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 singlephase loads, which
are balanced as evenly as possible across the 415 V threephase
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 singlephase voltage for a 415V threephase
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
Thus the current per kilowatt
will be
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 singlephase 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. 
Singlephase 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 
