CHAPTER 2 Fluid Statics


 Cory Harvey
 3 years ago
 Views:
Transcription
1 Chapter / Fluid Statics CHPTER Fluid Statics FEtype Eam Review Problems: Problems  to 9. (C). (D). (C).4 ().5 () The pressure can be calculated using: p = γ h were h is the height of mercury. p= γ h= (.6 980N/m ) (8.5 in m/in) = Pa Hg Since the pressure varies in a vertical direction, then: Hg p= p0 ρgh= Pa.00kg/m 9.8m/s 4000m = Pa p = p + γ h γ h = = 800 Pa w atm m m water w This is the gage pressure since we used p atm =0. Initially, the pressure in the air is pir, = γ H = (.6 980) 0.6 = 50 Pa. fter the pressure is increased we have: p = = 50 = H. H = m ir, The moment of force P with respect to the hinge, must balance the moment of 5 the hydrostatic force F with respect to the hinge, that is: ( ) P = F d 5 F = γ h= 9.8 kn/m m ( m )] F = 98. kn The location of F is at ( ) I. yp = y+ =.67 + =. m d =.. =. m y.67(. ). P = 98.. P =.7 kn The gate opens when the center of pressure is at the hinge:.6 (). + h I. + h b(. + h) / y = + 5. yp = y + = + = y (. + h) b(. + h) / This can be solved by trialanderror, or we can simply substitute one of the answers into the equation and check to see if it is correct. This yields h =.08 m. 4 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2 Chapter / Fluid Statics.7 (D).8 () The hydrostatic force will pass through the center, and so F H will be balanced by the force in the hinge and the force P will be equal to F. P= F = w+ 9.8 ( π. / 4) w= 00. w = 5.6 m. The weight is balanced by the buoyancy force which is given by F where is the displaced volume of fluid: = w. w = 6 m = γ The pressure on the plug is due to the initial pressure plus the pressure due to the acceleration of the fluid, that is: pplug = pinitial + γ gasolineδ Z where,.9 () a Δ Z =Δ g 5 p plug = (. ) = Pa 9.8 F = p = π 0.0 = 0.5 N. plug plug Pressure..4 Since p = γ h, then h = p/γ a) h = /980 = 5.5 m c) h = /(.6 980) =.874 m This requires that p = p ( γh) = ( γh) water Hg water Hg b) 980 h = (.6 980) 0.75 h = 0. m..6 Δ p = γδz Δp = (0,000) = 77 psf or 5.7 psi..8 From the given information the specific gravity is S =.0 + z/00 since S(0) = and S(0) =.. y definition ρ = 000 S, where ρ water = 000 kg/m. Using dp = γ dz then, by integration we write: p 0 z dp = 000( + z /00) gdz = 000g z p = ( 0 + ) = Pa or 0 kpa Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
3 Chapter / Fluid Statics Note: we could have used an average S: S avg =.05, so that ρ avg = 050 kg/m and so p = γ h p = = Pa.0 / From Eq. (.4.8): atm[( 0 ) / 0] g α p = p T α z T R = 00 [( )/88] 9.8/ = kpa ssuming constant density, then 00 p= p atm ρ gh= / = kpa % error = 00 = 0.05% Since the error is small, the density variation can be ignored over heights of 00 m or less. Eq..5. gives ρ gdh = d ρ or ρ dp = ρ ut, dp = ρgdh. Therefore, d ρ dρ ρ = T g dh Integrate, using ρ 0 =.00 slug/ft, and =,000 lb/in :..4 ρ h dρ g dh ρ =. 0. ft/s = h= 7. 0 ρ, 000 ( lb/in ) 44 in ft This gives ρ = h Now, h h g g 7 p = ρ gdh = dh ln( h) h = If we assume ρ = const: p = ρ gh =.0. h = 64.4h b) For h = 5000 ft: p accurate =,00 psf and p estimate =,000 psf., 000, 00 % error = 00 = 0.7 %, 00 Use Eq..4.8: p = 0( / 88) 7 h 9.8 z Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4 Chapter / Fluid Statics a) for z = 000 p = 69.9 kpa. c) for z = 9000 p = 0.6 kpa. Manometers Referring to Fig..7b, the pressure in the pipe is: p = γh = (.6 980) 0.5 = 50 Pa or.5 kpa. Referring to Fig..7a, the pressure in the pipe is p = ρgh. If p = 400 Pa, then 400 = ρgh = ρ 9.8h or 400 ρ = 9.8h 400 a) ρ = = 680 kg/m The fluid is gasoline c) ρ = = 999 kg/m The fluid is water See Fig..7b: The pressure in the pipe is given by p = γ h + γ H p = p p = γ 0. + γ H γ 0. water oil oil Hg water = psf or 4.5 psi = H Solving for H we get: H = 0.74 m or 7.4 cm ( ) p = γ 0.6 γ (0.0) γ (0.04) γ (0.0) water Hg water Hg water Using γ water p water = 5.6 kpa = 9.8 kn/m and γ = kn/m Hg p water = p oil With p water = 5 kpa, p oil = 4 kpa pgage = pair + γ water 4 where, pair = patm γ Hg H p = γ H + γ 4 gage Hg water p gage = = 7.89 kpa Note: we subtracted atmospheric pressure since we need the gage pressure. 7 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
5 Chapter / Fluid Statics.40 p = p = 587 Pa or 5.87 kpa. The distance the mercury drops on the left equals the distance along the tube that the mercury rises on the right. This is shown in the sketch. Oil (S = 0.87) Water Δh 0 cm 9 cm 7 cm Δh 40 o Mercury.4.44 From the previous problem we have: ( p) p γwater γhg γoil = sin =. kpa () For the new condition: ( ) = + γ ( +Δ ) γ γ ( Δ ) p p h h () water 0.07 HG 0.sin 40 oil 0. sin 40 where Δh in this case is calculated from the new manometer reading as: Δ h+δ h/ sin 40 = 9 cm Δ h= 0.78 cm Subtracting Eq.() from Eq.() yields: ( p ) ( p ) = γ ( Δh) γ γ ( Δh ) water HG oil 0.0sin 40 sin 40 Substituting the value of Δh gives: ( ) ( ) ( ) p = sin sin = 0.5 kpa a) Using Eq. (.4.6): ( ) ( ) p = γ z z + γh+ γ γ H where h= z5 z=7 6 = cm 4000 = 9800(0.6 0.) (0.0) + ( )H H = m or.76 cm 8 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6 Chapter / Fluid Statics.46 From No..0: p oil = 4.0 kpa From No..6: p oil = p water 9.8 (0.+Δz) (0. Δz) (0. Δz) Δz = m or 4.5 cm. Forces on Plane reas The hydrostatic force is calculated using: F = γ h where, h = 0 m, and = πr = π ( 0.5 m) hence, F π ( ) = = 694 N. For saturated ground, the force on the bottom tending to lift the vault is: F = p c = ( ) = N The weight of the vault is approimately: W = ρ g walls ( ) ( ) ( ) W = = N. The vault will tend to rise out of the ground..5 b) Since the triangle is horizontal the force is due to the uniform pressure at a depth of 0 m. That is, F = p, where p= γ h= = 98. kn/m The area of the triangle is = bh =.88 / =.88 m.54 F = = 77.4 kn. a) F = γh= π = 79.7 kn 4 I π /4 yp = y+ = 6 + = 6.67 m (, y) p = (0, 0.67) m y 4π 6 c) F = 9.8 (4 + 4/) 6 =.9 kn y p = /. 6 = 5.50 m y = /.5 =.5. = (, y) p = (0.975,.5) m y.56 F = γh = = N, or 77 kn 9 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
7 Chapter / Fluid Statics y = I p y + y = / 75. = m Σ = 0 ( ) 77 = 5 P P = 5 kn M Hinge The vertical height of water is h =. 0.4 =.4 m The area of the gate can be split into two areas: = + or = =.80 m Use forces: F = γ h = (..4) = 754 N.60.4 F = γ h = 980 (0.4.4) = 674 N The location of F is at y p = (.4) = m, and F is at y p I / 6 = y+ = + = m y 0.4 (.4 / ) (.4 / ).4 ΣM hinge = 0 : ( ).4P = 0 P = 46 N The gate is about to open when the center of pressure is at the hinge..6 b) b / yp =. + H = (.0/ + H) + ( + H)b H = m freebodydiagram of the gate and block is sketched. T Sum forces on the block: Σ F = 0 W = T + F y F stop 0 T.64 where F is the buoyancy force which is given by F = γ πr ( H) F H y p F W Take moments about the hinge: R T.5 = F ( y ) H p Ry where F H is the hydrostatic force acting on the gate. It is, using 0 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
8 Chapter / Fluid Statics h = = = FH.5 m and 6 m, ( )( ) = γ h= 9.8 kn/m.5 m 6 m = 88.9 kn From the given information, y p ( ) I / = y+ =.5 + = m y.5 6 ( ) 88.9 T = = 5. kn.5 F = W T = = kn. γπ R H = kn H = m =.55 m ( 9.8 kn/m ) π ( m) ( ).66 The dam will topple if there is a net clockwise moment about O. The weight of the dam consists of the weight of the rectangular area + a triangular area, that is: W = W+ W. The force F acting on the bottom of the dam can be divided into two forces: Fp due to the uniform pressure distribution and Fp due to the linear pressure distribution. W = = 56, 609 lb assume m deep b) W = / =, 9 lb W = 6.4 (60.86/) = 4, 794 lb F = =,0 lb F = =,0 lb F p = = 8,70 lb W F W F O F F p = / = 46,800 lb Σ M O : (,0)(0) + (8,70)(5) + (46,800)(0) (56,609)(),0(0/) (,9)(4) 4,794(.8) = 740,78 > 0. will tip. 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
9 Chapter / Fluid Statics Forces on Curved Surfaces Since all infinitesimal pressure forces pass thru the center, we can place the resultant forces at the center. Since the vertical components pass thru the bottom point, they produce no moment about that point. Hence, consider only horizontal forces: ( γ h) ( γ h) ( F ) = = 9.8 (4 0) = kn H water water ( F ) = = = 68.7 kn H oil oil ΣM: P = P = 66. kn freebodydiagram of the volume of water in the vicinity of the surface is shown. Force balances in the horizontal and vertical directions give: FH = F F = W + F where FH and F are the horizontal and vertical components of the force acting on the water by the surface. Hence, FH = F = 9.8 kn/m = 706. kn ( )( )( ) The line of action of F H is the same as that of F. Its distance from the surface is y p ( ) I 4 = y+ = 9 + = 9.07 m y 9 8 F. W F F F H. To find F we find W and F : π W = γ = ( 9.8 kn/m ) ( ) 4 =.7 kn 4 F ( ) = 9.8 kn/m 8 4 = 68 kn F = F + W = = 66 kn To find the line of action of F, we take moments at point : F = F d + W d where R d = m, and d = = =.55 m: 4 4 ( π) ( π) 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10 Chapter / Fluid Statics F d+ W d = = =.08 m F 66 Finally, the forces F H and F that act on the surface are equal and opposite to those calculated above. So, on the surface, F H acts to the right and F acts downward..7 Place the resultant FH + F at the center. F passes through the hinge. The moment of F H must equal the moment of P with respect to the hinge: (9.8 0) =.8 P. P = 70. kn. The resultant F + F H of the unknown liquid acts thru the center of the circular arc. F passes through the hinge. Thus, we use only F ). ssume m wide: ( )( ) F = γ h= γ R R = γ R H ( H.74 F = γ h= γ R R = γ R The horizontal force due to the water is ( )( ) The weight of the gate is W Sγw γw( πr ) w w w w = = 0. 4 Summing moments about the hinge: ( ) ( 4 π ) F R + W R = F R w H R R R R R a) π = γ R 4 π γ = 4580 N/m.76 The pressure in the dome is: a) p = = Pa or 4.87 kpa The force is F = p projected = (π ) 4.87 = 40.4 kn b) From a freebody diagram of the dome filled with oil: W F weld + W = p p Using the pressure from part (a): F weld F weld = π ( ) 4 π = 400 or.4 kn 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
11 Chapter / Fluid Statics uoyancy.78 Under static conditions the weight of the barge + load = weight of displaced water. (a) = 980 (6 d + d /). d + d 8.5 = 0 d =.7 m.80.8 The weight of the cars will be balanced by the weight of displaced water: = 6.4(5 00 Δd ) Δd = ft or 4.6 in T + F = W (See Fig.. c.) T = = 8804 N or kn t the limit of lifting: F = W+p where p is the pressure acting on the plug. (b) ssume h> 5 + Rand use the above equation with W F.86 R =. ft and h = 6.4 ft: ( ) F = γ = γ 0 πr = = 858 lb w w segment ( ) W + p= 500 lb π 4 = 857 lb Hence, the plug will lift for h >6.4 ft. θ p R h 5 (a) When the hydrometer is completely submerged in water:.88 W = γ w π 0.05 π (0.0 + m Hg)9.8 = m Hg = kg When the hydrometer without the stem is submerged in a fluid: W = γ S =.089 ( 0.05) π ( )9.8 = S Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12 Chapter / Fluid Statics Stability.90 With ends horizontal ( γ π ) 4 Io = π d 64 The displaced volume is = W γ water = d h/ 4 / 980 = γ d since h = d. 5 The depth the cylinder will sink is depth = γ d = = π d /4 5 γ d The distance CG is CG h 5 = 0. 0 γ d /. Then GM = I O 4 πd 64 CG = γ d / d γ d / > 0 This gives (divide by d and multiply by γ ): γ γ > 0. Consequently, γ > 868 N/m or γ < 46 N/m s shown, y = = 6.5 cm above the bottom edge γ γ Sγ 4 G = = 6.5 cm. 0.5γ 8 + γ 8 + S γ S = S. S =. The centroid C is.5 m below the water surface. CG =.5 m Using Eq..4.47: 8 / GM =.5 = = > The barge is stable. Linearly ccelerating Containers.96 (a) p ma = (0 4) 000 (9.8) (0 ) = Pa (c) p ma = (0 ).94 (. + 60) (0 6) = 470 psf or 7.5 psi.98 Use Eq..5.: b) = 000 a ( 8) 000 ( ).5 + 8a Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
13 Chapter / Fluid Statics 8a 60 = 8 a or a. = a 5. a +.44 = 0 a = 0.5, 4.8 m/s a.00 a) The pressure on the end (z is zero at ) is, using Eq..5., p(z) = ( 7.66) (z) = z 5. F = ( z) 4dz = N or 640 kn 0 b) The pressure on the bottom C is p() = ( 7.66) = FC = ( ) 4d = N or 6 kn 0 Use Eq..5. with position at the open end: b) p = (0.9 0) = 9000 Pa. z p = (0.9) ( 0.6) = 4 Pa.0 p C = ( 0.6) = 5886 Pa. e) p = = 64 psf. C p = = 4 psf. p C = = 0 psf. Rotating Containers Use Eq..6.4 with position at the open end: z a) p = (0 0.9 ) = Pa.04 p = = Pa p C = = 5886 Pa r C ω c) p = = 947 psf 6 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14 Chapter / Fluid Statics p = = 87 psf p C = = 0 psf The air volume before and after is equal. πrh 0 = π. 6.. rh 0 = h z r 0 (a) Using Eq..6.5: r 0 5 / = 9.8 h h = 0.48 m p = ( 0.7) r.06 = 849 Pa (c) For ω = 0, part of the bottom is bared. π Using Eq..6.5: ω 0 g π π = r0 h r h r ω r = h, = h g g g = h h ω ω h h = 9.8 or h lso, h h = h 0.64 = h = m, r = 0.08 m p = ( ) = 7,400 Pa h z r 0 r pr () = [0 (0.8 )] ρω r ρg h d = π r d r pr ( ) = 500ω r + 980(0.8 h) if h < 0.8 d r.08 pr () = ω ( r r ) if h > a) 0.6 π π ( ) = 6670 N 0 F = p rdr = r + r dr (We used h = 0.48 m) 7 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15 Chapter / Fluid Statics c) 0.6 π π (50 000( 0.08 ) = 950 N 0.08 F = p rdr = r r dr (We used r = 0.08 m) 8 0 Cengage Learning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
CHAPTER 2 Fluid Statics
Chapter / Fluid Statics CHPTER Fluid Statics FEtpe Exam Review Problems: Problems  to 9. (C) p = γ h = (.6 98) (8.5.54) = 96 6 Pa = 96.6 kpa Hg. (D) p = p ρ gh = 84. 9.8 4 = 44 76 Pa. (C) p = p + γ
More informationCHAPTER 2 Fluid Statics
Chater / Fluid Statics CHAPTER Fluid Statics FEtye Exam Review Problems: Problems  to 9. (C) h (.6 98) (8.5.54) 96 6 Pa Hg. (D) gh 84. 9.8 4 44 76 Pa. (C) h h. 98. 8 Pa w atm x x water w.4 (A) H (.6
More informationCHAPTER 2 Fluid Statics
Chater / Fluid Statics CHAPTER Fluid Statics FEtye Exam Review Problems: Problems  to 9. (C) h (.6 98) (8.5.54) 96 6 Pa Hg. (D) gh 84. 9.8 4 44 76 Pa. (C) h h. 98. 8 Pa w atm x x water w.4 (A) H (.6
More informationChapter 3 Fluid Statics
Chapter 3 Fluid Statics 3.1 Pressure Pressure : The ratio of normal force to area at a point. Pressure often varies from point to point. Pressure is a scalar quantity; it has magnitude only It produces
More informationMECHANICS of FLUIDS, 4 TH EDITION, SI
A STUDENT S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS, 4 TH EDITION, SI MERLE C. POTTER DAID C. WIGGERT BASSEM H. RAMADAN STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY MECHANICS of FLUIDS FOURTH EDITION,
More informationFluid Mechanics61341
AnNajah National University College of Engineering Fluid Mechanics61341 Chapter [2] Fluid Statics 1 Fluid Mechanics2nd Semester 2010 [2] Fluid Statics Fluid Statics Problems Fluid statics refers to
More informationstorage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface.
Hydrostatic Forces on Submerged Plane Surfaces Hydrostatic forces mean forces exerted by fluid at rest.  A plate exposed to a liquid, such as a gate valve in a dam, the wall of a liquid storage tank,
More informationP = ρ{ g a } + µ 2 V II. FLUID STATICS
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: 1. Pressure is constant
More informationEric G. Paterson. Spring 2005
Eric G. Paterson Department of Mechanical and Nuclear Engineering Pennsylvania State University Spring 2005 Reading and Homework Read Chapter 3. Homework Set #2 has been posted. Due date: Friday 21 January.
More informationIntroduction to Marine Hydrodynamics
1896 1920 1987 2006 Introduction to Marine Hydrodynamics (NA235) Department of Naval Architecture and Ocean Engineering School of Naval Architecture, Ocean & Civil Engineering Shanghai Jiao Tong University
More informationFluid Statics. Pressure. Pressure
Pressure Fluid Statics Variation of Pressure with Position in a Fluid Measurement of Pressure Hydrostatic Thrusts on Submerged Surfaces Plane Surfaces Curved Surfaces ddendum First and Second Moment of
More informationCHAPTER 4 The Integral Forms of the Fundamental Laws
CHAPTER 4 The Integral Forms of the Fundamental Laws FEtype Exam Review Problems: Problems 4 to 45 4 (B) 4 (D) 4 (A) 44 (D) p m ρa A π 4 7 87 kg/s RT 87 9 Refer to the circle of Problem 47: 757 Q A
More informationHydrostatics. ENGR 5961 Fluid Mechanics I: Dr. Y.S. Muzychka
1 Hydrostatics 2 Introduction In Fluid Mechanics hydrostatics considers fluids at rest: typically fluid pressure on stationary bodies and surfaces, pressure measurements, buoyancy and flotation, and fluid
More informationDIMENSIONS AND UNITS
DIMENSIONS AND UNITS A dimension is the measure by which a physical variable is expressed quantitatively. A unit is a particular way of attaching a number to the quantitative dimension. Primary Dimension
More informationFluid Mechanics Discussion. Prepared By: Dr.Khalil M. AlAstal Eng.Ahmed S. AlAgha Eng.Ruba M. Awad
Discussion Prepared By: Dr.Khalil M. AlAstal Eng.Ahmed S. AlAgha Eng.Ruba M. Awad 20142015 Chapter (1) Fluids and their Properties Fluids and their Properties Fluids (Liquids or gases) which a substance
More informationCHARACTERISTIC OF FLUIDS. A fluid is defined as a substance that deforms continuously when acted on by a shearing stress at any magnitude.
CHARACTERISTIC OF FLUIDS A fluid is defined as a substance that deforms continuously when acted on by a shearing stress at any magnitude. In a fluid at rest, normal stress is called pressure. 1 Dimensions,
More informationQ1 Give answers to all of the following questions (5 marks each):
FLUID MECHANICS First Year Exam Solutions 03 Q Give answers to all of the following questions (5 marks each): (a) A cylinder of m in diameter is made with material of relative density 0.5. It is moored
More informationCHAPTER 28 PRESSURE IN FLUIDS
CHAPTER 8 PRESSURE IN FLUIDS EXERCISE 18, Page 81 1. A force of 80 N is applied to a piston of a hydraulic system of crosssectional area 0.010 m. Determine the pressure produced by the piston in the hydraulic
More informationChapter 1 INTRODUCTION
Chapter 1 INTRODUCTION 11 The Fluid. 12 Dimensions. 13 Units. 14 Fluid Properties. 1 11 The Fluid: It is the substance that deforms continuously when subjected to a shear stress. Matter Solid Fluid
More informationM E 320 Professor John M. Cimbala Lecture 05
M E 320 Professor John M. Cimbala Lecture 05 Today, we will: Continue Chapter 3 Pressure and Fluid Statics Discuss applications of fluid statics (barometers and Utube manometers) Do some example problems
More informationANSWER KEY. 3. B 15. A 27. 3m 39. B 51. D 63. C 40. C 52. A 64. D
NSWE KEY CUMUIVE ES (578)_Solutions.. N/m. C 5. 7. C 49. 6. D. m 4. D 6. C 8. D 5. 6. D. 5. 7. m 9. 5. D 6. C 4. 6. 8. W 9 4. C 5. 64. D 5. C 7. C 9..559m 4. D 5. C 65. D 6. D 8. 4%. 4. a 4k M 54. 7.
More informationPressure in stationary and moving fluid Lab Lab On On Chip: Lecture 2
Pressure in stationary and moving fluid LabOnChip: Lecture Lecture plan what is pressure e and how it s distributed in static fluid water pressure in engineering problems buoyancy y and archimedes law;
More informationThe online of midtermtests of Fluid Mechanics 1
The online of midtermtests 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.
More informationChapter 2 Hydrostatics Buoyancy, Floatation and Stability
Chapter 2 Hydrostatics uoyancy, Floatation and Stability Zerihun Alemayehu Rm. E119 AAiT Force of buoyancy an upward force exerted by a fluid pressure on fully or partially floating body Gravity Archimedes
More informationMULTIPLECHOICE PROBLEMS:(Two marks per answer) (Circle the Letter Beside the Most Correct Answer in the Questions Below.)
MULTIPLECHOICE PROLEMS:(Two marks per answer) (Circle the Letter eside the Most Correct Answer in the Questions elow.) 1. The absolute viscosity µ of a fluid is primarily a function of: a. Density. b.
More informationCEE 342 Aut Homework #2 Solutions
CEE 4 Aut 005. Homework # Solutions.9. A freebody diagram of the lower hemisphere is shown below. This set of boundaries for the free body is chosen because it isolates the force on the bolts, which is
More informationFluid Mechanics. Forces on Fluid Elements. Fluid Elements  Definition:
Fluid Mechanics Chapter 2: Fluid Statics Lecture 3 Forces on Fluid Elements Fluid Elements  Definition: Fluid element can be defined as an infinitesimal region of the fluid continuum in isolation from
More informationSourabh V. Apte. 308 Rogers Hall
Sourabh V. Apte 308 Rogers Hall sva@engr.orst.edu 1 Topics Quick overview of Fluid properties, units Hydrostatic forces Conservation laws (mass, momentum, energy) Flow through pipes (friction loss, Moody
More informationME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 3. (Hydrostatic forces on flat and curved surfaces)
ME 262 BSIC FLUID MECHNICS ssistant Professor Neslihan Semerci Lecture 3 (Hydrostatic forces on flat and curved surfaces) 15. HYDROSTTIC PRESSURE 15.1 Hydrostatic pressure force on flat surfaces When a
More informationLagrangian description from the perspective of a parcel moving within the flow. Streamline Eulerian, tangent line to instantaneous velocity field.
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
More informationPressure in stationary and moving fluid. LabOnChip: Lecture 2
Pressure in stationary and moving fluid LabOnChip: Lecture Fluid Statics No shearing stress.no relative movement between adjacent fluid particles, i.e. static or moving as a single block Pressure at
More informationCivil Engineering Hydraulics Mechanics of Fluids. Pressure and Fluid Statics. The fastest healing part of the body is the tongue.
Civil Engineering Hydraulics Mechanics of Fluids and Fluid Statics The fastest healing part of the body is the tongue. Common Units 2 In order to be able to discuss and analyze fluid problems we need to
More informationMULTIPLECHOICE PROBLEMS :(Two marks per answer) (Circle the Letter Beside the Most Correct Answer in the Questions Below.)
Test Midterm 1 F2013 MULTIPLECHOICE PROBLEMS :(Two marks per answer) (Circle the Letter Beside the Most Correct nswer in the Questions Below.) 1. The absolute viscosity µ of a fluid is primarily a function
More informationCivil Engineering Hydraulics. Pressure and Fluid Statics
Civil Engineering Hydraulics and Fluid Statics Leonard: It wouldn't kill us to meet new people. Sheldon: For the record, it could kill us to meet new people. Common Units 2 In order to be able to discuss
More informationFE Exam Fluids Review October 23, Important Concepts
FE Exam Fluids Review October 3, 013 mportant Concepts Density, specific volume, specific weight, specific gravity (Water 1000 kg/m^3, Air 1. kg/m^3) Meaning & Symbols? Stress, Pressure, Viscosity; Meaning
More informationLiquids CHAPTER 13 FLUIDS FLUIDS. Gases. Density! Bulk modulus! Compressibility. To begin with... some important definitions...
CHAPTER 13 FLUIDS FLUIDS Liquids Gases Density! Bulk modulus! Compressibility Pressure in a fluid! Hydraulic lift! Hydrostatic paradox Measurement of pressure! Manometers and barometers Buoyancy and Archimedes
More informationMEB41 Lab 1: Hydrostatics. Experimental Procedures
MEB41 Lab 1: Hydrostatics In this lab you will do four brief experiments related to the following topics: manometry, buoyancy, forces on submerged planes, and hydraulics (a hydraulic jack). Each experiment
More informationFluids. Fluid = Gas or Liquid. Density Pressure in a Fluid Buoyancy and Archimedes Principle Fluids in Motion
Chapter 14 Fluids Fluids Density Pressure in a Fluid Buoyancy and Archimedes Principle Fluids in Motion Fluid = Gas or Liquid MFMcGrawPHY45 Chap_14HaFluidsRevised 10/13/01 Densities MFMcGrawPHY45 Chap_14HaFluidsRevised
More informationFluid Engineering Mechanics
Fluid Engineering Mechanics Chapter 3 Fluid Statics: ressure intensity and pressure head: pressure and specific weight relationship, absolute and gauge pressure, Forces on submerged planes & curved surfaces
More informationCHAPTER 13. Liquids FLUIDS FLUIDS. Gases. Density! Bulk modulus! Compressibility. To begin with... some important definitions...
CHAPTER 13 FLUIDS Density! Bulk modulus! Compressibility Pressure in a fluid! Hydraulic lift! Hydrostatic paradox Measurement of pressure! Manometers and barometers Buoyancy and Archimedes Principle! Upthrust!
More informationCHAPTER 2 Pressure and Head
FLUID MECHANICS Gaza, Sep. 2012 CHAPTER 2 Pressure and Head Dr. Khalil Mahmoud ALASTAL Objectives of this Chapter: Introduce the concept of pressure. Prove it has a unique value at any particular elevation.
More informationCHEN 3200 Fluid Mechanics Spring Homework 3 solutions
Homework 3 solutions 1. An artery with an inner diameter of 15 mm contains blood flowing at a rate of 5000 ml/min. Further along the artery, arterial plaque has partially clogged the artery, reducing the
More informationChapter 15  Fluid Mechanics Thursday, March 24 th
Chapter 15  Fluid Mechanics Thursday, March 24 th Fluids Static properties Density and pressure Hydrostatic equilibrium Archimedes principle and buoyancy Fluid Motion The continuity equation Bernoulli
More informationFluid Dynamics Exam #1: Introduction, fluid statics, and the Bernoulli equation March 2, 2016, 7:00 p.m. 8:40 p.m. in CE 118
CVEN 311501 (Socolofsky) Fluid Dynamics Exam #1: Introduction, fluid statics, and the Bernoulli equation March 2, 2016, 7:00 p.m. 8:40 p.m. in CE 118 Name: : UIN: : Instructions: Fill in your name and
More information( 4. AP Exam Practice Questions for Chapter 7. AP Exam Practice Questions for Chapter 7 1 = = x dx. 1 3x So, the answer is A.
AP Eam Practice Questions for Chapter 7 AP Eam Practice Questions for Chapter 7. e e So, the answer is A. e e ( ) A e e d e e e e. 7 + + (, ) + (, ) (, ) 7 + + + 7 + 7 + ( ) ( ),, A d + + + + + + + d +
More informationCHAPTER 1 Basic Considerations
CHAPTER Basic Considerations FEtype Exam Review Problems: Problems. to. Chapter / Basic Considerations. (C) m = F/a or kg = N/m/s = N s /m. (B) [μ] = [τ/(/dy)] = (F/L )/(L/T)/L = F. T/L. (A) 8 9.6 0 Pa
More information! =!"#$% exerted by a fluid (liquid or gas) !"#$ =!"# FUNDAMENTAL AND MEASURABLE INTENSIVE PROPERTIES PRESSURE, TEMPERATURE AND SPECIFIC VOLUME
FUNDAMENTAL AND MEASURABLE INTENSIVE PROPERTIES PRESSURE, TEMPERATURE AND SPECIFIC VOLUME PRESSURE, P! =!"#$%!"#! exerted by a fluid (liquid or gas) Thermodynamic importance of pressure One of two independent
More informationCE MECHANICS OF FLUIDS
CE60  MECHANICS OF FLUIDS (FOR III SEMESTER) UNIT II FLUID STATICS & KINEMATICS PREPARED BY R.SURYA, M.E Assistant Professor DEPARTMENT OF CIVIL ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SRI VIDYA COLLEGE
More informationDetermine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1 4a.
E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct
More informationCIVE HYDRAULIC ENGINEERING PART I Pierre Julien Colorado State University
CIVE 401  HYDRAULIC ENGINEERING PART I Pierre Julien Colorado State University Problems with and are considered moderate and those with are the longest and most difficult. In 2018 solve the problems with
More informationMTE 119 STATICS FINAL HELP SESSION REVIEW PROBLEMS PAGE 1 9 NAME & ID DATE. Example Problem P.1
MTE STATICS Example Problem P. Beer & Johnston, 004 by Mc GrawHill Companies, Inc. The structure shown consists of a beam of rectangular cross section (4in width, 8in height. (a Draw the shear and bending
More informationPhysics 107 HOMEWORK ASSIGNMENT #9
Physics 07 HOMEORK ASSIGNMENT #9 Cutnell & Johnson, 7 th edition Chapter : Problems 6, 8, 33, 40, 44 *6 A 58kg skier is going down a slope oriented 35 above the horizontal. The area of each ski in contact
More informationNew Website: M P E il Add. Mr. Peterson s Address:
Brad Peterson, P.E. New Website: http://njut009fall.weebly.com M P E il Add Mr. Peterson s Email Address: bradpeterson@engineer.com If 6 m 3 of oil weighs 47 kn calculate its If 6 m 3 of oil weighs 47
More information1 Fluid Statics. 1.1 Fluid Properties. Fluid
1 Fluid Statics 1.1 Fluid Properties Fluid A fluid is a substance, which deforms when subjected to a force. A fluid can offer no permanent resistance to any force causing change of shape. Fluid flow under
More informationChapter 9: Solids and Fluids
Chapter 9: Solids and Fluids State of matters: Solid, Liquid, Gas and Plasma. Solids Has definite volume and shape Can be crystalline or amorphous Molecules are held in specific locations by electrical
More informationFigure 3: Problem 7. (a) 0.9 m (b) 1.8 m (c) 2.7 m (d) 3.6 m
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)
More informationChapter 2 SOLUTION 100 = km = h. = h. ft s
Chapter.1. Convert the information given in the accompanying table from SI units to U.S. Customary units. Show all steps of your solutions. See Example.. km 1000 m.8 ft 1 mile 10 = 74.5 miles/h h 1 km
More informationChapter (6) Energy Equation and Its Applications
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
More informationCourse: TDEC202 (Energy II) dflwww.ece.drexel.edu/tdec
Course: TDEC202 (Energy II) Thermodynamics: An Engineering Approach Course Director/Lecturer: Dr. Michael Carchidi Course Website URL dflwww.ece.drexel.edu/tdec 1 Course Textbook Cengel, Yunus A. and Michael
More informationFluid Mechanics Testbank By David Admiraal
Fluid Mechanics Testbank By David Admiraal This testbank was created for an introductory fluid mechanics class. The primary intentions of the testbank are to help students improve their performance on
More informationF kg. Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal s principle.
14 Review roblems 185 A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. Assumptions 1 The cylinders of the lift are vertical.
More informationThe Bernoulli Equation
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 3D, unsteady (x, y, z, t; r,θ, z, t; etc) Let consider
More informationFluid Mechanics. Chapter 14. Modified by P. Lam 6_7_2012
Chapter 14 Fluid Mechanics PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman Lectures by James Pazun Modified by P. Lam 6_7_2012 Goals for Chapter 14 To study
More informationEF 152 Exam 1 Fall 2018 Page 1 Copy 165
EF 152 Exam 1 Fall 2018 Page 1 Copy 165 Name: Seat Assignment: Specify your EXAM ID on the right. Use 000 if you don t know your exam ID. Circle your TEAM SECTION 11:10 12:40 2:10 216 217 218 A216 Matt
More informationρ mixture = m mixture /V = (SG antifreeze ρ water V antifreeze + SG water ρ water V water )/V, so we get
CHAPTER 10 1. When we use the density of granite, we have m = ρv = (.7 10 3 kg/m 3 )(1 10 8 m 3 ) =.7 10 11 kg.. When we use the density of air, we have m = ρv = ρlwh = (1.9 kg/m 3 )(5.8 m)(3.8 m)(.8 m)
More informationPhysical Sciences 2: Assignments for Oct Oct 31 Homework #7: Elasticity and Fluid Statics Due Tuesday, Oct 31, at 9:30AM
Physical Sciences 2: Assignments for Oct. 24  Oct 31 Homework #7: Elasticity and Fluid Statics Due Tuesday, Oct 31, at 9:30AM After completing this homework, you should Be able to describe what is meant
More informationStatic Forces on SurfacesBuoyancy. Fluid Mechanics. There are two cases: Case I: if the fluid is above the curved surface:
Force on a Curved Surface due to Hydrostatic Pressure If the surface is curved, the forces on each element of the surface will not be parallel (normal to the surface at each point) and must be combined
More informationHydrostatic. Pressure distribution in a static fluid and its effects on solid surfaces and on floating and submerged bodies.
Hydrostatic Pressure distribution in a static fluid and its effects on solid surfaces and on floating and submerged bodies. M. Bahrami ENSC 283 Spring 2009 1 Fluid at rest hydrostatic condition: when a
More information11.1 Mass Density. Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an
Chapter 11 Fluids 11.1 Mass Density Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an important factor that determines its behavior
More information9. Pumps (compressors & turbines) Partly based on Chapter 10 of the De Nevers textbook.
Lecture Notes CHE 31 Fluid Mechanics (Fall 010) 9. Pumps (compressors & turbines) Partly based on Chapter 10 of the De Nevers textbook. Basics (pressure head, efficiency, working point, stability) Pumps
More informationb) (5) Find the tension T B in the cord connected to the wall.
General Physics I Quiz 6  Ch. 9  Static Equilibrium July 15, 2009 Name: Make your work clear to the grader. Show formulas used. Give correct units and significant figures. Partial credit is available
More informationSteven Burian Civil & Environmental Engineering September 25, 2013
Fundamentals of Engineering (FE) Exam Mechanics Steven Burian Civil & Environmental Engineering September 25, 2013 s and FE Morning ( Mechanics) A. Flow measurement 7% of FE Morning B. properties Session
More informationFormulae that you may or may not find useful. E v = V. dy dx = v u. y cp y = I xc/a y. Volume of an entire sphere = 4πr3 = πd3
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
More informationCHAPTER 1 Basic Considerations
CHAPTER 1 Basic Considerations FEtype Exam Review Problems: Problems 1.1 to 1.1. 1.1 (C) m F/a or kg N/m/s N. s /m. 1. (B) [µ] [τ//dy] (F/L )/(L/T)/L F. T/L. Chapter 1 / Basic Considerations 1. (A) 8
More informationEngineering Mechanics. Equivalent force systems: problems
Engineering Mechanics Equivalent force systems: problems A 36N force is applied to a wrench to tighten a showerhead. Knowing that the centerline of the wrench is parallel to the x axis. Determine the
More informationHomework Assignment on Fluid Statics
AMEE 0 Introduction to Fluid Mecanics Instructor: Marios M. Fyrillas Email: m.fyrillas@fit.ac.cy Homework Assignment on Fluid Statics 
More information1.4 Perform the following unit conversions: (b) (c) s. g s. lb min. (d) (e) in. ft s. m 55 h. (f) ft s. km h. (g)
1.4 Perform the following unit conversions: 0.05 ft 1 in. (a) 1L 61in. 1L 1ft (b) 1kJ 650 J 10 J 1Btu 1.0551kJ 0.616 Btu (c) 41 Btu/h 0.15 kw 1kW 1h 600 s 778.17 ft lbf 1Btu ft lbf 99.596 s (d) g 78 s
More informationChapter 12. Fluid Mechanics. A. The density ρ of a substance of uniform composition is defined as its mass M divided by its volume V.
Chapter 12 Fluid Mechanics 12.1 Density A. The density ρ of a substance of uniform composition is defined as its mass M divided by its volume V. That is,! = M V The density of water at 4 o C is 1000 kg/m
More informationFigure 1 Answer: = m
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
More informationAerodynamics. Basic Aerodynamics. Continuity equation (mass conserved) Some thermodynamics. Energy equation (energy conserved)
Flow with no friction (inviscid) Aerodynamics Basic Aerodynamics Continuity equation (mass conserved) Flow with friction (viscous) Momentum equation (F = ma) 1. Euler s equation 2. Bernoulli s equation
More informationAtmospheric pressure. 9 ft. 6 ft
Name CEE 4 Final Exam, Aut 00; Answer all questions; 145 points total. Some information that might be helpful is provided below. A Moody diagram is printed on the last page. For water at 0 o C (68 o F):
More informationCHAPTER 3 Introduction to Fluids in Motion
CHAPTER 3 Introduction to Fluids in Motion FEtpe Eam Review Problems: Problems 3 to 39 nˆ 0 ( n ˆi+ n ˆj) (3ˆi 4 ˆj) 0 or 3n 4n 0 3. (D) 3. (C) 3.3 (D) 3.4 (C) 3.5 (B) 3.6 (C) Also n n n + since ˆn
More informationFluid Mechanics Introduction
Fluid Mechanics Introduction Fluid mechanics study the fluid under all conditions of rest and motion. Its approach is analytical, mathematical, and empirical (experimental and observation). Fluid can be
More informationNicholas J. Giordano. Chapter 10 Fluids
Nicholas J. Giordano www.cengage.com/physics/giordano Chapter 10 Fluids Fluids A fluid may be either a liquid or a gas Some characteristics of a fluid Flows from one place to another Shape varies according
More informationHOMEWORK ASSIGNMENT ON BERNOULLI S EQUATION
AMEE 0 Introduction to Fluid Mechanics Instructor: Marios M. Fyrillas Email: m.fyrillas@frederick.ac.cy HOMEWORK ASSIGNMENT ON BERNOULLI S EQUATION. Conventional sprayguns operate by achieving a low pressure
More informationPhysics 201 Chapter 13 Lecture 1
Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal s Principle Archimedes Principle (Buoyancy) Fluid Dynamics Continuity Equation Bernoulli Equation 11/30/2009 Physics 201, UWMadison 1 Fluids Density
More informationTHERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
THERMOYNAMICS, FLUI AN PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. SELF ASSESSMENT EXERCISE No.1 FLUI MECHANICS HYROSTATIC FORCES
More informationPhysics 201 Chapter 13 Lecture 1
Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal s Principle Archimedes Principle (Buoyancy) Fluid Dynamics Continuity Equation Bernoulli Equation 11/30/2009 Physics 201, UWMadison 1 Fluids Density
More informationChapter 15. m. The symbolic equation for mass density is: ρ= m V. Table of Densities
Chapter 15 Density Often you will hear that fiberglass is used for racecars because it is lighter than steel. This is only true if we build two identical bodies, one made with steel and one with fiberglass.
More informationUnit C1: List of Subjects
Unit C: List of Subjects The elocity Field The Acceleration Field The Material or Substantial Derivative Steady Flow and Streamlines Fluid Particle in a Flow Field F=ma along a Streamline Bernoulli s
More informationFluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur. Lecture  9 Fluid Statics Part VI
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  9 Fluid Statics Part VI Good morning, I welcome you all to this session of Fluid
More informationPhysics 220: Classical Mechanics
Lecture 10 1/34 Phys 220 Physics 220: Classical Mechanics Lecture: MWF 8:40 am 9:40 am (Phys 114) Michael Meier mdmeier@purdue.edu Office: Phys Room 381 Help Room: Phys Room 11 schedule on course webpage
More informationME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 4. (Buoyancy and Viscosity of water)
ME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 4 (Buoyancy and Viscosity of water) 16. BUOYANCY Whenever an object is floating in a fluid or when it is completely submerged in
More informationChapter 14  Fluids. Archimedes, On Floating Bodies. David J. Starling Penn State Hazleton PHYS 213. Chapter 14  Fluids. Objectives (Ch 14)
Any solid lighter than a fluid will, if placed in the fluid, be so far immersed that the weight of the solid will be equal to the weight of the fluid displaced. Archimedes, On Floating Bodies David J.
More informationStevens High School AP Physics II Work for Notschool
1. (AP SAMPLE QUESTION) An ideal fluid is flowing with a speed of 12 cm/s through a pipe of diameter 5 cm. The pipe splits into three smaller pipes, each with a diameter of 2 cm. What is the speed of the
More informationPhysics 2c Lecture 9. Recap of Entropy. First part of chapter 18: Hydrostatic Equilibrium Measuring Pressure Pascal's Law Archimedes Principle
Physics 2c Lecture 9 Recap of Entropy First part of chapter 18: Hydrostatic Equilibrium Measuring Pressure Pascal's Law Archimedes Principle Defining Entropy Macroscopic Definition of entropy difference:
More informationCONCEPTS AND DEFINITIONS. Prepared by Engr. John Paul Timola
CONCEPTS AND DEFINITIONS Prepared by Engr. John Paul Timola ENGINEERING THERMODYNAMICS Science that involves design and analysis of devices and systems for energy conversion Deals with heat and work and
More informationChapter 15: Fluid Mechanics Dynamics Using Pascal s Law = F 1 = F 2 2 = F 2 A 2
Lecture 24: Archimedes Principle and Bernoulli s Law 1 Chapter 15: Fluid Mechanics Dynamics Using Pascal s Law Example 15.1 The hydraulic lift A hydraulic lift consists of a small diameter piston of radius
More informationAnNajah National University Civil Engineering Departemnt. Fluid Mechanics. Chapter [2] Fluid Statics
AnNajah National University Civil Engineering Deartemnt Fluid Mechanics Chater [2] Fluid Statics 1 Fluid Statics Problems Fluid statics refers to the study of fluids at rest or moving in such a manner
More informationHong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering Structural Mechanics. Chapter 1 PRINCIPLES OF STATICS
PRINCIPLES OF STTICS Statics is the study of how forces act and react on rigid bodies which are at rest or not in motion. This study is the basis for the engineering principles, which guide the design
More information