Page 364 - Air Pollution Control Engineering

07_Chap_Wang.qxd 05/05/2004 4:20 pm Page 341
Flare Process 341

Q = [(300 − 180)30,000]/582
f
Q = 6200 scfm of natural gas flow rate
f
3. Determine the flare gas flow rate, using Eq. (2):

Q = 30,000 scfm
e
Q = 6200 scfm
f
Q = Q + Q = 30,000 + 6200 scfm
flg e f
Q = 36,200 scfm
flg
4. Determine the flare gas heat content. Given h < 300 Btu/scf, then h = 300 Btu/scf.
e flg
5. Determine the maximum flare gas exit velocity, U , using Table 2. Given h < 300
max e
Btu/scf, use the equation in Table 2 to calculate U . Thus,
max
U = 3.28 [10 (0.0118h flg + 0.908) ]
max
= 3.28 [10 (0.00118 × 300 + 0.908) ]
U = 60 ft/s
max
6. Determine the flare gas exit velocity, using Eq. (3). Given
Q = 36,200 scfm
flg
T = 95°F
flg
D = 54 in.
tip
then

(5 766. ×10 −3 )( Q flg)( T flg + )
460
U flg = 2
D tip
−
3
,
U = (5 766. × 10 )(36 200 )(95 + 460)
flg 54 2
U = 39.7 = 40 ft s
flg
Because 0.03 ft/s < U = 40 ft/s < U = 60 ft/s, the required level of 98% DE can
flg max
be achieved under these conditions.
7. Determine the steam requirement using Eqs. (4) and (5). Given
Q = 36,200 scfm
flg
MW = 33.5 lb/lb-mol
e
MW = [(Q )(16.7) + (Q )(MW )]/Q
flg f s e flg
MW = [(6200)(16.7) + (30,000)(33.5)]/36,200
flg
= 30.6 lb/lb-mol
Then,
Q = 1.03 × 10 −3 (Q )(MW )
s flg flg
Q = 1.03 × 10 −3 (36,200)(30.6)
s
Q = 1,140 lb/min
s
Example 2

Assume that the flare gas exit velocity U = 40 ft/s (see Example 1); determine the flame
flg
angle of a stream-assisted elevated flare system shown in Fig. 1.