T x. T k x. is a constant of integration. We integrate a second time to obtain an expression for the temperature distribution:

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1 ME 336 Fall 8 HW solution Poblem - The geneal fom of the heat diffusion equation is: T cp = ( T) + eg t - one-dimensional conduction (along the x - diection only): = ˆi and T = T( x) x - steady state conditions: t - constant themal popeties: = constant - no heat geneation with the base plate mateial: e g Theefoe, the educed heat conduction equation is: T x T x Integatg with espect to x : T =, x whee is a constant of tegation. We tegate a second time to obta an expession fo the tempeatue distibution: T ( x) = x +, whee is anothe constant of tegation. To solve fo the two constants, we need to conside the bounday conditions this poblem. - B: the heat flux along the lowe suface is equal to q T = q x x = q = q - B: the conductive heat flux along the top suface is equal to the convective heat flux T x = h( T ( ) T ) x= = h + T + h = T h ( )

2 ME 336 Fall 8 HW solution Substitutg bac the geneal fom of the tempeatue distibution: q T( x) = + x+ T h Numeical applications: T = T() = T = 339 T = T( ) = T = 3 - T = 3 : the oute suface tempeatue is highe than the safe tempeatue of. The oute suface of the enge should be coveed with potective sulation to pevent fie hazad the event of oil leaage. Poblem In ode to deteme the side suface tempeatue of the funace T, the tempeatue distibution with the funace must be detemed. This tempeatue distibution depends on the oute suface tempeatue T( ), which can be detemed by consideg an enegy balance on the oute suface. Along that suface, heat is enteg though a unifom flux q. Heat is lost the same amount via convection and adiation. The enegy balance can then be witten: 4 4 ( ( ) ) ( ( ) su ) q = h T T + T T convection adiation 8 4 Numeical application: 5 T (73 + ) +.3( 5.67 ) T ( + 73) 4 Usg a numeical solve on the calculato: T = 594 K With the nowledge of this bounday condition, we can now deive an expession fo the tempeatue distibution the funace font. The geneal fom of the heat diffusion equation is: T cp = ( T) + eg t - one-dimensional conduction (along the x - diection only): = ˆi and T = T( x) x

3 ME 336 Fall 8 HW solution - steady state conditions: t - constant themal popeties: = constant - no heat geneation: e g Theefoe, the educed heat conduction equation is: T x T x Integatg with espect to x : T =, x whee is a constant of tegation. We tegate a second time to obta an expession fo the tempeatue distibution: T ( x) = x +, whee is anothe constant of tegation. To solve fo the two constants, we need to conside the bounday conditions this poblem. - B: the heat flux along the left suface is equal to q T = q x x = q = q - B: the tempeatue along the ight suface is T q + = T = T + Substitutg bac the geneal fom of the tempeatue distibution: q T ( x) = ( x) + T 5 Numeical application: T = T() = T = 598 K 5

4 ME 336 Fall 8 HW solution T = T( ) = T = 3 Poblem 3 The geneal fom of the heat diffusion equation spheical coodates is: T T s T q T gen c s s = t - one-dimensional conduction (along the - diection only): = and T = T() - steady state conditions: t - constant themal popeties: = constant - no heat geneation: q gen Theefoe, the educed heat conduction equation is: T = T = T = We tegate both sides with espect to : We tegate aga with espect to : T = T = T() = + - B: at =, q= q The heat tansfeed to the spheical contae is due to the tic heate, which geneates a heat flux along the e diection acoss a suface aea A= 4 : ˆ q Q = 4 Sce % of the tical powe is lost to the sulation, 9% is effectively tansfeed to the contae:

5 ME 336 Fall 8 HW solution q.9p = 4 Theefoe, the bounday condition on the oute suface of the contae is expessed as: T.9P =.9P.9P = 4 = 4 4 = - B: T ( ) = T T + = = T+.9P = T + 4 Substitutg bac to the tempeatue distibution:.9p T() = T Numeical application: T = T( ) T =.3 The ate of heat supply to the wate is calculated as: Q = mc T p Q m = c p T m.9p = c T T ( ) p w Numeical application:.98 m = 485 ( ) m.5 g/s = 7.74 g/h