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Chapter 10: Q59P (page 260)

(I) Engine oil (assume SAE 10, Table 10–3) passes through a fine \({\bf{1}}{\bf{.80}}\;{\bf{mm}}\)-diameter tube that is \({\bf{10}}{\bf{.2}}\;{\bf{cm}}\)long. What pressure difference is needed to maintain a flow rate of \({\bf{6}}{\bf{.2}}\;{\bf{mL/min}}\)?

Short Answer

Expert verified

The pressure difference required to maintain the flow rate is\(8.2 \times {10^3}\;{\rm{Pa}}\).

Step by step solution

01

Understanding Poiseuille’s Equation

The liquid's flow rate in a tube depends upon the radius of tube, pressure difference and length of the tube. The relation between the mentioned quantities is given by Poiseuille's equation.

02

Given Data

The diameter of tube is\(D = 1.80\;{\rm{mm}}\).

The length of tube is\(L = 10.2\;{\rm{cm}}\).

The flow rate of oil is \(Q = 6.2\;{\rm{mL/min}}\).

03

Calculate the pressure difference

The radius of tube can be given as,

\(R = \frac{D}{2}\)

Substitute the values in the above equation,

\(\begin{array}{c}R = \frac{{1.80\;{\rm{mm}}}}{2}\\ = 0.90\;{\rm{mm}}\end{array}\)

The expression for the Poiseuille’s equation,

\(Q = \frac{{\pi {R^4}\Delta P}}{{8\eta L}}\)

Substitute the values in the above equation,

\(\begin{array}{c}6.2\;{\rm{mL/min}} = \frac{{\left( {3.14} \right){{\left( {0.90\;{\rm{mm}}} \right)}^4}\Delta P}}{{8\left( {200 \times {{10}^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}}} \right)\left( {10.2\;{\rm{cm}}} \right)}}\\\Delta P = \frac{{\left( {6.2\;{\rm{mL/min}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\rm{L}}}}{{1\;{\rm{mL}}}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{{\rm{m}}^3}}}{{1\;{\rm{L}}}}} \right)\left( {\frac{{1\;\min }}{{60\;{\rm{s}}}}} \right)8\left( {200 \times {{10}^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}}} \right)\left( {10.2\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{m}}}}{{{\rm{100}}\;{\rm{cm}}}}} \right)}}{{\left( {3.14} \right){{\left( {0.90\;{\rm{mm}}} \right)}^4}{{\left( {\frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)}^4}}}\\\Delta P = 8.2 \times {10^3}\;{\rm{Pa}}\end{array}\)

Therefore, the pressure difference required to maintain the flow rate is \(8.2 \times {10^3}\;{\rm{Pa}}\).

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