6.2.3 -  Applied diversity

Apart from indicating that diversity and maximum demand must be assessed, the
Regulations themselves give little help. Suggestions of values for the allowances
for diversity are given in {Table 6.2}.
Distribution boards must not have diversity applied so that they can carry the total load connected to them.

Example 6.1
A shop has the following single-phase loads, which are balanced as evenly as possible across the 415 V three-phase supply.

2 x 6 kW and 7 x 3kw thermostatically controlled water heaters
2 x 3 kW instantaneous water heaters
2 x 6 kW and 1 x 4 kW cookers
12 kW of discharge lighting (Sum of tube ratings)
8 x 30 A ring circuits feeding 13 A sockets.

Calculate the total demand of the system, assuming that diversity can be applied. Calculations will be based on {Table 6.2}.

The single-phase voltage for a 415V three-phase system is 415/Ö 3 = 240 V.

All loads with the exception of the discharge lighting can be assumed to be at unity power factor, so current may be calculated from

l =	P
	U

Thus the current per kilowatt will be

1000 A =	4.17	A
	240

 

Table 6.2 - Allowance for diversity
Note the following abbreviations : X is the full load current of the largest appliance or circuit Y is the full load current of the second largest appliance or circuit Z is the full load current of the remaining appliances or circuits
Type of final circuit		Type of premises
	Households	Small shops, stores, offices	Hotels, guest houses
Lighting	66% total demand	90% total demand	75% total demand
Heating and power	100% up to 10 A + 50% balance	100%X + 75%(Y+Z)	100%X + 80%Y + 60%Z
Cookers	10 A + 30% balance + 5 A for socket	100%X + 80%Y + 60%Z	100%X + 80%Y + 60%Z
Motors (but not lifts)		100%X + 80%Y + 60%Z	100%X + 50%(Y+Z)
Instantaneous water heaters	100%X + 100%Y + 25%Z	100%X + 100%Y + 25%Z	100%X + 100%Y + 25%Z
Thermostatic water heaters	100%	100%	100%
Floor warming installations	100%	100%	100%
Thermal storage heating	100%	100%	100%
Standard circuits	100%X + 40%(Y+Z)	100%X + 50%(Y+Z)	100%X + 50%(Y+Z)
Sockets and stationary equip.	100%X + 40%(Y+Z)	100%X + 75%(Y+Z)	100%X + 75%Y + 40%Z

Water heaters (thermostatic)

No diversity is allowable, so the total load will be:

(2 x 6) + (7 x 3) kW = 12 + 21kw = 33kw

This gives a total single-phase current of I = 33 x 4.17 = 137.6 A

Water heaters (instantaneous)

100% of largest plus 100% of next means that in effect there is no allowable diversity.
Single-phase current thus	= 2 x 3 x 4.17 =	25.0 A

 

Cookers

100% of largest	= 6 x 4.17A =	25.0 A
80% of second	= 80 x 6 x 4.17A =	20.0 A
	=100
60% of remainder	= 60 x 4 x 4.17 A =	10.0 A
	=100
Total for cookers	=	55.O A

Discharge lighting

90% of total which must be increased to allow for power factor and control gear losses.
Lighting current	= 12 x 4.17 x 1.8 x 90 =	81.1 A
	100

Ring circuits

First circuit 100%, 50 current is 30 A
75% of remainder	= 7 x 30 x 75 =	157.5 A
	100
Total current demand for ring circuits =		187.5 A
Total single phase current demand =		486.2 A
Since a perfect balance is assumed, three phase line current =		486.2 A
		=3
	=	162 A