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The carrying capacity of a drain pipe is directly proportional to the area of its cross section.If cylindrical pipe drain can carry 36L/sec,determine the percentage increase in the diameter of the drain pipe necessary to enable it to carry 60 litres per second .

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Explanation:

x and y are directly proportional means that

y = kx

y grows with the same factor k applied to x.

the area of a circle (the cross section of a cylindrical pipe) is

pi × r²

r being the radius (which is half of the diameter).

so, the area in terms of the diameter is

pi × (d/2)² = pi × d²/4

so, we know

36 correlates to pi × (d small)²/4

60 correlates to pi × (d large)²/4

therefore (directly proportional),

d small² = 36×4/pi

d small = 12/sqrt(pi)

d large² = 60×4/pi

d large = sqrt(4×15×4/pi) = 4×sqrt(15/pi)

the difference between large and small diameter (increase from small to large) is then

4×sqrt(15/pi) - 12/sqrt(pi)

d small = 100%

1% = 100%/100 = 12/sqrt(pi) / 100 = 12/(100sqrt(pi))

the % of the diameter increase is then

increase/1% =

(4×sqrt(15/pi) - 12/sqrt(pi)) / 12/(100sqrt(pi)) =

= (400sqrt(pi)×sqrt(15/pi) - 1200sqrt(pi)/sqrt(pi)) / 12 =

= (400×sqrt(15) - 1200) / 12 = 100×sqrt(15)/3 - 100 =

= 100×(sqrt(15)/3 - 1) = 29.09944487...% ≈ 29.1%

the diameter has to increase by about 29.1%, so that the pipe can carry 60 liters per second.

please consider : only the area of the cross section and the carrying capacity are directly proportional.

the diameter of the cross section and the area of the cross section are NOT directly proportional.

they have the pi×(d/2)² relationship.

and so, while the capacity and the cycle area both increase by the same factor, the impact on the diameter (or radius) is different.

so, for capacity and area we have

60 = k×36

k = 60/36 = 5/3 = 1.66666666...

and therefore both increase by 66.66666...%.

but because of the square relation between diameter and cross section area, the diameter only increases by 29.1%.

User Kuldeep Saxena
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