1 The Bernoulli Equation The most used and the most abused equation in fluid mechanics. Newton s Second Law: F = ma In general, most real flows are 3-D, unsteady (x, y, z, t; r,θ, z, t; etc) Let consider a -D motion of flow along streamlines, as shown below. Velocity (V ): Time rate of change of the position of the particle. Streamlines: The lines that are tangent to the velocity vectors throughout the flow field. Note: For steady flows, each particle slide along its path, and its velocity vector is everywhere tangent to the path. Streamline coordinate: v S = S(t) ; V = ds / dt (the distance along the streamline can be decided by V and R(s) )
2 By chain rule, a s = dv = V ds = V V ; a dt S dt S where R is the local radius of curvature of the streamline, and S is the distance measured along the streamline from some arbitrary initial point. n = V R Figure 3. Isolation of small fluid particle in a flow field. V 0; S R (not straight line) Forces: Gravity & Pressure (important) v F = ma v Viscous, Surface tension (negligible) along a streamline
3 Figure 3.3 Free-body diagram of a fluid particle for which the important forces are those due to pressure and gravity.. Consider the small fluid particle (δ S δ n ), as show above. If the flow is steady. N. δ F S = δ m a S = δ mv V S = ρ δ V V S (3.) (Eq. (3.) is valid for both compressible and incompressible fluids) W. δ W S = δ W sin θ = γ δ sin θ (θ = 0 δ W S = 0)
4 δ P P. S P δ S S (first term of Taylor series expansion, because the particle is small; P + δ P S P δ P S ) Thus δ F P S δ F PS : the net pressure force on the particle in the streamline direction = (P δ P S )δ nδ y (P + δ P S )δ nδ y = δ P δ nδ y = P δ sδ nδ y S S = P δ S *Note: if pressure gradient is not zero ( P C ), then there is a net pressure force. P = P S v + S P n v n Net Force δ F S = δ W S + δ F PS ; ρ V V S = γ sin θ P S (3.4) Note: Force balancing consideration ρ V (mass flux), not ρ or V, is a key parameter for fluid mechanics, if ρ constant.
5 Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A-B in front of the sphere of radius a as shown in Fig. E3.1a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is
6 3 V = V (1 + a ) 0 x 3 Determinate the pressure variation along the streamline from point A far in front the sphere (x A = and V A = V 0 ) to point B on the sphere ( x B = a and V B = 0)
7 Figure E3.1 Solution: Since the flow is steady and inviscid, Eq. 3.4 is valid. In addition, since the streamline is horizontal, sin θ = sin 0 = 0 and the equation of motion along the streamline reduces to p = ρ V V s s (1) with the given velocity variation along the streamline, the acceleration term is V V s = V V x = V 0 (1 + 3 a 3 x 3 a a = 3V 0 (1 + ) x 3 x 4 3 )( 3V 0a 3 ) x
8 where we have replaced s by x since the two coordinates are
9 identical (within an additive constant) along streamline A-B. It follows that V V s < along the streamline. The fluid slows 0 down from V 0 far ahead of the sphere to zero velocity on the nose of the sphere (x = a). Thus according to Eq. 1, to produce the given motion the pressure gradient along the streamline is p 3 a 3 3ρ a V 0 (1 + x 3 ) x 4 () = x This variation is indicated in Fig. E3. 1b. It is seen that the pressure increase in the direction of flow ( p x > from point A to point B. The maximum pressure gradient 0) (0.610 ρ V a) 0 occurs just slightly ahead of the sphere (x = 1.05a). It is the pressure gradient that slows the fluid down from V A = V 0 to V B = 0. The pressure distribution along the streamline can be obtained by integrating Eq. from p = 0 (gage) at x = to pressure p at location x. The result, plotted in Fig. E3.1c, is a (a x) 6 3
10 p = ρ V ( ) + 0 x (Ans) The pressure at B, a stagnation point since V B = 0, is the
11 highest pressure along the streamline ( p B = ρ V 0 ). As shown in Chapter 9, this excess pressure on the front of the sphere (i.e. p B > 0 ) contributes to the net drag force on the sphere. Note that the pressure gradient and pressure are directly proportional to the density of the fluid, a representation of the fact that the fluid inertia is proportional to its mass. Since sin θ = dz dv 1 d(v ) ds, also V = ds ds streamline, n = const, so dn = 0., and along the Thus dp = p ds + s p dn = n dp ds ds Eq. (3.4) γ dz dp = 1 ρ d(v ) ds ds ds along a streamline (s) dp + 1 ρ d(v ) + γ dz = 0 dp 1 or + V + gz = const ρ (3.5) (3.6) Note: if ρ constant, then ρ = f ( p)
12 must be known. The well-known Bernoulli equation:
13 P + 1 ρ V + γ Z = constant Note: 4 assumptions 1) μ = 0 (inviscid) ) = 0 t 3) ρ = C (steady) (incompressible) 4) along a streamline (dn = 0 ) Example 3. Consider the flow of air around a bicyclist moving through still air with velocity V 0 as is shown in Fig. E3.. Determine the difference in the pressure between points (1) and (). Figure E 3. Solution: In a coordinate system fixed to the bike, it appears as through the air is flowing steadily toward the bicycle with speed
14 V 0. If the assumptions of Bernoulli s equation are valid (steady,
15 incompressible, inviscid flow), Eq. 3.7 can be applied as follows along the streamline that passes through (1) and () p ρ V + γ z 1 p + 1 ρ V + γ z 1 = We consider (1) to be in the free stream so that V 1 = V 0 and () to be at the tip of the bicyclist s nose and assume that z 1 = z and V = 0 (both of which, as is discussed in Section 3.4, are reasonable assumptions). It follows that the pressure at () is greater than at (1) by an amount p p 1 = 1 ρ V = 1 ρ V 1 0 (Ans) A similar result was obtained in Example 3.1 by integrating pressure gradient, which was known because the velocity distribution along the streamline, V (s), was known. The Bernoulli equation is a general integration of F = ma. To determine p p 1, knowledge of the detailed velocity distribution is not needed-only the boundary conditions at (1) and () are required. Of course, knowledge of the value of V along the streamline is needed to determine the speed V 0. As discussed in Section 3.5, this is the principle upon which many
16 velocity measuring devices are based. If the bicyclist were accelerating or decelerating, the flow would unsteady (i.e. V 0 constant) and the above analysis would be in correct since Eq. 3.7 is restricted to steady flow. v v F = ma normal to a streamline. Example: The devastating low-pressure region at the center of a tornado can be explained by applying Newton s nd law across the nearly circular streamlines of the tornado.. δ Fn = δ mv R. δ Wn = δ W cosθ = ρ δ V R = δ m a v = γ δ cos θ n (3.8) (θ = 90 ;δ W n = 0). δ Fpn = (P δ Pn)δ Sδ y (P + δ Pn)δ Sδ y = δ Pnδ Sδ y = P δ Sδ nδ y n. Thus, = P δ n δ Fn = δ Wn + δ Fpn = cos θ P (3.9) ( γ )δ n Eq. (3.8) & (3.9) &cosθ = dz dn
17 γ dz P = dn n ρ V R (3.10)
18 . if gravity is neglected or if the floor is horizontal P n ρ V γ = 0 dz = 0 dn = R This pressure difference is needed to be balance the centrifugal acceleration associated with the curved streamlines of the fluid motion.. Read example 3.3. Integrate across the streamline, using the factor that P n = dp dn if S = constant dp V ρ + dn + gz = constant (3.11) R across the streamline v V = V (S,n) ; R = R(S,n) p + ρ V dn + γ z = C R if 1) t ) μ = 0 3) ρ = C (3.1)
19 Example 3.3 Shown in Figs. E3.3a, b are two fields with circular streamlines. The velocity distributions are V (r) = C 1 r V (r) = C r for case (a) for case(b) and where C 1 and C are constant. Determine the pressure distributions, p = p(r), for each given that p = p 0 at r = r 0.
20 Figure E 3.3 Solution: We assume the flow are steady, inviscid, and incompressible with streamline in the horizontal plane ( dz dn = 0) Since the streamlines are circles, the coordinate n points in a direction opposite of that of the radial coordinate, n = r, and the radius of curvature is given by R = r. Hence, Eq becomes p ρ V = r For case (a) this gives r p = ρ C 1 r r p ρ C While for case (b) it gives = r r 3 For either the pressure increases as r increase since δ p δ r > 0 Integration of these equations with respect to r, starting with a known pressure p = p 0 at r = r 0, gives
21 p = 1 ρ C (r r 1 0 ) + p 0 for case (a) and (Ans) p = 1 ρ C ( 1 1 ) + p r r 0 0 (Ans) for case (b). These pressure distributions are sketched in Fig E3.3c. The pressure distributions needed to balance the centrifugal accelerations in cases (a) and (b) are not the same because the velocity distributions are different. In fact for case (a) the pressure increase without bound as r, while for case (b) the pressure approaches a finite value as r. The streamline patterns are the same for each case, however. Physically, case (a) represents rigid body rotation (as obtained in a can of water on a turntable after it has been spun up ) and case(b) represents a free vortex (an approximation to a tornado or the swirl of water in a drain, the bathtub vortex ).
22 Physical Interpretation. Along a streamline: P + 1 ρ V + γ z = C Across the streamline: P + ρ V R dn + γ z = c If: steady, inviscid, and incompressible (None is exactly true for real flows). Physics: Force balance; Bernoulli s Principle The work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle. (Newton s second law; first & second laws of thermodynamics). Example 3.4 Consider the flow of water from the syringe shown in Fig A force applied to the plunger will produce a pressure greater than atmospheric at point (1) within the syringe. The water flows from the needle, point (), with relatively high velocity and coasts up to point (3) at the top of its trajectory. Discuss the
23 energy of the fluid at points (1), (), and (3) by using the Bernoulli equation. Figure E 3.4 Solution: If the assumptions (steady, inviscid, incompressible flow) of the Bernoulli equation are approximately valid, it then follows that the flow can be explained in terms of the partition of the total energy of the water. According to Eq the sum of the three types of energy (kinetic, potential, and pressure) or heads (veloci ty, elevation, and pressure) must remain constant. The following table indicates the relative magnitude of each of these energies at the three points shown in the figure.
24 Energy Type Point Kinetic ρ V Potential γ z Pressure p 1 Small Zero Large Large Small Zero 3 Zero Large Zero The motion results in (or is due to) a change in the magnitude of each type of energy as the fluid flows from one location to another. An alternate way to consider this flow is as follows. The pressure gradient between (1) and () produces an acceleration to eject the water from the needle. Gravity acting on the particle between () and (3) produces a deceleration to cause the water to come to a momentary stop at the top of its flight. If friction (viscous) effects were important, there would be an energy loss between (1) and (3) and for the given p 1 the water would not be able to reach the height indicated in the figure. Such friction may arise in the needle (see chapter 8, pipe flow) or between the water stream and the surrounding air (see chapter
25 9, external flow). Example 3.5 Consider the inviscid, incompressible, steady flow shown in Fig. E3.5. From section A to B the streamlines are straight, while from C to D they follow circular paths. Describe the pressure variation between points (1) and () and points (3) and (4). Figure E 3.5 Solution: With the above assumption and the fact that R = for the portion from A to B, Eq becomes p + γ z = constant The constant can be determined by evaluating the known variables at the two locations using P = 0 (gage), Z 1 = h -1 to give
26 p 1 = p + γ ( z z 1 ) = p + γ h 1 (Ans) Note that since the radius of curvature of the streamline is infinite, the pressure variation in the vertical direction is the same as if the fluid were stationary. However, if we apply Eq between points (3) band point (4) we obtain (using dn = -dz) V p 4 + ρ Z 4 ( dz ) + Z 3 R γ z 4 = p 3 + γ z 3 With p 4 =0 and z 4 -z 3 =h 4-3 this becomes p h = γ Z dz ρ Z 4 V 3 R (Ans) To evaluate the integral we must know the variation of V and R with z. Even without this detailed information we note that the integral has a positive value. Thus, the pressure at (3) is less than the hydrostatic value, γ h 4, by an amount equal to 3 ρ dz Z V 4 Z 3 R. This lower pressure, caused by the curved streamline, is necessary to accelerate the fluid around the curved path.
27 Note that we did not apply the Bernoulli equation (Eq. 3.13)
28 across the streamlines from (1) to () or (3) to (4). Rather we used Eq As is discussed in section 3.6 application of the Bernoulli equation across streamlines (rather than along) them may lead to serious errors. 3.5 Static, Stagnation (Total), and Dynamic Pressure γ z - hydrostatic pressure not a real pressure, but possible due to potential energy variations of the fluid as a result of elevation changes. p - static pressure 1 ρ V -dynamic pressure stagnation point V = 0 P = P + 1 ρ V stagnation pressure static pressure+dynamic pressure (if elevation effect are neglected) Total pressure It represents the conversion of all of the kinetic energy into a pressure rise.
29 P T = γ Z ρ V P Total P Hydrostatic P Static P Dynamic P Figure 3.4 Measurement of static and stagnation pressures Pitot-static tube: simple, relatively in expensive.
30 Figure 3.6 The Pitot-static tube Figure 3.7 Typical Piotot-static tube designs. p + 1 ρ V + γ z = p T = constant along a streamline p = p + 1 ρ V center tube: 3 elevation is neglected. outer tube: p 4 = p 1 = p p p = 1 ρ V V 3 4 = ( p p ) / ρ 3 4
31 Figure 3.9 Typical pressure distribution along a Pitot-static tube. The cylinder is rotated until the pressures in the two sideholes are equal, thus indicating that the center hole points directly upstream-measure stagnation pressure. Side holes ( β = 9.5 ) measure static pressure. Figure 3.10 Cross section of a directional-finding Pitot-static tube.
32 3.6 Example of Use of the Bernoulli Equation steady ( t = 0), inviscid ( μ = 0 ), incompressible ( ρ = constant ) and along a streamline ( n = 0). Between two points: P + 1 ρ V 1 + γ Z 1 1 = P + 1 ρ V + γ Z Unknowns: P 1, P, V 1, V, Z 1, Z (6) If 5 unknowns are given, then determining the remaining one. Free Jets (Vertical flows from a tank) (1) - ():
33 Figure 3.11 Vertical flow from a tank (1) - ():
34 P 1 = 0 (gage pressure); P = 0 (Free jet) Z 1 = h; Z = 0; V 1 0; γ h = 1 ρ V V = γ h = ρ gh () and (4): atmospheric pressure (5): V 5 = g(h + H ) (3) and (4): P 3 = γ (h l) ; = 0 ; Z 3 P 4 1 V 3 =0 P 3 + γ l = P 4 + ρ V 4 = l ; Z 4 = 0, Q V 4 = gh Recall from Physics or dynamics 自由落體在一真空中 V = gh ; the same as the liquid leaving from the nozzle. All potential energy kinetic energy (neglecting the viscous effects).
35 Fig 3.1 Horizontal flow from a tank. Vena contracta effect: Fluid can not turn 90 ; Contraction coefficient: C = A /A ; c j h A j cross area of jet fluid column; A h cross area of the nozzle exit.
36 Figure 3.14 Typical flow patterns and contraction coefficients for various round exit configurations. Confined flows Figure 3.15 Steady flow into and out of a tank.
37 mass flow rate m = ρ Q (kg/s or slugs/s) where Q: volume flow rate (m 3 /s or ft 3 /s) Q = VA ; m = ρ Q = ρ VA From Fig. 3.15: ρ 1 A 1 V 1 = ρ A V Example 3.7 A stream of diameter d = 0.1m flows steadily from a tank of diameter D = 1.0m as shown in Fig E3.7a. Determine the flow rate, Q, needed from the inflow pipe if the water depth remains constant. h =.0m.
38 Figure E 3.7 Solution: For steady, inviscid, incompressible the Bernoulli equation applied between points (1) and () is p + 1 ρ V 1 + γ z = p ρ V + γ Z (1) With the assumptions that p 1 = p = 0, z 1 = h, and z = 0, Eq. 1 becomes 1 V + gh = 1 V () 1 Although the water level remains constant (h = constant), there is an average velocity, V 1, across section (1) because of the flow
39 from the tank. From Eq 3.19 for steady incompressible flow, conservation of mass requires Q 1 = Q = AV. Thus, A 1 V 1 = A V, or Hence, π D V = π d V d V 1 = ( ) V (3) D Equation 1 and 3 can be combined to give V = gh 1 (d / D) 4 Thus, with the given data V = (9.81m/s )(.0 m) 1 (0.1m/1m) 4 = 6.6 m/s and Q = AV = A V = π (0.1m) (6.6m/s) = 0.049m 3 /s (Ans) In this example we have not neglected the kinetic energy in the water in the tank (V 1 0). If the tank diameter is large compared to the jet diameter (D>>>d), Eq. 3 indicates that V 1 <<V and the assumption that V 1 0 would be reasonable. The error associated with this assumption can be seen by calculating the ratio of the flowrate assuming V 1 0, denoted Q, to that assuming V 1 = 0, denoted Q 0. This ratio, written as
40 Q V gh (1 (d D) 4 ) 1 = = = Q 0 V D= gh 1 (d D) 4 is plotted in Fig. E3.7b. With 0< d/d < 0.4 it follows that 1< Q/Q , and the error in assuming V 1 = 0 is less than 1%. Thus, it is often reasonable to assume V 1 = 0. Example 3.9 Water flows through a pipe reducer as is shown in Fig. E3.9. The static pressure at (1) and () are measured by the inverted U-tube manometer containing oil of specific gravity. SG. Less than one. Determine the manometer reading, h. FIGURE E 3.9 Solution: With the assumption of steady, inviscid, incompressible flow, the Bernoulli equation can be written as
41 p + 1 ρ V +γ z = p + 1 ρ V +γ z The continuity equation (Eq. 3.19) provides a second relationship between V 1 and V, if we assume the velocity profiles are uniform as those two locations and the fluid incompressible: Q = AV = AV 1 1 By combining these two equations we obtain p p = γ ( z ] z ) + 1 ρ V [1 A (1) 1 1 A1 This pressure difference is measured by the manometer and can be determined by using the pressure-depth ideas developed in Chapter. Thus. p γ ( z z ) l γ h + SGγ h +γ l = p 1 1 γ or p p 1 = γ ( z z 1 ) + (1 SG )γ h () As discussed in Chapter, the pressure difference is neither merely γh not γ(h+z 1 -z ). Equations 1 and can be combined to give the desired result as follows (1 SG )γ h = 1 ρ V [1 A ] A1
42 A 1 Q A or V =Q/A, thus, h = 1. (Ans) A g(1 SG ) The difference in elevation, z 1 -z, was not needed because the change in elevation term in the Bernoulli equation exactly cancels the elevation term in the manometer equation. However, the pressure difference, p 1 -p, depends on the angle θ, because of the elevation, z 1 -z, in Eq. 1. Thus, for a given flowrate, the pressure difference, p 1 -p, as measured by a pressure gage would vary with θ, but the manometer reading, h, would be independent of θ. In general, an increase in velocity is accompanied by a decrease in pressure. Air (gases): Compressibility (Ch 11) Liquids: Cavitation (Propeller etc.) Flow Rate Measurement z Ideal flow meter neglecting viscous and compressible effects. (loss some accuracy)
43 z Orifice meter; Nozzle meter: V, P ; Venturi meter Q = ( P P ) A ρ 1 A 1 1 A (3.0) 1 z Sluice gate: Q = z b g( z z ) z 1 z 1 (3.1) z Sharp crested weir: Q = C Hb gh = C b 1 1 gh 3 The Energy Line and the Hydraulic Grade Line z Bernoulli equation is an energy equation with four assumptions: inviscid, incompressible, steady flow, and along a streamline from pts ( 1) - (), respectively, and the sum of partition of energy remains constant from pts (1)-(). (Head) H = p + V + z = constant along a streamline. γ g
44 FIGURE 3.1 Representation of the energy line and the hydraulic grade line FIGURE 3. The energy line and hydraulic grade line for flow from a tank. Restrictions on the use of Bernoulli Equation Compressibility Effect
45 Gases: dp ρ not a constant A special case: for compressible flow Given - steady, inviscid, isothermal (T = C along the streamline) Sol: p = ρ RT ρ dp = RT ρ = p RT dp = RT ln p + c p V V RT p 1 + = + z ln 1 z + g g g p1 V 1 or + z + 1 g RT p V ln 1 = + z g p g Compare to the incompressible Bernoulli Equation. Q if p p p 1 = = 1 +ε, ε = p p 1 << 1 p p p p then ln 1 = ln(1 + ε ) ε p V RT p 1 + z + ( 1 1) = + z 1 g g p g V p V p + + V 1 z 1 + g 1 γ = + g z γ p p p = 1 = 1 1 p p 3.8. Unsteady Effect
46 V = V(s) steady; V = V(s, t) unsteady.
47 a s t = V +V V s (local + convective accelerations) z Repeating the steps leading to the Bernoulli Equation. ρ V ds + dp + 1 ρ d (V ) +γ dz = 0 t along a streamline z If incompressible, inviscid, p S v 1 ρ V +γ z = ρ 1 1 S ds + + ρ V +γ p 1 1 t z Use velocity potential to simplify the problem (Ch 6). FIGURE 3.5 Oscillation of a liquid column in a U-tube
48 3.8.3 Rotational Effect In general, the Bernoulli constant varies from streamline to streamline. However, under certain restrictions, this constant may be the same throughout the entire flow field, as ex.3.19 illustrates this effect Other restrictions μ = 0 (inviscid); Bernoulli Equation: A first integral of Newton s nd laws along a streamline.
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1. For the manometer shown in figure 1, if the absolute pressure at point A is 1.013 10 5 Pa, the absolute pressure at point B is (ρ water =10 3 kg/m 3, ρ Hg =13.56 10 3 kg/m 3, ρ oil = 800kg/m 3 ): (a)
The online of midterm-tests of Fluid Mechanics 1 1) The information on a can of pop indicates that the can contains 460 ml. The mass of a full can of pop is 3.75 lbm while an empty can weights 80.5 lbf.
Experiment (4): Flow measurement Introduction: The flow measuring apparatus is used to familiarize the students with typical methods of flow measurement of an incompressible fluid and, at the same time
Fluid Mechanics 1. Which is the cheapest device for measuring flow / discharge rate. a) Venturimeter b) Pitot tube c) Orificemeter d) None of the mentioned 2. Which forces are neglected to obtain Euler
CLASS SCHEDULE 2013 FALL Class # or Lab # 1 Date Aug 26 2 28 Important Concepts (Section # in Text Reading, Lecture note) Examples/Lab Activities Definition fluid; continuum hypothesis; fluid properties
Detailed Outline, M E 320 Fluid Flow, Spring Semester 2015 I. Introduction (Chapters 1 and 2) A. What is Fluid Mechanics? 1. What is a fluid? 2. What is mechanics? B. Classification of Fluid Flows 1. Viscous
CHAPTER 1 Properties of Fluids ENGINEERING FLUID MECHANICS 1.1 Introduction 1.2 Development of Fluid Mechanics 1.3 Units of Measurement (SI units) 1.4 Mass, Density, Specific Weight, Specific Volume, Specific
The angular momentum A control volume analysis can be applied to the angular momentum, by letting B equal to angularmomentum vector H. If O is the point about which moments are desired, the angular moment
Some Basic Plane Potential Flows Uniform Stream in the x Direction A uniform stream V = iu, as in the Fig. (Solid lines are streamlines and dashed lines are potential lines), possesses both a stream function
Homework 4 Solutions 1. (15 points) Bernoulli s equation can be adapted for use in evaluating unsteady flow conditions, such as those encountered during start- up processes. For example, consider the large
EXPERIMENT No.1 FLOW MEASUREMENT BY ORIFICEMETER 1.1 AIM: To determine the co-efficient of discharge of the orifice meter 1.2 EQUIPMENTS REQUIRED: Orifice meter test rig, Stopwatch 1.3 PREPARATION 1.3.1
Appendix FLUID MECHANICS Approximate physical properties of selected fluids All properties are given at pressure 101. kn/m and temperature 15 C. Liquids Density (kg/m ) Dynamic viscosity (N s/m ) Surface
Important Definitions: MECHANICAL PROPERTIES OF FLUIDS: Fluid: A substance that can flow is called Fluid Both liquids and gases are fluids Pressure: The normal force acting per unit area of a surface is
Chapter 5: Mass, Bernoulli, and Energy Equations Introduction This chapter deals with 3 equations commonly used in fluid mechanics The mass equation is an expression of the conservation of mass principle.
When the mass m of the control volume remains nearly constant, the first term of the Eq. 6 8 simply becomes mass times acceleration since 39 CHAPTER 6 d(mv ) CV m dv CV CV (ma ) CV Therefore, the control
IV. DIFFERENTIAL RELATIONS FOR A FLUID PARTICLE This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations and common
Topic: Fluid Properties 1. If 6 m 3 of oil weighs 47 kn, calculate its specific weight, density, and specific gravity. 2. 10.0 L of an incompressible liquid exert a force of 20 N at the earth s surface.
Chapter 2 Hydrostatics 2.1 Review Eulerian description from the perspective of fixed points within a reference frame. Lagrangian description from the perspective of a parcel moving within the flow. Streamline
Fluid Dynamics Problems M.Sc Mathematics-Second Semester Dr. Dinesh Khattar-K.M.College 1. (Example, p.74, Chorlton) At the point in an incompressible fluid having spherical polar coordinates,,, the velocity
Chapter (6) Energy Equation and Its Applications Bernoulli Equation Bernoulli equation is one of the most useful equations in fluid mechanics and hydraulics. And it s a statement of the principle of conservation
bernoulli_11 In which of the following scenarios is applying the following form of Bernoulli s equation: p V z constant! g + g + = from point 1 to point valid? a. 1 stagnant column of water steady, inviscid,
1. The x and y components of velocity for a two-dimensional flow are u = 3.0 ft/s and v = 9.0x ft/s where x is in feet. Determine the equation for the streamlines and graph representative streamlines in
skiladæmi 10 Due: 11:59pm on Wednesday, November 11, 015 You will receive no credit for items you complete after the assignment is due Grading Policy Alternative Exercise 1115 A bar with cross sectional
Part A: 1 pts each, 10 pts total, no partial credit. 1) (Correct: 1 pt/ Wrong: -3 pts). The sum of static, dynamic, and hydrostatic pressures is constant when flow is steady, irrotational, incompressible,
Introduction to Fluid Dynamics Roger K. Smith Skript - auf englisch! Umsonst im Internet http://www.meteo.physik.uni-muenchen.de Wählen: Lehre Manuskripte Download User Name: meteo Password: download Aim
Hydraulic Coefficient & Flow Measurements ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 3 1. Mass flow rate If we want to measure the rate at which water is flowing
Q1. Figure 1 shows a solid cylindrical steel rod of length =.0 m and diameter D =.0 cm. What will be increase in its length when m = 80 kg block is attached to its bottom end? (Young's modulus of steel
Shell Balances in Fluid Mechanics R. Shankar Subramanian Department of Chemical and Biomolecular Engineering Clarkson University When fluid flow occurs in a single direction everywhere in a system, shell
UNIT III FLOW THROUGH PIPES 1. List the types of fluid flow. Steady and unsteady flow Uniform and non-uniform flow Laminar and Turbulent flow Compressible and incompressible flow Rotational and ir-rotational
Lecture 3 The energy equation Dr Tim Gough: firstname.lastname@example.org General information Lab groups now assigned Timetable up to week 6 published Is there anyone not yet on the list? Week 3 Week 4 Week 5
COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour Basic Equations in fluid Dynamics Course teacher Dr. M. Mahbubur Razzaque Professor Department of Mechanical Engineering BUET 1 Description of Fluid
Department of Chemical Engineering B.E/B.Tech/M.E/M.Tech : Chemical Engineering Regulation: 2016 PG Specialisation : NA Sub. Code / Sub. Name : CH16304 FLUID MECHANICS Unit : I LP: CH 16304 Rev. No: 00
Neatly print your name: Signature: (Note that unsigned exams will be given a score of zero.) Circle your lecture section (-1 point if not circled, or circled incorrectly): Prof. Dabiri Prof. Wassgren Prof.
INTRODUCTION TO FLUID MECHANICS June 27, 2013 PROBLEM 3 (1 hour) A perfect liquid of constant density ρ and constant viscosity µ fills the space between two infinite parallel walls separated by a distance
CE30 Test 1 Solution Key Date: 26 Sept. 2017 COVER PAGE Write your name on each sheet of paper that you hand in. Read all questions very carefully. If the problem statement is not clear, you should ask
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BEE 5330 Fluids FE Review, Feb 24, 2010 1 A fluid is a substance that can not support a shear stress. Liquids differ from gasses in that liquids that do not completely fill a container will form a free
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1. Coordinate System Introduction to Turbomachinery Since there are stationary and rotating blades in turbomachines, they tend to form a cylindrical form, represented in three directions; 1. Axial 2. Radial
University of Engineering and Technology, Taxila Department of Civil Engineering Course Title: CE-201 Fluid Mechanics - I Pre-requisite(s): None Credit Hours: 2 + 1 Contact Hours: 2 + 3 Text Book(s): Reference
0 - FLUID MECHANICS Page Introduction Fluid is a matter in a state which can flow. Liquids, gases, molten metal and tar are examples of fluids. Fluid mechanics is studied in two parts: ( i ) Fluid statics
CHAPTER-10 MECHANICAL PROPERTIES OF FLUIDS QUESTIONS 1 marks questions 1. What are fluids? 2. How are fluids different from solids? 3. Define thrust of a liquid. 4. Define liquid pressure. 5. Is pressure
EXPERIMENT SEVEN: FLOW VISUALIZATION AND ANALYSIS I OBJECTIVE OF THE EXPERIMENT: Visualization of flow pattern over or around immersed objects in open channel flow. II THEORY AND EQUATION: Open channel:
Chapter 6 /60 The Impulse-Momentum Principle F F Chapter 6 The Impulse-Momentum Principle /60 Contents 6.0 Introduction 6. The Linear Impulse-Momentum Equation 6. Pipe Flow Applications 6.3 Open Channel
SPC 307 - Aerodynamics Course Assignment Due Date Monday 28 May 2018 at 11:30 1. The maximum velocity at which an aircraft can cruise occurs when the thrust available with the engines operating with the