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  • 8/8/2019 HW IV Solution

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    ME3333Thermodynamics

    HWIV_Solution,Fall2010

    232E

    Theworkrequiredtocompressaspringistobedetermined.

    Analysis Sincethereisnopreload,F=kx. Substitutingthisintotheworkexpressiongives

    [ ]

    Btu0.0107

    ftlbf8.33

    =

    =

    =

    =

    ====

    ftlbf169.778

    Btu1ft)lbf33.8(

    in12

    ft10)in1(

    2

    lbf/in200

    )(2

    22

    21

    22

    2

    1

    2

    1

    2

    1

    xxk

    xdxkkxdxFdsW

    238 Acaristoclimbahillin12s.Thepowerneededistobedeterminedforthreedifferentcases.Assumptions Airdrag,friction,androllingresistancearenegligible.AnalysisThetotalpowerrequiredforeachcaseisthesumoftheratesofchangesinpotentialandkineticenergies.Thatis,

    ga WWW&&& +=total

    (a) &Wa = 0 sincethevelocityisconstant.Also,theverticalriseish=(100m)(sin30)=50m.Thus,

    kW.047s)/(12/smkg1000

    kJ1m))(50m/skg)(9.81(1150/)(

    22

    212 =

    == tzzmgWg&

    and kW47.0=+=+= 0.470total ga WWW &&& (b) Thepowerneededtoaccelerateis

    ( )[ ] kW1.43s)/(12/smkg1000

    kJ10m/s30kg)(1150

    2

    1/)(

    2

    122

    221

    22 =

    == tVVmWa&

    and kW90.1=+=+= 1.430.47total ga WWW &&& (c) Thepowerneededtodecelerateis

    ( ) ( )[ ] kW5.57s)/(12/smkg1000

    kJ1m/s35m/s5kg)(1150

    2

    1/)(

    2

    122

    2221

    22 =

    == tVVmWa&

    and kW10.5=+=+= 1.475.57total ga WWW &&& (breakingpower)

    F

    x

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    251Afanistoacceleratequiescentairtoaspecifiedvelocityataspecifiedflowrate.Theminimumpowerthatmustbesuppliedtothefanistobedetermined.AssumptionsThefanoperatessteadily.Properties Thedensityofairisgiventobe=1.18kg/m3.AnalysisAfantransmitsthemechanicalenergyoftheshaft(shaftpower)tomechanicalenergyofair(kineticenergy).Foracontrolvolumethatenclosesthefan,theenergybalancecanbewrittenas

    0/

    energiesetc.potential,kinetic,internal,inchangeofRate

    (steady)0system

    massandwork,heat,bynsferenergy tranetofRate

    ==444 3444 2143421

    && dtdEEE outin outin EE && =

    2ke

    2out

    airoutairinsh,

    VmmW &&& ==

    wherekg/s62.10/s)m)(9kg/m18.1(

    33air === V&& m

    Substituting,theminimumpowerinputrequiredisdeterminedtobeW340==

    == J/s340

    /sm1

    J/kg1

    2

    m/s)8(kg/s).6210(

    2 22

    22out

    airinsh,

    VmW &&

    DiscussionTheconservationofenergyprinciplerequirestheenergytobeconservedasitisconvertedfromoneformtoanother,anditdoesnotallowanyenergytobecreatedordestroyedduringaprocess.Inreality,thepowerrequiredwillbeconsiderablyhigherbecauseofthelossesassociatedwiththeconversionofmechanicalshaftenergytokineticenergyofair.2-76 A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor

    efficiency is specified. The mechanical efficiency of the pump is to be determined.

    Assumptions1Theflowissteadyandincompressible.2Theelevationdifferenceacrossthepumpis

    negligible.PropertiesThedensityofoilisgiventobe=860kg/m3.AnalysisThenthetotalmechanicalenergyofafluidisthesumofthepotential,flow,andkineticenergies,andisexpressedperunitmassas 2/2mech VPvghe ++= .Todeterminethemechanicalefficiencyofthepump,weneedtoknowtheincreaseinthemechanicalenergyofthefluidasitflowsthroughthepump,whichis

    +=

    +==

    2)(

    2)(

    2)()(

    21

    22

    12

    21

    1

    22

    2inmech,outmech,fluidmech,

    VVPP

    VPv

    VPvmeemE V&&&&

    since vVV /&&& == m , and there is no change in the

    potential energy of the fluid. Also,

    m/s9.194/m)(0.08

    /sm1.0

    4/2

    3

    211

    1 ====DA

    VVV &&

    m/s84.84/m)(0.12

    /sm1.0

    4/2

    3

    222

    2 ====DA

    VVV &&

    PUMP

    Pumpinlet

    35kW

    Motor

    2

    1

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    Substituting, the useful pumping power is determined to be

    kW3.26

    m/skN1

    kW1

    m/skg1000

    kN1

    2

    m/s)(19.9)m/s84.8()kg/m860(kN/m400/s)m(0.1

    2

    22323

    fluidmech,upump,

    =

    +=

    = EW &&

    Then the shaft power and the mechanical efficiency of the pump become

    kW5.31kW)35)(90.0(electricmotorshaftpump, === WW &&

    83.6%==== 836.0kW31.5

    kW3.26

    shaftpump,

    upump,

    pumpW

    W

    &

    &

    Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies,

    which is 0.90.836 = 0.75.

    283CCarbondioxide(CO2),watervapor,andtraceamountsofsomeothergasessuchasmethaneandnitrogenoxidesactlikeablanketandkeeptheearthwarmatnightbyblockingtheheatradiatedfromtheearth.Thisisknownasthegreenhouseeffect.Thegreenhouseeffectmakeslifeonearthpossiblebykeepingtheearthwarm.Butexcessiveamountsofthesegasesdisturbthedelicatebalancebytrappingtoomuchenergy,whichcausestheaveragetemperatureoftheearthtoriseandtheclimateatsomelocalitiestochange.Theseundesirableconsequencesofthegreenhouseeffectarereferredtoasglobalwarming or global climate change. The greenhouse effect can be reduced by reducing the netproduction of CO2 by consuming less energy (for example, by buying energy efficient cars andappliances)andplantingtrees.