circular sump-oil drain

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14.10.2010 R0 First Issue SDC SIVEN P SKL

DATE REV. NO. DESCRIPTION esigne Checked Approved

REVISIONS

AREVA T & D

CLIENT

MSETCLSS1B/SS2B PACAKAGE

PROJECT :

220/132/33KV SUBSTATION

JOB No.5257PN902/903

TOTAL NO. OF PAGES TITLE :

NAME SIGN DATE

Design of Circular Water Sump (Circular type)

DSGN SDC 31.05.10

CHKD IVEN 01.06.10

APPD SKL 07.06.10

REV.

DOC. N 5 2 5 7 P N 9 0 2 - C M D - C - S Y D - C A LR0

- 0 7 1 5

RELEASED FOR PRELIMINARY TENDER INFORMATION P APPROVAL CONSTRUCTION

T h i s d o c u m e n t i s t h e p r o p e r t y o

f A R E V A T & D I N D I A L T D a n d m u s

t n o t b e p a s s e d o n t o a n y t h i r d p e r s o n o r f i r m

n o t a u t h o r i s e d

b y u s , n o r b e c o p i e d / m a d e u s e o f i n f u l l o r p a r t b y s u c h p e r s o n o r f i r m

w i t h o u t o

u r p r i o r p e r m i s s i o n w r i t i n g



DESIGN OF CIRCULAR SUMP



AREVA T& D INDIA LIMITED

PROJECT 400/220/132 KV SUBSTATION, Nawada ( FARIDABAD)DOCUMENT NO. DATE

5257PN088-NWD-C-SYD-CAL-0715 14.10.2010

TITLE Design of Circular Water Sump (Circular type)DES CHKD APP

SKL ASL SKL

Calculation Total Depth of the Sump

`

Maximum oil in largest unit = 50 MT

Density of Oil = 1

Volume of oil = 50 Cum

Add 25% of total oil to account Rain water = 25% x 50 = 12.5 Cum

Add 25% of total oil towards fire fighting water = 25% x 50 = 12.5 Cum

Net Capacity of the sump required = 50 + 12.5 + 12.5 = 75 Cum

Provide a circular sump of diameter (Internal) = 6.8 m

Elevation of Bottom of pipe at the entry of sump (h1) = 2.5 m

Net Depth required to store entire Oil (h2) = 75

0.79 x 6.8 x 6.8

= 2.07 m

Total depth of the sump from FGL = h1 + h2

= 2.5 + 2.07

= 4.57 m

Say 6.1 m



Cap(m3 Dia Area Ht req FB Hoop Bars Vertical steel

20 4 12.57 1.59 0.15 1.74 1.75 0.15 0.15 0.15

25 4 12.57 1.99 0.15 2.14 2.14 0.15 0.15 0.15

30 4 12.57 2.39 0.15 2.54 2.54 0.2 0.2 0.240 5 19.63 2.04 0.15 2.19 2.19 0.2 0.2 0.2

50 5 19.63 2.55 0.15 2.70 2.70 0.2 0.2 0.2

Total clear ht

Roundu p

Wallthicknes

RaftThickness

TopSlabThickness




PROJECT SS1B & SS2B PACKAGE MSETCLDOCUMENT NO. DATE

10/14/2010

TITLE Design Of circular water Tank DESIGNED CHECKED APPROVED

SDC SIVEN P SKL


INPUT DATA

1) Diameter of well (D) = 4 m

2) Density of soil, ϒ = 1.76

3) Thickness of wall ( t ) at bottom = 0.25 m

4) = 30 °

5) Depth of well. (H) = 1.8 m

6) Density of Oil 1.00 t/m3

7) Density of Water 1.00 t/m3

8) Internal radius of the well r = 2 m

9) Maximum Permissible stress in direct Tension (Concrete) = 1.3 As per IS 3370 table 1 clause 3.3.1

10) Maximum Permissible stress in bending Tension (Concrete) 1.8 As per IS 3370 table 1 clause 3.3.1

11) Maximum Permissible stress in steel (as per IS 3370) = 150 N/mm2

12) Unit weight of Concrete = 2.5

13) Grade of concrete M 25 N/mm2

14) Grade of reinforcement steel Fe 415 N/mm2

Design coefficient

15) Permissible bending stress in conc 9 N/mm2

16) Modular Ratio (m) 10.37

17) Nutral axis depth factor (k) 0.38

18) Lever arm ( j ) 0.87

A Design of Wall

A well lining should be design considering the well as a special case of a thick cylinder,

the principle stresses developed are hoop stress & radial stress. Both hoop stress and radial stressdeveloped in awell are principl stresses and compressive in nature. The hoop stresses devloped is the

maximum and is twice the radial stress.Hence it is considered for the purpose of design.

Wall of the tank with fixed base will be subjected to bending moments also in addition to hoop tension

Moment shall be maximum at the base of the tank and for coefficient table 10 IS 3370

Asuume total thickness of the wall at the base = 150mm

Effective thickness of the wall 9 cm

As bending stress will reduce from bottom to top, thickness of the wall can be reduced as it moves up.

provide thickness of wall from top to 2.0 m below FGL. = 150 mm

provide thickness of wall from 2.01 to 4.0 m below FGL. = 150 mm


H^2 = 3.06 = 5.1

D x t 4 x 0.15

Max Bending Moment = k1 * w * H^3 Kg-m/m

Refer Table 10 IS 3370 IV k1 = 0.01 (at Baseof the tank)

Max bending Moment = 0.01 x x 1000 x 5.36

= 42 Kgm/m

t/m3

Angle of repose (φ)

N/mm2

N/mm2

t/m3





10/14/2010


SDC SIVEN P SKL

Max stress due to bending = 6 * M 6 * 4200 =

b x d^2 100 x 81

= 3.11 Kg/sqcm < 18 O.K.

P P

P

Hoop stress , = P*D*k2

2

for k2 refer table 9 of IS 3370 part IV

When Tank is empty P shall be due to earth pressure

P = 0.33 x 0 x 175

= 0.1 Kg/Cm2

When Tank is Full and there is no Soil outside (Density of Water is considered for calculation purpose)

P = 1 x 0 x 175.0= 0.18 Kg/Cm2

Maximum Pressure = 0.18K g/Cm2

k2 for H^2/Dt of 14 = 0.477

Hoop tension (T) at Base = 0.18 x 400 x 0.477

2

= 16.7 Kg/cm

Max direct tension in conc (T/thk) = 16.7

9

= 1.86 Kg/sq.cm < 13 O.K.

Check for Acutal direct tension + Actual bending stress < 1

Per. Direct tension per. Bending stress

1.86 + 3.11 = 0.32 < 1 O.K.

13 18

Reinforcement Steel

στmax

r r1





10/14/2010


SDC SIVEN P SKL

1) Area of hoop reinforcement = 16.7 x 1001500

= 1.2 Cm2/ meter (both faces)

At each face 60 mm2

Provide reinf. both way @ 230 m c/c of dia 8 mm

Actual area of steel provided = 2.18 cm2 > 0.6 O.K.

2) Vertical reinforcement steel Ast' 420000

Ast = = 150 x 0.87 x 90

= 35.67 sq. mm

Minimum Reinforcement required = 0.3% on both the faces and hence 0.15% on each face.

= 0.15 x 9 = 1.35 cm2 per meter

Provide reinf. both way @ 210 mm c/c of dia 8 mm

Actual area of steel provided = 2.39 cm2 > 1.35 O.K. (On both the faces.)

B. Design of Base slab

Provide same thickness as of wall 15 cm

provide minimum reinforcement steel 0.24% on both faces

= 0.24 x 15 x 100100

= 3.6 cm2

Area of steel required at each face = 1.8 cm2

Provide bottom reinf. both way @ 193 mm c/c of dia 8 mm

Actual area of steel provided = 2.60 cm2 > 1.8 O.K. (On both faces and bothways)

B. Design of Roof slab

Assume thickness of roof slab = 15 cm

Load Calculations

1) Self weght of slab = 2500 x 0.15 = 375 Kgs

2) Plaster and finishes = 2400 x 0.015 = 36 Kgs

3) Live Load = = 350 Kgs

Total 761 Kgs

Say 7.5 KN

Maximum Radial Moment = 1 x W x r^2

16

B.M./(σst*j*d)





10/14/2010


SDC SIVEN P SKL

= 2.37 KN-m

Maximum Circumferential Moment = 2 x W x r^2

16

= 4.75 KN-m

Depth of the slab required = sqrt(Mmax/R.b)

= 73.69 mm

Provide total thickness of the slab = 150 mm

Effective thickness of the slab = 150 - 25 - 5 = 120 mm

Mu/bd2 (Radial) = 0.16Corrseponding % of steel (pt) = 0.05

Minimum % of steel pt,min 0.12

Area of steel required Ast 1.44 cm2


Area provided 2.51 > 1.44 O.K.

Provide 8 mm dia r/f @ 200 mm c/c bothways at bottom of slab

Mu/bd2 (Radial) = 0.33

Corrseponding % of steel (pt) = 0.09



Radial moment will become zero at sqrt(2/3xRadius) = 1.63 from center OR from edge = 2.25 - 1.63 = 0.62

Hence Point of inflection from face of support = 0.62 - ### = 0.37 m

Distance upto which -ve r/f shall be provided should be greater of the following

1) Development length (50 d) = 50 x 8 = 400

2) = ### + 150 = 517

3) ### + 96 = 463


Area provided 2.51 > 1.44 O.K. upto 750 mm from the face of support




PROJECT /220/132 KV SUBSTATION, Pimpalgaon (Nashik)DOCUMENT NO. DATE

10/14/2010


SDC SIVEN P SKL


INPUT DATA




4) = 30 °










Design coefficient




18) Lever arm ( j ) 0.87

A Design of Wall


the principle stresses developed are hoop stress & radial stress. Both hoop stress and radial stress

developed in awell are principl stresses and compressive in nature. The hoop stresses devloped is themaximum and is twice the radial stress.Hence it is considered for the purpose of design.









H^2 = 4.84 = 8.07

D x t 4 x 0.15




= 156 Kgm/m


t/m3


N/mm2

N/mm2

t/m3





10/14/2010


SDC SIVEN P SKL

b x d^2 100 x 81

= 11.56 Kg/sqcm < 18 O.K.

P

P P

PHoop stress , = P*D*k2

2



P = 0.33 x 0 x 220

= 0.13 Kg/Cm2


P = 1 x 0 x 220.0

= 0.22 Kg/Cm2


k2 for H^2/Dt of 14 = 0.575


2

= 25.3 Kg/cm


9

= 2.81 Kg/sq.cm < 13 O.K.



2.81 + 11.56 = 0.86 < 1 O.K.

13 18

Reinforcement Steel

στmax

r r1





10/14/2010


SDC SIVEN P SKL

1) Area of hoop reinforcement = 25.3 x 100

1500= 1.7 Cm2/ meter (both faces)

At each face 85 mm2




Ast = = 150 x 0.87 x 90

= 132.5 sq. mm


= 0.15 x 9 = 1.35 cm2 per meter






= 0.24 x 15 x 100

100= 3.6 cm2






Load Calculations




Total 761 Kgs

Say 7.5 KN


16

= 2.37 KN-m

B.M./(σst*j*d)





10/14/2010


SDC SIVEN P SKL


= 4.75 KN-m


= 73.69 mm




Corrseponding % of steel (pt) = 0.05Minimum % of steel pt,min 0.12









Radial moment will become zero at sqrt(2/3xRadius) = 1.63 from center

OR from edge = 2.25 - 1.63 = 0.62




2) = ### + 150 = 517

3) ### + 96 = 463







10/14/2010


SDC SIVEN P SKL


INPUT DATA




4) = 30 °










Design coefficient




18) Lever arm ( j ) 0.87

A Design of Wall












H^2 = 6.76 = 8.45

D x t 4 x 0.20




= 257 Kgm/m


t/m3


N/mm2

N/mm2

t/m3





10/14/2010


SDC SIVEN P SKL

b x d^2 100 x 196

= 7.87 Kg/sqcm < 18 O.K.

P

P P


2



P = 0.33 x 0 x 260

= 0.15 Kg/Cm2


P = 1 x 0 x 260.0

= 0.26 Kg/Cm2


k2 for H^2/Dt of 14 = 0.608


2

= 31.62 Kg/cm


14

= 2.26 Kg/sq.cm < 13 O.K.



2.26 + 7.87 = 0.61 < 1 O.K.

13 18

Reinforcement Steel

στmax

r r1





10/14/2010


SDC SIVEN P SKL

1) Area of hoop reinforcement = 31.62x 100


At each face 110 mm2




Ast = = 150 x 0.87 x 140

= ### sq. mm


= 0.15 x 14 = 2.1 cm2 per meter






= 0.24 x 20 x 100

100= 4.8 cm2






Load Calculations




Total 761 Kgs

Say 7.5 KN


16

= 2.37 KN-m

B.M./(σst*j*d)





10/14/2010


SDC SIVEN P SKL


= 4.75 KN-m


= 73.69 mm














OR from edge = 2.25 - 1.63 = 0.62




2) = ### + 150 = 517

3) ### + 96 = 463







10/14/2010


SDC SIVEN P SKL


INPUT DATA




4) = 30 °



8) Internal radius of the well r = 2.5 m







Design coefficient




18) Lever arm ( j ) 0.87

A Design of Wall












H^2 = 4.84 = 4.84

D x t 5 x 0.20




= 235 Kgm/m


t/m3


N/mm2

N/mm2

t/m3





10/14/2010


SDC SIVEN P SKL

b x d^2 100 x 196

= 7.19 Kg/sqcm < 18 O.K.

P

P P


2



P = 0.33 x 0 x 220

= 0.13 Kg/Cm2


P = 1 x 0 x 220.0

= 0.22 Kg/Cm2


k2 for H^2/Dt of 14 = 0.477


2

= 26.24 Kg/cm


14

= 1.87 Kg/sq.cm < 13 O.K.



1.87 + 7.19 = 0.54 < 1 O.K.

13 18

Reinforcement Steel

στmax

r r1





10/14/2010


SDC SIVEN P SKL



At each face 90 mm2




Ast = = 150 x 0.87 x 140

= ### sq. mm


= 0.15 x 14 = 2.1 cm2 per meter






= 0.24 x 20 x 100

100= 4.8 cm2






Load Calculations




Total 761 Kgs

Say 7.5 KN


16

= 3.54 KN-m

B.M./(σst*j*d)





10/14/2010


SDC SIVEN P SKL


= 7.09 KN-m


= 90.07 mm














OR from edge = 2.75 - 2.04 = 0.71




2) = ### + 150 = 609

3) ### + 96 = 555







10/14/2010


SDC SIVEN P SKL


INPUT DATA




4) = 30 °



8) Internal radius of the well r = 2.5 m







Design coefficient




18) Lever arm ( j ) 0.87

A Design of Wall












H^2 = 7.29 = 7.29

D x t 5 x 0.20




= 369 Kgm/m


t/m3


N/mm2

N/mm2

t/m3





10/14/2010


SDC SIVEN P SKL

b x d^2 100 x 196

= 11.3 Kg/sqcm < 18 O.K.

P

P P


2



P = 0.33 x 0 x 270

= 0.16 Kg/Cm2


P = 1 x 0 x 270.0

= 0.27 Kg/Cm2


k2 for H^2/Dt of 14 = 0.575


2

= 38.81 Kg/cm


14

= 2.77 Kg/sq.cm < 13 O.K.



2.77 + 11.3 = 0.84 < 1 O.K.

13 18

Reinforcement Steel

στmax

r r1





10/14/2010


SDC SIVEN P SKL



At each face 130 mm2




Ast = = 150 x 0.87 x 140

= ### sq. mm


= 0.15 x 14 = 2.1 cm2 per meter






= 0.24 x 20 x 100

100= 4.8 cm2






Load Calculations




Total 761 Kgs

Say 7.5 KN


16

= 3.54 KN-m

B.M./(σst*j*d)





10/14/2010


SDC SIVEN P SKL


= 7.09 KN-m


= 90.07 mm














OR from edge = 2.75 - 2.04 = 0.71




2) = ### + 150 = 609

3) ### + 96 = 555



circular sump-oil drain

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