- 1. KV MOMENT OF MOMENTUM {Rotodynamic Machines}
- 2. KV Identify the unique vocabulary used in the description and analysis of fluid flow with an emphasis on Moment of Momentum and Rotodynamic Machines Describe and discuss the flow of fluids through rotodynamic machines and the momentum of the fluids passing through such devices. Construct and resolve the velocity vectors associated with the flow of fluids through rotodynamic devices. Derive and apply the governing equations associated with fluid momentum when applied to rotodynamic devices OBJECTIVES
- 3. KV Consider a two arm sprinkler system as illustrated. The direction and magnitude of fluid flow through this sprinkler changes direction as it passes from the inlet to the outlet. The fluid discharging through the arms of the sprinkler exerts a torque on the sprinkler head and as a consequence causes it to rotated about the z axis. Provided that the flow is steady, equation 1 can be applied to the system. ∑ 𝑟 × 𝑈 𝑜𝑢𝑡𝑚 · 𝑜𝑢𝑡 − ∑ 𝑟 × 𝑈 𝑖𝑛𝑚 · 𝑖𝑛 = ∑ 𝑟 × 𝐹 (1) Eq.1 is known as the moment of moment equation MOMENTUM OF MOMENTUM
- 4. KV • The fluid enters the control volume of the sprinkler (figure 1) axially through the stem at section 1. • The fluid leaves the control volume through each of the two nozzles at section 2. • The component of 𝑟 × 𝑈 along the axis of rotation is zero. • The magnitude of the the axial component of 𝑟 × 𝑈 is 𝑟2 × 𝑈𝜃 where r2 is the radius from the axis of rotation to the nozzle centreline and 𝑈𝜃2 is the tangential component of the velocity of flow exiting the nozzle. • The fluid velocity measured relative to a fixed control surface is an absolute velocity, U • The velocity of the fluid exiting the nozzle is called the relative velocity, Uθ. • Absolute and relative velocities can be related by a vector diagram. Figure 1: Two arm sprinkler (Young, et al. 2012)
- 5. KV For the sprinkler system, it can be stated that; ∑ 𝑟 × 𝑈 𝑜𝑢𝑡𝑚 · 𝑜𝑢𝑡 − ∑ 𝑟 × 𝑈 𝑖𝑛𝑚 · 𝑖𝑛 𝑎𝑥𝑖𝑎𝑙 = −𝑟2 × 𝑈𝜃2 𝑚 · (2) For mass conservation, ṁ is the total mass flowrate through both nozzles. Given that the direction of rotation is opposite to that discharging jet, a negative sign is appropriate in the right hand portion of the equation. Noting that the mass flow rate ṁ times the velocity U equals force, then can be stated that ∑ 𝑟 × 𝑈 𝑚 · 𝑎𝑥𝑖𝑎𝑙 = ∑ 𝑟 × 𝐹 𝑐𝑣 𝑎𝑥𝑖𝑎𝑙 = 𝑇𝑠ℎ𝑎𝑓𝑡 (3) 𝑈2 = 𝑟2𝜔 𝑈𝜃2 𝑈𝜃2 𝑈𝜃 𝑤2 𝑤2 𝜔 𝑇𝑠ℎ𝑎𝑓𝑡 Control Volume Section (2) Section (1) Figure 2: Sprinkler control volume and velocities (Young, et al. 2012)
- 6. KV It follows from equations 2 & 3 that −𝑟2 × 𝑈𝜃2𝑚 · = 𝑇𝑠ℎ𝑎𝑓𝑡 (4) It is obvious that Tshaft is negative thereby the shaft torque opposes the rotation of the sprinkler. This is true for all turbine devices. The shaft power Ẇshaft associated with the shaft torque Tshaft can be determined 𝑊 · 𝑠ℎ𝑎𝑓𝑡 = 𝑇𝑠ℎ𝑎𝑓𝑡 × 𝜔 = −𝑟2𝑈𝜃2𝑚 · 𝜔 (5) Given that 𝑟2𝑈𝜃2 is the speed, U of each nozzle, eq 5 can be written as 𝑊 · 𝑠ℎ𝑎𝑓𝑡 = −𝑈2𝑈𝜃2𝑚 · or written per unit mass 𝑊 · 𝑠ℎ𝑎𝑓𝑡 = −𝑈2𝑈𝜃2 𝑈2 = 𝑟2𝜔 𝑈𝜃2 𝑈𝜃2 𝑈𝜃 𝑤2 𝑤2 𝜔 𝑇𝑠ℎ𝑎𝑓𝑡 Control Volume Section (2) Section (1) Figure 2: Sprinkler control volume and velocities (Young, et al. 2012)
- 7. KV LAWN SPRINKLER Example Problem Water enters a rotating lawn sprinkler through its base at the steady rate of 1000 ml ・s-1. The exit area of each of the two nozzles is 30 mm2 and the flow leaving each nozzle is in the tangential direction. The radius from the axis of rotation to the centerline of each nozzle is 200 mm. Determine (a) The resisting torque required to hold the sprinkler head stationary (b) The resisting torque associated with the sprinkler rotating with a constant speed of 500 rev・min-1. (c) The speed of the sprinkler if no resisting torque is applied.
- 8. KEITH VAUGH The fluid enters through the stem of the sprinkler and leaves through each of the two nozzles as illustrated. The torque can be determined by 𝑇𝑠ℎ𝑎𝑓𝑡 = −𝑟2𝑈𝜃2𝑚 · where 𝑈𝜃2 = 𝑈2 SOLUTION The mass flow rate can be determined 𝑚 · = 𝑉 · 𝜌 = 1000𝑚𝑙/𝑠 × 0.001𝑚3 /𝑙 × 999𝑘𝑔/𝑚3 1000𝑚𝑙/𝑙 = 0.999𝑘𝑔
- 9. KEITH VAUGH The velocity of the water issuing from the nozzle, relative to the nozzle must be determined. Assuming that the sprinkler head is stationary; 𝜕 𝜕𝑡 0 ∫ 𝜌𝑑𝑉 + ∑𝜌𝑜𝑢𝑡𝐴𝑜𝑢𝑡𝑈𝑜𝑢𝑡 − ∑𝜌𝑖𝑛𝐴𝑖𝑛𝑈𝑖𝑛 = 0 The time rate of changed of the mass of water is zero because the flow is steady and the control volume is filled with water. Because there is only one inlet (subscript 1) and two outlets (subscript 2 and 3), the above equation becomes 𝜌2𝐴2𝑈2 + 𝜌3𝐴3𝑈3 − 𝜌1𝐴1𝑈1 = 0 Given that outlet 2 and 3 are equal and given that the density of the fluid does not change; 2 𝐴2𝑈2 − 𝐴1𝑈1 = 0
- 10. KEITH VAUGH The volumetric flow rate is 𝑉 · = 𝐴𝑈 Then the velocity can be determined for the outlet: 𝑈2 = 𝑉 · 2𝐴2 𝑈2 = 1000𝑚𝑙/𝑠 × 0.001𝑚3 /𝑙 × 106 𝑚𝑚2 /𝑚2 1000𝑚𝑙/𝑙 × 2 × 30𝑚𝑚2 = 16.7𝑚/𝑠 (a) The resistance torque can be determined for the insistence when the sprinkler is not rotating; 𝑇𝑠ℎ𝑎𝑓𝑡 = 200 × 10−3 𝑚 × 16.7𝑚/𝑠 × 0.997𝑘𝑔/𝑠 = −3.34𝑁 − 𝑚
- 11. KEITH VAUGH (b) When the sprinkler rotates at a constant speed, the flow field is unsteady but cyclical. In such cases the flow field can be regarded as being steady in the mean. 𝑈𝜃 = 𝑊2 − 𝑈2 where 𝑊 = 16𝑚/𝑠 and 𝑈2 = 𝑟2𝜔 Applying the axial component of the moment of momentum equation 𝑈𝜃 = 16.7𝑚/𝑠 − 𝑟2𝜔 = 16.7𝑚/𝑠 − 200 × 10−3 𝑚 × 500𝑟𝑒𝑣/𝑚𝑖𝑛 × 2𝜋 60 𝑈𝜃 = 6.2𝑚/𝑠 The torque can now be determined for this case; 𝑇𝑠ℎ𝑎𝑓𝑡 = 200 × 10−3 × 6.2𝑚/𝑠 × 0.999𝑘𝑔/𝑠 = −1.24𝑁 − 𝑚 It can be observed from this result that the resisting torque associated with the head rotating is less that the torque required to hold it stationary.
- 12. KEITH VAUGH (c) When no resisting torque is applied, a maximum constant speed of rotation will occur. 𝑇𝑠ℎ𝑎𝑓𝑡 = −𝑟2 𝑊2 − 𝑟2𝜔 𝑚 · therefore 0 = −𝑟2 𝑊2 − 𝑟2𝜔 𝑚 · When the mass flow rate is constant 𝜔 = 𝑊2 𝑟2 = 16.7𝑚/𝑠 200 × 10−3𝑚 = 83.5𝑟𝑎𝑑/𝑠 𝜔 = 83.5𝑟𝑎𝑑/𝑠 × 60𝑠/𝑚𝑖𝑛 2𝜋𝑟𝑎𝑑/𝑟𝑒𝑣 = 797𝑟𝑝𝑚 The torque can now be determined for this case; 𝑇𝑠ℎ𝑎𝑓𝑡 = 200 × 10−3 × 6.2𝑚/𝑠 × 0.999𝑘𝑔/𝑠 = −1.24𝑁 − 𝑚 It can be observed from this result that the resisting torque associated with the head rotating is less that the torque required to hold it stationary.
- 13. KEITH VAUGH (d) Investigate varying ω between 0 and 1000 rpm and plot the results against the torque. -3.94 -3.38 -2.81 -2.25 -1.69 -1.13 -0.56 0.00 0.56 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Torque N-m Rotational Speed (rpm) Torque Vs. Rotation Speed
- 14. KV ω Runner blade u0 u1→ Absolute velocity vector. Velocity of the flow at the inlet. It’s direction is governed by the guide vane angle α v1→ Velocity of blade at inlet ur1→ The relative velocity vector of the jet at the inlet uf1→ The velocity of flow in the radial direction uw1→The component of the velocity of the jet u1 in horizontal/tangential direction α1→ Guide vane angle, or the angle between the velocity vector of the blade and the direction of the jet β1→ Blade angle at outlet. Angle between the relative velocity vector and the direction of motion of the blade at the inlet
- 15. KV ω Runner blade u0 u2→ Absolute velocity vector water leaving the runner. It’s direction is governed by the guide vane angle α and if it is in the radial direction, will be equal to uf2 v2→ Velocity of blade at exit ur2→ The relative velocity vector of the water leaving the impeller uf2→ The velocity of flow in the radial direction uw2→The component of the velocity for exit u2 in horizontal/tangential direction α2→ Guide vane angle, or the angle between the velocity vector of the blade and the direction of the u2 exiting β2→ Blade angle at outlet, Angle between the relative velocity vector and the direction of motion of the blade at the outlet 𝜶2 uf2 β2 u0
- 16. KV uf1 uf1 u1 uw1 ur1 uf1 u1 uf1 ur1 Velocity triangles Use trigonometry to solve for each vectors uw1 Two triangles within uw1 uw1 uw1 - v1 v1 - uw1
- 17. KV ϑ Opposite Adjacent 𝑠𝑖𝑛𝜃 = 𝑂𝑃𝑃 𝐻𝑌𝑃 𝑐𝑜𝑠𝜃 = 𝐴𝐷𝐽 𝐻𝑌𝑃 𝑡𝑎𝑛𝜃 = 𝑂𝑃𝑃 𝐴𝐷𝐽 Also remember that 𝑡𝑎𝑛𝜃 = 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑐𝑜𝑡𝜃 = 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 𝑠𝑒𝑐𝜃 = 1 𝑐𝑜𝑠𝜃 𝑐𝑠𝑐𝜃 = 1 𝑠𝑖𝑛𝜃 Recall from Trigonometry in Mathematics
- 18. KV 𝑃𝑜𝑤𝑒𝑟 = 𝜌 × 𝑔 × 𝑉 · × 𝐻 𝑘𝑔 𝑚3 × 𝑚 𝑠2 × 𝑚3 𝑠 × 𝑚 𝑘𝑔 𝑚3 × 𝑚5 𝑠3 𝑘𝑔 × 𝑚2 𝑠3 𝑘𝑔 × 𝑚 𝑠2 × 𝑚 𝑠 𝑁 × 𝑚 𝑠 = 𝐽 𝑠 = 𝑊𝑎𝑡𝑡𝑠

- To appreciate energy conversion such as hydro, wave, tidal and wind power a detailed knowledge of fluid mechanics is essential. During the course of this lecture, a brief summary of the basic physical properties of fluids is provided and the conservation laws of mass and energy for an ideal (or inviscid) fluid are derived. The application of the conservation laws to situations of practical interest are also explored to illustrate how useful information about the flow can be derived. Finally, the effect of viscosity on the motion of a fluid around an immersed body (such as a turbine blade) and how the flow determines the forces acting on the body of interest.
- Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinϑ, therefore u0A0 = u1 A1 sinϑ
- Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinϑ, therefore u0A0 = u1 A1 sinϑ
- Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinϑ, therefore u0A0 = u1 A1 sinϑ