Chemical Engineering - Oil - Sizing Heating Coil.pdf

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CHAPTER 13

Calculating heat loss and heater size for a tank 

Heat lo Heat loss ss an and d th thee he heat ater er su surfa rface ce ar area ea to co comp mpen ensa sate te forthe he heat at los losss ma may y be ca calc lcula ulate ted  d  using the following procedure: Step 1. Establish the bulk temperature for the tank contents. Fix the ambient air  temper tem peratu ature re andthe win wind d ve veloc locitynormalfor itynormalfor thearea in whi which ch thetank is to be sit sited. ed. Step 2. Calculate the inside film resistance to heat transfer between the tank contents and the tank wall. The following simplified equation may be used for this: hc

=

8.5(t /µ /µ)0.25

where h c = Inside film resistance to wall in Btu/hr sqft ◦ F. t  = Temperature difference between the tank contents and the wall in ◦ F. µ = The viscosity of the tank contents at the bulk temperature in cps. The heat loss calculation is iterative with assumed temperatures being made for  the tank wall. Step 3. Using the assumed wall temperature made in step 2 calculate the heat loss to atmosphere atmos phere by radiation using Figure 13.8. Then calculate calculate the heat loss from the tank wall to the atmosphere using Figure 13.9. Note the temperature difference in this case is that between the assumed wall temperature and the ambient air tem perature. Correct these figures by multiplying multi plying the radiation loss by the emissivity factor given in Figure 13.8. Then correct the heat loss by convection figure by the factors as described in item 4 below. Step 4. The value of  h co read from Figure 13.9 is corrected for wind velocity and for  shape (vertical or horizontal) by multiplying by the following shape factors: Ver erti tica call pl plat ates es Horizo Hor izonta ntall plate platess

1.3 1.3 2.0 (fac (facing ing up) 1.2 (facing down)

Correction for wind velocity use  F w

=

F 1

+  F 2

where  F w = wind correction factor   F 1 = wind factor @ 200◦ F calculated from:  F 1 = (MPH/1.47)0.61  F 2 = Read from Figure 13.10 Then the corrected  h co is: h co × shape correction × F   F w w.

SUPPORT SYSTEMS COMMON TO MOST REFINERIES

539

 Figure 13.8. Heat loss by radiation.

Step 5. The resistance of heat transferred from the bulk of the contents to the wall must equal the heat transferred from the wall to the atmosphere. Thus: Heat transferred from the bulk to the wall = ‘a’ =

h c from step 2 × t  in Btu/hr  · sqft.

where t  in this case is (bulk temp—  assumed wall temp)

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CHAPTER 13

 Figure 13.9. Heat loss to atmosphere by natural convection.

 Figure 13.10. Plot of ‘ F 2 ’v ersus surface temperature.

SUPPORT SYSTEMS COMMON TO MOST REFINERIES

541

Heat transferred from the wall to the atmosphere = ‘b’ =

(h co + hr ) × t  in Btu/hr sqft

where t  in this case is (assumed wall temp—  air temp). Step 6. Plot the difference between the two transfer rates against the assumed wall temperature. This difference (‘a’ − ‘b’) will be negative or positive but the wall temperature that is correct will be the one in which the difference plotted  = 0. Make a last check calculation using this value for the wall temperature. Step 7. The total heat loss from the wall of the tank is the value of ‘a’or b’calculated  ‘ in step 6 times the surface area of the tank wall. Thus: Q wall

=

hc

×

t  × (π Dtank  × tank height) in Btu/hr .

Step 8. Calculate the heat loss from the roof in the same manner as that for the wall described in steps 2–7. Note the correction for shape factor in this case will be for  horizontal plates facing upward, and the surface area will be that for the roof. Step 9. Calculate the heat loss through the floor of the tank by assuming the ground  temperature as 50 ◦ F and using; h f  = 1.5 Btu/hr sqft◦ F Step 10. Total heat loss then is: Total heat loss from tank  = Q wall + Q roof  + Q floor . Step 11. Establish the heating medium to be used. Usually this is medium pressure steam. Calculate the resistance to heat transfer of the heating medium to the outside of the heating coil or tubes. If steam is used then take the condensing steam value for h as 0.001 Btu/hr sqft ◦ F. Take value of steam fouling as .0005 and tube metal resistance as 0.0005 also. The outside fouling factor is selected from the following: Light hydrocarbon Medium hydrocarbon Heavy Hc such as fuel oils

0.0013 = 0.002 = 0.005 =

The resistance of the steam to the tube outside

=

1

h + R  where R = rsteam fouling + rtube metal + routside fouling . Step 12. Assume a coil outside temperature. Then using the same type of iterative calculation as for heat loss, calculate for ‘a’as the heat from the steam to the coil outside surface in Btu/hr sqft. That is ‘a’

=

h

×

ti

Calculate for ‘b’as the heat from the coil outside surface to the bulk of the tank  contents. Use Figure 13.11 to obtain ho and again ‘ b’is h o × to where the to is the temperature between the tube outside and that of the bulk tank contents. Make further assumptions for coil outside temperature until ‘a’ = ‘b ’. Step 13. Use ‘a’or ‘ b’from step 12 which is the rate of heat transferred from the heating medium in btu/hr sqft and divide this into the total heat loss calculated 

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CHAPTER 13

 Figure 13.11. Convection heat transfer coefficient.

in step 10. The answer is the surface area of the immersed heater required for  maintaining tank content’s bulk temperature. An example calculation using this technique is given as Appendix 13.1 at the end of  this chapter. Product blending facilities

Blending is the combining of two or more components to produce a desired end   product. The term in refinery practice usually refers to process streams being combined  to make a saleable product leaving the refinery. Generally these include gasolines, middle distillates such as: jet fuel, kerosene, diesel, and heating oil. Other blended