ρ v = 1.11
ρ l = 33.37
G f = 6,415.8 lb/h ft
2
Tower diameter will be designed to 80 % of flood. Then G A = 5,132.6 lb/h ft
2
Bubble section A B of tray at 80 % flood will be
57, 297
5, 132:6 ¼ 11:16 ft
2 .
Downcomer area (inlet and outlet) A dc :
Downcomer velocity will be 0.4 ft/s (see Table 13).
Then area of one downcomer will be 0.417 Ä 0.4 = 1.04 ft
2 .
Total downcomer area A dc = 2.08 ft
2
.
Waste area of tray A w will be 15 % (see Table 13).
Total tray area A s = A B + A dc + A w .
Then for the top section of the tower, the diameter will be:
A s – 0.15A s = 2.08 + 11.6 ft
2
A s = 15.58 ft
2 diameter = 4.45 f. say 4.5 ft
Similarly for the stripping side of the tower:
G f ¼ 1, 110√3:44 Â 38:9 À 3:44
ð
Þ
¼ 12, 259 lb=h ft
2 and at 80 % flood G a ¼ 9, 807 lb=h ft
2
Downcomer area = 1.715/0.4 = 4.29 ft
2 and A dc = 8.58 ft
2 .
A s = 24 ft
2 and diam is 5.53 ft, say 6 ft.
Tower Hydraulics and Downcomer Filling
Using the pressure drop equations defined earlier, the percentage of downcomer
filled by liquid is calculated. This calculation is based on the stripping section of the
tower only. A similar one will be completed for the rectifying section. Thus:
Clear Liquid Height, h cl
h cl ¼ 0:5  V L Ä N p  L o
À
Á
Â
à 2=3
(7)
V L = 768.4 GPM.
N p = 2 (liquid loading is relatively high so the option of a two pass tray is used).
L o = 58.8 in. Use the correlation given in Appendix 1 of the chapter entitled
“▶ Process Equipment for Petroleum Processing” in this handbook.
Then h cl = 1.74 in. of hot liquid.
Effective Dry Tray, ΔP
(a)
ΔP po ¼ 1:35t m :ρ m =ρ 1 þ K 1 : V o
2
À Á :ρ v =ρ l
(8)
224
D.S.J. Jones
ρ l = 33.37
G f = 6,415.8 lb/h ft
2
Tower diameter will be designed to 80 % of flood. Then G A = 5,132.6 lb/h ft
2
Bubble section A B of tray at 80 % flood will be
57, 297
5, 132:6 ¼ 11:16 ft
2 .
Downcomer area (inlet and outlet) A dc :
Downcomer velocity will be 0.4 ft/s (see Table 13).
Then area of one downcomer will be 0.417 Ä 0.4 = 1.04 ft
2 .
Total downcomer area A dc = 2.08 ft
2
.
Waste area of tray A w will be 15 % (see Table 13).
Total tray area A s = A B + A dc + A w .
Then for the top section of the tower, the diameter will be:
A s – 0.15A s = 2.08 + 11.6 ft
2
A s = 15.58 ft
2 diameter = 4.45 f. say 4.5 ft
Similarly for the stripping side of the tower:
G f ¼ 1, 110√3:44 Â 38:9 À 3:44
ð
Þ
¼ 12, 259 lb=h ft
2 and at 80 % flood G a ¼ 9, 807 lb=h ft
2
Downcomer area = 1.715/0.4 = 4.29 ft
2 and A dc = 8.58 ft
2 .
A s = 24 ft
2 and diam is 5.53 ft, say 6 ft.
Tower Hydraulics and Downcomer Filling
Using the pressure drop equations defined earlier, the percentage of downcomer
filled by liquid is calculated. This calculation is based on the stripping section of the
tower only. A similar one will be completed for the rectifying section. Thus:
Clear Liquid Height, h cl
h cl ¼ 0:5  V L Ä N p  L o
À
Á
Â
à 2=3
(7)
V L = 768.4 GPM.
N p = 2 (liquid loading is relatively high so the option of a two pass tray is used).
L o = 58.8 in. Use the correlation given in Appendix 1 of the chapter entitled
“▶ Process Equipment for Petroleum Processing” in this handbook.
Then h cl = 1.74 in. of hot liquid.
Effective Dry Tray, ΔP
(a)
ΔP po ¼ 1:35t m :ρ m =ρ 1 þ K 1 : V o
2
À Á :ρ v =ρ l
(8)
224
D.S.J. Jones