Question

An oil of relative density $0.9$ and viscosity $0.12kg/ms$ flows through a $2.5cm$ diameter pipe with a pressure drop of $38.4kN/{m^2}$ in a length of $30cm$. Determine the power required to maintain the flow.
A) $2.2W$
B) $3.84W$
C) $5.6W$
D) $9.3W$

Answer 1

Hint: Relative density is defined as the ratio of the densities of any two materials taken. It is calculated as the ratio of the density of the substance given and a reference substance. The reference substance taken is usually water. It is used to determine the density of unknown substances.

Complete step by step solution:
Step I: According to Poiseuille’s Law, flow of liquid is related to a number of factors like pressure gradient $\Delta P$, viscosity of fluid $(\eta )$, length of the tube $L$ through which it flows and the radius. It is written by the equation
$Q = \dfrac{{\Delta P \times \pi {r^4}}}{{8\eta L}}$---(i)

Step II: Given $\Delta P = 38.4 \times {10^3}N/{m^2}$
If diameter, $D = 2.5 \times {10^{ - 2}}m$
Then radius, $r = 1.25 \times {10^{ - 2}}$
$L = 30m$
$\eta = 0.12N/{m^2}$

Step III: Substitute all the values in the given formula of equation (i)
$Q = \dfrac{{38.4 \times {{10}^3} \times 3.14 \times {{(1.25 \times {{10}^{ - 2}})}^4}}}{{8 \times 0.12 \times 30}}$
$Q = \dfrac{{38.4 \times {{10}^3} \times 3.14 \times 2.44 \times {{10}^8}}}{{28.8}}$
$Q = 1 \times {10^{ - 4}}m/s$

Step IV: Power gained by a fluid from a pump is given by the formula:
$P = \Delta P \times Q$
Where $\Delta P$is the change in pressure and
$Q$ is the Rate of flow of volume of oil in the pipe
Substituting the value in the above formula and solving for power,
$ = 38.4 \times {10^3} \times 1 \times {10^{ - 4}}$
$P = 3.84W$

Step V:
Hence the minimum power required to maintain the flow of oil in the pipe will be $3.84W.$

Therefore, Option (B) is the right answer.

Note: It is to be noted that the particles of liquid have spaces between them and their position is not fixed. Therefore due to the movement of the particles, the liquids do not have a fixed shape. They take the shape of the container in which they are placed. But the change in shape does not affect the volume of the liquids. The volume remains the same.