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7/28/2019 Blow Off Pipe Sizing
1/13
MAIN STEAM LINE
SAFETY VALVE DATA TAG NO. 10LBA10
Valve set pressure = 2196.2 psia
Name plate flow = 74.559 lbm/sec. (3) Reaction Force at Discharge Elbow Exit
Valve actual flow = 74.56 lbm/sec. Reaction force:
Orifice size = 3.341 in.2
Valve inlet I.D. = 3 in. EQ. (3) Calculate F1 = 7951 lbf
Valve outlet I.D. = 6 in. Supply Valve F1 = 0 lbf
Valve discharge elbow = 6 in. ..
Elbow I.D. = 6.066 in. Max F1 = 7951 lbf
Seismic coefficient = 0.16 g
Nozzle material = A182F91
Allowable stress at T = 16030 psi (4) Bending Moments at Points (1) and (2)
Valve weight = 571 lb
Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2) due to
STEAM CONDITIONS moment arm "L" = 22.875 in.
weigth of valve "W" = 571 lb
Temperature = 1005.8 F "h1" = 22 in.
Pressure = 2196.2 psia "h2" = 12 in.
Enthalpy " ho" = 1471.44 Btu/lbm [Young's modulus at des.temp.]E = 25367094 psi
nozzle "Do" = 5.65 in.
nozzle "Di" = 3 in.
[Moment of inertia Nozzle] "I" = 46.0 in.
EQ. (4) T = 0.01332 sec
(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) [Valve rise time] to = 0.04 sec
(Analysis of Section 1) Ratio t0/T = 3.0
From Fig. 3-2, DLF = 1.21
W (actual) = 74.56 lbm/sec M1(1) = M1(2) =F1xLxDLF = 220085 in.-lb
A1 = 28.90 in.
a = 831 (B) Bending Moments at Points (1) and (2) due tb = 4.33
J = 778.2 ft-lbf/Btu Seismic force
gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 91.36 lbf
EQ. (1) P1 = 126 psia
EQ. (2) V1 = 2047 ft/sec Moment arm for Point (1)
Ms(1) =Fs x h1 2009.92 in.-lb
Moment arm for Point (2)
(2) Discharge Elbow Maximun Operating Pressure Ms(2) =Fs x h2 1096.32 in.-lb
Elbow I.D. = 6.066 in. (C) Combined Bending Moments at Point (1) and
Height W.N. Flange = 4 in.
L/D = 0.66 M(1) = M1(1) + Ms(1) = 222095 in.-lbM(2) = M1(2) + Ms(2) = 221182 in.-lb
Short Radius Elbow
L/D = 30 (5) Stress Intensification Factors at Point (1) a
Pipe (A) At Point (1), Branch Connection [For nomenc
Length Pipe = 8 in.
L/D = 1.32 [Outside Dia. RUN PIPE] Do = 16 in.
[Thickness RUN PIPE] Tr= 1.438 in.
31.98 [Radius medium RUN PIPE] Rm = 7.281 in.
[Outside radius BRANCH] rp = 2.825 in.
[Thickness BRANCH] T'b = 1.2 in.
= 0.013 [Radius medium BRANCH] r'm = 2.225 in.
k = 1.3 EQ. (5) i (1) = 1.61
(Lmax/D) = 0.416
(B) Stress Intensification Factors at Point (2) Bu
SAFETY VALVE INSTALLATIONS
ACCORDING TO ASME B31.1 APPENDIX II
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
2/13
MAIN STEAM LINE
(6) Predicted Stresses at Point (1) and (2) (7) Calculate the Maximun Operating Press
( Vent Pipe Analysis at Section 2 and Secti
(A) Predicted Stresses at Point (1), Branch Connection A Vent Pipe must estimated to start the ca
A good starting size is 3 pipes sizes larger
Do/tn [for run pipe] = 11.12656
Do/tn [for branch pipe] = 4.708333 Vent Pipe Size = DN 14
Max [Do/tn] = 11.12656 Vent Pipe I.D. = 13.123 in.A3 = 135.26 in.
Pressure stress(1)
P Do/4 tn = 6109 psi P3 = P1 ( A1 / A3 ) = 26.9 psia
tS = [lesser oftr or(i) tb] = 1.438 in.
rb = 2.225 in. 21.87
22.4 in.3
= 0.013
k = 1.3
Flexure stress(1)
0.75 i M(1) / Z(1) = 11969 psi (Lmax/D) = 0.284
Combined stress(1) = Pressure stress(1) + Flexure stress(1) From Chart 1, P/P* = 1.8
= 18078 psi P2 = P3 (P/P*) = 48.5 psia
(B) Predicted Stresses at point (2), buttweld (8) Check for Blowback From Vent Pipe
Calculate the velocity V2 that exists at the
Do = 5.65 in. (Para. 2.2.1.4)
tn = 1.2 in.
(Lmax/D) = 0.284
Pressure stress(2) V3 = V1 = 2047 ft/sec
P Do/4 tn = 2585 psi
From Chart 1, V / V* = 0.68
Di = 3.25 in. V2 = V3 (V / V*) = 1392 ft/sec
Check the inequality from Para. 2.3.1.2.
15.8 in.3
W ( V1 - V2 ) / gc = 1517
(P2-Pa) A2 - (P1-Pa) A1= 1321
EQ. (6) = O.K.
Flexure stress(2) (9) Calculate Forces Acting on Vent Pipe
0.75 i M(2) / Z(2) = 14027 psi
EQ. (3) F2 = 7755 lbf
Combined stress(2) = Pressure stress(2) + Flexure stress(2) EQ. (3) F3 = 6356 lbf
= 16612 psi
(C) Comparison of Predicted Stress with Allowable stress
Allowable stress of nozzle material at temperature NOTE
When the vent outlet is perpendicular to the axis of
Sh = 16030 psi the vent pipe. This results in a flow that is vertical.
[See ASME B31.1 Para.104.8] k = 1.2 When the vent outlet is beveled. This results in a
flow that is not vertical. To take this into account
k Sh = 19236 psi the force at the outlet is shown to act at an angle
of 20 with the axis of the vent pipe. This will
Combined stress(1) = 18078 psi introduce a horizontal component force at the outlet.
Combined stress(2) = 16612 psi O.K.
=p= rZ sb(1) t2
=p
= D
D-D
32Z
o
4i
4o
(2)
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
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HOT REHEAT
SAFETY VALVE DATA TAG N. 10LBB10
Valve set pressure = 551.359 psia
Name plate flow = 69.4823 lbm/sec. (3) Reaction Force at Discharge E
Valve actual flow = 69.48 lbm/sec. Reaction force:
Orifice size = 16 in.2
Valve inlet I.D. = 6 in. EQ. (3) Calculate F1 = 74
Valve outlet I.D. = 8 in. Supply Valve F1 =
Valve discharge elbow = 8 in. ..
Elbow I.D. = 7.98 in. Max F1 = 74
Seismic coefficient = 0.16 g
Nozzle material = A182F91
Allowable stress at T = 16115.5 psi (4) Bending Moments at Points (1)
Valve weight = 902 lb
Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) an
STEAM CONDITIONS moment arm "L" = 2
weigth of valve "W" = 9Temperature = 1004 F "h1" = 2
Pressure = 551.359 psia "h2" =
Enthalpy " ho" = 1520.977 Btu/lbm [Young's modulus at des.temp.]E = 2538160
nozzle "Do" = 7.
nozzle "Di" =
[Moment of inertia Nozzle] "I" = 113
EQ. (4) T = 0.0106
(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) [Valve rise time] to = 0.
(Analysis of Section 1) Ratio t0/T = 3
From Fig. 3-2, DLF = 1.
W (actual) = 69.48 lbm/sec M1(1) = M1(2) =F1xLxDLF = 20977
A1 = 50.01 in.
a = 823 (B) Bending Moments at Points (1) ab = 4.33
J = 778.2 ft-lbf/Btu Seismic force
gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 144.3
EQ. (1) P1 = 71 psia
EQ. (2) V1 = 2137 ft/sec Moment arm for Point (1)
Ms(1) =Fs x h1 3175.0
Moment arm for Point (2)
(2) Discharge Elbow Maximun Operating Pressure Ms(2) =Fs x h2 1731.8
Elbow I.D. = 7.98 in. (C) Combined Bending Moments at P
Height W.N. Flange = 4 in.
L/D = 0.50 M(1) = M1(1) + Ms(1) = 21294
M(2) = M1(2) + Ms(2) = 21150
Short Radius Elbow
L/D = 30 (5) Stress Intensification Factors a
Pipe (A) At Point (1), Branch Connection [
Length Pipe = 10 in.
L/D = 1.25 [Outside Dia. RUN PIPE] Do = 2
[Thickness RUN PIPE] Tr=
31.75 [Radius medium RUN PIPE] Rm = 12
[Outside radius BRANCH] rp = 3.8
[Thickness BRANCH] T'b = 0.
= 0.013 [Radius medium BRANCH] r'm = 3
k = 1.3
EQ. (5) i (1) = 2.
(Lmax/D) = 0.413
(B) Stress Intensification Factors at P
SAFETY VALVE INSTALLATIONS
ACCORDING TO ASME B31.1 APPENDIX II
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
4/13
HOT REHEAT
(6) Predicted Stresses at Point (1) and (2) (7) Calculate the Maximun Ope
( Vent Pipe Analysis at Section
(A) Predicted Stresses at Point (1), Branch Connection A Vent Pipe must estimated to
A good starting size is 3 pipes
Do/tn [for run pipe] = 26
Do/tn [for branch pipe] = 10.33333 Vent Pipe Size =
Max [Do/tn] = 26 Vent Pipe I.D. =
A3 = 1
Pressure stress(1)
P Do/4 tn = 3584 psi P3 = P1 ( A1 / A3 ) =
tS = [lesser oftr or(i) tb] = 1 in.
rb = 3.5 in. 23.7
38.5 in.3
=
k =
Flexure stress(1)
0.75 i M(1) / Z(1) = 12018 psi (Lmax/D) =
Combined stress(1) = Pressure stress(1) + Flexure stress(1) From Chart 1, P/P* =
= 15602 psi P2 = P3 (P/P*) =
(B) Predicted Stresses at point (2), buttweld (8) Check for Blowback From V
Calculate the velocity V2 that e
Do = 7.75 in. (Para. 2.2.1.4)
tn = 0.75 in.
(Lmax/D) =
Pressure stress(2) V3 = V1 =
P Do/4 tn = 1424 psi
From Chart 1, V / V* =
Di = 6.25 in. V2 = V3 (V / V*) =
Check the inequality from Para. 226.4 in.
3
W ( V1 - V2 ) / gc =
(P2-Pa) A2 - (P1-Pa) A1=
EQ. (6) = O.K.
Flexure stress(2) (9) Calculate Forces Acting on
0.75 i M(2) / Z(2) = 8021 psi
EQ. (3) F2 =
Combined stress(2) = Pressure stress(2) + Flexure stress(2) EQ. (3) F3 =
= 9445 psi
(C) Comparison of Predicted Stress with Allowable stress
Allowable stress of nozzle material at temperature NOTEWhen the vent outlet is perpendicular to th
Sh = 16115.5 psi the vent pipe. This results in a flow that is
[See ASME B31.1 Para.104.8] k = 1.2 When the vent outlet is beveled. This resu
flow that is not vertical. To take this into a
k Sh = 19338.6 psi the force at the outlet is shown to act at an
of 20 with the axis of the vent pipe. This w
Combined stress(1) = 15602 psi introduce a horizontal component force at
Combined stress(2) = 9445 psi O.K.
=p= rZ sb(1) t2
=p= D
D-D
32Z
o
4i4o(2)
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
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COLD REHEHEAT
SAFETY VALVE DATA TAG N. 10LBC10
Valve set pressure = 624 psia
Name plate flow = 104.03 lbm/sec. (3) Reaction Force at Discharge Elbow Exit
Valve actual flow = 104.03 lbm/sec. Reaction force:
Orifice size = 16 in.2
Valve inlet I.D. = 6 in. EQ. (3) Calculate F1 = 9508 lbf
Valve outlet I.D. = 8 in. Supply Valve F1 = 0 lbf
Valve discharge elbow = 8 in. ..
Elbow I.D. = 7.98 in. Max F1 = 9508 lbf
Seismic coefficient = 0.16 g
Nozzle material = SA 234 WPC
Allowable stress at T = 19831.2 psi (4) Bending Moments at Points (1) and (2)
Valve weight = 902 lb
Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2) due t
STEAM CONDITIONS moment arm "L" = 24 in.
weigth of valve "W" = 902 lbTemperature = 642.2 F "h1" = 17 in.
Pressure = 624 psia "h2" = 12 in.
Enthalpy " ho" = 1315.39 Btu/lbm [Young's modulus at des.temp.]E = 25863100 psi
nozzle "Do" = 7.75 in.
nozzle "Di" = 6 in.
[Moment of inertia Nozzle] "I" = 113.5 in.
EQ. (4) T = 0.00717 sec
(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) [Valve rise time] to = 0.04 sec
(Analysis of Section 1) Ratio t0/T = 5.6
From Fig. 3-2, DLF = 1.15
W (actual) = 104.03 lbm/sec M1(1) = M1(2) =F1xLxDLF = 262421 in.-lb
A1 = 50.01 in.
a = 823 (B) Bending Moments at Points (1) and (2) dueb = 4.33
J = 778.2 ft-lbf/Btu Seismic force
gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 144.32 lbf
EQ. (1) P1 = 89 psia
EQ. (2) V1 = 1795 ft/sec Moment arm for Point (1)
Ms(1) =Fs x h1 2453.44 in.-lb
Moment arm for Point (2)
(2) Discharge Elbow Maximun Operating Pressure Ms(2) =Fs x h2 1731.84 in.-lb
Elbow I.D. = 7.98 in. (C) Combined Bending Moments at Point (1) a
Height W.N. Flange = 4 in.
L/D = 0.50 M(1) = M1(1) + Ms(1) = 264875 in.-lbM(2) = M1(2) + Ms(2) = 264153 in.-lb
Short Radius Elbow
L/D = 30 (5) Stress Intensification Factors at Point (1
Pipe (A) At Point (1), Branch Connection [For nomen
Length Pipe = 8 in.
L/D = 1.00 [Outside Dia. RUN PIPE] Do = 24 in.
[Thickness RUN PIPE] Tr= 1 in.
31.50 [Radius medium RUN PIPE] Rm = 11.5 in.
[Outside radius BRANCH] rp = 3.875 in.
[Thickness BRANCH] T'b = 0.75 in.
= 0.013 [Radius medium BRANCH] r'm = 3.5 in.
k = 1.3EQ. (5) i (1) = 2.86
(Lmax/D) = 0.410
SAFETY VALVE INSTALLATIONS
ACCORDING TO ASME B31.1 APPENDIX II
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
6/13
COLD REHEHEAT
(6) Predicted Stresses at Point (1) and (2) (7) Calculate the Maximun Operating Pre
( Vent Pipe Analysis at Section 2 and Sec
(A) Predicted Stresses at Point (1), Branch Connection A Vent Pipe must estimated to start the c
A good starting size is 3 pipes sizes larg
Do/tn [for run pipe] = 24
Do/tn [for branch pipe] = 10.33333333 Vent Pipe Size = DN
Max [Do/tn] = 24 Vent Pipe I.D. = 12.09 in.A3 = 114.80 in.
Pressure stress(1)
P Do/4 tn = 3744 psi P3 = P1 ( A1 / A3 ) = 38.8 psia
tS = [lesser oftr or(i) tb] = 1 in.
rb = 3.5 in. 23.73863
38.5 in.3
= 0.013
k = 1.3
Flexure stress(1)
0.75 i M(1) / Z(1) = 14743 psi (Lmax/D) = 0.309
Combined stress(1) = Pressure stress(1) + Flexure stress(1) From Chart 1, P/P* = 1.56
= 18487 psi P2 = P3 (P/P*) = 60.6 psia
(B) Predicted Stresses at point (2), buttweld (8) Check for Blowback From Vent Pipe
Calculate the velocity V2 that exists at th
Do = 7.75 in. (Para. 2.2.1.4)
tn = 0.75 in.
(Lmax/D) = 0.309
Pressure stress(2) V3 = V1 = 1795 ft/se
P Do/4 tn = 1612 psi
From Chart 1, V / V* = 0.68
Di = 6.25 in. V2 = V3 (V / V*) = 1220 ft/se
Check the inequality from Para. 2.3.1.2.
26.4 in.3
W ( V1 - V2 ) / gc = 1856
(P2-Pa) A2 - (P1-Pa) A1= 1526
EQ. (6) = O.K.
Flexure stress(2) (9) Calculate Forces Acting on Vent Pipe
0.75 i M(2) / Z(2) = 10017 psi
EQ. (3) F2 = 9178 lbf
Combined stress(2) = Pressure stress(2) + Flexure stress(2) EQ. (3) F3 = 8536 lbf
= 11629 psi
(C) Comparison of Predicted Stress with Allowable stress
Allowable stress of nozzle material at temperature NOTE
When the vent outlet is perpendicular to the axis of
Sh = 19831.2 psi the vent pipe. This results in a flow that is vertical.
[See ASME B31.1 Para.104.8] k = 1.2 When the vent outlet is beveled. This results in a
flow that is not vertical. To take this into account
k Sh = 23797.44 psi the force at the outlet is shown to act at an angle
of 20 with the axis of the vent pipe. This will
Combined stress(1) = 18487 psi introduce a horizontal component force at the outlet.
Combined stress(2) = 11629 psi O.K.
=p= rZ sb(1) t2
=p
= D
D-D
32Z
o
4i
4o
(2)
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
7/13
STEAM DRUM SV-I
SAFETY VALVE DATA TAG NO. 10HAD11
Valve set pressure = 2407.84 psia
Name plate flow = 111.83867 lbm/sec. (3) Reaction Force at Discharge Elbow Exit
Valve actual flow = 111.84 lbm/sec. Reaction force:Orifice size = 3.976 in.
2
Valve inlet I.D. = 3 in. EQ. (3) Calculate F1 = 6264 lbf
Valve outlet I.D. = 6 in. Supply Valve F1 = 0 lbf
Valve discharge elbow = 6 in. ..
Elbow I.D. = 6.066 in. Max F1 = 6264 lbf
Seismic coefficient = 0.16 g
Nozzle material = SA 234 WPC
Allowable stress at T = 19440.47 psi (4) Bending Moments at Points (1) and (2)
Valve weight = 630.526 lb
Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2) due to
STEAM CONDITIONS moment arm "L" = 22.875 in.
weigth of valve "W" = 630.526 lb
Temperature = 662 F "h1" = 22 in.
Pressure = 2407.84 psia "h2" = 12 in.
Enthalpy " ho" = 718.4712 Btu/lbm [Young's modulus at des.temp.]E = 25671700 psi
nozzle "Do" = 6.102 in.
nozzle "Di" = 3 in.
[Moment of inertia Nozzle] "I" = 64.1 in.
EQ. (4) T = 0.01179 sec
(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) [Valve rise time] to = 0.04 sec
(Analysis of Section 1) Ratio t0/T = 3.4
From Fig. 3-2, DLF = 1.23
W (actual) = 111.84 lbm/sec M1(1) = M1(2) =F1xLxDLF = 176238 in.-lb
A1 = 28.90 in.
a = 291 (B) Bending Moments at Points (1) and (2) due to
b = 11
J = 778.2 ft-lbf/Btu Seismic force
gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 100.8842 lbf
EQ. (1) P1 = 110 psia
EQ. (2) V1 = 1010 ft/sec Moment arm for Point (1)
Ms(1) =Fs x h1 2219.452 in.-lb
Moment arm for Point (2)
(2) Discharge Elbow Maximun Operating Pressure Ms(2) =Fs x h2 1210.61 in.-lb
Elbow I.D. = 6.066 in. (C) Combined Bending Moments at Point (1) and
Height W.N. Flange = 4 in.
L/D = 0.66 M(1) = M1(1) + Ms(1) = 178457 in.-lb
M(2) = M1(2) + Ms(2) = 177449 in.-lb
Short Radius Elbow
L/D = 30 (5) Stress Intensification Factors at Point (1) a
Pipe (A) At Point (1), Branch Connection [For nomenc
Length Pipe = 8 in.
L/D = 1.32 [Outside Dia. RUN PIPE] Do = 66.77 in.
[Thickness RUN PIPE] Tr= 4.842 in.
31.98 [Radius medium RUN PIPE] Rm = 30.964 in.
[Outside radius BRANCH] rp = 3.051 in.
[Thickness BRANCH] T'b = 1.426 in.
= 0.013 [Radius medium BRANCH] r'm = 2.338 in.
k = 1.1
EQ. (5) i (1) = 0.32
(Lmax/D) = 0.416
(B) Stress Intensification Factors at Point (2), Butt
SAFETY VALVE INSTALLATIONS
ACCORDING TO ASME B31.1 APPENDIX II
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
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STEAM DRUM SV-I
(6) Predicted Stresses at Point (1) and (2) (7) Calculate the Maximun Operating Pressur
( Vent Pipe Analysis at Section 2 and Section
(A) Predicted Stresses at Point (1), Branch Connection A Vent Pipe must estimated to start the calcu
A good starting size is 3 pipes sizes larger tha
Do/tn [for run pipe] = 13.7897563
Do/tn [for branch pipe] = 4.27910238 Vent Pipe Size = DN 12
Max [Do/t
n] =
13.7897563 Vent Pipe I.D. = 12.09 in.A3 = 114.80 in.
Pressure stress(1)
P Do/4 tn = 8301 psi P3 = P1 ( A1 / A3 ) = 27.8 psia
tS = [lesser oftr or(i) tb] = 1.426 in.
rb = 2.338 in. 23.73863 in
24.5 in.3
= 0.013
k = 1.1
Flexure stress(1)
0.75 i M(1) / Z(1) = 7287 psi (Lmax/D) = 0.309
Combined stress(1) = Pressure stress(1) + Flexure stress(1) From Chart 1, P/P* = 1.45= 15588 psi P2 = P3 (P/P*) = 40.3 psia
(B) Predicted Stresses at point (2), buttweld (8) Check for Blowback From Vent Pipe
Calculate the velocity V2 that exists at the inle
Do = 6.102 in. (Para. 2.2.1.4)
tn = 1.426 in.
(Lmax/D) = 0.309
Pressure stress(2) V3 = V1 = 1010 ft/sec
P Do/4 tn = 2576 psi
From Chart 1, V / V* = 0.75
Di = 3.25 in. V2 = V3 (V / V*) = 758 ft/sec
Check the inequality from Para. 2.3.1.2.
20.5 in.3
W ( V1 - V2 ) / gc = 877
(P2-Pa) A2 - (P1-Pa) A1= 147
EQ. (6) = O.K.
Flexure stress(2) (9) Calculate Forces Acting on Vent Pipe
0.75 i M(2) / Z(2) = 8651 psi
EQ. (3) F2 = 5533 lbf
Combined stress(2) = Pressure stress(2) + Flexure stress(2) EQ. (3) F3 = 4975 lbf (
= 11227 psi
(C) Comparison of Predicted Stress with Allowable stressAllowable stress of nozzle material at temperature NOTE
When the vent outlet is perpendicular to the axis of
Sh = 19440.47 psi the vent pipe. This results in a flow that is vertical.
[See ASME B31.1 Para.104.8] k = 1.1 When the vent outlet is beveled. This results in a
flow that is not vertical. To take this into account
k Sh = 21384.517 psi the force at the outlet is shown to act at an angle
of 20 with the axis of the vent pipe. This will
Combined stress(1) = 15588 psi introduce a horizontal component force at the outlet.
Combined stress(2) = 11227 psi O.K.
=p= rZ sb(1) t2
=p
= D
D-D
32Z
o
4i
4o
(2)
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
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STEAM DRUM SV-II
SAFETY VALVE DATA TAG NO. 10HAD11
Valve set pressure = 2393.34 psia
Name plate flow = 111.83867 lbm/sec. (3) Reaction Force at Discharge Elbow Exit
Valve actual flow = 111.84 lbm/sec. Reaction force:Orifice size = 3.976 in.
2
Valve inlet I.D. = 3 in. EQ. (3) Calculate F1 = 8811 lbf
Valve outlet I.D. = 6 in. Supply Valve F1 = 0 lbf
Valve discharge elbow = 6 in. ..
Elbow I.D. = 6.066 in. Max F1 = 8811 lbf
Seismic coefficient = 0.16 g
Nozzle material = SA 234 WPC
Allowable stress at T = 19440.47 psi (4) Bending Moments at Points (1) and (2)
Valve weight = 630.526 lb
Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2) due to R
STEAM CONDITIONS moment arm "L" = 22.875 in.
weigth of valve "W" = 630.526 lb
Temperature = 662 F "h1" = 22 in.
Pressure = 2393.34 psia "h2" = 12 in.
Enthalpy " ho" = 1105.5638 Btu/lbm [Young's modulus at des.temp.]E = 25671700 psi
nozzle "Do" = 6.102 in.
nozzle "Di" = 3 in.
[Moment of inertia Nozzle] "I" = 64.1 in.
EQ. (4) T = 0.01179 sec
(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) [Valve rise time] to = 0.04 sec
(Analysis of Section 1) Ratio t0/T = 3.4
From Fig. 3-2, DLF = 1.23
W (actual) = 111.84 lbm/sec M1(1) = M1(2) =F1xLxDLF = 247921 in.-lb
A1 = 28.90 in.
a = 291 (B) Bending Moments at Points (1) and (2) due tob = 11
J = 778.2 ft-lbf/Btu Seismic force
gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 100.8842 lbf
EQ. (1) P1 = 152 psia
EQ. (2) V1 = 1394 ft/sec Moment arm for Point (1)
Ms(1) =Fs x h1 2219.452 in.-lb
Moment arm for Point (2)
(2) Discharge Elbow Maximun Operating Pressure Ms(2) =Fs x h2 1210.61 in.-lb
Elbow I.D. = 6.066 in. (C) Combined Bending Moments at Point (1) and
Height W.N. Flange = 4 in.
L/D = 0.66 M(1)
= M1(1)
+ Ms(1)
= 250140 in.-lb
M(2) = M1(2) + Ms(2) = 249132 in.-lb
Short Radius Elbow
L/D = 30 (5) Stress Intensification Factors at Point (1) a
Pipe (A) At Point (1), Branch Connection [For nomencl
Length Pipe = 8 in.
L/D = 1.32 [Outside Dia. RUN PIPE] Do = 66.77 in.
[Thickness RUN PIPE] Tr= 4.842 in.
31.98 [Radius medium RUN PIPE] Rm = 30.964 in.
[Outside radius BRANCH] rp = 3.051 in.
[Thickness BRANCH] T'b = 1.426 in.
= 0.013 [Radius medium BRANCH] r'm = 2.338 in.
k = 1.1
EQ. (5) i (1) = 0.32
(Lmax/D) = 0.416
(B) Stress Intensification Factors at Point (2) Butt
SAFETY VALVE INSTALLATIONS
ACCORDING TO ASME B31.1 APPENDIX II
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
10/13
STEAM DRUM SV-II
(6) Predicted Stresses at Point (1) and (2) (7) Calculate the Maximun Operating Pressu
( Vent Pipe Analysis at Section 2 and Sectio
(A) Predicted Stresses at Point (1), Branch Connection A Vent Pipe must estimated to start the calc
A good starting size is 3 pipes sizes larger t
Do/tn [for run pipe] = 13.7897563
Do/tn [for branch pipe] = 4.27910238 Vent Pipe Size = DN 12
Max [Do/t
n] =
13.7897563 Vent Pipe I.D. = 12.09 in.A3 = 114.80 in.
Pressure stress(1)
P Do/4 tn = 8251 psi P3 = P1 ( A1 / A3 ) = 38.3 psia
tS = [lesser oftr or(i) tb] = 1.426 in.
rb = 2.338 in. 23.73863
24.5 in. = 0.013
k = 1.1
Flexure stress(1)
0.75 i M(1) / Z(1) = 10215 psi (Lmax/D) = 0.309
Combined stress(1) = Pressure stress(1) + Flexure stress(1) From Chart 1, P/P* = 1.45
= 18466 psi P2 = P3 (P/P*) = 55.6 psia
(B) Predicted Stresses at point (2), buttweld (8) Check for Blowback From Vent Pipe
Calculate the velocity V2 that exists at the in
Do = 6.102 in. (Para. 2.2.1.4)
tn = 1.426 in.
(Lmax/D) = 0.309
Pressure stress(2) V3 = V1 = 1394 ft/sec
P Do/4 tn = 2560 psi
From Chart 1, V / V* = 0.75
Di = 3.25 in. V2 = V3 (V / V*) = 1046 ft/sec
Check the inequality from Para. 2.3.1.2.
20.5 in.
W ( V1 - V2 ) / gc = 1211
(P2-Pa) A2 - (P1-Pa) A1= 693
EQ. (6) = O.K.
Flexure stress(2) (9) Calculate Forces Acting on Vent Pipe
0.75 i M(2) / Z(2) = 12146 psi
EQ. (3) F2 = 8293 lbf
Combined stress(2) = Pressure stress(2) + Flexure stress(2) EQ. (3) F3 = 7523 lbf
= 14707 psi
(C) Comparison of Predicted Stress with Allowable stress
Allowable stress of nozzle material at temperature NOTE
When the vent outlet is perpendicular to the axis of
Sh = 19440.47 psi the vent pipe. This results in a flow that is vertical.
[See ASME B31.1 Para.104.8] k = 1.1 When the vent outlet is beveled. This results in a
flow that is not vertical. To take this into account
k Sh = 21384.517 psi the force at the outlet is shown to act at an angle
of 20 with the axis of the vent pipe. This will
Combined stress(1) = 18466 psi introduce a horizontal component force at the outlet.
Combined stress(2) = 14707 psi O.K.
=p= rZ sb(1) t2
=p
= D
D-D
32Z
o
4i
4o
(2)
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
11/13
SOOT BLOWER SV
SAFETY VALVE DATA TAG NO. 10HCB11
Valve set pressure = 449.802 psia
Name plate flow = 13.472694 lbm/sec. (3) Reaction Force at Discharge Elbow
Valve actual flow = 13.47 lbm/sec. Reaction force:
Orifice size = 2.853 in.2
Valve inlet I.D. = 4 in. EQ. (3) Calculate F1 = 833 lb
Valve outlet I.D. = 6 in. Supply Valve F1 = 0 lb
Valve discharge elbow = 6 in. ..
Elbow I.D. = 6.066 in. Max F1 = 833 lb
Seismic coefficient = 0.16 g
Nozzle material = SA 234 WPB
Allowable stress at T = 16470.8 psi (4) Bending Moments at Points (1) and
Valve weight = 284.39 lb
Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2)
STEAM CONDITIONS moment arm "L" = 21.24 in
weigth of valve "W" = 284.39 lb
Temperature = 671 F "h1" = 16 in
Pressure = 449.802 psia "h2" = 8 inEnthalpy " ho" = 1344.4201 Btu/lbm [Young's modulus at des.temp.]E = 25570200 p
nozzle "Do" = 5.118 in
nozzle "Di" = 4 in
[Moment of inertia Nozzle] "I" = 21.1 in
EQ. (4) T = 0.00857 s
(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) [Valve rise time] to = 0.04 s
(Analysis of Section 1) Ratio t0/T = 4.7
From Fig. 3-2, DLF = 1.18
W (actual) = 13.47 lbm/sec M1(1) = M1(2) =F1xLxDLF = 20878 in
A1 = 28.90 in.
a = 291 (B) Bending Moments at Points (1) and (2
b = 11
J = 778.2 ft-lbf/Btu Seismic force
gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 45.5024 lb
EQ. (1) P1 = 21 psia
EQ. (2) V1 = 1586 ft/sec Moment arm for Point (1)
Ms(1) =Fs x h1 728.0384 in
Moment arm for Point (2)
(2) Discharge Elbow Maximun Operating Pressure Ms(2) =Fs x h2 364.0192 in
Elbow I.D. = 6.066 in. (C) Combined Bending Moments at Point
Height W.N. Flange = 4 in.
L/D = 0.66 M(1) = M1(1) + Ms(1) = 21606 in
M(2) = M1(2) + Ms(2) = 21242 in
Short Radius Elbow
L/D = 30 (5) Stress Intensification Factors at Po
Pipe (A) At Point (1), Branch Connection [For n
Length Pipe = 8 in.
L/D = 1.32 [Outside Dia. RUN PIPE] Do = 6 in
[Thickness RUN PIPE] Tr= 0.431 in
31.98 [Radius medium RUN PIPE] Rm = 2.7845 in
[Outside radius BRANCH] rp = 2.559 in
[Thickness BRANCH] T'b = 0.559 in
= 0.013 [Radius medium BRANCH] r'm = 2.2795 in
k = 1.1
EQ. (5) i (1) = 5.44
(Lmax/D) = 0.416
(B) Stress Intensification Factors at Point
From Chart 1, P/P* = 1.54
SAFETY VALVE INSTALLATIONS
ACCORDING TO ASME B31.1 APPENDIX II
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
12/13
SOOT BLOWER SV
(6) Predicted Stresses at Point (1) and (2) (7) Calculate the Maximun Operating
( Vent Pipe Analysis at Section 2 and
(A) Predicted Stresses at Point (1), Branch Connection A Vent Pipe must estimated to start t
A good starting size is 3 pipes sizes
Do/tn [for run pipe] = 13.92111369
Do/tn [for branch pipe] = 9.155635063 Vent Pipe Size = DN
Max [Do/tn] = 13.92111369 Vent Pipe I.D. = 8.07
A3
=51.15
Pressure stress(1)
P Do/4 tn = 1565 psi P3 = P1 ( A1 / A3 ) = 11.8
tS = [lesser oftr or(i) tb] = 0.431 in.
rb = 2.2795 in. 35.56382
7.0 in.3
= 0.013
k = 1.1
Flexure stress(1)
0.75 i M(1) / Z(1) = 12527 psi (Lmax/D) = 0.462
Combined stress(1) = Pressure stress(1) + Flexure stress(1) From Chart 1, P/P* = 1.56
= 14092 psi P2 = P3 (P/P*) = 18.4
(B) Predicted Stresses at point (2), buttweld (8) Check for Blowback From Vent Pi
Calculate the velocity V2 that exists a
Do = 5.118 in. (Para. 2.2.1.4)
tn = 0.559 in.
(Lmax/D) = 0.462
Pressure stress(2) V3 = V1 = 1586
P Do/4 tn = 1030 psi
From Chart 1, V / V* = 0.85
Di = 4 in. V2 = V3 (V / V*) = 1348
Check the inequality from Para. 2.3.1.2.
8.3 in.3
W ( V1 - V2 ) / gc = 100
(P2-Pa) A2 - (P1-Pa) A1= 4
EQ. (6) = O.K.
Flexure stress(2) (9) Calculate Forces Acting on Vent P
0.75 i M(2) / Z(2) = 2575 psi
EQ. (3) F2 = 737
Combined stress(2) = Pressure stress(2) + Flexure stress(2) EQ. (3) F3 = 499
= 3604 psi
(C) Comparison of Predicted Stress with Allowable stress
Allowable stress of nozzle material at temperature NOTE
When the vent outlet is perpendicular to the axis o
Sh = 16470.8 psi the vent pipe. This results in a flow that is vertical
[See ASME B31.1 Para.104.8] k = 1.1 When the vent outlet is beveled. This results in a
flow that is not vertical. To take this into account
k Sh = 18117.88 psi the force at the outlet is shown to act at an angle
of 20 with the axis of the vent pipe. This will
Combined stress(1) = 14092 psi introduce a horizontal component force at the outl
Combined stress(2) = 3604 psi O.K.
=p= rZ sb(1) t2
=p= D
D-D32Z
o
4i4o(2)
=
=
D
Lmax
D
L
7/28/2019 Blow Off Pipe Sizing
13/13
NOMENCLATURE and FORMULAS
NOMENCLATURE
P = Absolute pressure, psia
ho = Enthalpy, Btu/lbm
V = Velocity, ft/sec
W = Mass rate of flow, lbm/sec
A = Cross sectional area, in.2
L = lenght, in.
D = Inside diameter, in.
F = Force, lbf
f = Friction fractor = .013
gc = acceleration given to unit mass by init force = 32.2 lbm-ft/lbf-sec
J = Mechanical equivalent of heat = 778.2 ft-lbf/Btu
Pa = Atmospheric press = 15 psia
k = Ratio of specific heats for steam (see table 1)
a = constant (see table 1)
b = constant (see table 1)
Steam Condition k a b
P*=1 to 1000 psia
Superheated 1.3 823 (see note 2) 4.33
P*=1000 to 2000 psia
831 (see note 2)
Saturated 1.1 291 11
NOTES:
1 - These constants are used to represent steam at the sonic velocity.
2 - Normally "P*" will fall in the range of 1 to 1000 psia and "a" will normally equal 823.
FORMULAS
EQ. (1)
EQ. (2)
(asterick denotes values at sonic velocity)
EQ. (3)
EQ. (4)
EQ. (5)
EQ. (6)
TABLE 1 (see note 1)
Section 1
Section 2
Section 3
F2
F3 (see note)
h2 h1
L
Point 2
Point 1
( ) ( )( )1-2bg
Ja-h2bA1-bW*P
c
o=
( )( )1-2b
a-hJg2*V oc=
( )AP-Pg
VWF a
c
+=
IE
h1W
0,1846T
3
=
=
pr
m
r
b
1/2
m
m
2/3
r
m r'T
T'
R
r'
T
R1,5i
( )( ) ( ) 1a12a2
c
21 AP-P-AP-Pg
V-VW>