00:01
Hello students, in the question first of all we calculate the discharge through each branch.
00:06
The total discharge q is given as 2 .0 meter cube per second.
00:14
Since the system is parallel, so we calculate the discharge in each branch calculating for branch 1 that is v1.
00:21
We have l equals to 500 meter, d equals to 0 .5 meter.
00:28
So we have q1 equals to q into a1 upon a total.
00:36
Now calculating for branch 2, we have for branch 2 q2 equals to q into a2 upon a total where a1, a2 is the cross sectional area of branch a1 and branch a2 and similarly for the other branches.
01:00
Calculating for branch b3, we have for b3 q3 equals to q into a3 upon a total.
01:12
Now we calculate area first of all.
01:17
We have area a1 equals to pi into d square by 4.
01:31
We have area a2 equals to pi into d2 square by 4 and we have area a3 equals to pi into d4 square by 4.
01:44
Now when we plug in the values and solve for the total area, we have a total equals to a1 plus a2 plus a3 which is equals to pi by 4 into d1 square plus d2 square plus d3 square which gives us area equals to 0 .196 square.
02:12
Now we have q1 equals to 2 into pi into 0 .5 square upon 4 whole upon 0 .196 which is equals to 3156 .8...