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IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Momentum Conservation n Newton’s Second Law u Force = Mass * Acceleration u Alternate method: From Reynold’s.

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Presentation on theme: "IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Momentum Conservation n Newton’s Second Law u Force = Mass * Acceleration u Alternate method: From Reynold’s."— Presentation transcript:

1 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Momentum Conservation n Newton’s Second Law u Force = Mass * Acceleration u Alternate method: From Reynold’s theorem n Fluid Flow u Force = Momentum flux + Momentum Accumulation rate Flux Accumulation Rate

2 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example n Straight Pipe A1V1A1V1 A2V2A2V2 n Steady State 0 n Assumption: No friction

3 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example n Flow in a straight pipe. Realistic case 1 2

4 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example n Problem: 5.32 d 2,V 2 n Steady State 0 h=58 cm d 1,V 1 d 1 =8cm d 2 =5cmV 1 =5m/s  1 =1g/cm 3 12

5 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example n Example 1 (bent pipe), page 47  1,V 1  2,V 2 Area=A 1 Area=A 2 =A 1 BxBx ByBy  Weight of Fluid x y

6 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example n Steady State u d/dt =0 n From Eqn of Conservation of Mass outinout in

7 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example V1 V2 Area=A1Area=A2=A1 BxBx ByBy 

8 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example P1,V1 P2,V2 F n Pipe with U turn Use Gage Pressure! In case of gas, use absolute pressure to calculate density

9 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example P1 P2 F n “L” bend Assume the force by the pipe on the fluid is in the positive direction E N What will the force be, if the flow is reversed (a) in a straight pipe? (b) in a L bend?

10 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Pbm & 5.19 VwVw V2V2 V1V1 11 22 XY Conservation of Mass P2P2 P1P1 Stationary CV

11 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Pbm VwVw V2V2 V1V1 11 22 XY P1P1 P2P2 Stationary CV V 1 =-3 m/s Vw=Velocity of Sound in water V 2 =0 m/s Pressure difference? ~ 41 atm

12 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Pbm. 5.1,5,3 V 1 =6 m/s V 2 =? 0.6 m1.2 m x y

13 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Momentum Conservation Angular Momentum In a moving system Torque = Angular Momentum flux + Angular Momentum Accumulation rate Flux Accumulation Rate

14 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Example 4 in book Find the torque on the shaft In a moving system Torque = Angular Momentum flux + Angular Momentum Accumulation rate 

15 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Approach-1. Find effective Force in X direction Find the moment of Force Assume: No frictional loss, ignore gravity, steady state, atmospheric pressure everywhere 

16 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Approach-2. Using conservation of angular momentum Stationary CV

17 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Consider a jet hitting a moving plate V noz water has entered into the CV Plate has moved by V plate In a control volume which moves with the plate, V noz -V plate water has entered the CV (and exited at the bottom) After 1 second

18 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example

19 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example n Consider plate half Vnoz, alpha = 180 degrees

20 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Pbm 5.24 Thickness of slit =t, vol flow rate =Q, dia of pipe=d, density given Ignore gravity effects Flux Accumulation Rate 3ft 6ft

21 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Pbm 5.31 P1, P2, density, dia, vol flow rate given 3ft 1 2 Calculate velocity at 1 (=2)

22 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Energy Conservation Heat Mechanical Work done by the system Work done by pressure force Friction Loss (Viscous)

23 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Energy Conservation No Frictional losses Incompressible Steady No heat, work No internal energy change

24 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Flow from a tank Dia = d1 Dia = d2 Pressure = atm at the top and at the outlet Velocity at 1 ~ 0 h1 h3 0 n Toricelli’s Law Sections 2 and 3

25 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example How long does it take to empty the tank? What if you had a pipe all the way upto level 3? Dia = d1 Dia = d h1 h3 0 section 2 != atm section 3 = atm

26 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example What if you had a pipe all the way upto level 3? Dia = d1 Dia = d h1 h3 0 More flow with the pipe Turbulence, friction Unsteady flow Vortex formation

27 IIT-Madras, Momentum Transfer: July 2005-Dec km/h Example Moving reference; Aircraft 60 km/h Height is known Find P and r (eg from tables) 231 Flight as Reference

28 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Flow through a siphon vs constriction h1 h2 h3 1 2

29 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Pbm. 6.4 Steady flow through pipe, with friction Friction loss head = 10 psi Area, vol flow rate given Find temp increase Assume no heat transfer

30 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Pbm Fluid entering from bottom, exiting at radial direction Steady, no friction P1=10 psig P2=atm t D2 D1 Find Q, F on the top plate h2

31 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example P1=10 psig D1 F y If the velocity distribution just below the top plate is known, then P can be found using Bernoulli’s eqn

32 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Modifications to Eqn Unsteady state, for points 1 and 2 along a stream line

33 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Draining of a tank We can obtain the time it takes to drain a tank (i) Assume no friction in the drain pipe (ii) Assume you know the relationship between friction and velocity L H R D 1 2 Le us take that the bottom location is 2 and the top fluid surface is 1 Incompressible fluid

34 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Draining of a tank: Quasi steady state Quasi steady state assumption Velocity at fluid surface at 1 is very small i.e. R >> D No friction : L is negligible P1 = P2 = Patm

35 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Draining of a tank: Quasi steady state Original level of liquid is at H = H 0 Integrating above equation from t=0, H=H 0 to t=t final, H =0, we can find the efflux time

36 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Draining of a tank: Unsteady state BSL eg At any point of time, the kinetic + potential energy of the fluid in tank is converted into kinetic energy of the outgoing fluid We still neglect friction Potential Energy of a disk at height z and thickness dz

37 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Draining of a tank: Unsteady state Also, using continuity equation Substituting, you get a 2 nd order non linear ODE with two initial conditions. Please refer to BSL for solution

38 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Draining of a tank (accounting for friction) What if the flow in the tube is laminar and you want to account for friction? Bernoulli’s eqn is not used (friction present) Continuity L H R D 1 2 Hagen-Poiseuille’s eqn

39 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Draining of a tank (accounting for friction) Substituting and re arranging, Integrating with limits Note: The answer is given in terms of diameter of tube, so that it is easier to compare with the answer given in the book

40 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example A1,A2, initial height h1 known A1 >> A2 1 2 L h1 3 Consider section 3 and 2 Pseudo Steady state ==> Toricelli’s law

41 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example Rearranging and solving, we get As t increases, the solution approaches the Toricelli’s equation

42 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Appendix:Example Oscillating fluid in a U-tube L3 h1 h2 1 2 Let h=h1-h2

43 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Appendix:Example Blood Flow in vessels Minimization of ‘work’ Murray’s Law: Laminar Flow, negligible friction loss (other than that due to viscous loss in laminar flow), steady Turbulent, pulsating flow Assume

44 IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Appendix:Example If the ratio of ‘smaller’ to larger capillary is constant And Metabolic requirement =m= power/volume Work for maintaining blood vessel Total work Optimum radius


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