Page 474 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

440 Applied Process Design for Chemical and Petrochemical Plants

0.60 1 k � I Vl ma.� = maximum mass flow at critical or choked
\ I I 2 conditions, lb/sec
1
059 Pc /P • (k+I) 1 I
\ Pc= Pcrit = critical flow throat pressure, psia = sonic =
0.58 Pc = Critical Pressure, psi a. choked pressure = maximum downstream
\ P • Total Absolute Pressure Upstream pressure producing maximum flow
0.57 of Safety Valve at Full Relief, _
\ Equals Set Pressure pkJs
Q56 Overpressure , psla. When/if the downstream pressure exceeds the critical
\ flow pressure, then sub-critical pressure will occur and the
0.55
\ equations for sub-critical flow should be used.
a. 0.54 I\ When the downstream pressure is less than ( or below)
.....
a." \ the critical or choked pressure, the velocity and fluid flow
• 0.53 rate at a restriction or throat will not/ cannot be increased
.s ' \
.. \ fluid velocity at the restriction or throat is the velocity of
� Q52 by lowering the downstream pressure further, and the
.. I sound at the conditions [29].
� 0.5
.t 0.50 I\ The critical or sonic ratio is conveniently shown on Fig-
\
.; � ure 7-21, but this does not eliminate the need for calcu-
:= 0.49 lating the PJP1 ratio for a more accurate result.
·.::: \
u
0118 -, Example 7-2: Flow through Sharp Edged Vent Orifice
0117 (adapted after Ref. [29])
I\
0116
'\ A small hole bas been deliberately placed in a vessel near
0!45 I\ the top to provide a controlled vent for a nitrogen
\ purge/blanket. The hole is 0.2-inch diameter with the vessel
0·4 \o I.I 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 operating at 150 psig at 100°F. Determine the flow through
Ratio of Specific Heats, k = Cp/Cv this vent hole. Assume it acts as a sharp edged orifice.
Figure 7-21. Critical backpressure ratio for vapors and gases.
k (for nitrogen) = 1.4
Pc/P1 = See Equation 7-7 [2/(k + l)Jk/k - 1
1
Pc= (150 + 14.7) (2/1.4 + ])l.4/1.4 - = 86.9 psia, critical
At critical conditions, the maximum flow through the pressure
2
2
nozzle or orifice is: [29] Hole area= A= 1t d /4 = 1t (0.2) /4 = 0.0314 sq in.
= 0.0002181 sq ft
Discharge coef, (\ = assumed = 1.0 (Note, could calculate
Re to verify.)
\V max (crirical now) = Co AP o (7- 9) Inside pressure= 150 psig + 14.7 = 164.7 psia
T 0 = 100 + 460 = 560°R

where M = mo! wt of vapor or gas
0
TO = temp of service, R v\lma, = [1.0 (0.0002181)(164.7)(144sq in./sq ft)]
Rg = ideal universal gas constant = 1544, ft lb/lb- [ [ (1.4)(32.12)(28)](-2-)(l. 4+l)/(l.-1-l) l
mol-0R, also= MR
C = discharge coefficient for sharp-edged orifice -� ( 1545 ) ( 560 ) 1.4 + 1
0
= 0.61 for Reynolds Number> 30,000 and
not sonic
C = 1.0 for sonic flow, C increases from 0.61 to 1.0. critical flow rate, Wma.x = 0.002518 lb/sec
0
0
Use 1.0 Lo be conservative [29] [36]. 5th Ed.]
A = area of opening, orifice, or hole, or nozzle, Orifice Area Calculations [68]
sq. ft.
P 1 = P = upstream pressure, lb/sq foot, abs, (psfa) Calculations of Orifice Flow Area for Conventional
0
g = 32.12 ft/sec/sec Pressure Relieving Valves, and Flow is Critical (sonic)
M = mo!. wt. of vapor or gas Through Part of Relieving System, i.e., backpressure is
� = ratio of orifice diameter/pipe diameter (or less than 55% of the absolute relieving pressure (includ-
nozzle inlet diameter) ing set pressure plus accumulation). See Figure 7-7A, use