Question

Glycerine flows steadily through a horizontal tube of length $1.5{\text{ m}}$ and radius $1.0{\text{ cm}}$. If the amount of glycerine collected per second at one end is $4.0 \times {10^{ - 3}}\;{\text{kg}}{{\text{s}}^{{\text{ - 1}}}}$, what is the pressure difference between the two ends of the tube ? (Density of glycerine $= 1.3 \times {10^3}\;{\text{kg}}{{\text{m}}^{{\text{ - 3}}}}$ and viscosity of glycerine $= 0.83\;{\text{Pa}}\;{\text{S}}$). [You may also like to check if the assumption of laminar flow in the tube is correct].

Answer 1

First we convert all the units in SI form, i.e. convert all centimetres to metres. Since, the radius is given, the diameter can be obtained. From Poiseuille's formula $V = \dfrac{{\pi P \times {r^4}}}{{8\eta l}}$, and then we calculate the value of pressure. To obtain Reynolds number we use the formula $R = \dfrac{{4\rho V}}{{\pi d\eta }}$

Complete step by step answer:
Given, length of the horizontal tube, $l = 1.5\;{\text{m}}$
Radius of the tube, $r = 1\;{\text{cm}}$
Glycerine is flowing at a rate of $4.0 \times {10^{ - 3}}\;{\text{kg}}{{\text{s}}^{{\text{ - 1}}}}$
$M = 4.0 \times {10^{ - 3}}\;{\text{kg/s}}$
Density of glycerine, $\rho = 1.3 \times {10^3}\;{\text{kg/}}{{\text{m}}^3}$
Viscosity of glycerine, $\eta = 0.83\;{\text{Pa}}\;{\text{S}}$
Convert all the units in SI form, i.e. convert all centimetres to metres.
Hence,
Radius of the tube, $r = 1\;{\text{cm}} = {\text{0}}{\text{.01}}\;{\text{m}}$
Diameter of the tube, $d = 2r = 0.02\;{\text{m}}$
Now, volume of the glycerine per second is obtained by dividing $M$
by density of glycerine.
i.e.

$$V = \dfrac{{4 \times {{10}^{ - 3}}}}{{1.3 \times {{10}^3}}} \\\ = 3.08 \times {10^{ - 6}}\;{{\text{m}}^{\text{3}}}{\text{/s}} \\\$$

For the relation of the rate of flow, apply Poiseuille's formula.
$V = \dfrac{{\pi P \times {r^4}}}{{8\eta l}}$; $P$ is the difference of pressure of the two ends of the tube.
Rearrange the above equation to obtain the value of $P$.

$$P = \dfrac{{V8\eta l}} {{\pi {r^4}}} \\\ = \dfrac{{3.08 \times {{10}^{ - 6}} \times 8 \times 0.83 \times 1.5}} {{\pi \times {{\left( {0.01} \right)}^4}}} \\\ = 9.8 \times {10^2}\;{\text{Pa}} \\\$$

Reynolds number is obtained by the relation,

$$R = \dfrac{{4\rho V}}{{\pi d\eta }} \\\ = \dfrac{{4 \times 1.3 \times {{10}^3} \times 3.08 \times {{10}^{ - 6}}}}{{\pi \times 0.02 \times 0.83}} \\\ = 0.3 \\\$$

Since, the Reynolds number is $0.3$.
Therefore, the flow is laminar.

So, the correct answer is “Option A”.

Additional Information:
The Reynolds number is the ratio of inertial forces and viscous forces inside a fluid that, owing to different fluid velocities, is subject to similar spatial movement. As a boundary plate, such as the flowing fluid in the interior of a container, an area in which these forces change behaviour is known.
It is also observed that if Reynolds number (based on pipe diameter) is less than $2100$ , the flow in a pipe is laminar and if it is greater than $4000$ it is turbulent.

Note:
The number of Reynolds is a dimensionless number that is used in various fluid contexts to distinguish between the turbulent as well as laminar flow of a stream and to anticipate how it will behave.