Chapter 10, Problem 41P

Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/s

Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4

Problem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…

The flow rate is 12.275 Liters/s and the diameter is 6.266 cm.

An experimental setup is being built to study the flow in a large water main (i.e., a large pipe). The water main is expected to convey a discharge (Qp). The experimental tube will be built at a length scale of 1/20 of the actual water main. After building the experimental setup, the pressure drop per unit length in the model tube (APm/Lm) is measured. Problem (20): Given the value of APm/Lm [kPa/m], and assuming pressure coefficient similitude, calculate the drop in the pressure per unit length of the water main (APP/Lp) in [Pa/m]. Givens: AP M/L m = 590.637 kPa/m meen Answers: ( 1 ) 59.369 Pa/m ( 2 ) 73.83 Pa/m (3) 95.443 Pa/m ( 4 ) 44.444 Pa/m *******

Find the reaction force in y if Ain = 0.169 m^2, Aout = 0.143 m^2, p_in = 0.552 atm, Q = 0.367 m^3/s, α = 31.72 degrees. The pipe is flat on the ground so do not factor in weight of the pipe and fluid.

Find the reaction force in x if Ain = 0.301 m^2, Aout = 0.177 m^2, p_in = 1.338 atm, Q = 0.669 m^3/s, and α = 37.183 degrees

Problem 5: Three-Force Equilibrium A structural connection at point O is in equilibrium under the action of three forces. • • . Member A applies a force of 9 kN vertically upward along the y-axis. Member B applies an unknown force F at the angle shown. Member C applies an unknown force T along its length at an angle shown. Determine the magnitudes of forces F and T required for equilibrium, assuming 0 = 90° y 9 kN A

Problem 19: Determine the force in members HG, HE, and DE of the truss, and state if the members are in tension or compression. 4 ft K J I H G B C D E F -3 ft -3 ft 3 ft 3 ft 3 ft- 1500 lb 1500 lb 1500 lb 1500 lb 1500 lb