The speedA horizontal pipe narrows from a radius of 0.250 m to 0.100 m. If the speed of the water in the pipe is 1.00 m/s in the larger-radius pipe, what is the speed in the small pipe?

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We can calculate the average speed with the following relationship:

d^2=4*V/pi*v

d=diameter of the pipe section

V=volume of water

pi=3.14

v= speed of water

Now, we'll write the relation for the section of the pipe which has the radius r1=0.250 m.

d1^2=(2*r1)^2=4*V/pi*v1

4*(0.250)^2=4*V/3.14*1

0.250*3.14/4=V

V=0.196 m^3

We have to know that the same volume of water will crosses the larger section of the pipe, also the smaller narrowed section of the pipe.

So, 4*r2^2*3.14*v2=0.196

4*0.100^2*3.14*v2=0.196

v2=0.196/0.040*3.14

v2=1.56 m/s

Being given the fact that the same volume of water flows through the pipeline, both in the narrow area, also in the wide area, water velocity in the narrowed area will increase.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

The pipe is horizontal and so the  the flow is horizontal. Therefore the flow of the  volume of water per second in the  larger radius part of the pipe = cross sectional area *speed of flow=p*(.25)^2*velocity of flow/sec = P*(0.25)^2(1.0) m^3/sec...................(1)

Let the speed in the narrow part of the pipe be v. Then the volume of water flow per second at this end = cross sectional area*v = p(0.1)^2*V.....................................(2)

But the volume of the flow at (1) and (2) must be equal. So,

p(0.1)^2*v = p(0.25)^2*(0.1). Therefore,

v =  0.25^2 * 0.1/ (0.1)^2 = 0.625m/s

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The pipe is horizontal and so the  the flow is horizontal. Therefore the flow of the  volume of water per second in the pipe larger radius part of the pipe = cross sectional area *speed of flow=p*(.25)^2*velocity of flow/sec = P*(0.25)^2(1.0)m^3/sec...................(1)

Let the speed in the narrow part of the pipe be v. Then the volume of water flow per second = cross sectional area*v = p(0.1)^2*V.....................................(2)

But the volume of the flow at (1) and (2) must be equal. So,

p(0.1)^2*v = p(0.25)^2*(0.1). Thherefore,

v =  0.25^2 * 0.1/ (0.1)^2 = 0.625m/s

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