david tamayo case study

8/6/2019 David Tamayo Case Study

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ThePennsylvaniaStateUniversity

DepartmentofMechanicalEngineering

CaseStudy:

PressureLossesAcrossanEvaporator

Dr.JamesGBrasseur

ME320

Section3

DateofSubmition:

December3,2010

Submittedby:

DavidTamayo

ID#927165567


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ABSTRACT

Thissystemanalysisisforthepurposeofanalyzinganevaporator.Manythingsare

takenintoconsiderationonhowanevaporatorworksandhowitfunctions.Firstoff,

pressuredropaffectsthewaytheevaporatorperforms.Thegeometry,andthematerial

usedtobuildtheevaporatorwillgreatlyaffectthewaythesystemwillperformaswell.

Onethingthatcanbetakenintoconsiderationwhendesigninganevaporatoriswhatkind

ofpumpcanbeusedtocausetheflowofthefluidthroughtheevaporator.Moreover,the

roughnessofthematerialusedintheevaporatormustbetakenintoconsideration,and

heatconductivityofthematerial.

Thedatausedforthisanalysiswascollectedduringanexperimentwithonetube,

whichhadelevenchambersthroughit.Theflowratewasmeasuredatthedivingheader,

andthepressurewascollectedbetweentheentranceandexitofthechannels.

Aftercollectingthedata,andperformingtheanalysis,themostimportantresults

fromthisanalysisisrealizingwhatkindofflowtouse,whichwillbemoreefficient.

Moreover,knowingwhatflowratestouseisaveryimportantfactorbecauseitalsoaffects

thetypeofflowthatisbeingused.Oneconclusionthatonecanmakeafterperformingthis

analysisisthatLaminarflowwillhavealowerpressuredroponasystemthanturbulent

flowonthesamesystem.

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Fig1. Schematic Diagram of the flow through the Evaporator


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INTRODUCTION

An evaporator is a heat exchanger in which a refrigerant liquid enters at low pressure and

Temperature (relative to atmospheric) and leaves as a vapor. During the vaporization process, the

refrigerant "boils," absorbing energy from the refrigerated space surrounding the evaporator, and

everything within. The fluid within the refrigerated space, typically air or water, is forced over

the exterior sides of the lateral tubes of the heat exchanger containing a volatile refrigerant (e.g.,

R134a is used in automobile cooling systems). Heat energy from the air/water flow enters the

lateral tubes of the heat exchanger by convection, then is conducted through the tube walls and

into the refrigerant (again by convection.) If the refrigerant is in a saturated state and sufficient

latent heat enters the refrigerant, it changes phase (i.e., it evaporates). The schematic diagram

in Fig 1 shows the geometry and the flow directions of the two fluids in the evaporator under

analysis. The top-dividing header distributes the liquid refrigerant to the lateral tubes and the

bottom-combining header collects the vaporized refrigerant, ultimately to be condensed again

in a condenser later in the cycle. The geometry of the flat lateral tubes used in this evaporator is

shown in Fig 2. Each lateral tube has 11 channels with the dimensions shown. In this case study

you shall analyze the pressure drop from the inlet to the outlet of a single lateral tube due to a

liquid water flow that does not change phase. The data were collected by Brasseur [ref. 1] as a

preliminary analysis of two-phase pressure drop in the full evaporator. In this case study we will

use water as the refrigerant that flows through the evaporator. Moreover, we are measuring thepressure change from the inlet of the channel to the outlet of the channel. We assume that the

water will not go through a phase change, therefore the viscosity, and density of the water will

remain constant.

Setup, Data, and Methods of Analysis

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The pressure drop between the inlet and exit of a tube is, of course, the same as the

pressure drop from the inlet to the exit of a channel in the tube. The pressure drop depends on the

Reynolds number of the flow in the channel. An appropriate Reynolds number for the inertia-

dominated flow within a channel is

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(eqn. 1)

where and are density and absolute viscosity of the liquid within the channel, Vis the

average velocity of the flow on a cross section, and DHis the hydraulic diameter of the channel.

The transition between laminar and turbulent flow is when the Reynolds number is much higher

than the critical Reynolds number. Although the critical Reynolds number,Recrit, for circular

tube is often quoted to be 2300 (with fully turbulent flow produced atRe roughly between

5000 to 10000),Recritcan vary with the cross-sectional shape of the channel, roughness, flow

conditions, etc.. In addition to Reynolds number, pressure drop is affected by inlet and exit

geometries, entrance length, and tube roughness. The pressure drop depending on different flow

rates where measured with an experiment.

EXPERIMENTAL ANALYSIS

The overall pressure drop across the evaporator (Fig. 1) is important to the overall

performance of the heat exchanger and will affect the choice of other components in the system,

such as the compressor. Although the overall pressure drop is the sum of that within the two

headers plus that within the lateral tubes, the greatest contribution is in the pressure drop across

each lateral tube in the evaporator. In the experiment you will analyze, the pressure drop across a

single lateral tube measured due to the flow of liquid water at room temperature. The tests were

conducted with the procedure below.

Test Section

The tests were performed on a single aluminum flat tube with 11 channels as shown in Fig.


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2. The channel length was 24 inches, and because the tubes were new and smooth when the tests

were performed, roughness effects were negligible. All data were collected at room temperature,

about 20C.

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setup

As illustrated in Fig. 3, the two ends of the flat tube were connected to plenum chambers each of

which was connected to a flexible hose. The cross section of the plenum is shown in Fig 4. The

two plenum chambers were connected to different sides of a differential pressure gage that

measures the pressure difference between the inlet and outlet of the flat tube,P1 - P2. A variable-

area flow meter was used to measure the flow rate of waterQ into the lateral tube.


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Table 1. This is the experimental volume flow rate and pressure drop found from the experiment.

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Q volume flow rate [gal/hr] P1 - P2 Pressure drop [psi]

1.0 0.075

2.0 0.08

3.0 0.15

4.0 0.325

5.0 0.4

6.0 0.525

7.0 0.625

9.5 0.85

12.5 1.23

15.6 1.675

18.6 1.8

21.6 2.15

24.7 2.6

26.2 2.9

27.7 2.95

30.7 3.6

33.8 4.2

35.3 4.65

36.8 5.1


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The data collected from the experiment was used to analyze the different factors that will

cause drop of pressure. I used volume flow rate, pressure drop, and the physical parameters of

the device used during the experiment to evaluate the different velocities through the each

channels, the Reynolds number, major and minor pressure drops, and entrance lengths. Each

different volume flow rate affected the results, and using the following approach and analysis

many different results were found.

ANALYSIS AND RESULTS

Part I: Analysis of the Pressure Drop Using the Fully-developed Approximation

(a) In order for us to apply the correlations of the data that was found using a circular pipe to a non-

circular pipe; we must first find an effective diameter. The effective diameter is found by finding

the hydraulic diameter, or as also referred to as Dh. To find this we must use equation 2. This

equation is specific for rectangular pipes, and it is four times the cross sectional area divided by

the parameter of the rectangular pipe.

(eqn. 2)

Where Ac is the cross sectional area, so for this we would use the length times

width, and p would be 2(L + W). Where L is the length and W is the width. For this

my result for the hydraulic diameter is: Dh = 1.285714 mm

(b)Assume the flow is fully developed and calculate the Reynolds number for each flow. Also

calculate the Reynolds number, and pressure drop assuming that the flow is laminar, and then

that the flow is turbulent. Then with the calculated data plot to compare P vs. Q, and P vs. Re.

In order to find the Reynolds number we must use eqn.1, For this we must use the different flow

rates. We know that the flow rates given are measured before the water enters the channels. We


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also know that there are 11 channels so in order to make the analysis simpler I divided the flow

rate by 11, and in order to get the Reynolds number, I converted the flow rate from Gal/hr to

m3/s. Since we have conservation of mass we can use eqn. 3 to solve for the average velocity of

the channel.

(eqn.3)

FromthisweknowtheQinandweknowQoutisthecrosssectionalofthechanneltimesthe

averagevelocityofthechannel;wealsoknowthatQin=11*Qout.SowemustalsodividetheQin

by11togettheaccurateaveragevelocity.Knowingtheaveragevelocity,andlookinginthe

frontofthebook,Ifoundthedensity,theviscosity,andfromthatIusedequation1tofind

theReynoldsnumber.Moreover,tofindthepressuredropIfirsthadtofindthefrictionfactor,

f.Forlaminarflowthefrictionfactorofentranceisfoundusingeqn.4,andexitbylookingat

table8-4[engel],andforturbulentflowthefrictionfactorisfoundusingtheColebrook

equation,whichiseqn.5.Thenonceyouhavethemajorlossfrictionfactoryouapplyittoeqn.

6.Moreover,table.2containstheresultsforReynoldsnumbers,[Pmajor]lam, [Pmajor]turb,

Pminor, [P]lam, [P]turb

(eqn.4)

(eqn.5)

(eqn.6)


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Table.2TabulationofchangeinPressure,Reynoldsnumber,andflowrate.

Q(Channel)

m^3/s

Re(Channel)

flaminarP Laminar

Fully

Developd

f turbulentP

turbulentFully

Developd

P majorturbulent

P majorlaminar

P min

9.63385E-08

68.67556929 0.931918012 0.09160403 0.309909336 0.03046292 0.03046292 0.09160403 0.000308281

1.92601E-07

137.2968276 0.466143327 0.183135616 0.1829628 0.071881336 0.071881336 0.183135616 0.00123215

2.88863E-07

205.9180859 0.3108032 0.274667203 0.141893334 0.125395894 0.125395894 0.274667203 0.002771607

3.85125E-07

274.5393443 0.233117771 0.366198789 0.120600127 0.189447678 0.189447678 0.366198789 0.004926652

4.81388E-07

343.1606026 0.186501596 0.457730376 0.107218894 0.263147048 0.263147048 0.457730376 0.007697285

5.7765E-07

411.7818609 0.155422096 0.549261962 0.097871901 0.34587947 0.34587947 0.549261962 0.011083506

6.73912E-07

480.4031192 0.13322145 0.640793549 0.090890602 0.437182686 0.437182686 0.640793549 0.015085314

9.06797E-07

646.4163421 0.099007398 0.862232998 0.079394334 0.691427272 0.691427272 0.862232998 0.027312872

1.19673E-06

853.097848 0.075020703 1.137918501 0.070513042 1.069546042 1.069546042 1.137918501 0.047570796

1.48666E-06

1059.779354 0.060389929 1.413604004 0.064566778 1.511375447 1.511375447 1.413604004 0.073413111

1.7766E-06

1266.46086 0.050534527 1.689289508 0.060228337 2.013338295 2.013338295 1.689289508 0.10483982

2.06653E-06

1473.142366 0.043444545 1.964975011 0.056881164 2.572706565 2.572706565 1.964975011 0.141850921

2.35647E-06

1679.823872 0.038099232 2.240660514 0.054195493 3.187300491 3.187300491 2.240660514 0.184446415

2.50143E-06

1783.164625 0.035891246 2.378503266 0.053036775 3.514732892 3.514732892 2.378503266 0.207838309

2.6464E-06

1886.505378 0.033925162 2.516346017 0.051977098 3.855320242 3.855320242 2.516346017 0.232626302

2.93633E-06

2093.186884 0.030575387 2.79203152 0.05010324 4.575242961 4.575242961 2.79203152 0.286390581

3.22627E-06

2299.86839 0.027827679 3.067717024 0.048492077 5.345755564 5.345755564 3.067717024 0.345739253

3.37124E-06

2403.209143 0.026631057 3.205559775 0.047766515 5.749618461 5.749618461 3.205559775 0.377507737

3.5162E-06

2506.549896 0.025533104 3.343402527 0.047086667 6.165708668 6.165708668 3.343402527 0.410672318


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(c)Usingthetabulateddatafromtable.2,Iplottedthedifferentpressuredropsvs.QandRe.We

assumethattheflowisfullydevelopedentirelythroughthetube.Wealsoassumethatthereis

minorandmajorpressureloss.Thispressurelossisrelatedtotheheadloss.Bothpressureloss

andheadlossaredirectlyrelatedtolossofenergy.WeknowthatReynoldsnumberisrelated

tofrictionlossandkineticenergy.SothehigherthefrictionlossthesmallertheReynolds

number.Onthefigure5andfigure6weseethatallthreelinesareverysimilar.The

experimentaldata,predictedlaminarandturbulentflowsallareverysimilarwiththesame

slope,andfollowthesametrendlineandverysimilarmagnitudesuntiltheyreacharound20

gal/h.Equallylookingatfig.7andfig.8youcanseethattheturbulentflowdivergesfromthe

laminarflowatReynoldsnumberofabout1100.Theexperimentaldataonfigures5,6,7,and8

showthatitstaysclosetothelaminarflowuntilaround30gal/horaReynoldsnumberofabout2000.Thiscouldmeanthatfromlookingatwhentheturbulentflowdivergesfromboth

thedataandlaminarflowisRecritical,1100,andthetransitionalflowisfrom1100Reuntilthe

datadivergesfromthepredictedlaminarflow,2000.Therefore,theRecriticalis1100,andthe

transitionalflowisbetween1100and2000.ThiscriticalReynoldsnumberismuchlowerthan

thenormalpredictedcriticalReynoldsnumberofabout2300.Thiscouldbeduetothefact

thattherectangularductwillhavemuchmorefrictionthanacircularduct.Sinceitisasmaller

Recriticalitisshowingthatitisamorefrictiondominatedflow.Sincethediameterissosmall,thentheLength/Diameterisbig,thereforetheflowisfrictiondominatedanditmakesthe

criticalReynoldsnumbermuchsmaller.ThecornersalsocauseabigeffectontheReynolds

number.TheycauseagreaterminorlosswhichlowersthecriticalReynoldsnumber,andthis

makestheReynoldsnumberwefoundtobesmallerthan2300,whichisfoundusingflowwith

nocorners.HavingnocornerswillhaveahighercriticalReynoldsnumberbecauseitisless

frictiondominated.


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Figure5.Comparisonbetweenthedatafoundfromtheexperiment,andthepredictedlaminarand

turbulentpressuredropsincludingmajorandminorpressuredropsversusflowrate.

Figure6.Thepredictionofthecomparisonbetweenthedatafoundfromtheexperiment,andthepredicted

laminarandturbulentpressuredropsincludingmajorandminorpressuredropsversusflowrate.


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Figure7.Reynoldsnumberversusthepredictionofpressuredropincludingmajorandminorpressuredrops.

Figure8.ThepredictionofReynoldsnumberversusthepredictionofpressuredropincludingmajorandminor

pressuredrops.


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(d)NowwiththedatafromTable.2wemusttabulateandplotPminor/ P. Once that is tabulated,

from Table.3, we plotted Pminor/ P vs. Q. From this looking at Figure 9, and Figure 10. From

these plots you see the difference in the percentage of the minor pressure drop versus the

total pressure drop. On the first sight of Figure 9, we can assume that Re critical can be 1000-

1100. This is due to the fact that before that the two lines are very close together; therefore,

they are both representing laminar flow since they both have a similar slope, which is nearly

equal to the laminar slope. However, once you get to around the .04 ofPminor/ Pthey

diverge having the laminar flow increasing at a steady state, and the turbulent flow

decreasing with flow rate until it reaches a limit of around 0.06 Pminor/ P. This could mean

that no matter how high the flow is for laminar flow, the minor loss will increase at a steady

slope. When the flow is turbulent, the minor loss reaches a limit and stops increasing.

Figure9.Thedifferencebetweenthecomponentofminimalpressurelossovertotalpressurelossforturbulent

andlaminarflow.


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Figure10.Thepredictionofthedifferencebetweenthecomponentofminimalpressurelossovertotalpressure

lossforturbulentandlaminarflow.

(e)Wemustfindthemajor,minor,andexperimentalfrictionalfactorsusingthedatawefound

above.Aftercomputingthedatafoundonpartd,Iusedthepercentageofminorpressuredrop

foundinthetotalpressuredrop,andmultipliedittimesthetotalpressuredroptogetthe

actualvaluesofminorpressuredrop.ThenIsubtractedtheminorpressuredropsthatIfound

fromthetotalpressuredropstofindthemajorpressuredrops.Itabulatedtheseresultsnextto

theReynoldsnumbersonTable3.OnceIhadthemajorpressuredropIappliedittoeqn.7.

(eqn.7)

FromthereIobtainedthefrictionfactorformajorpressureloss,whichisequivalenttothe

experimentalfrictionfactor.WithallofthisdataIplottedthemajor,minor,andexperimental

frictionfactorsvs.Re.Oneoftheplotswasonalogloggraph,andtheotherwasonalinear

graph.LookingatFigure11.ShowsusthatatReynoldsnumber1059thethreelinesintersect,

showingthataspredictedbefore,thatthecriticalReynoldsnumberisbetween1000and1100.

Comparingfigure11tofigure13,themoodydiagram,youseesimilaritiesinwhichthelaminar

flowislinear,andhasasimilarslopeasinfigure13.Figure12,thelinearplotoffvs.Reisnot


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veryeffectiveatfindingthecriticalReynoldsnumberbecausethevaluesaretooclosetogether

andyoucanttellwhenthelinescrosseachotherorevenwhentheydiverge.Thisshowsus

thatsometimesitisbettertousealogloggraphtofindvalues.Youcantellthatthethreelines

intersectat1059,butitismorevaguethanwhenyoulookatfigure11.Moreover,the

turbulentflowalsofollowsthesamepatternasintheMoodychart.Itiseasytoseethecritical

pointbecauseyoujusthavetolookatwherethethreelinesintersect,whichisthesameasI

thecriticalpointthatIstatedabove.Thevaluefoundforthisplot,1059,ismoreprecisethan

thevaluefoundonpartc.MynewestimatedcriticalReynoldsnumberis1059.

Figure 11.fvs Re. This includes the predicted laminar flow, turbulent flow, and the experimental flow in a loglog

scale to makefvs Re more similar to the Moody Diagram.


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Firguer 12.fvs Re. This includes the predicted laminar flow, turbulent flow, and the experimental flow in a linear

scale

Figure 13. Moody Diagram.


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Table 3. Percentage of min pressure drop to total pressure drop. Experimental major, and minor pressure drop.

Part II: Estimating Accuracy of the Fully Developed Approximation.

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Figure 10. Schematic entrance length.

On part one, we estimated the pressure loss assuming the fluid flow is fully developed

through the whole channel. However, we know that is not realistic since it does take time for the

fluid to develop. The since it takes time for the fluid to develop, it must travel a distance from the

Re (Channel) pmin/P (experimental) p-total p-minor (experimental) p-major (experimental)

68.67556929 0.003354081 0.075 0.000251556 0.074748444

137.2968276 0.006683112 0.08 0.000534649 0.079465351

205.9180859 0.009989977 0.15 0.001498497 0.148501503

274.5393443 0.013274898 0.325 0.004314342 0.320685658

343.1606026 0.016538091 0.4 0.006615236 0.393384764

411.7818609 0.019779772 0.525 0.01038438 0.51461562

480.4031192 0.023000153 0.625 0.014375096 0.610624904

646.4163421 0.030704288 0.85 0.026098645 0.823901355

853.097848 0.040127562 1.23 0.049356902 1.180643098

1059.779354 0.049369379 1.675 0.08269371 1.59230629

1266.46086 0.04949528 1.8 0.089091505 1.710908495

1473.142366 0.052255634 2.15 0.112349612 2.037650388

1679.823872 0.054703517 2.6 0.142229145 2.457770855

1783.164625 0.055831923 2.9 0.161912577 2.738087423

1886.505378 0.056905417 2.95 0.167870979 2.7821290212093.186884 0.058908303 3.6 0.21206989 3.38793011

2299.86839 0.060746652 4.2 0.255135937 3.944864063

2403.209143 0.061612528 4.65 0.286498257 4.363501743

2506.549896 0.062446552 5.1 0.318477416 4.781522584


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entrance, to a certain point where it is finally fully developed. When the fluid is fully developed,

its velocity profile only changes with the y direction, and it stays constant with the x direction.

The length, or distance in which the fluid is developing is called the entrance length. On figure

10. you can see the entrance length and it is noted as Le. Neglecting this, and assuming fully

developed fluid flow can cause errors in the analysis. On part two we will find the percent error

that was made on part two by finding a more precise pressure drop by including the changes of

the boundary layer. On figure 10. When the fluid reaches point e, the flow is now fully

developed, and it will remain constant until it reaches an exit, or a distortion of or change of the

pipe.2

(a)We must estimate the entrance length of the pipe knowing, and taking into consideration that the

flow is laminar. To do this we can use eqn.8, which is used to find the entrance length of a

laminar flow. This involved using the previously estimated Reynolds number, and the hydraulic

diameter that was found on Part I.

(eqn.8)

Then more over, after finding the entrance length,Le, we must find the percent it makes up from

the original length which is 24 inches. Since we found the entrance length using Dh, the entrance

length is in meters, so we must also convert the total length in meters. To calculte Le/L, we can

use eqn.9. On the bottom we have table 4. Which shows the values of the entrance length,

Reynolds number, and percent of total length that the entrance length consists of.

(eqn.9)


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Table 4. Reynolds number, Entrance length, and percent of total length that is equal to entrance length.

(b) At point e, on figure 10, the flow has now reached a fully developed flow. This is when the

boundary layers have joined, and it now is also called a parabolic flow. Since we know that the

average velocity in the entire channel is equal through x, due to conservation of mass, and

equation 2, then the maximum velocity, Vmax, will change as the fluid flows from point 1 to 2.

Here, on part b), we must find a relationship between average velocity at e, and maximum

velocity. We know that at point e the flow is now fully developed, so the flow is also parabolic.

Since we know that the flow is mainly laminar, then that at e the flow is parabolic; then we know

that the relation ship between Vmax and Vaverage from e to 2 will be a constant. Knowing that the

flow is parabolic and laminar then we know that Vmax/Vavg is a constant. This constant for this

case is 2 [Brasseur, assignment 7.3.] So the maximum velocity, which is Vemax, will be 2 times

bigger than average velocity at e.

Re (Channel) Le (Laminar) Le/L68.67556929 0.004414857 0.00724222

137.2968276 0.008826223 0.014478712

205.9180859 0.013237588 0.021715204

274.5393443 0.017648954 0.028951696

343.1606026 0.02206032 0.036188188

411.7818609 0.026471685 0.04342468

480.4031192 0.030883051 0.050661173

646.4163421 0.041555327 0.068168187

853.097848 0.054841992 0.089963898

1059.779354 0.068128658 0.111759609

1266.46086 0.081415323 0.13355532

1473.142366 0.094701988 0.15535103

1679.823872 0.107988653 0.177146741

1783.164625 0.114631986 0.188044597

1886.505378 0.121275319 0.198942452

2093.186884 0.134561984 0.220738163

2299.86839 0.147848649 0.242533874

2403.209143 0.154491982 0.253431729

2506.549896 0.161135315 0.264329584


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(c) It must be proven that the Bernoulli equation can be used from points 1 to e. Before we can

prove anything we must state the requirements for the application of Bernoulli equation. The

fluid must be inviscid, incompressible, and not in the boundary layer. In order to use the

Bernoulli equation from point e to 1, we must follow a streamline from point e to point 1. In

order to avoid the boundary layer, we must use a streamline that runs at D/2, so at the center of

the channel. Since there is no boundary layer, then there is no friction loss, so there is no head

loss. The Bernoulli equation is eqn. 10.

(eqn.10)

From eqn.10 we can solve forp1, which is Pe-P1. From this we can get p1 as a function of

Vavg. Note that at point one, since the velocity profile is flat and has almost not boundary layer,

the maximum velocity can be said to be equal to the average velocity. Moreover, at point e, thevelocity is taken to be the velocity at the center. We know that that is the maximum velocity, and

from part b we know that Vemax = 2Ve,average, So we can replace the Vemax with the Ve average.

We also know from conservation of mass that the average velocity remains the same throughout

the channel, so Ve,avg is qual to V1average. From this, as shown below, we have change of

pressure as a function of Vaverage.


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(d)p2 and [p]f.d must be found as a function of average velocity. We can apply conservation of

energy to find this. On the first place, to find p2, we can use eqn. 11, the conservation of

energy, taking into consideration head-loss. Since we know that the average velocities are equal

in point one and two, then they cancel out. Moreover, the kinetic correction factors are the same

since they are both for laminar fully developed flow, so the whole kinetic energy component of

this equation cancels out. Since both points are both on the same heights, then the potential

energy part of this equation cancels out as well. The only thing we have left is p2 = gHL. We

know that head loss is also equal to eqn.12. So we can make p2 equal to eqn.12. From this we

now have a function of average velocity.

eqn.11

eqn.12

For fully developed we can apply the same idea, only that instead of going from e to 2, we go

from 1 to 2. This should give us the same results as we found on part 1.


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(e) For this we must develop a mathematical relationship between the percent error and the flow

Reynolds number. We have eqn.13, which is to find error the fully developed assumption from

part 1. Moreover. We also have the drop of pressures as a function of average velocities. We see

on equation 1 that Reynolds number is made up of density, length, average velocity, and

viscosity. We can substitute the different average velocities we found into the Reynolds

numbers. We must include the different lengths, for example, entrance length, and length of fully

developed region. For the assumption of fully developed flow through the entire channel, we can

do the same thing. Once we have the Reynolds number forp, and [p]fd we can plug this into

eqn.13.

eqn.13


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Figure 11. Error of approximating only fully developed to including entrance length on the Laminar Reynolds

number range.

Re (Channel)delta p F.D(psi)

delta p (psi) Error %

68.67556929 0.091884587 0.091843123 0.00045146

137.2968276 0.184256965 0.184091242 0.000900222

205.9180859 0.277189577 0.276816798 0.001346661

274.5393443 0.370682423 0.370019794 0.001790795

343.1606026 0.464735505 0.463700228 0.002232643

411.7818609 0.55934882 0.5578581 0.002672221

480.4031192 0.65452237 0.652493411 0.003109548

646.4163421 0.887089872 0.883416318 0.004158349

853.097848 1.181211703 1.174813479 0.005446161

1059.779354 1.480415781 1.470541792 0.006714525

Table.5: The tabulation for Part e).


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From what we see the highest error, assuming the flow is laminar, as seen on table 5, and

figure.11 is 0.67%. This is a very low error to take into consideration. Taking entrance length

into consideration for this low Reynolds number does not make much of a difference.

(f) If turbulent flow is used instead of laminar flow, the relative error will be much greater. This is

due to the higher Reynolds number, and friction factor. The friction factor will increase as the

Reynolds number increases. Moreover, with laminar flow, the friction factor decreases as the

Reynolds number increases due to eqn.4. Since turbulent flows has higher Reynolds numbers,

than the entrance length will be higher too, so it will have a greater effect, and have a greater

error. By looking at figure 12, you can see the much higher errors with high Reynolds numbers

when turbulent pressure drop is computed using the same approach as for laminar flow and only

changing the friction factors.

Figure 12. Laminar Error vs Turbulent error.


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Discussion and Summary

(a) Itemized list of the new knowledge that I produced with this analysis.

Calculation of entrance length

Calculation of head loss including minor and major head loss

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Calculating pressure drop from head loss.

Calculating error due to different pressure drops.

More proficient at the Bernoullis equation and conservation of energy

More proficient at reading graphs and data.

Calculating critical Reynolds number, and transitional Reynolds number.

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The way an evaporator works, and why it is designed the way it is.

(b)High-pressure drops are undesirable in an evaporator. Ways in which the design can reduce the

pressure drop, but still maximizing aspect ration can be by making the entrance and exit edges

smooth or fillet them, so there is less of a minor head loss. More over keeping low flow rates canalso lower the pressure drop. Using higher temperature, and using the same length will lower the

pressure drop.

REFERENCES

(1) Cengel, Y & Cimbala, J.M. 2006Fluid Mechanics, McGraw-Hill, New York, NY

(2) Brasseur, J.G, & Habte,M. 2007.Pressure Losses Across Evaporators. The

Pennsylvania State University. University Park, PA.

(3) "Moody Chart." Wikipedia, the Free Encyclopedia. 28 Nov. 2010. Web. 30 Nov.

2010. .

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