4. The surface temperature of a planet is T, and the measured temperature is Tm (i.e. the temper-ature measured at the top-of-atmosphere based on the upwards flux Ft there). Assume radiativeequilibrium.(a) Show that the infrared optical depth of the planet's surface 7 is related to the surfacetemperature T, and the measured temperature Tm byCalculate the infrared optical depth at the surfaces of(a) Earth, where T, = 300 K and Tm = 250 K.(b) Venus, where T, = 750 K and Tm = 230 K.(c) Mars, where T, = 240 K and T, = 220 K.

calculat the planet s surface=τ0=? Earth, Tq=300 KTm=250 Kτ*0=2TgTm4-1 =23002504-1τ*0=2.1472 for…

ature measured at the top-of-atmosphere based on the upwards flux Ft there). Assume radiative equilibrium. (a) Show that the infrared optical depth of the planet's surface r is related to the surface temperature T, and the measured temperature Tm by Calculate the infrared optical depth at the surfaces of () Earth, where T. 300 Kand T 250 K

ature measured at the top-of-atmosphere based on the upwards flux Ft there). Assume radiative equilibrium. (a) Show that the infrared optical depth of the planet's surface r is related to the surface temperature T, and the measured temperature Tm by Calculate the infrared optical depth at the surfaces of () Earth, where T. 300 Kand T 250 K

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

7) Given a window of 20 ft facing east at 40° north latitude, estimate the cooling load due to solar radiation at 4:00 pm for a day in June. Assume the window does not have any interior shading and the shading coefficient is 0.2 (in Btu/hr) (a) 105 (d) 276 (b) 430 (c) 314 (c) 225
a. What does it mean when we say that a surface 'sees itself' in the context of radiative heat transfer? b. If heat transfer by conduction through a medium occurs under steady- state, will the temperature at a particular location vary with time? Will the temperature vary with location in the medium?
i need correct explanation my best wishes ton
A right circular cylinder has a height of 1 m and diameter of 75 cm. Find the view factor for the cylindrical to the base.
Attached is the question and the formula to solve it. I attached the formula directly from the book, the Blackbody radiation Pressure formula that is. Thank you :)
Solve fast
What is the solution to this problem?
A long, horizontal, cylindrical steel reactor, 1 m in diameter, has a surface temperature of 300ºC. The emissivity of the steel is 0.6, and the heat transfer coefficient for natural convection is 5 W m−2 K−1 . Heat is lost by convection to the air at 15ºC, and also by radiation to the surroundings, which can be considered to be a black body at 15ºC. a) Calculate the total heat loss per metre length of the reactor, and the proportions lost by convection and radiation b) The reactor is then insulated with a thin layer of insulation material to reduce the total heat loss to one-tenth of its original value. This causes the surface temperature of the steel to rise to 400ºC. The thermal conductivity of the insulation is 0.01 W m−1 K−1 , and its surface emissivity is 0.2. Show that the resulting surface temperature of the insulation is about 89ºC, and calculate the thickness of insulation required, stating any assumptions made. Specifically need help with part b
3. A star has a surface temperature of 5,500K. Determine the radiative heat transfer received by planet Z in Petawatts with diameter of 2100 km, temperature of surface of 300K and mean absorptivity of 0.8. Radiative energy loss due to radiation dispersion on outer space is 99%.
You have stated that the convective area is πr^2. Please explain this. πr^2 is the area for a circle, not a sphere. It should be 4πr^2 as this is equal to the surface area of the sphere as you have done correctly in the radiative area. A conv= 4πr^2*3*293-292 = 0.377 A rad = 4πr^2*0.2*5.67x10^-8*(292^4-291^4) = 0.144 I have tried to multiply the convective area by 4πr^2 instead of πr^2 but it is giving me an incorrect total heat transfer rate of 0.377+0.144 = 0.521 the correct answer is supposed to be 0.236