00:06
In this question, we have to repeat problem 8 .146 and to find the diameter of the pipe of the length 2l, which generate the same flow as the pipe of length out.
00:25
So we have to find the diameter that makes the flow rate in two pipes to be the same.
00:33
So first, for pipe with length out, we can apply the entity equation and we have that.
00:45
It is g, z1, minus z2, minus 1 over 2, v2 squared equals f from l over g, v2 square over 2 plus k, then trans v2 square over 2.
01:04
And we have that v2 is equal v and z1 minus z2 is equal h so v is the square root of 2g divided by f l over d plus k entrance plus 1 so give this equation 1 and the random number is equal to vd over new, give this equation 2.
01:36
So we must solve for the flow rate from the iteration and first we can make the gaze of a v of 1 meter per second and that we can find the rainbow number of 5 times 10 to the power of 4 and the friction factor that is corresponding to the value of random number is 0286.
02:19
And then we do the iteration.
02:22
And so at this f, we plug it to equation 1, and then we can find the value of 3.
02:30
So lastly, we can have that.
02:34
V is equal to 5 .36 meter per second for the f of 0 .066.
02:46
So, half that q is pi over 4 d square times v, where v is this value...