Heater, prinsip kerjanya, serta peran aktif atau aplikasinya dlm kehidupan
A ppt on mems microheater design.Full description
PCA Circular Tank DesignDescripción completa
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Design of a liquid storage tank per API 650 2013 Standard.
Tank Baffles formulas for Design and Calculation
designing procedure of rectangular steel tank and placing of stiffeners.Descripción completa
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SUPP SUPPOR ORT T SYST SYSTEM EMS S COMM COMMON ON TO MOST OST REFI REFINE NERI RIES ES
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types of tanks range from 30 to 250 psig. The maximum allowable is limited by tank size and code requirements. For a 1,000 bbl sphere, the maximum pressure is 215 psig, for a 30,000 bbl it is 50 psig. These pressure limits can be increased if the tank is stress relieved. Although it is possible to store material in tanks with pressure in excess of 250 psig normally when such storage is required refrigerated storage is usually a better alternate.
Heated storage tanks
Heated storage tanks are more common in the petroleum industry than most others. They are used to store material whose flowing properties are such as to restrict flow at normal ambient temperatures. In the petroleum industry products heavier than diesel oil, such as heavy gas oils, lube oil, and fuel oil are stored in heated tanks. Generally speaking tanks are heated by immersed heating coils or bayonet type immersed heaters. Steam is normally used as the heating medium because of its availability in petroleum complexes. Very often where immersed heating is used the tank is agitated usually by side located propeller agitators for large tanks. Where external circulating heating is used for tanks, the contents are mixed by means of jet mixing. Here the hot return stream is introduced into the tank via a specially designed jet nozzle as shown in Figure 13.7. External tank heating is used when there is a possibility of a hazardous situation occurring if an immersed heater leaks.
Figure 13.7. A jet mixing nozzle.
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CHAPTER 13
Calculating heat loss and heater size for a tank
Heat loss and the heater surface area to compensate forthe heat loss may be calculated using the following procedure: Step 1. Establish the bulk temperature for the tank contents. Fix the ambient air temperature andthe wind velocitynormalfor thearea in which thetank 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 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 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 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: Vertical plates Horizontal plates
1.3 2.0 (facing 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 w.
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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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Figure 13.9. Heat loss to atmosphere by natural convection.
Figure 13.10. Plot of ‘ F 2 ’v ersus surface temperature.
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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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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 termin refinerypracticeusuallyrefersto 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
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Appendix 13.1: Example calculation for sizing a tank heater
Problem It is required to calculate the surface area for a heating coil which will maintain the bulk temperature of fuel oil in a cone roofed tank at a temperature of 150◦ F. The ambient air temperature is an average 65 ◦ F and the wind velocity averaged over the year is 30 MPH. The fuel oil data is as follows:
= 36cps@150 ◦F Sg@150◦ F = 0.900
Viscosity (μ)
The tank is to be heated with 125 psig saturated steam. The tank dimensions are 60 ft diameter 180 ft high it is not insulated but is painted with non metallic color paint.
×
Solution 1.0 Calculating the heat loss from the wall. 1st trial . Assume wall temperature is 120 ◦ F hi
t o is temperature difference between assumed wall temp and the ambient air 120 65 55◦ F.
− =
h co
=
= 0.495 × 1.3 (from Figure 13.9)
Wind correction factor F w is as follows:
= (MPH/1.47)0.61 = 6.29 F 2 = 1.04 (From Figure 13.10) F w = 6.29 + 1.04 = 7.33. h co (corrected) = 0.495 × 1.3 × 7.33 = 4.20 Btu/hr. sqft.◦ F F l
Heat loss from wall dueto radiationh ro isfoundfromFigure13.8
= 1.18 Btu/hr.sqft.◦ F.
SUPPORT SYSTEMS COMMON TO MOST REFINERIES
Corrected for emissivity h ro
= 1.18 × .95 = 1.123 Btu/hr.sqft.◦ F
‘b’
= (hco + hro ) × t o = (4.21 + 1.12) × 55 = 293 Btu/hr.sqft ‘a’ −‘b’ = 244 − 293 = −49 Btu/hr.sqft 2nd trial. Assume wall temperature is 110◦ F Carrying out the same calculation procedure as for trial 1: ‘a’
− ‘b’ = +172 Btu/hr .sqft
3rd trial. Assume wall temperature is 115 ◦ F Again carrying out the calculation procedure as for trial 1: ‘a ’
− ‘b’ = +35
The results of the above trials are plotted linearly below:
‘a’and ‘ b’are close enough call total heat loss 275.7 Btu/hr.sqft.
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Surface area of wall
= circumference × height.
= π D × 180 ft = 33929 sqft Total heat loss through wall
= 275.7 × 33929 = 9.35 mm Btu/hr
2.0 Calculating heat loss through roof. Trial 1. Assume roof temperature is 116 ◦ F.
= 8.38 Btu/hr.sqft.◦F = 8.38 × (150 − 116)◦ F = 284.9 Btu/hr.sqft. h co (corrected) = (.470 × 2.0) × 1.04 × 6.29 hi ‘a’
( Note: the number read from Figure 13.9 is multiplied by 2.0 in this case as the roof is an upward facing plate.)
= 6.35 Btu/hr.sqft.◦ F h ro (corrected) = 1.165 × 0.95 = 1.11 Btu/hr.sqft.◦ F ‘b’ = (6.35 + 1.11) × (116 − 65)◦ F = 380 Btu/hr.sqft. ‘a’ − ‘b’ = −95 Btu/hr.sqft Trial 2. Assume a wail temperature of 110 ◦ F ‘a’
− ‘b’in this case = +23 which is within acceptable limits.
The heat loss is taken as an average of ‘a’and ‘ b’
= 338 Btu/hr.sqft. Total heat loss from the roof
= area of roof × 338
= 2827 sqft × 338 = 0.956 mm Btu/hr 3.0 Calculating the heat loss through the floor. Assume the ground temperature is 50◦ F and the heat transfer coefficient is 1.5 Btu/ hr.sqft. ◦ F. Then heat loss
= 1.5 × (150 − 50) × 2827 sqft = 0.424 mm Btu/hr
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4.0 Total heat loss from the tank. Heat loss from the Walls Heat loss from the Roof Heat loss from the Floor Total Heat Loss
= 9.350 mm Btu/hr = 0.956 mm Btu/hr = 0.424 mm Btu/hr = 10.730 mm Btu/hr
5.0 Calculating the tank heater coil surface area required. The heating medium is saturated 125 psig steam. Temperature of the steam
= 354◦F
Steam side calculations. Approx resistance of steam, h s Fouling factor on steam side, r1 Tube metal resistance r2 Outside fouling factor r3
= 2043 sq ft Appendix 13.2: Example calculation for sizing a relief value
A vessel containing naphtha C 5 –C8 range is uninsulated and is not fireproofed. The vessel is vertical and has a skirt 15 in length. Dimensions of the vessel are I/D 6 0 T-T 20 0 liquid height to HLL = 16 0. Calculate the valve size for fire condition relief. Set pressure is 120 psig. Latent heat of naphtha at 200 ◦ F is 136 Btu/lb Q A
= 21000 FA0.82 = Wetted area and is calculated as follows: Liquid height above grade = 15 + 16 ft = 31 ft
= H L
Therefore wetted surface of vessel need only be taken to 25 ft above grade which is 25 15 10 ft of vessel height.
− =
Wetted surface
= π D×10 for walls = 188.5 sq.ft
= Q= = Q / H L =
plus 28.3 sq.ft for bottom 216.8 sq.ft 21000 1.0 (216.8)0.82 1.729 Btu/hr 106 1.729 106 12713 lbs/hr 136
