1. 1 MECH 221 FLUID MECHANICS (Fall 06/07) Tutorial 5

2. 2 Outline 1. Equations of motion for inviscid flow 1. Conservation of mass 2. Conservation of momentum 2. Bernoulli Equation 1. Bernoulli equation for steady flow 2. Static, dynamic, stagnation and total pressure 3. Example

3. 3 1. Equations of Motion for Inviscid Flow  Conservation of Mass  Conservation of Momentum

4. 4 1.1. Conservation of Mass  Mass in fluid flows must conserve. The total mass in V(t) is given by:  Therefore, the conservation of mass requires that dm/dt = 0. where the Leibniz rule was invoked.

5. 5 1.1. Conservation of Mass  Hence: This is the Integral Form of mass conservation equation.

6. 6 1.1. Conservation of Mass  As V(t) → 0, the integrand is independent of V(t) and therefore, This is the Differential Form of mass conservation and also called as continuity equation.

7. 7 1.2. Conservation of Momentum  The Newton’s second law, is Lagrangian in a description of momentum conservation. For motion of fluid particles that have no rotation, the flow is termed irrotational. An irrotational flow does not subject to shear force, i.e., pressure force only. Because the shear force is only caused by fluid viscosity, the irrotational flow is also called as “inviscid” flow

8. 8 1.2. Conservation of Momentum  For fluid subjecting to earth gravitational acceleration, the net force on fluids in the control volume V enclosed by a control surface S is: where s is out-normal to S from V and the divergence theorem is applied for the second equality.  This force applied on the fluid body will leads to the acceleration which is described as the rate of change in momentum. Pressure force Body force

9. 9 1.2. Conservation of Momentum where the Leibniz rule was invoked.

10. 10 1.2. Conservation of Momentum  Hence: This is the Integral Form of momentum conservation equation.

11. 11 1.2. Conservation of Momentum  As V → 0, the integrands are independent of V. Therefore, This is the Differential Form of momentum conservation equation for inviscid flows.

12. 12 1.2. Conservation of Momentum  By invoking the continuity equation,  The momentum equation can take the following alternative form: which is commonly referred to as Euler’s equation of motion.

13. 13 2.1. Bernoulli Equation for Steady Flows  From differential form of the momentum conservation equation 1. g=-g V z 2. By vector identity,  Therefore, we get,

14. 14 2.1. Bernoulli Equation for Steady Flows  Assumption, Steady flow;  v and t are independent Irrotational flow;  V xv=0 =0 (Steady flow)=0 (irrotational flow)

15. 15 2.1. Bernoulli Equation for Steady Flows  Finally, we can get, Or where v=magnitude of velocity vector v, i.e. v=√(u 2 +v 2 +w 2 )

16. 16 2.1. Bernoulli Equation for Steady Flows  Since, for dr in any direction, we have:  For anywhere of irrotational fluids  For anywhere of incompressible fluids

17. 17 2.1. Bernoulli Equation for Steady Flows  Bernoulli Equation in different form: 1. Energy density: 2. Total head (H):

18. 18 2.2. Static, Dynamic, Stagnation and Total Pressure  Consider the Bernoulli equation,  The static pressure p s is defined as the pressure associated with the gravitational force when the fluid is not in motion. If the atmospheric pressure is used as the reference for a gage pressure at z=0. (for incompressible fluid)

19. 19  Then we have as also from chapter 2.  The dynamic pressure p d is then the pressure deviates from the static pressure, i.e., p = p d +p s. The substitution of p = p d +p s. into the Bernoulli equation gives 2.2. Static, Dynamic, Stagnation and Total Pressure

20. 20  The maximum dynamic pressure occurs at the stagnation point where v=0 and this maximum pressure is called as the stagnation pressure p 0. Hence,  The total pressure p T is then the sum of the stagnation pressure and the static pressure, i.e., p T = p 0 - ρgz. For z = -h, the static pressure is ρgh and the total pressure is p 0 + ρgh. 2.2. Static, Dynamic, Stagnation and Total Pressure

21. 21 2.3.1. Example (1)  Determine the flowrate through the pipe.

22. 22 2.3.1. Example (1)  Procedure: Choose the reference point From the Bernoulli equation  P, V, Z all are unknowns  For same horizontal level, Z 1 =Z 2  V = V(P 1, P 2 ) From the balance of static pressure  P = ρgh  Δh is given, ρ m, ρ water are known V = V(Δh, ρ m, ρ water ) Q = AV = π D 2 V/4

23. 23 2.3.1. Example (1)  From the Bernoulli equation,

24. 24 2.3.1. Example (1)  From the balance of static pressure,

25. 25 2.3.1. Example (1)  Volume flow rate (Q),

26. 26 2.3.2. Example (2)  A conical plug is used to regulate the air flow from the pipe. The air leaves the edge of the cone with a uniform thickness of 0.02m. If viscous effects are negligible and the flowrate is 0.05m 3 /s, determine the pressure within the pipe.

27. 27 2.3.2. Example (2)  Procedure: Choose the reference point From the Bernoulli equation  P, V, Z all are unknowns  For same horizontal level, Z 1 =Z 2 Flowrate conservation  Q=AV

28. 28 2.3.2. Example (2)  From the Bernoulli equation,

29. 29 2.3.2. Example (2)  From flowrate conservation,

30. 30 2.3.2. Example (2)  Sub. into the Bernoulli equation,

31. 31 The End