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Chapter 9 · 6 hours

Emergency and Backup Supply Systems

IOE past exam questions

Past questions and answers

18 questions set from this chapter, 5 of them more than once. Most asked first.

  • Asked 3 times
  • 2077 Chaitra · 2+6 marks
  • 2076 Bhadra · 8 marks
  • 2069 Bhadra · 8 marks

State the difference between Emergency power Supply and Back-up power supply. Explain on-line UPS and off-line UPS.

Answer

Emergency power supply vs back-up power supply

An emergency power supply keeps only the life-safety and essential loads working when normal supply fails, usually within seconds and for a limited time. A back-up power supply keeps normal production or business loads running during an outage so that work continues.

PointEmergency supplyBack-up supply
PurposeSafety of people, safe shutdownContinuity of production/service
LoadsEscape lighting, fire alarm, fire pumps, lifts, controlMachines, computers, process loads
Required bySafety codes/lawOwner's choice (economic)
Changeover timeVery short (≤ 0.5–15 s)Few seconds to minutes acceptable
DurationUsually 1–3 hAs long as needed
CapacitySmallLarge, may equal full load
SourceBatteries, central battery, small DGLarge DG sets, UPS, second feeder

On-line UPS (double conversion)

  • The load is always fed through the inverter. Mains AC → rectifier/charger → DC bus (battery floats on it) → inverter → load.
  • When mains fails, the battery supplies the DC bus with zero transfer time.
  • A static bypass switch transfers load directly to mains if the inverter fails or is overloaded.
  • Advantages: no break, isolates load from spikes, sags, frequency variations; pure sine output.
  • Disadvantages: higher cost, lower efficiency (≈ 90–94%), more heat.
  • Used for data centres, process control, hospitals.
         ┌────── static bypass ──────┐
 AC ──►[Rectifier]──►DC──►[Inverter]─┴─► Load
                     │
                  [Battery]

Off-line UPS (standby)

  • Normally the load is fed directly from mains through a transfer switch; the charger keeps the battery charged.
  • When mains fails, the transfer switch shifts the load to the inverter, which runs from the battery. Transfer time is about 4–10 ms.
  • Advantages: cheap, simple, high efficiency in normal mode.
  • Disadvantages: no protection against sags/spikes in normal mode, short break at changeover, often square/quasi-sine output.
  • Used for PCs and small office loads.
 AC ──────────────────────┐
  │                    [Transfer]──► Load
  └►[Charger]►[Battery]►[Inverter]┘
  • Asked 3 times
  • 2074 Bhadra · 4 marks
  • 2070 Bhadra · 8 marks
  • 2068 Jestha (old course)

Explain various types of battery charging system with sketch.

Answer

A battery charging system passes direct current through a secondary battery in the reverse direction to discharge, to restore its chemical energy. The main methods differ in how voltage and current are controlled.

1. Constant-current charging

  • Charging current is kept constant (about 1/10 of Ah capacity) by a series resistance or controlled rectifier while the battery voltage rises.
  • Simple; used for initial/first charge.
  • Drawback: overcharging and gassing near the end unless current is reduced (two-rate charging).
 DC ──[R variable]──(A)──┬──(+ Battery −)──┐
 supply                  └──────(V)────────┘

2. Constant-voltage (constant-potential) charging

  • A fixed voltage (≈ 2.3–2.4 V/cell for lead-acid) is applied. The current is high at first and falls automatically as the battery's EMF rises.
  • Fast and charging stops itself; a current-limiting resistor protects against initial surge.
  • Used in vehicles and modern chargers.
 I │\
   │ \___
   │     ‾‾‾───────
   └──────────────── t

3. Trickle charging

  • A very small continuous current (≈ 1–2% of capacity) is supplied to make up for self-discharge of an idle battery.
  • Keeps standby batteries (emergency lighting, alarms) always fully charged.

4. Float charging

  • Charger, battery and load are connected in parallel. The charger supplies the load and a small float current to the battery at ≈ 2.15–2.25 V/cell.
  • On mains failure the battery takes over the load without any break. Used in substations, telecom and UPS.
 AC→[Rectifier]──┬────────┬──► DC Load
                 │        │
             [Battery]    │

5. Boost (quick) charging

  • A high current for a short time to restore a deeply discharged battery quickly; followed by normal charging.

6. Equalising charge

  • A periodic controlled overcharge to bring all cells of a series string to equal specific gravity and voltage.

Modern chargers use constant-current then constant-voltage (CC–CV) with a final float stage.

  • Asked 2 times
  • 2076 Bhadra · 4 marks
  • 2071 Bhadra · 8 marks

Explain the trickle charging method for a battery charging system. Calculate the (i) ampere-hour efficiency and (ii) watt-hour efficiency of a secondary cell which is discharged at a uniform rate of 30 A for 6 hr. at an average terminal voltage of 2 V. It is then charged at the uniform rate of 40 A for 5 hr. to restore it to its original condition. The terminal voltage during charging is 2.5 V.

Answer

Trickle charging

Trickle charging is the method of giving a battery that is not in use a small, continuous charging current just large enough to make up for its self-discharge (internal losses), so it always stays fully charged.

  • Current is very small, about 1–2% of the Ah rating (e.g. 0.5–1 A for a 100 Ah battery), or about 1 mA per Ah.
  • The charger is permanently connected; the battery does not supply the load in normal times.
  • Used for standby batteries: emergency lighting, fire alarm, engine starting batteries of DG sets, switchgear tripping batteries.
  • Over-trickling causes gassing and water loss, so the current must be adjusted.
 AC ─►[Transformer+Rectifier]─►[R]─► + Battery −
                  (small current)

Numerical

Given: Discharge 30 A for 6 h at 2 V average; charge 40 A for 5 h at 2.5 V average.

(i) Ampere-hour efficiency

Ah on discharge = 30 × 6 = 180 Ah
Ah on charge    = 40 × 5 = 200 Ah
η_Ah = 180 / 200 × 100 = 90 %

(ii) Watt-hour efficiency

Wh on discharge = 180 × 2   = 360 Wh
Wh on charge    = 200 × 2.5 = 500 Wh
η_Wh = 360 / 500 × 100 = 72 %

(Check: η_Wh = η_Ah × V_dis/V_ch = 0.9 × 2/2.5 = 0.72.)

Answer: Ampere-hour efficiency = 90 %, watt-hour efficiency = 72 %.

  • Asked 2 times
  • 2075 Bhadra · 8 marks
  • 2068 Bhadra (old course) · 8 marks

Explain trickle charging method for battery charging system. A lead acid cell is discharged at a steady current of 4 A for 12 hours, the average terminal voltage being 1.2 volt. To restore it to its original state of charge, a steady current at 3 A for 20 hours is required the average terminal voltage being 1.44 volt. Calculate the ampere-hour efficiency and watt-hour efficiency in this particular case.

Answer

Trickle charging

Trickle charging means connecting an idle, fully charged battery permanently to a charger that supplies a very small current, equal to its self-discharge (local action) loss, so that the battery remains fully charged and ready.

  • Trickle current ≈ 1–2% of the Ah capacity (about 1 mA per Ah of capacity).
  • Normally the battery does not feed the load; it is called upon only when the mains fails.
  • Applications: emergency lights, fire alarm panels, DG set starting batteries, telephone exchanges, switchgear control batteries.
  • The charger is a small transformer–rectifier with a series resistance or regulator to set the current.
  • Too high current causes gassing and water loss; too low lets the battery sulphate. Periodic check of specific gravity is needed.
 230 V AC ─►[Step-down]─►[Rectifier]─►[R]─┐
                                          + │
                                      [Battery]
                                          − │
 ─────────────────────────────────────────┘

Numerical

Given: Discharge 4 A for 12 h, average voltage 1.2 V; charge 3 A for 20 h, average voltage 1.44 V.

Ampere-hour efficiency

Ah output = 4 × 12 = 48 Ah
Ah input  = 3 × 20 = 60 Ah
η_Ah = 48 / 60 × 100 = 80 %

Watt-hour efficiency

Wh output = 48 × 1.2  = 57.6 Wh
Wh input  = 60 × 1.44 = 86.4 Wh
η_Wh = 57.6 / 86.4 × 100 = 66.67 %

(Check: η_Wh = η_Ah × 1.2/1.44 = 0.8 × 0.8333 = 0.6667.)

Watt-hour efficiency is lower than Ah efficiency because the cell is charged at a higher voltage than it gives on discharge.

Answer: η_Ah = 80 %, η_Wh = 66.67 %.

  • Asked 2 times
  • 2074 Bhadra · 4 marks
  • 2073 Bhadra · 4 marks

Explain about emergency power supply system for an industry.

Answer

An emergency power supply system is an independent source that automatically supplies essential loads of an industry when the normal (utility) supply fails, to protect people, equipment and critical processes.

Need / loads served

  • Emergency and escape lighting, exit signs.
  • Fire alarm, fire pumps, smoke extraction, lifts.
  • Control and protection circuits (switchgear tripping, PLCs, instruments).
  • Critical process loads where a sudden stop causes danger or loss (furnace cooling, chemical stirring, compressors).
  • Communication systems and security systems.

Sources

  1. Battery systems – central battery or self-contained luminaires; instant supply.
  2. UPS – no-break supply for computers and controls.
  3. Diesel generator set with automatic mains failure (AMF) panel – starts in 10–15 s, supplies large loads for long periods.
  4. Second independent utility feeder with automatic changeover.
 Utility ──►[ATS]──► Essential bus ──► Emergency loads
              ▲
 DG set ──────┘
 Battery/UPS ───────► Critical loads (no break)

Main features

  • Automatic start and changeover (ATS/AMF).
  • Essential loads on a separate bus, separated from non-essential loads.
  • Batteries kept charged by float/trickle charger.
  • Regular testing and maintenance of DG and batteries.
  • 2080 Chaitra · 8 marks

Design a Battery Backup system with an inverter for the following loads for 1 days, if DC bus voltage is 48V.
DevicesQty.WattageBackup Time
Computer61004
Tube light24106
Camera16024
Backup Supply34006
Induction Cooker65505
Water Pump220006
Following are the battery available:
OptionVoltageCapacityDoD
Option 112V200Ah80% DoD
Option 22V1500Ah80% DoD

Answer

Assumptions: inverter efficiency 90%, battery depth of discharge 80% (given), all loads may run at the same time, DC bus 48 V, back-up for 1 day as listed.

Step 1: Load and daily energy

DeviceQtyWTotal WhWh/day
Computer610060042400
Tube light241024061440
Camera16060241440
Backup supply3400120067200
Induction cooker65503300516 500
Water pump220004000624 000
Total940052 980

Step 2: Inverter rating

Connected load        = 9400 W
With 25 % margin      = 11 750 W
At pf 0.8 → kVA       = 11 750/0.8 = 14.7 kVA

Choose a 15 kVA, 48 V DC input, pure sine-wave inverter (pumps need surge capability of about 3 × running current; a soft starter/VFD for pumps is advised).

Step 3: Battery energy and Ah

Energy from battery = 52 980 / (η_inv × DoD)
                    = 52 980 / (0.9 × 0.8)
                    = 73 583 Wh
Ah at 48 V          = 73 583 / 48 = 1533 Ah

Step 4: Battery bank options

Option 1: 12 V, 200 Ah

Series  = 48 / 12     = 4
Parallel = 1533 / 200 = 7.67 → 8 strings
Total   = 4 × 8       = 32 batteries
Bank    = 48 V, 1600 Ah
Usable  = 1600 × 48 × 0.8 × 0.9 = 55 296 Wh ≥ 52 980 ✓

Option 2: 2 V, 1500 Ah

Series   = 48 / 2      = 24 cells
Parallel = 1533 / 1500 = 1.02 → 2 strings
Total    = 48 cells, 48 V, 3000 Ah

One string (24 cells, 1500 Ah) gives 1500 × 48 × 0.8 × 0.9 = 51 840 Wh, about 2% short (enough only if inverter losses are ignored: 52 980/(0.8 × 48) = 1380 Ah).

Selection

Option 1 (32 × 12 V, 200 Ah, 4 series × 8 parallel) matches the need closely (1600 Ah vs 1533 Ah). Option 2 with one string is cheaper in wiring and 2 V tubular cells last longer, but slightly undersized; with two strings it is oversized (3000 Ah). Option 2 (single string) can be chosen only if the load hours are reduced a little.

Step 5: Charger

To recharge 1600 Ah in about 10 h: charging current ≈ 0.1 C = 160 A at 48 V (≈ 8–9 kW charger, or the inverter's built-in charger / solar charge controller).

AC ─►[Charger 48V,160A]─┬─►[15 kVA Inverter]─► Loads
                        │
              [48 V battery bank]

Answer: Daily energy 52.98 kWh; 15 kVA inverter; battery 1533 Ah at 48 V → 32 nos. 12 V 200 Ah (4S × 8P) recommended.

  • 2079 Chaitra · 4+4 marks

What are the main purposes of Installation of Emergency Lighting in Industry? Explain the Trickle Charging method for a battery charging system.

Answer

Main purposes of emergency lighting in industry

Emergency lighting is lighting that comes on automatically when the normal lighting fails. Its purposes are:

  1. Safe evacuation: light the escape routes, stairs and exits so people can leave quickly without panic.
  2. Show exits and direction: illuminated exit signs and direction arrows.
  3. Locate safety equipment: fire extinguishers, fire alarm call points, first-aid boxes.
  4. Safe shutdown of processes: operators can stop machines, close valves and furnaces safely (standby/high-risk task lighting).
  5. Prevent accidents and panic in areas with moving machinery, open pits or hazardous chemicals.
  6. Help rescue and fire-fighting teams.
  7. Meet legal and fire safety codes (minimum ≈ 1 lux on escape route floor, for 1–3 h).

Trickle charging method

Trickle charging keeps a standby battery fully charged by supplying a small continuous current equal to its self-discharge loss.

  • Current is about 1–2% of Ah capacity (≈ 1 mA per Ah).
  • The charger (step-down transformer, rectifier and series resistance or regulator) is permanently connected to the battery.
  • The battery normally does not feed the load; it supplies the emergency lights only when mains fails, after which a normal/boost charge is given and then trickle charging resumes.
  • Used for emergency lights, fire alarm panels, engine starting and switchgear batteries.
  • The current must be set correctly: too high → gassing and water loss; too low → sulphation.
 AC ─►[Transformer]─►[Rectifier]─►[R]─► + Battery −
                                     │
                       Emergency lamps (on mains fail)
  • 2078 Chaitra · 1+3 marks

What is an emergency lighting system? Explain about escape lightning system for escape routes.

Answer

Emergency lighting system

An emergency lighting system is lighting supplied from an independent source (battery or generator) that switches on automatically when the normal supply fails, so people can see to escape or to finish critical tasks safely.

It has three types: escape lighting, standby lighting (to continue normal work) and high-risk task area lighting.

Escape lighting for escape routes

Escape lighting makes sure that escape routes can be identified and used safely at all times.

  • Illuminance: at least about 1 lux along the centre line of the escape route (0.5 lux for open areas), with a uniformity (max:min) not more than 40:1.
  • Response time: 50% of the level within 5 s and full level within 60 s.
  • Duration: usually 1 h minimum; 3 h for premises occupied at night or where immediate re-occupation is likely.
  • Luminaire positions: near every exit door, at changes of direction, at stair landings (each flight should get direct light), near fire-fighting equipment and alarm call points, at changes of floor level, outside final exits.
  • Exit signs: illuminated "EXIT" and directional signs visible from any point on the route.
  • Supply: self-contained luminaires with their own batteries, or a central battery system with fire-resistant wiring.
  • Maintenance: monthly function test and yearly full-duration test.
 [Room]──►corridor──►stairs──►EXIT
    ●         ●        ●        ●
  (emergency luminaires at doors,
   turns, stairs and final exit)
  • 2078 Chaitra · 4+4 marks

Explain battery Charger with neat sketch circuit diagram. Explain any two methods for battery charging system.

Answer

Battery charger

A battery charger converts AC mains into controlled DC and drives current into a secondary battery in the direction opposite to discharge, restoring its chemical energy. A typical single-phase charger has:

  1. Step-down transformer – reduces 230 V to a suitable level and isolates the battery from mains.
  2. Rectifier – bridge of diodes or thyristors (SCRs) to convert AC to DC.
  3. Filter/choke – smooths the DC current.
  4. Regulator/control – series resistance, SCR phase control or switch-mode regulator to set constant current or constant voltage.
  5. Protection and metering – fuse/MCB, reverse-polarity diode, ammeter, voltmeter, overcharge cut-off.
230 V ─►[Fuse]─►[Step-down ]─►[Bridge   ]─►[Choke/ ]─┐
  AC            [transformer]  [rectifier]  [regul. ] │
                                                  (A) │
                                    ┌──(V)──┐         │
                                    └─+Batt−┘◄────────┘

The SCR-controlled version measures battery voltage and current and adjusts the firing angle to give the required charging profile.

Method 1: Constant-current charging

  • The charging current is held constant (about 1/10 of the Ah rating, e.g. 10 A for 100 Ah) by adjusting a series resistance or the rectifier output while battery voltage rises from about 2.0 to 2.6 V/cell.
  • Charging time ≈ Ah to be restored / current, plus about 10–20% for losses.
  • Two-rate variant: high current until gassing begins (~2.4 V/cell), then a lower finishing rate.
  • Advantages: simple, accurate control of charge given. Disadvantage: slow; overcharge and gassing if not stopped.

Method 2: Constant-voltage charging

  • A constant voltage of about 2.3–2.4 V per cell is applied. At first the current is high because the battery EMF is low; it falls automatically as the EMF rises, approaching a small value at full charge.
  • A current-limit resistor or electronic limit protects against the initial surge.
  • Advantages: faster charging (about 80% in a few hours), less risk of overcharging, no supervision.
  • Used in automobiles and modern CC–CV chargers.
Const. current      Const. voltage
 I │──────┐          I │\
   │      │            │ \__
   │      └─ t         │    ‾‾───── t
  • 2077 Chaitra · 8 marks

What are the electrical characteristics of the lead-acid cell? Explain the methods of battery charging system.

Answer

Electrical characteristics of a lead-acid cell

  1. EMF / voltage: about 2.0–2.1 V per cell when fully charged (open circuit); it depends on the specific gravity of the acid (EMF ≈ 0.84 + specific gravity).
  2. Discharge voltage: falls quickly to about 2.0 V, stays nearly flat, and must not go below 1.8 V (end-point); deeper discharge causes sulphation.
  3. Charge voltage: rises from about 2.1 V to 2.5–2.7 V at the end of charge; gassing starts at about 2.4 V/cell.
  4. Capacity (Ah): product of discharge current and time to reach 1.8 V; it decreases at higher discharge rates and at low temperature, and is usually quoted at the 10-hour rate.
  5. Specific gravity: 1.25–1.28 when fully charged, 1.10–1.15 when discharged – used to judge state of charge.
  6. Internal resistance: very low (a few milliohms), so it can supply large currents (engine starting); it rises as the cell discharges.
  7. Efficiency: ampere-hour efficiency 80–90%; watt-hour efficiency 70–80%.
  8. Self-discharge: about 2–5% per month.
V/cell
 2.6 ┤            ___ charge
 2.2 ┤  ______----
 2.0 ┤‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾\  discharge
 1.8 ┤                \
     └──────────────────── time

Methods of battery charging

  1. Constant-current charging: current held constant (≈ C/10) while voltage rises; may be two-rate (high then low) to limit gassing. Simple but needs supervision.
  2. Constant-voltage (constant-potential) charging: fixed voltage (≈ 2.3–2.4 V/cell); current falls automatically as EMF rises. Fast and self-limiting; used in vehicles.
  3. CC–CV (modified constant voltage): constant current until a set voltage, then constant voltage; used in most modern chargers.
  4. Trickle charging: small continuous current (1–2% of capacity) to balance self-discharge of a standby battery.
  5. Float charging: charger, battery and load in parallel at ≈ 2.2 V/cell; charger feeds load and keeps battery full; battery takes over at mains failure without a break (substations, telecom, UPS).
  6. Boost charging: high current for short time to quickly restore a discharged battery.
  7. Equalising charge: periodic small overcharge to equalise all cells of a string.
  • 2075 Baisakh · 8 marks

Explain the importance of emergency supply system in industrial plant? Differentiate between on-line and off-line UPS with proper schematic diagrams.

Answer

Importance of emergency supply in an industrial plant

An emergency supply system automatically supplies essential loads when the utility supply fails. It is important because:

  1. Safety of people: escape lighting, exit signs, fire alarm and fire pumps must work during a power cut or fire.
  2. Safe shutdown of processes: furnaces, boilers, chemical reactors and kilns need cooling, stirring or control to avoid explosion, fire or damaged products.
  3. Protection of equipment: lubrication pumps, cooling fans, control and switchgear tripping supplies must continue.
  4. Avoid production and data loss: PLCs, SCADA and computers need no-break power; restarting a continuous process can take hours.
  5. Communication and security: telephones, CCTV, access control.
  6. Legal requirement: fire and factory regulations require emergency lighting and supply.

Sources: batteries (central/self-contained), UPS, DG sets with AMF panel, and a second utility feeder.

On-line UPS vs off-line UPS

ON-LINE (double conversion)
        ┌───────── static bypass ────────┐
 AC ─►[Rectifier/]─► DC bus ─►[Inverter]─┴─► Load
      [ charger  ]     │
                   [Battery]

OFF-LINE (standby)
 AC ─────────────────────────────┐
  │                          [Transfer ]─► Load
  └►[Charger]─►[Battery]─►[Inverter]┘ switch
PointOn-line UPSOff-line UPS
Normal path of loadAlways through rectifier and inverterDirectly from mains
Transfer timeZero (no break)4–10 ms
Protection from sags, spikes, noiseCompleteVery little in normal mode
Output waveformPure sine, stable V and fMains in normal; often quasi-sine on battery
EfficiencyLower (≈ 90–94%)Higher (≈ 97–99%)
Cost and sizeHighLow
Inverter dutyContinuousOnly during outage
Typical useData centres, process control, hospitalsHome PCs, small offices
  • 2073 Magh · 2+3+3 marks

Explain trickle charging method for a battery charging system. A lead acid cell is discharged at a steady current of 4A for 12 hours, the average terminal voltage being 2.2V. To restore it to its original state of charge, steady current at 3A for 20 hours is required, the average terminal voltage being 2.44V. Calculate the ampere-hour efficiency and watt-hour efficiency in this particular case.

Answer

Trickle charging

Trickle charging keeps an idle battery fully charged by supplying a very small continuous current equal to its self-discharge loss.

  • Current ≈ 1–2% of Ah capacity (about 1 mA per Ah).
  • Charger (step-down transformer + rectifier + series resistance/regulator) remains permanently connected.
  • The battery feeds the load only when mains fails; after that it is recharged normally and trickle charging resumes.
  • Used for emergency lighting, fire alarm, DG starting and switchgear batteries.
AC ─►[Transformer]─►[Rectifier]─►[R]─► + Battery −

Numerical

Given: Discharge 4 A × 12 h at 2.2 V average; charge 3 A × 20 h at 2.44 V average.

Ampere-hour efficiency

Ah output = 4 × 12 = 48 Ah
Ah input  = 3 × 20 = 60 Ah
η_Ah = 48 / 60 × 100 = 80 %

Watt-hour efficiency

Wh output = 48 × 2.2  = 105.6 Wh
Wh input  = 60 × 2.44 = 146.4 Wh
η_Wh = 105.6 / 146.4 × 100 = 72.13 %

Answer: Ampere-hour efficiency = 80 %, watt-hour efficiency = 72.13 %.

  • 2073 Bhadra · 4 marks

What are different types of battery charging scheme? Describe each of them with their specific purposes. Which method is most effective?

Answer

Battery charging schemes differ in how current and voltage are controlled during charging.

  1. Constant-current charging – current fixed at about C/10. Purpose: first charge of new batteries, accurate restoring of known Ah; used in battery workshops.
  2. Constant-voltage charging – fixed voltage ≈ 2.3–2.4 V/cell; current falls as battery charges. Purpose: fast, self-limiting charging, e.g. vehicle alternator charging.
  3. Trickle charging – tiny continuous current equal to self-discharge. Purpose: keep idle standby batteries (emergency lights, alarms, DG starting) always full.
  4. Float charging – charger, battery and load in parallel at ≈ 2.2 V/cell. Purpose: no-break DC supply in substations, telecom and UPS; battery takes load at mains failure.
  5. Boost charging – high current for a short time. Purpose: quick recovery of a deeply discharged battery.
  6. Equalising charge – periodic controlled overcharge. Purpose: equalise voltages and specific gravity of all cells in a string.

Most effective method: the modified constant-voltage (CC–CV) method – constant current at first, then constant voltage, then float. It charges fast without overcharging or excessive gassing and gives the longest battery life, so it is used in most modern chargers. For standby duty, float charging is the most effective in practice.

  • 2072 Magh · 4 marks

Write down main purposes for installation of emergency lighting in industry.

Answer

Emergency lighting is lighting that operates automatically from an independent source when the normal lighting supply fails. Its main purposes in an industry are:

  1. Safe evacuation: light escape routes, corridors, stairs and exits so people can leave quickly and safely.
  2. Exit identification: keep exit and direction signs lit so the way out is clear.
  3. Safe shutdown: let operators stop machines, close valves and make processes safe (high-risk task lighting).
  4. Avoid panic and accidents near moving machinery, open pits, hot or hazardous areas.
  5. Locate fire-fighting and safety equipment: extinguishers, call points, first-aid boxes.
  6. Help rescue and fire-fighting teams during an emergency.
  7. Continue essential work (standby lighting) in control rooms, substations and medical rooms.
  8. Meet legal requirements of fire and factory safety codes (about 1 lux minimum on escape routes for 1–3 h).
  • 2072 Magh · 4 marks

Why UPS is required for industrial electrification? Explain about On-line UPS.

Answer

Why UPS is required in industrial electrification

A UPS (Uninterruptible Power Supply) gives continuous, clean AC power to critical loads during supply failure and disturbances.

  • PLCs, DCS/SCADA, computers and instruments lose data or trip on even a few milliseconds' interruption.
  • It protects equipment from voltage sags, spikes, surges, harmonics and frequency variation.
  • It bridges the 10–15 s gap until the DG set starts and takes load.
  • It enables safe, orderly shutdown of processes and avoids loss of production.
  • It supplies emergency lighting, communication, alarm and security systems.

On-line UPS

In an on-line (double-conversion) UPS, the load is always supplied by the inverter.

  1. Rectifier/charger converts mains AC to DC, feeds the inverter and keeps the battery charged (float).
  2. Battery is connected to the DC bus permanently.
  3. Inverter converts DC to regulated sine-wave AC for the load.
  4. Static bypass switch transfers the load to mains if the inverter fails or is overloaded.

When mains fails, the battery continues feeding the inverter, so there is zero transfer time.

        ┌────── static bypass ──────┐
 AC ─►[Rectifier]─►DC─►[Inverter]───┴─► Load
                    │
                [Battery]

Advantages: no break, complete isolation from mains disturbances, stable voltage and frequency. Disadvantages: higher cost and losses (efficiency ≈ 90–94%).

  • 2071 Magh · 8 marks

What do you mean by starting, lighting and ignition and station battery in an industry? What are the different methods of battery charging system?

Answer

Starting, lighting and ignition (SLI) battery

An SLI battery is a lead-acid battery used in vehicles and engine-driven sets to:

  • Start the engine – supply a very high current (hundreds of amperes) for a few seconds to the starter motor;
  • Light – supply head lamps and other lamps when the engine is off;
  • Ignition – supply the ignition system of petrol engines.

In an industry, SLI batteries are used in diesel generator sets, fire-pump engines, forklifts and plant vehicles. They have thin plates with large surface area for high current; they are kept charged by the engine's alternator or by trickle charging when idle. Usual ratings: 12 V or 24 V, 80–200 Ah.

Station battery

A station battery is a large stationary battery bank (usually 24, 48, 110 or 220 V DC) installed in substations, power stations and industrial switchrooms to supply:

  • Tripping and closing coils of circuit breakers,
  • Protection relays, control and indication,
  • Emergency lighting and communication.

It must be completely reliable, so it is kept on float charge with the charger, battery and DC load in parallel. Thick-plate tubular lead-acid or Ni-Cd cells with long life (15–20 years) are used.

Methods of battery charging

  1. Constant-current charging – current held at about C/10 while voltage rises; two-rate variant reduces gassing.
  2. Constant-voltage charging – fixed 2.3–2.4 V/cell; current falls automatically.
  3. CC–CV (modified constant voltage) – constant current then constant voltage; used in modern chargers.
  4. Trickle charging – small continuous current to make up self-discharge of idle batteries (e.g. DG starting batteries).
  5. Float charging – charger feeds load and battery together at ≈ 2.2 V/cell (station batteries).
  6. Boost charging – high current briefly after deep discharge.
  7. Equalising charge – occasional overcharge to equalise all cells.
Station battery – float scheme
 AC ─►[Charger]──┬──────────┬──► DC bus (trip,
                 │          │    control, lights)
            [Battery bank]  │
  • 2070 Magh · 8 marks

What is emergency supply system for industrial plant? Differentiate between Ah efficiency and Wh efficiency of battery.

Answer

Emergency supply system for an industrial plant

An emergency supply system is an independent, automatically started power source that supplies the essential loads of a plant when the normal supply fails, to protect life, equipment and critical processes.

Essential loads: escape and exit lighting, fire alarm, fire pumps, lifts, control and protection supplies, process loads needing safe shutdown (cooling, stirring, lubrication), communication and security.

Sources:

  1. Battery systems – self-contained or central battery; supply is instant.
  2. UPS – no-break supply for computers, PLCs and instruments.
  3. Diesel generator with AMF panel – starts automatically in about 10–15 s; feeds large loads for long periods.
  4. Second utility feeder with automatic transfer switch (ATS).
 Utility ─►[ATS]─► Essential bus ─► fire pump, lights
             ▲
 DG set ─────┘
 Battery/UPS ─────► PLC, alarm, control (no break)

Essential loads are kept on a separate bus, batteries are on float/trickle charge, and the system is tested regularly.

Ah efficiency vs Wh efficiency

η_Ah = Ah delivered on discharge / Ah supplied on charge
η_Wh = Wh delivered on discharge / Wh supplied on charge
     = η_Ah × (V_avg discharge / V_avg charge)
PointAh efficiencyWh efficiency
Based onQuantity of charge onlyEnergy (charge × voltage)
Considers voltageNoYes
Losses includedGassing, side reactionsAlso I²R and polarisation losses
Value (lead-acid)80–90%70–80%
RelationAlways higherLower, since V_charge > V_discharge
UseDeciding charging time/Ah neededEnergy cost and economics

Example: discharge 4 A × 12 h at 2.2 V, charge 3 A × 20 h at 2.44 V → η_Ah = 48/60 = 80%, η_Wh = 105.6/146.4 = 72.13%.

  • 2068 Magh (old course) · 8 marks

What are the methods commonly adopted for charging a lead acid battery from an A.C. single phase source. What are the principle components of charger of the above type?

Answer

Methods of charging a lead-acid battery from a single-phase AC source

Since the battery needs DC, the AC is first stepped down and rectified. The common methods are:

  1. Constant-current charging using a transformer, rectifier and variable series resistance (or SCR control) to hold current at about C/10. Two-rate charging lowers the current when gassing starts.
  2. Constant-voltage charging – rectifier output regulated at about 2.3–2.4 V/cell; current falls as the battery charges. A small series resistance limits the initial current (modified constant-voltage).
  3. Trickle charging – small rectifier with high series resistance gives a tiny continuous current to standby batteries.
  4. Float charging – a regulated rectifier connected in parallel with battery and load at ≈ 2.2 V/cell.
  5. Boost charging – high-current rectifier for short periods.

Rectifiers used: half-wave (small trickle chargers), centre-tap full-wave, or bridge full-wave (most common), with diodes or SCRs.

 230 V ─[F]─┬─────────┐
 1φ AC      │ Step-   │ ┌─D1─┬─D2─┐
            │ down    ├─┤    │    ├─[L]─[R]─(A)─┐
            │ transf. │ └─D3─┴─D4─┘             +│
 N ─────────┴─────────┘        │            [Battery]
                               └───────────────−─┘
                 (V) across battery

Principal components of such a charger

  1. Fuse/MCB and ON-OFF switch – protection on AC side.
  2. Step-down (isolating) transformer – lowers 230 V to the needed level and isolates the battery from mains; tappings allow coarse voltage adjustment.
  3. Rectifier – diode bridge (or SCR bridge for control) converting AC to DC.
  4. Filter / smoothing choke and capacitor – reduces ripple in the charging current.
  5. Current-control device – variable resistor, rheostat, SCR firing circuit or switch-mode regulator.
  6. Ammeter and voltmeter – to monitor charging current and battery voltage.
  7. Reverse-current/polarity protection – blocking diode so the battery cannot discharge back into the charger; reverse polarity alarm.
  8. Automatic cut-off or float controller – switches to float/trickle mode at full charge to prevent overcharging.
  9. Cooling and enclosure – heat sinks for rectifier, ventilated cabinet.

Questions from Old Question Collection (EE 653) (Scanned IOE exam papers from 2068 to 2080 (2068 papers from the older Industrial Electrification course)). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the IOE syllabus ↗