P = ρ{ g a } + µ 2 V II. FLUID STATICS


 Scot Thompson
 1 years ago
 Views:
Transcription
1 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 along a horizontal plane, 2. Pressure at a point is independent of orientation, 3. 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 prolems in fluid statics and lead to the following: 1. Two points at the same depth in a static fluid have the same pressure. 2. The orientation of a surface has no earing on the pressure at a point in a static fluid. 3. 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 with constant density and viscosity, we would otain the following: P = ρ{ g a } + µ 2 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 2 V ) Chs VI & VII Note: For prolems involving the effects of oth (1) fluid statics and (2) inertial effects, it is the net g " a " acceleration vector that controls oth the magnitude and direction of the pressure gradient. II1
2 This equation can e 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 P 2 P 1 = ρgdz 2 1 It is appropriate at this point to review the differences etween asolute, gage, and vacuum pressure. These differences are illustrated in Fig. 2.3 (shown elow) and are ased on the following definitions: Patm asolute pressure  Pressure measured relative to asolute zero. gage pressure  Pressure > Patm measured relative to Patm vacuum pressure  Pressure < Patm measured relative to Patm Patm  local asolute pressure due to the local atmosphere only. standard Patm at sea level = 1atm = kpa = 2116 lf/ft 2 P (kpa) Pressure aove atmospheric e.g P = 120 kpa asolute or P = 30 kpa gage P (relative to Patm) = Pgage = 30 kpa gage 0 P(relative to Patm) = P(vac) = 40 kpa P(as) = 50 kpa Local atmospheric pressure e.g at 1000 m Patm 90 kpa = 0 Pa gage = 0 Pa vacuum Pressure elow atmospheric e.g. P = 50 kpa as Asolute zero, perfect vacuum P(as) = 0, P(vac) = 90kPa, P(gage) =  90 kpa Fig. 2.3 Illustration of asolute, gage, and vacuum pressure II2
3 Hydrostatic Pressure in Liquids For liquids and incompressile fluids, the previous integral expression for P2  P1 integrates to Note: P 1 P 2 = ρg (Z 2 Z 1 ) z Free surface Pressure = Pa P 2 Z 2 Z 1 is positive for Z 2 aove Z 1. ut P 2 P 1 is negative for Z 2 aove Z 1. We can now define a new fluid parameter: γ = ρg specific weight of the fluid x Z 2 h = Z  Z 2 1 Z 1 P 1 y With this, the previous equation ecomes (for an incompressile, static fluid) P 2 P 1 =  γ (Z 2 Z 1 ) The most common application of this result is that of manometry. Consider the Utue, multifluid manometer shown on the right. If we first lael all intermediate points etween A & a, we can write for the overall pressure change II3
4 P A  P a = (P A  P 1) + (P 1  P 2 ) + (P 2  P a ) This equation was otained y adding and sutracting each intermediate pressure. The total pressure difference is now expressed in terms of a series of intermediate pressure differences. Sustituting the previous result for static pressure difference, we otain P A  P B =  ρ g(z A  Z 1) ρ g (Z 1 Z 2 ) ρ g (Z 2  Z B ) Again note: Z positive up and Z A > Z 1, Z 1 < Z 2, Z 2 < Z a. In general, follow the following steps when analyzing manometry prolems: 1. On the manometer schematic, lael points on each end of the manometer and at each intermediate point where there is a fluidfluid interface, e.g. A B. 2. Express the overall pressure difference in terms of appropriate intermediate pressure differences. P A  P B = (P A  P 1 ) + (P 1  P 2 ) + (P 2  P B ) 3. Express each intermediate pressure difference in terms of an appropriate product of specific weight (or ρ g) * elevations change ( and watch the signs). P A  P B = γ (z A  z 1 ) + γ (z 1  z 2 ) + γ (z 2  z B ) When developing a solution for manometer prolems, take care to: 1. Include all pressure changes. 2. Use correct Z and γ with each fluid. 3. Use correct signs with Z. If pressure difference is expressed as P A P 1, the elevation change should e written as Z A Z Watch units. II4
5 Manometer Example: Given the indicated manometer, determine the gage pressure at A. Pa = kpa. The fluid at A is Meriam red oil no. 3. ρg w = 9790 N/m 3 A S.G. = cm H 0 2 a 2 18 cm ρg A = S.G.*ρg w = 0.83*9790 N/m ρg A = 8126 N/m 3 ρg air = 11.8 N/m 3 With the indicated points laeled on the manometer, we can write P A  P a = P A (gage) = (P A  P 1 ) + (P 1 P 2 ) + (P 2  P a ) Sustituting the manometer expression for a static fluid, we otain P A (gage) =  ρg A (z A  z 1) ρg w (z 1 z 2 ) ρg a (z 2  z a ) Neglect the contriution due to the air column. Sustituting values, we otain P A (gage) = N/m 3 * 0.10 m 9790 N/m 3 * = N/m 2 Ans Note why: (z A  z 1 ) = 0.10 m and (z 1 z 2 ) = m, & we did not use P a Review the text examples for manometry. See Tale 2.1 for values of specific weight, γ, in oth B.G. and S.I. units. II5
6 Hydrostatic Pressure in Gases Since gases are compressile, density is a nonconstant variale in the previous expression for dp/dz. Assuming the gas is an ideal gas, we can write dp dz P 2 = ρg = RT g or dp = ln P 2 = g 2 P P 1 R For an isothermal atmosphere with T = T o, this integrates to P 2 = P 1 exp gz 2 z 1 RT o 1 ( ) 1 dz T Up to an altitude of approximately 36,000ft (l l,000 m), the mean atmospheric temperature decreases nearly linearly and can e represented y T T o  B z where B is the lapse rate The following values are assumed to apply for air from sea level to 36,000 ft: To = R = K (15 C) B = R/ft = K/m Sustituting the linear temperature variation into the previous equation and integrating we otain P = P a 1 Bz T o g /( RB) where g RB = 5.26 ( for air ) and R = 287 m 2 /(s 2 K) Review example 2.2 in the text. II6
7 Hydrostatic Forces on Plane Surfaces Consider a plane surface of aritrary shape and orientation, sumerged in a static fluid as shown: If P represents the local pressure at any point on the surface and h the depth of fluid aove any point on the surface, from asic physics we can easily show that the net hydrostatic force on a plane surface is given y (see text for development): F = PdA A = P cg A Thus, asic physics says that the hydrostatic force is a distriuted load equal to the integral of the local pressure force over the area. This is equivalent to the following: The hydrostatic force on one side of a plane surface sumerged in a static fluid equals the product of the fluid pressure at the centroid of the surface times the surface area in contact with the fluid. Also: Since pressure acts normal to a surface, the direction of the resultant force will always e normal to the surface. Note: In most cases, since it is the net hydrostatic force that is desired and the contriution of atmospheric pressure Pa will act on oth sides of a surface, the result of atmospheric pressure Pa will cancel and the net force is otained y F = ρ gh cg A F = P cg A Pcg is now the gage pressure at the centroid of the area in contact with the fluid. II7
8 Therefore, to otain the net hydrostatic force F on a plane surface: 1. Determine depth of centroid hcg for the area in contact with the fluid. 2. Determine the (gage) pressure at the centroid Pcg. 3. Calculate F = PcgA. The following page shows the centroid, and other geometric properties of several common areas. It is noted that care must e taken when dealing with layered fluids. The procedure essentially requires that the force on the part of the plane area in each individual layer of fluid must e determined separately for each layer using the steps listed aove. We must now determine the effective point of application of F. This is commonly called the center of pressure  cp of the hydrostatic force. Note: This is not necessarily the same as the c.g. Define an x y coordinate system whose origin is at the centroid, c.g, of the area. The location of the resultant force is determined y integrating the moment of the distriuted fluid load on the surface aout each axis and equating this to the moment of the resultant force aout that axis. Therefore, for the moment aout the x axis: Fy cp = A ypda Applying a procedure similar to that used previously to determine the resultant force, and using the definition (see text for detailed development), we otain Y cp = ρgsinθ I xx P cg A 0 where: I xx is defined as the Moment of Inertia, or the 2 nd moment of the area Therefore, the resultant force will always act at a distance ycp elow the centroid of the surface ( except for the special case of sin θ = 0 ). II8
9 PROPERTIES OF PLANE SECTIONS Geometry Centroid Moment of Inertia I x x Product of Inertia I x y Area y x L L 2, 2 L L y x 2R π R 4 0,0 0 π R 2 4 L y x 3, L 3 L L 2 72 L 2 a y x R 0,a = 4R 3π R 4 π π 0 π R 2 2 y x s L a = L 3 L 3 36 ( 2s)L L y x R a = 4R 3 π π π R π R 4 π R 2 4 y 1 x h a ) = h ( ( + 1 ) h 3 ( ( + 1 ) ) 0 ( + 1 ) h 2 Fluid Specific Weight Air 1f/ft N /m Seawater 3 1f /ft N /m ,050 Oil ,996 Glycerin ,360 Water Ethyl ,790 7,733 Mercury Caron ,100 15,570 II9
10 Proceeding in a similar manner for the x location, and defining I xy = product of inertia, we otain X cp = ρgsinθ I xy P cg A where Xcp can e either positive or negative since Ixy can e either positive or negative. Note: For areas with a vertical plane of symmetry through the centroid, i.e. the y  axis (e.g. squares, circles, isosceles triangles, etc.), the center of pressure is located directly elow the centroid along the plane of symmetry, i.e., X cp = 0. Key Points: The values X cp and Y cp are oth measured with respect to the centroid of the area in contact with the fluid. X cp and Y cp are oth measured in the inclined plane of the area; i.e., Y cp is not necessarily a vertical dimension, unless θ = 90 o. Special Case: For most prolems where (1) we have a single, homogeneous fluid (i.e. not applicale to layers of multiple fluids) and (2) the surface pressure is atmospheric, the fluid specific weight γ cancels in the equation for Y cp and X cp and we have the following simplified expressions: F = ρ gh cg A Y cp = I xx sinθ h cg A X cp = I xy sinθ h cg A However, for prolems where we have either (1) multiple fluid layers, or (2) a container with surface pressurization > P atm, these simplifications do not occur and the original, asic expressions for F, Y cp, and X cp must e used; i.e., take II10
11 care to use the approximate expressions only for cases where they apply. The asic equations always work. II11
12 Summary: 1. The resultant force is determined from the product of the pressure at the centroid of the surface times the area in contact with the fluid. 2. The centroid is used to determine the magnitude of the force. This is not the location of the resultant force. 3. The location of the resultant force will e at the center of pressure which will e at a location Ycp elow the centroid and Xcp as specified previously. 4. Xcp = 0 for areas with a vertical plane of symmetry through the c.g. Example 2.5 Given: Gate, 5 ft wide Hinged at B Holds seawater as shown Find: a. Net hydrostatic force on gate 15 Seawater 64 lf/ft 3 h c.g. 9 A. Horizontal force at wall  A c. Hinge reactions  B B θ c.g. 6 8 a. By geometry: θ = tan 1 (6/8) = o Neglect P atm Since the plate is rectangular, h cg = 9 ft + 3ft = 12 ft A = 10 x 5 = 50 ft 2 P cg = γ h cg = 64 lf/ft 3 * 12 ft = 768 lf/ft 2 F p = P cg A = 768 lf/ft 2 * 50 ft 2 = 38,400 lf II12
13 II13
14 . Horizontal Reaction at A Must first find the location, c.p., for F p y cp = ρ gsinθ I xx P cg A = I xx sinθ h cg A For a rectangular wall: I xx = h 3 /12 I xx = 5 * 10 3 /12 = 417 ft 4 Note: The relevant area is a rectangle, not a triangle. Bx Bz θ Fw y c.p. c.g. c.p. 8 ft θ P 6 ft Note: Do not overlook the hinged reactions at B. y cp = 417 ft = ft elow c.g. 12 ft 50 ft x cp = 0 due to symmetry M B =0 P ( ) 38,400 6 P = 0 P = 29,330 lf Bx Bz θ Fw y c.p. c.g. c.p. θ 6 ft 8 ft II14
15 c. F x = 0, B x + Fsinθ P = 0 B x + 38,400*0.629,330 = 0 B x = 6290 lf F z = 0, B z Fcosθ = 0 B z = 38,400 * 0.8 = 30,720 lf Note: Show the direction of all forces in final answers. Summary: To find net hydrostatic force on a plane surface: 1. Find area in contact with fluid. 2. Locate centroid of that area. 3. Find hydrostatic pressure P cg at centroid, typically = γ h cg (generally neglect P atm ). 4. Find force F = P cg A. 5. The location will not e at the c.g., ut at a distance y cp elow the centroid. y cp is in the plane of the area. II15
16 Plane Surfaces in Layered Fluids For plane surfaces in layered fluids, the part of the surface in each fluid layer must essentially e worked as a separate prolem. That is, for each layer: 1.) Identify the area of the plate in contact with each layer, 2.) Locate the c.g. for the part of the plate in each layer and the pressure at the c.g., and 3. Calculate the force on each layered element using F 1 = Pc.g 1 A 1. Repeat for each layer. Use the usual procedure for finding the location of the force for each layer. Review all text examples for forces on plane surfaces. II16
17 Forces on Curved Surfaces Since this class of surface is curved, the direction of the force is different at each location on the surface. Therefore, we will evaluate the x and y components of net hydrostatic force separately. Consider curved surface, a. Force alances in x & y directions yield Fh = F H Fv = Wair + W 1 + W 2 From this force alance, the asic rules for determining the horizontal and vertical component of forces on a curved surface in a static fluid can e summarized as follows: Horizontal Component, Fh The horizontal component of force on a curved surface equals the force on the plane area formed y the projection of the curved surface onto a vertical plane normal to the component. The horizontal force will act through the c.p. (not the centroid) of the projected area. h cg F h y cp a Projected vertical plane Curved cp surface a II17
18 Therefore, to determine the horizontal component of force on a curved surface in a hydrostatic fluid: 1. Project the curved surface into the appropriate vertical plane. 2. Perform all further calculations on the vertical plane. 3. Determine the location of the centroid  c.g. of the vertical plane. 4. Determine the depth of the centroid  h cg of the vertical plane. 5. Determine the pressure  Pcg = ρ g h cg at the centroid of the vertical plane. 6. Calculate Fh = P cg A, where A is the area of the projection of the curved surface into the vertical plane, ie. the area of the vertical plane. 7. The location of Fh is through the center of pressure of the vertical plane, not the centroid. Get the picture? All elements of the analysis are performed with the vertical plane. The original curved surface is important only as it is used to define the projected vertical plane. Vertical Component  Fv The vertical component of force on a curved surface equals the weight of the effective column of fluid necessary to cause the pressure on the surface. The use of the words effective column of fluid is important in that there may not always actually e fluid directly aove the surface. (See graphic that follows.) This effective column of fluid is specified y identifying the column of fluid that would e required to cause the pressure at each location on the surface. Thus, to identify the effective volume  Veff: 1. Identify the curved surface in contact with the fluid. 2. Identify the pressure at each point on the curved surface. 3. Identify the height of fluid required to develop the pressure. II18
19 4. These collective heights comine to form V eff. V eff P P a P Fluid aove the surface a V eff P P P fluid No fluid actually aove surface These two examples show two typical cases where this concept is used to determine V eff. The vertical force acts vertically through the centroid (center of mass) of the effective column of fluid. The vertical direction will e the direction of the vertical components of the pressure forces. Therefore, to determine the vertical component of force on a curved surface in a hydrostatic fluid: 1. Identify the effective column of fluid necessary to cause the fluid pressure on the surface. 2. Determine the volume of the effective column of fluid. 3. Calculate the weight of the effective column of fluid  Fv = ρgv eff. 4. The location of Fv is through the centroid of V eff. Finding the Location of the Centroid A second prolem associated with the topic of curved surfaces is that of finding the location of the centroid of V eff. Recall: II19
20 Centroid = the location where a point area, volume, or mass can e place to yield the same first moment of the distriuted area, volume, or mass, e.g. x cg V 1 = V1 xdv This principle can also e used to determine the location of the centroid of complex geometries. For example: If V eff = V 1 + V 2 then x cg V eff = x 1 V 1 + x 2 V 2 V 1 V 2 a or for the second geometry V T = V 1 + V eff V eff x T V T = x 1 V 1 + x cg V eff a V 1 fluid Note: In the figures shown aove, each of the x values would e specified relative to a vertical axis through since the cg of the quarter circle is most easily specified relative to this axis. II20
21 Example: Gate AB holds ack 15 ft of water. Neglecting the weight of the gate, determine the magnitude (per unit width) and location of the hydrostatic forces on the gate and the resisting moment aout B. 15 ft Water F H A F V Width  W B a. Horizontal component γ = ρg = 62.4 lf/ft 3 Rule: Project the curved surface into the vertical plane. Locate the centroid of the projected area. Find the pressure at the centroid of the vertical projection. F = P cg A p Note: All calculations are done with the projected area. The curved surface is not used at all in the analysis. P cg h cg a A B The curved surface projects onto plane a  and results in a rectangle, (not a quarter circle) 15 ft x W. For this rectangle: h cg = 7.5, P cg = γh cg = 62.4 lf/ft 3 * 7.5 ft = 468 lf/ft 2 F h = P cg A = 468 lf/ft 2 * 15 ft*w= 7020 W lf per ft of width Location: I xx = h 3 /12 = W * 15 3 /12 = W ft 4 y cp = I xx sinθ h cg A = W ft4 sin90 o 7.5 ft15w ft 2 = 2.5 ft The location is 2.5 ft elow the c.g. or 10 ft elow the surface, 5 ft aove the ottom. II21
22 . Vertical force: A C c.g. Rule: F v equals the weight of the effective column of fluid aove the curved surface (shown y the dashed diagonal lines). F v B Q: What is the effective volume of fluid aove the surface? What volume of fluid would result in the actual pressure distriution on the curved surface? Vol = Vol A  B  C V rec = V qc + V ABC, V ABC = V rec  V qc V ABC = V eff = 15 2 W  π 15 2 /4*W = W ft 3 F v = ρg V eff = 62.4 lf/ft 3 * ft 3 = 3013 lf per ft of width Note: F v is directed upward even though the effective volume is aove the surface. c. What is the location? A C Rule: F v will act through the centroid of the effective volume causing the force. F v c.g. B We need the centroid of volume ABC. How do we otain this centroid? Use the concept which is the asis of the centroid, the first moment of an area. Since: A rec = A qc + A ABC M rec = M qc + M ABC M ABC = M rec  M qc II22
23 Note: We are taking moments aout the left side of the figure, ie., point. WHY? (The c.g. of the quarter circle is known to e at 4 R/ 3 π w.r.t..) x cg A = x rec A rec  x qc A qc x cg { π*15 2 /4} = 7.5* {4*15/3/π}* π*15 2 /4 x cg = ft { distance to rt. of to the centroid } Q: Do we need a y location? Why? d. Calculate the moment aout B needed for equilirium. A c.g. C M B =0 clockwise positive. F H M B + 5F h + ( 15 x v )F v = 0 F v M B B M B W + ( ) 3013W = 0 M B + 35,100W +10,093.6W = 0 M B = 45,194W ft lf per unit width Why is the answer negative? (What did we assume for an initial direction of M B?) The hydrostatic forces will tend to roll the surface clockwise relative to B, thus a resisting moment that is counterclockwise is needed for static equilirium. II23
24 Always review your answer (all aspects: magnitude, direction, units, etc.) to determine if it makes sense relative to physically what you understand aout the prolem. Begin to think like an engineer. II24
25 Buoyancy An important extension of the procedure for vertical forces on curved surfaces is that of the concept of uoyancy. The asic principle was discovered y Archimedes. It can e easily shown that (see text for detailed development) the uoyant force F is given y: P atm F = ρ g V where V is the volume of the fluid displaced y the sumerged ody and ρ g is the specific weight of the fluid displaced. V c.g. F Thus, the uoyant force equals the weight of the fluid displaced, which is equal to the product of the specific weight times the volume of fluid displaced. The location of the uoyant force is through a vertical line of action, directed upward, which acts through the centroid of the volume of fluid displaced. Review all text examples and material on uoyancy. II25
26 Pressure Distriution in Rigid Body Motion All of the prolems considered to this point were for static fluids. We will now consider an extension of our static fluid analysis to the case of rigid ody motion, where the entire fluid mass moves and accelerates uniformly (as a rigid ody). The container of fluid shown elow is accelerated uniformly up and to the right as shown. From a previous analysis, the general equation governing fluid motion is P = ρ( g a ) + µ 2 V For rigid ody motion, there is no velocity gradient in the fluid, therefore µ 2 V = 0 The simplified equation can now e written as P = ρ( g a ) = ρg where G = g a the net acceleration vector acting on the fluid. II26
27 This result is similar to the equation for the variation of pressure in a hydrostatic fluid. However, in the case of rigid ody motion: * P = f {fluid density & the net acceleration vector G = g a } * P acts in the vector direction of G = g a. * Lines of constant pressure are perpendicular to G. The new orientation of the free surface will also e perpendicular to G. The equations governing the analysis for this class of prolems are most easily developed from an acceleration diagram. Acceleration diagram: For the indicated geometry: Free surface θ a a z θ = tan 1 a x g + a z dp ds = ρg where G = a 2 x + (g + a z ) 2 { } 1 2 a G θ g s a x P 2 P 1 and P 2 P 1 = ρg(s 2 s 1 ) Note: P 2 P 1 ρ gz ( 2 z 1 ) and s 2 s 1 is not a vertical dimension Note: s is the depth to a given point perpendicular to the free surface or its extension. s is aligned with G. II27
28 In analyzing typical prolems with rigid ody motion: 1. Draw the acceleration diagram taking care to correctly indicate a, g, and θ, the inclination angle of the free surface. 2. Using the previously developed equations, solve for G and θ. 3. If required, use geometry to determine s 2 s 1 (the perpendicular distance from the free surface to a given point) and then the pressure at that point relative to the surface using P 2 P 1 = ρ G (s 2 s 1 ). Key Point: Do not use ρg to calculate P 2 P 1, use ρ G. Example 2.12 Given: A coffee mug, 6 cm x 6 cm square, 10 cm deep, contains 7 cm of coffee. The mug is accelerated to the right with a x = 7 m/s 2. Assuming rigid ody motion and ρ c = 1010 kg/m 3, Determine: a. Will the coffee spill?. P g at a &. z θ a x 6 cm 7 cm 10 cm c. F net on left wall. a. First draw schematic showing the original orientation and final orientation of the free surface. ρ c = 1010 kg/m 3 a x = 7m/s 2 a z = 0 g = m/s 2 a We now have a new free surface at an angle θ where a θ = tan 1 a x g + a z θ θ = tan = 35.5 z = 3 tan 35.5 = 2.14 cm G g a II28
29 h max = = 9.14 cm < 10 cm Coffee will not spill.. Pressure at a &. P a = ρ G s a G = {a 2 x + g 2 } 1/2 = { } 1/2 z θ a x 6 cm G = m/s 2 10 cm s a = {7 + z} cos θ s a = 9.14 cm cos 35.5 = 7.44 cm θ s a 7 cm P a = 1010 kg/m 3 *12.05m/s 2 * m P a = 906 (kg m/s 2 )/m 2 = 906 Pa Note: P ρ gy G g a a Q: How would you find the pressure at, P? c. What is the force on the left wall? We have a plane surface, what is the rule? Find cg, P cg, F = P cg. A Vertical depth to cg is: z cg = 9.14/2 = 4.57 cm s cg = 4.57 cos 35.5 = 3.72 cm z θ cg a x 6 cm θ s cg 7 cm 10 cm P cg = ρ G s cg P cg = 1010 kg/m 3 *12.05 m/s 2 * m P cg = N/m 2 F = P cg A = N/m 2 *0.0914*0.06m 2 a F = 2.48 N II29
30 What is the direction? Horizontal, perpendicular to the wall; i.e. Pressure always acts normal to a surface. Q: How would you find the force on the right wall? II30
Hydrostatics. 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 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 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 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 informationCHAPTER 2 Fluid Statics
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=
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 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 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 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 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 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 informationHydrostatic forces (Cont.)
25/01/2017 Lecture 8 Hydrostatic forces (Cont.) Yesterday, we discussed about hydrostatic forces acting on one side of a submerged plane surface. The plane area centroid was discussed and also the concept
More informationragsdale (zdr82) HW7 ditmire (58335) 1 The magnetic force is
ragsdale (zdr8) HW7 ditmire (585) This printout should have 8 questions. Multiplechoice questions ma continue on the net column or page find all choices efore answering. 00 0.0 points A wire carring
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 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 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 informationThe general rules of statics (as applied in solid mechanics) apply to fluids at rest. From earlier we know that:
ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 2 Pressure This section will study the forces acting on or generated by fluids at rest. Objectives Introduce the concept
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 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 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 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 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 informationAll that begins... peace be upon you
All that begins... peace be upon you School of Mechanical Engineering Mechanics of Fluids & Engineering Computing SKMM 2313 Mechanics of Fluids 1 Pressure Distribution in a Fluid «An excerpt (mostly) from
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 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 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 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 informationSection 8.5. z(t) = be ix(t). (8.5.1) Figure A pendulum. ż = ibẋe ix (8.5.2) (8.5.3) = ( bẋ 2 cos(x) bẍ sin(x)) + i( bẋ 2 sin(x) + bẍ cos(x)).
Difference Equations to Differential Equations Section 8.5 Applications: Pendulums MassSpring Systems In this section we will investigate two applications of our work in Section 8.4. First, we will consider
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 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 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 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 informationFluid Mechanics. du dy
FLUID MECHANICS Technical English  I 1 th week Fluid Mechanics FLUID STATICS FLUID DYNAMICS Fluid Statics or Hydrostatics is the study of fluids at rest. The main equation required for this is Newton's
More informationApplications of Hydrostatics
Applications of Hydrostatics Pressure measurement with hydrostatics Mercury Barometer  This is a device used to measure the local atmospheric pressure, p a. As seen in the sketch, it is formed by inverting
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 informationa b a b ab b b b Math 154B Elementary Algebra Spring 2012
Math 154B Elementar Algera Spring 01 Stud Guide for Eam 4 Eam 4 is scheduled for Thursda, Ma rd. You ma use a " 5" note card (oth sides) and a scientific calculator. You are epected to know (or have written
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 informationPlanar Rigid Body Kinematics Homework
Chapter 2: Planar Rigid ody Kinematics Homework Chapter 2 Planar Rigid ody Kinematics Homework Freeform c 2018 21 Chapter 2: Planar Rigid ody Kinematics Homework 22 Freeform c 2018 Chapter 2: Planar
More informationPhysics 210: Worksheet 26 Name:
(1) A all is floating in. If the density of the all is 0.95x10 kg/m, what percentage of the all is aove the? (2) An with a density of 19.x10 kg/m and a mass m50 kg is placed into a cylinder which contains
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 informationFluid Mechanics. Spring Course Outline
Fluid Mechanics (Fluidmekanik) Course Code: 1TV024 5 hp Fluid Mechanics Spring 2011 Instruct: Chris Hieronymus Office: Geocentrum Dk255 Phone: 471 2383 email: christoph.hieronymus@geo.uu.se Literature:
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 informationSolution: (a) (b) (N) F X =0: A X =0 (N) F Y =0: A Y + B Y (54)(9.81) 36(9.81)=0
Prolem 5.6 The masses of the person and the diving oard are 54 kg and 36 kg, respectivel. ssume that the are in equilirium. (a) Draw the freeod diagram of the diving oard. () Determine the reactions at
More informationTotal energy in volume
General Heat Transfer Equations (Set #3) ChE 1B Fundamental Energy Postulate time rate of change of internal +kinetic energy = rate of heat transfer + surface work transfer (viscous & other deformations)
More informationIX. Mass Wasting Processes
IX. Mass Wasting Processes 1. Deris Flows Flow types: Deris flow, lahar (volcanic), mud flow (few gravel, no oulders) Flowing mixture of water, clay, silt, sand, gravel, oulder, etc. Flowing is liquefied
More informationHOW TO GET A GOOD GRADE ON THE MME 2273B FLUID MECHANICS 1 EXAM. Common mistakes made on the final exam and how to avoid them
HOW TO GET A GOOD GRADE ON THE MME 2273B FLUID MECHANICS 1 EXAM Common mistakes made on the final exam and how to avoid them HOW TO GET A GOOD GRADE ON THE MME 2273B EXAM Introduction You now have a lot
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 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 informationThermodynamic Systems
Thermodynamic Systems For purposes of analysis we consider two types of Thermodynamic Systems: Closed System  usually referred to as a System or a Control Mass. This type of system is separated from its
More informationPhysics 1A Lecture 10B
Physics 1A Lecture 10B "Sometimes the world puts a spin on life. When our equilibrium returns to us, we understand more because we've seen the whole picture. Davis Barton Cross Products Another way to
More informations and FE X. A. Flow measurement B. properties C. statics D. impulse, and momentum equations E. Pipe and other internal flow 7% of FE Morning Session I
Fundamentals of Engineering (FE) Exam General Section Steven Burian Civil & Environmental Engineering October 26, 2010 s and FE X. A. Flow measurement B. properties C. statics D. impulse, and momentum
More informationFluid Mechanics Chapter 1 Effects of pressure
Fluid Mechanics Chapter 1 Effects of pressure last edited eptember 10, 2017 1.1 Motivation 29 1.2 Concept of pressure 29 1.2.1 The direction of pressure 29 1.2.2 Pressure on an infinitesimal volume 30
More informationThe University cannot take responsibility for any misprints or errors in the presented formulas. Please use them carefully and wisely.
Aide Mémoire Suject: Useful formulas for flow in rivers and channels The University cannot take responsiility for any misprints or errors in the presented formulas. Please use them carefully and wisely.
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 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 information2.8 Hydrostatic Force on a Plane Surface
CEE 3310 Hydrostatics, Sept. 7, 2018 35 2.7 Review Pressure is a scalar  independent of direction. Hydrostatics: P = γ k. pressure, vertical motion: z P, z P. Horizontal motion in the same fluid  no
More informationFluid Mechanics. If deformation is small, the stress in a body is proportional to the corresponding
Fluid Mechanics HOOKE'S LAW If deformation is small, the stress in a body is proportional to the corresponding strain. In the elasticity limit stress and strain Stress/strain = Const. = Modulus of elasticity.
More informationJurong Junior College 2014 J1 H1 Physics (8866) Tutorial 3: Forces (Solutions)
Jurong Junior College 2014 J1 H1 Physics (8866) Tutorial 3: Forces (Solutions) Take g = 9.81 m s 2, P atm = 1.0 x 10 5 Pa unless otherwise stated Learning Outcomes (a) SubTopic recall and apply Hooke
More informationProperties of Sections
ARCH 314 Structures I Test Primer Questions Dr.Ing. Peter von Buelow Properties of Sections 1. Select all that apply to the characteristics of the Center of Gravity: A) 1. The point about which the body
More informationDynamical Systems Solutions to Exercises
Dynamical Systems Part 56 Dr G Bowtell Dynamical Systems Solutions to Exercises. Figure : Phase diagrams for i, ii and iii respectively. Only fixed point is at the origin since the equations are linear
More informationChapter 2 Pressure Distribution in a Fluid
Chapter Pressure Distribution in a Fluid.1 For the twodimensional stress field in Fig. P.1, let xx 000 psf 000 psf xy yy 500 psf Find the shear and normal stresses on plane AA cutting through at 0. Solution:
More informationAMME2261: Fluid Mechanics 1 Course Notes
Module 1 Introduction and Fluid Properties Introduction Matter can be one of two states: solid or fluid. A fluid is a substance that deforms continuously under the application of a shear stress, no matter
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 information5.1 Fluid momentum equation Hydrostatics Archimedes theorem The vorticity equation... 42
Chapter 5 Euler s equation Contents 5.1 Fluid momentum equation........................ 39 5. Hydrostatics................................ 40 5.3 Archimedes theorem........................... 41 5.4 The
More informationP = 1 3 (σ xx + σ yy + σ zz ) = F A. It is created by the bombardment of the surface by molecules of fluid.
CEE 3310 Thermodynamic Properties, Aug. 27, 2010 11 1.4 Review A fluid is a substance that can not support a shear stress. Liquids differ from gasses in that liquids that do not completely fill a container
More informationPhysics 106 Lecture 13. Fluid Mechanics
Physics 106 Lecture 13 Fluid Mechanics SJ 7 th Ed.: Chap 14.1 to 14.5 What is a fluid? Pressure Pressure varies with depth Pascal s principle Methods for measuring pressure Buoyant forces Archimedes principle
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 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 informationME 323 Examination #2
ME 33 Eamination # SOUTION Novemer 14, 17 ROEM NO. 1 3 points ma. The cantilever eam D of the ending stiffness is sujected to a concentrated moment M at C. The eam is also supported y a roller at. Using
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 informationSports biomechanics explores the relationship between the body motion, internal forces and external forces to optimize the sport performance.
What is biomechanics? Biomechanics is the field of study that makes use of the laws of physics and engineering concepts to describe motion of body segments, and the internal and external forces, which
More informationCh. 5: Distributed Forces
234 5.0 Outline 234 Introduction 235 Center of Mass 236 Centroids (Line, Area, Volume) 241 Composite Bodies and Figures 257 Theorem of Pappus 265 Fluid Statics 272 5.0 Outline 5.0 Introduction Ch. 5: Distributed
More informationENGI Multiple Integration Page 801
ENGI 345 8. Multiple Integration Page 801 8. Multiple Integration This chapter provides only a very brief introduction to the major topic of multiple integration. Uses of multiple integration include
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 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 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 informationV (r,t) = i ˆ u( x, y,z,t) + ˆ j v( x, y,z,t) + k ˆ w( x, y, z,t)
IV. DIFFERENTIAL RELATIONS FOR A FLUID PARTICLE This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations and common
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 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 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 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 informationTOPICS. Density. Pressure. Variation of Pressure with Depth. Pressure Measurements. Buoyant ForcesArchimedes Principle
Lecture 6 Fluids TOPICS Density Pressure Variation of Pressure with Depth Pressure Measurements Buoyant ForcesArchimedes Principle Surface Tension ( External source ) Viscosity ( External source ) Equation
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 informationE 490 FE Exam Prep. Engineering Mechanics
E 490 FE Exam Prep Engineering Mechanics 2008 E 490 Course Topics Statics Newton s Laws of Motion Resultant Force Systems Moment of Forces and Couples Equilibrium Pulley Systems Trusses Centroid of an
More informationPROPERTIES OF FLUIDS
Unit  I Chapter  PROPERTIES OF FLUIDS Solutions of Examples for Practice Example.9 : Given data : u = y y, = 8 Poise = 0.8 Pas To find : Shear stress. Step  : Calculate the shear stress at various
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 informationME3560 Tentative Schedule Fall 2018
ME3560 Tentative Schedule Fall 2018 Week Number 1 Wednesday 8/29/2018 1 Date Lecture Topics Covered Introduction to course, syllabus and class policies. Math Review. Differentiation. Prior to Lecture Read
More informationNumber Plane Graphs and Coordinate Geometry
Numer Plane Graphs and Coordinate Geometr Now this is m kind of paraola! Chapter Contents :0 The paraola PS, PS, PS Investigation: The graphs of paraolas :0 Paraolas of the form = a + + c PS Fun Spot:
More informationFluid Mechanics Indian Institute of Technology, Kanpur Prof. Viswanathan Shankar Department of chemical Engineering. Lecture No.
Fluid Mechanics Indian Institute of Technology, Kanpur Prof. Viswanathan Shankar Department of chemical Engineering. Lecture No. # 05 Welcome to this fifth lecture on this nptel course on fluid mechanics
More informationTorque. Physics 6A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Physics 6A Torque is what causes angular acceleration (just like a force causes linear acceleration) Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque
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 informationExploring the relationship between a fluid container s geometry and when it will balance on edge
Exploring the relationship eteen a fluid container s geometry and hen it ill alance on edge Ryan J. Moriarty California Polytechnic State University Contents 1 Rectangular container 1 1.1 The first geometric
More information2. a) Explain the equilibrium of i) Concurrent force system, and ii) General force system.
Code No: R21031 R10 SET  1 II B. Tech I Semester Supplementary Examinations Dec 2013 ENGINEERING MECHANICS (Com to ME, AE, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions
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 informationSecond Moments or Moments of Inertia
Second Moments or Moments of Inertia The second moment of inertia of an element of area such as da in Figure 1 with respect to any axis is defined as the product of the area of the element and the square
More information