,Radiative Heat Transfer, 4e Michael Modest, Sandip Mazumder
Solution Manual all Chapters
CHAPTER 1
1.1 Solar energy impinging on the outer layer of earth’s atmosphere (usually called “solar constant”) has been
measured as 1367 W/m2. What is the solar constant on Mars? (Distance earth to sun = 1.496 × 1011 m,
Mars to sun = 2.28 × 1011 m).
Solution
The total energy emitted from the sun Q = 4πR2 σT4 goes equally into all directions, so that
S sun
2
4 RS
qSC(R) = Q/4πR2 = σTsun R
, !2
q R 2 R 2 2 1.8×1011
SC mars S S RE
= = = = 0.4305
qSC earth RM RE RM 2.28×1011
qSC mars = 0.4305 × 1367 = 588 W/m2
2
, CHAPTER 1 3
1.2 Assuming 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 = qsolπR2E
Qem = σT4 A = 4πR2 σT4
E E E
Thus
!1/4
qsol 1/4 1367 W/m2
TE = =
4σ 4×5.670×10−8 W/m2K4
TE = 279 K = 6◦C
, 4 RADIATIVE HEAT TRANSFER
1.3 Assuming 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 earth RE = 6.37 × 106 m).
Solution
Performing an energy balance on earth:
Q˙ = Qemitted − Qabsorbed = σT4 A − qsol Aproj
E
= 4πR2 σT4 − qsolπR2 = πR2 4σT4 − qsol
E E E E E
6 2 W W
= π × 6.37×10 m 4×5.670×10−8 × 3004 K4 − 1367
m2K4 m2
Q˙ = 6.00 × 1016 W
Solution Manual all Chapters
CHAPTER 1
1.1 Solar energy impinging on the outer layer of earth’s atmosphere (usually called “solar constant”) has been
measured as 1367 W/m2. What is the solar constant on Mars? (Distance earth to sun = 1.496 × 1011 m,
Mars to sun = 2.28 × 1011 m).
Solution
The total energy emitted from the sun Q = 4πR2 σT4 goes equally into all directions, so that
S sun
2
4 RS
qSC(R) = Q/4πR2 = σTsun R
, !2
q R 2 R 2 2 1.8×1011
SC mars S S RE
= = = = 0.4305
qSC earth RM RE RM 2.28×1011
qSC mars = 0.4305 × 1367 = 588 W/m2
2
, CHAPTER 1 3
1.2 Assuming 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 = qsolπR2E
Qem = σT4 A = 4πR2 σT4
E E E
Thus
!1/4
qsol 1/4 1367 W/m2
TE = =
4σ 4×5.670×10−8 W/m2K4
TE = 279 K = 6◦C
, 4 RADIATIVE HEAT TRANSFER
1.3 Assuming 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 earth RE = 6.37 × 106 m).
Solution
Performing an energy balance on earth:
Q˙ = Qemitted − Qabsorbed = σT4 A − qsol Aproj
E
= 4πR2 σT4 − qsolπR2 = πR2 4σT4 − qsol
E E E E E
6 2 W W
= π × 6.37×10 m 4×5.670×10−8 × 3004 K4 − 1367
m2K4 m2
Q˙ = 6.00 × 1016 W
