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6. Momentum Theory

6.1. Derivation of the Linear Momentum Equation. Consider a fluid mass contained in a closed (control) surface S at an instant of time t. After a small time interval dt, t dt the fluid mass is moved and has the surface S' at t dt in Fig. (6.1) . ? Fig. 6.1. The intersection of S and S' divides the both partial surface S1 and S2. V1 is the fluid volume moved during the small time interval dt from S1 towards S' and V3 that from S2 towards S'.

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6. Momentum Theory

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    1. 6. Momentum Theory 6.1. Derivation of the Linear Momentum Equation 6.2. Applications 6.3 Derivation of the moment of Momentum Equation

    2. 6.1. Derivation of the Linear Momentum Equation Consider a fluid mass contained in a closed (control) surface S at an instant of time t. After a small time interval dt, t+dt the fluid mass is moved and has the surface S at t+dt in Fig. (6.1)

    6. 6.2. Applications 6.2.1. Free jets 6.2.2. Sudden Change of Duct Cross Section : Carnots Mixing Loss 6.2.3. Mixing Process in Ducts of Constant Cross Section

    7. Example #1 A jet with a uniform velocity of U hits a bucket and is turned through an angle (180o ) symmetrically. To calculate is the force on the bucket acting by the free jet.

    17. 6.2.2. Sudden Change of Duct Cross Section : Carnots Mixing Loss

    21. 6.2.3.Mixing Process in Ducts of Constant Cross Section

    26. 6.3 Derivation of the Moment of Momentum Equation

    29. 6.4 Applications Consider problems with the simplifications : a) 1-D flow b) Steady or steady-in-the mean cyclical flow c) The component of Eq. (6.29) resolved along the axis of rotation is existing.

    30. 6.4.1. Rotating Water Sprinkler

    33. 6.4.2. Torque and Power of Rotating Machine (Turbomachines) : Euler Turbine Equation

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