Let subsript "1" denote the original inside pipe dimensions a "2" denote the final dimensions
A2/A1 = (D2/D1)^2 = 0.8
D2/D1 = 0.894
D2 = D1 - 2 (mm)
0.894 D1 = D1 - 2
0.106 D1 = 2
D1 = 18.87 mm
10 mm is not correct
When mineral deposits formed a coating 1 mm thick on the inside of a pipe, the area through which fluid can flow was reduced by 20%. Find the original inside diameter of the pipe.
I got 10 mm. Is this correct?
2 answers
Answer I got was 18.94
A2/A1 = (D2/D1)^2 = 0.8
(R1 -1)^2 /R^2 = .8
5(R1 -1)^2) = 4R^2
R^2 - 10R + 5 = 0
Solving for positive values of R
The radius R = 9.47
and The diameter D = 18.94
A2/A1 = (D2/D1)^2 = 0.8
(R1 -1)^2 /R^2 = .8
5(R1 -1)^2) = 4R^2
R^2 - 10R + 5 = 0
Solving for positive values of R
The radius R = 9.47
and The diameter D = 18.94
1 answer