Question

(II) What must be the pressure difference between the two ends of a 1.6-km section of pipe, 29 cm in diameter, if it is to transport oil (p= 950 kg/m³,n = 0.20 Pa. s) at a rate of \(650\;{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/s}}\)?

Short answer

The pressure difference between two ends of a pipe is \(1198.206\;{\rm{Pa}}\).

Step-by-step solution

Concepts

The pressure difference in a pipe flow is calculated using Hagen-Poiseuille’s Law.

The expression for the pressure change is given by,

\(\Delta P = \frac{{128\eta QL}}{{\pi {D^4}}}\)

Given data

The length of the pipe is\(L = 1.6\;{\rm{km}} \times \frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}} = 1600\;{\rm{m}}\).

The diameter of the pipe is\(D = 29\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 0.29\;{\rm{m}}\).

The density of the transport oil is\(\rho = 950\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).

The viscosity of the transport oil is heat added is\(\eta = 0.20\;{\rm{Pa}} \cdot {\rm{s}}\).

The volume rate of the transport oil is \(Q = 650\;{\rm{c}}{{\rm{m}}^3}{\rm{/s}} \times \frac{{{{10}^{ - 6}}\;{{\rm{m}}^3}{\rm{/s}}}}{{1\;{\rm{c}}{{\rm{m}}^3}{\rm{/s}}}} = 650 \times {10^{ - 6}}\;{{\rm{m}}^3}{\rm{/s}}\).

Calculation

The expression for the pressure difference between the two ends of the pipe is given by,

\(\Delta P = \frac{{128\eta QL}}{{\pi {D^4}}}\)

Substituting the values in the above equation,

\(\begin{array}{l}\Delta P = \frac{{128 \times 0.20\;{\rm{Pa}} \cdot {\rm{s}} \times 650 \times {{10}^{ - 6}}\;{{\rm{m}}^3}{\rm{/s}} \times 1600\;{\rm{m}}}}{{\pi \times {{\left( {0.29\;{\rm{m}}} \right)}^4}}}\\\Delta P = 1198.206\;{\rm{Pa}}\end{array}\)

Thus, the pressure difference in the pipe is \(1198.206\;{\rm{Pa}}\).