Hello, I was putting together some information to calculate water flow rates from fire hydrants and hoses. I have attached a copy in case anyone would be interested to have a look and point out any issues. We use the Freeman formula for this which makes it pretty easy. Q=AV Q is quantity of water, A is area of outlet, V is velocity. To calculate velocity we use Bernoulli's Equation V=[tex]\sqrt{2gh}[/tex] g=gravity h=head or height of water source above outlet. As we normally use pressure not head we would substitute gh for P in kPA i.e. 10m of head pressure is equal to 9.91 x 10 = 98.1kPa. I was wondering if we could calculate the velocity of the water using P=F/A F=ma? Question: If we had a large square hose that was 1m x 1m with a cap on the end. There is a pressure gauge in the cap that reads 100kPa. Assume there is a massive dam 10m above the hose outlet, this will provide a constant water supply at 100kPa (although when water is flowing the pressure will be nil. If we were to remove the cap, what would be the velocity of the water as it leaves the hose? Can we use P=F/A F=ma? Thanks for any assistance.
No, you cannot find equivalent force on an element of water. P = F/A, where F is lift. In the large reservoir, we assume F equals gravity; however it is not, or water wont flow. BTW, Q is quatity rate of water.
I suspect if you tried to derive yourself an equation starting with f=ma, you'd just end up deriving Bernoulli's equation. So just use that.