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

Hint: 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.