empty and contains 600 L of water at 20°C. Pipes 1 and 2 have D = 6 cm and Q = 300 m3/hr. What should the scale reading W be, in newtons? ... Then the...

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Solutions Manual • Fluid Mechanics, Fifth Edition

3.66 The tank in Fig. P3.66 weighs 500 N empty and contains 600 L of water at 20°C. Pipes 1 and 2 have D = 6 cm and Q = 300 m3/hr. What should the scale reading W be, in newtons? Solution: Let the CV surround the tank, cut through the two jets, and slip just under the tank bottom, as shown. The relevant jet velocities are V1 = V2 =

Fig. P3.66

Q (300/3600) m 3 /s = ≈ 29.5 m/s A (π /4)(0.06 m)2

The scale reads force “P” on the tank bottom. Then the vertical force balance is & 2 v2 − m & 1v1 = m[0 & − (−V1 )] å Fz = P − Wtank − Wwater = m æ 300 ö Solve for P = 500 + 9790(0.6 m 3 ) + 998 ç ÷ (29.5) ≈ 8800 N è 3600 ø

Ans.

3.67 Gravel is dumped from a hopper, at a rate of 650 N/s, onto a moving belt, as in Fig. P3.67. The gravel then passes off the end of the belt. The drive wheels are 80 cm in diameter and rotate clockwise at 150 r/min. Neglecting system friction and air drag, estimate the power required to drive this belt.

Fig. P3.67

Solution: The CV goes under the gravel on the belt and cuts through the inlet and outlet gravel streams, as shown. The no-slip belt velocity must be rev rad 1 min ù m é Vbelt = Voutlet = ΩR wheel = ê150 2π (0.4 m) ≈ 6.28 ú min rev 60 s û s ë

Chapter 3 • Integral Relations for a Control Volume

183

Then the belt applies tangential force F to the gravel, and the force balance is & out uout − m & in u in , but u in = 0. å Fx = Fon belt = m 650 kg öæ mö & out = æç Then Fbelt = mV ÷ç 6.28 ÷ = 416 N sø è 9.81 s øè

The power required to drive the belt is P = FVbelt = (416)(6.28) ≈ 2600 W

3.68 The rocket in Fig. P3.68 has a supersonic exhaust, and the exit pressure pe is not necessarily equal to pa. Show that the force F required to hold this rocket on the test stand is F = ρeAeVe2 + Ae(pe − pa). Is this force F what we term the thrust of the rocket?

Ans.

Fig. P3.68

Solution: The appropriate CV surrounds the entire rocket and cuts through the exit jet. Subtract pa everywhere so only exit pressure ≠ 0. The horizontal force balance is & e ue − m & f uf − m & o uo , but u f = u o = 0, m & e = ρe Ae Ve å Fx = F − (pe − pa )Ae = m

Thus F = ρe Ae V e2 + (pe − pa )Ae

(the thrust ) Ans.

3.69 A uniform rectangular plate, 40 cm long and 30 cm deep into the paper, hangs in air from a hinge at its top, 30-cm side. It is struck in its center by a horizontal 3-cmdiameter jet of water moving at 8 m/s. If the gate has a mass of 16 kg, estimate the angle at which the plate will hang from the vertical. Fig. P3.69

Solution: The plate orientation can be found through force and moment balances, æπ ö å Fx = Fj = − m& inuin = −(998) ç ÷ (0.032 )(82 ) = 45.1 N è 4ø

å M B = 0 = −(45)(0.02)(sinθ ) + (16)(9.81)(0.02)(cosθ ) θ = 16°