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Negative if into control volume Positive if out of control volume In simple unidirectional flow casesIn general Unit normal pointing out from control volume.

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Presentation on theme: "Negative if into control volume Positive if out of control volume In simple unidirectional flow casesIn general Unit normal pointing out from control volume."— Presentation transcript:

1 Negative if into control volume Positive if out of control volume In simple unidirectional flow casesIn general Unit normal pointing out from control volume Magnitude of velocity—square root of velocity dot product Applied forces include pressure force Always points into control volume in direction opposite to Outward pointing normal 1 2 3 In 2 and 3 if control volume moves put yourself on moving volume and use relative velocities Voller’s cheat sheet

2 A B Voller 1 Water flows through a constant diameter pipe of cross section 0.005 m 2. The discharge in the pipe is Q = 0.005 m 3 /s. The water exits from the pipe, into the atmosphere by turning through a 180 o bend. The pipe is supported on a frictionless base, in the x-y plane (i.e., gravity points into the page). Given that the head loss from the pipe joint (point A) to the end of the pipe (point B) is 1 m determine the force on the joint to hold the pipe. (  = 9810 N/m 3,  = 1000 kg/m 3,  = 1) V=Q/A=1 m/s F joint = -59 N

3 A block (mass 1 kg) is moving along a table top at a constant velocity of 5 m/s. The block is moved by water (  kg/m 3 ) issuing from a nozzle at 10 m/s and is deflected through an angle of 150 0 by a vane on the cart. The cross-section of the water jet is 0.0012 m 2. Calculate A. the resistive force acting on the cart in the x-direction. B. the total normal force N acting in the y direction 10 m/s 5 m/s X-direction -1000*(10-5)*(10-5)*.0012-1000*(10-5)*(10-5)*cos(30)*.0012 = F x = -56 N -1000*(10-5)*0*.0012+1000*(10-5)*(10-5)*sin(30)*.0012 = -1*g + N Y-direction N=24.81 N


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