Selecting a 70-kva DG Set

How did you arrive at the size of the generator? Seems a bit large for a residence.

1. Have an AMF Panel installed with the DG

2. Have a Manual Changeover also installed for the possibility of AMF malfunction work.

3. Look for higher fuel efficiency and longer hours between subsequent maintenance.

Happy Buying..

Rgds

Sanjeev

Can the AMF be expalined in detail.Thanks for prompt response and guiding.

Automatic Mains Failure (AMF) Panel is for sensing the mains / grid failure, starting the DG and transferring the load to DG Supply... and the reverse process of DG Switch OFF in case of mains / grid restoration.

rgds

Sanjeev

Thank you very much.

70 KVA DG SET IS TOO BIG FOR A RESIDENCE AS PER MY EXPERIENCE. BUT BEFORE THAT PLEASE CONFIRM WHATS THE LOAD & MAX DEMAND AFTER CONSIDERING THE DIVERSITY FACTOR. THIS IS IMPORTANT FOR CALCULATING THE SIZE OF THE SET. FOR A DUPLEX VILLA YOU NEED A 25 KVA DG SET TO MEET THE DEMAND INCLUDING THE AIR CONDITIONERS & GEYSERS.

HOWEVER CALCULATE THE LOAD FIRST.

Thank you very much and I would be checking the details once again.The transformer rating selected is 100 kva and 70% loading of it is considered for DG set.However we would check the loads once again.

I appreciate if you can brief the meaning of diversity factor.

Dear pokalasrao

This paper may can helping you to calculate of DG loading, its found from the website, but I forgot the web address...

Good Luck...

Rgds

Sis

Tables of contents

1. Legislation, statutory regulation in Hong Kong & relevant standard for electrical services

All of building services system is designed in accordance to legislation, statutory regulation in Hong Kong , relevant standards and guidance listed in the followings.

2.Electrical load estimation

( a ). For flats

According to 'Code of practice for electricity (wiring) regulation), the electrical load demand for each flat is calculated as the following table:

The main switch for each flat shall be 100A single phase or 30A three phase. However, according to HEC's supply rule, 30A three phase is recommanded as the main switch rather than 100A single phase because meter for 100A single phase is not available.

The electrical load for all flat = 89.14A/3 per flat x 220V x 64 flats

= 418.4kVA

( b ). For shops and kindergarten

In reference to 'COMMERCIAL CODE OF PRACTICE 215 LOAD ASSESSMENT PROCEDURE (REVISION 2)' issued by China Light & Power Company Ltd., the electrical load is estimated as below:

*1 means Usable Floor Area (U.F.A)

*2 means After Diversified Maximum Demand (ADMD) based on COP 215 by CLP

*3 Shops is assumed as light metal retail stores.

*4 ADMD is included central A/C.

( c ). For landlord Supply

*5 Data is based on Maximum Allowable Lighting Power Density extracted from Code of Practice for Energy Efficiency of Lighting Installation.

( d ). For essential Supply

Overall electrical load = (418.4 + 204.72 + 350.7 + 175)kVA

= 1138.82kVA

However, two factors shall be considered as below:

Total usable floor area = 7600m²

Total demand density = 0.196kVA/m²

So, 1 No of 1500kVA HEC's transformer is required for the proposed residential building.

3. Generator loading

An emergency generator set is required to handle the following essential loads in case of power failure & fire conditions. If the equipment is started simultaneously, the starting current will be larger. So, in the building, the equipment of essential load is started in sequence and control by BMS to reduce the starting current so as to select a smaller generator.

The essential load estimated in (1) = 175kVA

Taking a safety factor of 10% for overload and a factor of 30% for future expansion,

The total essential load = 175kVA x 1.1 x 1.3

= 250kVA

So, a 300kVA generator set is selected. The generator should be operated normally within 15 seconds in case and the capacity of fuel tank should be capable to operate the generator for 6 hours.

Based on some manufacturer's catalogue, the capacity of fuel tank is about 400L.

4. Lightning hazards

Based on BS 6651:1992, Part 2, the overall lightning is calculated as the following to determine whether or not lightning protection is needed.

Probability of being struck, P = A_c x N_g x K_a x K_b x K_c x K_d x K_e x 10^-6 ,where

Collection area, A_c = Area of roof + Area of perimeter of building + Area of four

rounded corners formed by quarter circles of building height

= LW + 2LH + 2WH + pH²

= 46x12 + 2x46x73 + 2x16x73 + px73²

= 26.3 x 10³ m²

No. of flash per km2 per year, N_g = 1.1

Factor for use of structure, K = 1.2

Factor for type of construction, K_b = 0.4

Factor for contents (or consequential effects), K_c = 0.3

Factor for degree of isolation, K_d = 0.4

Factor for type of terrain, K_e = 0.3

So, P = 26.3 x 10³ x 1.1 x 1.2 x 0.4 x 0.3 x 0.4 x 0.3 x 10^-6

= 5.0 x 10^-3

The probability of risk (5.0x10^-3) is greater than the critical risk (10^-5) so that lightning protection is necessary. And, the actual design of the installation is illustrated by drawing no. EE-EP-RF

5. Earthing system

In accordance with BS 7671, the earthing loop impedance shall not be greater than 0.5 W Due to the site constraint, the plate electrode and tape electrode cannot be used as the earthing conductor. The most suitable and economic way is to use the rod electrode. The following calculation is to determine the no. of rod electrode and the size.

Resistance of rod electrode, where L = length of rod,

d = diameter of rod,

r = soil resistivity

Code 12C of COP electricity (wiring) regulation stated that the diameter of rod electrode should not be less than 12.5mm. The standard size of rod electrode in the market is 12.5mm & 15mm.

After site visit, the soil in the Wan Chai is marshland. The soil resistivity is around 4 W-m.

Consider a rod electrode of 4.5m length and 15mm diameter,

the resistance = = 0.9596 W

so that 2 rod electrodes in parallel is required to maintain 0.5 W earth loop impedance. And, the spacing between the earth pits shall be more than 9m (2x4.5m) to avoid the return loop.

6. Cable sizing

In this design report, the cable sizing procedure is referring to Code of Practice for Energy Efficiency of Electrical Installation (Draft).

• To determine the design current I_b, nominal rating of protective device In
• To calculate the min. tabulated value of current I_t(min) as the below formula:

,

C_a = Correction factor for ambient temperature

(Assume the ambient temp = 35 ⁰C)

C_g = Corection factor for grouping

C_i = Correction factor for thermal insulation

Note that all correction factor is referring to Appendix 4 of IEE Wiring Regulation (16^th Edition)

• To select a suitable size It of conductor which satisfy I_t
• To find effective current-carrying capacity I_z (I_z = I_t x C_a x C_g x C_i) and ensure that
• To calculate the voltage drop along the conductor and determine whether or not cable size selected is acceptable. For sub-main circuit, the max. voltage drop is 1.5% and for final circuit, the max. voltage drop is 2.5%.

V.D. = r x p.f. x k x I_b x L , where

r = Voltage drop per ampere per metre at the conductor operating temperature

p.f. = Power factor (assume 0.85)

k = correction factor of operating temperature

, where t_l = operating temperature

t_p = max. permitted conductor operating temp

, t_a = expected ambient temp. (assume 35 ⁰C)

• To calculate the percentage copper loss with respect to the total active active power transmitted.

The following table (EE-Table 1) shows the cable size.

7. Vertical transportation

( a ). Lift Performance

In accordance to the CIBSE Guide D, the lift performance is calculated as the following table:

The lift performance is also simulated by computer software HEVACOMP (Results please refer to )

From the CIBSE Guide D, the recommended arrival rate and waiting interval are 5-7% and

40-90s for flats.

For the proposed building, the performance is satisfactory.

( b ). Escalator performance

According to CIBSE Guide D,

the escalator handling capacity, Cp = 60 v k s , where

v = horizontal speed (m/s)

k = average occupant density (people/step)

s = number of step (steps/m)

Due to architectural constraint, the physical data of escalator:

inclination of escalator = 30⁰

height of escalalator = 4m

length of escalator = 6.9m

no. of step = 6900mm/400m = 17.25

width of step = 800mm

k = 1.5

h = height of step = 4000/17.25 = 231mm

s = 1000mm/231mm = 4.33

the speed = 0.5m/s (max. for 30⁰)

the handling capacity = 60 x 0.5 x 1.5 x 4.33

= 194.85 person/min

From the CIBSE Guide D, the handling capacity for kindergarten is 92 person/min.

For the proposed building, the performance is satisfactory.

8. Lighting system

( a ). Shopping arcade corridor for G/F-2/F

Physical data of corridor, width = 2m, height = 4m

Data of luminaire: Manufacture = BEGA

Model no. = 6743

Type = Wall mounted discharge lamp with asymmetrical reflector

Lamp = 250W metal halide lamp (MBI-E)

Lumen output = 17000lm

Mounting height of luminaire = 3.5m

The angle of illuminating area = 30⁰-0⁰

The following calculation is based on technical data of BEGA 6743

the spacing between luminaires = 4m (obtained from cone diagram)

the average illuminance = 265 lx (obtained from cone diagram)

the luminous intensity (angle = 30⁰) = 360 cd/klm x 17 lm = 6120 cd

the horizontal illuminance at the opposite side of corridor = 6120cd/(2²+3.5²)x0.8

= 302 lx

the luminous intensity (angle = 0⁰) = 300 cd/klm x 17 lm = 5100 cd

the horizontal illuminance below the light fitting = 5100cd/(3.5)²x0.8

= 333 lx

(Assume the light loss factor = 0.8 & neglect the reflectance of walls due to open corridor)

( b ). Canopy

Physical data of corridor, width = 3m, height = 4m

Data of luminaire: Manufacture = BEGA

Model no. = 6180

Type = Recessed ceiling downlight c/w aluminium refelctor

Lamp = 80W metal halide lamp (MBF)

Lumen output = 4000lm

Mounting height of luminaire = 4m

The angle of illuminating area = -20⁰-20⁰

The following calculation is based on technical data of BEGA 6743

the spacing between luminaires = 4m (obtained from cone diagram)

the average illuminance = 80 lx (obtained from cone diagram)

the luminous intensity (angle = 30⁰) = 360 cd/klm x 4 lm = 1440 cd

the horizontal illuminance at the side of canopy = 1440cd/(1.5²+4²)x0.8

= 63 lx

the luminous intensity (angle = 0⁰) = 600 cd/klm x 4 lm = 2400 cd

the horizontal illuminance below the lighting fitting = 2400cd/(4)²x0.8

= 120 lx

(Assume the light loss factor = 0.8 & neglect the reflectance of walls due to open corridor)

( c ). Open atrium/corridor for L1-L5

Data of luminaire: Manufacture = THORN

Model no. = DLC 250/400

Type = Narrow bam medium bay lighting

Lamp = 250W/400W high pressure sodium lamp (SON-E)

Lumen output = 25500/45000 lm

Power input = 260W/410W per luminaire

The illuminance level is simulated by OPTILUME INTERIOR V3.03 by THORN LIGHTING LTD. And the results is summarised as below tables

( d ). Shops

Physical data of typical shop: 9m(L) x 6m(W) x 4m(H)

Data of luminaire: Manufacture = SIEMENS

Model no. = CR370SA-cent04-S70T

Type = Sealed-frame luminaire with dished triumph lens

Lamp = 70W high pressure sodium lamp (SON-T)

Lumen output = 5900lm

Mounting height of luminaire = 3.3m

By using lumen method,

According to CIBSE Lighting Guide LG1(The Industrial Environment), the standard illuminance is 500lx.

Assume UF = 0.5 & LLF = 0.7

nos. required

So, the decision is made to use N=15 luminaires. (3 columns of 5 pcs each).

The spacing between each luminaires = 1800mm

The power input = 15x87W = 1305W

Lighting power density = 24W/m²

The results is also simulated by manufacturer software SILICHT V4.1.

Appendix

Appendix - P.

Thanks a lot for giving a lot of informatin.