Fluid Mechanics II. Newton s second law applied to a control volume


 Jodie Moore
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1 Fluid Mechanics II Stead flow momentum equation Newton s second law applied to a control volume Fluids, either in a static or dnamic motion state, impose forces on immersed bodies and confining boundaries. In the engineering contet these forces are clearl of major relevance and importance for rational design requirements. It is frequentl possible to estimate the stead overall forces applied to either immersed bodies or contained boundaries using basic momentum considerations epressed in Newton s second law. The present section of the course develops the relevant background theor for this simple but powerful and effective methodolog with eamples of its practical application. Newton s second law states that: The net force acting on a bod or sstem of bodies in a particular direction is equal to the rate of change of momentum of the bod or sstem of bodies in that direction. The mathematical formulation of Newton s second law is: F = d I d t, (1) where F is the net force acting on the sstem and I momentum all particles of the sstem: I = m i V i. i is the total linear Note that both F and I are vector quantities and can be represented via the components in a suitable coordinate sstem. For eample, in two dimensions for the and directions we have F = d I d t ; F = d I d t. 1
2 P (pressure) Control volume D (drag force) R (reaction on fluid) τ (shear stresses) Fig. 1: Fig. : The net force acting on the sstem is equal to the sum of all eternal forces acting on the individual particles of the sstem. Internal forces, that is the forces acting between different particles of the sstem, do not contribute to the change of the momentum of the sstem because b Newton s third law for each such a force there is an equal opposite reaction (figure 1). For a fluid it is convenient to appl phsical principles to a sstem of fluid particles occuping a certain suitable volume called a control volume. The force on fluid particles within the control volume include the net force acting on the bounding surface of the control volume, forces on solid surfaces of immersed bodies and gravit or inertial forces. Main applications of the momentum principle include evaluation of forces on immersed objects produced b a flowing stream. B Newton s third law there must be an equal and opposite reaction on the fluid. We wish to appl the law to a flowing fluid in order to deduce the magnitude of the force F which must have acted on the fluid within the control volume in order to produce the observed rate of change of momentum. The force F will then be the total force on the fluid inside the control volume arising from all sources which will include not onl the reaction from the bod drag D, but also due to an pressure or shear forces over the bounding surface of of the control volume (figure ) plus an weight components. Let us consider a stream tube of a small cross section such that the fluid velocit can be assumed constant across the tube. It is convenient to chose a part of the stream tube between two cross section 1 and as a control volume (region abcd on figure 3). The areas of the cross section 1 and are A 1, and the corresponding fluid velocit vectors are V 1,. During a small time interval t the fluid in the region a add enters the control volume, and the corresponding mass gain is m 1 = ρ A 1 V 1 t.
3 3 I I a c V 1 c 1 A 1 d V 1 t a d b b V A V t Fig. 3: Similarl, the fluid in the region bcc b eits the control volume, and the mass loss is m = ρ A V t. According to the continuit principle, the mass flow rates across the surfaces 1 and are equal to each other, and we can write m 1 = m = ρ Q t, where Q is the volumetric flow rate through the stream tube. Note that V 1, are the velocit components normal to the surfaces 1,, and for the case when the velocit vectors are perpendicular to that surfaces (as on figure 3) V 1, are the absolute values of the corresponding velocit vectors: V 1, = V 1,. The momenta associated with masses m 1, are I 1, = ρ Q V 1, t. Note the vector values in both sides of the equation. Therefore, during the time period t the control volume gains the momentum I 1 = ρ Q V 1 t due to the fluid flow into the control volume trough the cross section 1, and the corresponding loss of the momentum due to the flow out of the control volume trough the cross section is I = ρ Q V t. For a stead flow the momentum of the fluid inside the control volume does not change: I = 0. This implies that certain force should act on the fluid inside the control volume to compensate the change of momentum due to the fluid flow in and out of the control volume. According to Newton s second law (1) the change of momentum caused b an action of such a force during the time period t
4 4 is: I F = F t, and the overall balance of the change of momentum in the control volume is I = ρ Q V 1 t ρ Q V t + F t = 0. From this equation the total force on the fluid in the control volume is given b F = ρ Q V ρ Q V 1 = ρ Q ( V V 1 ). () Verball, equation () sas: Momentum principle: For a stead flow the total force on the fluid in the control volume is equal to the momentum flow rate out of the control volume minus the momentum flow rate into the control volume. Equation () is written in a vector form. Equation for an specific direction can be obtained from () b taking the projection on this direction. For eample, in the case of 3 dimensions we have 3 coordinate ais (,, z). The corresponding components of force and velocit vectors are (F, F, F z ) and (u, v, w) respectivel. The vector equation () is then equivalent to three equations in each of three coordinate directions: F = ρ Q (u u 1 ); F = ρ Q (v v 1 ); F z = ρ Q (w w 1 ). For a collection of stream tubes forming a larger control volume with perhaps nonuniform velocit distributions across the inlet and outlet boundaries we ma integrate equation () to obtain the general stead flow momentum equation F = ρ V Vn da ρ V Vn da, S where the integrals are taken over the inlet and outlet surfaces S 1 and S, and V n is the component of velocit normal to an area element da of the corresponding integration surface. S 1 Applications of stead flow momentum equation General approach 1. Draw the flow sstem.. Make and justif necessar assumptions.
5 5 3. Chose a coordinate sstem and a suitable control volume. 4. Define velocities on the boundaries of the control volume and their components in the chosen sstem of coordinates. 5. Define and label forces on bodies, ducts, etc. and their reactions to the fluid. 6. Assess total force on fluid in control volume including an drag of internal immersed bodies, boundar pressure and shear forces, gravit forces. Consider components of the total force in the coordinate sstem. 7. Assess momentum flows into and out control volume and appl the momentum principle in chosen directions. Assumptions The following assumptions are used in this section, if not stated otherwise: 1. Flow is twodimensional.. Flow is frictionless. 3. Gravit effects are neglected. 4. Flow in ducts and jets are assumed parallel and uniform with a constant velocit over the cross section. Nomenclature Selected notations used in this section:, Coordinates. V Velocit vector. V Absolute value of velocit. u, v Velocit components along aes and respectivel. F, F Components of the net force acting along aes and on fluid inside a control volume. X, Y Force components acting along aes and on solid surfaces (bodies, ducts, etc.)
6 6 V A V1 U F V3 Impact of jets on surfaces Normal plate Formulation: Fig. 4: A jet of water of cross section A and velocit V 1 strikes a flat plate perpendicular to the direction of the jet and moving with a constant velocit U < V 1 in the jet direction (figure 4). Find the force applied b the jet to the plate. Note: Stead flow momentum principle can be applied not onl to a stationar control volume but also to a control volume moving with a constant velocit. In this case fluid motion is considered in the coordinate sstem moving together with the control volume. Solution: We chose a control volume and coordinate aes (, ) moving together with the plate (figure 4). The pressure in the jet is equal to an atmospheric pressure. Therefore, the pressure on the boundar of the control volume is constant and does not make a contribution into the total force acting on the control volume, which means that in this particular case the net force on the fluid is equal to the reaction from the wall (F = X; F = Y ). It is useful to note that the flow has a horizontal smmetr ais.
7 7 V θ A V1 F U F V3 Fig. 5: Flow rate through the moving boundar: Q = A (V 1 U) Flow smmetr: V = V 3 ; Q = Q 3 Velocit components in the moving coordinate sstem: Momentum principle: direction: u 1 = V 1 U; u = u 3 = 0 direction: v 1 = 0; v = v 3 direction: F = ρ Q u + ρ Q 3 u 3 ρ Q u 1 = ρ Q (V 1 U) direction: F = ρ Q v + ρ Q 3 v 3 ρ Q v 1 = 0 Force components acting on the plate: X = F = ρ A (V 1 U) ; Y = F = 0. The walue of X is positive, which means that the force acts in the positive direction (from left to right).
8 8 A F F V U V θ Fig. 6: Inclined plate Formulation: The jet has an angle θ to the plate normal. The plane is moving with velocit U < V 1 in the jet direction (figure 5). Find the force applied b the jet to the plate. Solution: It is convenient to consider aes oriented as shown. Assuming no friction, no forces can be applied to the fluid b the plane in the direction: F = 0. Flow rate: Q = A (V 1 U) Velocit components in the direction: u 1 = (V 1 U) cos(θ); u = u 3 = 0 Momentum principle in the direction: F = ρ Q u + ρ Q 3 u 3 ρ Q u 1 = ρ Q (V 1 U) cos(θ) Force components acting on the plate: Curved vane Formulation: X = F = ρ A (V 1 U) cos(θ); Y = F = 0. A curved vane deflects a jet of water of diameter A through angle θ. (figure 6). The mean velocit of water across the jet is V 1. The vane is moving in the jet direction Assuming that on inlet the jet is tangential to the vane surface
9 A F A P V1 P V Fig. 7: find the force eerted b the jet on the vane. Solution: Flow rate: Q = A (V 1 U) For frictionless flow the magnitudes of jet velocities relative to the vane are the same at inlet and outlet. This gives the following velocit components: Momentum principle: direction: u 1 = V 1 U; u = (V 1 U) cos(θ) direction: v 1 = 0; v = (V 1 U) sin(θ) direction: F = ρ Q (u u 1 ) = ρ Q (V 1 U)( cos(θ) 1) direction: Force components acting on the plate: F = ρ Q (v v 1 ) = ρ Q (V 1 U) sin(θ) X = F = ρ A (1 cos(θ))(v 1 U) ; Y = F = ρ A sin(θ)(v 1 U). Flows in contractions and nozzles Straight contraction Formulation: Water flows through a pipe with a straight contraction which cross section graduall redices from A 1 to A. The mean flow velocit at the inlet of the contraction is V 1 and pressure here is P 1. Determine the force eerted b the
10 A P V1 F A P = 0 V Fig. 8: water on the contraction Solution: Continuit: Q = A 1 V 1 = A V V = A 1 V 1 /A. Assuming frictionless uniform flow we can relate pressures at inlet and outlet b Bernoulli equation: P = P ρ V 1 1 ρ V. Pressures at inlet and outlet are different, which means the there is pressure contribution into a net force on the control volume. Therefore the total force on the fluid in the control volume includes the pressure component and the reaction from the contraction. Momentum principle: Force on the contraction: F = X + (P 1 A 1 P A ) = ρ Q (V V 1 ). X = ρ Q (V V 1 ) + (P 1 A 1 P A ). Nozzle Formulation: A pipe with cross section area A 1 terminates in a conical nozzle from which a jet of water is discharged to atmosphere. The cross section of the jet is A and the gauge pressure before the nozzle is P 1. Find the total force acting
11 11 on the nozzle. Solution: Pressure across the jet is constant and equal to the atmospheric pressure (gauge pressure is 0). Therefore, the gauge pressure on the right boundar of the control volume is constant and equal to the atmospheric pressure: P = 0. Continuit: Bernoulli: A 1 V 1 = A V = Q (3) P ρ V 1 = P + 1 ρ V. (4) With known P 1 and P = 0 simultaneous solution of (3) and (4) gives values of V 1 and V. The total force on the control volume includes the pressure component and the reaction from the nozzle. Momentum principle: Force on the nozzle: F = X + P 1 A 1 = ρ Q (V V 1 ). X = ρ Q (V V 1 ) + P 1 A 1. Pipe bend Formulation: A pipe with a flowing water bends through an angle θ and reduces cross section from A 1 to A. The pressure and velocit before the bend are P 1 and A 1 respectivel. Find the magnitude and direction of the force eerted on the bend b the water. Solution: Continuit: Q = A 1 V 1 = A V V = A 1 V 1 /A. Bernoulli: P = P ρ V 1 1 ρ V. Velocit components: direction: u 1 = V 1 ; u = V cos(θ) direction: v 1 = 0; v = V sin(θ)
12 1 A V θ F P A P 1 1 V1 F Fig. 9: Total force components on the control volume: direction: F = X + P 1 A 1 P A cos(θ) direction: F = Y P A sin(θ) Momentum principle: direction: F = ρ Q (u u 1 ) direction: F = ρ Q (v v 1 ) Force components on the bend: direction: X = P 1 A 1 P A cos(θ) ρ Q (V cos(θ) V 1 ) direction: Y = P A sin(θ) ρ Q V sin(θ) Force magnitude: R = X + Y Force direction (angle from ais): α = sin 1 (Y/R) Bod drag Formulation: A clindrical model is placed across a working section of a wind tunnel of height 4d (figure 10). Air flow at the inlet of the working section can be assumed parallel and uniform with velocit V 1. At the outlet of the working section, where the flow is also parallel, the velocit profile is measured. It consists of a viscous wake of thickness d with profile V w = V ( 1 + (/d) (/d) 4 )/,
13 13 V 1 V =d =d P 1 D V wake =V() 1 = d = d Fig. 10: and a uniform flow outside of the wake with velocit V. Assuming the flow to be twodimensional, find the drag per one meter of span of the model. Solution: Continuit: Flow rates through inlet and outlet are the same: Q = 4 d V 1 = d V + V d ( 1 + (/d) (/d) 4 )/ d. After taking the integral we find the value of V : V = V 1. d Bernoulli: We can appl Bernoulli equation in the inviscid region outside of the viscous wake to specif the pressures at the outlet. The flow here is parallel, and the pressure will be the same across the whole section. For the stream line 1 we have: P 1 + ρ V 1 = P + ρ V, and substituting the known values of velocities we have the following pressure difference: P 1 P = 8 ρ V Momentum: We neglect friction on the walls of the wind tunnel. Then the total force on the fluid inside the control volume includes the pressure forces at the inlet and outlet of the control volume and the reaction due to drag of the model R = D. The flow at the outlet is not uniform, therefore, we have to appl
14 14 the momentum principle in the general integral form. Momentum flow rate at inlet: I 1 = ρ d d V 1 d = 4d ρ V 1. At the outlet the integral for momentum flow rate consists of tree parts: I = ρ d d d V d + ρ d V ( 1 + (/d) (/d) 4 ) /4 d + ρ d d V d, After taking the integrals and substituting the value for V this gives gives: I = d ρ V 1. Now we can appl the momentum principle: and substituting all values we obtain: F = 4 d (P 1 P ) D = I I 1, D = d ρ V 1
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