Building Design and Construction Handbook - Merritt - Ventech!

15.44 SECTION FIFTEEN
The voltage drop between phase wires can be obtained from Eq. (15.28), with
the area of No. 8 wire taken as 16,510 circular mils and length of wire as 110 ft:
�3 � 10.7 � 26.4 � 110
Vd � � 3.24 V
16,510
3.24 � 100
% voltage drop � � 1.56%
208
2. Feeder to Meter Bank on Sixth Floor.
Total load (24 apartments) � 24 � 5700 W � 136,800 W
Demand load: first 15,000 W at 100% � 15,000
Balance, 121,800 W at 50% � 60,900
Total demand load � 75,900 W
Assume a power factor of 90% and apply Eq. (15.26):
75,900
Current per phase � � 234 A
3 � 120 � 0.9
Minimum-size type RH wire for this current is 250 kcmil. This has to be tested
for voltage drop. For a one-way run of 150 ft from service switch to sixth-floor
meter bank and an area of 250,000 circular mils, Eq. (15.28) gives:
�3 � 10.7 � 234 � 150
Vd � � 2.60 V
250,000
The correction factor for ac resistance of 250-kcmil wire is 1.06. Application of
this factor yields a corrected voltage drop of 1.06 � 2.60 � 2.76 volts.
2.76 � 100
% voltage drop � � 1.33%
208
Then, the total voltage drop from service switch to apartment panel A is
1.56 � 1.33% � 2.89%
which exceeds the 2.5% maximum voltage drop allowed by the local code. It is
necessary, therefore, to increase the size of the meter bank feeder over the minimum
size required for the current.
To bring the total drop down to 2.5% the meter bank feeder drop must be
reduced to 0.94% or 1.96 volts. The required wire size may be found by proportion:
2.76
Required area � � 250,000 � 352,000 circular mils
1.96
The nearest larger wire size is 400 kcmil.
For computing the size of the neutral for carrying 234 A, the local code allows
a demand factor of 70% on lighting loads over 200 A. Hence, the net current in
the neutral is 200 � 34 � 0.70 � 223.8 A. Minimum-size wire is No. 4/0 RH.
And the size of conduit required for the feeder is computed as follows with the aid
of tables in the NFPA ‘‘National Electrical Code Handbook’’: