Explanation Given: The initial velocity of the water is ( v 1 ) x = 12 m/s . The diameter of the nozzle is 50 mm . The mass of the cart is m c = 20 kg . The density of water is ρ w = 1000 kg/m 3 . Draw the free body diagram of the water stream in control volume as shown in Figure 1. Draw the free body diagram of the cart as shown in Figure 2. Refer Figure (1). Write the formula for cross sectional area of nozzle. A = π d 4 2 (I) Here, d is the nozzle diameter. Write the formula for discharge of water ( Q ) . Q = v A (II) Here, v is the velocity and A is the cross sectional area. Write the formula for mass flow rate of the water. d m d t = ρ w Q (III) Here, ρ w is the density of the water and Q is the discharge. Refer Figure (1). Write the steady flow equation of forces along x and y − a x e s . ∑ F x = d m d t [ ( v 2 ) x − ( v 1 ) x ] (IV) ∑ F y = d m d t [ ( v 2 ) y − ( v 1 ) y ] (V) Here, d m d t is the rate of mass flow, v 1 is the initial velocity, v 2 is the final velocity, ∑ F x is the net forces along x − a x i s , ∑ F y is the net forces along y − a x i s . Conclusion: Calculate the cross sectional area of the nozzle ( A ) . Substitute 50 mm for d in Equation (I). A = π ( 50 mm × 1 m 1000 mm ) 2 4 = 1.9635 × 10 − 3 m 2 Calculate the discharge of the water ( Q ) . Substitute 12 m/s for v and = 1.9635 × 10 − 3 m 2 for A in Equation (II). Q = ( 12 m/s ) ( 1.9635 × 10 − 3 m 2 ) = 0.02356 m 3 /s Calculate the mass flow rate of water. Substitute 1000 kg/m 3 for ρ w and 0.02356 m 3 /s for Q in Equation (III). d m d t = 1000 kg ( 0.02356 m 3 /s ) = 23.5619 kg/s Consider the free body diagram of the water stream in control volume as shown in Figure (1). From Figure (1). The net force along x − a x i s is ∑ F x = − F x Here, v 1 = v 2 = 12 m/s . The initial velocity is ( v 1 ) x = v 1 = 12 m/s The final velocity (exit velocity) of the water is ( v 2 ) x = ( v 1 ) cos 45 ° = 12 m/s ( cos 45 ° ) = 8

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Chapter 15.9, Problem 125P

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Chapter 15.9, Problem 124P

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