is: H
= N * I * k = 50 amps * 60 cells * 0.0158 ft3 /cell/hour = 47.4 ft3 /hour Step 2: Calculate Room Volume Room having a volume of 1000 cubic feet Step 3: Determine Critical Volume Critical
volume, based on 1 percent by volume = 1000 ft3 * 0.01 = 10 ft3 Step 4: Time to reach Critical Level of Hydrogen Concentration
1% Time
to produce critical level of 1 percent hydrogen (10 cubic feet) in the 1000
cubicfeet battery room is: t= CV/H = 10 ÷ 47.4 = 0.21 hour (12.6 minutes)
The
ventilation system must move 1000 cubic feet (the room volume), with the 10
cubic feet of hydrogen contained within, before the 0.21 hour (12.6 minutes)
elapses. Step 5: Determining the Ventilation Rate Q = H/(60*PC%) = 47.4 / (60*0.01) = 79 ft3 Step 6: Fan Sizing QA
= Q x FS QA
= 79 x 1.25 QA = 98.75 ft3 Step
7: Determine Air changes per hour Room Volume = 1000 sq.ft. ACH = QA * 60/RV ACH = 99.38 * 60 / 1000 = 5.96 say 6 airchanges per hour. Step 8: calculate time required for one air change Time required for one air change = 60 minute/ ACH = 60/6 =10
minutes Step 9: compare results from step 4 and step 8 Step 4 time = 12.6 minutes Step 8 time = 10 minutes So, step 8 result < step 4 result, Then the ventilation is quicker than the required and ventilation system is ok. Critical hydrogen concentration will not be reached with
continuous operation of the ventilation system. Example #3 Per manufacturer specification, one fully charged leadacid battery cell at 77°F will pass 0.24 amperes of floating current for every 100 amperehour cell capacity when subject to an equalizing potential of 2.33 volts. Each cell has a nominal 1,360amphere hour’s capacity at the 8hour rate. Calculate the ventilation rate for a battery room consisting of 182cell battery and 3 battery banks. Assume the battery room has dimensions of 20’ (l) x 15’ (w) x 10’ (h). Solution: First, we need to calculate the Charging current I = FC*AH/100 Where, I = Charging current, amperes FC
= Float current per 100 amperehour. FC varies with battery types, battery
condition, and electrolyte temperature. Ah = Rated capacity of the battery in Ampere hours. I = FC*AH/100 = 0.24*1360/100= 3.264 amps Step 1: Calculate Hydrogen Release N = 182 cells /battery * 3 battery banks = 546 cells k = 0.0158 ft3/amp hr cell H = 3.264 * 0.0158 * 546 H = 28.16 ft3/hr Step 2: Calculate Room Volume Room Volume = 20 x 15 x 10 = 3,000 cubic feet Step 3: Determine Critical Volume Critical
volume, based on 1 percent by volume = 3000 ft3 * 0.01 = 30 ft3 Step 4: Time to reach Critical Level of Hydrogen Concentration 1% t= CV/H = 30 ÷ 28.16 = 1.07 hour (64.2
minutes) Step 5: Determining the Ventilation Rate Q = H/(60*PC%) = 28.16 / (60*0.01) = 46.93 ft3 Step 6: Fan Sizing QA
= Q x FS QA = 46.93 * 1.25 = 58.66 ft3/min The ventilation system should be capable of extracting 58.66
cubic feet per minute. Step 7: Determine Air changes per hour ACH
= QA * 60/RV = 58.66*60/3000 = 1.17 say 2
airchanges per hour Step 8: calculate time required for one air change Time required for one air change = 60 minute/ ACH = 60/2 =30
minutes Step 9: compare results from step 4 and step 8 Step 4 time = 64.2 minutes Step 8 time = 30 minutes So, step 8 result < step 4 result, Then the ventilation is quicker than the required and ventilation system is ok. Critical hydrogen concentration will not be reached with
continuous operation of the ventilation system. Notes for example#3:
Example#
4 In example#3, how much time it will take to build 1% concentration of gas with no ventilation? Assume 1000 cuft of room volume is covered by battery equipment. Solution: Volume
of the room = 3000 cu ft Volume
taken by the batteries infrastructure = 1000 cu ft Net
volume = 2000 cu ft Step 3: Determine Critical Volume Critical
volume, based on 1 percent by volume = 2000 ft3 * 0.01 = 20 ft3 Step 4: Time to reach Critical Level of Hydrogen Concentration 1% t= CV/H = 20 ÷ 28.16 = 0.71 hour (42 minutes) 
Second method: Metric Units per British Standard BS
6133 The following is an extract from BS 6133:1985 – Safe operation of leadacid stationary cells and batteries:
The
following method may be used to calculate the ventilation requirements of a
battery room. 26.8Ah input to a fully charged cell will liberate 8 g of
oxygen and 1 g of hydrogen. One (1) g of hydrogen occupies a volume of 12
liters at 20°C and at a pressure of one standard atmosphere. Therefore 26.8Ah
input will evolve 12 liters of hydrogen. Therefore the volume of hydrogen
evolved from a battery per hour: H
= no. of cells * charge current *12 liter /26.8 H
= no. of cells * charge current *0.45 liter Since
1 liter = 0.001 m3 H = no. of cells * charge current *0.45 * 0.001 H
= no. of cells * charge current *0.00045 Where:
The
volume of hydrogen found by the above calculation can be expressed as a
percentage of the total volume of the battery room, and from this, the number
of changes of air per hour to keep the concentration of hydrogen below 1% can
be calculated. Example#
5 Consider
a battery of 120 cells with charge current of 17 amperes. The battery is
installed in a double tier, double row terraced arrangement in a room of 4m x
2m x 3m. Determine the ventilation rate to limit hydrogen concentration to
less than 1%. Solution: Rate
of hydrogen produced, H H
= no. of cells * charge current *0.00045 H = 120 x 17 x 0.00045 = 0.92 m3/hr Maximum
permissible concentration of hydrogen, PC = 1% Ventilation required, Q Q = H/(60*PC%) m3/min Q
= H/PC m3/h Q
= 0.92/0.01 = 92 m3/hr Fan
Capacity QA
= Q x 1.25 QA
= 92 x 1.25 = 115 m3/hr Room
Volume, RV
= 4 x 2 x 3 = 24 m3. Air changes per hour ACH
= QA/RV ACH = 115/24 = 4.79 Therefore
to keep the concentration of hydrogen gas at a maximum of 1%, the air in the
room will need to be changed 4.79 times per hour, or about five times per
hour. 
Requirement
of Air Conditioner for UPS 
UPS
system produces heat, which must be removed to prevent the UPS temperature
from rising to an unacceptable level. Selection of air conditioner for UPS
room requires an understanding of the amount of heat produced by the UPS.
Heat is energy and is commonly expressed in Joules, BTU, tons, or calories. common
measures of heat output rate for equipment are BTU per hour, Tons per day,
and Joules per second (Joules per second is equal to Watts)
The
selection of UPS for air conditioner has to be calculated based on:

Sizing of Air Conditioner Step
1: Calculate the basic capacity in BTU/Hr required for the UPS room Multiply
the length of UPS room by its width, which will gives us the total area of
the room. Based on the below table, the basic capacity in BTU/Hr required for
the UPS room can be calculated
Step
2: Arrive at the no of person who will work in the UPS room Generally
the UPS room is unmanned apart from the time when the technician visits to
service the UPS or during the visit of maintenance engineer. It is ideal to
consider 600 BTU/Hr per person to arrive at the air conditioner capacity of
the UPS room. Step
3: Heat Loss of UPS To arrive at the capacity of the air conditioner required for UPS, we need to calculate the heat loss of UPS in KW using the formula: Heat
Loss of UPS (in KW) = Input Power in KW – Output Power in KW Heat
Loss of UPS (in KW) = Output Power in KW / Efficiency of UPS  Output Power
in KW Heat
Loss of UPS = heat loss of UPS in KW x 3412.14 BTU/Hr (in BTU/Hr) As a thumb rule, 7% of the UPS capacity can be considered as heat loss which gives the thumb rule formula of Heat
Loss in BTU/Hr = 7% x Capacity of UPS in KW X 3412.14 BTU/hr Step
4: Insulation Loss As
a thumb rule, a 10% of insulation loss can be considered in the calculation Step 5: calculate required tons of Air Conditioner Sum
total heat losses, then Required tons of Air Conditioner = total heat losses / 12000 Round
the result to the nearest standard value. Step 6: calculate the required air flow Required
air flow = RT (from step 5) * 400 CFM Example#6:
UPS
capacity 200KVA Area
of UPS Room 9 sq m (3X3X3,WXDXH in m) Heat
loss of UPS 7% Heat
Loss of Other Loads 1% No
of Person in UPS Room 1 Insulation
Loss 10% Solution: From
Step 1: Area
of UPS Room = 9 sq m (3X3X3,WXDXH in m)Capacity in From
Table , Basic Cooling Capacity in BTU/Hr
= 5,000 BTU/Hr From
Step 2: No
of Person in UPS Room = 1 Person Consider
600 BTU/Hr per person From
Step 3: Heat loss of UPS = 7% * 200 KVA * 3412.14 BTU/hr = 47,770 BTU/Hr Heat
Loss of Other Loads Assumption
of 1% of UPS capacity = 1%*200 KVA*3412.14 BTU/hr = 6,824 BTU/Hr From Step 4: Insulation
Loss Consider
10% of UPS, Other loads and person heat = 10%* (600+47,770+6,824) = 5,500
BTU/Hr From Step 5: Total Losses = 600 + 47,770 + 6,824 + 5,500 = 65,694 BTU/ Hr Required tons of Air Conditioner = total heat losses / 12000 = 65,694 / 12000 = 5.47 RT Round
the result to the nearest standard value = 6 RT From Step 6: Required
air flow = RT (from step 5) * 400 CFM = 6 * 400 = 2400 CFM 
In the next Article, we will
explain the Installation and testing of UPS. So, please keep following.
Subject
Of Pervious Article 
Article 
Applicable Standards for UPS Systems What is a UPS? Why do we need a UPS? UPS Rating Classification of UPS: 1Voltage range, 2No. of phases, 3 Mobility, 4 Technological design, 

5 Physical Size/capacity, 6 Form factor/ configurations: 6.1 “N” System Configuration 

6.2 “N+1” System Configuration, which
includes:
6.3 Parallel Redundant with Dual Bus Configuration
(N+1 or 1+1) 
Classification and Types of UPS –
Part Three 
6.4 Parallel Redundant with STS Configuration
6.5 System plus System 2(N+1), 2N+2, [(N+1) + (N+1)],
and 2N 
Classification and Types of UPS –
Part Four 
7 According to UPS Topology 7.1
Offline or Standby UPS, 7.2 Line
Interactive UPS, 7.3
StandbyFerro UPS, 7.4 Online
Double Conversion UPS, 7.5 The
Delta Conversion OnLine UPS. 
Classification and Types of UPS –
Part Five 
8 According to UPS Distribution Architecture 8.1 Centralized UPS Configuration, 8.2 Distributed (Decentralized) UPS
Configuration, 8.2.1 Distributed UPSZonewise
Configuration 8.3 Hybrid UPS Configuration. Conventional (Monolithic) Vs Modular UPS System:

Classification and Types of UPS –
Part Six 
Three Basic Configurations Of Mains
And Bypass For A UPS System:
9According to Use of transformers
with the UPS:

Classification and Types of UPS –
Part Seven 
Transformer Arrangements in Practical UPS Systems: 1Transformer options for the “single mains”
configuration 2Transformer Options for the “Dual Mains”
Configuration 
Classification and Types of UPS –
Part Eight 
3 Transformer options for “single mains without
bypass” 

Components of Online Double Conversion UPS: 1 Rectifier, 2 Inverter, 3 Energy Storage system: 3.1 Battery 
Components of
Online Double Conversion UPS– Part One

3.1.1 Battery Configurations Serial Strings, Parallel Strings. 3.1.2 Battery Size and Location 3.1.3 Battery Transition Boxes 3.1.4 Battery Monitoring 3.2 Energy Storage System – Flywheel 3.3 Energy Storage system – Super Capacitors 3.4 Hydrogen Fuel Cells 4 Static switch Earthing Principles of UPS Systems 
Components of Online Double
Conversion UPS – Part Two 
Evaluation Criteria for Selecting an UPS 
Evaluation Criteria
for Selecting an UPSPart One

Example: Selecting an Uninterruptible Power Supply
(UPS) UPS System Ratings and Service Conditions First: from IEC 601464 Second: according to American
standards 
Evaluation Criteria for Selecting an
UPSPart Two 
The UPS sizing calculations steps 
Stationary UPS
Sizing Calculations – Part One

2 Rectifier/Charger Sizing Calculations 3 Inverter sizing calculations & Static Switch
Sizing 4 The Battery sizing calculations First: The Manufacturers’ methods, which include:

Stationary UPS Sizing Calculations –
Part Two 
Second: The IEEE methods of Battery Sizing Calculations
which includes:

Stationary UPS Sizing Calculations
Part Three 
 UPS Backup time calculation  Selection and sizing of UPS protective devices (CBs or Fuses) 
Stationary UPS Sizing Calculations – Part Four 
Selection of UPS Cables Sizing of UPS Cables:
Sizing a generator set for UPS system 
Stationary UPS Sizing Calculations – Part Five 
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