Spring 2010 Solutions - FSU Physics Department

Use Bernoulli’s equation to work backward from the top of the column of water.1 2 1 2P3 + ρgy3 +2ρv3 = P4 + ρgy4+2ρv41 2ρ gy2ρ v =3 42v = 2gy3 4,which is just the speed required to project and object to height y 4 . Solving for thevelocity in terms of known parameters gives:3 42( )( )v = 2gy = 2 9.8 m / s 140mv3= 52.4 m/s(b) What is the flow rate of the water just as it leaves the nozzle (in liters/sec)?The flow rate is volume/time, or V/Δt = AΔx /Δt = A (Δx /Δt) = Av which is just theterms on each side of the continuity equation.−2 2( ) ⎡πrate = v1A1 = 52.4 m / s ⎣ (2.5x10 m)⎤⎦3rate = 0.103 m / s = 103 liters / s(c) What pump pressure is required?Working backwards, we have1 1P2 + gy2 +2ρv2 = P3+ gy3+22 2ρ ρ ρv321 ⎛v3⎞ 12+2ρ=3+2ρ2P ⎜ ⎟ P v 3⎝ 9 ⎠1 2 1v3P2 = Patm+2ρv3−2ρ ⎛ ⎜ ⎞⎟⎝ 9 ⎠1 2 12P1802= P + ρ 1− v3= P + ρ v3atmP = P + gy2 ( 81) atm 2 ( 81)80( ) ρ2 atm 81 4Finally, we can obtain the pressure at the pump, P 1 , simply by adding this pressuredue to the 6 meter column of water (since v 1 = v 2 ),1 2 1 2P1+ ρgy1+ 2ρv1 = P2 + ρgy2+2ρv2P+ ρgy = P + ρgy 1P = P + ρg y − y1 1 2( )1 2 2 1280( ) ( )P1 = Patm + ρg⎡ 81y4y2 y ⎤⎣+ −1 ⎦