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Elementary Mechanics of Fluids

Elementary Mechanics of Fluids. CE 319 F Daene McKinney. Bernoulli Equation. Euler Equation. Fluid element accelerating in l direction & acted on by pressure and weight forces only (no friction) Newton’s 2 nd Law. Flow. l. 30 o. Ex (5.1).

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Elementary Mechanics of Fluids

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  1. Elementary Mechanics of Fluids CE 319 F Daene McKinney Bernoulli Equation

  2. Euler Equation • Fluid element accelerating in l direction & acted on by pressure and weight forces only (no friction) • Newton’s 2nd Law

  3. Flow l 30o Ex (5.1) • Given: Steady flow. Liquid is decelerating at a rate of 0.3g. • Find: Pressure gradient in flow direction in terms of specific weight.

  4. vertical A 1 m B EX (5.3) • Given: g = 10 kN/m3, pB-pA=12 kPa. • Find: Direction of fluid acceleration.

  5. HW (5.7) • Ex (5.6) What pressure is needed to accelerate water in a horizontal pipe at a rate of 6 m/s2?

  6. Ex (5.10) V1/2=(80+30)/2 ft/s = 55 ft/s • Given: Steady flow. Velocity varies linearly with distance through the nozzle. • Find: Pressure gradient ½-way through the nozzle dV/dx =(80-30) ft/s /1 ft = 50 ft/s/ft

  7. HW (5.11)

  8. Bernoulli Equation • Consider steady flow along streamline • s is along streamline, and t is tangent to streamline

  9. Ex (5.47) Point 1 • Given: Velocity in outlet pipe from reservoir is 6 m/s and h = 15 m. • Find: Pressure at A. • Solution: Bernoulli equation Point A

  10. Point 1 Point A Example • Given: D=30 in, d=1 in, h=4 ft • Find: VA • Solution: Bernoulli equation

  11. D D d 2 1 3 Example – Venturi Tube • Given: Water 20oC, V1=2 m/s, p1=50 kPa, D=6 cm, d=3 cm • Find: p2 and p3 • Solution: Continuity Eq. • Bernoulli Eq. Nozzle: velocity increases, pressure decreases Diffuser: velocity decreases, pressure increases Similarly for 2  3, or 1  3 Pressure drop is fully recovered, since we assumed no frictional losses Knowing the pressure drop 1  2 and d/D, we can calculate the velocity and flow rate

  12. Ex (5.48) • Given: Velocity in circular duct = 100 ft/s, air density = 0.075 lbm/ft3. • Find: Pressure change between circular and square section. • Solution: Continuity equation • Bernoulli equation Air conditioning (~ 60 oF)

  13. Ex (5.49) • Given: r = 0.0644 lbm/ft3V1= 100 ft/s, and A2/A1=0.5, gm=120 lbf/ft3 • Find: Dh • Solution: Continuity equation • Bernoulli equation Heating (~ 170 oF) • Manometer equation

  14. HW (5.51)

  15. Stagnation Tube

  16. Pipe 2 Flow 1 Stagnation Tube in a Pipe

  17. Pitot Tube

  18. V 1 z1-z2 2 l y Pitot Tube Application (p.170)

  19. HW (5.69)

  20. HW (5.75)

  21. HW (5.84)

  22. HW (5.93)

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