Explanation Given: The gauge pressure of the water at C is P = 40 lb/in . 2 The velocity of water at A is v A = 12 ft/s . The velocity of water at B is v B = 25 ft/s . The diameter of the pipe at A and B is d A = d B = 0.5 in . The diameter of the pipe at C is d C = 0.75 in . . The density of water is ρ w = 62.4 lb/ft 3 . Draw the free body diagram of the elbow as shown in Figure 1. Write the formula for mass flow rate of the water. d m d t = ρ w Q (I) Here, ρ w is the density of the water and Q is the discharge. Write the relation between the discharge of water ( Q ) , area ( A ) and velocity ( v ) . Q = v A (II) v = Q A (III) Here, v is the velocity and A is the cross sectional area. Write formula for cross sectional area of the pipe. A = π d 2 4 (IV) Refer Figure (1). Write the steady flow equation of forces along x and y − a x e s . ∑ F x = d m B d t ( v B ) x + d m A d t ( v A ) x − d m C d t ( v C ) x (V) ∑ F y = d m B d t ( v B ) y + d m A d t ( v A ) y − d m C d t ( v C ) y (VI) Here, ∑ F x is the net forces along x − a x i s , ∑ F y is the net forces along y − a x i s . Here, d m d t is the rate of mass flow, v A is the velocity of the water at A , v B is the velocity of the water at B , v C is the velocity of the water at C , ∑ F x net forces along x − a x i s , ∑ F y net forces along y − a x i s and the suffixes x , y indicates the velocity of the water along x and y axes. Write the formula for the force exerted by the pressure ( P ) . F P = P A C (VII) Here, P is the pressure of the water and A C is the cross sectional area at C . Conclusion: Refer Equation (IV) Calculate the cross sectional area at C . A C = π d C 2 4 Substitute 0.75 in . for d C . A C = π ( 0.75 in . × 1 ft 12 in . ) 2 4 = 0.0009765 π Calculate the cross sectional area at A . A A = π d A 2 4 Substitute 0.5 in . for d A . A A = π ( 0.5 in . × 1 ft 12 in . ) 2 4 = 0.0004340 π Here, A A = A B = 0.0004340 π . Refer Equation (II) and (IV) Calculate the discharge of water at C . Q C = v C A C Substitute 0.0009765 π for A C . Q C = v C ( 0.0009765 π ) = v C ( 0.0009765 π ) ft 3 /s Calculate the discharge of water at A . Q A = v A A A Substitute 12 ft/s for v A and 0.0004340 π for A A . Q A = 12 ft/s ( 0.0004340 π ) = ( 0.0052083 π ) ft 3 /s Calculate the discharge of water at B . Q B = v B A B Substitute 25 ft/s for v B and 0.0004340 π for A B . Q B = 25 ft/s ( 0.0004340 π ) = ( 0.01085 π ) ft 3 /s Refer Figure (I). The flow of water is divided into two discharges at A and B . Hence the discharge of water at A is equal to the sum of the discharge of water at A and B . Q C = Q A + Q B Substitute v C ( 0.0009765 π ) ft 3 /s for Q C , ( 0.0052083 π ) ft 3 /s for Q A and ( 0.01085 π ) ft 3 /s for Q B . v C ( 0.0009765 π ) ft 3 /s = ( 0.0052083 π ) ft 3 /s + ( 0.01085 π ) ft 3 /s v C = 0.0160583 π 0.0009765 π v C = 16.4444 ft/s Refer Equation (I). Calculate the mass flow rate of water at A . d m A d t = ρ w Q A Substitute 62.4 lb/ft 3 for ρ w and ( 0.0052083 π ) ft 3 /s for Q A . d m A d t = ( 62.4 lb/ft 3 × 1 32.2 ft/s 2 ) [ ( 0

Engineering Mechanics: Dynamics (14th Edition)

14th Edition

ISBN: 9780133915389

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.9, Problem 121P

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The horizontal and vertical force components exerted on the elbow.

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Chapter 15.9, Problem 120P

Chapter 15.9, Problem 122P

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