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Section 27

FACT Method of Cost Estimation

INTRODUCTION Whenever you start to look at a new design or modifying an old design, one of the first questions that is likely to be asked is “What will it cost”. This presents a dilemma because the ques-tion is always asked when very little, if any, design work has been completed. Since cost estimates are almost always based on factoring the cost of major equipment items in the design, some method is required to approximate the size and cost of these major equipment items very early in the design phase. By way of definition, major materials means any uniquely identifi-able item of equipment such as distillation columns, pressure vessels, reactors, heat exchangers, pumps, compressors, fired

heaters and the like. To resolve this dilemma Olin Engineering has developed the First Approximation Costing Technique or FACT method for developing order of magnitude sizing and costs for major items of capital equipment. The costs derived by the FACT method are based on the equipment at the vendor’s works. To arrive at a total installed cost estimate of the process facilities we add the following ‘Fac-tors” to the sum of the cost of all of the major equipment items.

FIG. 27- 1

Installed Cost Factors for Purchased Equipment Factor Item Subtotals

A. Major Equipment Costs 100 B. Instrumentation and Control Systems 15% of A 15 C. Major Equipment including I&C 115 D. Minor Material (piping, electrical , etc) 60% of C 69 E. Total Equipment Cost 184 F. Freight Insurance and Handling 15% of E 28 G. Engineering 20% of E 37 H. Construction Labor 60% of E 110 I. Construction Equipment 10% of E 18 J. Construction Supervision 10% of E 18 K. Total Installation Costs 211 L. Total Capital Cost E + K 395 M. Owners Costs 8% of L 32 N. Total Installed Cost of ISBL L + M 427

This is the cost of the process facilities, also known as Inside Battery Limits (ISBL) facilities. These facilities cannot exist in isolation. They will need to be supported by Outside Battery Limit (OSBL) facilities such as storage tanks, feedstock receiving and product shipping, cool-ing water, steam, fire water, instrument air, flare stack, envi-

ronmental treating and other facilities. Since it is only possible to estimate the amount of tankage at this point in time, Olin Engineering uses a factor of 30% of the ISBL cost to account for the OSBL facilities other than storage tanks

.FIG. 27-2

Nomenclature

A = Area (square feet) ACFS = Actual Cubic Feet per Second (Ft3/sec) c.a. = Corrosion allowance (inches) ISBL = Inside Battery Limits. Equipment directly used in the manufacturing process. GPM = Gallons per Minute. HP = Horsepower = 550 ft.-lbs./sec. L = Liquid flow rate in a tower. (Lb. moles/hour) L/D = Length ? Diameter of a tank. OSBL = Outside Battery Limits. Storage tanks, all buildings and equipment not directly involved in manufacturing.

P = Pressure (PSI) (Either Absolute or Gauge) R/D = Reflux Ratio. Reflux/Distillate (Overhead Product). t = Thickness of a pressure vessel wall (inches) T = Temperature (oR) V = Vapor flow rate in a tower. (Lb. moles/hour) W = Weight (pounds) Z = Compressibility (Videal ? Vactual @ inlet conditions) ? = Relative Volatility. Ratio of vapor pressures of light key to heavy key at operating temperatures. ? = Density (Lbs./cubic foot) ? = Cp/Cv

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The cost of storage tanks is included at $0.50/US gallon. Olin Engineering also adds a contingency factor that should be applied at this point to account for items not included and for the inevitable cost growth that accompanies the identifi-cation of additional equipment and facilities as the design process progresses. Thus, the Total Installed Cost of the entire facility is made up as follows based on the ISBL Cost = 100: ISBL 100 Storage Tanks* 15 OSBL 35 Contingency Allowance at 40%Of ISBL + OSBL 60 Total Project Cost 210 *This number is used as an example. For an actual estimate, the storage tank cost would be estimated by multiplying the total storage tank volume in gallons by $0.50/gallon. On the following pages, we show Olin Engineering’s FACT method for estimating the cost of individual equipment items.

PRESSURE VESSELS The cost of a carbon steel pressure vessel is the sum of the metal cost (55%) and the fabrication cost (45%). For vessels where the shell thickness is less than 1.25 inches, the cost is related to vessel weight as follows:

? ? 62.080$ weight?? where weight is in pounds For vessels where the shell thickness is 1.25 inches or greater, the cost per pound is higher due to more expensive fabrication requirements. For these heavy wall vessels, the cost is related to the weight as follows:

? ? 7.056$ weight?? where weight is in pounds For materials other than mild steel, see Figure 27-3 below. Estimating the weight of a vessel: Most vessels contain a gas and a liquid phase. Assume a liq-uid holdup of 5 minutes with the vessel half full. Use a length to diameter ratio of L/D = 2.5. Calculate the cylinder area. Add 20% to the calculated area for the heads at each end.

The thickness of the metal is given by:

t = Pressure x Diameter/25,000* where P is in psig

D is in inches *Allowable Stress (for steel)

Add a minimum corrosion allowance of 0.125 inches. Note Olin Engineering’s practice is to specify that all pressure vessels have a minimum design pressure of 50 psig. If you have a vessel that operates under vacuum, then designing for 50 psig will almost always provide sufficient shell thickness to be able to withstand vac-uum conditions. . One square foot of steel one inch thick weighs 40 lbs. Steel plate comes in plate thickness increments of 1/16 of an inch. Example: A flash drum is required to handle 10, 000 ACFS of vapor and 300 gpm of liquid at 100psig. Volume = 600* gpm x 5 minutes x 1 cu ft./ 7.5 gallons = 400 cu. ft. * 600 gpm to account for half full.

? ? LDV ??? 2

4? Substitute 2.5 D = L and

? ? DDV 5.24

2 ??? ? = 400 or approx. D3 = 200

D = say 6 feet; L = 2.5 x D = 15 feet A = p x 6 x 15 = 282 square feet. Add 20% for heads = 339 say 340 square feet. t = 100 x 6 x 12 /25,000 = 0.29 inches. t + c.a. = .415 inches Adjust up to nearest steel plate thick-ness of 7/16 inch. Weight = 40 x 7/16 x 340 = 5950 lbs. say 6000

? ? 600,17$600080$ 62.0 ???Cost For the same vessel in stainless steel the cost will be:

Materials = .55 x 17,600 x 5.0 = 48,400 Labor = .45 x 17,600 x 1.4 = 11,100 Total = 59,500

FIG. 27-3 Adjustments for different materials of construction:

Cost of Carbon Steel Low Chrome Steel Stainless Steel Hastelloy etc. Materials 1.0 2.5 5.0 7.5 Labor 1.0 1.3 1.4 1.5

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STAGED SEPARATION COLUMNS Find length, diameter and wall thickness and calculate weight as for pressure vessels except add a 15 foot long column skirt of 3/8 “ carbon steel plate and add 15 % of calculated weight for heads and appurtenances. Length is number of trays x 2.5 feet plus 5 feet for disengag-ing space at the top and 10 feet for a liquid reservoir at the bottom. (This 15 feet does not include the skirt) You can assume that the cost of trays and other internals is equal to the cost of the column shell. That is to say a column plus the internals (trays or packing) costs twice the shell cost. Distillation Columns: Use the following table to estimate number of actual distillation trays Product Purity <95% <98% <99% 99.5% # of actual trays: 17/? -1 19/? -1 22/? -1 25/? -1 Use the following table to estimate the reflux ration Value of ? -1 >1 +/- 0.5 +/- 0.15 Reflux Ratio R/D 1 2.5 6 Stripping Columns and Absorber Columns: Number of actual trays = 20. The liquid and vapor rates are those at the V & L inlets. Example A distillation column is required to separate A and B. The feed is 1000 mols/hr of A and 850 mols/hr of B. The rela-tive volatility of A to B is 1.4. The MW of A is 40. The liquid density of A is 50 lbs./cubic foot at the top tray conditions. The column operates at 100 psig. and 150 ºF at the column top. Product purity top and bottom is 98%. Number of trays = 19/(1.4-1) = 48, Column length = 15 + 48 x 2.5 = 140 feet. From above, reflux ratio = 2.5 Use the method found in Section 19 (Figure 19-14) of the GPSA Engineering Data Book to find the column diameter. The following information will be needed: At the column top L = 2.5 x 1000 = 2500 mols/hr 2500 mols/hr x 40 lb/mol ? 50 lbs./ft3 = 2,000 cu ft./hr = 250 gpm

At the column top; V = Overhead Product + Reflux ? V = 1000 mols/hr. + 2.5* x 1000 = 3500 mols/hr V = 3500 x 359 cu ft/mol x (14.7/115) x (460+150/460+32) x 1 hr/3600 sec = 55.3 ACFS Vapor Flow Rate = 3500 x 40 ÷ 3600sec/hr = 39.8 lb/sec Vapor Density = 38.9 lbs/sec/ 55.3cu ft./sec = 0.70lbs/cu ft

v

acfsVloadl ????

?v

Vload = 55.3 x v(.70)/ (50 - 0.70) = 7.1 From Figure 19-14, the column diameter is 5'6" Thickness = 100 x 5.5 x 12/25,000 = 0.26 inches Add 0.125” for corrosion allowance: t = 0.385" Adjust to standard plate thickness: t = 7/16" (0.437") W = 40 x p x 6 x 140 x 7/16 + 40 x p x 6 x 15 x 3/8 = say 48,000 lbs. Add 15% for heads and appurtenances so weight = 55,000 lbs.

? ? 62.0000,5580$ ??Cost Therefore cost = say $69,500 Double the shell cost to include the cost of the trays Cost of column and trays = $ 140,000

HEAT EXCHANGERS Heat exchanger costs are based on square feet of heat trans-fer surface For shell and tube heat exchangers use the following: Heat exchanger design pressure in psig:

<300 300 – 600 >600 Cost = $8000 + $25/ft2 $30/ ft2 $35/ ft2

The cost of an air cooler is = ? ? 87.0110$ A Where A = heat transfer area in square feet To approximate "A", divide the heat duty by the appropriate heat flux rate as shown in FIG. 27-4 next page:

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FIG. 27 - 4 Heat Flux in various exchangers

Flux Rates (Btu/hr-sq. ft). Tubular Exchangers Air Coolers

Gas to Gas 3,000 3,500 Liquid to Liquid 8,000 9,000 Phase Change 10,000 11,000 Steam Condensing 12,000 N/A

Example: A heat exchanger is exchanging heat between reactor feed and reactor effluent. There is a phase change on both sides. That is evaporation on one side and condensation on the other. The heat duty is 20 MM btu./hr. Heat exchanger surface A = 20 MM Btu/hr / 10,000 Btu/hr/sq. ft. = 2000 sq. ft. Cost = 8,000 + (25 x 2000) = $58,000 For the same exchanger in high pressure (>600 psig) service: Cost = 8,000 +(35 x 2000)= $78,000

FIRED HEATERS Fired heaters are made up of convection surface and radiant surface all contained in a firebox. Assume the heat flux in the radiant section is 12,000 Btu/hr/sq. ft. and 3000 Btu/hr/sq. ft. in the convection section. Assume that two thirds of the heat is absorbed in the radiant section and one third in the convection section. If there is steam raised in the convection section use a flux rate of 20,000 Btu/hr/sq. ft for this section. Calculate the cost of each section separately and then add the cost of the firebox. Firebox cost is $50,000 + $50/sq. ft. of installed heat exchange surface

Coil Material $/sq. ft. of Surface

Carbon Steel 100 Stainless Steel 250 Steam Reformer (HK 40) 500

Example:

Basis a steam reformer where the radiant process duty is 120 MM Btu/hr and the convection process duty is 60 MM Btu/hr and the steam generation duty is 30 MM btu/hr. The radiant coil is HK 40, the other coils are carbon steel.

Radiant surface = 120,000,000/12,000 = 10,000 sq. ft. Cost of radiant coil 10,000 sq.ft. x $500/sq. ft. = $5, 000,000

Convection surface = 60,000,000/3000 = 20,000 sq. ft. Cost of convection coil 20,000 sq. ft. x $ 100 = $2,000,000

Steam Generation Surface 30,000,000/20,0000 = 1,500 Cost of steam generation coil = 1500 x 100 = $150,000

Total coil cost = $7,150,000

Firebox cost = 50,000 +50 x (10,000 + 20,000 +1500) = $1,575,000

Total cost = $8,725,000

PUMPS

Pump horsepower is calculated by the following formula:

HP = GPM x specific gravity x Head (in feet)/3960/ effi-ciency

For most services 75% is a good approximation of efficiency so we can restate this as:

HP = G.P.M. x s.g. x head/3000

Multistage pumps are used when the head is > 450 feet

FIG. 27-5 Cost of Various Pumps

Pump HP Cost of pump and motor Cost of pump and motor

for single stage pumps for multi stage pumps Up to 10 $5,000 10 – 50 $5,000 + $100/HP 50 – 500 $10,000 + $115/HP $25,000 +$125/HP 500 – 1000 $20,000 + $125/HP $30,000 + $150/HP Over 1000 $25,000 + $150/HP $50,000 + $175/HP

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COMPRESSORS

kk

PP k

k

ZRTHead 1

112

1

????

?

?

???

?

??

?

?

where Z = compressibility (@ inlet cond.) T = inlet temperature in ºR ? = Cp/Cv P1= inlet pressure (PSIA) P2= outlet pressure (PSIA) H = head (feet)

Theoretical Horsepower = Head x mass rate ( lb/second) ? 550 ft lb/sec/HP Divide by the efficiency (typically 77 %) to get Actual HP

Compressor Costs:

Discharge $/HP for compressor and Pressure(atm) motor driver Less than 5 $ 100/HP 5 – 15 $ 1250/HP

Greater than 15 $ 1500/ HP

Example:

Compress 5,000 mols/hour of air from 1 atm to 4 atm

T = 70 F k = 1.4 Z = 1.0 k-1/k = 0.29 R = 1544/MW = 53

Head = 1 x 53 x 530 x ((40.29 – 1)/0.29) ? 50,000 feet

W = 5,000 mols/hr x 29lbs/mol/3600 sec/hr = 41 lb/sec

HP = W x Head/550 = 41 x 50,000/550 = 3700 HP

Compressor and driver cost = $100/HP x 3700 HP ? $3,700,000

ATMOSPHERIC PRESSURE STORAGE TANKS Installed cost is $.50/gallon or $132/cubic meter of capacity. This includes foundations, dykes, sewers, and interconnect-ing piping.

A. M. Center School of Chemical Engineering Cornell University