met 330: fluid mechanics 0. ayala spring 2017 … · applied fluid mechanics 7 th edition by robert...

Team 7 Continental AG proposal

MET 330: Fluid Mechanics

0. Ayala

Spring 2017 Project

Tasks 1-9 submission

Team 7

Eric Stone

Sylvia Homes-Caldwell

Nathan Williamson

links

Task l

Purpose: According to the problem description we are required to have three different tanks; an operating tank, a storage tank, and a dirty fluid tank.

Diagrams: See following overhead tank view.

Sources: Mott, R. Untener, J.A., "Applied Fluid Mechanics", 7 t h ed. Pearson Education, Inc, (2015).

http://www.engineeringtoolboxxoni/fuel-oil"Storage-tanks~dimensioiis-d 1585 .html

Design Considerations: Since we are designing this system to work in the real world, we decided to try and find a supplier for most of our parts. We found our tank dimensions from the referenced website.

After finding this website we agreed to make each tank a cylindrical shaped tank.

Data/variables:

1,000 gal. operating tank: Diameter = 48in Length = 10ft lOin

15,000 gal. holding tank: Diameter = 120in Length = 25ft 6in

7,500 gal. dirty fluid tank: Diameter = 96in Length = 19ft 8in

Procedure: With placing each tank we thought it would be best to locate them where they will be put to use. The storage tank was placed by the railroad tracks, the operating tank was placed close to the machining area, and the dirty fluid tank was placed close to the driveway.

Calculations: The dimensions for the given size tanks are supplied from the manufacturer

Summary: The listed tanks have been chosen as the minimum sizes necessary for operation of the system.

Materials: 1,000 gal. operating tank, 15,000 gal. holding tank, 7,500 gal. dirty fluid tank

Analysis: Since we were allowed some freedom with the building plans, we moved the driveway towards the right side of the building where it was originally placed in figure 1 of our project packet. We then assumed a building size of 500 ft x 1,000 ft.

Group Project

Task 2

Purpose

One of the first steps to designing a good pipe system is selecting the tank material and specify

the wall thickness of the storage tanks.

Sources

Applied Fluid Mechanics 7 t h edition by Robert L . Mott and Joseph A. Untener

Design Considerations

The locations of the tanks and how they are place around the building. While also taking in to

account how far apart each is and if they are grouped together. These design considerations were

determined in task one.

Data and Variables

Tank Size Dimensions

Cylindrical lOOOgal operating tank:

Diameter = 48in Length = 1 Oft 1 Oin

Cylindrical 7500gal dirty fluid tank:

Diameter = 96in Length = 19ft 8in

Cylindrical 15000gal storage tank:

Diameter = 120in Length = 25ft 6inariables

Tank Tank Material Wall Thickness

Operating Tank Stainless Steel 0.02in Storage Tank Stainless Steel 0.02in Dirty Fluid Tank Stainless Steel 0.02 in

Procedure

The first step was to look over where we decide to place each tank on the layout provided in the

problem descriptions. Then to measure out how far apart each tank would be apart to get a rough

estimate of how much pipe would have to be added to the system. Then considering that coolant

would be the fluid flowing through the pipes we chose a time limit. There was also the fact that

the shifts only lasted 8 hours so we had to have a flow rate that allowed all tanks to be emptied

and refilled within that window.

Calculations

The Equation to use to Calculate the Wall Thickness

t = pD - [ 2( SE + pY)]

Operating Tank

p = yh = 0.033944 #/inA3 X h

P = 0.033944 #/inA3 X 48in = 1.63 psig

D = 48 in or 4 ft

S = 23,300 psi

E= 0.85

Y = 0.40

t = p X 48in - [ 2( 23,300psi X 0.85 + p X 0.40)]

t = p X 48in - [2(19805psig X 1.7 +p X 0.40]

t = 1.63 psig X 48in - [2(19806pisg+ 1.63psig X 0.40]

t = 78.24pisgxin- 39613.3 = 1.9xlA-3

Dirty Fluid Tank

t = pD + [ 2( SE + P Y ) ]

p = yh = 0.033944 #/inA3 X h = 0.033944 #/inA3 X 96 in =3.26 psig

D = 96 in or 8 ft

S = 23,300 psi

E=0.85

Y = 0.40

t = p X 96in + [ 2( 23,300psi X 0.85 + p X 0.40)]

t = p X 96in -s- [2(19805psi +p X 0.40)]

t = 3.26pisg X 96in + [2(19805psi + 3.26 pisg X 0.40] = 312.96pisgxin •*• 39612.698psig

t = 7.9xlOA-3in

Storage Tank

t = pD + [ 2(SE + pY)]

p = yh = 0.0339 #/inA3 X h = 0.0339 #/inA3 X 120 in = 4.76 pisg

D = 120 in or 10 ft

S = 23,300 psi

E= 0.85

Y = 0.40

t = p X 96in - [ 2( 23,300psi X 0.85 + p X 0.40)]

t = p X 96in - [2( 19805psi +p X 0.40]

t= 4.76 pisg X 120 in * [ 2(19805psi + 4.76 pisg X 0.40]

t=571.2pisg.in + 39613.8pisg = 0.014in

Drawings & Diagrams

Summary

To calculated the wall thickness of the tanks you have consider pressure, dimensions and use

references for the allowable tension, longitudinal joint quality factor and the correction factor.

With this information, you can determine the wall thickness which will have a major impact on

the longevity of your tank.

Materials

Operating Tank, Storage Tank, Dirty Fluid Tank

Analysis

Tanks do not require thick walls like homes in buildings because steel is a very strong and

durable material. When determining the wall thickness is it important not to waste material or

space in the tank for the fluid being held. That is way wall thickness is very small. We will be

using a wall thickness of 0.02in for our project instead of the wall thickness calculated. That's

why we will have uniformity and all the tanks will have the required thickness to remain in

service.


Task 3

Purpose: To calculate the required thickness of the 24 in. nominal diameter blind flange for the physical access port to the 7,500 gal. dirty coolant holding tank.

24 inch NPS 1 1/4 in x 20

26 inches

Diagrams:


Academia.edu, https://www.academia.edu/6520039/Blind Flange Thickness calculation .

Standard of New England, LLC, "Number and Size of bolts for flanged joints - ASME B16.5", http://www.standardne.com/STD%20NE%20BOLT%20CHART.pdf

Design Considerations:

• Incompressible fluids • Constant properties • Temperature range from -20 F — 105 F • Inner Diameter measurements, Dg, from appropriate inner diameter of schedule 40, 24 in NPS. • Academia.edu resource references ASME-B-31.3, from the clause 304.5.3. I believe these are

boiler code calculations. General enough for my purposes, but due to such low pressures this system is working with, required thickness is expected to be low.

• I will assume a quality factor, E, and weld joint strength, W, to be one. The materials for this project will be of the highest quality.

• Sum of allowances variable, C, will be taken as 3 mm for corrosion allowances. • Design pressure, P, and stress value of material need not be completely converted to SI units, as

these terms' units cancel in the equation to be used, thus the ratio remains the same. • Material: ASTM A 105 - low carbon steel for piping applications. • The referenced table from Standard of New England, LLC denotes use of (20) 1 % in. diameter

bolts as standard for this blind.

tm= + C

1 6 S E W

[Equation]:


Data/variables:

• P = Static pressure of fluid in a full tank, at the centroid of the flat, vertical, blind flange. • P = (GammaCooiant)(Diameter of tank/2) = 58.656 #/ft3 (4 ft) = 234.624 #/ft 2(lft/12 in) 2 = [1.63 psi] • S (ATSM A105 steel in tension) = 23.3 ksi = [23,300 psi] • Inner Diameter, Dg = [574.7 mm] • E = quality factor = (1) = weld joint strength = W

Procedure: Importing values for variables, I will calculate the required thickness using the equation referenced above to show that class 150 standards will be more than required.

Calculations:

Summary: The minimum thickness required for this low pressure static system is only 5.1 mm

Materials: Raised face blind flange, Class 150. Secured with 20 standard steel bolts of 1 % in. diameter.

Analysis: According to parts listings for class 150 blind flanges on the website for Coastal, precision engineered flanges, the thickness of their 24 NPS blind flange is 1.88 in. = 47.8 mm thickness. This is well above requirements shown above.

Another class for even lower pressures may be appropriate, but I am not familiar with any.

t)l\ U5

NOTES:

. THE LEFT WALL OF THE BUILDING IS SO FT FROM THE RAILROAD TRACKS

• THE BOTTOM WALL OF THE BUILDING IS 200 FT FROM THE HIGHWAY

15000 GAL STORAGE TANK • •

IS FT 1000 GAL OPERATING TANK

T

50C FT

7500 GAL DIRTY FLUID TANK

DRAWN Nate CHECKED

2/9/2017

QA

MFC lank Qwjtfnh. SEE [ IDWGNO 1 REV

D [ |Tank Locations Building Plajns

APPROVED lank Qwjtfnh.

SEE [ IDWGNO 1 REV

D [ |Tank Locations Building Plajns l/60o| [SHEET 1 OFl

Group Project

Task 4

Purpose

To determine the time required to fill and empty each tank in the system while considering how many

gallons of fluid the tank would have to hold. Once the time limit is chosen we use that number to

calculate the desired flow rate to fill and empty all tanks.

Sources



We must take make sure the time it takes to empty and fill all the takes falls within the 8-hour window of

the workers have to complete the job. The sizes of the three tanks must be reviewed as well as the length

of the pipe which can have a major impact on the flow rate.

Data and Variables

Tank Time to Empty Tank Desired flow Rate

Operating Tank 0.5 hrs 33.33 gal/min Storage Tank 4 hrs 62.5 gal/min

Dirty Fluid Tank 2 hrs 62.5 gal/min

Tank size dimensions:

Cylindrical 1 OOOgal operating tank: Diameter = 48in Length = 1 Oft 1 Oin

Cylindrical 7500gal dirty fluid tank: Diameter = 96in Length = 19ft 8in

Cylindrical 15000gal storage'tank: Diameter = 120in Length = 25ft 6in

Procedure

The first step is to set the time required to empty and fill each tank. Then take the measurements of the

tanks which was determined in the first task of the project. With this data we are able to use the equation

to find the flow rate.

Calculations

Operating Tank: Diameter 48in or 4 ft, Time to empty/Fill 30 min, holds lOOOgal

Flow rate= 1000gal/30min = 33.33 gal/min

Storage Tank: Diameter 120 in or 10 ft, Time to empty/fill 4 hrs or 240 min, holds 15000gal

Flow rate = 15000gal/240 min = 62.5 gal/min

Dirty Fluid Tank: Diameter 96 in or 8 ft, Time to empty/fill 2 hrs or 120 min, holds 7500gal

Flow rate = 7500 gal/120 min = 62.5 gal/min

Summary

With these calculation, we were able to calculate a flow rate that was high enough considering the size of

the tank and the time constraints the problem description gave.

Materials

The materials involved with this tasks were the tanks, coolant, and the flange.

Analysis

With the flow rates provided we will have an appropriate estimate of the time it takes to empty and fill the

tanks. Providing the client with the necessary information to schedule out his deliveries if more coolant is

needed or a tank needs to be emptied.

Vtpt

Task 5

Purpose: To determine the pipe layout and sizes.

Diagrams: See following pipe layout views, overhead and side.


Design Considerations: We thought we should run most of the piping along the side of the building in the thought of attaching the supports needed to the wall.

We assumed the inlets to each tank were at the tops of each tank and the outlets located on the back sides towards the ground.

From class, we noticed a trend in the use of schedule 40 steel pipes so we decided to go with what we have seen work the most.

Data/variables: Volumetric flow-rate (Q-high) = 62.5 gal/min = 0.1392 ftA3/s

Volumetric flow rate (Q-low) = 33.3 gal/min = 0.0743 ftA3/s

Critical velocity = 3 m/s = 9.843 ft/s

Procedure: Using the critical velocity we can solve for the approximate required flow area, comparing it to Appendix F (properties of schedule 40 steel pipe).

Calculations: Area (high rate) = Q-high/critical velocity = 0.01414 => 1.5" pipe.

Area (low rate) = Q-low/critical velocity = 0.007545 => 1.25" pipe.

Summary: The beginning section of pipe going into the storage tank and the ending section of pipe going out of the dirty fluid tank we decided to make with a 1V2 in diameter. Every other section of the system, from the outlet of the storage tank to the inlet of the dirty fluid tank, we decided to make 1 % in diameter.

Materials: 30 ft - 1.5" sch. 40 steel pipe, 208 ft - 1.25" sch 40 steel pipe.

Analysis: In total, we will need 238 ft of piping to complete our system, along with a 50 ft length of smooth plastic flexible engaging lock hose to attach to the train and reclamation truck.

Group Project

Task 6

Purpose

To provide information on the number of valves, elbows and fitting that will be in our pipe system as well

as the type of valve, material and size it will be. This is not a system driven by gravity so it will not be a

pumped system.

Drawings and Diagrams

See following pipe and valve layout, overhead and side views

Sources



The number of valves and their specific sizes was determined from the calculations and design choses

made in task one through five. We also researched the safety codes and procedures to follow when

choosing the material to use for our system. Al l the elbows in our project will be long radius elbows Data and Variables

Type of Valve Size Material

Connect flanges (x2) 1.5 in Steel

Gate Valves (x2) 1.5 in Brass

90 Degree elbows (x3) 1.5 in Schedule 40 Steel

Gate Valves (x2) 1.25 Brass

90 Degree elbows (x7) 1.25 Schedule 40 Steel

III Pfe-

Procedure

To look at the layout of our pipe system. To decide what material would work best with the amount of

fluid and the chemical composition of our fluid. Then to choose what size and material pipes to have

between each tank to connect every part of the system. Then to calculate the flow rate and wall thickness

to determine the size of the values and the types we could use. The next step was to use this data to

calculated the minor losses we would have to add to our system because of the valves and elbows.

Calculations

The calculations for the sizes of the elbows and valves were found using the calculations done in task five

when determining the appropriate pipe sections.

Summary

Task six was to determine the number of valves, elbows and fittings were needed in the system that would

limit the number of minor losses and not cause the desired flow rate to be lost. The gates allow the tank to

have maintenance work do on it if technical issues happen in the future and give the supervisor the option

to drain the tank if necessary.

Materials

The materials that were used to complete this task were: tanks, pipes, valves, elbows, gates and the fluid

that flowed through the pipe system the coolant.

Analysis

With the number of elbows and valves added to the pipe system it also always the client to replace the

pipe without shutting down the entire system for the day if the pipes burst or become damaged. The

valves also allow the workers to control the flow in the pipe system in case of emergencies. With the

gates changes, could be made to the pipe system if a better rate of flow is required in the future.

Hi V Team 7 Continental AG proposal

Task 9

Purpose: To determine the minimum pipe wall thickness needed to handle the pressures in the system.

Diagrams:

Sources: Mott, R . Untener, J.A., "Applied Fluid Mechanics", 7 t h ed. Pearson Education, Inc, (2015).


• Constant properties

• Incompressible fluids • For cylindrical Metal shell use the formula: t = P(R+c)/(SE - 0.6P)

t = cylindrical wall thickness, inches P = Design pressure, psi c = corrosion allowance, inches R = Inside cylindrical radius, inches S = Stress value of the metal material, psi

E = Joint efficiency of the welded sections, expressed as a fraction

Data/variables:

• P D = 7.65 psig

• PB = 10psig • c = 0.0125 inches • RD-I sin = 1.61/2 = 0.805 inches • RB-i.2Sin = 1.38/2 = 0.69 inches • SASTM A105steel = 23,300 psi • E = 1

Procedure: Minimum required thickness will be calculated using the appropriate maximum pressure from the hydraulic analysis for both sizes of schedule 40 steel pipe.

Calculations:

t D-i.5,n = P D ( R D + C ) / ( S E - 0.6PD) = [7.65psig (0.805in + 0.0125in)] / [23,300psig - 0.6(7.65psig)] = 0.00027 in.

t B - i . 2 5in = P B ( R B + C ) / ( S E - 0.6PB) = [lOpsig (0.69in + 0.0125in)] / [23,300psig - 0.6(10psig)] = 0.0003 in.

Summary: The required thickness of the 1.5 in. steel pipe is 0.27 m-in., and for the 1.25 in. = 0.3 m-in.

Materials: 1.5 in. and 1.25 in. schedule 40 steel pipe made of ASTM A105 low carbon steel for pipe apps.

Analysis: The actual thickness of schedule 40 steel pipe is orders of magnitude greater than these calculated requirements, even considering lower than 1 (fractional) joint-weld strength. This suggests that PVC may have been more appropriate for use in this system.

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Task 7

Purpose: To develop the hydraulic analysis of all portions of the Team 7 system.

Diagrams: All section diagrams have been flattened into equivalent systems, for ease of view.

Train Car -> Sec. A -> 15,000 gal -> Sec. B -> 1,000 gal -> Sec. C -> 7,500 gal -> Sec. D -> Truck

Section diagrams will appear at the top of appropriate calculations page.



• Incompressible fluids • Constant properties • Temperature range from-20 F — 105 F • Pipe sizes are fixed, using critical velocity of 9.84 ft/s (3 m/s) • Sections A and D use 1.5 in. schedule 40 steel pipe and fittings, coupled with a 1.5 in. plastic roll-

up quick connect hose for industrial applications. • Sections B and C use 1.25 in. schedule 40 steel pipe and fittings. • This analysis uses only two volumetric flow rates. Sections A,D, using Qhigh and B,C, using Oj o w . • Vhigh and V|0W have been calculate from the appropriate Q value. • Two Reynolds numbers will be computed for the appropriate section, using 5 degrees F. • Since the kinematic viscosity value rises as temperature lowers, the Reynolds number will also

lower. Thus, the most energy losses due to friction will occur at lowest temperature values. • This value will be taken as 5 degrees F since appropriate insulation will prevent the freezing of

coolant at extreme temperature lows. • Inlets to the respective pipe section from tank will be taken as squared edge inlet, K = 0.5

Data/variables:

• Di25= 0.1150 ft • ReHi8h(5F) = Vhigh D L S = (9.84 ft/s)(0.1342 ft) = 4.7 x 10 4

• ReLow(5F) = Vio„ Di.25 = (7.15 ft/s)(0.1150 ft) = 3 x 10 4

Procedure: Using given values and properties of known materials I can obtain relevant f-values from the Moody chart and values of fT for new, clean schedule 40 steel of the appropriate size.

• Computing all energy losses within each section is a straightforward calculation. • Using Bernoulli's equation to solve for the energy contributed to the fluid by the pump, h A . • Using a point at the outlet for the relevant section's pump, I will use Bernoulli's equation again

to determine the maximum pressure in the line for each section.

Qh,gh =62.5 gal/min Ojow =33.3 gal/min Di.5 = 0.1342 ft

0.1392 ft 3/s 0.0743 ft3/s


Sec hon A

- ~ IRl

is'u&^tfe Y :

r 5 f f

Otit* fife' 1.5" *A HO $4e* (

Calculations, section A: Car lioll^a hnlc. /\/\^ <ru\

J i ^ L - cj?!L ' - <t sop* k , - K¥ = OS(1.5*35fi) = 0 . * < t ~7Z7

5?

/, r \6ois.4H- & o< #


" j ^ r m/vfc ^ 7 ^ Lm -rmdhii

l>pc •'

Calculations, section B: f-/o//̂ T^H/C 71* ^£r# «L ^ £ ^775J - 5 ? CO^f^Jl

, 7 7 # W -fir* - o M % ~ ro.^sx^J * <JL

r , - 10f si J


Sech'on C

k — n < — M ®

H5&

Calculations, section C: foh Ofenkj A, J A dirty f/u.J J ^ L ~ 0> 7?3f^

P3 5 *7J£f<y


1.5" StL Ho 5 he I f>if><

(Cti-

C a l c u l a , i o n , section D: ff,, Uy M U T4 Uck .^-MSg J^ll ,<^lc

faict •J coiled

* 7

7-7 ' ^

P3 - 7'65 psij


Summary: Maximum pressure for the system occurs in section B, where the max pressure at the outlet of pump B is 10 psig.

Materials:

• Tanks, 15,000 gal cylindrical, 1,000 gallon cylindrical, 7,500 gallon cylindrical • 20 ft 1.5 in. schedule 40 steel pipe • (2) 50 ft sections of 1.5 in. roll-able industrial strength plastic quick-connect hose • (2) 1.5 in. inlet/outlet pumps • (2) 1.5 in. quick connect flanges (entrance and exit of entire system) • (2) 1.5 in. brass gate valves • (3) 1.5 in. schedule 40 steel 90 degree elbows • 208 ft 1.25 in. schedule 40 steel pipe • (2) 1.25 in. inlet/outlet pumps • (2) 1.25 in. brass gate valves • (7) 1.25 in. schedule 40 steel 90 degree elbows

Analysis: Another area of interest would be at the maximum pressure for section D. Although the max pressure in this section is only 7.65 psig, the industrial strength plastic quick-connect pipe must be able to withstand this max pressure, as well as the second quick-coupler.


Task 8

Purpose: To determine the required pump head needed to maintain flow rate for appropriate section.

Diagrams: /- — ~ —̂ W


Holland, Applied Technologies, https://hollandaptblog.com/2014/02/07/what-is-pump-head-

how-is-it-different-than-pressure/


• Constant properties • Incompressible fluids • Evaluation at temperature of 5 degrees F (max friction)

• Pressure differential between pump outlet and system exit is known from hydraulic analysis.

Data/variables:

• [S.G.] = 0.94

• Psect.onA = 7.15 psig • PsectionB = 10 psig • PsectionC = 7.66 psig • PsectlonD = 7.65 p S i g

Procedure: Required pump head is the developmental head the pump must support without static pressure behind it. It is related to pressure by the equation H=2.31p/SG, where p= pressure (psi), and H=head (ft). Using this simplistic equation, I will solve for required head using the pressure differential between the known max pressure at the outlet of each section pump and the exit for that particular section.

Calculations:

Section A: H=2.31p/SG = 2.31 (7.15 psig) / [0.94] = 17.6 ft, where QA = 62.5 gal/min

Section B: H=2.31p/SG = 2.31 (10 psig) / [0.94] = 24.6 ft, where QB = 33.3 gal/min

Section C: H=2.31p/SG = 2.31 (7.66 psig) / [0.94] = 18.8 ft, where QC = 33.3 gal/min

Section D: H=2.31p/SG = 2.31 (7.65 psig) / [0.94] = 18.8 ft, where QD = 62.5 gal/min

Summary: The pump head requirements and desired flow rate are shown.

Materials: 4 pumps, tailored to the given specified minimum requirements.

Analysis: It may be possible, because of the close range of Head values, to use the same pump for each of the two same diameter pumps, A and D, as well as B and C. This will be explored in Task 15.

met 330: fluid mechanics 0. ayala spring 2017 … · applied fluid mechanics 7 th edition by robert...

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