00:02
First, we need to convert vapor pressure to head, which would be meters of water.
00:07
So let's let gamma be row g, which is about 1 ,000 times 9 .81, which is 9 ,810 newtons per cubic meter.
00:20
Then the pressure piece of v over gamma would be negative 98 ,700 over 98 ,10.
00:34
And that's negative 10 .06 meters.
00:37
Now, bernoulli from free surface a, p sub a over gamma plus z sub a equals p sub b over gamma plus z sub b plus v squared over 2g plus h sub l.
00:58
So p sub b over gamma equals negative the difference of z sub b and z sub a minus v squared over 2g minus h sub l.
01:09
To avoid cavitation anywhere in the hose, that would require p.
01:14
Sub b over gamma to be greater than or equal to piece of v over gamma, which is negative 10 .06.
01:24
So negative 3 minus v squared over 2g minus h subl has got to be greater than equal to negative 10 .06, which means 3 plus v squared over 2g minus h subal is got to be greater than equal to negative 10 .06, which means 3 plus v squared over 2g plus h sub l's less than are equal to 10 .06 head loss in the hose h sub l would be f times l over d times v squared over 2g l over d is 5 over 0 .05 which is a hundred so f times l over d would be 0 .03 times 100 which is 3 so h sub l is 3 times v squared over 2g...