Page 499 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

Applied Process Design 465

1.13. Disk burst pressure to be 10 psig. Nli-\WP of vessel is Disk Area
50 psig. Discharge line back pressure is 1 psig.
( � )
Heal Input Flow is subcritical, since pl > �
pi
Using ASME flanged and dished heads (F&D) from
Appendix Tables of Blanks, the circle size is 152 in. for a
12-ft diameter tank. Then add 3-in. straight flange which
becomes 158 in. which is 158/12 = 13.166 fl diameter.
Area of this diameter for surface area of head= 136.14 sq
ft equivalent surface area of one head. For a horizontal 'vV = 106,000 lbs/hr
vessel there are two heads possibly exposed to fire. Kn= 0.62
Pb= 10 + 14.7 = 24.7 psia (10) (l.lO see Fig. 7-llA) + 14.7
External Surface Area: cylindrical area + surface area = 25.7 psia
heads (2) P2 = 15.7 psia
M = 154
= rt (12) (36) + ( l:'5§) 2 ( � ) 2 T = 460 + 202 = 662° R (B.P. of CCL 1@ 10 psig)
k=l.13
\ 12 4
1357 + 274 = 1629 sq ft (approximate) F 2 = 0.691 interpolated from Figure 7-29

Area calculated substituting in above relation:
Total Heat Input (From Figure 7-30)
A = 43.52 sq in.
6
Q = 9.0 X 10 BTU/hr. at area of 1629 sq ft
8-inch Sch. 40 pipe has a cross-sectional area of 50.0 sq in.
Solving equation: Q = 0 21,000 F A0.82 so an 8-inch frangible rupture disk will be satisfactory.
82 Disk material to be lead or lead covered aluminum.
Q = 0 21,000 ( 1.0) (1629) O = 9,036,300
BTU/hr
Example 7-6: Rupture Disk for Vapors or Gases; Non-
Fire Condition
Quantity of Vapor Released
Determine the rupture disk size required to relieve the
Latent Heat CC1 4 = 85 BTU/lb pressure in a process vessel with the following conditions:
k = 1.4.
9,000,000 .
·w = = 106, 000 lb I hr (rounded) Vessel: MAWP = 85 psig; also = disk set pressure
85
Vapor flow to relieve: 12,000 std. cu fl/min@ 60°F and
Critical Flow Pressure 14.7 psia
Flowing temperature: 385°F
Vapor mol. wt: 28

( 2 ·,k./•:k-l) ( 2 Jl.13/(1.13-1) Backpressure on discharge of disk: 30 psig
P = P --J = 24. 7 ---
c k + 1 1.13 + 1
Determine if conditions on rupLure are critical or non-
critical, based on 10% overpressure for primary relief.
Pc =27.7(0.573)=15.8psia,or
P (30+14.7)
�= =44.7/108.2=0.413
P,
- " actual= ( 14.7+1.0 J = 0.610 pl [(85)(1.10)+ 14.7]
pl (10)(1.1)+ 14.7

[--2--] 3 . . . [ 2 I:� I _
]
� = 1. · = 0.573 Critical pressure ratio: = Pc IP = -- , = 0.::>28
P ( 1.13 + l) ( l.3 - I) 1 1.4 + 1
1