MAIN STEAM LINE

SAFETY VALVE INSTALLATIONS

ACCORDING TO ASME B31.1 APPENDIX II SAFETY VALVE DATA Valve set pressure = Name plate flow = Valve actual flow = Orifice size = Valve inlet I.D. = TAG NO. 2196.2 74.559 74.56 3.341 3 10LBA10 psia lbm/sec. lbm/sec. in.2 in. .. Sch. Max F1 = 7951 lbf

(3) Reaction Force at Discharge Elbow Exit Reaction force: EQ. (3) Calculate F1 = Supply Valve F1 = 7951 lbf 0 lbf

Valve outlet I.D. = 6 in. Valve discharge elbow = 6 in. Elbow I.D. = 6.066 in. Seismic coefficient = 0.16 g Nozzle material = A182F91 Allowable stress at T = 16030 psi Valve weight = 571 lb Valve rise time = 0.04 sec. STEAM CONDITIONS Temperature = Pressure = Enthalpy " ho" =

(4) Bending Moments at Points (1) and (2) (A) Bending Moment at Points (1) and (2) due to Reaction atPoint (1): moment arm "L" = 22.875 in. weigth of valve "W" = 571 lb "h1" = 22 in. "h2" = 12 in. [Young's modulus at des.temp.]E = 25367094 psi nozzle "Do" = 5.65 in. nozzle "Di" = [Moment of inertia Nozzle] "I" = EQ. (4) T = [Valve rise time] to = Ratio t0/T = From Fig. 3-2, DLF =
M1(1) = M1(2) =F1xLxDLF =

20.528

1005.8 F 2196.2 psia 1471.44 Btu/lbm

(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) (Analysis of Section 1) W (actual) = A1 = a= b= J= gc = EQ. (1) P1 = EQ. (2) V1 = 74.56 28.90 831 4.33 778.2 32.2 lbm/sec in.2

3 in. 4 46.0 in. 0.01332 sec 0.04 sec 3.0 1.21 220085 in.-lb

(B) Bending Moments at Points (1) and (2) due to Seismic Loading ft-lbf/Btu lbm-ft/lbf-sec2 Seismic force
Fs = mass x acceleration

91.36 lbf

126 psia 2047 ft/sec Moment arm for Point (1) Ms(1) =Fs x h1 Moment arm for Point (2) Ms(2) =Fs x h2 2009.92 in.-lb

(2) Discharge Elbow Maximun Operating Pressure Elbow I.D. = Height W.N. Flange = L/D = Short Radius Elbow L/D = Pipe Length Pipe = L/D =
L Lmax = = D D

1096.32 in.-lb

6.066 in. 4 in. 0.66

(C) Combined Bending Moments at Point (1) and (2) M(1) = M1(1) + Ms(1) = M(2) = M1(2) + Ms(2) = 222095 in.-lb 221182 in.-lb

30

(5) Stress Intensification Factors at Point (1) and (2) (A) At Point (1), Branch Connection [For nomenclature see FIG. 7-2]

8 in. 1.32 31.98

[Outside Dia. RUN PIPE] Do = [Thickness RUN PIPE] Tr =

[Radius medium RUN PIPE] Rm =

16 in. 1.438 in. 7.281 in. 2.825 in. 1.2 in. 2.225 in. 1.61 1.61 1.793

[Outside radius BRANCH] rp = [Thickness BRANCH] T'b = = k= (Lmax/D) = From Chart 1, P/P* = P1a = P1 (P/P*) = 0.013 1.3 0.416
[Radius medium BRANCH] r'm =

EQ. (5) i (1) =

(B) Stress Intensification Factors at Point (2), Butt-weld 1.64 207 psia i (2) = 1

Pagina 1 di 13

MAIN STEAM LINE

(6) Predicted Stresses at Point (1) and (2) (A) Predicted Stresses at Point (1), Branch Connection Do/tn [for run pipe] = 11.12656 Do/tn [for branch pipe] = 4.708333 Max [Do/tn] = 11.12656 Pressure stress(1) P Do/4 tn = tS = [lesser of tr or (i) tb] = rb = 6109 psi 1.438 in. 2.225 in. 22.4 in.
3

(7) Calculate the Maximun Operating Pressure for Vent Pipe ( Vent Pipe Analysis at Section 2 and Section 3 ) A Vent Pipe must estimated to start the calculation. A good starting size is 3 pipes sizes larger than the safety valve outlet. Vent Pipe Size = Vent Pipe I.D. = A3 = P3 = P1 ( A1 / A 3 ) =
L Lmax = D D

DN 13.123 in. 2 135.26 in.

14

Sch.

40

26.9 psia

21.87 0.013 1.3 0.284 1.8 48.5 psia

includes 30bend 2 Nos

Z (1) = p rb t s =
2

= k= (Lmax/D) = From Chart 1, P/P* = P2 = P3 (P/P*) =

Flexure stress(1) 0.75 i M(1) / Z(1) = 11969 psi

Combined stress(1) = Pressure stress(1) + Flexure stress(1) = 18078 psi

(B) Predicted Stresses at point (2), buttweld Do = tn = Pressure stress(2) P Do/4 tn = Di = 2585 psi 3.25 in. 5.65 in. 1.2 in.

(8) Check for Blowback From Vent Pipe Calculate the velocity V2 that exists at the inlet to the vent pipe (Para. 2.2.1.4) (Lmax/D) = V3 = V 1 = From Chart 1, V / V* = V2 = V3 (V / V*) = 0.284 2047 ft/sec 0.68 1392 ft/sec

Z(2) =

p D o 4 - Di 4 = 32 Do

Check the inequality from Para. 2.3.1.2.

3 15.8 in.

W ( V1 - V2 ) / g c = 1517 (P2-Pa) A2 - (P1-Pa) A1= 1321 EQ. (6) = O.K. (9) Calculate Forces Acting on Vent Pipe 14027 psi EQ. (3) F2 = 7755 lbf 6356 lbf (see note) EQ. (3) F3 =

Flexure stress(2) 0.75 i M(2) / Z(2) =

Combined stress(2) = Pressure stress(2) + Flexure stress(2) = 16612 psi (C) Comparison of Predicted Stress with Allowable stress Allowable stress of nozzle material at temperature Sh = [See ASME B31.1 Para.104.8] k = k Sh = Combined stress(1) = Combined stress(2) = 16030 psi 1.2 19236 psi 18078 psi 16612 psi O.K.

NOTE
When the vent outlet is perpendicular to the axis of the vent pipe. This results in a flow that is vertical. When the vent outlet is beveled. This results in a flow that is not vertical. To take this into account the force at the outlet is shown to act at an angle of 20 with the axis of the vent pipe. This will introduce a horizontal component force at the outlet.

Pagina 2 di 13

HOT REHEAT

SAFETY VALVE INSTALLATIONS

ACCORDING TO ASME B31.1 APPENDIX II SAFETY VALVE DATA Valve set pressure = Name plate flow = Valve actual flow = Orifice size = Valve inlet I.D. = Valve outlet I.D. = Valve discharge elbow = Elbow I.D. = Seismic coefficient = Nozzle material = Allowable stress at T = Valve weight = Valve rise time = STEAM CONDITIONS Temperature = 1004 F Pressure = 551.359 psia Enthalpy " ho" = 1520.977 Btu/lbm TAG N. 551.359 69.4823 69.48 16 6 8 8 7.98 0.16 A182F91 16115.5 902 0.04 10LBB10 psia lbm/sec. lbm/sec. in.2 in. in. in. in. g psi lb sec. .. Sch. Max F1 = 7407 lbf

(3) Reaction Force at Discharge Elbow Exit Reaction force: EQ. (3) Calculate F1 = Supply Valve F1 = 7407 lbf 0 lbf

(4) Bending Moments at Points (1) and (2) (A) Bending Moment at Points (1) and (2) due to Reaction atPoint (1): moment arm "L" = 24 weigth of valve "W" = 902 "h1" = 22 "h2" = 12 [Young's modulus at des.temp.]E = 25381600 nozzle "Do" = 7.75 nozzle "Di" = 6 [Moment of inertia Nozzle] "I" = EQ. (4) T = [Valve rise time] to = Ratio t0/T = From Fig. 3-2, DLF =
M1(1) = M1(2) =F1xLxDLF =

in. lb in. in. psi

20.528

(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) (Analysis of Section 1) W (actual) = A1 = a= b= J= gc = EQ. (1) P1 = EQ. (2) V1 = 69.48 50.01 823 4.33 778.2 32.2 lbm/sec in.2

in. in. 4 113.5 in. 0.01066 sec 0.04 sec 3.8 1.18 209773 in.-lb

(B) Bending Moments at Points (1) and (2) due to Seismic Loading ft-lbf/Btu lbm-ft/lbf-sec2 Seismic force
Fs = mass x acceleration

144.32 lbf

71 psia 2137 ft/sec Moment arm for Point (1) Ms(1) =Fs x h1 Moment arm for Point (2) Ms(2) =Fs x h2 3175.04 in.-lb

(2) Discharge Elbow Maximun Operating Pressure Elbow I.D. = Height W.N. Flange = L/D = Short Radius Elbow L/D = Pipe Length Pipe = L/D =
L Lmax = = D D

1731.84 in.-lb

7.98 in. 4 in. 0.50

(C) Combined Bending Moments at Point (1) and (2) M(1) = M1(1) + Ms(1) = M(2) = M1(2) + Ms(2) = 212948 in.-lb 211504 in.-lb

30

(5) Stress Intensification Factors at Point (1) and (2) (A) At Point (1), Branch Connection [For nomenclature see FIG. 7-2]

10 in. 1.25 31.75

[Outside Dia. RUN PIPE] Do = [Thickness RUN PIPE] Tr =

[Radius medium RUN PIPE] Rm =

26 in. 1 in. 12.5 in. 3.875 in. 0.75 in. 3.5 in. 2.90 2.90 modified for calc purpose only

[Outside radius BRANCH] rp = [Thickness BRANCH] T'b = = k= (Lmax/D) = From Chart 1, P/P* = P1a = P1 (P/P*) = 0.013 1.3 0.413
[Radius medium BRANCH] r'm =

EQ. (5) i (1) =

(B) Stress Intensification Factors at Point (2), Butt-weld 1.64 116 psia i (2) = 1

Pagina 3 di 13

HOT REHEAT

(6) Predicted Stresses at Point (1) and (2) (A) Predicted Stresses at Point (1), Branch Connection Do/tn [for run pipe] = 26 Do/tn [for branch pipe] = 10.33333 Max [Do/tn] = 26 Pressure stress(1) P Do/4 tn = tS = [lesser of tr or (i) tb] = rb = 3584 psi 1 in. 3.5 in.
3 38.5 in.

(7) Calculate the Maximun Operating Pressure for Vent Pipe ( Vent Pipe Analysis at Section 2 and Section 3 ) A Vent Pipe must estimated to start the calculation. A good starting size is 3 pipes sizes larger than the safety valve outlet. Vent Pipe Size = Vent Pipe I.D. = A3 = P3 = P1 ( A1 / A3 ) =
L Lmax = D D

DN 12.09 in. 2 114.80 in.

12

Sch.

30

30.9 psia

23.73863 0.013 1.3 0.309 1.56 48.2 psia

includes 30bend 2 Nos

Z (1) = p rb t s =
2

= k= (Lmax/D) = From Chart 1, P/P* = P2 = P3 (P/P*) =

Flexure stress(1) 0.75 i M(1) / Z(1) = 12018 psi

Combined stress(1) = Pressure stress(1) + Flexure stress(1) = 15602 psi

(B) Predicted Stresses at point (2), buttweld Do = tn = Pressure stress(2) P Do/4 tn = Di = 1424 psi 6.25 in. 7.75 in. 0.75 in.

(8) Check for Blowback From Vent Pipe Calculate the velocity V2 that exists at the inlet to the vent pipe (Para. 2.2.1.4) (Lmax/D) = V3 = V1 = From Chart 1, V / V* = V2 = V3 (V / V*) = 0.309 2137 ft/sec 0.68 1453 ft/sec

Z(2) =

p D o 4 - Di 4 = 32 Do

Check the inequality from Para. 2.3.1.2.

3 26.4 in.

W ( V1 - V2 ) / gc = 1476 (P2-Pa) A2 - (P1-Pa) A1= 1014 EQ. (6) = O.K. (9) Calculate Forces Acting on Vent Pipe 8021 psi EQ. (3) F2 = 6946 lbf 6435 lbf (see note) EQ. (3) F3 =

Flexure stress(2) 0.75 i M(2) / Z(2) =

Combined stress(2) = Pressure stress(2) + Flexure stress(2) = 9445 psi (C) Comparison of Predicted Stress with Allowable stress Allowable stress of nozzle material at temperature Sh = [See ASME B31.1 Para.104.8] k = k Sh = Combined stress(1) = Combined stress(2) = 16115.5 psi 1.2 19338.6 psi 15602 psi 9445 psi O.K.

NOTE
When the vent outlet is perpendicular to the axis of the vent pipe. This results in a flow that is vertical. When the vent outlet is beveled. This results in a flow that is not vertical. To take this into account the force at the outlet is shown to act at an angle of 20 with the axis of the vent pipe. This will introduce a horizontal component force at the outlet.

Pagina 4 di 13

COLD REHEHEAT

SAFETY VALVE INSTALLATIONS

ACCORDING TO ASME B31.1 APPENDIX II SAFETY VALVE DATA Valve set pressure = Name plate flow = Valve actual flow = Orifice size = Valve inlet I.D. = Valve outlet I.D. = Valve discharge elbow = Elbow I.D. = Seismic coefficient = Nozzle material = Allowable stress at T = Valve weight = Valve rise time = STEAM CONDITIONS Temperature = Pressure = Enthalpy " ho" = 642.2 F 624 psia 1315.39 Btu/lbm TAG N. 624 104.03 104.03 16 6 8 8 7.98 0.16 SA 234 WPC 19831.2 902 0.04 10LBC10 psia lbm/sec. lbm/sec. 2 in. in. in. in. in. g psi lb sec. .. Sch. Max F1 = 9508 lbf

(3) Reaction Force at Discharge Elbow Exit Reaction force: EQ. (3) Calculate F1 = Supply Valve F1 = 9508 lbf 0 lbf

(4) Bending Moments at Points (1) and (2) (A) Bending Moment at Points (1) and (2) due to Reaction atPoint (1): moment arm "L" = 24 in. weigth of valve "W" = 902 lb "h1" = 17 in. "h2" = 12 in. [Young's modulus at des.temp.]E = 25863100 psi nozzle "Do" = 7.75 in. nozzle "Di" = [Moment of inertia Nozzle] "I" = EQ. (4) T = [Valve rise time] to = Ratio t0/T = From Fig. 3-2, DLF =
M1(1) = M1(2) =F1xLxDLF =

20.528

(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) (Analysis of Section 1) W (actual) = A1 = a= b= J= gc = EQ. (1) P1 = EQ. (2) V1 = 104.03 50.01 823 4.33 778.2 32.2 lbm/sec in.2

6 in. 4 113.5 in. 0.00717 sec 0.04 sec 5.6 1.15 262421 in.-lb

(B) Bending Moments at Points (1) and (2) due to Seismic Loading ft-lbf/Btu lbm-ft/lbf-sec2 Seismic force
Fs = mass x acceleration

144.32 lbf

89 psia 1795 ft/sec Moment arm for Point (1) Ms(1) =Fs x h1 Moment arm for Point (2) Ms(2) =Fs x h2 2453.44 in.-lb

(2) Discharge Elbow Maximun Operating Pressure Elbow I.D. = Height W.N. Flange = L/D = Short Radius Elbow L/D = Pipe Length Pipe = L/D =
L Lmax = = D D

1731.84 in.-lb

7.98 in. 4 in. 0.50

(C) Combined Bending Moments at Point (1) and (2) M(1) = M1(1) + Ms(1) = M(2) = M1(2) + Ms(2) = 264875 in.-lb 264153 in.-lb

30

(5) Stress Intensification Factors at Point (1) and (2) (A) At Point (1), Branch Connection [For nomenclature see FIG. 7-2]

8 in. 1.00 31.50

[Outside Dia. RUN PIPE] Do = [Thickness RUN PIPE] Tr =

[Radius medium RUN PIPE] Rm =

24 in. 1 in. 11.5 in. 3.875 in. 0.75 in. 3.5 in. 2.86 2.86 modified for calc purpose only

[Outside radius BRANCH] rp = [Thickness BRANCH] T'b = = k= (Lmax/D) = From Chart 1, P/P* = P1a = P1 (P/P*) = 0.013 1.3 0.410
[Radius medium BRANCH] r'm =

EQ. (5) i

(1)

(B) Stress Intensification Factors at Point (2), Butt-weld 1.64 146 psia i (2) = 1

Pagina 5 di 13

COLD REHEHEAT

(6) Predicted Stresses at Point (1) and (2) (A) Predicted Stresses at Point (1), Branch Connection Do/tn [for run pipe] = Do/tn [for branch pipe] = Max [Do/tn] = Pressure stress(1) P Do/4 tn = tS = [lesser of tr or (i) tb] = rb = 3744 psi 1 in. 3.5 in.
3 38.5 in.

(7) Calculate the Maximun Operating Pressure for Vent Pipe ( Vent Pipe Analysis at Section 2 and Section 3 ) A Vent Pipe must estimated to start the calculation. A good starting size is 3 pipes sizes larger than the safety valve outlet. Vent Pipe Size = Vent Pipe I.D. = A3 = P3 = P1 ( A1 / A3 ) =
L Lmax = D D

24 10.33333333 24 DN 12.09 in. 2 114.80 in. 38.8 psia 12 Sch. 30

23.73863 0.013 1.3 0.309 1.56 60.6 psia

includes 30bend 2 Nos

Z (1) = p rb 2 t s =
Flexure stress(1) 0.75 i M(1) / Z(1) =

= k= (Lmax/D) = From Chart 1, P/P* = P2 = P3 (P/P*) =

14743 psi

Combined stress(1) = Pressure stress(1) + Flexure stress(1) = 18487 psi

(B) Predicted Stresses at point (2), buttweld Do = tn = Pressure stress(2) P Do/4 tn = Di = 1612 psi 6.25 in. 7.75 in. 0.75 in.

(8) Check for Blowback From Vent Pipe Calculate the velocity V2 that exists at the inlet to the vent pipe (Para. 2.2.1.4) (Lmax/D) = V3 = V1 = From Chart 1, V / V* = V2 = V3 (V / V*) = 0.309 1795 ft/sec 0.68 1220 ft/sec

Z(2) =

p D o 4 - Di 4 = 32 Do

Check the inequality from Para. 2.3.1.2.

3 26.4 in.

W ( V 1 - V2 ) / g c = 1856 (P2-Pa) A2 - (P1-Pa) A1= 1526 EQ. (6) = O.K. (9) Calculate Forces Acting on Vent Pipe 10017 psi EQ. (3) F2 = 9178 lbf 8536 lbf (see note) EQ. (3) F3 =

Flexure stress(2) 0.75 i M(2) / Z(2) =

Combined stress(2) = Pressure stress(2) + Flexure stress(2) = 11629 psi (C) Comparison of Predicted Stress with Allowable stress Allowable stress of nozzle material at temperature Sh = [See ASME B31.1 Para.104.8] k = k Sh = Combined stress(1) = Combined stress(2) = 19831.2 psi 1.2 23797.44 psi 18487 psi 11629 psi O.K.

NOTE
When the vent outlet is perpendicular to the axis of the vent pipe. This results in a flow that is vertical. When the vent outlet is beveled. This results in a flow that is not vertical. To take this into account the force at the outlet is shown to act at an angle of 20 with the axis of the vent pipe. This will introduce a horizontal component force at the outlet.

Pagina 6 di 13

STEAM DRUM SV-I

SAFETY VALVE INSTALLATIONS

ACCORDING TO ASME B31.1 APPENDIX II SAFETY VALVE DATA Valve set pressure = Name plate flow = Valve actual flow = Orifice size = Valve inlet I.D. = TAG NO. 2407.84 111.83867 111.84 3.976 3 10HAD11 psia lbm/sec. lbm/sec. in.2 in. .. Sch. Max F1 = 6264 lbf

(3) Reaction Force at Discharge Elbow Exit Reaction force: EQ. (3) Calculate F1 = Supply Valve F1 = 6264 lbf 0 lbf

Valve outlet I.D. = 6 in. Valve discharge elbow = 6 in. Elbow I.D. = 6.066 in. Seismic coefficient = 0.16 g Nozzle material = SA 234 WPC Allowable stress at T = 19440.47 psi Valve weight = 630.526 lb Valve rise time = 0.04 sec. STEAM CONDITIONS Temperature = Pressure = Enthalpy " ho" =

(4) Bending Moments at Points (1) and (2) (A) Bending Moment at Points (1) and (2) due to Reaction atPoint (1): moment arm "L" = 22.875 in. weigth of valve "W" = 630.526 lb "h1" = 22 in. "h2" = 12 in. [Young's modulus at des.temp.]E = 25671700 psi nozzle "Do" = 6.102 in. nozzle "Di" = [Moment of inertia Nozzle] "I" = EQ. (4) T = [Valve rise time] to = Ratio t0/T = From Fig. 3-2, DLF =
M1(1) = M1(2) =F1xLxDLF =

20.528

662 F 2407.84 psia 718.4712 Btu/lbm

(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) (Analysis of Section 1) W (actual) = A1 = a= b= J= gc = EQ. (1) P1 = EQ. (2) V1 = 111.84 28.90 291 11 778.2 32.2 lbm/sec in.2

3 in. 4 64.1 in. 0.01179 sec 0.04 sec 3.4 1.23 176238 in.-lb

(B) Bending Moments at Points (1) and (2) due to Seismic Loading ft-lbf/Btu lbm-ft/lbf-sec2 Seismic force
Fs = mass x acceleration 100.8842 lbf

110 psia 1010 ft/sec Moment arm for Point (1) Ms(1) =Fs x h1 2219.452 in.-lb Moment arm for Point (2) Ms(2) =Fs x h2

(2) Discharge Elbow Maximun Operating Pressure Elbow I.D. = Height W.N. Flange = L/D = Short Radius Elbow L/D = Pipe Length Pipe = L/D = 6.066 in. 4 in. 0.66

1210.61 in.-lb

(C) Combined Bending Moments at Point (1) and (2) M(1) = M1(1) + Ms(1) = M(2) = M1(2) + Ms(2) = 178457 in.-lb 177449 in.-lb

30

(5) Stress Intensification Factors at Point (1) and (2) (A) At Point (1), Branch Connection [For nomenclature see FIG. 7-2]

8 in. 1.32

[Outside Dia. RUN PIPE] Do = [Thickness RUN PIPE] Tr =

[Radius medium RUN PIPE] Rm =

66.77 in. 4.842 in. 30.964 in. 3.051 in. 1.426 in. 2.338 in. 0.32 1.00

L Lmax 31.98 = D D
= k= (Lmax/D) = 0.013 1.1 0.416

[Outside radius BRANCH] rp = [Thickness BRANCH] T'b =

[Radius medium BRANCH] r'm =

EQ. (5) i (1) =

(B) Stress Intensification Factors at Point (2), Butt-weld From Chart 1, P/P* = P1a = P1 (P/P*) = 1.54 170 psia i (2) = 1

Pagina 7 di 13

STEAM DRUM SV-I

(6) Predicted Stresses at Point (1) and (2) (A) Predicted Stresses at Point (1), Branch Connection Do/tn [for run pipe] = 13.7897563 Do/tn [for branch pipe] = 4.27910238 Max [Do/tn] = 13.7897563 Pressure stress(1) P Do/4 tn = tS = [lesser of tr or (i) tb] = rb = 8301 psi 1.426 in. 2.338 in. 24.5 in.
3

(7) Calculate the Maximun Operating Pressure for Vent Pipe ( Vent Pipe Analysis at Section 2 and Section 3 ) A Vent Pipe must estimated to start the calculation. A good starting size is 3 pipes sizes larger than the safety valve outlet. Vent Pipe Size = Vent Pipe I.D. = A3 = P3 = P1 ( A1 / A 3 ) = DN 12.09 in. 2 114.80 in. 27.8 psia 12 Sch. 30

Z (1) = p rb 2 t s =
Flexure stress(1) 0.75 i M(1) / Z(1) =

L Lmax 23.73863 = D D
= k= 0.013 1.1 0.309 1.45 40.3 psia

includes 30bend 2 Nos

7287 psi

(Lmax/D) = From Chart 1, P/P* = P2 = P3 (P/P*) =

Combined stress(1) = Pressure stress(1) + Flexure stress(1) = 15588 psi

(B) Predicted Stresses at point (2), buttweld Do = tn = Pressure stress(2) P Do/4 tn = Di = 2576 psi 3.25 in. 6.102 in. 1.426 in.

(8) Check for Blowback From Vent Pipe Calculate the velocity V2 that exists at the inlet to the vent pipe (Para. 2.2.1.4) (Lmax/D) = V3 = V 1 = From Chart 1, V / V* = V2 = V3 (V / V*) = 0.309 1010 ft/sec 0.75 758 ft/sec

Z(2)

p Do 4 - Di 4 20.5 in.3 = = 32 Do

Check the inequality from Para. 2.3.1.2. W ( V1 - V2 ) / g c = 877 (P2-Pa) A2 - (P1-Pa) A1= 147 EQ. (6) = O.K. (9) Calculate Forces Acting on Vent Pipe

Flexure stress(2) 0.75 i M(2) / Z(2) = 8651 psi

EQ. (3) F2 = Combined stress(2) = Pressure stress(2) + Flexure stress(2) = 11227 psi (C) Comparison of Predicted Stress with Allowable stress Allowable stress of nozzle material at temperature Sh = [See ASME B31.1 Para.104.8] k = k Sh = Combined stress(1) = Combined stress(2) = 19440.47 psi 1.1 21384.517 psi 15588 psi 11227 psi O.K. EQ. (3) F3 =

5533 lbf 4975 lbf (see note)

NOTE
When the vent outlet is perpendicular to the axis of the vent pipe. This results in a flow that is vertical. When the vent outlet is beveled. This results in a flow that is not vertical. To take this into account the force at the outlet is shown to act at an angle of 20 with the axis of the vent pipe. This will introduce a horizontal component force at the outlet.

Pagina 8 di 13

STEAM DRUM SV-II

SAFETY VALVE INSTALLATIONS

ACCORDING TO ASME B31.1 APPENDIX II SAFETY VALVE DATA Valve set pressure = Name plate flow = Valve actual flow = Orifice size = Valve inlet I.D. = TAG NO. 2393.34 111.83867 111.84 3.976 3 10HAD11 psia lbm/sec. lbm/sec. 2 in. in. .. Sch. Max F1 = 8811 lbf

(3) Reaction Force at Discharge Elbow Exit Reaction force: EQ. (3) Calculate F1 = Supply Valve F1 = 8811 lbf 0 lbf

Valve outlet I.D. = 6 in. Valve discharge elbow = 6 in. Elbow I.D. = 6.066 in. Seismic coefficient = 0.16 g Nozzle material = SA 234 WPC Allowable stress at T = 19440.47 psi Valve weight = 630.526 lb Valve rise time = 0.04 sec. STEAM CONDITIONS Temperature = Pressure = Enthalpy " ho" =

(4) Bending Moments at Points (1) and (2) (A) Bending Moment at Points (1) and (2) due to Reaction atPoint (1): moment arm "L" = 22.875 in. weigth of valve "W" = 630.526 lb "h1" = 22 in. "h2" = 12 in. [Young's modulus at des.temp.] E = 25671700 psi nozzle "Do" = 6.102 in. nozzle "Di" = [Moment of inertia Nozzle] "I" = EQ. (4) T = [Valve rise time] to = Ratio t0/T = From Fig. 3-2, DLF =
M1(1) = M1(2) =F1xLxDLF =

20.528

662 F 2393.34 psia 1105.5638 Btu/lbm

(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) (Analysis of Section 1) W (actual) = A1 = a= b= J= gc = EQ. (1) P1 = EQ. (2) V1 = 111.84 28.90 291 11 778.2 32.2 lbm/sec in.2

3 in. 4 64.1 in. 0.01179 sec 0.04 sec 3.4 1.23 247921 in.-lb

(B) Bending Moments at Points (1) and (2) due to Seismic Loading ft-lbf/Btu lbm-ft/lbf-sec2 Seismic force
Fs = mass x acceleration

100.8842 lbf

152 psia 1394 ft/sec Moment arm for Point (1) Ms(1) =Fs x h1 2219.452 in.-lb Moment arm for Point (2) Ms(2) =Fs x h2

(2) Discharge Elbow Maximun Operating Pressure Elbow I.D. = Height W.N. Flange = L/D = Short Radius Elbow L/D = Pipe Length Pipe = L/D = 6.066 in. 4 in. 0.66

1210.61 in.-lb

(C) Combined Bending Moments at Point (1) and (2) M(1) = M1(1) + Ms(1) = M(2) = M1(2) + Ms(2) = 250140 in.-lb 249132 in.-lb

30

(5) Stress Intensification Factors at Point (1) and (2) (A) At Point (1), Branch Connection [For nomenclature see FIG. 7-2]

8 in. 1.32 31.98

[Outside Dia. RUN PIPE] Do = [Thickness RUN PIPE] Tr =

[Radius medium RUN PIPE] Rm =

66.77 in. 4.842 in. 30.964 in. 3.051 in. 1.426 in. 2.338 in. 0.32 1.00

L Lmax = D D

[Outside radius BRANCH] rp = [Thickness BRANCH] T'b = = k= (Lmax/D) = From Chart 1, P/P* = P1a = P1 (P/P*) = 0.013 1.1 0.416
[Radius medium BRANCH] r'm =

EQ. (5) i (1) =

(B) Stress Intensification Factors at Point (2), Butt-weld 1.54 235 psia i (2) = 1

Pagina 9 di 13

STEAM DRUM SV-II

(6) Predicted Stresses at Point (1) and (2) (A) Predicted Stresses at Point (1), Branch Connection Do/tn [for run pipe] = Do/tn [for branch pipe] = Max [Do/tn] = Pressure stress(1) P Do/4 tn = tS = [lesser of tr or (i) tb] = rb = 8251 psi 1.426 in. 2.338 in. 24.5 in.
3

(7) Calculate the Maximun Operating Pressure for Vent Pipe ( Vent Pipe Analysis at Section 2 and Section 3 ) A Vent Pipe must estimated to start the calculation. A good starting size is 3 pipes sizes larger than the safety valve outlet. Vent Pipe Size = Vent Pipe I.D. = A3 = P 3 = P1 ( A1 / A 3 ) =
L Lmax = D D

13.7897563 4.27910238 13.7897563 DN 12.09 in. 2 114.80 in. 38.3 psia 12 Sch. 30

23.73863 0.013 1.1 0.309 1.45 55.6 psia

includes 30bend 2 Nos

Z (1) = p rb t s =
2

= k= (Lmax/D) = From Chart 1, P/P* = P2 = P3 (P/P*) =

Flexure stress(1) 0.75 i M(1) / Z(1) = 10215 psi

Combined stress(1) = Pressure stress(1) + Flexure stress(1) = 18466 psi

(B) Predicted Stresses at point (2), buttweld Do = tn = Pressure stress(2) P Do/4 tn = Di = 2560 psi 3.25 in. 6.102 in. 1.426 in.

(8) Check for Blowback From Vent Pipe Calculate the velocity V2 that exists at the inlet to the vent pipe (Para. 2.2.1.4) (Lmax/D) = V 3 = V1 = From Chart 1, V / V* = V2 = V3 (V / V*) = 0.309 1394 ft/sec 0.75 1046 ft/sec

Z(2) =

p D o 4 - Di 4 = 32 Do

Check the inequality from Para. 2.3.1.2.

3 20.5 in.

W ( V1 - V2 ) / gc = (P2-Pa) A2 - (P1-Pa) A1= EQ. (6) = O.K.

1211 693

Flexure stress(2) 0.75 i M(2) / Z(2) = 12146 psi

(9) Calculate Forces Acting on Vent Pipe EQ. (3) F2 = 8293 lbf 7523 lbf (see note)

Combined stress(2) = Pressure stress(2) + Flexure stress(2) = 14707 psi (C) Comparison of Predicted Stress with Allowable stress Allowable stress of nozzle material at temperature Sh = [See ASME B31.1 Para.104.8] k = k Sh = Combined stress(1) = Combined stress(2) = 19440.47 psi 1.1 21384.517 psi 18466 psi 14707 psi O.K.

EQ. (3) F3 =

NOTE
When the vent outlet is perpendicular to the axis of the vent pipe. This results in a flow that is vertical. When the vent outlet is beveled. This results in a flow that is not vertical. To take this into account the force at the outlet is shown to act at an angle of 20 with the axis of the vent pipe. This will introduce a horizontal component force at the outlet.

Pagina 10 di 13

SOOT BLOWER SV

SAFETY VALVE INSTALLATIONS

ACCORDING TO ASME B31.1 APPENDIX II SAFETY VALVE DATA Valve set pressure = Name plate flow = Valve actual flow = Orifice size = Valve inlet I.D. = Valve outlet I.D. = Valve discharge elbow = Elbow I.D. = Seismic coefficient = Nozzle material = Allowable stress at T = Valve weight = Valve rise time = STEAM CONDITIONS Temperature = Pressure = Enthalpy " ho" = 671 F 449.802 psia 1344.4201 Btu/lbm TAG NO. 449.802 13.472694 13.47 2.853 4 6 6 6.066 0.16 SA 234 WPB 16470.8 284.39 0.04 10HCB11 psia lbm/sec. lbm/sec. in.2 in. in. in. in. g psi lb sec. .. Sch. Max F1 = 833 lbf

(3) Reaction Force at Discharge Elbow Exit Reaction force: EQ. (3) Calculate F1 = Supply Valve F1 = 833 lbf 0 lbf

(4) Bending Moments at Points (1) and (2) (A) Bending Moment at Points (1) and (2) due to Reaction atPoint (1): moment arm "L" = 21.24 in. weigth of valve "W" = 284.39 lb "h1" = 16 in. "h2" = 8 in. [Young's modulus at des.temp.]E = 25570200 psi nozzle "Do" = 5.118 in. nozzle "Di" = [Moment of inertia Nozzle] "I" = EQ. (4) T = [Valve rise time] to = Ratio t0/T = From Fig. 3-2, DLF =
M1(1) = M1(2) =F1xLxDLF =

20.528

(1) Pressure and Velocity at Discharge Elbow Exit (Para. 2.2.1) (Analysis of Section 1) W (actual) = A1 = a= b= J= gc = EQ. (1) P1 = EQ. (2) V1 = 13.47 28.90 291 11 778.2 32.2 lbm/sec in.2

4 in. 4 21.1 in. 0.00857 sec 0.04 sec 4.7 1.18 20878 in.-lb

(B) Bending Moments at Points (1) and (2) due to Seismic Loading ft-lbf/Btu lbm-ft/lbf-sec2 Seismic force
Fs = mass x acceleration

45.5024 lbf

21 psia 1586 ft/sec Moment arm for Point (1) Ms(1) =Fs x h1 728.0384 in.-lb Moment arm for Point (2) Ms(2) =Fs x h2 364.0192 in.-lb (C) Combined Bending Moments at Point (1) and (2) M(1) = M1(1) + Ms(1) = M(2) = M1(2) + Ms(2) = 21606 in.-lb 21242 in.-lb

(2) Discharge Elbow Maximun Operating Pressure Elbow I.D. = Height W.N. Flange = L/D = Short Radius Elbow L/D = Pipe Length Pipe = L/D = 6.066 in. 4 in. 0.66

30

(5) Stress Intensification Factors at Point (1) and (2) (A) At Point (1), Branch Connection [For nomenclature see FIG. 7-2]

8 in. 1.32 31.98

[Outside Dia. RUN PIPE] Do = [Thickness RUN PIPE] Tr =

[Radius medium RUN PIPE] Rm =

6 in. 0.431 in. 2.7845 in. 2.559 in. 0.559 in. 2.2795 in. 5.44 5.44 modified for calc purpose only

L Lmax = = D D
= k= (Lmax/D) = From Chart 1, P/P* = P1a = P1 (P/P*) =

[Outside radius BRANCH] rp = [Thickness BRANCH] T'b = 0.013 1.1 0.416

[Radius medium BRANCH] r'm =

EQ. (5) i

(1)

(B) Stress Intensification Factors at Point (2), Butt-weld 1.54 32 psia i (2) = 1

Pagina 11 di 13

SOOT BLOWER SV (6) Predicted Stresses at Point (1) and (2) (A) Predicted Stresses at Point (1), Branch Connection Do/tn [for run pipe] = Do/tn [for branch pipe] = Max [Do/tn] = Pressure stress(1) P Do/4 tn = tS = [lesser of tr or (i) tb] = rb = 1565 psi 0.431 in. 2.2795 in. 7.0 in.
3

(7) Calculate the Maximun Operating Pressure for Vent Pipe ( Vent Pipe Analysis at Section 2 and Section 3 ) A Vent Pipe must estimated to start the calculation. A good starting size is 3 pipes sizes larger than the safety valve outlet. Vent Pipe Size = Vent Pipe I.D. = A3 = P3 = P1 ( A1 / A3 ) = DN 8.07 in. 2 51.15 in. 11.8 psia 8 Sch. 30

13.92111369 9.155635063 13.92111369

Z (1) = p rb 2 t s =
Flexure stress(1) 0.75 i M(1) / Z(1) =

L Lmax 35.56382 = D D
= k= 0.013 1.1 0.462 1.56 18.4 psia

includes 30bend 2 Nos

12527 psi

(Lmax/D) = From Chart 1, P/P* = P2 = P3 (P/P*) =

Combined stress(1) = Pressure stress(1) + Flexure stress(1) = 14092 psi

(B) Predicted Stresses at point (2), buttweld Do = tn = Pressure stress(2) P Do/4 tn = Di = 1030 psi 4 in. 5.118 in. 0.559 in.

(8) Check for Blowback From Vent Pipe Calculate the velocity V2 that exists at the inlet to the vent pipe (Para. 2.2.1.4) (Lmax/D) = V3 = V1 = From Chart 1, V / V* = V2 = V3 (V / V*) = 0.462 1586 ft/sec 0.85 1348 ft/sec

Z(2)

p Do 4 - Di 4 8.3 in. = = 32 Do

Check the inequality from Para. 2.3.1.2.

3

W ( V 1 - V 2 ) / gc = (P2-Pa) A2 - (P1-Pa) A1= EQ. (6) = O.K.

100 4

Flexure stress(2) 0.75 i M(2) / Z(2) = 2575 psi

(9) Calculate Forces Acting on Vent Pipe EQ. (3) F2 = 737 lbf 499 lbf (see note)

Combined stress(2) = Pressure stress(2) + Flexure stress(2) = 3604 psi (C) Comparison of Predicted Stress with Allowable stress Allowable stress of nozzle material at temperature Sh = [See ASME B31.1 Para.104.8] k = k Sh = Combined stress(1) = Combined stress(2) = 16470.8 psi 1.1 18117.88 psi 14092 psi 3604 psi O.K.

EQ. (3) F3 =

NOTE
When the vent outlet is perpendicular to the axis of the vent pipe. This results in a flow that is vertical. When the vent outlet is beveled. This results in a flow that is not vertical. To take this into account the force at the outlet is shown to act at an angle of 20 with the axis of the vent pipe. This will introduce a horizontal component force at the outlet.

Pagina 12 di 13

NOMENCLATURE and FORMULAS NOMENCLATURE

F3
(see note)

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-sec2 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) Point 2 b = constant (see table 1) Point 1 TABLE 1 (see note 1) Steam Condition k a P*=1 to 1000 psia Superheated 1.3 823 (see note 2) P*=1000 to 2000 psia 831 (see note 2) Saturated 1.1 291 b 4.33
L

Section 3

F2

Section 2 Section 1 h2
h1

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)

P* =

W (b - 1) Ab

2(h o - a ) J g c (2b - 1)

EQ. (2)

V*=

2 gc J (ho - a) (2b - 1)

(asterick denotes values at sonic velocity)

EQ. (3)

F=

WV + (P - Pa ) A gc

EQ. (4)

T = 0,1846
Rm i = 1,5 T r
2/3

W h13 EI
r'm R m
1/2

EQ. (5)

T' b r'm T r r p

EQ. (6)

W (V1 - V2 ) > (P2 - Pa ) A 2 - (P1 - Pa ) A1 gc