ρ water = 1000 kg/m 3 = 1.94 slugs/ft 3 γ water = 9810 N/m 3 = 62.4 lbs/ft 3

Transcription

1 CEE 34 Aut 004 Midterm # Anwer all quetion. Some data that might be ueful are a follow: ρ water = 1000 kg/m 3 = 1.94 lug/ft 3 water = 9810 N/m 3 = 6.4 lb/ft 3 1 kw = 1000 N-m/ 1. (10) A 1-in. and a 4-in. hole are drilled next to each other in the ide of a reervoir. Through which hole will water exit with a higher velocity? Will the tream get wider, thinner, or remain the ame diameter a they fall to the ground? Explain your reaoning for each part in one or at mot two entence. The velocity i generated by the elevation head, which i the ame for the two hole. Therefore, the velocitie will be the ame. Thi reult can alo be derived from the ernoulli equation. Since the preure i zero at the top of the reervoir and in the exiting tream, the equation i the ame when applied between thoe two location, for both hole: p 1 + v1 z + p 1 = g v + z + g ( ) v = g z z 1 A the water fall, the magnitude of it velocity increae. Since the flow rate i contant, continuity require that the cro-ectional area decreae, o the tream will get thinner.. (5) The pump, uction pipe, dicharge pipe, and nozzle in the ytem hown below are all welded together a a ingle unit. Calculate the horizontal component of the force (magnitude and direction) exerted by the water on the unit when the pump i adding.5 m of head to the water that pae through it. Ignore friction. 1

2 d 0 o 0.3 m d 1. m 0.75 m d 1.8 m water 1.5 m We can apply the energy equation to determine the velocity of the dicharge water. For thi calculation, the mot convenient control volume i between the water urface and the dicharge from the pump, ince the preure i equal (and zero, if expreed a a gage preure) at thoe point. The velocity at the water urface i alo approximately zero. Since we are ignoring friction, the energy equation i: p 1 v1 p + z1 + + hpump = g v + z + g ( pump ) ( )( ( )) v = g h + z z = 9.81 m/.5 m + 3 m = m/ 1 The dicharge flow rate i: [ ] m m m Q va = = π = The net horizontal force on the water can then be computed uing the momentum equation. Thi equation tate that the horizontal force equal the rate of change in horizontal momentum. The water ha zero horizontal velocity, and therefore zero horizontal momentum, until it goe through the pump and i directed toward the right. When it exit the pump, it horizontal velocity i given by: v x m = = o co m

3 The force on the water i therefore: 3 kg m m m Fx = ρq( v v1) = ,360 N 5.36 kn 3 = = m The force on the water i to the right, o the force that the water exert on the tructure i to the left. 3. (5) Determine the horizontal component of the anchoring force required to hold the luice gate in the poition hown. The gate i 4 ft wide (into the page). 4 ft/ 6 ft 30 o 4 ft Conider a CV from a point uptream to a point downtream of the gate, where the flow i horizontal at both location. y continuity, the velocity downtream of the gate i: A ft 4 ft x 6 ft ft = = 4 = 6 4 ft x 4 ft 1 v v1 A The preure force on the water i h c at each location, where h c = y/. At both location, the preure i applied to a vertical plane of the water, o the force i horizontal. Alo, both v 1 and v are horizontal velocitie. Therefore, defining the force exerted by the gate on the water to be poitive to the left, the horizontal force on the water i given by: ( ) p A p A F = Qρ v v 1 1 gate 1 y y A A F = v A ρ v v ( ) ( ) 1 1 gate y y Fgate = A A v A ρ v v ( ) ( )

4 F gate 6 ft lb 4 ft lb = ( 4 ft ) ( 16 ft ) ft ft ft lug ft ft 4 ( 4 ft ) ft F gate = 14 lb The force on the gate i therefore 14 lb to the right, and the anchoring force mut be 14 lb to the left. 4. (35) The two-dimenional corrugated ramp hown below i being ued to generate friction and diipate energy in the dicharge from a cooling tower. The ramp i olid, o that all the water flow over it. The flow rate i 15 m 3 /, and at point A, the energy line i 5.35 m above the bae of the water column. Note: If you get tuck on one part of the problem and need that value to olve another part of the problem, make a reaonable aumption for the unknown value and continue. a. (10) Sketch on the diagram the energy line and the hydraulic grade line from plane A to plane. b. (10) Indicate on the figure how high water would rie in the pitot tube pointing directly into the flow at plane C. Would the water level in the tube rie, fall, or not change elevation if the long part of the tube remained vertical, but the hort ection at the bottom were bent o that it wa horizontal. Explain your reaoning very briefly. c. (15) How high above the datum i the energy line at point? 4

5 EL Pitot tube 5.35 m 0.9 m Corrugated ramp Datum A C a. A the water goe over the ramp, the energy line drop continuouly due to friction. The line would probably drop omewhat more rapidly at the beginning of the ramp becaue the water velocity i greatet there. Since the (gage) preure i zero at all location, the hydraulic grade line follow the urface of the water. b. If the pitot tube point directly into the flow, the water rie in the pitot tube to the elevation of the energy line at that location. A the tip of the pitot tube i moved away from facing into the flow, the contribution of the velocity head to the preure at the tip of the tube decline, o the level of the water in the tube fall. If the tip point perpendicular to the flow, the water rie only to the HGL. 5

6 EL EL Water level in pitot tube 5.35 m HGL 0.9 m A C c. The velocity at point A can be determined, ince the ditance from the hydraulic grade line to the energy line i v /g: y va y = g EL, A HGL, A m m va = g( yel yhgl ) = 9.81 ( 5.35 m ) = 9.65 y continuity, the velocity at point i therefore: AA m 0.6m x w m v = va = 9.65 = 6.44 A 0.9m x w 6

7 The height of the energy line at point i therefore: y m 6.44 v = y + = 1.5m + = 3.61 m g m 9.81 EL, HGL, 7