The proposed expression is:

Cp(soln) = W_{1} Cp(solid) + W_{2} Cp(water) | Eq. (1) |

where:

Cp(soln) = Heat capacity of the solution

Cp(solid) = Heat capacity of the dissolved solid

Cp(water) = Heat capacity of water

W_{1} = Weight percent of dissolved solid

W_{2} = Weight percent of water

The following charts are used in the calculation to follow. Table 1 shows Engish units while Table 2 displays the values in Metric units.

Table 1: Heat Capacity of Selected Element, English Units | Table 2: Heat Capacity of Selected Element, Metric Units |

Step-by-Step Example

For an example problem, calculate the heat capacity of a 20% by weight solution of Na_{2}CO_{3} at 150 Â°F.

Step 1

Look up the heat capacity of the solid or calculate this value from Kopp's Rule. If literature data is not available for the dissolved solid, it can be estimated from the elemental heat capacities with Kopp's Rule:

Cp(Na_{2}CO_{3}) = 2 Cp(Na) + 1 Cp© + 3 Cp(O) | Eq. (2) |

From Table 1 above, we read the values at 150 ^{Â°}F. Notice that the heat capacity for oxygen is given as O_{2} (it's natural form). This value must be divided by two to get the heat capacity for one atom of oxygen.

Cp(Na_{2}CO_{3}) = 2 (6.8247) + 1 (2.7805) + 3 (1/2 (6.7198)) = 26.51 Btu/lb mole ^{Â°}F

Dividing by the molecular weight of the solid yields:

(26.51 Btu/lb mole ^{Â°}F) / (105.9 lb/lb mole) = 0.25 Btu/ lb ^{Â°}F

Step 2

Determine the heat capacity of the solution at a given temperature. Figure 1 below shows how the heat capacity of water varies with temperature in English units. To convert Btu/lb ^{0}F to kJ/kg K, multiple by 4.1867.

Now, recall from Equation 1:

Figure 1: Heat Capacity of Water as a Function of Temperature |

Cp(soln) = W_{1} Cp(solid) + W_{2} Cp(water)

Cp(soln) = 0.20 Cp(Na_{2}CO_{3}) + 0.80 Cp (water)

At 150 ^{0}F, the heat capacity of water is 0.9975 Btu/lb ^{0}F.

Cp(soln) = 0.20 (0.25 Btu/lb ^{0}F) + 0.80 (0.9975 Btu/lb ^{0}F)

*Cp(soln) = 0.848 Btu/lb ^{0}F at 150 ^{0}F (literature data is 0.850)*

Other Methods

A quick alternative to the above procedure was proposed by Vosseller in 1973:

Cp(soln) = 1 - 0.70 W_{1} (valid for English units only)

For the example shown here, the result would be 0.86 Btu/lb ^{0}F

For aqueous organic solutions, Vosseller proposed:

Cp(soln) = 1 - 0.45 W_{1} (valid for English units only)

Download the MS Excel spreadsheet with the heat capacity data for the elements shown in Tables 1 and 2.

]]>When I think of such rules, two authors come to my mind, Walas and Branan. Dr. Walas' book, *Chemical Process Equipment: Selection and Design* has been widely used in the process industry and in chemical engineering education for years. Mr. Branan has either helped write or edit numerous books concerning this topic. Perhaps his most popular is *Rules of Thumb for Chemical Engineers.* Here, I'll share some of these rules with you along with some of my own. Now, be aware that these rules are for estimation and are not necessary meant to replace rigorous calculations when such calculations should be performed. But at many stages of analysis and design, these rules can save you hours and hours.

Download these rules of thumb in MS Excel format.

Physical Properties

Property | Units | Water | Organic Liquids | Steam | Air | Organic Vapors |

Â | Â | Â | Â | Â | Â | Â |

Heat Capacity | KJ/kg 0C | 4.2 | 1.0-2.5 | 2.0 | 1.0 | 2.0-4.0 |

Â | Btu/lb 0F | 1.0 | 0.239-0.598 | 0.479 | 0.239 | 0.479-0.958 |

Density | kg/m3 | 1000 | 700-1500 | Â | 1.29@STP | Â |

Â | lb/ft3 | 62.29 | 43.6-94.4 | Â | 0.08@STP | Â |

Latent Heat | KJ/kg | 1200-2100 | 200-1000 | Â | Â | Â |

Â | Btu/lb | 516-903 | 86-430 | Â | Â | Â |

Thermal Cond. | W/m 0C | 0.55-0.70 | 0.10-0.20 | 0.025-0.070 | 0.025-0.05 | 0.02-0.06 |

Â | Btu/h ft 0F | 0.32-0.40 | 0.057-0.116 | 0.0144-0.040 | 0.014-0.029 | 0.116-0.35 |

Viscosity | cP | 1.8 @ 0 0C | **See Below | 0.01-0.03 | 0.02-0.05 | 0.01-0.03 |

Â | Â | 0.57 @ 50 0C | Â | Â | Â | Â |

Â | Â | 0.28 @ 100 0C | Â | Â | Â | Â |

Â | Â | 0.14 @ 200 0C | Â | Â | Â | Â |

Prandtl Number | Â | 1-15 | 10-1000 | 1.0 | 0.7 | 0.7-0.8 |

** Viscosities of organic liquids vary widely with temperature

Liquid densities vary with temperature to this approximation:

Eq. (1) |

Gas densities can be calculated by:

Eq. (2) |

The boiling point of water can be approximated as a function of pressure by:

Eq. (3) |

Â

Materials of Construction

Material | Advantage | Â | Disadvantage |

Carbon Steel | Low cost, easy to fabricate, abundant, most common material. Resists most alkaline environments well. | Â | Very poor resistance to acids and stronger alkaline streams. More brittle than other materials, especially at low temperatures. |

Stainless Steel | Relatively low cost, still easy to fabricate. Resist a wider variety of environments than carbon steel. Available is many different types. | Â | No resistance to chlorides, and resistance decreases significantly at higher temperatures. |

254 SMO (Avesta) | Moderate cost, still easy to fabricate. Resistance is better over a wider range of concentrations and temperatures compared to stainless steel. | Â | Little resistance to chlorides, and resistance at higher temperatures could be improved. |

Titanium | Very good resistance to chlorides (widely used in seawater applications). Strength allows it to be fabricated at smaller thicknesses. | Â | While the material is moderately expensive, fabrication is difficult. Much of cost will be in welding labor. |

Pd stabilized Titanium | Superior resistance to chlorides, even at higher temperatures. Is often used on sea water application where Titanium's resistance may not be acceptable. | Â | Very expensive material and fabrication is again difficult and expensive. |

Nickel | Very good resistance to high temperature caustic streams. | Â | Moderate to high expense. Difficult to weld. |

Hastelloy Alloy | Very wide range to choose from. Some have been specifically developed for acid services where other materials have failed. | Â | Fairly expensive alloys. Their use must be justified. Most are easy to weld. |

Graphite | One of the few materials capable of withstanding weak HCl streams. | Â | Brittle, very expensive, and very difficult to fabricate. Some stream components have been know to diffusion through some types of graphites. |

Tantalum | Superior resistance to very harsh services where no other material is acceptable. | Â | Extremely expensive, must be absolutely necessary. |

Compressors and Vacuum Equipment

A. The following chart is used to determine what type of compressor is to be used:

Figure 1: Range Chart for Various Types of Compressors |

B. Fans should be used to raise pressure about 3% (12 in water), blowers to raise to less than 2.75 barg (40 psig), and compressors to higher pressures.

C. The theoretical reversible adiabatic power is estimated by:

Power = m z_{1} R T_{1} [({P_{2} / P_{2}}a - 1)] / a | Eq. (4) |

where:

T_{1} is the inlet temperature

R is the gas constant

z_{1} is the compressibility

m is the molar flow rate

a = (k-1)/k

k = Cp/Cv

D. The outlet for the adiabatic reversible flow, T_{2} = T_{1} (P_{2} / P_{1})a

E. Exit temperatures should not exceed 204 Â°C (400 Â°F).

F. For diatomic gases (Cp/Cv = 1.4) this corresponds to a compression ratio of about 4

G. Compression ratios should be about the same in each stage for a multistage unit, the ratio = (Pn / P1) 1/n, with n stages.

H. Efficiencies for reciprocating compressors are as follows:

65% at compression ratios of 1.5

75% at compression ratios of 2.0

80-85% at compression ratios between 3 and 6

I. Efficiencies of large centrifugal compressors handling 2.8 to 47 m3/s (6000-100,000 acfm) at suction is about 76-78%

J. Reciprocating piston vacuum pumps are generally capable of vacuum to 1 torr absolute, rotary piston types can achieve vacuums of 0.001 torr.

K. Single stage jet ejectors are capable of vacuums to 100 torr absolute, two stage to 10 torr, three stage to 1 torr, and five stage to 0.05 torr.

L. A three stage ejector requires about 100 lb steam/lb air to maintain a pressure of 1 torr.

M. Air leakage into vacuum equipment can be approximated as follows:

Leakage = k V(2/3)

where:

k =0.20 for P >90 torr, 0.08 for 3 < P < 20 torr, and 0.025 for P < 1 torr

V = equipment volume in cubic feet

Leakage = air leakage into equipment in lb/h

Download these rules of thumb in MS Excel format.

Cooling Towers

A. With industrial cooling towers, cooling to 90% of the ambient air saturation level is possible.

B. Relative tower size is dependent on the water temperature approach to the wet bulb temperature:

T_{water}-T_{wb} (Â°F) | Relative Size |

5 | 2.4 |

15 | 1.0 |

25 | 0.55 |

C. Water circulation rates are generally 2-4 GPM/ft^{2} (81-162 L/min m^{2}) and air velocities are usually 5-7 ft/s (1.5-2.0 m/s){parse block="google_articles"}

D. Countercurrent induced draft towers are the most common. These towers are capable of cooling to within 2 Â°F (1.1 Â°C) of the wet bulb temperature. A 5-10 Â°F (2.8-5.5 Â°C) approach is more common.

E. Evaporation losses are about 1% by mass of the circulation rate for every 10 Â°F (5.5 Â°C) of cooling. Drift losses are around 0.25% of the circulation rate. A blowdown of about 3% of the circulation rate is needed to prevent salt and chemical treatment buildup.

Conveyors

A. Pneumatic conveyors are best suited for high capacity applications over distances of up to about 400 ft. Pneumatic conveying is also appropriate for multiple sources and destinations. Vacuum or low pressure (6-12 psig or 0.4 to 0.8 bar) is used for generate air velocities from 35 to 120 ft/s (10.7-36.6 m/s). Air requirements are usually in the range of 1 to 7 cubic feet of air per cubic foot of solids (0.03 to 0.5 cubic meters of air per cubic meter of solids).

B. Drag-type conveyors (Redler) are completed enclosed and suited to short distances. Sizes range from 3 to 19 inches square (75 to 480 mm). Travel velocities can be from 30 to 250 ft/min (10 to 75 meters/min). The power requirements for these conveyors is higher than other types.

C. Bucket elevators are generally used for the vertical transport of sticky or abrasive materials. With a bucket measuring 20 in x 20 in (500 mm x 500 mm), capacities of 1000 cubic feet/hr (28 cubic meters/hr) can be reached at speeds of 100 ft/min (30 m/min). Speeds up to 300 ft/min (90 m/min) are possible.

D. Belt conveyors can be used for high capacity and long distance transports. Inclines up to 30Â° are possible. A 24 in (635 mm) belt can transport 3000 ft^{3}./h (85 m^{3}/h) at speeds of 100 ft/min (30.5 m/min). Speeds can be as high as 600 ft/min (183 m/min). Power consumption is relatively low.

E. Screw conveyors can be used for sticky or abrasive solids for transports up to 150 ft (46 m). Inclines can be up to about 20Â°. A 12 in (305 mm) diameter screw conveyor can transport 1000-3000 ft^{3}./h (28-85 m^{3}/h) at around 40-60 rpm.

Crystallization

A. During most crystallizations, C/C_{sat} (concentration/saturated concentration) is kept near 1.02 to 1.05

B. Crystal growth rates and crystal sizes are controlled by limiting the degree of supersaturation.

C. During crystallization by cooling, the temperature of the solution is kept 1-2 Â°F (0.5-1.2 Â°C) below the saturation point at the given concentration.

D. A generally acceptable crystal growth rate is 0.10 - 0.80 mm/h

Drivers and Power Recovery

A. Efficiencies: 85-95% for motors, 40-75% for steam turbines, 28-38% for gas engines and turbines.

B. Electric motors are nearly always used for under 100 HP (75 kW). They are available up to 20,000 HP (14,915 kW).

C. Induction motors are most popular. Synchronous motors have speeds as low as 150 rpm at ratings above 50 HP (37.3 kW) only. Synchronous motors are good for low speed reciprocating compressors.

D. Steam turbines are seldom used below 100 HP (75 kW). Their speeds can be controlled and they make good spares for motors in case of a power failure.

E. Gas expanders may be justified for recovering several hundred horsepower. At lower recoveries, pressure let down will most likely be through a throttling valve.

Download these rules of thumb in MS Excel format.

Drum Type Vessels

A. Liquid drums are usually horizontal. Gas/Liquid separators are usually vertical.

B. Optimum Length/Diameter ratio is usually 3, range is 2.5 to 5.

C. Holdup time is 5 minutes for half full reflux drums and gas/liquid separators. Design for a 5-10 minute holdup for drums feeding another column.{parse block="google_articles"}

D. For drums feeding a furnace, a holdup of 30 minutes is a good estimate.

E. Knockout drum in front of compressors should be designed for a holdup of 10 times the liquid volume passing per minute.

F. Liquid/Liquid separators should be designed for settling velocities of 2-3 inches/min

G. Gas velocities in gas/liquid separators:

Eq. (5) |

where:

k is 0.35 with horizontal mesh de-entrainers and 0.167 with vertical mesh deentrainers

k is 0.1 without mesh de-entrainers

velocity is in ft/s

?_{L} is liquid density (lb/ft^{3})

?_{V} is vapor density (lb/ft^{3})

H. A six inch mesh pad thickness is very popular for such vessels.

I. For positive pressure separations, disengagement spaces of 6-18 inches before the mesh pad and 12 inches after the pad are generally suitable.

Drying of Solids

A. Spray dryer have drying times of a few seconds. Rotary dryers have drying times ranging from a few minutes to up to an hour.

B. Continuous tray and belt dryers have drying times of 10-200 minutes for granular materials or 3-15 mm pellets.

C. Drum dryers used for highly viscous fluids use contact times of 3-12 seconds and produce flakes 1-3 mm thick. Diameters are generally 1.5-5 ft (0.5 - 1.5 m). Rotation speeds are 2-10 rpm and the maximum evaporation capacity is around 3000 lb/h (1363 kg/h).

D. Rotary cylindrical dryers operate with air velocities of 5-10 ft/s (1.5-3 m/s), up to 35 ft/s (10.5 m/s). Residence times range from 5-90 min. For initial design purposes, an 85% free cross sectional area is used. Countercurrent design should yield an exit gas temperature that is 18-35 Â°F (10-20 Â°C) above the solids temperature. Parallel flow should yield an exiting solids temperature of 212 Â°F (100 Â°C). Rotation speeds of 4-5 rpm are common. The product of rpm and diameter (in feet) should be 15-25.

E. Pneumatic conveying dryers are appropriate for particles 1-3 mm in diameter and in some cases up to 10 mm. Air velocities are usually 33-100 ft/s (10-30 m/s). Single pass residence time is typically near one minute. Size range from 0.6-1.0 ft (0.2-0.3 m) in diameter by 3.3-125 ft (1-38 m) in length.

F. Fluidized bed dryers work well with particles up to 4.0 mm in diameter. Designing for a gas velocity that is 1.7-2 times the minimum fluidization velocity is good practice. Normally, drying times of 1-2 minutes are sufficient in continuous operation.

Electric Motors and Turbines

A. Efficiencies range from 85-95% for electric motors, 42-78% for steam turbines 28-38% for gas engines and turbines.

B. For services under 75 kW (100 hp), electric motors are almost always used. They can be used for services up to about 15000 kW (20000 hp).

C. Turbines can be justified in services where they will yield several hundred horsepowers. Otherwise, throttle valves are used to release pressure.

D. A quick estimate of the energy available to a turbine is given by:

Eq. (6) |

where:

?H = actual available energy, Btu/lb

C_{p} = heat capacity at constant pressure, Btu/lb Â°F

T_{1} = inlet temperature, Â°R

P_{1} = inlet pressure, psia

P_{2} = outlet pressure, psia

x = C_{p}/C_{v}

Evaporation

A. Most popular types are long tube vertical with natural or forced circulation. Tubes range from 3/4"" to 2.5"" (19-63 mm) in diameter and 12-30 ft (3.6-9.1 m) in length.

B. Forced circulation tube velocities are generally in the 15-20 ft/s (4.5-6 m/s) range.

C. Boiling Point Elevation (BPE) as a result of having dissolved solids must be accounted for in the differences between the solution temperature and the temperature of the saturated vapor.

D. BPE's greater than 7 Â°F (3.9 Â°C) usually result in 4-6 effects in series (feed-forward) as an economical solution. With smaller BPE's, more effects in series are typically more economical, depending on the cost of steam.

E. Reverse feed results in the more concentrated solution being heated with the hottest steam to minimize surface area. However, the solution must be pumped from one stage to the next.

F. Interstage steam pressures can be increased with ejectors (20-30% efficient) or mechanical compressors (70-75% efficient).

Download these rules of thumb in MS Excel format.

Filtration

A. Initially, processes are classified according to their cake buildup in a laboratory vacuum leaf filter : 0.10 - 10.0 cm/s (rapid), 0.10-10.0 cm/min (medium), 0.10-10.0 cm/h (slow).

B. Continuous filtration methods should not be used if 0.35 sm of cake cannot be formed in less than 5 minutes.

C. Belts, top feed drums, and pusher-type centrifuges are best for rapid filtering.

D. Vacuum drums and disk or peeler-type centrifuges are best for medium filtering.{parse block="google_articles"}

E. Pressure filters or sedimenting centrifuges are best for slow filtering.

F. Cartridges, precoat drums, and sand filters can be used for clarification duties with negligible buildup.

G. Finely ground mineral ores can utilize rotary drum rates of 1500 lb/day ft^{2} (7335 kg/day m^{2}) at 20 rev/h and 18-25 in Hg (457-635 mm Hg) vacuum.

H. Course solids and crystals can be filtered at rates of 6000 lb/day ft^{2} (29,340 kg/day m^{2}) at 20 rev/h and 2-6 in Hg (51-152 mm Hg) vacuum.

Heat Exchangers

A. For the heat exchanger equation, Q = UAF (LMTD), use F = 0.9 when charts for the LMTD correction factor are not available.

B. Most commonly used tubes are 3/4 in. (1.9 cm) in outer diameter on a 1 in triangular spacing at 16 ft (4.9 m) long.

C. A 1 ft (30 cm) shell will contains about 100 ft2 (9.3 m2)

A 2 ft (60 cm) shell will contain about 400 ft2 (37.2 m2)

A 3 ft (90 cm) shell will contain about 1100 ft2 (102 m2)

D. Typical velocities in the tubes should be 3-10 ft/s (1-3 m/s) for liquids and 30-100 ft/s (9-30 m/s) for gases.

E. Flows that are corrosive, fouling, scaling, or under high pressure are usually placed in the tubes.

F. Viscous and condensing fluids are typically placed on the shell side.

G. Pressure drops are about 1.5 psi (0.1 bar) for vaporization and 3-10 psi (0.2-0.68 bar) for other services.

H. The minimum approach temperature for shell and tube exchangers is about 20 Â°F (10 Â°C) for fluids and 10 Â°F (5 Â°C) for refrigerants.

I. Cooling tower water is typically available at a maximum temperature of 90 Â°F (30 Â°C) and should be returned to the tower no higher than 115 Â°F (45 Â°C)

J. Shell and Tube heat transfer coefficient for estimation purposes can be found in many reference books or an online list can be found at one of the two following addresses:

http://www.cheresources.com/uexchangers.shtml

K. Double pipe heat exchangers may be a good choice for areas from 100 to 200 ft2 (9.3-18.6 m2)

L. Spiral heat exchangers are often used to slurry interchangers and other services containing solids

M. Plate heat exchanger with gaskets can be used up to 320 Â°F (160 Â°C) and are often used for interchanging duties due to their high efficiencies and ability to "cross" temperatures. More about compact heat exchangers can be found at: http://www.virginiaheattransfer.com/

Mixing and Agitation

A. Mild agitation results from superficial fluid velocities of 0.10-0.20 ft/s (0.03-0.06 m/s). Intense agitation results from velocities of 0.70-1.0 ft/s (0.21-0.30 m/s).

B. For baffled tanks, agitation intensity is measured by power input and impeller tip speeds:

Â | Â | Power Requirements | Tip Velocity | ||||||||||

Â | HP/1000 gal | kW/m^{3} | ft/s | m/s | |||||||||

Blending | 0.2 | - | 0.5 | 0.033 | - | 0.082 | -------------------- | ------------------ | |||||

Homogeneous Reaction | 0.5 | - | 1.5 | 0.082 | - | 0.247 | 7.5 | - | 10.0 | 2.3 | - | 3.1 | |

Reaction w/ Heat Transfer | 1.5 | - | 5.0 | 0.247 | - | 0.824 | 10.0 | - | 15.0 | 3.1 | - | 4.6 | |

Liquid-Liquid Mixture | 5.0 | 0.824 | 15.0 | - | 20.0 | 4.6 | - | 6.1 | |||||

Liquid-Gas Mixture | 5.0 | - | 10.0 | 0.824 | - | 1.647 | 15.0 | - | 20.0 | 4.6 | - | 6.1 | |

Slurries | 10.0 | 1.647 | -------------------- | ------------------ |

C. Various geometries of an agitated tank relative to diameter (D) of the vessel include:

Liquid Level = D

Turbine Impeller Diameter = D/3

Impeller Level Above Bottom = D/3

Impeller Blade Width = D/15

Four Vertical Baffle Width = D/10

D. For settling velocities around 0.03 ft/s, solids suspension can be accomplished with turbine or propeller impellers. For settling velocities above 0.15 ft/s, intense propeller agitation is needed.

E. Power to mix a fluid of gas and liquid can be 25-50% less than the power to mix the liquid alone.

Download these rules of thumb in MS Excel format.

Pressure and Storage Vessels

Pressure Vessels

A. Design Temperatures between -30 and 345 Â°C (-22 to 653 Â°F) is typically about 25 Â°C (77 Â°F) above maximum operating temperature, margins increase above this range.

B. Design pressure is 10% or 0.69 to 1.7 bar (10 to 25 psi) above the maximum operating pressure, {parse block="google_articles"}whichever is greater. The maximum operating pressure is taken as 1.7 bar (25 psi) above the normal operation pressure.

C. For vacuum operations, design pressures are 1 barg (15 psig) to full vacuum.

D. Minimum thicknesses for maintaining tank structure are:

6.4 mm (0.25 in) for 1.07 m (42 in) diameter and under

8.1 mm (0.32 in) for 1.07-1.52 m (42-60 in) diameter

9.7 mm (0.38 in) for diameters over 1.52 m (60 in)

E. Allowable working stresses are taken as 1/4 of the ultimate strength of the material.

F. Maximum allowable working stresses:

Temperature | -20 to 650 Â°F | 750 Â°F | 850 Â°F | 1000 Â°F |

Â | -30 to 345 Â°C | 400 Â°C | 455 Â°C | 540 Â°C |

CS SA203 | 18759 psi | 15650 psi | 9950 psi | 2500 psi |

Â | 1290 bar | 1070 bar | 686 bar | 273 bar |

302 SS | 18750 psi | 18750 psi | 15950 psi | 6250 psi |

Â | 1290 bar | 1290 bar | 1100 bar | 431 bar |

G. Thickness based on pressure and radius is given by:

Eq. (7) |

where pressure is in psig, radius in inches, stress in psi, corrosion allowance in inches.

**Weld Efficiency can usually be taken as 0.85 for initial design work

H. Guidelines for corrosion allowances are as follows: 0.35 in (9 mm) for known corrosive fluids, 0.15 in (4 mm) for non-corrosive fluids, and 0.06 in (1.5 mm) for steam drums and air receivers.

Storage Vessels

I. For less than 3.8 m3 (1000 gallons) use vertical tanks on legs

J. Between 3.8 m3 and 38 m3 (1000 to 10,000 gallons) use horizontal tanks on concrete supports.

K. Beyond 38 m3 (10,000 gallons) use vertical tanks on concrete pads.

L. Liquids with low vapor pressures, use tanks with floating roofs.

M. Raw material feed tanks are often specified for 30 days feed supplies.

N. Storage tank capacity should be at 1.5 times the capacity of mobile supply vessels. For example, 28.4 m3 (7500 gallon) tanker truck, 130 m3 (34,500 gallon) rail cars.

Piping

A. Liquid lines should be sized for a velocity of (5+D/3) ft/s and a pressure drop of 2.0 psi/100 ft of pipe at pump discharges. At the pump suction, size for (1.3+D/6) ft/s and a pressure drop of 0.4 psi/100 ft of pipe. **D is pipe diameter in inches.

B. Steam or gas lines can be sized for 20D ft/s and pressure drops of 0.5 psi/100 ft of pipe.

C. Limits on superheated, dry steam or line should be 61 m/s (200 ft/s) and a pressure drop of 0.1 bar/100 m or 0.5 psi/100 ft of pipe.

D. For turbulent flow in commercial steel pipes, use the following:

Eq. (8) |

where:

ΔP = fricitional pressure drop (psi) / 100 equivalent feet of pipe

M = mass flow, lb/hr

μ = viscosity, cP

ρ = density, lb/ft^{3}D = pipe inside diameter, in.

** for smooth heat exchanger steel tubes, replace 20,000 with 23,000

E. For two phase flow, an estimate often used is Lockhart and Martinelli:

First, the pressure drops are calculated as if each phase exist alone in the pipe, then

Eq. (9) |

now, the total pressure drop can be calculated by one of the following:

Eq. (10) |

where:

Y_{L} = 4.6X^{-1.78} + 12.5X^{-0.68} + 0.65

Y_{G} = X^{2 }Y_{L}

F. Control valves require at least 0.69 bar (10 psi) pressure drop for sufficient control.

G. Flange ratings include 10, 20, 40, 103, and 175 bar (150, 300, 600, 1500, and 2500 psig).

H. Globe valves are most commonly used for gases and when tight shutoff is required. Gate valves are common for most other services.

I. Screwed fitting are generally used for line sizes 2 inches and smaller. Larger connections should utilize flanges or welding to eliminate leakage.

J. Pipe Schedule Number = 1000P/S (approximate) where P is the internal pressure rating in psig and S is the allowable working stress of the material is psi. Schedule 40 is the most common.

Pumps

A. Power estimates for pumping liquids:

kW = (1.67) [Flow (m3/min)] [Pressure drop (bar)] / Efficiency | Eq. (11) |

hp = [Flow (gpm)] [Pressure drop (psi)] / 1714 (Efficiency) | Eq. (12) |

**Efficiency expressed as a fraction in these relations

B. NPSH is defined as:

NPSH = (pressure at impeller eye-vapor pressure) / (density*gravitational constant) | Eq. (13) |

Common range is 1.2 to 6.1 m (4-20 ft) of liquid.

C. An equation developed for efficiency based on the GPSA Engineering Data Book is:

Efficiency = 80-0.2855F+.000378FG-.000000238FG^{2}+.000539F^{2}-.000000639(F^{2})G+ 0.0000000004(F^{2})(G^{2}) | Eq. (14) |

where Efficiency is in fraction form, F is developed head in feet, G is flow in GPM

Ranges of applicability are F=50-300 ft and G=100-1000 GPM

Error documented at 3.5%

D. Centrifugal pumps: Single stage for 0.057-18.9 m3/min (15-5000 GPM), 152 m (500 ft) maximum head;

For flow of 0.076-41.6 m3/min (20-11,000 GPM) use multistage, 1675 m (5500 ft) maximum head;

Efficiencies of 45% at 0.378 m3/min (100 GPM), 70% at 1.89 m3/min (500 GPM), 80% at 37.8 m3/min (10,000 GPM).

E. Axial pumps can be used for flows of 0.076-378 m3/min (20-100,000 GPM). Expect heads up to 12 m (40 ft) and efficiencies of about 65-85%.

F. Rotary pumps can be used for flows of 0.00378-18.9 m3/min (1-5000 GPM). Expect heads up to 15,200 m (50,000 ft) and efficiencies of about 50-80%.

G. Reciporating pumps can be used for 0.0378-37.8 m3/min (10-100,000 GPM). Expect heads up to 300,000 m (1,000,000 ft). Efficiencies: 70% at 7.46 kW (10 hp), 85% at 37.3 kW (50 hp), and 90% at 373 kW (500 hp).

Download these rules of thumb in MS Excel format.

Tray Towers

A. For ideal mixtures, relative volatility can be taken as the ratio of pure component vapor pressures.

B. Tower operating pressure is most often determined by the cooling medium in condenser or the maximum allowable reboiler temperature to avoid degradation of the process fluid.

C. For sequencing columns: {parse block="google_articles"}

- Perform the easiest separation first (least trays and lowest reflux.
- If relative volatility nor feed composition vary widely, take products off one at time as the overhead.
- If the relative volatility of components do vary significantly, remove products in order of decreasing volatility.
- If the concentrations of the feed vary significantly but the relative volatility do not, remove products in order of decreasing concentration.

D. The most economic reflux ratio usually is between 1.2Rmin and 1.5Rmin.

E. The most economic number of trays is usually about twice the minimum number of trays. The minimum number of trays is determined with the Fenske-Underwood Equation.

F. Typically, 10% more trays than are calculated are specified for a tower.

G. Tray spacings should be from 18 to 24 inches, with accessibility in mind.

H. Peak tray efficiencies usually occur at linear vapor velocities of 2 ft/s (0.6 m/s) at moderate pressures, or 6 ft/s (1.8 m/s) under vacuum conditions.

I. A typical pressure drop per tray is 0.1 psi (0.007 bar).

J. Tray efficiencies for aqueous solutions are usually in the range of 60-90% while gas absorption and stripping typically have efficiencies closer to 10-20%.

K. The three most common types of trays are valve, sieve, and bubble cap. Bubble cap trays are typically used when low-turn down is expected or a lower pressure drop than the valve or sieve trays can provide is necessary.

L. Seive tray holes are 0.25 to 0.50 in. diameter with the total hole area being about 10% of the total active tray area.

M. Valve trays typically have 1.5 in. diameter holes each with a lifting cap. 12-14 caps/square foot of tray is a good benchmark. Valve trays usually cost less than seive trays.

N. The most common weir heights are 2 and 3 in and the weir length is typically 75% of the tray diameter.

O. Reflux pumps should be at least 25% overdesigned.

P. The optimum Kremser absorption factor is usually in the range of 1.25 to 2.00.

Q. Reflux drums are almost always horizontally mounted and designed for a 5 min holdup at half of the drum's capacity.

R. For towers that are at least 3 ft (0.9 m) is diameter, 4 ft (1.2 m) should be added to the top for vapor release and 6 ft (1.8 m) should be added to the bottom to account for the liquid level and reboiler return.

S. Limit tower heights to 175 ft (53 m) due to wind load and foundation considerations.

T. The Length/Diameter ratio of a tower should be no more than 30 and preferrably below 20.

U. A rough estimate of reboiler duty as a function of tower diameter is given by:

Q = 0.5 D^{2} for pressure distillation

Q = 0.3 D^{2} for atmospheric distillation

Q = 0.15 D^{2} for vacuum distillation

where Q is in Million Btu/hr and D is tower diameter in feet.

Packed Towers

A. Packed towers almost always have lower pressure drop than comparable tray towers.

B. Packing is often retrofitted into existing tray towers to increase capacity or separation.

C. For gas flowrates of 500 ft^{3}/min (14.2 m^{3}/min) use 1 in (2.5 cm) packing, for gas flows of 2000 ft^{3}/min (56.6 m^{3}/min) or more, use 2 in (5 cm) packing.

D. Ratio of tower diameter to packing diameter should usually be at least 15.

E. Due to the possibility of deformation, plastic packing should be limited to an unsupported depth of 10-15 ft (3-4 m) while metallatic packing can withstand 20-25 ft (6-7.6 m).

F. Liquid distributor should be placed every 5-10 tower diameters (along the length) for pall rings and every 20 ft (6.5 m) for other types of random packings.

G. For redistribution, there should be 8-12 streams per sq. foot of tower area for tower larger than three feet in diameter. They should be even more numerous in smaller towers.

H. Packed columns should operate near 70% flooding.

I. Height Equivalent to Theoretical Stage (HETS) for vapor-liquid contacting is 1.3-1.8 ft (0.4-0.56 m) for 1 in pall rings and 2.5-3.0 ft (0.76-0.90 m) for 2 in pall rings

J. Design pressure drops should be as follows:

Service | Â | Pressure drop (in water/ft packing) |

Absorbers and Regenerators | Â | |

Â | Non-Foaming Systems | 0.25 - 0.40 |

Â | Moderate Foaming Systems | 0.15 - 0.25 |

Fume Scrubbers | Â | |

Â | Water Absorbent | 0.40 - 0.60 |

Â | Chemical Absorbent | 0.25 - 0.40 |

Atmospheric or Pressure Distillation | 0.40 - 0.80 | |

Vacuum Distillation | 0.15 - 0.40 | |

Maximum for Any System | 1.0 |

Â

Reactors

A. The rate of reaction must be established in the laboratory and the residence time or space velocity will eventually have to be determined in a pilot plant.

B. Catalyst particle sizes: 0.10 mm for fluidized beds, 1 mm in slurry beds, and 2-5 mm in fixed beds.

C. For homogeneous stirred tank reactions, the agitor power input should be about 0.5-1.5 hp/1000 gal (0.1-0.3 kW/m3), however, if heat is to be transferred, the agitation should be about three times these amounts.

D. Ideal CSTR behavior is usually reached when the mean residence time is 5-10 times the length needed to achieve homogeneity. Homogeneity is typically reached with 500-2000 revolutions of a properly designed stirrer.

E. Relatively slow reactions between liquids or slurries are usually conducted most economically in a battery of 3-5 CSTR's in series.

F. Tubular flow reactors are typically used for high productions rates and when the residence times are short. Tubular reactors are also a good choice when significant heat transfer to or from the reactor is necessary.

G. For conversion under 95% of equilibrium, the reaction performance of a 5 stages CSTR approaches that of a plug flow reactor.

H. Typically the chemical reaction rate will double for a 18 Â°F (10 Â°C) increase in temperature.

I. The reaction rate in a heterogeneous reaction is often controlled more by the rate of heat or mass transfer than by chemical kinetics.

J. Sometimes, catalysts usefulness is in improving selectivity rather than increasing the rate of the reaction.

Download these rules of thumb in MS Excel format.

]]>Torispherical heads are made up of two parts.Â The rounded end has a spherical shape.Â It has a radius that is described with the numerical value â€œfâ€ which is the ratio of the dish radius to the vessel shell diameter. {parse block="google_articles"}So, if the diameter of the vessel, D, is 48 inches and the value of â€œfâ€ is 1.0 then the radius of the dish is 48 inches.Â Unless f=0.5, there would be an abrupt angle between the shell and head unless a transition piece is installed. This is the second part of the torispherical head, a transition piece shaped like a donut or torus.Â It is called the â€œknuckleâ€ and there is a knuckle radius that again is related to the vessel diameter with the factor, k. Valid values for f are anything greater than 0.5; k must be greater or equal to 0 and less than or equal to 0.5.

Torispherical heads are also called â€œF&Dâ€ or â€œflanged and dishedâ€.Â The standard F&D head has f = 1.0 and kD = 3 times the metal thickness of the head.Â ASME F&D heads require a dish radius no greater than the diameter (f <= 1.0) and knuckle radius no less than 6% of the diameter (k >= 0.06) or three times the metal thickness, whichever is greater.

Figure 1: Torispherical Head |

When calculating the volume of fluid in a torispherical head, the head is divided into three parts.Â The first is the lower portion occupied by the knuckle (liquid height 0 to h1 as seen in the diagram).Â The second part is the portion covered by the spherical dish section, which also includes the knuckle at the perimeter.Â The third portion is that which is entirely within the knuckle but above the spherical section.

Each of the three parts are calculated separately and they each require solving an integral.Â Iâ€™ll use the integral for the first part for an example in this article.

Eq. (1) |

where

Â Â Â Â Â Â Â Â Â Â Â k = knuckle radius

Â Â Â Â Â Â Â Â Â Â Â D = vessel diameter

Â Â Â Â Â Â Â Â Â Â Â h = fill height of the vessel

Â Â Â Â Â Â Â Â Â Â Â n = R â€“ kD + (k^{2}D^{2} â€“ x^{2})^0.5, where R = vessel radius (= D/2)

Â Â Â Â Â Â Â Â Â Â Â w = R â€“ h

It looks complicated, but solving in Excel is straightforward if it is done sequentially as described below.

Integrals that are too complex to express implicitly are solved using â€œnumerical methodsâ€.Â A number of numerical methods have been used for integrals with Simpsonâ€™s Rule being one of the best.Â I chose to use Simpsonâ€™s Rule for this problem.Â (Other methods include the trapezoidal rule, Riemann sums, Romberg integration, Gaussian quadratures and the Monte Carlo method).Â There are many webpages that give the derivation of Simpsonâ€™s Rule and explain it in detail; I wonâ€™t repeat that work here.

Simpsonâ€™s rule requires that the integral be broken into intervals.Â Since the integral is evaluating the area under a curve, from x=0 to x= (2kDh-h^{2})^.5, the method first calculates the maximum value for x, divides that by the number of intervals, and then evaluates the function for each value of x.Â In other words, if the maximum value of x was 10 and there were 10 intervals, then the function would be evaluated with x = 0, 1, 2, 3, 4, etc. Â Notice that the term called â€œnâ€ above includes x in its formula.

An even number of intervals is required.Â The more the better. Â I found through trial and error that a good number to use for this particular problem is 1000 intervals.

Implementation could be done in a tabular form on an Excel worksheet.Â However, it is much more elegant to solve Simpsonâ€™s Rule in a Visual Basic for Applications function subroutine. Â Listing 1 gives the function.Â This is located in a VBA Module within TankVolume.Â Each place the calculation is required, the function is called using a cell formula of the form:

= Toris_V1(k, D, h)

where the variables k, D, and h are as defined above.Â Since recalculation takes time, which can be noticable, I put the function call in a conditional statement so it is only used when the tank is horizontal with torispherical heads.Â Assume the variable â€œhead_typeâ€ refers to the type of head and a value of head_type=4 refers to torispherical, the conditional function call becomes:

= if(head_type=4,Toris_V1(k, D, h), 0)

In this case, the function is called only if the heads are torispherical.Â Otherwise, a value of â€œ0â€ is returned.Â This is perfectly fine since the result of this cell in the spreadsheet is only used for torispherical heads.

Function Toris_V1(k, D, H) As Double ' ' Integral solution using Simpson's Rule ' Dim interval As Double Dim i As Integer Dim n As Double, n1 As Double, n2 As Double Dim X As Double, xmax As Double Dim steps As Integer Dim func As Double, sumfunc As Double ' On Error GoTo err_TorisV1 steps = 1000 ' '----------------------------------------------------------- ' Procedure: ' evaluate the maximum value for x (minimum value=0) ' for each value of x from min to max, at each interval, 'Â Â calculate n 'Â Â calculate the function 'Â Â sum results according to Simpson's Rule ' calculate result and return ' ' Derived function (arcsin is not an intrinsic function) 'Â Â Â Arcsin(x) = Atn (x / Sqr(-x * x + 1)) '----------------------------------------------------------- ' ' maximum value of x xmax = (2 * k * D * H - H ^ 2) ^ 0.5 R = D / 2 w = R - H interval = xmax / steps sumfunc = 0 X = 0 n1 = R - k * D n2 = k ^ 2 * D ^ 2 For i = 0 To steps Â Â Â n = n1 + Sqr(n2 - X ^ 2) Â Â Â func = Sqr(n ^ 2 - w ^ 2) / n Â Â Â func = n ^ 2 * Atn(func / Sqr(-func * func + 1)) - w * Sqr(n ^ 2 - w ^ 2) Â Â Â sumfunc = sumfunc + func Â Â Â If i > 0 And i < steps Then Â Â Â Â Â Â Â If (i / 2) = Int(i / 2) Then sumfunc = sumfunc + func Else sumfunc = sumfunc + 3 * func Â Â Â End If Â Â Â X = X + interval Â Â Â If X > xmax Then X = xmax Next i ' Toris_V1 = sumfunc * interval / 3 Exit Function ' err_TorisV1: Toris_V1 = 0 ' End Function |

This works surprisingly well and is very easy to do. **All measurements are in inches. The formula is invalid for any other units. Also, the discharging pipe must be full of liquid.**

Figure 1: Schematic Diagram |

Using a yardstick, measure any convenient distance X across the top of the liquid discharge (Figure 1).Â Â Then measure the corresponding drop distance Y from the end of X to the top of the liquid. Measure the diameter of the pipe in inches.

Eq. (1) |

For example â€“ measure a distance, X, of 24 inches across the top of the flowing discharge from the end of a 2 inch diameter pipe. Then measure vertically to the top of the liquid discharge, Y, say it is 25 inches. Then by the formula, the flow in the pipe in gallons per minute is:

2.56 times 24 times 2 squared divided by the square root of 25 =

2.56 times 96 divided by 5 =

49.15 say 50 GPM

References

- "Engineering Handbook of Conversion Factors" by Combustion Engineering, 1977, pg.142.