RADIATIVEHEATTRANSFER

RADIATIVEHEATTRANSFER

Fourth Edition Michael F. Modest Sandip Mazumder

SOLUTIONMANUAL 1 / 4

CHAPTER1

1.1Solar energy impinging on the outer layer of earth’s atmosphere (usually called “solar constant”) has been measured as 1367 W/m 2 . What is the solar constant on Mars? (Distance earth to sun=1.496×10 11 m, Mars to sun=2.28×10 11 m).Solution The total energy emitted from the sunQ=4πR 2 S σT 4 sungoes equally into all directions, so that q

SC(R)=Q/4πR

2 =σT 4 sun θ RS R ι2 q SC mars q SC earth = θ RS RM ι2 , θ RS RE ι2 = θ RE RM ι2 =

1.8×10

11

2.28×10

11 !2

=0.4305

q SC mars=0.4305×1367=588 W/m 2

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CHAPTER 1 3

1.2Assuming Earth to be a blackbody, what would be its average temperature if there was no internal heating from the core of earth?Solution Without internal heating an energy balance for earth gives Qabsorbed=Qemitted Qabs=qsol×Aproj=qsolπR 2 E Qem=σT 4 E

A=4πR

2 E σT 4 E Thus TE= θ qsol 4σ ι 1/4 =

1367 W/m 2

4×5.670×10

−8 W/m 2 K 4 !1/4

TE=279 K=6

◦

C 3 / 4

4 RADIATIVE HEAT TRANSFER

1.3Assuming Earth to be a black sphere with a surface temperature of 300 K, what must earth’s internal heat generation be in order to maintain that temperature (neglect radiation from the stars, but not the sun) (radius of the earthRE=6.37×10 6 m).Solution

Performing an energy balance on earth:

˙Q=Qemitted−Qabsorbed=σT 4 E A−qsolAproj =4πR 2 E σT 4 E −qsolπR 2 E =πR 2 E ζ 4σT 4 E −qsol ȷ =π× ζ

6.37×10

6 m ȷ 2 κ

4×5.670×10

−8 W m 2 K 4 ×300 4 K 4

−1367

W m 2 λ

˙Q=6.00×10

16 W

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