lecture 03 - piping specifications

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Piping Specifications

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Page 1: Lecture 03 - Piping Specifications

11/20/2013

1

Department of Chemical EngineeringCollege of EngineeringUniversity of the Philippines DilimanSecond Semester, AY 2013-2014

ChE 142: ChemicalEngineering Plant Design

LECTURE 03LECTURE 03Piping Specifications

• Introduction• Line Identification

Outline of Lecture

Line Identification• Line Sizing Criteria

o Single Phaseo Two Phase (liquid-vapor)

• Insulation• P&ID Update• Line List

Introduction Line Identification

A”-BBB-CCC-DDD-EEE-F

• A = Nominal pipe size (NPS)• BBB = Fluid code• CCC = Unit number (100, 200, etc)• DDD = P&ID number (001, 002, etc)• EEE = Line number (001, 002, etc)• F = Insulation type (H, C, P, A)

Page 2: Lecture 03 - Piping Specifications

11/20/2013

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P Process stream

Line IdentificationFluid Codes

BFW Boiler feed waterCWS Cooling water supplyCWR Cooling water returnDMW Demineralized waterFG Fuel gasHPS High pressure steamLPS Low pressure steamLPS Low pressure steamLPC Low pressure condensateMPS Medium pressure steamMPC Medium pressure condensateN NitrogenPA Plant airWW Waste water

Line Sizing CriteriaIntroduction

There are some guidelines in selecting pipes

Line Sizing CriteriaIntroduction

of certain sizes and schedules:

• Piping smaller than ¾” shall not be used.• The following pipe sizes shall not be used:

1¼”, 2½”, 3½, 5”, and 9”• As initial assumptions,

o Schedule 80 for sizes 1½” and smallero Schedule 40 for sizes 2” and larger.

Line Sizing CriteriaSingle Phase

Given: Mass flowrate, density, and viscosity as listed on the HMB table.

1. Assume an NPS.2. Solve for the Reynolds number.

Page 3: Lecture 03 - Piping Specifications

11/20/2013

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Line Sizing CriteriaSingle Phase

3. Determine the friction factor from the Reynolds number.

Line Sizing CriteriaSingle Phase

3. Determine the friction factor from the Reynolds number.

4. Calculate ∆P (in psi) for a length of 100 ft.5. Change NPS based on maximum ∆P and

velocity criteria.

Line Sizing CriteriaSingle Phase

FluidMaximum

P DMaximum

V l itPressure Drop Velocity

Boiling liquids at pump suction

0.25 psi/100 ft 4 ft/s

Subcooled liquids at pump suction

1.0 psi/100 ft 8 ft/s

Boiling liquids at pump discharge

4.0 psi/100 ft 15 ft/sp p g

Subcooled liquids at pump discharge

4.0 psi/100 ft 15 ft/s

Atmospheric orhigher pressure gases

2.0 psi/100 ft -

Vacuum gases 0.5 psi/100 ft -

Line Sizing CriteriaSingle Phase

What line sizing criteria shall be applied on the following lines?

• Bottoms from tower to pump/reboiler• Bottoms from pump to cooler• Bottoms from cooler to storage• Reboiler outlet to tower• Cooling water supply/return

Page 4: Lecture 03 - Piping Specifications

11/20/2013

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For two-phase flow, the recommended

Line Sizing CriteriaTwo-Phase (Liquid and Vapor)

maximum pressure drop is 4 psi/100 ft.

Calculate the pressure drop for two-phase flow using the Duckler method (see handout for the procedure).

Insulation is done on equipment and piping mainly for temperature maintenance and

i f i d l

Insulation

protection of equipment and personnel.

• Heat conservation• Cold conservation• Personnel protection• Antisweat protection

Fi f i• Fireproof protection• Explosion protection• Acoustic protection

Limits the loss of heat for process reasons

InsulationHeat Conservation

where the temperature is to be kept either constant or higher than ambient.

Limit the loss of heat for process reasons

InsulationCold Conservation

where the temperature is to be kept below ambient.

Page 5: Lecture 03 - Piping Specifications

11/20/2013

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For lines where heat/cold conservation is not

InsulationPersonnel Protection

required, install insulation for those operating greater than 50ºC or less than 0ºC in order to avoid direct personnel contact.

When the piping surface is cold enough,

InsulationAntisweat Protection

airborne moisture may condense on the surface. Corrosion is promoted if left as is.

Insulation

Common insulation materials include mineral wool, calcium silicate, and polystyrene.

Insulation

An optimum insulation thickness exists.• Thin insulation = High cost due to energy lossesg gy• Thick insulation = High cost due to insulation

Selection of insulation material and thickness is the job of piping engineers. Hence, only the identification of insulation type (H, C, P, A) will be done on this course.

Page 6: Lecture 03 - Piping Specifications

11/20/2013

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P&ID Update P&ID Update

Line List

Line Size Fluid Code

Unit Number

P&ID Number

Line Number

Insulation Type

Pressure Temp From To

Department of Chemical EngineeringCollege of EngineeringUniversity of the Philippines DilimanSecond Semester, AY 2013-2014

ChE 142: ChemicalEngineering Plant Design

LECTURE 03LECTURE 03Piping Specifications

Page 7: Lecture 03 - Piping Specifications

Lecture 03: Piping Specifications Pressure Drop Calculation for Two-Phase Flow Engr. Kenneth Robert de Guzman

The pressure drop of two-phase (vapor-liquid) flow is calculated using the Duckler method:

1. Calculate the superficial gas and liquid velocities in ft/s. (Volumetric flow in ft3/s, pipe diameter in inches)

576⁄ , 576⁄

2. Calculate the no-slip mixture velocity

3. Calculate the no-slip liquid holdup

4. Calculate the no-slip mixture density (in lb/ft3)

1

5. Calculate the no-slip mixture viscosity (in cP) 1

6. Calculate the dimensionless Froude number. Acceleration due to gravity

g is 32.2 ft/s2

12⁄

Steps 7 to 11 involve an iterative procedure to simultaneously solve for Reynolds number and liquid holdup.

7. Assume a value for the dimensionless Reynolds number 8. Calculate the dimensionless Hughmark liquid holdup parameter

⁄ ⁄

9. Calculate the dimensionless Bank off-flow parameter If Z < 8 log 0.7045 1.99305 log 0.20572 0.0068 If Z > 8 log 0.0887 log 0.19634

10. Calculate the dimensionless Hughmark liquid holdup 1 1

11. Calculate a new dimensionless Reynolds number

1241

Repeat Steps 8 to 11 until the assumed and calculated Reynolds numbers agree to within 1% on successive iterations.

12. Calculate the two-phase mixture density 11

13. Calculate the dimensionless two-phase Reynolds number

,124

14. Calculate the single-phase Fanning friction factor

0.00140.125

,.

15. Calculate the two-phase Fanning friction factor

1 ln1.281 0.4781 ln 0.444 ln 0.094 ln 0.00843 ln

16. Calculate the friction pressure drop per length (in psi/100 ft). Correction

factor gc is 32.174 ft-lb/lbf-s2.

∆16.67