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Example of Groundwater Primer - Yours will be fluid mechanics primer – see homework assignment sheet. http://www.cee.vt.edu/program_areas/environmental/teach/gwprimer/gw-sourc.html. LINEAR MOMENTUM EQUATION - APPLICATION TO PRESSURE (PIPE) FLOW. Chapter 6 (p. 190) of Text.
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Example of Groundwater Primer - Yours will be fluid mechanics primer – see homework assignment sheet http://www.cee.vt.edu/program_areas/environmental/teach/gwprimer/gw-sourc.html
LINEAR MOMENTUM EQUATION - APPLICATION TO PRESSURE (PIPE) FLOW Chapter 6 (p. 190) of Text
Linear momentum equation • Total force acting on fluid in x-direction equals rate of change of momentum in x-direction (6.6)
What is Ftot? • Total Force represents all forces acting on the fluid mass in the control volume, including gravity forces, shear forces, and pressure forces
Linear momentum equationBasic approach • Apply a control volume such that all inlets and outlets cross the control volume perpendicularly • Use continuity and Bernoulli and Continuity to calculate unknown pressures/velocities • Apply the linear momentum equation allowing for all forces acting on the fluid • Allows calculation of unknown force exerted on the fluid within the control volume (from this, calculate reaction force).
p2A2 v2 v1 p1A1 y x Linear momentum equation- Flow through a pipe bend Fy Fx
Example 1 • A liquid of relative density 0.75 flows at a rate of 7.5 kg/s through a pipe that contains a 90 bend in the horizontal plane. Fluid enters at a pressure of 25 kPa. The bend has a centerline radius of 200 mm and a constant circular cross-section of diameter 60 mm. The loss in head through the bend due to friction is 0.1 times the inlet velocity head. Calculate: • a)the mean velocity of the flow; • b)the pressure at exit from the bend; • c)the resultant force exerted by the fluid on the bend and its direction. • Why would the resultant force differ if the bend were located in the vertical plane? Describe how you would take account of this when calculating the resultant force exerted by the fluid on the bend.
(p1 = 25 kPa) R 200 mm Example 1 60 mm = 750 kg/m3
CV Resultant Force exerted on fluid and its direction p1A1 y x p2A2
Application of Newton’s Law Apply linear momentum to fluid in control volume in the x-direction: Force acting on Fluid Fx
Force acting on fluid in y-direction Apply linear momentum to fluid in control volume in the y-direction:
95.91 N 97.24 N Example 1…part c These are the forces acting on the fluid, therefore equal and opposite forces act on the bend:
If in Vertical Plane p2 will be changed because z1 z2. Also, the mass of the fluid now appears in the z-momentum equation.