When substituting this equation for a the continuity

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When substituting this equation for A, the Continuity Equation becomes:π ∙D124∙v1=π ∙D224∙v2Dividing both sides of the equation by π, and multiplying each side by 4 simplifies the equation:D12∙v1=D22∙v2Solving for the value of D2, the equation becomes:D2=D1v1v2An objective of the experiment is to verify that the relationship between v2and the height of the fluid above the orifice, h, is true, which would also verify that Bernoulli’s equation truly describes the fluid flow in this system. This process involves measuring v2at several fluid heights, h. v2is found by treating the fluid coming out of the orifice as a projectile. A projectile with an initial velocity v2in the horizontal direction, the measured range, R, is related to the velocity v2by the following equation:R=v22Hgwhich, when solving for v2, is the same as:v2=Rg2H(4)This equation of the relation between v2and R can be proved with the following equations for height, H, and range, R:H=12gt2andR=v2t2where, for the value of height, H:t=Rv2Substituting this value for time, t, into the equation for H, we get:H=12g(Rv2)2Multiplying both sides by 2, we get:2H=gR2v22
4Solving for v2, we get:v2=Rg2HDataTable 1. Measurements of the systemWaterlevel (h)Shootingheight (H)Time increment(∆t)Shootingrange (R)V1(∆h/∆t)V2=Rg/2HV22D1D2D2=D1v1/v21731.3 cm10.11 s36 cm0.0989cm/s142.4cm/s20278cm/s214.25cm 0.5cm0.376cm165.50 s35 cm0.1818cm/s138.5cm/s19182cm/s20.516cm156.19 s34 cm0.1616cm/s134.5cm/s18090cm/s20.494cm146.79 s33 cm0.1473cm/s130.6cm/s17056cm/s20.479cm136.57 s32 cm0.1522cm/s126.6cm/s16028cm/s20.494cm127.09 s31 cm0.1410cm/s122.7cm/s15055cm/s20.483cm117.19 s29.8 cm0.1391cm/s117.9cm/s13900cm/s20.489cm107.65 s27.9 cm0.1307cm/s110.4cm/s12188cm/s20.490cm97.56 s26.5 cm0.1323cm/s104.8cm/s10983cm/s20.506cm810.24 s25 cm0.0977cm/s98.9cm/s9781cm/s20.448cm79.86 s23.5 cm0.1014cm/s93.0cm/s8649cm/s20.471cm611.32 s21.8 cm0.0883cm/s86.3cm/s7448cm/s20.456cm
5513.23 s19.9 cm0.0756

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