# Flow of liquid in T shaped pipe

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In this how they have written the equation .

Applying the equation of continuity, $(2A)v=2(A)u$ - ie the velocities $v,u$ in the vertical and horizontal sections are equal.

The total volume of liquid is $V=(2A)H+2(AH)=4AH$ so the total mass is $M=\rho V$. The initial height of liquid in the vertical tube is $H$. When this height has fallen by $x$ a volume of $2Ax$ has moved into the horizontal sections. The height of the centre of this section of liquid above the horizontal section is $H-\frac12x$. The loss of GPE is therefore $mg(H-\frac12x)$ where $m=2Ax\rho$.

The whole of the fluid is moving with speed $v$ after this liquid falls. The loss in GPE equals the gain in the KE of the fluid so
$(2Ax\rho)g(H-\frac12x)=\frac12Mv^2=\frac12(4AH\rho)v^2$
$x(H-\frac12x)g=Hv^2$.

When $x=\frac12H$ then
$Hv^2=\frac12H(H-\frac14H)g$
$v=\sqrt{\frac38gH}$.

Differentiating wrt $x$ we get
$(H-\frac12x)g+x(-\frac12)g=(H-x)g=2Hv\frac{dv}{dx}=2H\frac{dx}{dt}\frac{dv}{dx}=2Ha$
where $a=\frac{dv}{dt}$ is acceleration.

When $x=0$ the initial acceleration of the liquid is $a=\frac12g$.

The mass of liquid to the right of P is $m'=\rho A\frac{H}{2}$. The initial acceleration of this liquid is $a=\frac12g$ so the resultant force on it is $m'a=\frac14\rho AHg$. The force on this liquid comes from the difference in pressure at the two ends, which is the gauge pressure $P$ since the right side of this liquid is at atmospheric pressure. Therefore
$PA= \frac14\rho AHg$
$P=\frac14\rho Hg$.

answered May 4, 2017 by (28,876 points)
selected May 4, 2017 by koolman
How can we apply equation of continuity at three different points at the same time .

I thought that we  can only apply it for two points .
No, you can apply it at any number of points. Continuity equation conserves mass (and also volume if the liquid is incompressible). It is similar to Kirchhoff's Current Law : the sum of currents entering a section of a circuit equals the sum of currents leaving that section.
Very good answer.one clarification why you have taken the kinetic energy of the whole mass to be same.I mean by principle of continuity the kinetic energy should have been $1/2(2A)p(H-x)v^2+1/2p(A)(2H+2x)(4v^2)=1/2pA(2(H-x)+8(H-x))v^2$
@Green The whole mass of liquid moves at the same speed, as stated in my 1st paragraph.
@sammygerbil thanks I got it