p64.mw

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Find the steady temperature of a white ball (of =0.2 and =0.8) and a black ball (of ==0.9), at sea level, and above the atmosphere, considered isothermal and exposed to solar radiation. In particular, find:
a) Temperature on ground, surrounded by air at 15 ºC with a natural convection coefficient of h=10 W/(m2·K) and solar irradiance of 1000 W/m2.
b) Temperature in outer space, far from the Earth.
c) Steady temperature at 300 km altitude over the subsolar point.
d) Steady temperature at the opposite side of this LEO.

 > su:="":dat:=[Tair=288.15*K_,h=10*W_/(m_^2*K_),E=1360*W_/m_^2,Es=1000*W_/m_^2];dat:=op(dat),Const,SI2,SI1:datW:=alpha=0.2,epsilon=0.8;datB:=alpha=0.9,epsilon=0.9;   a) Temperature on ground, surrounded by air at 15 ºC with a natural convection coefficient of h=10 W/(m2·K) and solar irradiance of 1000 W/m2.

On ground, the energy balance is:

 > eqEB:=m*c*dT/dt=Qs-Qair;Qs:=alpha*Pi*R^2*Es;Qair:=4*Pi*R^2*h*(T-Tair);eqEB:=0=Qs-Qair;Tground:=solve(eqEB,T);T_B_:=evalf(expand(subs(datB,dat,%)));'T_B_'=TKC(%);T_W_:=evalf(expand(subs(datW,dat,Tground)));'T_W_'=TKC(%);         (1)

i.e. the black ball attains 37.5 ºC, and the white ball 20 ºC. Notice that we have assumed no other heat transfer than convection to ambient air (radiation to the environment is usually included in the h-value).

b) Temperature in outer space, far from the Earth.

Outside the Earth's atmosphere there is no convection. If the equilibrium is established just between solar radiation and backgroun radiation (at T=2.7 K, that we approximate as 0):

 > eqEB:=m*c*dT/dt=Qs-Qinf;Qs:=alpha*Pi*R^2*E;Qinf:=epsilon*4*Pi*R^2*sigma*(T^4-Tinf^4);eqEB:=0=Qs-subs(Tinf=0,Qinf);Tspace0:=sqrt(sqrt((alpha*E/(4*epsilon*sigma))));T_B_:=evalf(subs(datB,dat,SI0,Tspace0))*K_;'T_B_'=TKC(%);T_W_:=evalf(subs(datW,dat,SI0,Tspace0))*K_;'T_W_'=TKC(%);         (2)

i.e. far from the Earth, the black ball would reach 5 ºC and the white ball -76 ºC. But, if the Earth is not so far, being so large (RE=6400 km), it may provide additional heat inputs by planet IR emission, and by solar reflection on the planet (albedo).

c) Steady temperature at 300 km altitude over the subsolar point.

At the subsolar point:

 > eqEB:=m*c*dT/dt=Qs+Qp+Qa-Qinf;Qp:=epsilon*4*Pi*R^2*Fsp*sigma*epsilon[p]*Tp^4;Qa:=alpha*4*Pi*R^2*Fsp*rho*E;eqEB:=0=Qs+Qp+Qa-Qinf;    (3)

 > eqF12:=Fsp=(1-sqrt(1-1/h^2))/2;eqh:=h=(H+R)/R;eqh_:=evalf(subs(H=300,R=6400,%));eqF12_:=subs(eqh_,eqF12);    (4)

and, with Earth's properties, Tp=288 K, emissivity ep=0.61, and albedo rho=0.3:

 > T_B_:=fsolve(subs(Tp=Tair,Tinf=0,epsilon[p]=0.61,rho=0.3,eqF12_,datB,dat,SI0,expand(eqEB/R^2)),T=100..1000)*K_;'T_B_'=TKC(%);T_W_:=fsolve(subs(Tp=Tair,Tinf=0,epsilon[p]=0.61,rho=0.3,eqF12_,datW,dat,SI0,expand(eqEB/R^2)),T=100..1000)*K_;'T_W_'=TKC(%);    (5)

i.e. with IR emission and maximum albedo (Sun-ball-Earth aligned) the black ball reaches 43 ºC, and the white ball -28 ºC.

d) Steady temperature at the opposite side of this LEO.

Now there is no solar radiation; only the planet IR input.

 > eqEB:=0=Qp-Qinf;T_B_:=fsolve(subs(Tp=Tair,Tinf=0,epsilon[p]=0.61,rho=0.3,eqF12_,datB,dat,SI0,expand(eqEB/R^2)),T=100..1000)*K_;'T_B_'=TKC(%);T_W_:=fsolve(subs(Tp=Tair,Tinf=0,epsilon[p]=0.61,rho=0.3,eqF12_,datW,dat,SI0,expand(eqEB/R^2)),T=100..1000)*K_;'T_W_'=TKC(%);     (6)

i.e. with only the planet emission, the two balls attain the same temperature, -77 ºC.

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