VAA3001 ELECTRICAL POWER SYSTEMS Assignment Help
Loads on submains
Lighting and power load on each hospital floor are calculated from given data as follows:
Lighting and power non essential load = 28 VA / m2
Lighting and power non essential load per floor = 1650 * 28 / 1000 = 46.2 kVA
Lighting and power essential load = 24 VA / m2
Lighting and power essential load per floor = 1650 * 24 / 1000 = 39.6 kVA
There are some additional essential and non essential load as per given in the problem. Details of load in kVA are given in Figure 1. We calculate here maximum demand current for each floor as per kVA loads.
Essential Loads
Roof
Total kVA load on the Roof = 295 kVA
Total current required per phase for a 3 phase supply = A
Similar calculation for other floors can be done.
L5
Total kVA load = 39.6 kVA
Total current demand per phase = 57.15 A
L4
Total kVA load = 39.6 kVA
Total current demand per phase = 57.15 A
L3
Total kVA load = 39.6 kVA
Total current demand per phase = 57.15 A
L2
Total kVA load = 39.6 kVA
Total current demand per phase = 57.15 A
L1
Total kVA load = 39.6 kVA
Total current demand per phase = 57.15 A
B1
Total kVA load = 3.96 kVA
Total current demand per phase = 5.71 A
B2
Total kVA load = 78.96 kVA
Total current demand per phase = 113.96 A
B3
Total kVA load = 3.96 kVA
Total current demand per phase = 5.71 A
Non Essential Load
Roof
Total kVA load = 500 kVA
Total current demand per phase = 721.68 A
L5
Total kVA load = 166 kVA
Total current demand per phase = 239.88 A
L4
Total kVA load = 46.2 kVA
Total current demand per phase = 66.68 A
L3
Total kVA load = 46.2 kVA
Total current demand per phase = 66.68 A
L2
Total kVA load = 46.2 kVA
Total current demand per phase = 66.68 A
L1
Total kVA load = 46.2 kVA
Total current demand per phase = 66.68 A
B1
Total kVA load = 9.24 kVA
Total current demand per phase = 13.33 A
B2
Total kVA load = 9.24 kVA
Total current demand per phase = 13.33 A
B3
Total kVA load = 9.24 kVA
Total current demand per phase = 13.33 A
Sizing of Cables
Sizing is required for three cables i.e.
Mains
Total current for the essential & non-essential loads is 2157.70A (from section 1)
Derating factor for circuit of multi core cables insulated in perforated air = 0.98 (from Table-24 AS /NZS 3008.1)
Derating factor of ambient air temperature = 0.88 (from Table-27 AS/NZS 3008.1)
So the total derating factor = 0.98×0.88 = 0.8624
Let’s select 4 cables in parallel in each phase, then,
Current in a single conductor cable in each phase = 2157.70/4 = 539.425
Tabulated current capacity required = 539.425/0.8624 = 625.49Amp
Hence the cable conductor size for the mains will be 500 mm2 (from Table-10 AS /NZS 3008.1)
Conductor Size as per Voltage Drop across its length (which should not be more than 2.5 %)
Resistance of 500mm2 circular conductor multi-core cable at 45 ⁰C temperature R= 0.0486 Ω /km (as per AS/NZS 3008.1)
Reactance of 500mm2 circular conductor multi-core cable X= 0.0666 Ω/km
Cable length L = 20m (as per given data)
Let us assume that load power factor = 0.8
Then, Voltage drop
V= 1.473V
For a 400V system, the voltage drop is = (1.473/400)*100 = 0.36 %, which is less than 2.5 % and acceptable.
Lift cable size
Lift power = 120 kVA (given)
Required current of the lift = 1000x = 173.21A
Derating factor for circuit of single core cables insulated in perforated trays or air = 0.97 (from Table-23 AS /NZS 3008.1)
Derating factor of ambient air temperature = 0.88 (from Table-27 AS/NZS 3008.1)
Total derating factor = 0.97×0.88 = 0.8536
Let’s select single cables in parallel in each phase, then,
Current in a single conductor cable in each phase = 173.21A
Tabulated current capacity required = 173.21/0.8536 = 202.92Amp
So the conductor size for the mains will be 95 mm2 (From Table-20 AS /NZS 3008.1)
Conductor Size as per Voltage Drop across its length (which should not be more than 2.5 %)
As per AS/NZS 3008.1
Resistance of 95 sq-mm circular conductor multi-core cable at 45 ⁰C temperature R= 0.214 Ω / km
AC Reactance of 95 sq-mm circular conductor multi-core cable X= 0.0904 Ω / km
Cable length L = 45.2 m
Let us assume that load power factor = 0.8
Then, Voltage drop,
= 3.58V
For a 400 V system the voltage drop = 3.58/400*100 = 0.89 %, which is under 2.5 %
Carpark B2
In section1, I have already derived total current for essential load for park level B2 = 117.72A
Derating factor for circuit of single core cables insulated in perforated trays or air = 0.97 (from Table-23 AS /NZS 3008.1)
Derating factor of ambient air temperature = 0.88 (from Table-27 AS/NZS 3008.1)
Total derating factor = 0.97×0.88 = 0.8536
Let’s select single cables in parallel in each phase, then,
Current in a single conductor cable in each phase = 117.72A
Tabulated current capacity required = 117.72/0.8536 = 137.91A
So the conductor size for the carpark level B2 will be 50 mm2 (From Table-20 AS /NZS 3008.1)
Conductor Size as per Voltage Drop across its length (which should not be more than 2.5 %)
As per AS/NZS 3008.1
50 sq-mm circular conductor multi-core cable at 45 ⁰C temperature
R= 0.426 Ω / km
X= 0.0829 Ω / km
Slab to slab height of each floor = 4.2m (given)
Cable length L for the lift will be = 4.2 * 6 + 20 = 24.2 m
Let us assume that load power factor = 0.8
Then, Voltage drop,
= 1.927V
For a 400 V system the drop is 1.927/400*100 = 0.481 %, which is less than 2.5 %
Maximum demand calculation and fault level in parking area
Following calculations are required:
Maximum demand in parking area
Fault level at B2
Maximum demand calculations for the Car park level
Parking area has major loads of lighting and typical load is 3-10 VA/m2 depending on selection of illumination level and types of light used.
Let’s assume power consumption in parking area = 8.0 VA/
Total power consumption in parking area = 8.0×1650 = 13.2kVA/ floor
Total power consumption for parking area = 13.2 * 3 + 75 = 114.6 kVA
Let’s assume parking area have 30% essential and 70% non essential load,
Then non-essential power consumption = 13.2 * 70 / 100 = 9.24 kVA
And, essential power consumption = 13.2 * 30 / 100 = 3.96 kVA
Fault level at DB B2
Electrical Diagram for Fault at parking level B2]
Length of the cable from MSB to DB = 4.2 + 20 = 24.2meter
For a 3 phase system fault level current in each phase If =
Vphase = Phase Voltage
Z = Line Impedance
Vphase = 230.94 V
Lets assign conductor temperature = 75 DegC (maximum operating temperature)
For 6 square mm copper conductor,
Resistance Rs = 3.75 &
Reactance Xs = 0.104
Then
R = Rs x = 3.75 x 0.0242 = 0.09075 = 90.75 mΩ and
X = 0.104 x = 0.104 x 0.0242 =0.00251= 2.51 mΩ
Z =
Fault Current in each Phase If=
Battery sizing calculation
UPS load rating kVA = 120 kVA
Backup time = 4 hour
Let’s assume load power factor on UPS p.f. = 0.9
Let’s assume efficiency of inverter part in UPS = 0.96
And total UPS efficiency η= 0.93
Now, real power required from battery during backup time,
Total discharged energy from batteries = 112.5 * 4 = 450 kWh
Therefore, Total energy required to charge the battery = 450 kWh
Let us assume that as per the load shading conditions in hospital, total time available to charge the batteries = 15 hours
Power required for charging the batteries = 450 / 15 = 30 kW
Let’s assume 16 block of 12V batteries in series are used for storage.
Total battery voltage for series combination = 16*12 = 192V
For a AVR45-33 battery from Deka East Penn Manuf,
Watt per cell when discharging the battery in 4 hours = 299 W
Power from the battery = 299 * 6 = 1794 W
Number of strings required to provide total 450 kWh =
Therefore, total 4 strings of 16 no of AVR45-33 (12V) batteries in each string can be used.
