流体力学与工程Chapter 6 Momentum and forces in fluid flow.ppt

6.1 Momentum Equation According to the law of momentum conservation ( Newton’s second law) Example 6 In the following Fig.，the inflow and outflow radii are 25 and 15cm, respectively; the inflow and outflow angles with respect to the horizontal line (θ1andθ2) are 450 and 300, respectively; q is 50 L/s; the area average inlet and outlet pressure are 8.5 and 5.83kPa; and the total fluid weight in the pipe is 2.0N. Find the horizontal and vertical force required to hold the pipe in place. Solution The momentum exchange and force vectors are drawn in the following figure. The magnitudes of Fp1 and Fp2 are calculated as The magnitudes of v and qm are calculated as follows According to the momentum equation, the components of F are calculated as Substitution of parameters and solving for Fx and Fy gives The negative signs indicate that the assumed directions for Fx and Fy are incorrect. 6.2 Moment-of-Momentum Equation According to the law of moment of momentum, For a steady flow The moment-of-momentum equation can be applied to only the inertial coordinate system. Example 11 The horizontal sprinkler of the following figure discharges 0.001 m3/s through each nozzle. Neglecting friction, find its speed of rotation. The area of each opening nozzle is 0.0001m2. Solution Still, according to the moment-of-momentum equation, 6.2.2 Basic Equations of Turbomachinery In the figure, the axis of centrifugal pump or blower is horizontal. The revolving impeller is driven by its shaft. And the fluid runs outward in radial direction. Analysis (1) Steady flow——Under steady running condition. (2) Mass force——Its resultant moment about the axis is zero because of symmetry . (3) Surface force—— 1) The resultant moment about the axis is zero because the pressure at the inlet is in the radial direction. 2) The magnitude of resultant moment about the axis, which is exerted on the fluid by vanes, equals what the shaft passes the impeller. Velocity——