What Is the Net Thermal Radiation Transfer Rate for a Stretched Cylinder?

[VitaX]

Homework Statement

A solid cylinder of radius r1 = 2.5 cm, length h1 = 5.1 cm, emissivity 0.90, and temperature 28°C is suspended in an environment of temperature 46°C. (a) What is the cylinder's net thermal radiation transfer rate P1? (b) If the cylinder is stretched until its radius is r2 = 0.51 cm, its net thermal radiation transfer rate becomes P2. What is the ratio P2/P1?

Homework Equations

P = σεA(Tenv^4 - T^4)

The Attempt at a Solution

My answer to part A is correct. I got P = (5.67E-8)(.9)(.0119)(319^4 - 301^4) = 1.304 W

But my answer to part B is wrong apparently. I don't know what I did wrong. But the formula for Area of Solid Cylinder = 2*pi*r^2 + 2*pi*r*h

I used r1 = .025 m and h1 = .051 m for Area in part a. But in part B the only thing I changed was r2 = .0051 m to find the new area. Then got P2 and divided it by P1 to get the ratio. But it says part B is wrong. What exactly did I do wrong with the area in part b? Am I to assume h changes as well? My exact formula for part B Area is this: A = 2*pi*(.0051)^2 + 2*pi*(.0051)(.051) = .0018 m^2

 

[gneill]

But my answer to part B is wrong apparently. I don't know what I did wrong. But the formula for Area of Solid Cylinder = 2*pi*r^2 + 2*pi*r*h

I used r1 = .025 m and h1 = .051 m for Area in part a. But in part B the only thing I changed was r2 = .0051 m to find the new area.

If the cylinder was stretched, presumably it stretched in some direction...
Also presumably, the overall mass of the cylinder didn't change (it's still got the same amount of "stuff") so what does that say about the volume of the stretched cylinder?

 

[VitaX]

If the cylinder was stretched, presumably it stretched in some direction...
Also presumably, the overall mass of the cylinder didn't change (it's still got the same amount of "stuff") so what does that say about the volume of the stretched cylinder?

So you are saying use V = pi*r^2*h to find the volume in part A. Then use that same volume and radius 2 in part B to find height 2. Then find the new area of part B using the new height and radius?

 

[gneill]

So you are saying use V = pi*r^2*h to find the volume in part A. Then use that same volume and radius 2 in part B to find height 2. Then find the new area of part B using the new height and radius?

Sounds like a plan!

 

[rwisz]

In order to find the net thermal radiation transfer rate in part B, you need to use the new values for the radius and length of the cylinder (r2 = 0.051 cm and h2 = 5.1 cm). This is because the cylinder has been stretched, so its dimensions have changed. Using the new values, the correct formula for the area would be A = 2*pi*(0.0051)^2 + 2*pi*(0.0051)(0.051) = 0.00018 m^2. Using this value for the area, you can then solve for P2 and find the ratio P2/P1.