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## 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