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Question: A venturi meter is connected to two points in the mains where its radii are \(20\,cm\) and \(10\,cm\...

A venturi meter is connected to two points in the mains where its radii are 20cm20\,cm and 10cm10\,cm . And the levels of the water column in the tubes differ by 10cm10\,cm. How much water flows through the pipe per minute?

Explanation

Solution

In fluid mechanics, Venturimeter is a mechanical device which measures how much of fluid is flowing through a given time or we can say the rate of flow of a fluid through a pipe is measured by venturi meter and basically it measures in units of litremin1litre{\min ^{ - 1}}.

Complete step by step answer:
Let us suppose the area of first radii section which is 20cm20\,cm is denoted by A1=π(20)2cm2{A_1} = \pi {(20)^2}\,c{m^2} and let the area of second radii section is denoted by A2=π(10)2cm2{A_2} = \pi {(10)^2}\,c{m^2}
So now we have,
A1=π(20)2cm2{A_1} = \pi {(20)^2}\,c{m^2}
A2=π(10)2cm2\Rightarrow {A_2} = \pi {(10)^2}\,c{m^2}
Acceleration due to gravity is g=980cmsec2g = 980\,cm{\sec ^{ - 2}}
And it’s given that the height up to which the water rises is h=10cmh = 10cm
Volume flowing per second through the venturi meter is given by
V=A1A22ghA12A22V = {A_1}{A_2}\sqrt {\dfrac{{2gh}}{{{A_1}^2 - {A_2}^2}}}

In order to convert this flow of water in second to minutes and the cubic centimetres to litre,
Just multiply by 6060 we get,
V=A1A22ghA12A22×60litremin1V = {A_1}{A_2}\sqrt {\dfrac{{2gh}}{{{A_1}^2 - {A_2}^2}}} \times 60\,litre\,{\min ^{ - 1}}
Putting the values of all known parameters we will get:
V=(227)2(20)2(10)22×980×10300(3.14)×60litremin1V = {(\dfrac{{22}}{7})^2}{(20)^2}{(10)^2}\sqrt {\dfrac{{2 \times 980 \times 10}}{{300(3.14)}}} \times 60litre{\min ^{ - 1}}
V=(227)2240000019600942litremin1\Rightarrow V = {(\dfrac{{22}}{7})^2}2400000\sqrt {\dfrac{{19600}}{{942}}} litre{\min ^{ - 1}}
V=2726.58litremin1\therefore V = 2726.58\,litre\,{\min ^{ - 1}}

Hence, the rate of flow of water measured by venturi meter in litres per minute is V=2726.58litremin1V = 2726.58\,litre{\min ^{ - 1}}.

Note: It should be remembered that, the basic units of conversions like: acceleration due to gravity is g=9.8msec2g = 9.8m{\sec ^{ - 2}} into g=980cmsec2g = 980cm{\sec ^{ - 2}} and litre is the most standard unit of volume which is used while measure large amount of fluids and its relation with cubic centimetre is given as 1litre=1000cm31litre = 1000c{m^3}.