Title HPGE 26 Solutions.pdf ✅

26 Hydraulics: Orifice Flow Solutions SITUATION 1. A cylindrical water tank discharges 8937.6 N of water in one minute through an orifice as shown. The diameter of the orifice is 6 cm. ▣ 1. Determine the coefficient of discharge, C. [SOLUTION] The coefficient of discharge is given by: C = Qactual Qtheoretical Compute for the actual discharge Qactual = 8937.6 N 9810 N m3 60 s Qactual = 0.0152 m3 s Compute for the theoretical discharge Qtheoretical = A√2gH Qtheoretical = π 4 (0.06 m) 2√2 (9.81 m s 2 ) (3 m) Qtheoretical = 0.0217 m3 s
Compute for the coefficient of discharge C = 0.0152 m3 s 0.0217 m3 s C = 0.70 ▣ 2. Determine the coefficient of velocity, Cv. [SOLUTION] The coefficient of velocity, Cv. Is given by: Cv = vactual vtheoretical Compute for the theoretical velocity vtheoretical = √2gH vtheoretical = √2 (9.81 m s 2 ) (3 m) vtheoretical = 7.672 m s Compute for the actual velocity based on its trajectory y = x tan θ − gx 2 2v 2 cos2 θ −4.5 m = 5.3 m tan 0° − (9.81 m s 2 ) (5. 3m) 2 2v 2 cos2 0° v = 5.53 m s Compute for the coefficient of velocity: Cv = 5.533 m s 7.672 m s Cv = 0.721 ▣ 3. Determine the coefficient of contraction, Cc . [SOLUTION]
C = CcCv Cc = C Cv Cc = 0.7 0.721 Cc = 0.97 SITUATION 2. A jet is issued from the side of a tank under a constant head as shown. The side of the tank has an inclination of 1H to 1V. The total depth of water in the tank is h1 = 6.70 m and the orifice is located h2 = 3.70 m above the bottom of the tank. Neglecting air resistance and assuming C = 1.0, determine the following: ▣ 4. The maximum height to which the jet will rise above the orifice. [SOLUTION] Compute for the velocity of the water at the orifice v = √2gH v = √2 (9.81 m s 2 ) (6.7 m − 3.7 m) v = 7.672 m s Compute for the maximum height vy = v sin θ vy = (7.672 m s ) sin 45° vy = 5.425 m s
hmax = v0y 2 2g hmax = (5.425 m s ) 2 2 (9.81 m s 2 ) hmax = 1.50 m ▣ 5. The point “x” the jet will strike a horizontal plane 1.20 m below the bottom of the tank. [SOLUTION] y = x tan θ − gx 2 2v 2 cos2 θ −4.9 m = x tan 45° − (9.81 m s 2 ) x 2 2 (7.672 m s ) 2 cos2 45° x = 9.196 m ▣ 6. The time it takes for the jet to strike a horizontal plane 1.20 m below the bottom of the tank. [SOLUTION] x = v0xt 9.196 m = (7.672 m s cos 45°) t t = 1.6953 s SITUATION 3. A 2-m diameter cylindrical vessel contains three types of immiscible liquids as shown. A 100- mm diameter circular orifice is located at the bottom of the tank with C = 0.7.