Study guide for fluid mechanics textbook

I FLUID MECHANICS I.1 Basic Concepts & Definitions: Fluid Mechanics - Study of fluids at rest, in motion, and the effects of fluids on boundaries Note: This definition outlines the key topics in the study of fluids: (1) fluid statics (fluids at rest), (2) momentum and energy analyses (fluids in motion), and (3) viscous effects and all sections considering pressure forces (effects of fluids on boundaries) Fluid - A substance which moves and deforms continuously as a result of an applied shear stress The definition also clearly shows that viscous effects are not considered in the study of fluid statics Two important properties in the study of fluid mechanics are: Pressure and Velocity These are defined as follows: Pressure - The normal stress on any plane through a fluid element at rest Key Point: The direction of pressure forces will always be perpendicular to the surface of interest Velocity - The rate of change of position at a point in a flow field It is used not only to specify flow field characteristics but also to specify flow rate, momentum, and viscous effects for a fluid in motion | ▲ ▲ I-1 | e-Text Main Menu | Textbook Table of Contents | Study Guide I.4 Dimensions and Units This text will use both the International System of Units (S.I.) and British Gravitational System (B.G.) A key feature of both is that neither system uses gc Rather, in both systems the combination of units for mass * acceleration yields the unit of force, i.e Newton’s second law yields S.I Newton (N) = kg m/s2 B.G lbf = slug ft/s2 This will be particularly useful in the following: Concept Expression Units kg/s * m/s = kg m/s2 =N & mV momentum slug/s * ft/s = slug ft/s2 = lbf ρgh manometry kg/m3*m/s2*m = (kg m/s2)/ m2 =N/m2 slug/ft3*ft/s2*ft = (slug ft/s2)/ft2 = lbf/ft2 dynamic viscosity µ N s /m2 = (kg m/s2) s /m2 = kg/m s lbf s /ft2 = (slug ft/s2) s /ft2 = slug/ft s Key Point: In the B.G system of units, the mass unit is the slug and not the lbm and slug = 32.174 lbm Therefore, be careful not to use conventional values for fluid density in English units without appropriate conversions, e.g., ρw = 62.4 lb/ft3 For this case the manometer equation would be written as ∆P=ρ g h gc | ▲ ▲ I-2 | e-Text Main Menu | Textbook Table of Contents | Study Guide Example: Given: Pump power requirements are given by W& p = fluid density*volume flow rate*g*pump head = ρ Q g hp For ρ = 1.928 slug/ft3, Q = 500 gal/min, and hp = 70 ft, Determine: The power required in kW W& p = 1.928 slug/ft3 * 500 gal/min*1 ft /s /448.8 gpm*32.2 ft/s2 * 70 ft W& p = 4841 ft–lbf/s * 1.3558*10-3 kW/ft–lbf/s = 6.564 kW Note: We used the following: lbf = slug ft/s2 to obtain the desired units Recommendation: Properties of the velocity Field Two important properties in the study of fluid mechanics are Pressure and Velocity The basic definition for velocity has been given previously, however, one of its most important uses in fluid mechanics is to specify both the volume and mass flow rate of a fluid I-3 | ▲ ▲ 1.5 In working with problems with complex or mixed system units, at the start of the problem convert all parameters with units to the base units being used in the problem, e.g for S.I problems, convert all parameters to kg, m, & s; for BG problems, convert all parameters to slug, ft, & s Then convert the final answer to the desired final units | e-Text Main Menu | Textbook Table of Contents | Study Guide Volume flow rate: & = V ⋅ n dA = Q ∫ ∫ Vn dA cs cs where Vn is the normal component of velocity at a point on the area across which fluid flows Key Point: Note that only the normal component of velocity contributes to flow rate across a boundary Mass flow rate: & = ∫ ρV ⋅n d A = ∫ ρV d A m n cs cs NOTE: While not obvious in the basic equation, Vn must also be measured relative to any flow area boundary motion, i.e., if the flow boundary is moving, Vn is measured relative to the moving boundary This will be particularly important for problems involving moving control volumes in Ch III | ▲ ▲ I-4 | e-Text Main Menu | Textbook Table of Contents | Study Guide 1.6 Thermodynamic Properties All of the usual thermodynamic properties are important in fluid mechanics P - Pressure (kPa, psi) T- Temperature ( C, F) ρ ñ Density (kg/m3, slug/ft3) o o Alternatives for density γ - specific weight = weight per unit volume (N/m3, lbf/ft3) γ=ρg H2O: γ = 9790 N/m3 = 62.4 lbf/ft3 Air: γ = 11.8 N/m3 = 0.0752 lbf/ft3 S.G - specific gravity = ρ / ρ (ref) where: ρ (ref) = ρ (water at atm, 20˚C) for liquids = 998 kg/m3 = ρ (air at atm, 20˚C) for gases = 1.205 kg/m3 Example: Determine the static pressure difference indicated by an 18 cm column of fluid (liquid) with a specific gravity of 0.85 ∆P = ρ g h = S.G γ h = 0.85* 9790 N/m3 0.18 m = 1498 N/m2 = 1.5 kPa I.7 Transport Properties Certain transport properties are important as they relate to the diffusion of momentum due to shear stresses Specifically: µ ≡ coefficient of viscosity (dynamic viscosity) {M / L t } ν ≡ kinematic viscosity ( µ / ρ ) {L /t} | ▲ ▲ I-5 | e-Text Main Menu | Textbook Table of Contents | Study Guide This gives rise to the definition of a Newtonian fluid Newtonian fluid: A fluid which has a linear relationship between shear stress and velocity gradient dU dy The linearity coefficient in the equation is the coefficient of viscosity µ τ =µ Flows constrained by solid surfaces can typically be divided into two regimes: a Flow near a bounding surface with significant velocity gradients significant shear stresses This flow region is referred to as a "boundary layer." b Flows far from bounding surface with negligible velocity gradients negligible shear stresses significant inertia effects This flow region is referred to as "free stream" or "inviscid flow region." An important parameter in identifying the characteristics of these flows is the Reynolds number = Re = ρV L µ This physically represents the ratio of inertia forces in the flow to viscous forces For most flows of engineering significance, both the characteristics of the flow and the important effects due to the flow, e.g., drag, pressure drop, aerodynamic loads, etc., are dependent on this parameter | ▲ ▲ I-6 | e-Text Main Menu | Textbook Table of Contents | Study Guide II Fluid Statics From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: Pressure is constant along a horizontal plane, Pressure at a point is independent of orientation, Pressure change in any direction is proportional to the fluid density, local g, and vertical change in depth These concepts are key to the solution of problems in fluid statics, e.g Two points at the same depth in a static fluid have the same pressure The orientation of a surface has no bearing on the pressure at a point in a static fluid Vertical depth is a key dimension in determining pressure change in a static fluid If we were to conduct a more general force analysis on a fluid in motion, we would then obtain the following: ∇ P = ρ{ g − a} + µ ∇ V Thus the pressure change in fluid in general depends on: effects of fluid statics (ρ g), Ch II inertial effects (ρ a), Ch III viscous effects ( µ ∇ V ) Chs VI & VII Note: For problems involving the effects v of v both (1) fluid statics and (2) inertial effects, it is the net g − a acceleration vector that controls both the magnitude and direction of the pressure gradient | ▲ ▲ II-1 | e-Text Main Menu | Textbook Table of Contents | Study Guide This equation can be simplified for a fluid at rest (ie., no inertial or viscous effects) to yield ∇p = ρ g ∂p ∂x = 0; ∂p ∂y =0; ∂p ∂z = dp dz = −ρ g P2 − P1 = − ∫ ρ g d Z Free surface Pressure = Pa For liquids and incompressible fluids, this integrates to z P2 P1 – P2 = -ρg (Z2 – Z1) h =Z Z2 Note: P Z Z 1 y Z2 – Z1 is positive for Z2 above Z1 but x P2 – P1 is negative for Z2 above Z1 We can now define a new fluid parameter useful in static fluid analysis: γ = ρg ≡ specific weight of the fluid With this, the previous equation becomes (for an incompressible, static fluid) P2 – P1 = - γ (Z2 – Z1) The most common application of this result is that of manometry | ▲ ▲ II-2 | e-Text Main Menu | Textbook Table of Contents | Study Guide Consider the U-tube, multifluid manometer shown on the right If we first label all intermediate points between A & a, we can write for the overall pressure change PA - Pa = (PA- P1) + (P1 - P2) + (P2 - Pa ) This equation was obtained by adding and subtracting each intermediate pressure The total pressure difference now is expressed in terms of a series of intermediate pressure differences Substituting the previous result for static pressure difference, we obtain PA - PB = - ρ g(ZA- Z1) – ρ g (Z1 – Z2) – ρ g (Z2 - ZB ) Again note: Z positive up and ZA > Z1 , Z1 < Z2 , Z2 < Za In general, follow the following steps when analyzing manometry problems: On manometer schematic, label points on each end of manometer and each intermediate point where there is a fluid-fluid interface: e.g., A – – - B Express overall manometer pressure difference in terms of appropriate intermediate pressure differences PA - PB = (PA- P1) + (P1 – P2) + (P2 - PB ) Express each intermediate pressure difference in terms of appropriate product of specific weight * elevation change (watch signs) PA - PB = - ρ g(zA- z1) – ρ g (z1 – z2) – ρ g (z2 - zB ) Substitute for known values and solve for remaining unknowns | ▲ ▲ II-3 | e-Text Main Menu | Textbook Table of Contents | Study Guide When developing a solution for manometer problems, take care to: Include all pressure changes Use correct ∆Z and γ with each fluid Use correct signs with ∆ Z If pressure difference is expressed as PA – P1, the elevation change should be written as ZA – Z1 Watch units Manometer Example: a Given the indicated manometer, determine the gage pressure at A Pa = 101.3 kPa The fluid at A is Meriam red oil no S.G = 83 H A 18 cm ρgw = 9790 N/m 10 cm ρg A = S.G.*ρgw = 0.83*9790 N/m ρg A = 8126 N/m3 ρgair = 11.8 N/m3 With the indicated points labeled on the manometer, we can write PA - Pa = (PA- P1) + (P1 – P2) + (P2 - Pa ) Substituting the manometer expression for a static fluid, we obtain PA - Pa = - ρgA(zA- z1) – ρgw(z1 – z2) – ρga(z2 - za ) Neglect the contribution due to the air column Substituting values, we obtain PA - Pa = - 8126 N/m3 * 0.10 m – 9790 N/m3 * -0.18 = 949.6 N/m2 Note why: (zA- z1) = 0.10 m and (z1 – z2) = -0.18 m, & did not use Pa Review the text examples for manometry ▲ ▲ II-4 | | 2 e-Text Main Menu | Textbook Table of Contents | Study Guide Likewise for the outer radius r2 we have the following: a the circumferential velocity due to the impeller rotation u = r2 ω blade tip speed at outer radius b relative flow velocity tangent to the blade w2 tangent to the blade angle V2 w2 β2 Vn2 α2 Vt u2 β2 These again combine to yield the absolute outlet velocity V2 at angle α2 The exit absolute velocity can also be resolved into two absolute velocity components: Normal ( radial ) component: Vn2 = V2 sin α = w2 sin β2 = Q π r2 b2 Note that Q is the same as for the inlet flow rate Absolute tangential velocity: Vt = V2 cos α = u - w2 cos β2 Vt = u - Vn2 tan β = u2 - Q π r2 b tan β2 Q = A1 Vn1 = π r1 b1 Vn1 = A Vn2 = π r2 b Vn where Again, each of the above expressions follows easily from the velocity diagram, and the student should draw and use the diagram with each pump theory problem | ▲ ▲ XI - | e-Text Main Menu | Textbook Table of Contents | Study Guide We can now apply moment - of – momentum equation { T = ρ Q r2 * Vt − r1 * Vt1 } (again Vt1 is zero for the ideal design) For a sign convention, we have assumed that Vt1 and Vt2 are positive in the direction of impeller rotation The “ ideal” power supplied to the fluid is given by { Pw = ω T = ρ Q ω r2 Vt2 − ω r1 Vt1 or { } } Pw = ω T = ρ Q u Vt − u1 Vt1 = ρ Qg H Since these are ideal values, the shaft power required to drive a non-ideal pump is given by BHP = Pw ηp The head delivered to the fluid is H= ρ Q{u Vt2 − u1 Vt1 } ρQg u { = Vt − u Vt1 } g For the special case of purely radial inlet flow H* = u Vt g | ▲ ▲ XI - | e-Text Main Menu | Textbook Table of Contents | Study Guide From the exit velocity diagram, substituting for Vt2 we can show that u ωQ H= − g π b g tan β 2 has the form C1 - C2 Q shutoff head, the head produced at zero flow, Q = u2 where: C1 = g Example: A centrifugal water pump operates at the following conditions: speed = 1440 rpm, r1 = in, r2 = in, β1 = 30o, β2 = 20o, b1 = b2 = 1.75 in Assuming the inlet flow enters normal to the impeller (zero absolute tangential velocity): find: (a) Q, (b) T, (c) Wp, (d) hp, (e) ∆P ω = 1440 rev π rad = 150.8 60 s Calculate blade tip velocities: u1 = r1 ω = rad ft ft150.8 = 50.3 12 s s u = r2 ω = Since design is ideal, at inlet rad ft ft150.8 = 88 12 s s V1 = Vn1 w1 α1 = 90 , Vt1 = o Vn1 = U1 tan 300 = 50.3 tan 30o = 29.04 ft/s 30Þ 90Þ Q = π r1 b1 Vn1 r1 • | ▲ ▲ XI - | e-Text Main Menu | Textbook Table of Contents | Study Guide 30Þ u1 ft ft Q = π ft1.75 ft 29.04 = 8.87 s 12 s ft gal gal s Q = 8.87 60 7.48 = 3981 ft s Repeat for the outlet: ft 8.87 Q s Vn2 = = π r2 b 2 π ft 1.75 ft 12 12 ft Vn2 = 16.6 s Vn2 16.6 ft/s ft w2 = = = 48.54 s sin 20 o sin 20 o V2 w2 20ị 20ị u2 r2 ã Vt = u - w cos β2 = 88 − 48.54 cos 20 o = 42.4 ft s We are now able to determine the pump performance parameters Since for the centrifugal pump, the moment arm r1 at the inlet is zero, the momentum equation becomes Ideal moment of momentum delivered to the fluid: { T = ρ Q r2 * Vt2 } slug ft ft = 1.938 8.87 ft 42.4 = 425.1ft − lbf ft s s 12 Ideal power delivered to the fluid: P = ω T = 150.8 rad ft − lbf 425.1ft − lbf = 64,103 = 116.5 hp s s | ▲ ▲ XI - | e-Text Main Menu | Textbook Table of Contents | Study Guide Head produced by the pump (ideal): H= P 64,103 ft − lbf/s = = 115.9 ft lbf ft ρ gQ 62.4 8.87 ft s Pressure increase produced by the pump: ft ∆ P = ρ g H = 62.4 115.9 ft = 7226 psf = 50.2 psi s Pump Performance Curves and Similarity Laws Pump performance results are typically obtained from an experimental test of the given pump and are presented graphically for each performance parameter • Basic independent variable - Q {usually gpm or cfm } • Dependent variables typically – head pressure rise, in some cases ∆P H BHP – input power requirements (motor size) η – pump efficiency • These typically presented at fixed pump speed and impeller diameter Typical performance curves appear as | ▲ ▲ XI - 10 | e-Text Main Menu | Textbook Table of Contents | Study Guide Fig 11.6 Typical Centrifugal Pump Performance Curves at Fixed Pump Speed and diameter These curves are observed to have the following characteristics: hp is approximately constant at low flow rate hp = at Qmax BHP is not equal to at Q = BHP increases monotonically with the increase in Q ηp = at Q = and at Qmax Maximum pump efficiency occurs at approximately Q* = 0.6 Qmax This is the best efficiency point BEP At any other operating point, efficiency is less, pump head can be higher or lower, and BHP can be higher or lower At the BEP, Q = Q*, hp = hp*, BHP = BHP* Measured Performance Data Actual pump performance data will typically be presented graphically as shown in Fig 11.7 Each graph will usually have curves representing the pump head vs flow rate for two or more impeller diameters for a given class/model of pumps having a similar design The graphs will also show curves of constant efficiency and constant pump power (BHP) for the impeller diameters shown All curves will be for a fixed pump impeller speed | ▲ ▲ XI - 11 | e-Text Main Menu | Textbook Table of Contents | Study Guide Fig 11.7 Measured performance curves for two models of a centrifugal water pump | ▲ ▲ XI - 12 | e-Text Main Menu | Textbook Table of Contents | Study Guide How to Read Pump Performance Curves Care must be taken to correctly read the performance data from pump curves This should be done as follows: (1) For a given flow rate Q (2) Read vertically to a point on the pump head curve h for the impeller diameter D of interest (3) All remaining parameters ( efficiency & BHP) are read at this point; i.e., graphically interpolate between adjacent curves for BHP to obtain the pump power at this point Note that the resulting values are valid only for the conditions of these curves: (1) pump model and design, (2) pump speed – N, (3) impeller size – D, (4) fluid (typically water) Thus for the pump shown in Fig 11.7a with an impeller diameter D = 32 in, we obtain the following performance at Q = 20,000 gpm: Q = 20,000 gpm, D = 32 in, N = 1170 rpm H ≅ 385 ft, BHP ≅ 2300 bhp, ηp ≅ 86.3 % Note that points that are not on an h vs Q curve are not valid operating points Thus for Fig 11.7b, the conditions Q = 22,000 gpm, BHP = 1500 bhp, hp = 250 ft not correspond to a valid operating point because they not fall on one of the given impeller diameter curves However, for the same figure, the point Q = 20,000 gpm, BHP = 1250 bhp is a valid point because it coincidentally also falls on the D = 38 in impeller curve at hp = 227 ft | ▲ ▲ XI - 13 | e-Text Main Menu | Textbook Table of Contents | Study Guide Net Positive Suction Head - NPH One additional parameter is typically shown on pump performance curves: NPSH = head required at the pump inlet to keep the fluid from cavitating NPSH is defined as follows: P NPSH = i + ρg Vi P − v 2g ρg where Pi = pump inlet pressure Pv = vapor pressure of fluid Pump inlet Considering the adjacent figure, write the energy equation between the fluid surface and the pump inlet to obtain the following: P NPSH = i + ρg zi Pa Pi z=0 Vi P P P − v = a − Z i − h f,a−i − v 2g ρg ρ g ρg For a pump installation with this configuration to operate as intended, the righthand-side of the above equation must be > the NPSH value for the operating flow rate for the pump Example: A water supply tank and pump are connected as shown Pa = 13.6 psia and the water is at 20 o C with Pv = 0.34 psia The system has a friction loss of 4.34 ft Will the NPSH of the pump of Fig 11.7a at 20,000 gpm work? a 10 ft i | ▲ ▲ XI - 14 | e-Text Main Menu | Textbook Table of Contents | Study Guide Applying the previous equation we obtain NPSH = Pa P − Z i − h f,a−i − v ρg ρg 13.6 − 0.34) lbf/in2 *144 in /ft ( NPSH = − (−10 ft) − 4.34 ft 62.4 lbf/ft NPSH = 36.26 ft The pump will work because the system NPSH as shown in Fig 11.7a is 30 ft which provides a 6.3 ft safety margin Conversely, the pump could be located as close as 3.7 ft below the water surface and meet NPSH requirements Pump Similarity Laws Application of the dimensional analysis procedures of Ch V will yield the following three dimensionless performance parameters: Dimensionless flow coefficient: CQ = Q ω D3 Dimensionless head coefficient: CH = gH ω D2 Dimensionless power coefficient: CP = BHP ρω D5 where ω is the pump speed in radians/time and other symbols are standard design and operating parameters with units that make the coefficients dimensionless How are these used? These terms can be used to estimate design and performance changes between two pumps of similar design | ▲ ▲ XI - 15 | e-Text Main Menu | Textbook Table of Contents | Study Guide Stated in another way: If pumps and are from the same geometric design family and are operating at similar operating conditions, the flow rates, pump head, and pump power for the two pumps will be related according to the following expressions: Q2 N2 = Q1 N1  D2     D1  H  N2   D2  =    H1  N1   D1  Use to predict the new flow rate for a design change in pump speed N and impeller diameter D Used to predict the new pump head H for a design change in pump speed, N and impeller diameter D BHP2  ρ2   N   D2  =      BHP1  ρ1   N1   D1  Used to predict the new pump power BHP for a design change in fluid, ρ, pump speed N and impeller diameter D Example It is desired to modify the operating conditions for the 38 in diameter impeller pump of Fig 11.7b to a new pump speed of 900 rpm and a larger impeller diameter of 40 in • H(ft) BEP1• Determine the new pump head and power for the new pump speed at the BEP Q(gpm) | ▲ ▲ XI - 16 | e-Text Main Menu | Textbook Table of Contents | Study Guide BEP For the D = 38 in impeller of Fig 11.7b operating at 710 rpm, we read the best efficiency point (BEP) values as Q* = 20,000 gpm, H* = 225 ft, BHP * = 1250 hp Applying the similarity laws for N2 = 900 rpm and D2 = D1 = 38 in, we obtain 3 Q2 N2  D2  900  40  = = 1.478   = 710  38  Q1 N1  D1  Q2 = 20,000*1.478 = 29,570 gpm ans 2 H  N2   D2  900   40   =    = = 1.78 710   38  H1  N1   D1  H2 = 225*1.78 = 400.5 ft ans 5 BHP2  ρ2   N   D2  900   40   =       = (1) = 2.632 710   38  BHP1  ρ1   N1   D1  BHP2 = 3290 hp ans Thus, even small changes in the speed and size of a pump can result in significant changes in flow rate, head, and power It is noted that every point on the original 38 in diameter performance curve exhibits a similar translation to a new operating condition The similarity laws are obviously useful to predict changes in the performance characteristics of an existing pump or to estimate the performance of a modified pump design prior to the construction of a prototype | ▲ ▲ XI - 17 | e-Text Main Menu | Textbook Table of Contents | Study Guide Matching a Pump to System Characteristics The typical design/sizing requirement for a pump is to select a pump which has a pump head which matches the required system head at the design/operating flow rate for the piping system Key Point hp = hsys at Qdes It is noted that pump selection should occur such that the operating point of the selected pump should occur on the pump curve near or at the BEP From the energy equation in Ch VI, the system head is typically expressed as h sys V2 P2 − P1 V2 − V1 L   = + + Z − Z + f + ∑ K i   D  2g ρg 2g Thus the selection of a pump for a piping system design should result in a pump for which the pump head hp at the design flow rate Qdes is equal ( or very close) to the head η p hp Hdes requirements hsys of the piping system at the same flow rate, and this should occur at or near the point of maximum efficiency for the chosen pump • hsys Q(gpm) Qdes Other operating and performance requirements (such as NPSH) are obviously also a part of the selection criteria for a pump | ▲ ▲ XI - 18 | e-Text Main Menu | Textbook Table of Contents | Study Guide Pumping Systems: Parallel and Series Configurations For some piping system designs, it may be desirable to consider a multiple pump system to meet the design requirements Two typical options include parallel and series configurations of pumps Specific performance criteria must be met when considering these options Given a piping system which has a known design flow rate and head requirements, Qdes, hdes The following pump selection criteria apply Pumps in Parallel: Assuming that the pumps are identical, each pump must provide the following: Q(pump) = 0.5 Qdes h(pump) = hdes Pumps in Series: Assuming that the pumps are identical, each pump must provide the following: Q (pump) = Qdes h(pump) = 0.5 hdes For example, if the design point for a given piping system were Qdes = 600 gpm, and hsys = 270 ft, the following pump selection criteria would apply: Single pump system Q(pump) = 600 gpm, hp = 270 ft Parallel pump system Q(pump) = 300 gpm, hp = 270 ft for each of the two pumps Series pump system Q(pump) = 600 gpm, hp = 135 ft for each of the two pumps | ▲ ▲ XI - 19 | e-Text Main Menu | Textbook Table of Contents | Study Guide StudyGuide for Fluid Mechanics Preface The following materials are provided as a study guide for the text Fluid Mechanics by Frank White A brief summary of the key concepts and theory is presented for each chapter along with the final form of basic equations (without detailed derivations) used in the various analyses being presented In most cases, a detailed explanation for the physical significance of each term in a fundamental governing equation is given (e.g., linear momentum, pg III-8) to assist the student in identifying when a given term should be included in the analysis Example problems are provided for major sections In each case, the starting general equation used in the solution is given followed by any necessary simplifications and the resulting complete solution Where appropriate, the control volume and coordinate system used in the analysis are shown with the problem schematic In many cases, an explanation is given with the final numerical answer to help the student understand the engineering significance of the answer (e.g., forces on curved surfaces, pg II–18) In selected cases, computer based solutions to example problems are provided as an example to the student in the use of computer based problem solving techniques (e.g., parallel pipe sections, pg VI-23) For problems areas involving multiple steps in the solution, a summary of the steps used in a typical problem solution sequence is provided and enclosed in a boxed border (e.g., rigid body motion, pg II-22) Areas where the author’s experience has shown that mistakes in the analysis can easily occur are noted as Key Points (e.g., laminar flat plate boundary layer, pg VII-5) throughout the material Finally, the author of this study guide appreciates the opportunity to contribute to the instructional materials provided with one of the leading texts in the area of fluid mechanics and to collaborate with an educator with whom he has long has the highest respect and had the privilege to further his education in fluid mechanics while a student at Georgia Tech | ▲ ▲ Jerry R Dunn, P.E Associate Professor, Department of Mechanical Engineering Texas Tech University | e-Text Main Menu | Textbook Table of Contents | Study Guide ... Main Menu | Textbook Table of Contents | Study Guide II Fluid Statics From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous,... {L /t} | ▲ ▲ I-5 | e-Text Main Menu | Textbook Table of Contents | Study Guide This gives rise to the definition of a Newtonian fluid Newtonian fluid: A fluid which has a linear relationship... properties in the study of fluid mechanics are Pressure and Velocity The basic definition for velocity has been given previously, however, one of its most important uses in fluid mechanics is to