What diameter must a \({\bf{15}}{\bf{.5}}\;{\bf{m}}\)-long air duct have if the ventilation and heating system is to replenish the air in a \({\bf{8}}{\bf{.0}}\;{\bf{m \times 14}}{\bf{.0}}\;{\bf{m \times 4}}{\bf{.0}}\;{\bf{m}}\) room every \({\bf{15}}{\bf{.0}}\;{\bf{min}}\)? Assume the pump can exert a gauge pressure of \({\bf{0}}{\bf{.710 \times 1}}{{\bf{0}}^{\bf{3}}}\;{\bf{atm}}\).

Poiseuille’s equation gives the relationship between the flow rate, radius of pipe, pressure difference and length of the pipe.

The equation can be expressed as,

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

Here, \(Q\)is the flow rate, \(R\)is the radius, \(\Delta P\)is the pressure difference, \(\eta \)is the viscosity and \(l\)is the length.

The length of air duct is\(L = 15.5\;{\rm{m}}\).

The volume of room is\(V = 8.0\;{\rm{m}} \times 14.0\;{\rm{m}} \times 4.0\;{\rm{m}}\).

The time taken to ventilate is\(t = 15.0\;\min \).

The gauge pressure of pump is \(\Delta P = 0.710 \times {10^3}\;{\rm{atm}}\).

Substitute the known values in the equation,

\(\begin{array}{c}Q = \frac{{\left( {8.0\;{\rm{m}} \times 14.0\;{\rm{m}} \times 4.0\;{\rm{m}}} \right)}}{{\left( {15.0\;\min } \right)\left( {\frac{{60\;{\rm{s}}}}{{1\;\min }}} \right)}}\\ = 0.4978\;{{\rm{m}}^{\rm{3}}}{\rm{/s}}\end{array}\)

According to Poiseuille’s equation,

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

Rearrange the above equation for radius,

\(\begin{array}{c}{R^4} = \frac{{8\eta lQ}}{{\pi \Delta P}}\\R = {\left( {\frac{{8\eta lQ}}{{\pi \Delta P}}} \right)^{1/4}}\end{array}\)

The diameter of duct can be given as,

\(D = 2R\)

Substitute the known expression in the equation,

\(D = 2{\left( {\frac{{8\eta lQ}}{{\pi \Delta P}}} \right)^{1/4}}\)

Substitute the known values in the equation,

\(\begin{array}{c}D = 2{\left( {\frac{{8\left( {0.018 \times {{10}^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}}} \right)\left( {15.5\;{\rm{m}}} \right)\left( {0.4978\;{{\rm{m}}^{\rm{3}}}{\rm{/s}}} \right)}}{{\left( {3.14} \right)\left( {0.710 \times {{10}^{ - 3}}\;{\rm{atm}}} \right)\left( {\frac{{1.013 \times {{10}^5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}}} \right)}}} \right)^{1/4}}\\ = 2\left( {0.047\;{\rm{m}}} \right)\\ = 0.094\;{\rm{m}}\end{array}\)

Therefore, the diameter of the duct must be \(0.094\;{\rm{m}}\).