Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 7 (Q.No. 21)
21.
The total pressure on the top of a closed cylindrical vessel competely filled up with a liquid is
Discussion:
16 comments Page 1 of 2.
Kuldeep Gupta said:
4 years ago
@Ajaypal.
Pressure force = pressure * area.
So, the pressure force and total pressure are same.
Pressure force = pressure * area.
So, the pressure force and total pressure are same.
Asma said:
4 years ago
Thanks for the explanation @Ajaypal.
Anonymous said:
5 years ago
Total pressure P=WhA.
h=ω^2r^2/2g.
A=πr^2.
So, P=r^4.
h=ω^2r^2/2g.
A=πr^2.
So, P=r^4.
Ramesh said:
5 years ago
@Muhhmad Azkar.
How the W = v * ρ?
How the W = v * ρ?
Abbas Ali said:
6 years ago
@Muhammad Azkar.
It is unit weight not a simple weight.
It is unit weight not a simple weight.
Muhammad Azkar said:
6 years ago
The total pressure on the top of the liquid-filled cylindrical vessel is given by -
P = π.W.ω^2.r^2 / 4.g
As we know, W = V x ρ = πr^2h.ρ
P = π.πr^2h.ρ.ω^2.r^2 / 4.g
P = π^2.h.ρ.ω^2.r^4 / 4.g
P ∝ r^4
C is the correct answer.
P = π.W.ω^2.r^2 / 4.g
As we know, W = V x ρ = πr^2h.ρ
P = π.πr^2h.ρ.ω^2.r^2 / 4.g
P = π^2.h.ρ.ω^2.r^4 / 4.g
P ∝ r^4
C is the correct answer.
(1)
Ganesh said:
6 years ago
@Ajaypal.
Here m^2 and m^2 will be cancelled. It should not be multiply.
Here m^2 and m^2 will be cancelled. It should not be multiply.
(1)
Mizanur said:
7 years ago
Thanks @Ajaypal.
(1)
Anirudha said:
7 years ago
It is πww^2r^4/4g.
Rajesh Rar said:
8 years ago
@Ajaypal.
If total pressure =pressure*area.
The Unit will become as N/mm4.
If total pressure =pressure*area.
The Unit will become as N/mm4.
(1)
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