substation design example

7/29/2019 Substation Design Example

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-: Design Example :-

Design a Single Phase Bus Support for a Substation inNagpur, given the following information,

Height of Bus Centerline above foundation = 5.5 m

Schedule 40 aluminum bus = 100 mm (mass = 5.51 kg/m)

Maximum Short Circuit force = 550 N/m

Short Circuit reduction factor = 0.66

Bus Support Spacing = 6.0 m

Insulator Height (hi) = 2.0 m

Insulator Diameter (Di) = 0.28 m

Insulator Weight (Wi) = 140 kg

Basic Wind Speed (Vb) = 33 m/sec (Zone = 1) Reliability level = 2 (Return period of design loads 150 yrs)

Terrain Category = 2


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LATTICE STRUCTURE DETAILS:-

Main Leg = ISA65x65x6 @ 5.8kg/mBracing,

Inclined = ISA45x45x5 @ 3.4kg/m

Plan = ISA45x45x5 @ 3.4kg/m

Part of Structure = 04

Each Part Length = 850 mm

Inclined Length = 931 mm

Back to Back of Str. = 380 mm


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Short-Circuit Loading:-

Fsc = 6.0 m x 0.66 x 550 N/m Fsc = 2178 N

Mom @ Base = 5.5 m x 2178 N

Mom @ Base = 11979 N.m


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Wind Loading :- IS 802 (Part 1 / Sec 1) :1995,

Basic wind speed Vb = 33 m/sec

Metrological Reference wind speed VR is, VR = Vb / K0 , where K0 = 1.375 (cl. 8.2 pg 3)

VR = 33 / 1.375 = 24 m/sec

Design wind speed Vd = VR x K1 x K2, Where K1 = Risk Coeff. (cl. 8.3.1)

K1 = 1.08 (Table 2)

Where K2 = Terrain roughness coeff. (cl.8.3.2)

K2 = 1.00 (Table 3)

Design wind speed Vd = 24 x 1.08 x 1.0 = 25.92 m/sec


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Wind Loading :-Design wind pressure Pd = 0.6 Vd2 (cl. 8.4)

Pd = 0.6 x 25.92 x 25.92 = 403.11 N/sq.m

==============================================

Wind load on Conductor Fwc (cl 9.2)

Fwc = Pd x Cdc x L x d x Gc,

Where,

Cdc = Drag coeff, taken as 1.0 for conductor

L = wind span, being sum of half the span on either side of

supporting point in meters

d = diameter of cable / tube

Gc = gust response factor (Table 7) = 1.83

Fwc = 403.11 x 1.0 x 6 x 0.1 x 1.83

Fwc = 442.615 N (737.7 N/sq.m)


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Wind Loading :-

Wind load on Insulator Strings (Fwi) :- (cl 9.3)

Fwi = Cdi x Pd x Ai x Gi

Where,

Cdi = drag coeff to be taken as 1.2

Ai = 50 % of the area of insulator string projected ona plane which is parallel to the longitudinal axis of

the string

Gi = Gust response factor (Table 6) = 1.92 Fwi = 1.2 x 403.11 x 0.5 x 2 x 0.28 x 1.92

Fwi = 260.1 N (928.8 N/sq.m)


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Wind on structure Fwt = Pd x Cdt x Ae x Gt

Where,

Cdt = Drag coeff for panel under considerationagainst which the wind is blowing . Values of Cdtfor different solidity ratios are given in Table 5.

Solidity ratio () is equal to the effective area of

a frame normal to the wind direction divided bythe area enclosed by the boundary of the framenormal to the wind direction.

Solidity ratio () = Aeff / Agross

Ae = Total net surface area of the legs, bracing,cross arms and secondary members of thepanel projected normal to the face in m2

Gt = Gust response factor, Table 6

Wind Loading :-


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Solidity Ratio () = Ae / Ag

= 0.678 / 1.292 = 0.525

Drag Coeff Cdt = 2.0 (Table 5)

Gt = 1.92 (Table 6)Fwt = 403.11 x 2 x 0.678 x 1.92

Fwt = 1049.50 N (1547.92 N/sq.m)

0.678Total Net Surface Area Ae =

0.0680.0450.3843,5,7 & 9

1.292Gross Surface Area Ag = 3.4 m x 0.38 m

0.1680.0450.93142,4,6 & 8

0.4420.0653.421

Area

(sq.m)

Width

(m)

Length

(m)

NosMember


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Wind load summary :-

5363.001752.22Totals

1784.151.71049.5Wind on Structure (Fwt)

1144.444.4260.1Wind on Insulator (Fwi)

2434.415.5442.62Wind on Bus (Fwc)

Moment @

Base (N.m)

Lever

Arm (m)

Force

(N)

Description


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Earthquake Loading:-

IS 1893 (Part 1) : 2002

Design Horizontal Acceleration Coeff (Ah),

Ah = Z I / 2R * Sa/g

Z = Zone factor = Zone II = 0.10 (Table 2)I = Importance factor = 1.5 (Table 6)

R = Response reduction factor = 4 (Table 7)

Sa/g = Avg. response acceleration coeff for rock or soil

sites as given by fig 2 & Table 3 based onappropriate natural periods and damping of thestructure.

Fundamental Natural Period Ta = 0.085 h0.75

Ta = 0.085 x 3.40.75 = 0.213 secSa/g = 2.5

Hence, Ah = 0.10 x 1.5 x 2.5 / 2 x 4 = 0.047


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Earthquake Loading:-IS 1893 (Part 1) : 2002

Design Seismic Base Shear VB = Ah W

W = Seismic weight of the structure

508.92162.78Totals

141.081.7180kg x 9.81 x 0.047 =

82.99

EQ on Structure

284.024.4140kg x 9.81 x 0.047 =

64.55

EQ on Insulator

83.825.56m x 5.51 kg/m x 9.81

x 0.047 = 15.24

EQ on Bus

Moment @

Base (N.m)

L.A

(m)

Force (N)Description


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Results:-

SCF + WL :-

Force = 2178 + 1752.22 = 3930.22 N

Moment @ Base = 11979 + 5363 = 17342 N

SCF + EQ :-

Force = 2178 + 162.78 = 2340.78 N

Moment @ Base = 11979 + 508.92 = 12487.92 N

The combined loading of wind and short-circuit forcesproduce the greatest forces and moment at the base

design for this condition.Therefore seismic forces are not critical for thisstructure.


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Forces in the Member:-

Moment at the base causes tension andcompression in the chord angles.

C = Tensile or Compressive force

C = 17342 N.m / [ 2(0.380 2 x 0.0181)]

C = 25221.1 N per leg

P = Applied loadP + C = [140 x 9.81 + 6 x 54] / 4 + 25221.1

P + C = 25645.45 N per leg

Forces in bracing member shall be 25 % of the legmember.


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Thank you..

substation design example

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