Answer to Question #208845 in Chemical Engineering for Sajad Asem

Question #208845

t is required to heat 10000 bbl/ day of crude oil (as given in case study of Lecture3) in a

fired still heater from 220 0F to 622 0F and vaporize (gasoline, naphtha, kerosene, light gasoil,

heavy gasoil) of the crude oil using refining gas, (NHV=50000 Btu /lb). In the radiant section

contains 1500 ft2 of projected area, and the tubes (5 in. outside diameter) are spaced at a center-

to-center distance of 10 in. there is only one row of radiant tubes, and they are 40 ft long. The

ratio of air to fuel is 17.

F, average latent heat of vaporization 0 Given: Average Specific heat of the crude oil = 2.3 Btu/ lb

is 120, 113, 100, 95, 90 Btu/lb for (gasoline, naphtha, kerosene, light gasoil, heavy gasoil)

respectively. and the furnace efficiency is 75% calculate:

a) The mass flow rate of the fuel required?

b) How many Btu are absorbed per hour in radiant section?

c) How many Btu are absorbed per hour through each squar

Expert's answer

Part a

"m_{air}*C_1*T_1 +\\dot{m} Q= (m_{air}+m_{f})C_p*T_2"

"13.24*1005*377.594 +\\dot{m} *139.4*12.2= (13.24+5.35)1005*600"

"5024341.2828 +\\dot{m} *139.4*12.2=11209770"

"\\dot{m} = \\frac{11209770-5024341.2828}{139.4*12.2} =3637.03 g\/s =3.6kg\/s"

Part b

"1 kg\/s = 33.5 btu\/hr"

"3.6 kg\/s = \\frac{33.5*3.6}{1}=120.btu\/hr"

Part c

"1500 ft^2 = 120.6 btu\/hr"

"1 ft^2 = \\frac{120.6}{1500}=0.0804 btu\/hr"

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