blow off pipe sizing

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


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


3/13

HOT REHEAT

SAFETY VALVE DATA TAG N. 10LBB10


Name plate flow = 69.4823 lbm/sec. (3) Reaction Force at Discharge E


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 =


Elbow I.D. = 7.98 in. Max F1 = 74


Nozzle material = A182F91

Allowable stress at T = 16115.5 psi (4) Bending Moments at Points (1)


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


gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 144.3

EQ. (1) P1 = 71 psia


Ms(1) =Fs x h1 3175.0


(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


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



=

=

D

Lmax

D

L


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 =


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


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


=p= rZ sb(1) t2

=p= D

D-D

32Z

o

4i4o(2)

=

=

D

Lmax

D

L


5/13

COLD REHEHEAT

SAFETY VALVE DATA TAG N. 10LBC10

Valve set pressure = 624 psia



Orifice size = 16 in.2






Nozzle material = SA 234 WPC

Allowable stress at T = 19831.2 psi (4) Bending Moments at Points (1) and (2)


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.



nozzle "Di" = 6 in.


EQ. (4) T = 0.00717 sec





A1 = 50.01 in.

a = 823 (B) Bending Moments at Points (1) and (2) dueb = 4.33



EQ. (1) P1 = 89 psia


Ms(1) =Fs x h1 2453.44 in.-lb



Elbow I.D. = 7.98 in. (C) Combined Bending Moments at Point (1) a


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.





k = 1.3EQ. (5) i (1) = 2.86

(Lmax/D) = 0.410



=

=

D

Lmax

D

L


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)


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


= 18487 psi P2 = P3 (P/P*) = 60.6 psia


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



Di = 6.25 in. V2 = V3 (V / V*) = 1220 ft/se


26.4 in.3

W ( V1 - V2 ) / gc = 1856

(P2-Pa) A2 - (P1-Pa) A1= 1526

EQ. (6) = O.K.


0.75 i M(2) / Z(2) = 10017 psi

EQ. (3) F2 = 9178 lbf


= 11629 psi




Sh = 19831.2 psi the vent pipe. This results in a flow that is vertical.



k Sh = 23797.44 psi the force at the outlet is shown to act at an angle




=p= rZ sb(1) t2

=p

= D

D-D

32Z

o

4i

4o

(2)

=

=

D

Lmax

D

L


7/13

STEAM DRUM SV-I

SAFETY VALVE DATA TAG NO. 10HAD11



Valve actual flow = 111.84 lbm/sec. Reaction force:Orifice size = 3.976 in.

2








Valve weight = 630.526 lb

Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2) due to


weigth of valve "W" = 630.526 lb

Temperature = 662 F "h1" = 22 in.




nozzle "Di" = 3 in.


EQ. (4) T = 0.01179 sec





A1 = 28.90 in.

a = 291 (B) Bending Moments at Points (1) and (2) due to

b = 11



EQ. (1) P1 = 110 psia


Ms(1) =Fs x h1 2219.452 in.-lb





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


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.






k = 1.1

EQ. (5) i (1) = 0.32

(Lmax/D) = 0.416

(B) Stress Intensification Factors at Point (2), Butt



=

=

D

Lmax

D

L


8/13

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



Max [Do/t

n] =

13.7897563 Vent Pipe I.D. = 12.09 in.A3 = 114.80 in.

Pressure stress(1)



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


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




Di = 3.25 in. V2 = V3 (V / V*) = 758 ft/sec


20.5 in.3

W ( V1 - V2 ) / gc = 877

(P2-Pa) A2 - (P1-Pa) A1= 147

EQ. (6) = O.K.


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









=p= rZ sb(1) t2

=p

= D

D-D

32Z

o

4i

4o

(2)

=

=

D

Lmax

D

L


9/13

STEAM DRUM SV-II

SAFETY VALVE DATA TAG NO. 10HAD11



Valve actual flow = 111.84 lbm/sec. Reaction force:Orifice size = 3.976 in.

2









Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2) due to R



Temperature = 662 F "h1" = 22 in.




nozzle "Di" = 3 in.


EQ. (4) T = 0.01179 sec





A1 = 28.90 in.

a = 291 (B) Bending Moments at Points (1) and (2) due tob = 11



EQ. (1) P1 = 152 psia


Ms(1) =Fs x h1 2219.452 in.-lb





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


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.






k = 1.1

EQ. (5) i (1) = 0.32

(Lmax/D) = 0.416

(B) Stress Intensification Factors at Point (2) Butt



=

=

D

Lmax

D

L


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



Max [Do/t

n] =

13.7897563 Vent Pipe I.D. = 12.09 in.A3 = 114.80 in.

Pressure stress(1)



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


= 18466 psi P2 = P3 (P/P*) = 55.6 psia


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




Di = 3.25 in. V2 = V3 (V / V*) = 1046 ft/sec


20.5 in.

W ( V1 - V2 ) / gc = 1211

(P2-Pa) A2 - (P1-Pa) A1= 693

EQ. (6) = O.K.


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

EQ. (3) F2 = 8293 lbf


= 14707 psi











=p= rZ sb(1) t2

=p

= D

D-D

32Z

o

4i

4o

(2)

=

=

D

Lmax

D

L


11/13

SOOT BLOWER SV

SAFETY VALVE DATA TAG NO. 10HCB11


Name plate flow = 13.472694 lbm/sec. (3) Reaction Force at Discharge Elbow


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


Elbow I.D. = 6.066 in. Max F1 = 833 lb


Nozzle material = SA 234 WPB

Allowable stress at T = 16470.8 psi (4) Bending Moments at Points (1) and


Valve rise time = 0.04 sec. (A) Bending Moment at Points (1) and (2)

STEAM CONDITIONS moment arm "L" = 21.24 in


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



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


gc = 32.2 lbm-ft/lbf-sec Fs = mass x acceleration 45.5024 lb

EQ. (1) P1 = 21 psia


Ms(1) =Fs x h1 728.0384 in


(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


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



=

=

D

Lmax

D

L


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 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


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


= 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



Di = 4 in. V2 = V3 (V / V*) = 1348


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



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





Combined stress(1) = 14092 psi introduce a horizontal component force at the outl


=p= rZ sb(1) t2

=p= D

D-D32Z

o

4i4o(2)

=

=

D

Lmax

D

L


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>

blow off pipe sizing

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