Electrical Design

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  • Published : December 17, 2012
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Electrical Design
Mini Project–Stage2

Part A :ElectricalLoading estimation
Total Tenants load of flat : 41x8X6.5K=2132KVA
Load of car park :1000+0.03K=30KVA
Total loading of landlord supply
Loading| no| Total Power| Designed current for one component | Total Designed current | MCB rated Current| Passenger Lifts| 2| 30kW| 15k/(0.9 X 0.85 X 380 X √3)= 29.79A | 59.58A| 63A| Fireman lift| 1| 15kW| 15k/(0.9 X 0.85 X 380 X √3)= 29.79A| 29.79A| 32A| Fresh water up-feed pumps| 2| 74kW| 37k/(0.9 X 0.85 X 380 X √3)= 73.48 A| 146.97A| 160A| Flush water up-feed pumps| 2| 22kW| 11k/(0.9 X 0.85 X 380 X √3)= 21.85 A| 43.69A| 60A| Fresh water booster pumps| 2| 8kW| 4k/(0.9 X 0.85 X 380 X √3)= 7.95A| 15.89A| 20A| Sprinkler pumps| 2| 44kW| 22k/(0.9 X 0.85 X 380 X √3)= 43.70A| 87.39A| 100A| Sprinkler jockey pumps| 1| 2.2kW| 2.2k/(0.9 X 0.85 X 380 X √3)= 4.37A| 4.37A| 6A| Lighting| --| 20kW| 20k/(0.9 X 0.85 X 380 X √3)=39.72A| 39.72A| 40A| Small power| --| 10kW| 10k/(0.9 X 0.85 X 380 X √3)= 19.86A| 19.86A| 20A| Fixed fire pumps| 2| 37kW| 37k/(0.9 X 0.85 X 380 X √3)= 73.49A| 146.97A| 160A| Fixed fire jockey pumps| 1| 2.2kW| 2.2k/(0.9 X 0.85 X 380 X √3)= 4.37A| 4.37A| 6A| Total loading of landlord supply

Assume the power factor is 0.8 and efficiency is 0.9 and overall demand factor is 0.8: Thus, the total loading of landlord is : [45k+74k+22k+8k+44k+2.2k+74k+2.2k+10k)(0.8x0.9)+30K](0.8) =186.0864KVA Therefore, the total electrical loading estimation is 186.0864+2132k=2318.0864KVA SO, two 1500 KVA transformer is required.

Part B : Cable Sizing
For the R/F
The Fresh Booster pump room:
* Assume: height:5.5m,Ca = 0.71(ambient temp 50o c), Cg = 0.1 , total length=120m The current demand Ib= 7.9A
Soothe fuse rating is In=10A
It = 4k/(0.9 X 0.85 X 380 X √3X0.71X1)=11.19A
From Table A6 (4),
From Table A6 (4), 1.5mm2 4/c PVC/SWA/PVC cu. Cable should be selected Checking the voltage drop 11.19x120x25x0.001 = 33.57V> 380 x 0.04= 15.2V ,thus it is not acceptable.

To choose another cable ,From Table A6 (4), 2.5mm2 4/c PVC/SWA/PVC cu. Cable should be selected Checking Voltage drop = 11.19 x 120 x 15 x 0.001= 20.14V>15.2V , thus it is not acceptable.

To choose another cable , From Table A6 (4), 4mm2 4/c PVC/SWA/PVC cu. Cable should be selected CheckingVoltage drop = 11.19 x 120 x 9.5 x 0.001= 12.76V <15.2V ,thus it is acceptable , 4mm2 4/c PVC/SWA/PVC cu. Cable should be selected

The Lift Machine Room
There are 2 passenger lift and one fireman lift
* Assume: height: 5.5m, Ca = 0.71(ambient temp 50o c), Cg = 0.1 , total length=120m The current demand Ib= 59.58+29.79 =87.37A
So, the fuse rating is In=100A
It = 45k/(0.9 X 0.85 X 380 X √3X0.71X1)=125.88A
From Table A6 (4), 35mm2 4/c PVC/SWA/PVC cu. Cable should be selected Checking Voltage drop = 125.88 x 120 x 0.81 x 0.001= 12.23V <15.2V ,thus it is acceptable,35mm2 4/c PVC/SWA/PVC cu. Cable should be selected.

Generator Room:
* Assume: height: 5.5m, Ca = 0.71(ambient temp 50o c), Cg = 0.1 , total length=120m The current demand Ib = 492.11A
So, the fuse rating is In=500A
It = 247.78k/(0.9 X 0.85 X 380 X √3X0.71X1)=697.67 A
From Table A6 (5), 240mm2 Single core XLPE insulated cables should be selected Checking Voltage drop = 697.67 x 120 x 0.22x 0.001= 18.41V > 380 x 0.04= 15.2V, thus it is not acceptable. To choose another cable, From Table A6 (5), 300mm2 Single core XLPE insulated cables should be selected Checking Voltage drop = 697.67 x 120 x 0.195 x 0.001= 16.33V >15.2V, thus it is acceptable ,

To choose another cable , From Table A6 (5), 400mm2 Single core XLPE insulated cables should be selected CheckingVoltage drop = 697.67 x 120 x 0.175 x 0.001= 14.65V >15.2V, thus it is acceptable , 400mm2 Single core XLPE insulated...
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