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HOW TO CHOOSE A GENERATOR

發布日期 2009-11-21
Diesel generator set is usual emergency power supply in civilian use area, it’s a part of the electricity net design of civilian building. So what’s the key of pruchasing diesel generator set?

1 Principle
Accoding to the building level(every country got different standard for the classification,like function of the building, here we use Chinese national standard) to set load and power source.

1) in level 1 building, if the electricity grid can supply 2 ways of power supply, Diesel generator set is not necessay, but if there are equipments with high importance, generator set is still required for emergency.

2)when the grid can only offer 1 way power supply. Usually, generator set is always needed as standby or emergency power source.

3)for ensuring normal running of Big-and-middle-sized commercial building after power cutting. Generator set is important.and as city grid is connect with others, error in one place can pause of the whole grid, Even a building with 2 ways power supply, for reliable supply of fire extinguishing equipments and other important equipments like communication euipment and smart terminal, diesel generator is still necessary.



2. power capacity

(一)Design stage

Power capacity of generator for self-provided should be 10-20% of the total capacity of the transformer.

(二)project fulfilling stage

There are 3 types load in a building,

first,sercurity type. The load of epuipments for man’s safety. i.e. fire service pump, elevant, exhuast system etc.

second, basic type. Base equipments for ensuring normal operation of the whole building, like lighting in workign area, main elevant and pass channel.

Third, ordinary type. Load excepts the former 2 types. Like air conditioner, pump, and other illumination lights.

When caculating the power capacity of the generator, No matter in what condition, the first type load has to be included. Second type is according to the function of the building and the grid performance.suppose the basic load is necessary, these 2 types of load is a big amount together. The best solution is caculate the first type load and the second type load independently, and choose a generator according to the bigger one. Because distinguishing system and the basic illumination system would not run together.

After getting the exact load of related equipments, set an actual requiement factor(normally 0.85-0.95), got a caculated capacity Pj=KxP∑,so you can compute the power load of the generator set by the following fomula P=kPj/η

P-power capacity of standby generator set

Pj—load of equipments

P∑—total load

η—generator parallelling running asymmerty

k—steady factor, normally 1.1

2) count generator capacity with the single maximum electricity consumer

P=(P∑-Pm)/η∑+ PmKCcosψm(KW)

Pm—electricity for start the maximum power consumer.(kw)

η∑一total power computing factor, usually 0.85

cosΨm —start factor for electric motor, usually 0.4

K—start multiple of the electic motor
C—start wit h full voltage C=1.0, Y—△ start c=0.67, start under auto-transformer 50% tapping, c =0.25. 65%tapping, c=0.42. 80%tapping, c=0.64

3) count generator capacity according to the normal fallen of the generatrix voltage. P=PnKCXd″(1/△E-1)(kw)

P- power capacity of electic motor which causing the maximum fallen of the generatix voltage (kw)

K—multiple of the start motor current

Xd″—tranient reactence of the generator, normally 0.25

E-Bearable sudden voltage fallen of generatrix. 0.20 with elevant, 0.25 without elevant.



case study:

A project plan build 12 floors, with building area of 10000㎡. Level 2 building. Fire fighting load(191kw) is the main load in security type capacity, spary pump(37kw) is the biggest power consumer, start with aotu-transformer 80% tapping.

1) certain caculating load P=kPj/η=1.1×191/1kw=210.1 kw

2)count generator capacity with the single maximum electricity consumer

P=(P∑-Pm)/η∑ +PmKCcosΨm =(191—37)/0.85+37×6×0.64×0.4 =238.0 kw
3) power capacity of electic motor which causing the maximum fallen of the generatix voltage. P=PnKCXd″(1/△E-1)
=37×6×0.64×0.25(1/0.20-1)
=142.08 kw

Based on these reckoning, we should choose generator set ≥238.0 kw,so 250kw is the right one. 


HOW TO CHOOSE A GENERATOR
HOW TO CHOOSE A GENERATOR




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