Test Bank for Fluid Mechanics Fundamentals and Applications 4th Edition By Yunus Cengel, John Cimbala (All Chapters, 100 Original Verified, A+ Grade) 2

FluidfMechanics

TestfProblems


1. Afnylonfstringfhasfafdiameterfoff2fmm,fpulledfbyfafforcefoff100fN.fDeterminefthefstress.

Given:
𝑭 = 𝟏𝟎𝟎𝑵
𝒅 = 𝟐𝒎𝒎 = 𝟎. 𝟎𝟎𝟐𝒎
𝒓 = 𝟏𝒎𝒎 = 𝟎. 𝟎𝟎𝟏𝒎

Solution:
𝑨 = 𝝅𝒓𝟐
𝑨 = (𝟑. 𝟏𝟒)(𝟎. 𝟎𝟎𝟏)𝟐
𝑨 = 𝟑. 𝟏𝟒 × 𝟏𝟎−𝟔 𝒎𝟐
𝑭 𝟏𝟎𝟎𝑵
𝝈= =
𝑨 𝟑. 𝟏𝟒 × 𝟏𝟎−𝟔 𝒎𝟐
𝝈 = 𝟑𝟏. 𝟓 × 𝟏𝟎𝟔 𝑵⁄𝒎𝟐

2. fAfstringf4fmmfinfdiameterfhasforiginalflengthf2fm.fThefstringfisfpulledfbyfafforcefoff200fN.fIff
theffinalflengthfoffthefspringfisf2.02fm,fdeterminefthefstress.

Given:
𝒅 = 𝟒𝒎𝒎 = 𝟎. 𝟎𝟎𝟒𝒎
𝒓 = 𝟐𝒎𝒎 = 𝟎. 𝟎𝟎𝟐𝒎
𝑭 = 𝟐𝟎𝟎𝑵

Solution:
𝑨 = 𝝅𝒓𝟐
𝑨 = (𝟑. 𝟏𝟒)(𝟎. 𝟎𝟎𝟐)𝟐
𝑨 = 𝟏𝟐. 𝟓𝟔 × 𝟏𝟎−𝟔 𝒎𝟐
𝑭 𝟐𝟎𝟎𝑵
𝝈= =
𝑨 𝟏𝟐. 𝟓𝟔 × 𝟏𝟎−𝟔 𝒎𝟐
𝝈 = 𝟏𝟓. 𝟗𝟐 × 𝟏𝟎𝟔 𝑵⁄𝒎𝟐

3. Afconcretefhasfafheightfoff5fmetersfandfhasfunitfareafoff3f𝑚2 fsupportsfafmassfoff30,000fkg
.fDeterminefthefstress.

Given:
𝑨 = 𝟑f 𝒎 𝟐
𝒎 = 𝟑𝟎, 𝟎𝟎𝟎𝒌𝒈

Solution:
𝑾 = 𝒎𝒈 = (𝟑𝟎, 𝟎𝟎𝟎𝒌𝒈)(𝟗. 𝟖𝟏 𝒎⁄𝒔𝟐 )
𝑾 = 𝟐𝟗𝟒, 𝟑𝟎𝟎𝑵
𝑭 𝟐𝟗𝟒, 𝟑𝟎𝟎𝑵
𝝈= =
𝑨 𝟑f 𝒎 𝟐
𝝈 = 𝟗. 𝟖𝟏 × 𝟏𝟎𝟒 𝑵⁄𝒎𝟐

,ProblemfSetfNo.f1 Title:fFluidfMechanics PagefNo.:f1
off2foff121f
f




4. Afhollowfsteelftubefwithfanfinsidefdiameterfoff100fmmfmustfcarryfaftensilefloadfoff400fkN.f
Determinefthefoutsidefdiameterfofftheftubefiffthefstressfisflimitedftof120f𝑀𝑁/𝑚2 .

Given:
𝒅𝒊𝒏 = 𝟏𝟎𝟎𝒎𝒎
𝑷 = 𝟒𝟎𝟎𝒌𝑵

fffffffffffff 𝝈 = 𝟏𝟐𝟎f𝑴𝑵/𝒎𝟐
ffffffffffffff



ffffffffffffff Solution:



𝟏 𝟐 𝟏
𝑨= 𝝅𝒅 − (𝟏𝟎𝟎𝟐 )
𝟒 𝟒
𝟏
= 𝝅(𝒅𝟐 − 𝟏𝟎, 𝟎𝟎𝟎)
𝟒
𝟏
𝟒𝟎𝟎, 𝟎𝟎𝟎 = 𝟏𝟐𝟎 [ 𝝅(𝒅𝟐 − 𝟏𝟎, 𝟎𝟎𝟎)]
𝟒
𝟒𝟎𝟎, 𝟎𝟎𝟎 = 𝟑𝟎𝝅𝒅𝟐 − 𝟑𝟎𝟎, 𝟎𝟎𝟎𝝅
𝟒𝟎𝟎, 𝟎𝟎𝟎 + 𝟑𝟎𝟎, 𝟎𝟎𝟎𝝅
𝒅𝟐 =
𝟑𝟎𝝅
𝒅 = 𝟏𝟏𝟗. 𝟑𝟓𝒎𝒎

5. fFindfthefstressfiffafforcefoff50fNfisfactingfonfanfareafoff5fmm2?

Given:
𝑭 = 𝟓𝟎𝑵
𝑨 = 𝟓𝒎𝒎𝟐 = 𝟓 × 𝟏𝟎−𝟔 𝒎𝟐

Solution:
𝑭 𝟓𝟎𝑵
𝝈= =
𝑨 𝟓 × 𝟏𝟎−𝟔 𝒎𝟐
𝝈 = 𝟏𝟎 × 𝟏𝟎𝟔 𝑵⁄𝒎𝟐

6. fAnfelasticfspringfisfgivenfafforcefoff1000fNfoverfanfareafoff0.2fm2.fCalculatefthefstress?

Given:
𝑭 = 𝟏𝟎𝟎𝟎𝑵
𝑨 = 𝟎. 𝟐𝒎𝟐

Solution:
𝑭 𝟏𝟎𝟎𝟎𝑵
𝝈= =
𝑨 𝟎. 𝟐𝒎𝟐
𝝈 = 𝟓𝟎𝟎𝟎 𝑵⁄𝒎𝟐

,ProblemfSetfNo.f1 Title:fFluidfMechanics PagefNo.:f2
off2foff121f
f




7. AfstringfisfstretchedfdueftofafWeightfoff20fNfwithfafcross-
sectionalfareafoff2fm.fFindfthefStress.

Given:
𝑾 = 𝟐𝟎𝑵
𝑪𝒓𝒐𝒔𝒔 − 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍f𝒂𝒓𝒆𝒂 = 𝟐𝒎
𝒈 = 𝟗. 𝟖𝐦/𝐬²

Solution:
𝑭 𝑾⁄𝒈 𝑾
𝝈= = =
𝑨 𝑨 𝑨𝒈
𝟐𝟎𝑵
𝝈=
(𝟐𝒎)(𝟗. 𝟖 𝒎⁄𝒔𝟐 )
𝝈 = 𝟏. 𝟎𝟐 𝑵⁄𝒎𝟐
8. Afmetalfwiref1fmflongfandfoff2fmmfdiameterfisfstretchedfbyfafloadfoff40kg
.fFindfthefStress.

Given:
𝒍 = 𝟏𝒎
𝒅 = 𝟐𝒎𝒎
𝒓 = 𝟏𝒎𝒎 = 𝟎. 𝟎𝟎𝟏𝒎
𝑾 = 𝟒𝟎𝒌𝒈

Solution:
𝑭 𝒎𝒈
𝝈= =
𝑨 𝝅𝒓𝟐
(𝟒𝟎𝒌𝒈)(𝟗. 𝟖 𝒎⁄𝒔𝟐 )
𝝈=
𝝅(𝟎. 𝟎𝟎𝟏𝟐 𝒎)
𝝈 = 𝟏. 𝟐𝟓 × 𝟏𝟎𝟖 𝑵⁄𝒎𝟐


9. Afmildfsteelfwirefoffradiusf0.5fmmfandflengthf3fmfisfstretchedfbyfafforcefo
ff49fN.fFindfthefStress.

Given:
𝒓 = 𝟎. 𝟓𝒎𝒎
𝒍 = 𝟑𝒎
𝑭 = 𝟒𝟗𝑵

Solution:
𝑭 𝑭
𝝈= =
𝑨 𝝅𝒓𝟐
𝟒𝟗𝑵
𝝈=
𝝅(𝟎. 𝟎𝟎𝟎𝟓𝟐 𝒎)
𝝈 = 𝟔. 𝟐𝟒 × 𝟏𝟎𝟕 𝑵⁄𝒎𝟐

, ProblemfSetfNo.f1 Title:fFluidfMechanics PagefNo.:f3
off2foff121f
f

10. Afcylindricalfspecimenfoffsteelfhavingfanforiginalfdiameterfoff12.8f𝑚𝑚fisftensileftestedffftoff
racturefandffoundftofhavefanfengineeringffracturefstrengthfoff460f𝑀𝑃𝑎.fIffitsfcross-
sectionalfdiameterfatffracturefisf10.7f𝑚𝑚,fdeterminefthefloadfatffractureffromftheffracturef
strength.

Given:
𝒅𝟎 = 𝟏𝟐. 𝟖f𝒎𝒎 𝝈 = 𝟒𝟔𝟎f𝑴𝑷𝒂ff 𝒅𝒇 = 𝟏𝟎. 𝟕f𝒎𝒎f

Solution:

𝐅
𝛔=
𝐀
𝐅 = 𝛔𝐀 𝟎

𝟏𝟐. 𝟖f𝐦𝐦 𝟐 𝟏𝐦𝟐
𝐅 = (𝟒𝟔𝟎f𝐌𝐏𝐚f) ( ) 𝛑( 𝟔 )
𝟐 𝟏𝟎 𝐦𝐦𝟐

𝐅 = 𝟓𝟗𝟐𝟎𝟎f𝐍

11. Afvacuumfgagefconnectedftofafchamberfreadsf5.8fpsifatfaflocationfwherefthefatmospheri
cfpressurefisf14.5fpsi.fDeterminefthefabsolutefpressurefinfthefchamber.

Given:
𝑷𝒈𝒂𝒈𝒆 = 𝟓. 𝟖𝒑𝒔𝒊
f𝑷𝒂𝒕𝒎 = 𝟏𝟒. 𝟓𝒑𝒔𝒊




Solution:
𝑷𝒂𝒃𝒔 = 𝑷𝒂𝒕𝒎 − 𝑷𝒈𝒂𝒈𝒆
𝑷𝒂𝒃𝒔 = 𝟏𝟒. 𝟓𝒑𝒔𝒊 − 𝟓. 𝟖𝒑𝒔𝒊
𝑷𝒂𝒃𝒔 = 𝟖. 𝟕𝒑𝒔𝒊
12. Afmanometerfisfusedftofmeasurefthefpressurefinfaftank.fTheffluidfusedfhasfafspecificfgravi
tyfoff0.85,fandfthefmanometerfcolumnfheightfisf55fcm.fIfftheflocalfatmosphericfpressurefisf
96fkPa,fdeterminefthefabsolutefpressurefwithinftheftank.

Given:
S.G=0.85 h=55cm=0.55m
𝑷𝒂𝒕𝒎 = 𝟗𝟔𝒌𝑷𝒂

Solution:

𝝆 = 𝑺𝑮(𝝆𝑯𝟐 𝒐 ) = 𝟎. 𝟖𝟓(𝟏𝟎𝟎𝟎 𝒌𝒈⁄𝒎𝟑 )
𝝆 = 𝟖𝟓𝟎 𝒌𝒈⁄𝒎𝟑
𝑷 = 𝑷𝒂𝒕𝒎 + 𝝆𝒈𝒉
𝟏𝑵 𝟏𝒌𝑷𝒂
𝑷 = 𝟗𝟔𝒌𝑷𝒂 + (𝟖𝟓𝟎 𝒌𝒈⁄𝒎𝟑 )(𝟗. 𝟖 𝒎⁄𝒔𝟐 )(𝟎. 𝟓𝟓𝒎) ( 𝟐
)( )
𝟏𝒌𝒈 ∙ 𝒎⁄𝒔 𝟏𝟎𝟎𝟎 𝑵⁄𝒎𝟐
𝑷 = 𝟒. 𝟔𝒌𝑷𝒂
13. Determinefthefatmosphericfpressurefatfaflocationfwherefthefbarometricfreadingfisf740fm
mfHgfandfthefgravitationalfaccelerationfisf𝑔 =