This tutorial discusses on the following topics of electrical design:
1. VOLTAGE SELECTION
2. TRANSFORMER SELECTION
3. H.V CABLE SELECTION
4. GENERATOR SELECTION
5. L.V CABLE SELECTION
6. INSTRUMENT TRANSFORMER AND METER SELECTION
7. GENERATOR PROTECTION AND TRANSFORMER PROTECTION
8. CAPACITOR BANK SELECTION
9. MSB BUSBAR SELECTION
10. FAULT CALCULATION AND EARTHING DESIGN
11. SINGLE LINE DIAGRAM DESIGN
For easiness, let us explain with the help of a detailed load scheme.
Consider you need to feed a factory which consists of the following electrical loads:
|
SL.NO. |
EQUIPMENT |
NO. OF EQUIPMENTS |
LOAD (HP) |
TOTAL LOAD(KW) |
LOAD (KVA) |
|
1 |
MOTOR |
1 |
200 |
149.2 |
165.78 |
|
2 |
MOTOR |
3 |
150 |
335.7 |
373 |
|
3 |
MOTOR |
5 |
100 |
373 |
414.44 |
|
4 |
MOTOR |
10 |
50 |
373 |
414.44 |
|
5 |
MOTOR |
10 |
25 |
186.5 |
207.22 |
|
6 |
MOTOR |
25 |
10 |
186.5 |
207.22 |
|
7 |
LIGHTING & POWER LOAD |
|
|
20 |
22.22 |
|
|
TOTAL |
2150 |
1623.9 |
1804.34 |
Let 10% be the future load, consider the demand factor as 2 and transformer efficiency as 80%, Percentage impedance of transformer= 5%, fault MVA as 250 MVA, and soil resistivity as 100ohm-meter, Let all the motor starters shall be Star-Delta, except the largest motors.
1 VOLTAGE SELECTION
Calculate
the total Maximum demand to select the voltage range.
Total load= 1804.34 +180.434= 1984.774KVA.
Maximum Demand= Total connected load/demand factor
=1804.34/2
=902.17 KVA
|
Supply
voltage |
Maximum
connected load |
Maximum
Contract Demand |
|
240v
(single phase) |
5
KW |
|
|
415v
(three phase) |
100
KVA |
100
KVA |
|
11KV |
|
3000
KVA |
|
22KV |
|
6000KVA |
|
33KV |
|
12000KVA |
|
66KV |
|
20000KVA |
|
110KV |
|
40000KVA |
|
220KV |
|
>40000KVA |
As the M.D is less than 3000KVA the supply voltage shall be 11KV
Voltage selection is the first step to Electrical Design
DO YOU NEED SINGLE PHASE 240V OR THREE PHASE 415V FOR YOUR HOUSE?
2 TRANSFORMER SELECTION
In
order to select a transformer, we need to go through 2 steps, and
Among which gives you the highest rating we have to select that rating for the transformer.
Step 1: SELECTION BASED ON MAXIMUM DEMAND
future load= 10% of the total load = 180.43.
Total Maximum Demand (M.D): = (1804.34+180.434)/2= 992.39 KVA
Required Demand: M.D/transformer efficiency = (992.39 KVA/80) *100
= 1240.49KVA
Selected Transformer: 1250 KVA.
Step II: SELECTION BASED ON MAXIMUM MOTOR CAPACITY
If motors
with larger capacity are available, then we have to consider the largest motor
into account in the selection of transformer.
KVA of the required transformer is (K*M+L) *Z/7
“K” is the motor starter constant, and the value depends on the type of starter used. The value is given
in the chart.
|
VALUE
OF K |
|
|
DOL |
7 |
|
STAR
DELTA |
2.5 |
|
AUTOTRANSFORMER/ROTOR
RESISTANCE |
2 |
|
SOFT
STARTER/VFD/VVVF |
1.5 |
M= Motor
capacity in HP = 200HP
L = Base Load = Required demand- largest motor capacity
= 992.39 - (165.78/efficiency)
= 992.39 - (165.78/.9)
= 808.19KVA
Z = % impedance of transformer= 5%
KVA of the
required transformer =(K*M+L) *Z/7
= (2.5*200+808.19) *5/7=934.42KVA
The available transformer sizes are 36,100,160,200,250, 315, 400, 500, 630, 750, 1000,1250, 1600, 2000 KV
So, the maximum
demand gives you a higher KVA value than the largest motor method. So, the selected
the transformer shall be 1250KVA, 3 phase, %Z = 5%, Delta-Star (Dyn11), ONAN, indoor, Offload tap changer.
3. H.V CABLE SELECTION
Selected
transformer capacity 1250KVA
Now find out the full load current, KVA=√3*V*I
= √3*11*I = 1250
I= 65.6A
The
short circuit current carrying capacity of 11 KV XLPE cable having aluminium
conductors can be
Calculated
from the following formula. A=11.1*Is*√t
A=
Area of the conductors, Is= Short circuit current, t= Duration of short circuit
current in seconds (1S)
How
to calculate the “Is”, get the fault MVA, which is 100MVA as given in the
question.
Is= Fault MVA/Line voltage= 250*1000/ (√3*11) =13122A
A =11.1*Is*√t=11.1*13122*√1
=145.65
~ 150sq.mm cable.
so, the selected cable shall be 150 sq.mm,1R,3C, Al, XLPE (A2XFY)
cable.
4. GENERATOR SELECTION
In generator
selection, we can follow the same steps as in transformer selection
Step I: SELECTION BASED ON MAXIMUM DEMAND
Maximum Demand
(M.D) = 1240.49KVA
So, as we did in
the transformer selection, based on M.D we can select 1250KVA
Step II: Selection Based on Maximum Motor capacity
Minimum
generator capacity = K*Largest motor capacity in HP
K= Motor starter
constant, largest motor capacity= 200HP
Minimum generator capacity = 2.5*200= 500KVA
So, considering
step I & II, can conclude that the selected generator size shall be
1250KVA,3 phase,
50Hz, 415V, Diesel Generator.
5. L.V CABLE SELECTION
Selected
transformer size: 1250KVA
Secondary
Current = (1250*1000)/ (433*1.732) = 1667 A
Note: select bus
bar trunking above 630 KVA
The current
density of the Aluminium bar shall not exceed 0.8A/sq.mm and that of copper is
1.2A/sq.mm
so, 1667/1.2
=1389, so the selected bus bar size is
2Rx75x10 Copper bus bar
As the generator
is of the same size, the selected busbar for LV cable shall be used for the
generator too
LV CABLE SELECTION, ELECTRICAL DESIGN-05
6. PROTECTION
Transformer primary protection
The protection includes VCB/GCB with 2 O/C (over current) relays with
high set elements, one instantaneous earth fault relay, 1 Buchholz relay, oil,
and winding temperature high alarm and trip.
As the selected transformer is 1250KVA, let us select 800A, 12KV, 3
Pole VCB for primary protection. (800A VCB/GCB to be used for transformers with capacity 1000KVA,
1250KVA,1600KVA).
Transformer secondary protection
The protection includes ACB draw-out type with 3 O/C and 1 E/F relay,
standby law set E/F protection using CT at the neutral earthing conductor. REF
protection relay with the primary trip. If the breaker is having
microprocessor-based programmable releases having different time settings
separate O/C and E/F relay may not be provided. But law set E/F relay and REF
relay to be provided.
The secondary current of the transformer is 1667A, so the selected ACB
is 2000A, 415V, 25KA,4 pole, with inbuilt 3O/C, 1 E/F release, standby law set
E/F protection using CT at the neutral earthing conductor (the size of neutral
C.T shall be 20% of the secondary current so 0.2 X 1667=335,
So the selected C.T shall be 400/5A).
Generator protection
ACB with thermal overload, voltage-controlled O/C relay, over voltage, under
voltage, negative sequence, law set stand by earth fault relays and
REF/differential relay with fuel shutoff facilities. Over speed protection
shall be provided for the engine.
2000A, 415V, 25KA,
4pole ACB, with inbuilt 3O/C, 1 E/F release, and standby law set E/F protection
using CT at the neutral earthing conductor (the size of neutral C.T shall be
20% of the secondary current
So, 0.2 X 1667=335,
So, the selected C.T shall be 400/5A.
8. Metering
For metering it is essential to install the C.T in both primary and
secondary side of the transformer and P.T shall be provided in the primary side
only. C.T selection for 11KV side,
Primary side current = 65.6A but considering the maximum demand (902.17KVA) so the current demand in primary side will be 902.17x1000/1.732x433
=
9. Fault Calculation and Earthing Design
The parameters
we require for the Fault calculation and Earthig Design are:
Soil resistivity (ρ), voltage (v), Fault MVA
For the Transformer:
Fault Current Calculation for 3 seconds = (Fault MVA / Voltage)
= 250*106/√3*11*103
=
13.12KA
Permissible
current density = 7.57*1000/ √(ρ*t),
t= time of the fault
=
437 A/sq.m
Area of plate required at 11 KV side,
= 13.12*103/437
=30
sq.m
Available Size of plate electrode = 1.2 M *1.2 M * 2
=
2.88 Sq.m
Total no. of plate electrodes required = 30/2.88
= 11 Nos
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