IBR Calculation
IBR Calculation
IBR Calculation
Rev. No. 00
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IBR CALCULATIONS
AS PER INDIAN BOILER REGULATIONS, 1950
INDIAN BOILERS ACT, 1923, Edition 2010
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DESIGN DATA
PWHT : YES
Head------------Full.
Head : 1.00
Shell : 1.00
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Material Specifications :
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Rev. No. 00
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1. SHELL 6
2. DISH 7
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Regulation: 270
Where,
= 1562.37 1124.912
E = Efficiency factor for welded shells = 1.00 for full radiographic examinations (ASME Sec VIII,
Division 1, Table UW-12)
0.69 = [(220.6320T-168.121)/(893.638+T)]
616.61+0.69T = 220.632T-168.121
219.942T = 784.931
T = 3.567 mm
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Regulation: 278
W.P = (2*f*(T-C))/(D*K);
T = 10 mm.
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NOZZLE N5 (3")
Regulation: 573
t = 0.04D + 2.5
= (0.04x88.9) + 2.5
t = 6.056 mm
t = 6.056/0.875 = 6.92 mm
NOZZLE N2 (2")
t = 0.04D + 2.5
= (0.04x60.3) + 2.5
t = 4.912 mm
t = 4.912/0.875 = 5.613 mm
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NOZZLE N3 (2")
t = 0.04D + 2.5
= (0.04x60.3) + 2.5
t = 4.912 mm
t = 4.912/0.875 = 5.613 mm
NOZZLE N4 (2")
t = 0.04D + 2.5
= (0.04x60.3) + 2.5
t = 4.912 mm
t = 4.912/0.875 = 5.613 mm
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t = 0.04D + 2.5
= (0.04x457.2) + 2.5
t = 20.788 mm
t = 20.788/0.875 = 23.737 mm
t = 0.04D + 2.5
= (0.04x60.3) + 2.5
t = 4.912 mm
t = 4.912/0.875 = 5.613 mm
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Rev. No. 00
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t = 0.04D + 2.5
= (0.04x60.3) + 2.5
t = 4.912 mm
t = 4.912/0.875 = 5.613 mm
NOZZLE N9 (2")
t = 0.04D + 2.5
= (0.04x60.3) + 2.5
t = 4.912 mm
t = 4.912/0.875 = 5.613 mm
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t = 0.04D + 2.5
= (0.04x48.3) + 2.5
t = 4.432 mm
t = 4.432/0.875 = 5.065 mm
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COMPENSATION CALCULATIONS
iii) Width of opening for 2" Nozzle = 60.3mm, Hence compensation not
required.
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Rev. No. 00
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C = Constant = 0.762
= 2179.92 mm2
= (10-4.768)x2(75+10)
= 889.44 mm2
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C = 2(100+Ts)[Tn-(T1+T2)]S2/S1 . Eqn. 2
= (W.P x D/2fe)+0.762
= (0.9806x894.4/2x118.0x1)+0.762
T2 = 4.478mm
= 10 x 10
= 100 mm2
iv) Compensation due to pad plate
Z = 2 x w x Tc x S3/S1 . Eqn. 3
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Where,
W = Width of compensation plate in mm = 210 mm
Tc = Thickness of compensating plates in mm = 10 mm
S3 = Minimum tensile strength of compensation plate(SA 516 Gr.
60) in N/mm2 = 485 N/mm2
S1 = Minimum tensile strength of Shell plate (SA 516 Gr. 60)
in N/mm2 = 485 N/mm2
Z = 2x210x10x(485/485)
Z = 4200 mm2
= 889.44+3207.886+100+4200
= 8397.326 mm2
Compensation required:-
= 2179.92 mm2
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C = Constant = 0.762
= 423.875 mm2
= (10-4.768)x2(75+10)
= 889.44 mm2
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C = 2(100+Ts)[Tn-(T1+T2)]S2/S1 . Eqn. 2
= (W.P x D/2fe)+0.762
= (0.9806x894.4/2x118.0x1)+0.762
T2 = 4.478mm
= 10 x 10
= 100 mm2
iv) Compensation due to pad plate
Z = 2 x w x Tc x S3/S1 . Eqn. 3
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Rev. No. 00
. Page No. Page 18 of 19
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Where,
W = Width of compensation plate in mm = 45 mm
Tc = Thickness of compensating plates in mm = 10 mm
S3 = Minimum tensile strength of compensation plate(SA 516 Gr.
60) in N/mm2 = 485 N/mm2
S1 = Minimum tensile strength of Shell plate (SA 516 Gr. 60)
in N/mm2 = 485 N/mm2
Z = 2x45x10x(485/485)
Z = 900 mm2
= 889.44+583.682+100+900
= 2473.122 mm2
Compensation required:-
= 423.875 mm2
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