Question

Oil (ρ = 925 kg/m3) is flowing through a pipeline at a constant speed when it encounters a vertical bend in the pipe raising it 4.0 m. The cross-sectional area of the pipe does not change.
a) What is the speed of the fluid at Point B?
b) What is the difference in pressure (PB – PA) in the portions of the pipe before and after the rise?

Answer 1

(a)

According to equation of continuity,

$$ A_{1} v_{1}=A_{2} v_{2} $$

Velcoity at one end having cros sectional area \(\left(A_{1}\right)\) is \(v_{1}\)

Velcoity at another end having crossectional area \(\left(A_{2}\right)\) is \(v_{2}\) Given that both have the same crossectional areas \(\left(A_{1}\right)=\left(A_{2}\right)=A\) Hence, velocity at the both end is als same.

$$ v_{2}=v_{1} $$

(b) pressure difference \((\Delta p)=\rho g h\) Here, density of mercury \((\rho)=925 \mathrm{~kg} / \mathrm{m}^{3}\) Acceleration due to gravity \((g)=9.8 \mathrm{~m} / \mathrm{s}^{2}\) raise in height \((h)=4.0 \mathrm{~m}\)

$$ \begin{aligned} (\Delta p) &=\rho g h \\ &=\left(925 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(4.0 \mathrm{~m}) \\ &=36260 \mathrm{~Pa} \\ & \approx 3.63 \times 10^{4} \mathrm{~Pa} \end{aligned} $$