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278 Chapter 6

16000 Ib 7.5 min 1 ft 3 1 h
3
V= 2 ————— ———— ————— ———— =64.12 ft 3 (1.816m )
1 h 1 62.38 Ib 60 min
From Equation 6.5.4, select an average L/D ratio.

L/D = 4.25

Substitute this ratio into Equation 6.5.3, and solve for D 3 to obtain

V

1.06371 + 2 fuv

Calculate the design pressure from Equation 6.3.1. Because the pressure is
atmospheric, the gage pressure P = 0, and therefore the design pressure is 25 psig
0
(1.72 barg) According to step 3 in Table 6.6, select a torispherical head because
the design pressure is less than 150 psig (10.3 barg). Thus, from Equation 6.5.5,
3
f~Hv = 0.0778. From the above equation for D , we obtain.
64.12
3
3
3
D = ——————————————— = 18.34 ft (0.5194 m )
1.063 (3.142)+ 2 (0.0778)
D = 2.637 ft (31.64 in, 0.803 m)

Because the drum diameter is greater than 30 in (0.762 m) but less than 36 in
(0.914 m), round off D to the highest 6 in increment, which is 36 in (0.914 m).
From Equation 6.5.4, L = 4.25 (3.0) = 12.75 ft (3.89 m). This length requires no
rounding.
Now, calculate the head thickness following the procedure outlined in Table
6.4. From Table 6.1, with no X-ray inspection, the weld efficiency for the weld
joining the head to the shell is 0.80. Because of the acetic acid present in the dis-
tillate, we select SS 316, which has an allowable stress of 15,200 psi (1.04xl05
kPa). For the moment, neglect the corrosion allowance. From Equations 6.3.1 and
6.3.3 for a torispherical head, the head thickness

1.104 (25) (36)
t H = ——————————————— = 0.04086 in (1.04 mm)
2 (0.80) (15200) - 0.2 (25)

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