Question

What must be the pressure difference between the two ends of a 2.6 km section of pipe, 36 cm in diameter, if it is to transport oil (rho = 950 kg/m^3, η = 0.20 Pa*s) at a rate of 900 cm^3/s ? Express your answer to two significant figures and include the appropriate units. Δ P = 1.3 x 10^3Pa

Answer 1

Answer: $1.13(10)^{3} Pa$

Explanation:

This problem can be solved by the following equation:

$\Delta P=\frac{8 \eta L Q}{\pi r^{4}}$

Where:

$\Delta P$ is the pressure difference between the two ends of the pipe

$\eta=0.20 Pa.s$ is the viscosity of oil

$L=2.6 km=2600 m$ is the length of the pipe

$Q=900 \frac{cm^{3}}{s} \frac{1 m^{3}}{(100 cm)^{3}}=0.0009 \frac{m^{3}}{s}$ is the Rate of flow of the fluid

$d=36 cm=0.36 m$ is the diameter of the pipe

$r=\frac{d}{2}=0.18 m$ is the radius of the pipe

Soving for $\Delta P$:

$\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}$

Finally:

$\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa$