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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?
Thanks in advance for your help.
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 301m²
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.
3.1 answer:
Total load :
o Socket outlets -Floor area 301 m² = (5 kw for first 100 m²) + (1kw for second 100m²) +( 1kw for part of third 100m²) 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.
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?
Thanks in advance for your help.
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 301m²
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.
3.1 answer:
Total load :
o Socket outlets -Floor area 301 m² = (5 kw for first 100 m²) + (1kw for second 100m²) +( 1kw for part of third 100m²) 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.
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