ME 321: FLUID MECHANICS-I

Transcription

1 6/07/08 ME 3: LUID MECHANI-I Dr. A.B.M. Toufique Hasan Professor Department of Mechanical Engineering Bangladesh Universit of Engineering & Technolog (BUET), Dhaka Lecture- 4/07/08 Momentum Principle teacher.buet.ac.bd/toufiquehasan/ Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) ecall TT TT (enolds Transport Theorem) relates between the sstem approach with finite control volume (C) approach for a sstem propert: DB Dt ss t C ρ ρb d b nˆ da B = an etensive propert (such as mass, momentum, energ etc.) b = an intensive propert per unit mass (such as mass per mass, momentum per mass, etc.) Conservation of linear momentum (L-05) M(momentum ) B M (momentum), b M (mass) TT takes the form of D M Dt ss t C d nˆ da Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08)

2 6/07/08 ecall Conservation of linear Momentum Newton s second law of motion for a sstem is Time rate of change of the linear momentum of the sstem D M Dt ss ss = Sum of eternal forces acting on the sstem contents of the control volume Since when a control volume coincident with a sstem at an instant of time, the forces acting on the sstem and the forces acting on the contents of the coincident control volume are instantaneousl identical. or a control volume (C) which is fied and nondeforming, the Newton s second law of motion takes the following form: contents of the ρ d nˆ da control volume t C or fluid dnamics: orce contents of the Time rate of change of the control volume (C) = linear momentum of the + contents of the control volume (C) Net rate of linear momentum through the control surface () ρ d n contents of the s B control volume t C ˆ da ** ector equation Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 3 Momentum Principle The above epression could be simplified considerabl if a flow sstem has onl one entrance and one eit and if the flow is stead: C C C ρ d t C nˆ da Using continuit: Then: A A m A A C m nˆ da nˆ (outflow ) nˆ (inflow ). Note that the momentum equation is a vector equation which represents three scalar equations: : : z : z m m m z z C. This is the fundamental principle which drives the turbomachiner (propulsion nozzle in jet engine, turbine, compressor cascade etc.) Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 4

3 6/07/08 Problem #: Water from a stationar nozzle strikes a flat plate as shown. The water leaves the nozzle at 5 m/s; the nozzle area is 0.0 m. Assuming water is directed normal to the plate, and flows along the plate, determine the horizontal force ou need to resist to hold it in place. 5m/s A 0.0m Solution: Consider an appropriate C and the equation of momentum for this C is: C nˆ da : m m 0.5 kn A 5m/s A 0.0m This.5 kn force will act on hand as reaction force in the right hand side m 4 3 m m cv Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 5 Problem #: Water flows steadil through the 90 reducing elbow shown in fig. At the inlet of the elbow, the absolute pressure is 0 kpa and the cross-sectional area is 0.0 m. At the outlet, the cross-sectional area is m and the velocit is 6 m/s. The elbow discharges to the atmosphere. Determine the force required to hold the elbow in place. Solution: from stead flow momentum principle: nˆ da C : m p A m p m 0 A A A p 0kPa abs 9 kpa gage cv Now, from continuit equation: Q Q A (0.0) A (0.005)(6) 4 m/s p 0kPa abs 0 kpa gage Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 6 3

4 6/07/08 Problem #: Then p Now, along -ais: 3 (9 0 )(0.0) (4).35 kn A A m m 0 A ; ve : m ( )(6) 0.64 kn So, the magnitude of resultant force is kn - tan 5.37 with ve ais p 0kPa abs 9 kpa gage p 0kPa abs θ 0 kpa gage cv Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 7 Problem #3: Determine the magnitude and direction of the anchoring force needed to hold the horizontal elbow and nozzle combination shown in figure in place. Atmospheric pressure is 00 kpa (abs) The gage pressure at section () is 00 kpa. At section (), the water eits to the atmosphere. Solution: from stead flow momentum principle; nˆ da : C m p A m p A A ; ve A Now, from continuit equation: Q Q A A d 4 ( ) d 4 d d 7.03 m/s ( ) Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 8 4

5 6/07/08 Problem #3: Then, : p A p A A A m 3 (00 0 ) kn Now, along -ais: m : m So, the magnitude of resultant anchoring force is kN - tan 0 with ve ais Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 9 Problem #4: A deflector turns a sheet of water through an angle of 30 as shown in ig. What force components are necessar to hold the deflector in place if mass flow rate is 3 kg/s? Solution: m 3 A ( ) 40 m/s : m 3(40 cos 30 40) 7 N : m 3(40 sin 30 0) 640 N Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 0 5

6 6/07/08 Problem # 5 A converging elbow (shown in figure) turns water through an angle of 35 in a vertical plane. The flow cross-section diameter is 400 mm at the elbow inlet, section (a), and 00 mm at the elbow outlet, section (). The water volume flowrate is 0.4 m 3 /s and the elbow inlet and outlet pressures are 50 kpa and 90 kpa. Calculate the horizontal (-direction) and vertical (z-direction) anchoring forces required to hold the elbow in place. (Neglect bod force) Dr. A.B.M. Toufique Hasan (BUET) L-3 T-, Dept. of ME ME 3: luid Mechanics-I (Jan. 08) 6