### Selection of Various Types of Inverter-(Part-3)

**How to select Right Inverter**

- Before buying an inverter for, It is very important to understand what is the right inverter for our requirement for that we do understand the basics criteria of Inverter.
- In order to make a good estimate of your power needs, you’ll need to take a look at all of the devices you plan on plugging into your new inverter.
- If we only need to use one device at a time, then that’s the only one you’ll need to look at. However, you’ll need to add together the numbers from multiple devices (like an LCD screen and a video game system) if you plan on using them at the same time.

**(A) Power Requirement:**

- One of the most important factors that we must know before buying an inverter is Power requirement.
- Power requirement means all electrical appliances (like fan, tube lights, television,Pumps, CFL etc.), we want to run at the time of power failure. The power requirement is simply addition of the power consumed by various electrical equipments.
- The thing we must understand that Inverter is not a Generator. Inverter has its own limitations. If power requirement is more than our estimation, then an inverter alone cannot create demands effectively.
- High power inverter can run our refrigerator and air conditioners, but battery will not last more than few hours. Hence it better to estimate Inverter Load Carefully
- The selection of correct size Inverter is very important, If we need to power small appliances like energy efficient light bulbs, we do not need to buy a 2000W power inverter because it will consume more power even in standby mode and work very inefficiently with small appliances. On the other hand, if we connect a coffee machine to a 150W inverter we will quickly blow a fuse, Therefore power estimation is important thing.
**The size of Inverter depends on the watts (or amps) of what we want to run. It is recommend that we choose at least 10% to 20% more than our requirement.**- Suppose you want 1No Fans, 1No Tube lights, 1No CFL and 1No television to operate at the time of power failure. Therefore the total power requirement to be (1×90 + 1×50 + 1×25 + 1×120) = 285 watts. Here Total Load is 285 watts.

**(B) Surge Power (Starting Power):**

- Starting and running power requirement of all electric appliances are different.
- Starting power of Electrical Equipment is several times greater than their normal working power.
- An 18 Watts CFL takes around 25 Watts power to start and after few seconds it works on 18 Watts. Some appliances like Refrigerator, Washing Machine etc take almost double power to start as compared to the normal running power.
- For example Gridding Machine have normal working power 1000W, their starting power is higher than 4000W, so inverters with continuous power 2000W are not suitable because their peak power is limited by 4000W.
- For selecting right Inverter, Always take into account starting power requirements of your equipment, especially devices with electric motors, Electronics Choke, Capacity, inductive
- The Size of the inverter should be chosen based on the power consumption of your load.
**Resistive loads:**All resistive loads like toaster, coffee maker, electric range, iron, Incandescent lamps, flood lighting. Laser Printers use Nichrome resistance wire in their heating**The Starting Power or Surge Power rating should be 6 times the Watt rating****Inductive Loads:**All Inductive Load like induction motor, reciprocating pumps and compressors, refrigeration, air-conditioning, Oxygen Concentrators have more starting current.**The Starting current (LRA) should be 5 times the full load current (FLA).****Capacitive Loads:**Switched Mode Power Supplies (SMPS), electronic equipment like battery chargers, computers, audio and video devices, radio etc.**The Starting Power or Surge Power rating should be 3 times the Watt rating****Microwaves :****The Initial power consummation by the microwave will be 2 times the cooking power.****A Water Supply Pump:****The starting surge can be 3 times the normal running rating of the pump**- The inverter should, therefore, be sized adequately to withstand the high inrush current.

| |

Type of Device | Starting Factor |

Air conditioner | 5 |

Refrigerator / Freezer | 5 |

Air Compressor | 4 |

Sump Pump / Well Pump / Submersible Pump | 3 |

Dishwasher | 3 |

Clothes Washer | 3 |

Microwave | 2 |

Furnace Fan | 3 |

Industrial Motor | 3 |

Portable Kerosene / Diesel Fuel Heater | 3 |

Circular Saw | 3 |

Bench Grinder | 3 |

Incandescent / Halogen / Quartz Lamps | 3 |

Laser Printer / Other Devices using Quartz Lamps for heating | 4 |

Switched Mode Power Supplies | 3 |

Photographic Strobe / Flash Lights | 4 |

** **** (C) Rating (VA rating) of the inverter:**

- It stands for the Volt ampere rating. It is the voltage and current supplied by the inverter to the equipments.
- If an inverter operates with 100% efficiency, then the power requirement of the electrical items and power supplied by inverter is same. But 100% or ideal conditions does not exist in real. Most inverters have the efficiency range from 60 % to 80%. This efficiency is also called power factor of an inverter and is simply the ratio of power required by the appliances to power supplied by an inverter. Power factor of most inverters ranges from 0.6 to 0.8.
**Rating of inverter (VA) =Power consumed by equipments (watts) / Power factor (efficiency).**- Power of inverter (VA) = 535/0.7 = 765 VA
- In the market 800 VA inverters are available. So Size of an inverter with
**800 VA.**

** ****(D) Type of Battery**

- The performance and Life of an inverter depends on the battery Quality.
- Battery stores the energy needed to run appliances during power cut.
- There are three Type of Battery
- Lead Acid Battery
- Tubular Battery
- Maintenance Free Battery
- Normally high power Lead Acid batteries are used to power inverters.

** ****(E) Battery Capacity (Ah):**

- Capacity of a battery is expressed in terms of Ampere Hour (Ah). It is decide the back up hours of Inverter.
- It indicates the rate of current a battery can supply for a given duration. If the capacity of a battery is 100 Ah, that battery can supply 100 Ampere current for 1 Hour or 1 Ampere Current for 100 Hrs
**Battery capacity = Power requirement (watts) x Back up hours (hrs) / Battery Voltage (volts) x Battery Efficiency x No of Battery**- Example: If Inverter Load is 390 Watt, We need 3 hours back up and Battery Voltage is 12V, Battery Efficiency is 90%
- Battery Capacity = (390 x3) / 12×0.9×1 =
**109 Ah** - Therefore a battery capacity of 120 Ah.
- In reality battery performance degrades with usage, so you are recommended to buy 5-10% higher capacity battery.
- Therefore a battery with a capacity of 150 Ah is suitable for our requirement.
- If we use series – parallel battery combination for backup is
**Backup capacity(hours) = N series X Volts X Battery Ah X N parallel / (UPS capacity in VA)**- Where, N series = no. of batteries in series and N parallel = no. of parallel battery banks

** ****(F) Backup Time (Hr):**

**Backup (hours) = Battery Ah X Battery Voltage X Inverter Efficiency / Total Load (Watt)**

**(G) Battery Voltage (12V, 24V or 48V):**

- We can choose Battery Voltage of 12V, 24Vor 48V for inverter. The selection is very important for Voltage drop and Cable size of Inverter.
- To control load current within limit, as load increase, we should select higher size of Battery Voltage.
- Let’s take example.

Watt | Volt | Amp | Cable Resistance(Ω) | Voltage Drop (V) |

1000 | 12 | 83 | 0.01 | 0.8 |

2000 | 24 | 83 | 0.01 | 0.8 |

4000 | 48 | 83 | 0.01 | 0.8 |

- As per above table, as the load increase, by selecting higher voltage,load current within limit.
- Let’s take example of selecting different voltage for same load (Watt)

Watt | Volt | Amp | Cable Resistance(Ω) | Voltage Drop (V) |

4200 | 12 | 350 | 0.01 | 3.50 |

4200 | 24 | 175 | 0.01 | 1.75 |

4200 | 48 | 87.5 | 0.01 | 0.88 |

- At Higher Load (Watt), by selecting lower voltage (12V).The load current is high. The higher the current needs the bigger the components. High currents require large diameter cables and fuses, which are expensive and the entire system is not safe.
- By only doubling the voltage we can get double the power (Watt) at the same current.
- 12 Volt used to be a standard for extra low voltage power systems. Today, mostly systems are 24V or 48V and include a 230V AC inverter. This means the wiring of the house does not have to be different from any other grid-connected household and cabling cost is greatly reduced.
- The Battery Cost for higher Voltage is not more for same rating of Battery. For 12KW Load, Instead of buying a battery which has a capacity of 1000AH at 12V, (12KWH) we can rearrange the battery as a 24V 500AH (12KWH) or 48V 250AH (12KWH). The total power available from the battery is exactly the same, but with lower losses in the cabling and inverter.
- A higher volt inverter will be more expensive than a lower volt one, and for small setups, that extra expense is unnecessary. If we are going to have high load demands, it makes more sense to go with a higher-volt inverter for a variety of reasons.

**(H) Battery Charger:**

- The battery capacity is limited to the size of the charger. Usually, the battery capacity should be no more than 12x the maximum charge current i.e. a 5A charger can only accommodate 60AH of batteries (5 x 12 =60).
- If discharges are expected to be less frequent than once in every 10 days, We may use formula of 20x the maximum charge current.

**(I) Discharge Rate:**

- Lead acid batteries have few Amps hour if the discharge rate is fast.
- Generally, the lead acid battery discharge rate is slow. It is 20 hours.
- At high discharge rate, the capacity of the battery drops steeply. Suppose the battery is 10 Ah and its discharge rate is 1C. One hour discharge of the battery at the rate of 1C (10 Amps in 1 hour), the capacity reduces to 5 Ah in one hour.
- Suppose we want to run a load at 20 Amps for 1hour. Then the capacity of the battery is
**20A x 1 Hour = 20Ah.** - Keep discharge rate to maximum 80%. Then the battery capacity should be
**20Ah / 0.8 = 25 Ah** - So a 25 Ah battery can give 20 Amps current for 1 hour to run the load.
- But it is better to drain the battery to 50% only then,
**25 Ah / 0.5 = 50 Ah.** - Hence it is better to use a 50Ah battery to run the load at 20 Amps per Hour to keep 50% charge in the battery.

Battery Capacity | Hours of Discharge |

100 | 20 |

90 | 10 |

87 | 8 |

83 | 6 |

80 | 5 |

70 | 3 |

60 | 2 |

50 | 1 |