Two students, A and B, are discussing a heat transfer problem regarding what the surface temperature of a cartridge electrical heater would be when exposed to a water flow (CASE 1) and an air flow (CASE 2).
hw = 5000 W / (m2 * K); Tsurr = 20ºC
ha = 50 W / (m2 * K); Tsurr = 20ºC
Student A insists that the surface temperature of the cartridge will be larger in CASE 2 than in CASE 1, while student B states the opposite.
The cartridge is shaped as a solid cylinder of length L=200 mm and diameter D=20 mm, which is known to dissipate 2 kW irrespective of the case. Find the surface temperature of the cartridge in each case to find out if student A is right or wrong.
qtot = qconv + qrad
qtot = A * q''conv + A * q''rad
I assumed that since [itex]\alpha[/itex] wasn't mentioned, [itex]\epsilon[/itex] = [itex]\alpha[/itex] and I could use this form:
qtot = 2[itex]\pi[/itex]RL * h(Ts-Tsurr) + 2[itex]\pi[/itex]RL * [itex]\epsilon[/itex] [itex]\sigma[/itex] (Ts4 - Tsurr4)
The Attempt at a Solution
So really, given the above equations, this should be a "plug and chug" problem by saying that qtot = 2 kW and solving the equation for Ts. However, I cannot do that because the surface emissivity (or absorption) is not given. So I started to think whether or not the cartridge had any radiation at all. I looked through my notes and the textbook and could not find a way to quantify [itex]\epsilon[/itex] using the given information. That's when I came here. Am I wrong to assume that there is any radiation? Or am I doing something completely wrong?
Thanks for any help!