Example: Convection? Radiation? Or Both?PowerPoint PresentationSlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Example: Convection? Radiation? Or Both?Heat transfer takes place between objects with different temperatures and all three modes of heat transfer exist simultaneously. However, there are many situations when one mode dominates over the others. Question: how do we know which modes to neglect in order to simplify the calculation? The following is an example to provide a general guideline to determine the relative importance between convection and radiation. Neglecting the less important usually leads to a simpler solution without significant error.Example: An electronic chip can dissipation between 0.1 W to 200 W of power depending on its operating configuration. Determine the operating temperature of the chip under three different power settings: (a) 0.1 W, (b) 200 W, (c) 5 W. Assume the convection heat transfer coefficient (h) over the chip is 10 W/(m K), the ambient temperature surrounding the chip is 27°C, the surface emissivity () of the chip is 0.3 and the chip has a surface area of 0.0004 m2.convectionradiationQuestion: what happens to conduction?high. is re temperatu the whenradiationneglect However,low. is re temperatu the whenconvectionneglect :2 thumbof Rule tell?one canhow :Question less. thescontribute that one eneglect th :1 thumbof Ruleone? which:Question .expression hesimplify t tomodes theof oneneglect can weMaybe.calculator leprogrammaba hout easily wit solved benot can that equationorder fourth A)300)(108.6()300(004.0)300)(0004.0)(1067.5)(3.0()300)(0004.0)(10()()(441244844SSSSsurrSSradiationconvTTqTTqTTATThAqqqconvectionconvectionradiationQuestion: what do you mean by high or low temperature?Unfortunately, there is no easy answer to this question. It will depend on your engineering judgement. The following calculation may help to answer this question.12 4 4( ) 0.1 : 0.1 0.004( 300) (6.8 10 )( 300 )It might be reasonable that this will give a low temperature (?) neglect the radiation term0.1 0.004( 300), 325 , not really that low but let'sS SS Sa W Power q T TT T K       12 4 4radiation check.Based on this temperature, the radiation contribution will beq (6.8 10 )( 300 ) 0.021( )and it is small compared to 0.1 W so it is reasonable to neglect the radiation sinceit doesST W    not contribute significantly to the heat loss. However, we can stillconsider this contribution as shown in the following example.ature.new temper the erecalculat and load its of some relief can now we and workmuch toodo toradiationask we,convection neglectingBy 1.8q200qit? estimate wecan However,exact.not is etemperatur calculated thesinceknow not do We term?radiation theisWhat term.radiation the tocompared as small also isit and)(1.8)300(004.0qbe willoncontributi convection there, temperatu thison Basedindeed. highvery ,2329),300)(108.6(200 termconvection eneglect th (?) re temperatuhigha assume to time thisreasonable bemight It )300)(108.6()300(004.0200:200)(convconv44124412radradSSSSSqqWTKTTTTqPowerWb!match! numbers thoseuntil re temperatusurface theerecalculatandnumber thissubstituteJust problem. No answer?better evena get want toandist perfectiona is one if happens what However,!enough! good be shouldThat )(02.8)3002305(004.0qconvectionCheck now. scontribute alsoconvection since shouldit as decrease does re temperatusurface The2305),300)(108.6(8.1-200q-qconv4412convWKTTSSWhy? one.dominant more heprobably t is radiation theTherefore,1550),300(0.0045only convection930),300)(108.6(5only radiation :Checkother. over the dominate still should One really.Not terms??both keep tohave weDolow. nor too high oneither to bemight etemperatur thesince troubleus givemight thisexperience previousour From)300)(108.6()300(004.05:5)(44124412KTTKTTTTqPowerWcSSSSSS...converges.it until process correction sRepeat thi09.2)300823(004.0)300)((0.004)(q823)(),300)(108.6()(q-593.1)300782(004.0)300)((0.004)(q:procedure Repeat the782)(),300)(108.6()(q-5:re temperatu theerecalculat and ionconsiderat into thisTake52.2)300930(004.0)300)((0.004)(qlevel at this is re temperatu theif convection by the oncontributi theDetermine930)(),300)(108.6(5T determine and convectionNeglect 33conv344122conv22conv244121conv11conv14412SSSSSSSSSSTKTTTKTTTKTT814K is solutionExact !enough! good be