A fire hose exerts a force on the person holding it. This is because the water accelerates as it goes from the hose through the nozzle.
How much force is required to hold a 6.7-cm-diameter hose delivering 490L/min through a 0.80-cm-diameter nozzle?
A1V1 = A2V2
P1 + .5ρV1^2 + ρgY1 = P2 + .5ρV2^2 + ρgY2
The Attempt at a Solution
First I converted the diameters into meters, divided it by 2 to get the radius and then found the areas of the hose and nozzle. A1 (the hose) = .0036 m^2 A2 (the nozzle) = 5.0x10^-5 m^2. Then i converted V1 (the velocity in the hose) to m^3/s. V1 = .0082 m^3/s.
Next I must find V2.
A1V1 = A2V2
V2 = A1V1/A2 = (.0036)(.0082)/(5.0x10^-5) = .59 m^3/s
Then I must use bernoulli's equation to find the pressure in the hose and multiply by the area to find the force. the part of the hose that connects to the nozzle and the nozzle are at the same height so Y1 and Y2 = 0. So I am left with
P1 + .5ρV1^2 = P2 + .5ρV2^2
This is where I am stuck. I know the density of water is around 1000 kg/m^3 and I know both of the velocities , but i still have two unknowns (P1 and P2). since the hose/nozzle is a confined space, can I say that P1=P2 or does that only work for non moving fluid?
anyway after that I would have to find P1 (pressure in hose) and multiply by A1 to find the force the hose puts on the person, which would be equal in magnitude to the force needed to hold it. How does my solution look so far?