An ideal fluid flows at 12 m/s in a horizontal pipe. If the pipe narrows to half its original radius, what is the flow speed in the narrower section?
A) 12 m/s
B) 24 m/s
C) 36 m/s
D) 48 m/s
2 Answers | Add Yours
The rate of flow of a liquid in terms of volume per second is equal to the flow (linear) speed of the liquid multiplied by cross sectional area of the pipe.
In the the given example let us say that:
The rate of flow of the liquid is V m^3 per second
The initial flow speed is s1 m per second, and
The initial area of pipe is a1 m^2.
Then: V = s1*a1
But s1 is given to be equal to 12 m/s. Substituting this value of s1 in above equation we get
V = 12a1
When the radius of the pipe is narrowed town to half of original, let us say:
The flow speed changes to s2, and
area reduces to a2.
However as V remains same, in this case:
V = s1*a2
Therefore: 12a1 = s2*a2
Therefore: s2 = 12*(a1/a2)
Area of cross section of a pipe is proportional to square of its diameter. Therefor if radius of is reduced to 1/2 of the original radius the cross section area gets reduced to (1/2)^2 = 1/4 th of the original area.
Therefore a1/a2 = 4
Substituting this value of a1/a2 in above equation for s2 we get:
S2 = 12*4 = 48 m/s
Therefore option D0 is correct.
Since the same volume of fluid flows through the sections at any given time,
the volume of flow per second in the normal cross sectional area of radius r is pr^2*12 m^3.
= the volume of flow per second through the (1/2) radius cross sectional area= p(r/2)^2* v m^3, where , r is the radius of the normal cross section and v is the speed of the liquid per second at the cross section with r/2 radius. Therefore,
pr^2*12 = p(r/2)^2*v
Solving for v, we get:
v = pr^2*12/[p(r/2)^2] = 4*12 = 48m/s is the speed at the cross section with radius r/2.
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