1. ## Estimating load with known diversity factors

Good day all.

I am going through old installation rules exam papers with no answers provided, and I attempmted to do one of the calculations as per topic title.

Please could you have a look at this for me, and comment. I am a bit stumped by part two of the question as well, I cannot see another method of checking the result in the code.
Also the max allowable loading of the supply transformer is 90% correct?

3.1 Calculate from the following information, the maximum amount of water heaters of 1kW each that may be installed according to the regulations:
o 10kVa/220 v single phase transformer using a power factor of 0.95
o Floor area 301mÂ²
o Two 0.75kW single phase motors
o One electric oil heater of 500w
o 20 lights

3.2 Test your answer by making use of any other alternative method.

o Socket outlets -Floor area 301 mÂ² = (5 kw for first 100 mÂ²) + (1kw for second 100mÂ²) +( 1kw for part of third 100mÂ²) TOTAL = 7kw x diversity of .5 = 3.5kw
o Motors = (.75kW x 2 ) = 1.5kW x diversity factor of 1 =1.5kW
o Electric oil heater = 0.5kW x diversity of .5 = .25kW
o 20 lights ( 20 x 60watts ) = 1200watts = 1.2kw x diversity of .5 = .6kW
o
o TOTAL LOAD ( 3.5kW+1.5kW+0.25kW+.6kW) = 5.85kW

Therefore total current required = 5.85kW / 220v = 5850 /220 = 26.6A

Max supply from transformer at full load (10000 *.95) /220 = 43.18 amps.

43.18 max load x 90% = 38.86amps allowable.

Therefore remaining amperage is 38.86-26.6 = 12.26Amps.

Each heater requires: 1kw
Therefore amps required per is 1kw/220 1000/220= 4.54 amps x diversity factor of 1 = 4.54 Amps.

Therefore total amount of water heaters allowed is 12.26amps/4.54 amps = 2.7.

To check total load with two connected : 26.6amps + (4.54 x 2) = 35.68
To check total load with three connected 26.6amps + (4.53 x 3) = 40.19

Therefore to enable the transformer to be loaded at 90% max only two may be added.  Reply With Quote

2. ## Thanks given for this post:

mikilianis (12-Oct-10)

In other papers they ask you to use the current (I) method to test your answer, so I am thinking that you need to do the calculations in KW by not converting to amps, and then convert to amps to check.  Reply With Quote

4. Can I respectfully enquire from where you derive/determine the diversity factors used in your calculations?  Reply With Quote

5. I get the factors from page 302 of the sans 10142  Reply With Quote

6. I could well imagine cases where most, if not all of the power-points are operating together. No sense in using 0.5 diversity at those times, I'd expect. Surely that situation could/would leave you with an under-designed system?  Reply With Quote

7. The standards state, and the way this question is phrased state that you should take the loading of the socket outlets if unknown, to be as follows (annexD)

5kw for the first 100m sq and thereafter 1kw per 100m sq or part thereof.

then mutliply the result by the diversity factor to get potential load.  Reply With Quote

8. Ok the lecturer got back to me. This is the way it must be answered, In table form.
NOTE I got the calculation of the sockets wrong there is an extra 100m sq to be factored in ( for the 1m)  Reply With Quote

9. Morning Maccie. Pls foward me infprmation on handling and installation of cables not exceeding 33kv if u have. Peace  Reply With Quote