Civil Engineering Review 2 - Chapter 1 Fluid Mechanics PDF

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CE Review IIChapter 1Fluid MechanicsChapter 1Fluid MechanicsIntroductionThis module deals with the review of Fluid Mechanics as a preparation for the Civil Engineering Licensure Examination.Specific ObjectivesAt the end of the lesson, the students should be able to: Review the basics of Fluid Mechan...

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CE Review II Chapter 1

Fluid Mechanics

Chapter 1

Fluid Mechanics Introduction This module deals with the review of Fluid Mechanics as a preparation for the Civil Engineering Licensure Examination. Specific Objectives At the end of the lesson, the students should be able to: -

Review the basics of Fluid Mechanics. Solve problems usually encountered on Civil Engineering Licensure Examination. Solve problems faster and easier.

Duration Chapter 1:

Fluid Mechanics

= 6 hours lecture = 9 hours laboratory

Lesson Proper

PROPERTIES OF FLUIDS Fluid Mechanics Fluid Mechanics is a physical science dealing with the action of fluids at rest or in motion. Fluid mechanics can be subdivided into two major areas, fluid statics, which deals with fluids at rest, and fluid dynamics, dealing with fluids in motion. Gas and Liquid are both considered as Fluid. TYPES OF FLUID Ideal Fluid • Assumed to have no viscosity (and hence, no resistance to shear) • Incompressible • Have uniform velocity when flowing • No friction between moving layers of fluid • No eddy currents or turbulence Real Fluid • Exhibit infinite viscosities • Non-uniform velocity distribution when flowing • Compressible • Experience friction and turbulence in flow Common Fluid Properties 1. Mass Density,  The mass density or commonly called “density” of fluid is the ratio of its mass per unit volume and is given by the formula 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑, 𝑀 𝜌= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑, 𝑉 Mass density is represented by the Greek symbol  (rho). It has units of kilograms per cubic meter (kg/m3) or pounds-mass per cubic foot (lbm/ft3). The mass density of water at 4°C is 1000 kg/m3 (62.4 lbm/ft3). 2. Specific Weight,  The gravitational force per unit volume of fluid, or simply the weight per unit volume, is defined as specific weight. It is represented by the Greek symbol  (gamma). 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑, 𝑊 𝛾= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑, 𝑉 A commonly used value of specific weight of water on Earth at 4°C is 9.81 kN/m3 (62.4 lb/ft3). Specific weight and density are related by 𝛾 = 𝜌𝑔

3. Specific Gravity, s Specific gravity is a dimensionless ratio of a fluid’s density to some standard reference density. For liquid and solid, the reference density is water. Also, specific gravity can be derived from the ratio of a liquid’s unit weight to the unit weight of water (reference). 𝛾𝑙𝑖𝑞𝑢𝑖𝑑 𝜌𝑙𝑖𝑞𝑢𝑖𝑑 𝑠= 𝑜𝑟 𝑠 = 𝛾𝑤𝑎𝑡𝑒𝑟 𝜌𝑤𝑎𝑡𝑒𝑟 In gases, the standard reference to calculate the specific gravity is the density of air 𝜌𝑔𝑎𝑠 𝑠= 𝜌𝑎𝑖𝑟 4. Specific Volume, Vs It is the volume occupied by a unit mass of fluid or simply the reciprocal of mass density 𝑉 1 𝑜𝑟 𝑉𝑠 = 𝑉𝑠 = 𝜌 𝑀 Commonly used units of specific volume are cm3/gram, m3/kg, ft3/lbm.

PRINCIPLES OF HYDROSTATICS Unit Pressure or Pressure, p Pressure is the force per unit area exerted by a liquid or gas on a body or surface, with the force acting at right angles to the surface uniformly in all directions. 𝒑=

𝑭𝒐𝒓𝒄𝒆 𝑭 , 𝑨𝒓𝒆𝒂 𝑨

Blaise Pascal, a French scientist observed that the pressure in a fluid at rest is the same at all points provided they are at the same height. Therefore, we can say that like other types of stress, pressure is not a vector quantity. No direction can be assigned to it. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area. Some units for pressure give a ratio of force to area. Newtons per square meter of area, or pascals (Pa), is the SI unit. The traditional units include psi, which is pounds-force per square inch, and psf, which is pounds-force per square foot. Absolute and Gage Pressures Gage Pressure (Relative Pressure) Gage pressures are the pressures above or below the atmosphere and can be measured by pressure gauges or manometers.

Atmospheric Pressure It is the pressure that we experience in the surface of the earth in our everyday lives. It is the pressure at any point on the earth’s surface due to the weight of air above it. Under Normal conditions at sea level: Patm = 1.0 atm = 101.325 kPa = 14.70 psi = 760 mm-Hg = 29.92 in-Hg = 2116 psf Absolute Pressure It is the actual pressure at any given point. This is the sum of atmospheric and gage pressures. 𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑃𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑒 + 𝑃𝑔𝑎𝑔𝑒 𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔

VARIATION OF PRESSURE WITH DEPTH The pressure varies depending on the location. As you go deeper below any liquid surface, the pressure increases because of the added weight of liquid above.

𝒑 = 𝜸𝒉

For any given point below a liquid surface, the gage pressure is given by the formula where: p = gage pressure γ = unit weight of the liquid h = depth of the point below the liquid surface

Two or more points below the same liquid surface having the same depth will have the same pressure. See the illustration.

PA = PB = PC = PD = PE = PF = PG = unit weight of water x height A, B, C, D, E, F, and G are points below the water surface having the same depth. They have the same gage pressures.

For any two points 1 and 2 below any liquid surface having different depth, the relation in their pressures is given by:

𝒑𝟐 = 𝒑𝟏 + 𝜸𝒉

PRESSURE BELOW LAYERS OF DIFFERENT LIQUIDS FOR CLOSED TANK

Consider the closed tank shown to be filled with liquids of different unit weights and with the air at the top under a gage pressure of 𝒑𝑨 , the pressure at the bottom of the tank is: 𝒑𝟏 = ∑ 𝜸𝒉 + 𝒑𝑨

𝒑𝟏 = 𝜸𝟏 𝒉𝟏 + 𝜸𝟐 𝒉𝟐 + 𝜸𝟑 𝒉𝟑 + 𝒑𝑨

FOR OPEN TANK

𝒑𝟏 = ∑ 𝜸𝒉

𝒑𝟏 = 𝜸𝟏 𝒉𝟏 + 𝜸𝟐 𝒉𝟐 + 𝜸𝟑 𝒉𝟑

PRESSURE HEAD Pressure head is the height “h” of a column of homogeneous liquid of unit weight γ that will produce an intensity pressure p. 𝒉=

𝒑 𝜸

Derived from the formula of pressure p = γh, h is expressed in meters or feet.

To Convert Pressure head (height) of liquid A to liquid B 𝒉𝑩 =

𝜸𝑨 𝒔𝑨 𝝆𝑨 𝒉 𝒉𝑨 , 𝒐𝒓 𝒉𝑩 = 𝒉𝑨 , 𝒐𝒓 𝒉𝑩 = 𝒔𝑩 𝑨 𝝆𝑩 𝜸𝑩

To Convert pressure head (height) of any liquid to water, just multiply its height by its specific gravity 𝒉𝒘𝒂𝒕𝒆𝒓 = 𝒔𝒍𝒊𝒒𝒖𝒊𝒅 𝒉𝒍𝒊𝒒𝒖𝒊𝒅

MANOMETER A manometer is a tube, usually bent in a form of U, containing a liquid of known specific gravity, the surface of which moves proportionally to changes of pressure. It is used to measure pressure or pressure differences. Types of Manometer 1. Open Type It has an atmospheric surface in one leg and is capable of measuring gage pressure.

2. Differential Type It has no atmospheric surface and capable of measuring only pressure differences.

3. Piezometer The simplest form of open manometer. It is a tube tapped into a wall of a container or conduit for the purpose of measuring pressure. The fluid in the container or conduit rises in this tube to form a free surface.

Steps in Solving Manometer Problems: 1. Decide on the fluid in feet or meter, of which the heads are to be expressed, (water is most advisable). 2. Starting from an end point, number in order, the interface of different fluids. 3. Identify the points of equal pressure (taking into account that for a homogeneous fluid at rest, the pressures along the same horizontal plane are equal). Label these points with the same number. 4. Proceed from level to level, adding (if going down) or subtracting (if going up) pressure heads as the elevation decreases or increases, respectively with due regard for the specific gravity of the fluids.

TOTAL HYDROSTATIC FORCE ON SURFACES Hydrostatic Force on a Plane Surface When a surface is submerged in a fluid, forces develop on the surface due to the fluid. The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic structures. For fluids at rest we know that the force must be perpendicular to the surface since there are no shearing stresses present. We also know that the pressure will vary linearly with depth as shown in Figure 1.0 if the fluid is incompressible. For a horizontal surface, such as the bottom of a liquid-filled tank (Fig. 1.0a), the magnitude of the resultant force is simply, 𝐹𝑅 = 𝑝𝐴 where p is the uniform pressure on the bottom and A is the area of the bottom. For the open tank shown, 𝑝 = 𝛾ℎ. Note that if atmospheric pressure acts on both sides of the bottom, as is illustrated, the resultant force on the bottom is simply due to the liquid in the tank. Since the pressure is constant and uniformly distributed over the bottom, the resultant force acts through the centroid of the area as shown in Fig. 1.0a. As shown in Fig. 1.0b, the pressure on the ends of the tank is not uniformly distributed. Determination of the resultant force for situations such as this is presented below.

In the case of an inclined or vertical plane submerged in a liquid, the total hydrostatic force can be found by the following: For Vertical Plane Submerged under a Liquid Surface 𝑭 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 ℎ𝑦𝑑𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒𝑎 𝜸 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑞𝑢𝑖𝑑  𝑖𝑠 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑙𝑖𝑞𝑢𝑖𝑑 𝒉 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑨 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝒄𝒈 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝒄𝒑 𝑖𝑠 𝑡ℎ𝑒 center of pressure 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑓𝑜𝑟𝑐𝑒 𝐹 𝑎𝑐𝑡𝑠 𝒆 𝑖𝑠 𝑡ℎ𝑒 Eccentricity - the distance between cg and cp  𝒚 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑙𝑖𝑞𝑢𝑖𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑛𝑑 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒𝑎

where:

In the case of vertical planes submerged in a liquid, ℎ = 𝑦

𝐸𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦, 𝑒 = where:

𝐼𝑔 𝐴𝑦

𝑰𝒈 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

For Inclined Plane Submerged under a Liquid Surface

For the case of inclined areas, ℎ 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑦  and 𝑦 forms a triangle with the liquid the ℎ surface The formula for 𝑭 𝒂𝒏𝒅 𝒆 is the same to that of the vertical plane.

Geometric properties of some common shapes.

Hydrostatic Force on a Curve Surface Consider the curved surface AB in Fig. 2.0a. The goal is to represent the pressure distribution with a resultant force that passes through the center of pressure. One approach is to integrate the pressure force along the curved surface and find the equivalent force. However, it is easier to sum forces for the free body shown in the upper part of Fig. 2.0b. The lower sketch in Fig. 2.0b shows how the force acting on the curved surface relates to the force F acting on the free body. Using the FBD and summing forces in the horizontal direction shows that 𝐹𝑥 = 𝐹𝐴𝐶 The line of action for the force 𝐹𝐴𝐶 is through the center of pressure for side AC. The vertical component of the equivalent force is 𝐹𝑦 = 𝑊 + 𝐹𝐶𝐵 where 𝑊 is the weight of the fluid in the free body and 𝐹𝐶𝐵 is the force on the side CB. The force 𝐹𝐶𝐵 acts through the centroid of surface CB, and the weight acts through the center of gravity of the free body. The line of action for the vertical force may be found by summing the moments about any convenient axis.

The magnitude of the resultant is obtained from the equation 𝑭 = √(𝑭𝒙 )𝟐 + (𝑭𝒚 )

𝟐

BUOYANCY Archimedes’ Principle A principle discovered by the Greek scientist Archimedes that states that “any object immersed in a fluid is acted upon by an upward force (buoyant force) equal to the weight of the displaced fluid”. This principle, also known as the law of hydrostatics, applies to both floating and submerged bodies and to all fluids. Consider the figure below. The ship of total weight W floats on the ocean. In order to maintain its position, the sea water exerts an upward force equal to the weight of the ship. This upward force is the Buoyant Force. When solving problems involving buoyancy, identify the forces acting and apply conditions of equilibrium ∑ 𝐹𝑥 = 0 ∑ 𝐹𝑦 = 0 ∑𝑀 = 0

when an object floats in any liquid, a portion of it is submerged under the liquid surface. This submerged portion is equal to the volume of the displaced liquid as shown in the figure. (a) the ship floats on the ocean having a portion of it is below water surface (b) shows the volume of the displaced liquid caused by the ship. The volume of the displaced liquid (VD) is equal to the volume of the submerged portion of the object.

The weight of the displaced liquid is also equal to the Buoyant Force

Based on the above explanation, we derive the formulas 𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒐𝒃𝒋𝒆𝒄𝒕 = 𝑩𝒖𝒐𝒚𝒂𝒏𝒕 𝑭𝒐𝒓𝒄𝒆,

𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒅 𝒍𝒊𝒒𝒖𝒊𝒅 = 𝑩𝒖𝒐𝒚𝒂𝒏𝒕 𝑭𝒐𝒓𝒄𝒆,

𝑾 = 𝑩𝑭

𝑾𝑫 = 𝑩𝑭

where: 𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒅 𝒍𝒊𝒒𝒖𝒊𝒅 𝑾𝑫 = 𝜸𝒍𝒊𝒒𝒖𝒊𝒅 𝒙 𝑽𝑫 For homogeneous solid body of volume V “floating” in a homogeneous fluid at rest: 𝑽𝑫 =

𝜸𝒃𝒐𝒅𝒚 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒈𝒓𝒂𝒗𝒊𝒕𝒚 𝒐𝒇 𝒃𝒐𝒅𝒚 𝑽= 𝑽 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒈𝒓𝒂𝒗𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅 𝜸𝒍𝒊𝒒𝒖𝒊𝒅

If the body of height H has a constant horizontal cross-sectional area such as vertical cylinders, blocks, etc.:

𝑫=

𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒈𝒓𝒂𝒗𝒊𝒕𝒚 𝒐𝒇 𝒃𝒐𝒅𝒚 𝜸𝒃𝒐𝒅𝒚 𝑯 𝑯= 𝜸𝒍𝒊𝒒𝒖𝒊𝒅 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒈𝒓𝒂𝒗𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅

If the body is of uniform vertical cross-sectional area A, the area submerged As is: 𝑨𝒔 =

𝜸𝒃𝒐𝒅𝒚 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒈𝒓𝒂𝒗𝒊𝒕𝒚 𝒐𝒇 𝒃𝒐𝒅𝒚 𝑨 𝑨= 𝜸𝒍𝒊𝒒𝒖𝒊𝒅 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒈𝒓𝒂𝒗𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅

RELATIVE EQUILIBRIUM OF LIQUIDS Relative equilibrium of liquid is a condition where the whole mass of liquid including the vessel in which the liquid is contained, is moving at uniform accelerated motion with respect to the earth, but every particle of liquid have no relative motion between each other. There are two cases of relative equilibrium that will be discussed in this section: linear translation and rotation. Note that if a mass of liquid is moving with constant speed, the conditions are the same as static liquid in the previous sections. Rectilinear Translation – Moving Vessel Horizontal Motion If a mass of fluid moves horizontally along a straight line at constant acceleration a, the liquid surface assume an angle θ with the horizontal, see figure below.

For any value of a, the angle θ can be found by considering a fluid particle of mass m on the surface. The forces acting on the particle are the weight W = mg, inertia force or reverse effective force REF = ma, and the normal force N which is the perpendicular reaction at the surface. These three forces are in equilibrium with their force polygon shown to the right. From the force triangle

𝑅𝐸𝐹 𝑀𝑎 = 𝑊 𝑀𝑔 𝑎 tan 𝜃 = 𝑔

tan 𝜃 =

Inclined Motion Consider a mass of fluid being accelerated up an incline α from horizontal. The horizontal and vertical components of inertia force REF would be respectively, x = mah and y = mav.

𝑥 𝑊+𝑦 𝑚𝑎 cos 𝛼 tan 𝜃 = 𝑚𝑔 + 𝑚𝑎 sin 𝛼 𝑎 cos 𝛼 tan 𝜃 = 𝑔 + 𝑎 sin 𝛼 but 𝑎 cos 𝛼 = 𝑎ℎ and 𝑎 sin 𝛼 = 𝑎𝑣 , hence 𝑎ℎ tan 𝜃 = 𝑔 + 𝑎𝑣

From the force triangle above

tan 𝜃 =

𝒂𝒉 𝒈 ± 𝒂𝒗 Use (+) sign for upward motion and (-) sign for downward motion 𝐭𝐚𝐧 𝜽 =

Vertical Motion The figure shown is a mass of liquid moving vertically upward with a constant acceleration a. The forces acting to a liquid column of depth h from the surface are weight of the liquid W = γV, the inertia force REF = ma, and the pressure F = pA at the bottom of the column. ∑ 𝐹𝑣 = 0

𝐹 = 𝑊 + 𝑅𝐸𝐹 𝑝𝐴 = 𝛾𝑉 + 𝑚𝑎 𝑝𝐴 = 𝛾𝑉 + 𝜌𝑉𝑎 𝛾 𝑝𝐴 = 𝛾(𝐴ℎ) + (𝐴ℎ)𝑎 𝑔 𝛾 𝑝 = 𝛾ℎ + ℎ𝑎 𝑔 𝑎 𝑝 = 𝛾ℎ (1 + ) 𝑔

𝒂 𝒑 = 𝜸𝒉 (𝟏 ± ) 𝒈 Use (+) sign for upward motion and (-) sign for downward motion. Also note that a is positive for acceleration and negative for deceleration

Rotation (Rotating Vessel) When at rest, the surface of mass of liquid is horizontal at PQ as shown in the figure. When this mass of liquid is rotated about a vertical axis at constant angular velocity ω (in radians per second), it will assume the surface ABC which is parabolic. Every particle is subjected to a centrifugal force (or reversed normal effective force) 𝐶𝐹 = 𝑚𝜔2 𝑥 which produces centripetal acceleration towards the center of rotation. Other forces that acts are gravity force W = mg and normal force N.

From the force polygon, 𝐶𝐹 𝑊 𝑚𝜔2 𝑥 tan 𝜃 = 𝑚𝑔 𝜔2 𝑥 tan 𝜃 = 𝑔 where tan 𝜃 is the slope at the surface of paraboloid at any distance x from the axis of rotation tan 𝜃 =

From Calculus, y’ = slope, thus 𝑑𝑦 = tan 𝜃 𝑑𝑥 𝑑𝑦 𝜔2 𝑥 = 𝑔 𝑑𝑥 𝜔2 𝑥 𝑑𝑥 𝑑𝑦 = 𝑔 𝜔2 𝑥 ∫ 𝑑𝑦 = ∫ 𝑑𝑥 𝑔 𝜔2 𝑥 2 𝑦= 2𝑔

For cylindrical vessel of radius r revolved about its vertical axis, the height h of paraboloid is 𝜔2 𝑟 2 ℎ = 2𝑔 Other Formulas By squared-property of parabola, the relationship of y, x, h and r is defined by 𝑟2 𝑥 2 = 𝑦 ℎ Volume of paraboloid of revolution 1 𝑉 = 𝜋𝑟 2 ℎ 2 Important conversion factor

1 𝑟𝑝𝑚 =

1 𝜋 𝑟𝑎𝑑/𝑠𝑒𝑐 30

Liquid Surface Conditions For open cylindrical containers more than half-full of liquid, rotated about its vertical axis (h>H/2):

For closed cylindrical containers more than half-full of liquid, rotated about its vertical axis (h>H/2):

For closed cylindrical containers completely filled with liquid:

Solved Problems: 1. A reservoir of glycerin has a mass of 1,200 kg and a volume of 0.952 m3. Determine its (a) weight, W, (b) unit weight, , (c) mass density, , (d) specific gravity, s, (e) specific volume, Vs. Solution: a. Weight, 𝑊 = 𝑀𝑔

= (1200 𝑘𝑔)(9.81𝑚/𝑠 2 )

Weight, 𝑊 = 𝟏𝟏, 𝟕𝟕𝟐 𝑵 𝒐𝒓 𝟏𝟏. 𝟕𝟕𝟐 𝒌𝑵

b. Unit weight, 𝛾 =

𝑊 𝑉

11.772 𝑘𝑁 0.952 𝑚3 𝒌𝑵 Unit weight, 𝛾 = 𝟏𝟐. 𝟑𝟔𝟔 𝟑 𝒎 =

c. mass density, 𝜌 =

𝑀 𝑉

1200 𝑘𝑔 = 0.952 𝑚3 𝒌𝒈 mass density, 𝜌 = 𝟏𝟐𝟔𝟎. 𝟓 𝟑 𝒎

d. specific gravity, 𝑠 =

𝛾𝑙𝑖𝑞𝑢𝑖𝑑

𝛾𝑤𝑎𝑡𝑒𝑟

12.366 𝑘𝑁/𝑚3 9.81 𝑘𝑁/𝑚3 specific gravity, 𝑠 = 𝟏. 𝟐𝟔 =

e. specific volume, 𝑉𝑠 =

1

𝜌

1 = 1260.5 𝑘𝑔/𝑚3 specific volume, 𝑉𝑠 = 𝟎. 𝟎𝟎𝟎𝟕𝟗 𝒎𝟑 /𝒌𝒈

2. If the specific volume of a certain gas is 0.7848 m3/kg, what is its specific weight? Solution: 1 𝜌 1 𝜌= 𝑉𝑠

𝑉𝑠 =

1 0.7848 𝑚3 /𝑘𝑔 𝜌 = 1.27421 𝑘𝑔/𝑚3

𝜌=

Specific weight 𝛾 = 𝜌𝑔 𝛾 = (1.27421 𝑘𝑔/𝑚3 )(9.81 𝑚/𝑠 2 ) 𝜸 = 𝟏𝟐. 𝟓 𝑵/𝒎𝟑 3. What is the water pressure at a depth of 35 ft in the tank shown?

Solution: 𝒑 = 𝜸𝒘𝒂𝒕𝒆𝒓 𝒉 𝑙𝑏 𝑝 = 62.4 𝑓𝑡 3 𝑥 35𝑓𝑡 𝑝 = 𝟐𝟏𝟖𝟒

𝒍𝒃

𝒇𝒕𝟐

𝒐𝒓 𝟐𝟏𝟖𝟒 𝒑𝒔𝒇 ------ gage pressure

4. If a depth of liquid of 1 m causes a pressure of 7 KPa, what is the specific gravity of the liquid? Solution: 𝒑 = 𝜸𝒍𝒊𝒒𝒖𝒊𝒅 𝒉 𝒑 = 𝒔(𝜸𝒘𝒂𝒕𝒆𝒓)𝒉 𝑠=

𝐾𝑁

7 2 𝑚 𝐾𝑁

(9.81 3 )(1𝑚) 𝑚

s = 0.714

5. Oil with a specific gravity of 0.80 forms a layer 0.90 m deep in an open tank that is otherwise filled with water. The total depth of water and oil is 3 m. What is the gage pressure at the bottom of the tank?

Solution: 𝒑𝟑 = (𝜸𝒐𝒊𝒍)(𝒉𝒐𝒊𝒍 ) + (𝜸𝒘𝒂𝒕𝒆𝒓 )(𝒉𝒘𝒂𝒕𝒆𝒓) 𝑝3 = 𝑠(𝛾𝑤𝑎𝑡𝑒𝑟 )(ℎ𝑜𝑖𝑙 ) + (𝛾𝑤𝑎𝑡𝑒𝑟 )(ℎ𝑤𝑎𝑡𝑒𝑟 ) 𝑘𝑁 𝑘𝑁 𝑝3 = 0.8 (9.81 3 ) (0.90𝑚) + (9.81 3 ) (2.10𝑚) 𝑚 𝑚 𝑘𝑁 𝑘𝑁 𝑝3 = 7.0632 2 + 20.601 𝑚2 𝑚 𝑘𝑁 𝒑𝟑 = 27.664 2 𝑜𝑟 𝟐𝟕. 𝟔𝟔𝟒 𝐤𝐏𝐚 𝑚 6. A point 5 meters below the water surface has a certain amount of pressure. How deep below the surface of the oil will have the same pressure reading? Use specific gravity of water = 1 and for oil, s = 0.8 Solution: 𝑠 ℎ𝑤𝑎𝑡𝑒𝑟 ℎ𝑜𝑖𝑙 = 𝑤𝑎𝑡𝑒𝑟 𝑠 ℎ𝑜𝑖𝑙 =

1

𝑜𝑖𝑙

0.8

(5𝑚)

𝒉𝒐𝒊𝒍 = 𝟔. 𝟐𝟓 𝒎

7. An open tank is filled with different types of liquid. Determine (a) the gage pressure between layer of water and mercury; (b) the gage pressure at point 1; (c) absolute pressure at point 1. Solution: a. pressure between water and mercury 𝑝 = 𝛾𝑜𝑖𝑙 ℎ𝑜𝑖𝑙 + 𝛾𝑤𝑎𝑡𝑒𝑟 ℎ𝑤𝑎𝑡𝑒𝑟 𝑝 = (0.8)(9.81)(4) + (9.81)(3) 𝒑 = 𝟔𝟎. 𝟖𝟐𝟐 𝑲𝑷𝒂 − − − −𝑎𝑛𝑠𝑤𝑒𝑟 b. the gage pressure at point 1 𝑝 = 𝛾𝑜𝑖𝑙 ℎ𝑜𝑖𝑙 + 𝛾𝑤𝑎𝑡𝑒𝑟 ℎ𝑤𝑎𝑡𝑒𝑟 + 𝛾𝑚𝑒𝑟𝑐𝑢𝑟𝑦 ℎ𝑚𝑒𝑟𝑐𝑢𝑟𝑦 𝑝 = (0.8)(9.81)(4) + (9.81)(3) + 13.6(9.81)(2) 𝒑 = 𝟑𝟐𝟕. 𝟔𝟓𝟒 𝑲𝑷𝒂 − − − −𝑎𝑛𝑠𝑤𝑒𝑟 c. absolute pressure at point 1 𝑝𝑎𝑏𝑠 = 𝑝𝑎𝑡𝑚 + 𝑝𝑔𝑎𝑔𝑒 𝑝𝑎𝑏𝑠 = 101.325 𝐾𝑃𝑎 + 327 .654 𝐾𝑃𝑎 𝒑𝒂𝒃𝒔 = 𝟒𝟐𝟖. 𝟗𝟕𝟗 𝑲𝑷𝒂 − − − −𝑎𝑛𝑠𝑤𝑒𝑟 8. If the three liquids in sample problem 2 are to be replace by gasoline having s = 0.68, how deep is the tank should be t...