Heating of Crude Oil

undefined
 
Lect./6                                                   Petroleum refinery
 
Heating of Crude Oil
 
Pipe Still Heater
 
In most of the furnaces, the direction of tubes is horizontal as
in all box type heaters and vertical in cylindrical stills. Radiant
walls also use horizontal tubes; however tubes can be placed
vertically also.
The radiant section design is based on 
Stefan's law of
radiation
:
 
Heating of Crude Oil
 
A: area of radiating surface, ft
2
b : 1.72x10
9 
Btu/ 
o
F ft
2 
hr. at black bod conditions.
T: absolute temperature of the surface,
o
F
 
Heating of Crude Oil
Design of a furnace radiation section is based on
Hottle, Wilson method
  and radiant heat absorption is given
as:-
 
 
R= % heat absorbed in radiant section.
G= Air /fuel ratio (wt. basis).
α = Factor to convert actual exposed surface to cold surface:
                  0.986 for 
two
  rows at spacing 2 OD.
                  0.88  for 
one 
 rows at spacing 2 OD.
If
   Q in Kj / hr      S=14200           Area in m
2
     
Q  in Btu/hr     S=4200             Area in ft
2
    Q in Kcal/hr    S=6930             Area in m
2
 
 
A
cp
  = wall area.
α 
A
cp
= equivalent cold plane surface ft
2
 
Heating of Crude Oil
 
L= length.
C= Center to center spacing.
N= Number of tube per row.
 
 
A= Projected area
D= Tube diameter ( in )
 
n= no. of rows
 
RQ = Aq
q= rate of heat absorption per square foot of projected tube area
 
 
 
 
Heating of Crude Oil
For a most commercial case D/C= 0.5 , n=2
 
Heating of Crude Oil
 
Example:-
A petroleum stock at a rate of 1200 bbl/hr. of sp. gr. 0.8524 is
passed through a train of heat exchangers and is allowed to
enter directly the radiant section of box type heater at 220 
o 
C .
The heater is designed to burn 3500 kgs per hour of refinery off
gases as fuel.
The net heating value of fuel is 47.46x10
3
 Kj per kg.
The radiant section contains 150 sq. meters of projected area of
one row of tubes (10.5 cm, 12 m long and spaced at 2 OD).
>>
 
Find the outlet temperature of the petroleum stock?
Where,
 
α=0.88
Air fuel ratio= 25
Average Specific heat of stock=2.268 Kj/Kg 
o
 C.
 
Heating of Crude Oil
 
Heating of Crude Oil
 
 
Solution
Total heat liberated (Q) = m 
fuel 
* NHV
                                      = 47.46*10
3 
* 3500 = 1.66*10
8
 Kj/hr.
Projected area of one tube (L * D) = 12 * 0.105
No. of tubes = 150 / ( 12 * 0.105 ) = 120 tubes
 α A 
cp
= 0.88 * 120 * 0.105 * 2 * 12 = 266 Sq. m.
 
Heat absorption %(R)
 
 
 
Heating of Crude Oil
 
Outlet temperature of the stock:-
     Q = m Cp Δt
0.44*1.66*10
8 
= 1200 * 200 * 0.8524* 2.268 *  Δt
  
 ؞
Δt =157 
o
 C
So the outlet temperature is equal to 157 + 220 = 377 
o
 C
 
Heating of Crude Oil
 
Example
A pipe still uses 7110 lb
/hr
 of a cracked gas (Net Heating Value
(NHV) 20560 Btu/lb). The radiant section contains 1500 sq ft of
projected area, and the tube (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
(21 -
 
30 percent excess air).
a) 
What percentage of the heat liberation is absorbed in the
radiant section?
b) 
How many Btu are absorbed per hour through each square foot
of projected area?
 
Heating of Crude Oil
 
Solution
T
o
t
a
l
 
h
e
a
t
 
l
i
b
e
r
a
t
e
d
(
Q
)
=
 
m
 
f
u
e
l
 
*
 
N
H
V
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
=
 
7
1
1
0
 
*
 
2
0
5
6
0
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
=
1
4
6
0
0
0
0
0
0
 
B
t
u
/
h
r
 
A 
= 1500
 
 
 
N
 
=
 
n
u
m
b
e
r
 
o
f
 
t
u
b
e
s
 
 = 40 * 90 *10 / 12 = 3000
 
= 90
 
α A 
cp
= 0.88 * 3000 = 2640 sq ft
 
 
Heating of Crude Oil
 
Heat absorption %(R)
 
Heat absorption in radiant section = 0.458 *146 *10
6
                                                       = 66900000 Btu/hr
 
Heat absorbed / sq ft projected area = q = 66900000/1500
                                                                = 44500 Btu/hr.ft
2
 
Heating of Crude Oil
 
 
Example
A furnace is to be designed for a heat duty of 50x10
6
 Btu/hr if the
overall efficiency of the furnace is 80% and an oil fuel with a
NHV = 17130 Btu/lb is to be fired with 25% excess air (17.5
lb
 
air/lb fuel ) with the air being preheated to 400 
o
 F .
 Steam is used for atomizing at a rate of 0.3 lb/lb of fuel at 190 
0
F.
Furnace tubes are of 5 in OD., 38.5 ft length and 10 in spacing
arranged in a single row. 1500 ft
2
 of projected area is available.
H
air 
( 400 
o
 F ) = 82 Btu/lb
H
 steam
 ( 190 
o
 F ) = 95 Btu/lb
H ( flue gases at 1730 
o
 F ) = 148 Btu/hr
 
Heating of Crude Oil
 
Calculate:-
1) The no. of tube required in radiation section.
2) % heat absorbed in convection section assuming wall losses
     of  5 %.
3) The heat rate available per unit projected area.
 
 
Solution:-
 
 
A = n 
* 
L *
 
N 
* 
D/12
A = 1500 ft
2
 
1
)
 
N
 
=
 
1
5
0
0
 
*
 
1
2
 
/
 
(
1
 
*
 
3
8
.
5
 
*
 
5
)
 
Heating of Crude Oil
 
N = 94 tube/row
A
 cp
= L * C /12 * N
      = 38.5 * 10/12 * 94 = 3015.8 ft
2
α A
 cp
= 0.88 * 3015.8 = 2653.9 ft
2
Q
 total 
= Q
 comb.
 + Q
 steam
 + Q
 air
Q
 comb.
= m
 fuel
 * NHV
m
 fuel
 = 6.25 * 10
7
/17130 = 3.648 * 10
3
 
 
m
 steam 
= 0.3 * m
 fuel
m
 steam 
= 0.3 * 3.648 * 10
3
 = 1.0944 * 10
3 
lb/hr
Q
 steam
 = m 
steam
 * H
 steam
           =  1.0944 * 10
3
 * 95 = 1.03968 * 10
5 
Btu/hr
0.3 lb steam / 1 lb fuel
 
Heating of Crude Oil
17.5 lb air / 1 lb fuel
 
m
 air 
= 17.5 * m
 fuel
m
 air 
= 17.5 * 3.648 *10
3
 = 6.384 * 10
4
 lb/hr
 
m
 air 
= 1.25 * 6.384 * 10
4
 = 7.98 * 10
4
 lb/hr
Q 
air 
= m
 air
 * H
air
Q 
air 
= 7.98 * 10
4 
* 82= 6.5436 * 10
6
 Btu/hr
m
 flue gases 
= m
 fuel 
+ m
 air 
+ m
 steam
m
 flue gases 
= 3.648 *10
3 
+ 7.98 *10
4
 + 1.0944 * 10
3
 = 8.454 * 10
4
 lb/hr
Q
 flue gases
 = m
 flue gases
 * H 
flue gases
 = 8.454 * 10
4
 * 148
               = 1.25 * 10
7
 Btu/hr
Excess air = 25 %
 
Heating of Crude Oil
 
Q 
total  
= Q 
comb.
 + Q 
steam
 + Q 
air
          = 6.25 * 10
7
 + 1.03968 * 105 + 6.5436 * 10
6
          = 6.9147 *10
7
  Btu/hr
 
% stack loss = Q
 flue gases 
/ Q 
total
  * 100
                    = 1.25 * 10
7
 /6.9147 * 10
7 
* 100 = 18 %
 
H
e
a
t
 
a
b
s
o
r
p
t
i
o
n
 
%
(
R
)
 
 
 
 
 
 
Heating of Crude Oil
 
2) 
% convection = 100 - % R - % stack loss - % wall loss
    % convection = 100 - 59.88 – 18 – 5 = 17.12 %
 
3) 
q = RQ/A = 0.5988 * 6.78 * 10
7
/1500
     q = 2.7 x 10
4
 Btu/hr.ft
2
 
Heating of Crude Oil
H.W (1)
A furnace is to be designed for a heat duty of 30 x 10
6
 Btu/hr
and efficiency of 75%.
The furnace is fired with gaseous fuel at a rate of 17 lb air/lb
fuel (NHV = 17000 Btu/lb).
The tube are arranged in two rows and are of 5 in OD., 40 ft
length and 2x OD.
Spacing, heat rate of 35000 Btu/hr of projected area is
recommended.
calculate:
1) 
% heat absorbed in radiation section (R %).
2) 
Heat absorbed in the convection section.
   
 
(State any
 
assumptions used).
3) 
The number of tubes in the radiation section.
 
Heating of Crude Oil
H.W (2)
7000 lb/hr of cracked gas of 20560 Btu/lb NHV is used as a
fuel in a furnace.
The radiant section absorbed 44500 Btu/hr ft
2
 of projected
area. The tubes are 5 in. OD. , 10 in. spacing, and 20 ft long.
They arranged in two rows. The air to fuel ratio is 21- 0.
Calculate :
1)
the number of tubes in the radiation section.
 2) the amount of heat absorbed in this section.
Slide Note
Embed
Share

The process of heating crude oil in petroleum refineries involves various types of furnaces and heat distribution methods. The furnaces can be categorized into box, cylindrical, and radiant wall types, each with separate radiation and convection sections. Design considerations include heat transfer percentages for convection and radiant heat, losses, stack losses, and methods for calculating heat absorption in the radiant section. Various factors such as air/fuel ratio, wall area, tube arrangement, and heat absorption rate per square foot are crucial in designing an efficient heating system.


Uploaded on May 10, 2024 | 1 Views


Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

E N D

Presentation Transcript


  1. Lect./6 Petroleum refinery Heating of Crude Oil

  2. Heating of Crude Oil Pipe Still Heater Pipe heaters can be categorized into three types: 1) Box/ Rectangular. 2) Cylindrical. 3) Radiant Wall All these furnaces have got separate radiation section and convection section. The most universal classification is based on direction of tubes as well as shape of furnace and mode of application of heat.

  3. Heating of Crude Oil In most of the furnaces, the direction of tubes is horizontal as in all box type heaters and vertical in cylindrical stills. Radiant walls also use horizontal tubes; however tubes can be placed vertically also. The radiant section design is based on Stefan's law of = = 4 4 radiation: 4 ( ) Q bAT bA T T r G B A: area of radiating surface, ft2 b : 1.72x109 Btu/ oF ft2 hr. at black bod conditions. T: absolute temperature of the surface,oF

  4. Heating of Crude Oil For a satisfactory design, the following schedule of heat distribution may be employed: Type of heat Percent Convection heat transfer 30-50% Radiant heat transfer 45-60% Losses (Furnace) 5% Stack losses 12% Design of a furnace radiation section is based on Hottle, Wilson method and radiant heat absorption is given as:-

  5. 1 Q = 100 x R / G A cp + 1 S R= % heat absorbed in radiant section. G= Air /fuel ratio (wt. basis). = Factor to convert actual exposed surface to cold surface: 0.986 for two rows at spacing 2 OD. 0.88 for one rows at spacing 2 OD. If Q in Kj / hr S=14200 Area in m2 Q in Btu/hr S=4200 Area in ft2 Q in Kcal/hr S=6930 Area in m2

  6. Heating of Crude Oil C Acp= wall area. Acp= equivalent cold plane surface ft2 Acp= LN 12 L= length. C= Center to center spacing. N= Number of tube per row. D A = LnN 12 A= Projected area D= Tube diameter ( in )

  7. Heating of Crude Oil n= no. of rows D = A nA cp C C A Acp= n D RQ = Aq q= rate of heat absorption per square foot of projected tube area ( / ) nA D C q Aq = = cp Q R R

  8. Heating of Crude Oil 1 = 100 x R q n D G R C + 1 S 2 1 ( ) D n R S = 2 ( ) ( ) q C R G For a most commercial case D/C= 0.5 , n=2 2 1 ( ) R S = ' 2 . 1 014 ( ) xq R G C a q = ' . 1 014 x x q D n

  9. Heating of Crude Oil Example:- A petroleum stock at a rate of 1200 bbl/hr. of sp. gr. 0.8524 is passed through a train of heat exchangers and is allowed to enter directly the radiant section of box type heater at 220oC . The heater is designed to burn 3500 kgs per hour of refinery off gases as fuel. The net heating value of fuel is 47.46x103Kj per kg. The radiant section contains 150 sq. meters of projected area of one row of tubes (10.5 cm, 12 m long and spaced at 2 OD). >> Find the outlet temperature of the petroleum stock? Where, =0.88 Air fuel ratio= 25 Average Specific heat of stock=2.268 Kj/KgoC.

  10. Heating of Crude Oil Solution Total heat liberated (Q) = m fuel * NHV = 47.46*103 * 3500 = 1.66*108Kj/hr. Projected area of one tube (L * D) = 12 * 0.105 No. of tubes = 150 / ( 12 * 0.105 ) = 120 tubes A cp= 0.88 * 120 * 0.105 * 2 * 12 = 266 Sq. m. 1 Q = 100 x Heat absorption %(R) / G A cp + 1 S

  11. Heating of Crude Oil 1 = = 44 % 8 . 1 66 * 10 25 * 266 + 1 14200 Outlet temperature of the stock:- Q = m Cp t 0.44*1.66*108 = 1200 * 200 * 0.8524* 2.268 * t t =157 oC So the outlet temperature is equal to 157 + 220 = 377 oC

  12. Heating of Crude Oil Example A pipe still uses 7110 lb/hr of a cracked gas (Net Heating Value (NHV) 20560 Btu/lb). The radiant section contains 1500 sq ft of projected area, and the tube (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 (21 - 30 percent excess air). a) What percentage of the heat liberation is absorbed in the radiant section? b) How many Btu are absorbed per hour through each square foot of projected area?

  13. Heating of Crude Oil Solution Total heat liberated(Q)= m fuel * NHV = 7110 * 20560 =146000000 Btu/hr D A = LnN 12 A = 1500 N = number of tubes 1500 = = 90 40 * 5 / 12 C Acp= LN = 40 * 90 *10 / 12 = 3000 12 A cp= 0.88 * 3000 = 2640 sq ft

  14. Heating of Crude Oil 1 = 100 x Heat absorption %(R) * / G Q A cp + 1 S 1 = = 45 8 . % 6 . 1 46 10 x 21 * 2640 + 1 4200 Heat absorption in radiant section = 0.458 *146 *106 = 66900000 Btu/hr Heat absorbed / sq ft projected area = q = 66900000/1500 = 44500 Btu/hr.ft2

  15. Heating of Crude Oil Example A furnace is to be designed for a heat duty of 50x106Btu/hr if the overall efficiency of the furnace is 80% and an oil fuel with a NHV = 17130 Btu/lb is to be fired with 25% excess air (17.5 lb air/lb fuel ) with the air being preheated to 400oF . Steam is used for atomizing at a rate of 0.3 lb/lb of fuel at 1900F. Furnace tubes are of 5 in OD., 38.5 ft length and 10 in spacing arranged in a single row. 1500 ft2of projected area is available. Hair( 400oF ) = 82 Btu/lb Hsteam( 190oF ) = 95 Btu/lb H ( flue gases at 1730oF ) = 148 Btu/hr

  16. Heating of Crude Oil Calculate:- 1) The no. of tube required in radiation section. 2) % heat absorbed in convection section assuming wall losses of 5 %. 3) The heat rate available per unit projected area. Solution:- 6 50 * 10 heatduty = = = 7 . 6 25 * 10 / Qcomb Btu hr . 8 . 0 efficiency A = n * L * N * D/12 A = 1500 ft2

  17. Heating of Crude Oil 1) N = 1500 * 12 / (1 * 38.5 * 5) N = 94 tube/row Acp= L * C /12 * N = 38.5 * 10/12 * 94 = 3015.8 ft2 Acp= 0.88 * 3015.8 = 2653.9 ft2 Qtotal = Qcomb.+ Qsteam+ Qair Qcomb.= mfuel* NHV mfuel= 6.25 * 107/17130 = 3.648 * 103 0.3 lb steam / 1 lb fuel msteam = 0.3 * mfuel msteam = 0.3 * 3.648 * 103= 1.0944 * 103 lb/hr Qsteam= m steam* Hsteam = 1.0944 * 103* 95 = 1.03968 * 105 Btu/hr

  18. Heating of Crude Oil 17.5 lb air / 1 lb fuel mair = 17.5 * mfuel mair = 17.5 * 3.648 *103= 6.384 * 104lb/hr Excess air = 25 % mair = 1.25 * 6.384 * 104= 7.98 * 104lb/hr Q air = mair* Hair Q air = 7.98 * 104 * 82= 6.5436 * 106Btu/hr mflue gases = mfuel + mair + msteam mflue gases = 3.648 *103 + 7.98 *104+ 1.0944 * 103= 8.454 * 104lb/hr Qflue gases= mflue gases* H flue gases= 8.454 * 104* 148 = 1.25 * 107Btu/hr

  19. Heating of Crude Oil Q total = Q comb.+ Q steam+ Q air = 6.25 * 107+ 1.03968 * 105 + 6.5436 * 106 = 6.9147 *107Btu/hr % stack loss = Qflue gases / Q total* 100 = 1.25 * 107/6.9147 * 107 * 100 = 18 % 1 = 100 x Heat absorption %(R) * / G Q A cp + 1 S 1 = = 59 88 . % 7 9147 . 6 10 x 17 5 . * 2653 9 . + 1 4200

  20. Heating of Crude Oil 2) % convection = 100 - % R - % stack loss - % wall loss % convection = 100 - 59.88 18 5 = 17.12 % 3) q = RQ/A = 0.5988 * 6.78 * 107/1500 q = 2.7 x 104Btu/hr.ft2

  21. Heating of Crude Oil H.W (1) A furnace is to be designed for a heat duty of 30 x 106Btu/hr and efficiency of 75%. The furnace is fired with gaseous fuel at a rate of 17 lb air/lb fuel (NHV = 17000 Btu/lb). The tube are arranged in two rows and are of 5 in OD., 40 ft length and 2x OD. Spacing, heat rate of 35000 Btu/hr of projected area is recommended. calculate: 1) % heat absorbed in radiation section (R %). 2) Heat absorbed in the convection section. (State any assumptions used). 3) The number of tubes in the radiation section.

  22. Heating of Crude Oil H.W (2) 7000 lb/hr of cracked gas of 20560 Btu/lb NHV is used as a fuel in a furnace. The radiant section absorbed 44500 Btu/hr ft2of projected area. The tubes are 5 in. OD. , 10 in. spacing, and 20 ft long. They arranged in two rows. The air to fuel ratio is 21- 0. Calculate : 1) the number of tubes in the radiation section. 2) the amount of heat absorbed in this section.

Related


More Related Content

giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#