pipe lengths: external inner vs. outer

Qaz

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Jul 24, 2009
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for a 3.8" OD 90 degree pipe bend, how do you calculate the the amount the outside/outer edge of a pipe exceeds the inside/inner outer edge? My answers come out to be 50% of the "correct" anwer when I tried--

(assuming pipe's cross-sectional (OD) = 3.8", 1/4 x pi x D:

.25 x 3.1417 x 3.8" = 2.985"

(correct answer = 5.969). I would have thought a 90 degree bend required dividing by 4; a 180 degree bend, dividing by 2, etc.
 
Do you mean what is the distance from the outside of the pipe to the inside at a 90 degree angle?.

We can use x=3.8sin(45)=5.37\displaystyle x=\frac{3.8}{sin(45)}=5.37

Unless I am misinterpreting. See the diagram.

Or are you meaning area?. I am not sure.

If you mean the circumference, that is given by πD=π(3.8)=11.938\displaystyle {\pi}D={\pi}(3.8)=11.938

Then, half the circumference of the pipe would be π(3.8)2=5.969\displaystyle \frac{{\pi}(3.8)}{2}=5.969
 

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Suppose the bend radius is "r". Then

the inner (quarter ) circumference is

Ci=π2r\displaystyle C_i \, = \, \frac{\pi}{2}\cdot r

the outer (quarter ) circumference is

Co=π2(r+3.8)\displaystyle C_o \, = \, \frac{\pi}{2}\cdot (r + 3.8)

CoCi=π2(r+3.8)π2(r)=π23.8=5.969\displaystyle {C_o} \, - {C_i} \, = \, \frac{\pi}{2}\cdot (r + 3.8) \, - \, \frac{\pi}{2}\cdot (r) \, = \frac{\pi}{2}\cdot 3.8 \, = \, 5.969
 
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