00:05
In this question, question 8 .155, we have to repeat the previous problem, problem 8 .154.
00:16
But now the pipe is vertical.
00:19
So q, we have to find q for three types of entrances.
00:26
So first we have to apply the energy equation between the tank surface and the pipe exit.
00:32
So it becomes gz1, the apportion.
00:36
0 .1 is the free surface and 0 .2 is the exit point and we have that gc1 minus v square over 2 is f time l over d plus k entrance time v square over 2.
00:56
So we can solve for v it is the equal to square root of 2g h plus l, divided by 1 plus k entrance plus f l over d.
01:19
This is equation 1 and equation 2 is the renal number which is vd or new.
01:28
That is the equation 2.
01:30
So first for the reentrance the k is 0 .78 so first we have to make a guess of v and we do the iteration to find the value of v and at f is 0 .01 we can have that v is 3 .85 meter per second from equation 1 and we have that the rentero number will be 9 .67 times 10 to the power 4.
02:10
So we do the iteration to find the value of f that corresponds to this value of rentero number and and in this head it is 0388 and we do the same for f equals 0 .0388v v is 8 .35 times 10 to the power sorry v is equal to 3 .32 peter per second and the random number we have is 8 .35 times 10 to the power fall and so we have that f that corresponds to this random number is 0389...