Water is conveyed through a uniform tube of 8 | vedclass

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(A) Given: Diameter $d = 8 	ext{ cm} = 0.08 	ext{ m}$,so radius $r = 0.04 	ext{ m} = 4 	imes 10^{-2} 	ext{ m}$.Length $l = 3140 	ext{ m}$.Flow r

Vedclass · Accepted Answer

$6.25 	imes 10^3 	ext{ N/m}^2$

Vedclass · Answer

(A) Given: Diameter $d = 8 	ext{ cm} = 0.08 	ext{ m}$,so radius $r = 0.04 	ext{ m} = 4 	imes 10^{-2} 	ext{ m}$.Length $l = 3140 	ext{ m}$.Flow rate $Q = 2 	imes 10^{-3} 	ext{ m}^3/	ext{s}$.Viscosity $\eta = 10^{-3} 	ext{ SI units}$.According to Poiseuille's equation for laminar flow in a pipe:$Q = \frac{\pi P r^4}{8 \eta l}$Rearranging for pressure difference $P$:$P = \frac{8 \eta l Q}{\pi r^4}$Substituting the values:$P = \frac{8 	imes 10^{-3} 	imes 3140 	imes 2 	imes 10^{-3}}{3.14 	imes (4 	imes 10^{-2})^4}$$P = \frac{16 	imes 3140 	imes 10^{-6}}{3.14 	imes 256 	imes 10^{-8}}$$P = \frac{16 	imes 1000 	imes 10^{-6}}{256 	imes 10^{-8}} = \frac{16000 	imes 10^2}{256} = \frac{1600000}{256} = 6250 	ext{ N/m}^2$$P = 6.25 	imes 10^3 	ext{ N/m}^2$.

Water is conveyed through a uniform tube of $8 \text{ cm}$ in diameter and $3140 \text{ m}$ in length at the rate of $2 \times 10^{-3} \text{ m}^3/\text{s}$. The pressure required to maintain the flow is (Viscosity of water $= 10^{-3} \text{ SI units}$):

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