Question

A cylindrical water-storage tank has a height of 8.1 m and a radius of 6.4 m. The storage tank tank is 30 m off the ground. A 12 cm diameter pipe runs vertically down from the tank and goes 1.9 m underground before turning horizontal. The water flow in the horizontal pipe is 77 L/s. full of water but is vented to the atmosphere. The bottom of the EXAMPLE оТES IMAGES DISCUSS UNITS STATS HELP Answer Description Chk History rt What is the water pressure (gauge) in the horizontal pipe underground? (include units with answer) 8.33 pts 100% tries: 0 Show Details Formet Check 2% try penalty Hints: 1,1 How fast does the water level in the 8.33 pts.100 % tries: 0 Show Details tank drop (include units with answer) Formet Check 2% try penalty Hints: 1.0 bUTch Jones drills a 6.2 mm diameter hole near the bottom of the tank. How fast does the water shoot out? (include units with answer) 8.33 pts.100% #tries 0 Show Datails Format Check 2% try penalty Hints: 2,0 What volume flow of water is lost wha hole? 8.33 pts.100 % tries: 0 Show Details Format Check (include units with answer) 2% try penalty Hints: 1.0

Answer 1

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Solution: (a The water in the pipe flow due to the horizontal presssure. If P is the pressure or the 1 pressure energy per unit volume that pushes the water in the pipe then P 2 p the density of water and v the velocity of flow. Given the water flow in the horizontal pipe dV Av, 'A' area of cross-scction dt 771s 77x 10m*/s. We know the rate of change of flow of the pipe and v the velocity of flow. Therefore, 77x10'm /s = n(6x102)vv= 1 x1000x (6.8 2 1 = 23120 N/m2 of Pa. Hence the pressure in the pipe P = 25 dV = Av. The volume of water that dt (b). Let dV/dı is the rate of fiow through the pipe, then- decreases from the tank is Agh, is Ao area of cross-section of the tank and h level of water and dV -- Av. 'A' area of cross-section of the pipe and v the velocity of flow. Hence di d(Ah dh Av > A v. Thus the rate of dropping of water level in the tank Ap dt 2 x(0.06) d(A,h) dh A -x 6,8 0.02448m /s 2.45cm s.Moreover, we can = Av dt (3.2) dt 77x10ms d(Ah) 77x10m/s dh Av > dt 2.4 cm s calculate it by = T(3.2) A0 dt (c). The hole is made below the surface at depth 38.1 m. If v is the velocity of flow then /2 V2gh 27.33m/s x 9.8x 38.1 =v (d). Thc volume of water flowing through the hole dV - x(3.1x10°)'. /s 8.25 cm 27.33 825.11x Av dt