
Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to \[\dfrac{a}{2}\] and the pressure increases to 2P, then what happens to the rate of flow?
A. 4Q
B. Q
C. \[\dfrac{Q}{4}\]
D. \[\dfrac{Q}{8}\]
Answer
163.2k+ views
Hint: Poiseuille's theorem states that the velocity of a liquid flowing through a capillary is directly proportional to the pressure of the liquid and the fourth power of the radius of the capillary tube and is inversely proportional to the viscosity of the liquid and the length of the capillary tube.
Formula Used:
Poiseuille’s formula is,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Where, \[\eta \] is dynamic viscosity, P is pressure, l is length of the pipe and r is radius of the pipe.
Complete step by step solution:
Suppose water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow is Q. If the radius is reduced to \[\dfrac{a}{2}\] and the pressure increased to 2P, then we need to find the rate of flow.
Here,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Q is proportional to P and the fourth power of r. In the given problem it is specifically mentioned that pressure as well as the radius of the capillary tube.
We can write,
\[\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{P_2}}}{{{P_1}}}{\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)^4}\]
Given that, \[{P_1} = P\] , \[{P_2} = 2P\], \[{r_2} = \dfrac{r}{2}\] and \[{r_1} = r\]
Substitute the value in above equation we obtain,
\[\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{2P}}{P}{\left( {\dfrac{{\dfrac{r}{2}}}{r}} \right)^4} \\ \]
\[\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{2}{1}{\left( {\dfrac{1}{2}} \right)^4} \\ \]
\[\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{2}{1} \times \dfrac{1}{{16}} \\ \]
\[\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{1}{8} \\ \]
\[\therefore {Q_2} = \dfrac{Q}{8}\]
Therefore, the flow rate of water is \[\dfrac{Q}{8}\].
Hence, Option D is the correct answer.
Note:Poiseuille number is a non-dimensional number which characterizes steady, fully-developed, laminar flow of a constant-property fluid through a duct of arbitrary, but constant, cross section.
Formula Used:
Poiseuille’s formula is,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Where, \[\eta \] is dynamic viscosity, P is pressure, l is length of the pipe and r is radius of the pipe.
Complete step by step solution:
Suppose water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow is Q. If the radius is reduced to \[\dfrac{a}{2}\] and the pressure increased to 2P, then we need to find the rate of flow.
Here,
\[Q = \dfrac{{\pi {{\Pr }^4}}}{{8\eta l}}\]
Q is proportional to P and the fourth power of r. In the given problem it is specifically mentioned that pressure as well as the radius of the capillary tube.
We can write,
\[\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{P_2}}}{{{P_1}}}{\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)^4}\]
Given that, \[{P_1} = P\] , \[{P_2} = 2P\], \[{r_2} = \dfrac{r}{2}\] and \[{r_1} = r\]
Substitute the value in above equation we obtain,
\[\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{2P}}{P}{\left( {\dfrac{{\dfrac{r}{2}}}{r}} \right)^4} \\ \]
\[\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{2}{1}{\left( {\dfrac{1}{2}} \right)^4} \\ \]
\[\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{2}{1} \times \dfrac{1}{{16}} \\ \]
\[\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{1}{8} \\ \]
\[\therefore {Q_2} = \dfrac{Q}{8}\]
Therefore, the flow rate of water is \[\dfrac{Q}{8}\].
Hence, Option D is the correct answer.
Note:Poiseuille number is a non-dimensional number which characterizes steady, fully-developed, laminar flow of a constant-property fluid through a duct of arbitrary, but constant, cross section.
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