Civil Engineering - GATE Exam Questions - Discussion

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

[A]. 1/r5
[B]. 1/r3
[C]. 1/r4
[D]. 1/r

Answer: Option A

Explanation:

No answer description available for this question.

Mohan said: (Dec 13, 2013)  
From Fluid kinematics. For radial flow the normal acceleration is An = dvn/ds+Vr2/r,

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

Ram said: (Sep 24, 2014)  
When the acceleration was take place on frist line.

Dileep Barnwal said: (Dec 19, 2014)  
Acceleration a = v*(dv/ds).
=> ao < 1/r5.

Chandra said: (Jul 6, 2015)  
I cannot understand this concept. Can anybody elaborate me please?

Aparna said: (Nov 13, 2015)  
v2/r but; v = 1/r2 so v = 1/r4+1; v = 1/r5.

Yoga said: (Dec 18, 2015)  
Acceleration a = V2/r = (1/r2)2/r = 1/r5.

Deepak said: (Jun 14, 2016)  
I agree with you @Aparna.

Teja said: (Sep 8, 2016)  
Nice @Aparna.

Pran said: (Sep 17, 2016)  
Please give me conceptual & answer.

Md Shams Zeya said: (Nov 21, 2016)  
I can not understand this, Please help me to get it.

Madhu Kumar said: (Feb 7, 2017)  
I agree with your answer @Aparna.

Nathan said: (Feb 8, 2017)  
What condition to apply r/2 in V in the question?

Please anyone tell me.

Tulasi Sivaprasad said: (Mar 9, 2017)  
Please explain clearly.

Kumari Juhi said: (Mar 13, 2017)  
a = V2/r = (1/r2)2/r = 1/r5.

Vishal Babu Jaiswal said: (Jun 5, 2017)  
For radial flow : a =v2/r.

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

Deepanchakaravarthy said: (Jun 7, 2017)  
Thanks @Aparna.

Maha said: (Aug 23, 2017)  
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.

Swapan said: (Oct 11, 2017)  
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.

M.Naresh said: (Dec 17, 2017)  
We know the radial acc is=V2/r .(1/(r2)2*r=1/r5.

Niki said: (Feb 8, 2018)  
I agree @Maha.

Nanu said: (Mar 11, 2018)  
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).

Maha said: (May 5, 2018)  
Thanks for explaining the answer.

Phani said: (May 6, 2018)  
Simple V = d/t,

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

Desalegn Degu said: (Feb 26, 2019)  
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.

Aki said: (Sep 21, 2019)  
Well explained, Thanks @Desalegn Degu.

Blackie said: (Jan 5, 2021)  
Yes, 1/r5 is correct.

Gourab said: (Aug 10, 2021)  
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.

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