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Chapter 4 Fluid Kinematics Solutions Manual for Fluid Mechanics: Fundamentals and Applications by Cengel & Cimbala CHAPTER 4 FLUID KINEMATICS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws.

By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party.

No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. 4-1 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics Introductory Problems 4-1C Solution We are to define and explain kinematics and fluid kinematics. Analysis Kinematics means the study of motion.

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Fluid kinematics is the study of how fluids flow and how to describe fluid motion. Fluid kinematics deals with describing the motion of fluids without considering (or even understanding) the forces and moments that cause the motion. Discussion Fluid kinematics deals with such things as describing how a fluid particle translates, distorts, and rotates, and how to visualize flow fields. 4-2 Solution We are to write an equation for centerline speed through a nozzle, given that the flow speed increases parabolically. Assumptions 1 The flow is steady. 2 The flow is axisymmetric. The water is incompressible. Analysis A general equation for a parabola in the x direction is u = a + b ( x ? c) General parabolic equation: 2 (1) We have two boundary conditions, namely at x = 0, u = uentrance and at x = L, u = uexit. By inspection, Eq. 1 is satisfied by setting c = 0, a = uentrance and b = (uexit - uentrance)/L2. Thus, Eq. 1 becomes u = uentrance + Parabolic speed: ( uexit ? uentrance ) L2 x2 (2) Discussion You can verify Eq. 2 by plugging in x = 0 and x = L. 4-3 Solution location. For a given velocity field we are to find out if there is a stagnation point.

If so, we are to calculate its Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity field is V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j (1) At a stagnation point, both u and v must equal zero. At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = 0. 5 + 1. 2 x v = ? 2. 0 ? 1. 2 y (2) x = ? 0. 4167 y = ? 1. 667 (3) Setting these to zero yields Stagnation point: 0 = 0. 5 + 1. 2 x 0 = ? 2. 0 ? 1. 2 y So, yes there is a stagnation point; its location is x = -0. 17, y = -1. 67 (to 3 digits). Discussion If the flow were three-dimensional, we would have to set w = 0 as well to determine the location of the stagnation point. In some flow fields there is more than one stagnation point. 4-2 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-4 Solution location. For a given velocity field we are to find out if there is a stagnation point.

If so, we are to calculate its Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity field is ( )( ) V = ( u, v ) = a 2 ? ( b ? cx ) i + ? 2cby + 2c 2 xy j 2 (1) At a stagnation point, both u and v must equal zero. At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = a 2 ? ( b ? cx ) 2 v = ? 2cby + 2c 2 xy (2) b? a c y=0 (3) Setting these to zero and solving simultaneously yields Stagnation point: 0 = a 2 ? ( b ? cx ) 2 x= v = ? 2cby + 2c xy So, yes there is a stagnation point; its location is x = (b – a)/c, y = 0. Discussion If the flow were three-dimensional, we would have to set w = 0 as well to determine the location of the stagnation point. In some flow fields there is more than one stagnation point. Lagrangian and Eulerian Descriptions 4-5C Solution We are to define the Lagrangian description of fluid motion. Analysis In the Lagrangian description of fluid motion, individual fluid particles (fluid elements composed of a fixed, identifiable mass of fluid) are followed. Discussion

The Lagrangian method of studying fluid motion is similar to that of studying billiard balls and other solid objects in physics. 4-6C Solution We are to compare the Lagrangian method to the study of systems and control volumes and determine to which of these it is most similar. Analysis The Lagrangian method is more similar to system analysis (i. e. , closed system analysis). In both cases, we follow a mass of fixed identity as it moves in a flow. In a control volume analysis, on the other hand, mass moves into and out of the control volume, and we don’t follow any particular chunk of fluid.

Instead we analyze whatever fluid happens to be inside the control volume at the time. Discussion to a point. In fact, the Lagrangian analysis is the same as a system analysis in the limit as the size of the system shrinks 4-3 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-7C Solution description. We are to define the Eulerian description of fluid motion, and explain how it differs from the Lagrangian

Analysis In the Eulerian description of fluid motion, we are concerned with field variables, such as velocity, pressure, temperature, etc. , as functions of space and time within a flow domain or control volume. In contrast to the Lagrangian method, fluid flows into and out of the Eulerian flow domain, and we do not keep track of the motion of particular identifiable fluid particles. Discussion The Eulerian method of studying fluid motion is not as “natural” as the Lagrangian method since the fundamental conservation laws apply to moving particles, not to fields. -8C Solution We are to determine whether a measurement is Lagrangian or Eulerian. Analysis Since the probe is fixed in space and the fluid flows around it, we are not following individual fluid particles as they move. Instead, we are measuring a field variable at a particular location in space. Thus this is an Eulerian measurement. Discussion If a neutrally buoyant probe were to move with the flow, its results would be Lagrangian measurements – following fluid particles. 4-9C Solution We are to determine whether a measurement is Lagrangian or Eulerian. Analysis

Since the probe moves with the flow and is neutrally buoyant, we are following individual fluid particles as they move through the pump. Thus this is a Lagrangian measurement. Discussion If the probe were instead fixed at one location in the flow, its results would be Eulerian measurements. 4-10C Solution We are to determine whether a measurement is Lagrangian or Eulerian. Analysis Since the weather balloon moves with the air and is neutrally buoyant, we are following individual “fluid particles” as they move through the atmosphere. Thus this is a Lagrangian measurement.

Note that in this case the “fluid particle” is huge, and can follow gross features of the flow – the balloon obviously cannot follow small scale turbulent fluctuations in the atmosphere. Discussion When weather monitoring instruments are mounted on the roof of a building, the results are Eulerian measurements. 4-11C Solution We are to determine whether a measurement is Lagrangian or Eulerian. Analysis Relative to the airplane, the probe is fixed and the air flows around it. We are not following individual fluid particles as they move. Instead, we are measuring a field variable at a particular location in space relative to the moving airplane.

Thus this is an Eulerian measurement. Discussion The airplane is moving, but it is not moving with the flow. 4-4 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-12C Solution We are to compare the Eulerian method to the study of systems and control volumes and determine to which of these it is most similar. Analysis The Eulerian method is more similar to control volume analysis.

In both cases, mass moves into and out of the flow domain or control volume, and we don’t follow any particular chunk of fluid. Instead we analyze whatever fluid happens to be inside the control volume at the time. Discussion In fact, the Eulerian analysis is the same as a control volume analysis except that Eulerian analysis is usually applied to infinitesimal volumes and differential equations of fluid flow, whereas control volume analysis usually refers to finite volumes and integral equations of fluid flow. 4-13C Solution flow. We are to define a steady flow field in the Eulerian description, and discuss particle acceleration in such a

Analysis A flow field is defined as steady in the Eulerian frame of reference when properties at any point in the flow field do not change with respect to time. In such a flow field, individual fluid particles may still experience non-zero acceleration – the answer to the question is yes. Discussion ( a = dV / dt ) Although velocity is not a function of time in a steady flow field, its total derivative with respect to time is not necessarily zero since the acceleration is composed of a local (unsteady) part which is zero and an advective part which is not necessarily zero. 4-14C Solution

We are to list three alternate names for material derivative. Analysis The material derivative is also called total derivative, particle derivative, Eulerian derivative, Lagrangian derivative, and substantial derivative. “Total” is appropriate because the material derivative includes both local (unsteady) and convective parts. “Particle” is appropriate because it stresses that the material derivative is one following fluid particles as they move about in the flow field. “Eulerian” is appropriate since the material derivative is used to transform from Lagrangian to Eulerian reference frames. Lagrangian” is appropriate since the material derivative is used to transform from Lagrangian to Eulerian reference frames. Finally, “substantial” is not as clear of a term for the material derivative, and we are not sure of its origin. Discussion All of these names emphasize that we are following a fluid particle as it moves through a flow field. 4-5 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 4 Fluid Kinematics 4-15 Solution We are to calculate the material acceleration for a given velocity field. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The velocity field is V = ( u , v ) = (U 0 + bx ) i ? byj (1) The acceleration field components are obtained from its definition (the material acceleration) in Cartesian coordinates, ? u ?u ?u ?u +u +v +w = 0 + (U 0 + bx ) b + ( ? by ) 0 + 0 ?t ?x ?y ?z ?v ?v ?v ?v ay = + u + v + w = 0 + (U 0 + bx ) 0 + ( ? by )( ? b ) +0 ?t ?x ?y ?z ax = (2) here the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is twodimensional. Eq. 2 simplifies to ax = b (U 0 + bx ) ay = b2 y (3) a = b (U 0 + bx ) i + b 2 yj Material acceleration components: (4) In terms of a vector, Material acceleration vector: Discussion For positive x and b, fluid particles accelerate in the positive x direction. Even though this flow is steady, there is still a non-zero acceleration field. 4-16 Solution particle. For a given pressure and velocity field, we are to calculate the rate of change of pressure following a fluid Assumptions 1 The flow is steady. The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The pressure field is P = P0 ? Pressure field: ?? 2U 0 bx + b 2 ( x 2 + y 2 ) ? 2? ? (1) By definition, the material derivative, when applied to pressure, produces the rate of change of pressure following a fluid particle. Using Eq. 1 and the velocity components from the previous problem, DP ? P ?P ?P = +u +v + Dt ?t ?x ?y Steady ( w ?P ?z (2) Two-dimensional ) ( = (U 0 + bx ) ? ?U 0 b ? ? b 2 x + ( ? by ) ? ? b 2 y ) where the unsteady term is zero since this is a steady flow, and the term with w is zero since the flow is two-dimensional.

Eq. 2 simplifies to the following rate of change of pressure following a fluid particle: ( ) DP 2 = ? ? ? U 0 b ? 2U 0 b 2 x + b3 y 2 ? x 2 ? ? ? Dt (3) Discussion The material derivative can be applied to any flow property, scalar or vector. Here we apply it to the pressure, a scalar quantity. 4-6 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-17 Solution

For a given velocity field we are to calculate the acceleration. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity components are Velocity components: u = 0. 20 + 1. 3 x + 0. 85 y v = ? 0. 50 + 0. 95 x ? 1. 3 y (1) The acceleration field components are obtained from its definition (the material acceleration) in Cartesian coordinates, ? u ?u ?u ?u +u +v +w = 0 + ( 0. 20 + 1. 3 x + 0. 85 y )(1. 3) + ( ? 0. 50 + 0. 95 x ? 1. 3 y )( 0. 85 ) + 0 ? t ?x ?y ?z ?v ?v ?v ?v + u + v + w = 0 + ( 0. 20 + 1. 3 x + 0. 85 y )( 0. 95 ) + ( ? 0. 50 + 0. 95 x ? 1. y )( ? 1. 3 ) +0 ay = ?t ?x ?y ?z ax = (2) where the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is twodimensional. Eq. 2 simplifies to Acceleration components: ax = ? 0. 165 + 2. 4975 x a y = 0. 84 + 2. 4975 y (3) At the point (x,y) = (1,2), the acceleration components of Eq. 3 are Acceleration components at (1,2): ax = 2. 3325 ? 2. 33 a y = 5. 835 ? 5. 84 Discussion The final answers are given to three significant digits. No units are given in either the problem statement or the answers. We assume that the coefficients have appropriate units. -7 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-19 Solution We are to generate an expression for the fluid acceleration for a given velocity. Assumptions 1 The flow is steady. 2 The flow is axisymmetric. 3 The water is incompressible. Analysis In Problem 4-2 we found that along the centerline, u = uentrance + Speed along centerline of nozzle: ( uexit ? uentrance ) x2 (1) ?u ?u ?u ?u +u +v +w ?t ?x y ?z (2) L2 To find the acceleration in the x-direction, we use the material acceleration, ax = Acceleration along centerline of nozzle: The first term in Eq. 2 is zero because the flow is steady. The last two terms are zero because the flow is axisymmetric, which means that along the centerline there can be no v or w velocity component. We substitute Eq. 1 for u to obtain Acceleration along centerline of nozzle: ax = u ( uexit ? uentrance ) 2 ? ( uexit ? uentrance ) ?u ? = ? uentrance + x ? ( 2) x ? ? ?x ? L2 L2 ? (3) or ax = 2uentrance Discussion ( uexit ? uentrance ) L2 x+2 ( uexit ? uentrance )

L4 2 x3 (4) Fluid particles are accelerated along the centerline of the nozzle, even though the flow is steady. 4-20 Solution We are to write an equation for centerline speed through a diffuser, given that the flow speed decreases parabolically. Assumptions 1 The flow is steady. 2 The flow is axisymmetric. Analysis A general equation for a parabola in x is General parabolic equation: u = a + b ( x ? c) 2 (1) We have two boundary conditions, namely at x = 0, u = uentrance and at x = L, u = uexit. By inspection, Eq. 1 is satisfied by setting c = 0, a = uentrance and b = (uexit - uentrance)/L2. Thus, Eq. becomes Parabolic speed: Discussion u = uentrance + ( uexit ? uentrance ) L2 x2 (2) You can verify Eq. 2 by plugging in x = 0 and x = L. 4-8 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-21 Solution We are to generate an expression for the fluid acceleration for a given velocity, and then calculate its value at two x locations. Assumptions 1 The flow is steady. 2 The flow is axisymmetric. Analysis

In the previous problem, we found that along the centerline, u = uentrance + Speed along centerline of diffuser: ( uexit ? uentrance ) 2 L x2 (1) To find the acceleration in the x-direction, we use the material acceleration, Acceleration along centerline of diffuser: ax = ?u ?u ?u ?u +w +u +v ?z ?t ?x ?y (2) The first term in Eq. 2 is zero because the flow is steady. The last two terms are zero because the flow is axisymmetric, which means that along the centerline there can be no v or w velocity component. We substitute Eq. 1 for u to obtain Acceleration along centerline of diffuser: ( uexit ? uentrance ) x 2 ? ( uexit ? entrance ) x ?u ? = ? uentrance + ax = u ? ( 2) ? ?x ? L2 L2 ? ? or ax = 2uentrance ( uexit ? uentrance ) 2 L x+2 ( uexit ? uentrance ) 2 4 L x3 (3) At the given locations, we substitute the given values. At x = 0, Acceleration along centerline of diffuser at x = 0: ax ( x = 0 ) = 0 (4) At x = 1. 0 m, Acceleration along centerline of diffuser at x = 1. 0 m: ax ( x = 1. 0 m ) = 2 ( 30. 0 m/s ) ( ? 25. 0 m/s ) ( ? 25. 0 m/s ) 3 (1. 0 m ) + 2 (1. 0 m ) 2 4 ( 2. 0 m ) ( 2. 0 m ) 2 (5) = -297 m/s 2 Discussion ax is negative implying that fluid particles are decelerated along the centerline of the diffuser, even though the flow is steady.

Because of the parabolic nature of the velocity field, the acceleration is zero at the entrance of the diffuser, but its magnitude increases rapidly downstream. 4-9 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics Flow Patterns and Flow Visualization 4-22C Solution We are to define streamline and discuss what streamlines indicate. Analysis A streamline is a curve that is everywhere tangent to the instantaneous local velocity vector.

It indicates the instantaneous direction of fluid motion throughout the flow field. Discussion If a flow field is steady, streamlines, pathlines, and streaklines are identical. 4-23 Solution For a given velocity field we are to generate an equation for the streamlines. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. The steady, two-dimensional velocity field of Problem 4-15 is Analysis V = ( u , v ) = (U 0 + bx ) i ? byj Velocity field: (1) For two-dimensional flow in the x-y plane, streamlines are given by Streamlines in the x-y plane: dy ? v = dx ? along a streamline u (2) We substitute the u and v components of Eq. 1 into Eq. 2 and rearrange to get dy ?by = dx U 0 + bx We solve the above differential equation by separation of variables: ?? dy dx = by ? U 0 + bx Integration yields 1 1 1 ? ln ( by ) = ln (U 0 + bx ) + ln C1 b b b (3) where we have set the constant of integration as the natural logarithm of some constant C1, with a constant in front in order to simplify the algebra (notice that the factor of 1/b can be removed from each term in Eq. 3). When we recall that ln(ab) = lna + lnb, and that –lna = ln(1/a), Eq. 3 simplifies to Equation for streamlines: y= C

U 0 + bx ) ( (4) The new constant C is related to C1, and is introduced for simplicity. Discussion Each value of constant C yields a unique streamline of the flow. 4-10 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-24E Solution For a given velocity field we are to plot several streamlines for a given range of x and y values. 3 Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis From the solution to the previous problem, an equation for the streamlines is 1 Streamlines in the x-y plane: y= C (U 0 + bx ) (1) y0 (ft) Constant C is set to various values in order to plot the streamlines. Several streamlines in the given range of x and y are plotted in Fig. 1. The direction of the flow is found by calculating u and v at some point in the flow field. We choose x = 1 ft, y = 1 ft. At this point u = 9. 6 ft/s and v = –4. 6 ft/s. The direction of the velocity at this point is obviously to the lower right. This sets the direction of all the streamlines. The arrows in Fig. indicate the direction of flow. Discussion -1 -2 -3 0 1 2 3 x (ft) 4 5 The flow is type of converging channel flow. FIGURE 1 Streamlines (solid blue curves) for the given velocity field; x and y are in units of ft. 4-25C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a snapshot of dye streaks in water, each streak shows the time history of dye that was introduced earlier from a port in the body. Thus these are streaklines. Since the flow appears to be steady, these streaklines are the same as pathlines and streamlines. Discussion

It is assumed that the dye follows the flow of the water. If the dye is of nearly the same density as the water, this is a reasonable assumption. 4-26C Solution We are to define pathline and discuss what pathlines indicate. Analysis A pathline is the actual path traveled by an individual fluid particle over some time period. It indicates the exact route along which a fluid particle travels from its starting point to its ending point. Unlike streamlines, pathlines are not instantaneous, but involve a finite time period. Discussion If a flow field is steady, streamlines, pathlines, and streaklines are identical. -27C Solution We are to define streakline and discuss the difference between streaklines and streamlines. Analysis A streakline is the locus of fluid particles that have passed sequentially through a prescribed point in the flow. Streaklines are very different than streamlines. Streamlines are instantaneous curves, everywhere tangent to the local velocity, while streaklines are produced over a finite time period. In an unsteady flow, streaklines distort and then retain features of that distorted shape even as the flow field changes, whereas streamlines change instantaneously with the flow field.

Discussion If a flow field is steady, streamlines and streaklines are identical. 4-11 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-28C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a snapshot of dye streaks in water, each streak shows the time history of dye that was introduced earlier from a port in the body.

Thus these are streaklines. Since the flow appears to be unsteady, these streaklines are not the same as pathlines or streamlines. Discussion It is assumed that the dye follows the flow of the water. If the dye is of nearly the same density as the water, this is a reasonable assumption. 4-29C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a snapshot of smoke streaks in air, each streak shows the time history of smoke that was introduced earlier from the smoke wire. Thus these are streaklines.

Since the flow appears to be unsteady, these streaklines are not the same as pathlines or streamlines. Discussion It is assumed that the smoke follows the flow of the air. If the smoke is neutrally buoyant, this is a reasonable assumption. In actuality, the smoke rises a bit since it is hot; however, the air speeds are high enough that this effect is negligible. 4-30C Solution We are to determine what kind of flow visualization is seen in a photograph. Analysis Since the picture is a time exposure of air bubbles in water, each white streak shows the path of an individual air bubble.

Thus these are pathlines. Since the outer flow (top and bottom portions of the photograph) appears to be steady, these pathlines are the same as streaklines and streamlines. Discussion It is assumed that the air bubbles follow the flow of the water. If the bubbles are small enough, this is a reasonable assumption. 4-31C Solution We are to define timeline and discuss how timelines can be produced in a water channel. We are also to describe an application where timelines are more useful than streaklines. Analysis A timeline is a set of adjacent fluid particles that were marked at the same instant of time.

Timelines can be produced in a water flow by using a hydrogen bubble wire. There are also techniques in which a chemical reaction is initiated by applying current to the wire, changing the fluid color along the wire. Timelines are more useful than streaklines when the uniformity of a flow is to be visualized. Another application is to visualize the velocity profile of a boundary layer or a channel flow. Discussion Timelines differ from streamlines, streaklines, and pathlines even if the flow is steady. 4-32C Solution For each case we are to decide whether a vector plot or contour plot is most appropriate, and we are to explain our choice.

Analysis In general, contour plots are most appropriate for scalars, while vector plots are necessary when vectors are to be visualized. (a) A contour plot of speed is most appropriate since fluid speed is a scalar. (b) A vector plot of velocity vectors would clearly show where the flow separates. Alternatively, a vorticity contour plot of vorticity normal to the plane would also show the separation region clearly. (c) A contour plot of temperature is most appropriate since temperature is a scalar. (d) A contour plot of this component of vorticity is most appropriate since one component of a vector is a scalar.

Discussion There are other options for case (b) – temperature contours can also sometimes be used to identify a separation zone. 4-12 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-33 Solution For a given velocity field we are to generate an equation for the streamlines and sketch several streamlines in the first quadrant. Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane.

Analysis The velocity field is given by V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j (1) For two-dimensional flow in the x-y plane, streamlines are given by dy ? v = ? dx ? along a streamline u Streamlines in the x-y plane: (2) We substitute the u and v components of Eq. 1 into Eq. 2 and rearrange to get dy ? 2. 0 ? 1. 2 y = dx 0. 5 + 1. 2 x We solve the above differential equation by separation of variables: dy dx = ?2. 0 ? 1. 2 y 0. 5 + 1. 2 x > dy dx ? ? 2. 0 ? 1. 2 y = ? 0. 5 + 1. 2 x Integration yields ? 1 1 1 ln ( ? 2. 0 ? 1. 2 y ) = ln ( 0. 5 + 1. 2 x ) ? ln C1 1. 2 1. 2 1. 2 here we have set the constant of integration as the natural logarithm of some constant C1, with a constant in front in order to simplify the algebra. When we recall that ln(ab) = lna + lnb, and that –lna = ln(1/a), Eq. 3 simplifies to Equation for streamlines: y= 5 y 4 3 2 C ? 1. 667 1. 2 ( 0. 5 + 1. 2 x ) 1 The new constant C is related to C1, and is introduced for simplicity. C can be set to various values in order to plot the streamlines. Several streamlines in the upper right quadrant of the given flow field are shown in Fig. 1. The direction of the flow is found by calculating u and v at some point in the flow field.

We choose x = 3, y = 3. At this point u = 4. 1 and v = -5. 6. The direction of the velocity at this point is obviously to the lower right. This sets the direction of all the streamlines. The arrows in Fig. 1 indicate the direction of flow. Discussion 6 (3) 0 0 1 2 3 4 5 x FIGURE 1 Streamlines (solid black curves) for the given velocity field. The flow appears to be a counterclockwise turning flow in the upper right quadrant. 4-13 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation.

If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-34 Solution For a given velocity field we are to generate a velocity vector plot in the first quadrant. Scale: 6 Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis 5 y4 The velocity field is given by V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j 3 (1) 2 At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = 0. 5 + 1. 2 x 10 m/s v = ? 2. 0 ? 1. 2 y 1 0 (2) 0

To plot velocity vectors, we simply pick an (x,y) point, calculate u and v from Eq. 2, and plot an arrow with its tail at (x,y), and its tip at (x+Su,y+Sv) where S is some scale factor for the vector plot. For the vector plot shown in Fig. 1, we chose S = 0. 2, and plot velocity vectors at several locations in the first quadrant. 1 2 3 4 5 x FIGURE 1 Velocity vectors for the given velocity field. The scale is shown by the top arrow. Discussion The flow appears to be a counterclockwise turning flow in the upper right quadrant. 4-35 Solution For a given velocity field we are to generate an acceleration vector plot in the first quadrant.

Assumptions 1 The flow is steady. 2 The flow is two-dimensional in the x-y plane. Analysis The velocity field is given by V = ( u , v ) = ( 0. 5 + 1. 2 x ) i + ( ? 2. 0 ? 1. 2 y ) j (1) At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1, Velocity components: u = 0. 5 + 1. 2 x v = ? 2. 0 ? 1. 2 y Scale: (2) 6 The acceleration field is obtained from its definition (the material acceleration), Acceleration components: ?u ?u ?u ?u ax = +u +v +w = 0 + ( 0. 5 + 1. 2 x )(1. 2 ) + 0 + 0 ?t ?x ?y ?z ?v ?v ?v ?v ay = + u + v + w = 0 + 0 + ( ? 2. 0 ? 1. 2 y )( ? 1. 2 ) +0 t ?x ?y ?z 5 4 y 3 2 (3) 1 0 0 where the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is two-dimensional. Eq. 3 simplifies to Acceleration components: ax = 0. 6 + 1. 44 x a y = 2. 4 + 1. 44 y 10 m/s2 (4) 1 2 3 4 5 x FIGURE 1 Acceleration vectors for the velocity field. The scale is shown by the top arrow. To plot the acceleration vectors, we simply pick an (x,y) point, calculate ax and ay from Eq. 4, and plot an arrow with its tail at (x,y), and its tip at (x+Sax,y+Say) where S is some scale factor for the vector plot. For the vector plot shown in Fig. , we chose S = 0. 15, and plot acceleration vectors at several locations in the first quadrant. Discussion Since the flow is a counterclockwise turning flow in the upper right quadrant, the acceleration vectors point to the upper right (centripetal acceleration). 4-14 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-36 For the given velocity field, the location(s) of stagnation point(s) are to be determined.

Several velocity Solution vectors are to be sketched and the velocity field is to be described. Assumptions 1 The flow is steady and incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or v with z. Analysis (a) The velocity field is Scale: V = ( u , v ) = (1 + 2. 5 x + y ) i + ( ? 0. 5 ? 1. 5 x ? 2. 5 y ) j (1) 5 Since V is a vector, all its components must equal zero in order for V itself to be zero. Setting each component of Eq. 1 to zero, Simultaneous equations: x = -0. 421 m 4 3 u = 1 + 2. 5 x + y = 0 v = ? 0. 5 ? 1. 5 x ? 2. y = 0 y 2 We can easily solve this set of two equations and two unknowns simultaneously. Yes, there is one stagnation point, and it is located at Stagnation point: 10 m/s y = 0. 0526 m 1 0 (b) The x and y components of velocity are calculated from Eq. 1 for several (x,y) locations in the specified range. For example, at the point (x = 2 m, y = 3 m), u = 9. 00 m/s and v = -11 m/s. The magnitude of velocity (the speed) at that point is 14. 21 m/s. At this and at an array of other locations, the velocity vector is constructed from its two components, the results of which are shown in Fig. . The flow can be described as a counterclockwise turning, accelerating flow from the upper left to the lower right. The stagnation point of Part (a) does not lie in the upper right quadrant, and therefore does not appear on the sketch. -1 0 1 2 3 4 5 x FIGURE 1 Velocity vectors in the upper right quadrant for the given velocity field. Discussion The stagnation point location is given to three significant digits. It will be verified in Chap. 9 that this flow field is physically valid because it satisfies the differential equation for conservation of mass. 4-15 PROPRIETARY MATERIAL. 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 4 Fluid Kinematics 4-37 For the given velocity field, the material acceleration is to be calculated at a particular point and plotted at Solution several locations in the upper right quadrant. Assumptions 1 The flow is steady and incompressible. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or v with z. Analysis (a) The velocity field is

V = ( u , v ) = (1 + 2. 5 x + y ) i + ( ? 0. 5 ? 1. 5 x ? 2. 5 y ) j (1) Using the velocity field of Eq. 1 and the equation for material acceleration in Cartesian coordinates, we write expressions for the two non-zero components of the acceleration vector: ax = ?u ?u +u ?t ?x +v ?u ?y +w ?u ?z Scale: = 0 + (1 + 2. 5 x + y )( 2. 5 ) + ( ? 0. 5 ? 1. 5 x ? 2. 5 y )(1) + 0 10 m/s2 5 4 and ay = ?v ?v +u ?t ?x +v ?v ?y +w ?v ?z = 0 + (1 + 2. 5 x + y )( ? 1. 5 ) + ( ? 0. 5 ? 1. 5 x ? 2. 5 y )( ? 2. 5 ) + 0 3 y 2 1 At (x = 2 m, y = 3 m), ax = 11. 5 m/s2 and ay = 14. 0 m/s2. b) The above equations are applied to an array of x and y values in the upper right quadrant, and the acceleration vectors are plotted in Fig. 1. Discussion The acceleration vectors plotted in Fig. 1 point to the upper right, increasing in magnitude away from the origin. This agrees qualitatively with the velocity vectors of Fig. 1 of the previous problem; namely, fluid particles are accelerated to the right and are turned in the counterclockwise direction due to centripetal acceleration towards the upper right. Note that the acceleration field is non-zero, even though the flow is steady. 0 -1 0 1 2 3 4 5 x

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