538 CHAPTER 13

Calculating heat loss and heater size for a tank

Heat loss and the heater surface area to compensate for the heat loss may be calculated

using the following procedure:

Step 1. Establish the bulk temperature for the tank contents. Fix the ambient air

temperature and thewind velocity normal for the area inwhich the tank is to be sited.

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)0.25

where

hc = 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 by radiation using Figure 13.8. Then 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 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 hco read from Figure 13.9 is corrected for wind velocity and for

shape (vertical or horizontal) by multiplying by the following shape factors:

Vertical plates 1.3

Horizontal plates 2.0 (facing up)

1.2 (facing down)

Correction for wind velocity use

Fw = F1 + F2

where

Fw =wind correction factor

F1 =wind factor @ 200◦F calculated from:

F1 = (MPH/1.47)0.61

F2 =Read from Figure 13.10

Then the corrected hco is:

hco× shape correction ×Fw.

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’

= hc from step 2×!t in Btu/hr · sqft.

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

540 CHAPTER 13

Figure 13.9. Heat loss to atmosphere by natural convection.

Figure 13.10. Plot of ‘F2’v ersus surface temperature.

SUPPORT SYSTEMS COMMON TO MOST REFINERIES 541

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

= (hco + 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:

Qwall = hc ×!t × (πDtank × 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;

hf = 1.5 Btu/hr sqft◦F

Step 10. Total heat loss then is:

Total heat loss from tank = Qwall + Qroof + Qfloor.

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 = 0.0013

Medium hydrocarbon = 0.002

Heavy Hc such as fuel oils = 0.005

The resistance of the steam to the tube outside =1

h + Rwhere 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 ho ×!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

542 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 refinerypractice usually refers to process streamsbeing 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