Physics Help Forum Finding time for velocity ( water hammer related)

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Mar 26th 2017, 09:23 AM #1 Junior Member   Join Date: Feb 2017 Posts: 4 Finding time for velocity ( water hammer related) Hello, The problem is given: I first found v0(final) from: $\displaystyle v0=sqrt(\frac {2*g*h}{lambda*\frac {l}{d}+1})$ $\displaystyle v0=sqrt(\frac {2*9.81*18}{0.03*\frac {400}{2}+1})$ v0 is 7.1 m/s 90% of v0 is v= 6.39 m/s. Time to reach v=6.39 m/s: $\displaystyle t=\frac {l*v0}{2*g*H}*log(\frac {v0+v}{v0-v})$ $\displaystyle t=\frac {400*7.1}{2*9.81*18}*log(\frac {7.1+6.39}{7.1-6.39})$ $\displaystyle t=10.28$ s But the answer key says it is 1 mins 20 secs Where may I go wrong? Thanks a lot
 Mar 26th 2017, 11:07 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 Don't know where you got your equations, but the way to solve this is to balance forces (via pressure heads) for the accelerating fluid. With your setup the main balance equation is Assuming the tank head is a constant H ( = 18m) At any time t let the velocity of flow in the pipe be v. Then if H is the constant head producing flow, H = Friction head + velocity head at exit + acceleration head + loss at valve + shock loss If L is the lenght of pipe and a the cross sectional area then mass of liquid in pipe = waL/g; where w is the specific weight. If the velocity is (v + dv) at time (t + dt) the acceleration of liquid in pipe = dv/dt For for this acceleration = (waL/g) dv/dt Head required for acceleration = (L/g) dv/dt Head required toovercome friction = (4fL/d)(v^2/2g) Loss of head at valve = (4f *25/d ) (v^2/2g) Loss of head at entry = 0.5v^2/2g Velocity head at exit = v^2/2g I'm not sure whether you are meant to include all of these or ignore some.

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