00:04
This is point a and this is point b.
00:09
Now writing bernoulli equation for point a and b, so we can write pa by rho g plus va square by 2g plus h .a equals to pb by rho g plus vb square by 2g plus hb.
00:35
Now, pa equals to pb equals to patm, that is atmospheric pressure.
00:45
Because here at the surface, the pressure is also atmospheric, and when the liquid comes out of this nozzle or outflow siphon, at that point also the pressure will be atmospheric.
01:01
And the another point is, at point a, the velocity, the velocity.
01:07
Is almost negligible so we can write va equals to 0 and pa equals to pb that means we can write pa by row g minus pb by row g plus va square by 2g plus ha minus hbb xb equals to b b b square by 2g now here we conclude that pa and pb are equal so this will cancel out 0 va that is velocity at point a is 0 so it will be also cancel out and ha minus hb is basically h so we can write vb square by 2g is equals to h so vb will be equals to root under 2g h now putting the value 2 into 9 .8 and h equals to 1 meter we will get velocity at point point b vb will be equals to 4 .42 meter per second.
02:22
Now for solving or understanding the second part of the question, let's suppose there is a point c somewhere here.
02:34
Now writing the bernoulli equation for point a and point c we can write pa by row g plus va square by 2g plus h .a.
02:47
H .a equals to pb by row g plus vb square by 2g plus hc.
02:56
Sorry, here it will be pc by rho g and vc square by 2g plus hc.
03:10
That is sum of all pressure at point c or all head.
03:17
This is the pressure head, this is the velocity head, this is the height...