Civil Engineering - GATE Exam Questions - Discussion

Discussion Forum : GATE Exam Questions - Section 1 (Q.No. 5)
In a steady radial flow into an intake, the velocity is found to vary as (1/r2), where r is the radial distance. The acceleration of the flow is proportional to
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27 comments Page 1 of 3.

Nanu said:   6 years ago
dr/dt = - constant / r*r.

To find the acceleration, we want to differentiate the velocity with respect to time. That is:
a = d/dt(dr/dt). = - constant * d/dt(1/r*r)
Now, if f is a function of r, then d/dt(f) = df/dr * dr/dt and so;
a = - constant * d/dt(1/r*r) = - constant * ( -2 / r*r*r) * (-constant) / r*r.
a = -2 * velocity^2 / r = - 2 * constant^2 / r*r*r*r*r.
So the acceleration is proportional to 1 / (r*r*r*r*r).

Mohan said:   1 decade ago
From Fluid kinematics. For radial flow the normal acceleration is An = dvn/ds+Vr2/r,

Acceleration is proportionate to V2/r, therefore 1/r5.

Swapan said:   7 years ago
Acceleration a= v2/r--- eq-01
Given velocity v=1/r2,
putting the value v in eq-. no 01,
therefore,a = v.v/r= ((1/r2).(1/r2))/r =(1/r4)/r = 1/r4 x 1/r = 1/r5.

Gourab said:   3 years ago
a= v(dv/ds) + dv/dt

As steady flow is taking place, dv/dt=0.
a= v (dv/ds).
now, v= k/r2.
dv/ds= dv/dr= -2k/ r3.
therefore, a varies as 1/r5.

Maha said:   7 years ago
For radial flow,

a = V^2/r.
Here, V = 1/r^2 hence,
a = [(1/r^2)^2]/r,
= (1/r^4)/r,
a = 1/r^5.

Vishal babu jaiswal said:   7 years ago
For radial flow : a =v2/r.

V=k/r2, hence put the value in above equation, then a = k2/r5.

Desalegn Degu said:   5 years ago
For this
a = V^2/r.

Here, V = 1/r^2 hence,
a = [(1/r^2)^2]/r,
= (1/r^4)/r.

a = 1/r^5.

Phani said:   6 years ago
Simple V = d/t,

Given V = 1/r^2 and d = r,
Substitute t = r^3,
Now a = V/t = 1/r^5.

NATHAN said:   7 years ago
What condition to apply r/2 in V in the question?

Please anyone tell me.

Chandra said:   9 years ago
I cannot understand this concept. Can anybody elaborate me please?

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