1)
\Given Data :
\Orifice diameter d = 52 mm
\do = 52/1000 = 0.052 m
\Pressure head H = 4.5 m
\Diameter of the vena contracta dv = 41 mm
\dv = 41/1000 = 0.041 m
\The horizontal distance of the water jet x = 2.15 m
\The vertical distance of the water jet y = 327 mm
\y = 327/1000 = 0.327 m
\The theoretical flow velocity vt = √(2gH)
\Substitute H = 4.5 m and g = 9.8
\vt = √(2*9.8*4.5)
\vt = 9.39 m/s
\Orifice area A = do²/4
\Ao= (0.052)²/4
\Ao = 0.000676 m²
\Area at vena contracta Av = dv²/4
\Av = 0.041²/4
\Av = 0.00042 m²
\The coefficient of velocity Cv = Av / Ao
\Cv = 0.00042/ 0.000676
\Cv = 0.621
\The coefficient of contraction Cc = x / √(4yH) = va / vt
\Cc = x / √(4yH)
\Substitute x = 2.15 and y = 0.327 m
\Cc = 2.15 / √(4*0.327*4.5)
\Cc = 0.8862
\Theoretical delivery of water Qt = Aovt
\Substitute Ao = 0.000676 and vt = 9.39
\Qt = (0.000676)(9.39)
\Qt = 0.00635
\The coefficient of delivery Cd = Cv Cc = Qa / Qt
\Cd = Cv Cc
\Cd = 0.621*0.8862
\Cd = 0.621*0.8862
\Cd = 0.55
\The coefficient of delivery Cd = Qa / Qt
\The actual delivery of the water Qa = Cd * Qt
\Qa = 0.55 * 0.0635
\Qa = 0.035
\The coefficient of contraction Cc = va / vt
\Actual flow velocity va = Cc * vt
\va =0.8862 * 9.39
\va = 8.32 m/s
\Solution : \ \
\vt = 9.39 m/s
\