Page 248 - Chemical Process Equipment - Selection and Design

218 HEAT TRANSFER AND HEAT EXCHANGERS
EXAMPLE 8.15-(continued) With four shield tubes, equilateral spacing and 3 in. distance
to walls,
radiant rate of 10,000 lead to an initial specification of 87 tubes, but
90 were taken. The final results are quite close to the estimates, 104’907(12) = 0.325 lb/sec sqft.
being 77.1% to the radiant zone and 9900Btu/(hr)(sqft) with 90 G= 3600(38.5)(27.98)
tubes. If the radiant rate comes out much different from the desired
value, the number of tubes is changed accordingly. 13. The 90 radiant tubes are arranged as shown on Figure 8.22:
Because of the changing temperature of the process stream, the 4 shields, 14 at the ceiling, and 36 on each wall. Dimensions
heat flux also deviates from the average value. This variation is of the shell are shown.
estimated roughly from the variation of the quantity 14. A, = (38.5)(1)(90 - 4) = 3311 sqft.
15. Inside surface of the shell is
/3 = 1730(T: - Ti) + 7.0(T, - TL),
A, = 2[20(37 + 38.5) + 37(38.5)] = 5869 sqft.
where the gas temperature Tg, in the radiant zone is constant and Refractory surface,
TL is the temperature of the process stream, both in “R. In
comparison with the average flux, the effect is a slightly increased A, = 5869 - 3311 = 2558 sqft.
preheat rate and a reduced flux in the reaction zone. The inside skin
temperature also can be estimated on the reasonable assumptions of 16. (Center-to-center)/OD = 12/6.625 = 1.81,
heat transfer film coefficients of more than 100 before cracking
starts and more than 200 at the outlet. For the conditions of this a = 0.917, single rows of tubes [Eq. (5)].
example, with Q/A = 9900 and T, = 2011”R, these results are
obtained:
17. Effective absorptivity:
T,W 8/fh T,,inFW “AR = 4(38.5)(1) + 0.917(3311) = 3190 sqft,
547 1.093 >IO0 <655 A,/aA, = 2558/3190 = 0.8018.
724 1 ,100 <823
900 1.878 >200 <943 18. Mean beam length:

The equation numbers cited following are from Table 8.16. The L = (2/3)(20 X 37 X 38.5)l” = 20.36.
step numbers used following are the same as those in Table 8.17:
19. From Eq. (6), with 25% excess air,
1. Flow rate = 195,394/3600(0.9455)(62.4) = 0.9200 cfs, P = 0.23.
velocity = 5.08 fps in 6-5/8 in. OD Schedule 80 pipe.
2. Short radius return bends have 12 in. center-to-center. 20. PL = 0.23(20.36) = 4.68 atm ft.
3. q =0.75. 21. Mean tube wall temp: The stream entering the radiant
4. Fraction excess air = 0.25. section has absorbed 25% of the total heat.
5. From the API data book and a heat of cracking of
332 Btu/(lb gas + gasoline):
H, = 248 + 0.25(77.14)(E6)/195,394 = 346.7,
TI = 565“F,
H,, = 0.9(590) + 0.08(770) + 0.02(855) = 609.6 Btu/lb, T, = 100 + (565 + 900)/2 = 832.5.
Qtotal = 195,394(609.6 - 248) + 19,539(332) = 77.14(E6).
22-24. Input data are summarized as:
6. Heat released:
PL = 4.68,
Q, = 77.14/0.75 = 102.86(E6) Btu/lb. D, = 0.8018,
D2 = 0.25,
7. Radiant heat absorption: = 832.5,
Q, = Q,/aAR = 102.86(E6)/3190 = 32,245.
QR = 0.75(77.14)(E6) = 57.86(E6).
From program “FRN-1”,
8. (Q/A) rad = 10,000 Btu/(hr)(sqft), average.
9. Radiant surface: Tg = 1553.7,
QR= ~A,F[I~~O[(-)~- (-r]
F=0.6496 [Eq. (9)],
A = 57.86(E6)/10,000 = 5786 sqft.

11. Tube length = 5786/1.7344 = 3336 ft; 40 foot tubes have an
exposed length of 38.5 ft; N = 3336/38.5 = 86.6, say 92 + 7(Tg - TI}
radiant tubes.
12. From Eq. (11) the flue gas rate is = 3190(0.6496)(28,679) = 59.43(E6).

Gf = 102.85(1020) = 104,9071b/hr. Compared with estimated 57.86(E6) at 75% heat absorption