# Civil Engineering - GATE Exam Questions - Discussion

Discussion Forum : GATE Exam Questions - Section 1 (Q.No. 5)

5.

In a steady radial flow into an intake, the velocity is found to vary as (1/

*r*^{2}), where*r*is the radial distance. The acceleration of the flow is proportional toDiscussion:

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).

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+Vr

Acceleration is proportionate to V

^{2}/r,Acceleration is proportionate to V

^{2}/r, therefore 1/r^{5}.
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.

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.

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.

(1)

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.

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.

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.

a = V^2/r.

Here, V = 1/r^2 hence,

a = [(1/r^2)^2]/r,

= (1/r^4)/r.

a = 1/r^5.

(8)

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.

Given V = 1/r^2 and d = r,

Substitute t = r^3,

Now a = V/t = 1/r^5.

(7)

NATHAN said:
7 years ago

What condition to apply r/2 in V in the question?

Please anyone tell me.

Please anyone tell me.

Chandra said:
9 years ago

I cannot understand this concept. Can anybody elaborate me please?

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