Sample Canadian Calculation System

7/30/2019 Sample Canadian Calculation System

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Synopsis:

Prepared by: Fazlul Karim, P. Eng.

Date: Title:Sr. Mechanical EngineerRCMT Canada Corp.

Verified by:

Date:

Approved by:

Date:

N-TMP-10020-R007 (Microsoft 2007), Engineering Calculation/ReportAssociated with N-PROC-MP-0044, Engineering Calculations


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Executive Summary

The addition of the strainers to the suction side of the ALWMS booster pumps (P2054& P2055) will improve the pumps operational status, efficiency and providemaintenance flexibility of the system.

In order to accommodate this requirement, the design calculation is to provide allengineering activities associated with the modification requested as per Reference: EC# 113833, to provide strainers to the inlet of the two booster pumps to collect anydebris carried by the waste water from the tanks or from other sources, to run thebooster pumps within its design capacity.

The pumps require frequent maintenance due to the clogging at the inlet of the pumps,and as a result the pumps have reduced efficiency. Both of the booster pumps arecritical from an operational and safety point of view.

The flange insert assembly was designed as Class 6 and has been assessed in thisdocument for the system performance and pressure boundary integrity.

The piping system is capable of handling 100 gpm of waste water at existing designpressure of 100 psi. This system complies with requirements of ASME B31.1-2004.

This modification that will be implimented under EC # 113833 does not impact on anyinterfacing system.

N-TMP-10020-R007 (Microsoft 2007)


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List of Revisions

RevisionNumber

PageNumber Description of Revisions

Preparedby

Verifiedby

Approvedby

R000 All Original Issue FazlulKarim

JasonRachele

PeterLindsay



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Table of Contents

Page

Executive Summary................................................................................................................... 2List of Revisions......................................................................................................................... 3

INTRODUCTION AND DESCRIPTION OF PROBLEM......................................6

ASSUMPTIONS, SIMPLIFICATIONS, AND LIMITATIONS................................6

DESIGN INPUTS....................................................................................................................... 6

SYSTEM PARAMETERS............................................................................7

CODE CLASSIFICATION.......................................................................................................... 7

MATERIAL................................................................................................................................. 7SYSTEM CATEGORIZATION...................................................................................................8APPLICABLE CODES AND STANDARDS................................................................................8TYPE OF CONNECTION...........................................................................................................8TEST 8

ANALYSIS METHOD.................................................................................8

CALCULATION METHODS.........................................................................8

ASSESSMENT.........................................................................................8

CONCLUSIONS......................................................................................10

RECOMMENDATIONS.............................................................................11

REFERENCE AND DESIGN INPUTS...........................................................11



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List of Tables and Figures

Figure 1: Arrangement of Strainer and Valve connection to the modification of suction line ---- 7

Table 1: Minor Loses ---------------------------------------------------------------------------------------------11

Figure 2: Isometric Sketch ----------------------------------------------------------------------------------------12



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INTRODUCTION AND DESCRIPTION OF PROBLEM

The addition of the strainers to the suction side of the ALWMS booster pumps willimprove the pumps operational status and efficiency. The addition of the strainers willalso provide maintenance flexibility to the system.

The strainers at the inlet of the two booster pumps will collect the debris carried bywaste water from the tanks or from other sources, and run the booster pumps at itsdesign efficiency.

The pumps required frequent maintenance due to the clogging at the inlet of thepumps, and as a result it reduces the pumps efficiency. Both of the booster pumps arecritical from an operational and safety point of view. The pumps can be shutdown onlyfor 24 hours.

The booster pumps 018-79210-P2054 and 018-79210-P2055 frequently requiremaintenance work request due to low suction. This factor is causing the pumps toperform poorly. For each maintenance time, the Common Services isolate each pump

under Personal Protective Tag (PPT), and find debris in the pump suction.To protect the pumps, a design modification is required. OPG proposed an inlinestrainer added upstream of the pumps. The design of the modification should be insuch a way that a new spool piece will have the same dimensions as the original toensure quick exchange in the field, because the pump cant be isolated for more thanone shift (24 hours).

The flange insert assembly is designed as Class 6 and has been assessed in thisdocument for the system performance and pressure boundary integrity.

The piping system is capable of handling 100 gpm of waste water at existing designpressure of 100 psi. This system complies with requirements of ASME B31.1-2004.

ASSUMPTIONS, SIMPLIFICATIONS, AND LIMITATIONS

Due to modifications of the upstream line of the ALWMS Booster Pumps, a pressuredrop calculation is required to confirm that the pump NPSHa is meeting its existingdesign condition (NPSHr).( see appendix A & B)

The modification is only at the upstream of the booster pump suction side, and if thepressure drop satisfies the existing NPSH available, then it is not required torecalculate the downstream (pump discharge side) pressure drop.

Assume that the line L2045 2 (Ref. DWG. Number NK30-FEH-79210-0001 Rev29, see appendix H) is receiving 100 gpm of waste water flow all the time and satisfiedthe pump design.

DESIGN INPUTS

The 2 , suction line to the booster pump is collecting debris and restricted the flow tothe suction side of the pump. To overcome the problem, OPG proposed to install astrainer to collect the debris before it is going to the pumps. The location of the



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upstream suction line is very congested and it is not possible to install individualstrainer at each suction line of the pump, the only option of the modification can bedone as shown in the Fig.1 (the main suction header line connected to the pumps).

Ball valve Y Strainer Ball Valve (system 1)

Ball valve Y Strainer Ball Valve (system 2)

Fig:1 Arrangement of strainer and valve connection to the modification of suction line

There are two booster pumps connected in parallel with the main line. The individualpipe connections to the pump P2054 and P2055 are very crowded with valves and

other control systems so it is not possible to install an additional strainer on each line.Therefore, it was decided to install two strainers in parallel to the main header at theinlet of the pumps as shown in Fig. 1. This arrangement will provide the continuousoperation of the pump without interruption, only required operator attention to switchover from system 1 to system 2. This option also allows maintenance to clean thestrainer time to time without the interruption of the pumps operation.

SYSTEM PARAMETERS

CODE CLASSIFICATION

The code classification is Class 6 as defined by CSA N285.0-95.

MATERIAL

The material used in the fabrication of the Pipe and the valves are Stainless SteelANC matching with existing pipe line (stainless Steel 304L) according to N-STM-01110-10002.



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SYSTEM CATEGORIZATION

The system is not a special safety system and is not required to be seismicallyqualified.

APPLICABLE CODES AND STANDARDS

CSA N285.0-95, CSA B51-97ASME B31.1-2004

TYPE OF CONNECTION

Welded, Flanged and threaded connections are used as per OPGN standards N-STM-01110-10002.

TEST

The fabricated spool shall be tested as per ASME B31.1-2004 standard for weldingand hydro test.

The installation test will be operational leak test as per ASME B31.1-2004 section137.1 & 137.2.

ANALYSIS METHOD

The analysis was carried out by classical methods supported by handcalculation.

The rules of ASME B31.1 are used to assess the pipes

No specific analysis is performed on the process pipe support plate, since the

stresses are bounded by those in the fillet weld between each process pipeinsert and the support plate.

No thermal expansion loads are considered in the design since the operationdoes not go through temperature cycles.

CALCULATION METHODS

Hand calculation method.

ASSESSMENT

Design consideration

Design input:

Design pressure: P= 100 psi

Flow: Q= 100 gpm = 13.37 ft3/min

Design Temperature 1000FN-TMP-10020-R007 (Microsoft 2007)


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Fluid pumping: Water

As the piping system is a complicated arrangement with the booster pump; as a resultfor calculation of suction behaviour, consider the pumps are drawing from an openended tank and the pumps are at a lower level as shown on the drawing NK30-DRAW-79219-10002 the marked up line as a reference (see appendix H & Fig. 2).

Pipe size: 2 Pipe ID =d= 2.469 = 2.469/12= 0.20575~0.206Area of flow: A= 3.14xd2/4 = 3.14x(2.469/12)2/4 = 0.033 ft2

Velocity of water: V =Q/A = 13.37/0.33= 402.32ft/min= 6.705ft/sec

Re = vxd/, where = kinematics viscosity = 0.738x10-5 (Re = Reynolds number)

Re= 6.705x0.206/0.738x10-5 = 186.11x103 : Therefore the flow is turbulent (Re>2000).

Consider equivalent length method for fittings pressure drop calculation.

Descriptionof fittings

Quantity Equivalentlength

Totalequivalentlength

Frictionfactor (fT)

Totalpressuredrop (ft)

Comments

4 900 longelbow

6 30f T 180xfT 0.018 3.24 f T from Cranetechnical paperTable sheet 1of 4, Appendix-F

2 900

long elbow

4 30f T 120xfT 0.018 2.16 f T from Crane

technical paperAppendix -F

2 Flowthrough Tee

2 60f T 120fT 0.018 2.16 f T from Cranetechnical paper

Appendix -F2 normalTee

1 30f T 30 fT 0.018 0.54 f T from Cranetechnical paper

Appendix -F2 x 3reducer

1 K(v2/2g)0.26(6.722/2X32.2)

0.26 0.18 From AppliedFluidMechanics Fig.

10.9 (Attachedappendix C)

2 Ballvalve

2 3fT 6 fT 0.018 0.108 f T from Cranetechnical paper

Appendix -E

Straight pipe 1 24 24 ft 0.018 0.432 f T from Cranetechnical paperTable sheet 1of 4, Appendix



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-D

Total 8.82

Table 1: Minor loses

Pressure drop for 2 strainer 1.61 ft ( from vendor data )

Total fitting loss= hf= 1.61+8.82 = 10.43 ft

hvp = 2.6 ft (from hvp= 2.6 ft @ 1000F (From Fig. 15.36 attached

see appendix B)

Elevation=241.5

Elevation = 228

Fig 2: Isometric sketch

For the installation of additional strainers at the suction side of the booster pumps are onlyrequired to calculate the required NPSH (Net Positive Suction Head).

Equation: NPSH

NPSH= hsp+hs-hf-hvp

Where hsp = static pressure head (ft), hs= Elevation difference (ft), hf= friction loss in suctionpipe (ft), hvp = vapour pressure of the liquid at the pumping temperature (ft)

hs = (228-241.6)=-13.5 ft

hf= 10.43 ft

hvp = 2.6 ft

Hsp = 14.7x144/62.4 = 33.92

Therefore NPSH = 33.92-13.5 -10.43-2.6 = 7.39 ft.

CONCLUSIONS

From the pressure drop calculation it shows that the NPHSr (required) is less thanNPHSa (available). It satisfies the pumps design suction head requirement. Thesystem will work as per existing design condition with this modification.



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RECOMMENDATIONS

Provide two strainers in a parallel connection to the main suction line of the pumpswith two isolation valves as shown on the fig1. These will facilitate flexiblemaintenance to the strainer without interrupt the operation of the booster pumps.

REFERENCE AND DESIGN INPUTS

1. NK30-DRAW-79210-10006-Purification Process & Equipment System P&ID

2. Applied Fluid Mechanics

3. Piping and support design hand book

4. N-STM-01110-10002- Design and Construction Requirements Manual



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Appendix A: Net Positive Suction Head

Ref book: (Applied Fluid Mechanics)



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Appendix B: Fluid Vapour Pressure with Temperature




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Appendix C: Resistance Coefficient for Reducer




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Appendix D : Pipe Friction Data

Ref: Crane Flowing of Fluid Hand Book



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Appendix E : Resistance Coefficient (K) for Valves and Fittings




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Appendix F : Resistance Coefficient for elbow & Tee




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Appendix G: Pipe Selection Calculation

Process Insert Pipe Properties

The wall thickness of the pipe shall be as per Equation 3 of ASME B31.1 -2004 Section 104.1.2

istm = PD0 /2(SE+Py) + A (A = Corrosion allowance)

From table A3, A213

Where y = 0.4 from table 104.1.2 (A), S= 16700 psi (from table A3), E=1

D0 = 2.875 inch, P= 100 psi (design condition given). A= 0.065 (Consider Corrosion allowances)

tm = 100x2.875/(2(16700x1+ 100x0.4)) + 0.065 = 0.0735

Consider pipe schedule 40S. (From pipe data 40S thickness is 0.203)

Design Conditions with Mechanical Load

The arrangement and loading of the piping system is shown in Fig.3 including support locations.

The effects of pressure, weight, and other sustained mechanical load must meet therequirements of the following equation (as per ASME B31.1) (page 70 Pipe design hand book)

PD0/4tn + 0.75iMA/Z< 1.0Sh

Where P = Internal design pressure psi,

D0 = outside diameter of the pipe, in

tn = nominal wall thickness of the pipe, in

MA = Resultant moment loading on cross section due to weight & other sustained loads, in.lb

Z = section modulus of pipe, in3

i = stress intensification factor (consider 1)

Sh = basic material allowable stress at maximum temperature.



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Fig: 3:- Pipe load distribution diagram

Given:

Pipe schedule 40S

Pipe OD= 2.875 in, Pipe ID= 2.469 in, Pipe moment of inertia (I)= 1.53 in4

Pipe Section Modulus = =(2 I/OD)= 1.064 in, Empty pipe unit wt = 5.79 lbs/ft,

with water= 7.86 lbs/ft, Pipe vertical length 2 ft, horizontal length = 45 in

Ball valve = 36 lbs, Strainer= 32 lbs

Calculation:Weight of vertical pipe = (d2l/4x144)x 62.4+ 2x5.79 = 15.72= 16 lbs

Taking moment at R1 and Ma=0 to determine the pipe stress. For simplification of calculationconsider reaction force are equal x,y,& z axies.

Fx=Fy=Fz, Ma=Mx=My=Mz

Ma = 82.73x40 = 3309.2 (R1 = 82.73 lb calculated)

Resultant moment = MR= 3309.22+3309.22+3309.22 = 5731.70 psi

PD0/4tn + 0.75iMR/Z ---- 1

Now putting the values in equation 1

= (100x2.875/4x0.203)+ (0.75x1x5731.70 /1.064) = 4394.26 psi

Allowable Sh = 16,700 psi ( from table A3 of ASME 31.1-2004 for 304L A 213 material)

The stresses due to sustained loads are within the code allowable stress limit. The pipeselection is OK.


R1 R2

16lbs 32lbs36lbs 16lbs36lbs

5 8.9 8.6 8.68.9

5

29.47lbs


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The reaction forces:

We can find the reaction forces from force analysis from the fig 3.

We have M=0 & F = 0.

MR1= -16x5+36x8.9+32x17.5+29.47x17.5+36x26.1-R2x35+16x40

R2x35= 2895.72

R2 = 82.73 lbs

F = 0 = 16+36+32+29.47+36+16-R1-R2

R1= 165.47-82.73= 82.73 lbs

Supports Calculations:

Consider: 3x3x3/8 angle for support, ( See fig: 4)

Base plate considered carbon steel plate, pipe clamping U bolt threaded steelrod.

From structural steel angle section (4x4x3/8) properties we have

Section area =A-2.11 in2, Ixx= 1.76 in4, pipe load= 82.73 lbs.

As shown in the pipe support drawing the load on the column is compressive rather than tensiletherefore, consider only compression load. The compressive capacity of a column is dependedon its slenderness ratio, which is define as

Slenderness ratio = Kl/r --------- (2)

Where K = a constant dependent on boundary conditions

L = un-braced length of column in

r = least radius of gyration of the member = I/A

I = moment of inertia of cross section, in4

A = area of cross section, in2

The classification of the column depends on column slenderness ratio as well as the material

property. The column classification is made by comparing the slenderness ratio to the materialparameter Cc, where Cc is define as

Cc = 22E/Sy, where E = modulus of elasticity, psi & Sy = material yield strength, psi

If Kl/r < Cc then the column is short, in this case the allowable compressive stress is

Fa= {1-(Kl/r)2/(2 Cc 2)} Sy /[5/3+(3Kl/r)/(8Cc) (Kl/r)3 /(8Cc3)] ------ (3)

E for carbon steel = 29x106 psi, Sy= 30 ksi for ASTM Grade 30. K = 0.5 from, l = 24



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By putting the values in the equation 2 & 3

Slenderness ratio= Sl = .05x20/( 1.76/2.11) = 1.09, Cc = 22x29x106/30x103 = 138.13

Therefore Sl < Cc so it is a short compression member so we put the value to equation (3)

Fa = {1-[(1.09)2/ 2x138.132]} 30,000 /[ 5/3+(3x1.09)/(8x138) {(1.092) /8x138.133}]

= 17788.41 psi is the allowable compressive force.

Force applied on the column = Fc = w/area = F/A = 165.46/2.86 = 57.83 psi.

So the angle selection is ok.

For u bolt is ok because the load capacity is 1,130 lbs ( from table 121.7.3 of ASME B 31.1)

Weld Check

5x5 plates are welding with base checker plate and welding size is 1/4.

The capacity of a weld is a function of the weld pattern properties, the size of the weld throat,and the types of electrode and the base metal used in the weld.

The forces in the weld are calculated as follows:

f1 = Faxial /Aw + MBx/Swx + MBy/Swy

f2 = Fx /Aw + MT (Y)/Jw

f3 = Fy /Aw + MT(X)/Jw

Where f1,f2,f3= mutually perpendicular forces per unit length developed in weld, lb/in

Aw = area of weld, in

Swx, Swy = section modulus of weld about the x and y axes, respectively, in2

Jw = polar moment of inertia of weld, in3

Faxial = force perpendicular to plane of weld, lb

Fx, Fy = shear forces along x and y axis, respectively, lb

MBx, MBy = bending moment about the x and y axis, respectively, in.lb

MT= torsional moment on weld, in.lb

x,y = distance to pint of interest from neutral axis, along the x and y axis, respectively, in.

The resultant force on the weld fw is found by taking the square root of the sum of the squares ofthe three perpendicular forces acting on the weld:

fw = f12+f22+f32 lb/in

For weld strength from Omer W. Blodgett we have



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Fw = 0.707xx0.3x70000 ( = weld size, from E70 electrode we have 70000 psi)

If Fw> fw then the weld is ok.

For simplicity is consider the forces along x, y, z axis are equal, fx=fy=fz.

The total reaction force on the weld is (82.73 lbs+ 82.73 lbs) 165.46 lb. Therefore, fx=fy=fz =

165.46 lb.Similarly for bending moment on weld MBx=MBy = (165.46x2.5) 413.65 in.lb

Swx = bxd + d2/3 =(5x0.25+ 0.252/3) = 1.27 in2 (b = 5 in, d= 1/4, weld size)

MT =Torsional momentum is zero, as there is no torsional forces.

Aw= (5+5+5+5) 20 in.

f1 = Faxial /Aw + MBx/Swx + MBy/Swy = 165.46/20+ 496.38/1.27+496.38/1.27 = 659.68 lb

f2 = Fx /Aw = 8.27

f3 = Fy /Aw = 8.27

Therefore, fw = f12+f22+f32 = 659.68 2+8.272+8.272 = 659.78 lb/in

Fw = 0.707x0.25x0.3x70000 = 3711.75 lb/in ( = 1/4 in).

Therefore, Fw> fw so weld is ok.



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Fig 4: Pipe Support



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Appendix H: ISOMETRIC DRAWING



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Appendix I: Modified Spool

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