00:01
Hello students in this question.
00:02
It is given that water under ideal flow as shown the figure is been discussed with the value of density of water, row to be equal to 1000 kilogram per meter cube, acceleration due to gravity, g to be equal to 9 .8 meter per second square, and here at the lower point, pressure p1 is taken to be 1 .75 into 10 rise to 4 pascal, diameter b1 ,000 ,000 ,000 ,000 ,000, is taken to be 6 into 10 raise to minus 2 meter and height h1 is taken to be 0 meter.
00:37
On the hand at the upper end pressure p2 is taken to be 1 .2 into 10 raise to 4 pascal diameter b2 is taken to be 3 10 rise to minus 2 meter and height h2 is taken to be 0 .25 meter.
00:55
Now in the first part the question we need to find out the velocity at the lower point which is taken as v1.
01:01
And here by using the equation of continuity, we get area a1 into velocity v1 to be equal to area a2 into velocity v2, where this area a1 will be equal to 5 into d1 square divided by 4 and area a2 will be equal to 5 into d2 square divided by 4.
01:22
So by substituting the value of a1 and a2 in the above equation and rearranging, the equation changes as d1 square into v1 will be equal to d2 square into v2 and here by substituting the values in this equation and undergoing simplification for equation changes as velocity v2 to be equal to 4 times the value of velocity v1 let this be equation number 1 and here by using the bernolice equation you can write that pressure p1 divided by density row into acceleration due to gravity g plus velocity v1 square divided by 2 into g plus height h1 will be equal to pressure p2 divided by row g plus v2 squared by 2g plus height h2.
02:09
And here by rearranging this equation, our equation changes as b2 square minus b1 square divided by 2g will be equal to pressure p1 minus pressure p2 divided by row into g plus height h1 minus height h2...