Question

PRACTICE IT Use the worked example above to help you solve this problem. A large pipe with a cross-sectional area of 1.24 m2 descends 4.80 m and narrows to 0.500 m2, where it terminates in a valve at point (see figure). If the pressure at point is atmospheric pressure, and the valve is opened wide and water allowed to flow freely, find the speed of the water leaving the pipe. 4.43 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s X EXERCISE HINTS: GETTING STARTED I'M STUCK! Water flowing in a horizontal pipe is at a pressure of 1.55 X 109 Pa at a point where its cross-sectional area is 5.50 m2. When the pipe narrows to 0.400 m2, the pressure drops to 1.16 X 10' Pa. (a) Find the water's speed in the wider pipe. m/s (b) Find the water's speed in the narrower pipe. m/s

Answer 1

Son The figure for the given condition is: - s ov Here, A = 1.24 m² A = 0.500 m² From continuity equation : - AV - AV . [Here, &, and we are velocities] Now, as pressure at 1 and 2 both are atmospheric pressure open to air). Thus, consider, P = P = Po Here, po atmospheric pressure Use Bernoulli's equation : - 8 + 1 sv2 +994, = P2 + 1 2 2 + 3 9 42. - %. + 19, 2+994, lot 1 ve ²+ 9942 - 1,-1 , JS = sy(4-y.) - $ $1,2 [:- )J = 8gh v = Jagh J - LA

Put 19:-2 the known values and solve ! - 2 x 9.8 m/s² x 4.80m. V oi soum) v = 10.6 m/s Exercise Here, the pipe is horizontal P = 155xiospa A = 5.some P = 1-16x 10spa A₂ = 0.400 m² from the continuity equation: - AV - A - a) Now, from Bernoulli's equation – P + 1 PV? +994, P + 1 +egge [As pipe is horizontal 7 -> P-P Pi- >> P- - 18(V-V-) IA) ] VA)-] 249,- ) V10-

Put the known values and solve: - Pg : 3 V- 2 (ISSNiospa - 1.16610spa) 1000 kg/m3 ils so moor) -] - 0.4553876 m/s. = 0.455 m/s =) v b) Calculate the water's speed in the narrower pipe : - AV - Ava – 5.50 m - x 0.455 m/s 0.400 m2 [ V2 = 6.256 m/s/
​​​​