00:01
Hello student, here volumetric flow rate q in the pipe is constant since the fluid is incompressible.
00:12
So, you can write a1 v1 equals to a2 v2 where a1 is the cross -sectional area of the first pipe or the small pipe and v1 is the velocity of the fluid in that pipe and a2 is the cross -sectional area of the large pipe and v2 is the velocity of the fluid in that pipe.
00:33
So, v2 equals to a1 v1 divided by a2.
00:41
This is pi r1 square v1 divided by pi r2 square or r1 minus r2 square into v1 which is 3 .5 divided by 14 square into v1.
01:06
V1 is 6 meter per second.
01:14
So, this is equals to 0 .375 meter per second.
01:25
So, this is the velocity of fluid in the large pipe is 5 meter per second.
01:38
Now, from bernoulli's theorem we have p1 plus half rho v1 square plus rho g h1 equals to p2 plus half rho v2 square plus rho g h2 where h1 and h2 are the elevation of the pipe small pipe and large pipe respectively.
02:08
So, h1 equals to h2 here...