ENGR. BON RYAN ANIBAN Properties of Fluid •Mass Density •Specific Volume •Unit Weight and Specific weight •Specific Gravity •Viscosity •Surface Tension •Capillarity •Compressibility •Pressure Disturbances •Property Changes in Ideal Gas Principles of Hydrostatic •Unit Pressure •Pascal’s Law •Absolute and Gauge Pressures •Variations and Pressure •Pressure Head •Manometers Total Hydrostatic Force on Surfaces •Total Hydrostatic Force on Plane and Curved Surfaces •Dams •Buoyancy •Statical stability of Floating Bodies •Thin-walled Pressure Vessels Relative Equilibrium of Liquid •Rectilinear Translation •Rotation FLUID MECHANICS FLUID STATICS FLUID IDEAL FLUID FLUID MECHANICS • Is a branch of fluid mechanics that studies Fluid Statics fluid at rest. It embraces the study of the conditions under which fluids are at rest in stable equilibrium. Fluid Mechanics It deals with the fluids ( liquid and gasses) in motion • • • Assumed to have no viscosity Incompressible Have uniform velocity when flowing No turbulence REAL FLUID • • • • Infinite viscosities Non-uniform velocity Compressible Experience friction and turbulence ENGR. BON RYAN ANIBAN MASS DENSITY (ρ) – mass per unit volume mass π kg g slugs π= = = 3= = Volume π m cm3 ft 3 IDEAL GAS DENSITY (ρ) π π= π π P = absolute pressure of gas in Pa recall: absolute pressure = gauge pressure + atmospheric pressure R = gas constant For Air : R = 287 Joule/kg-K (SI) R = 1716 lb-ft/slug-R (English) T = absolute temperature K = °C + 273 R = °F + 460 Densities of Common Fluid Fluid π in kg/m3 Air (STP) 1.29 Alcohol 790 Ammonia 602 Gasoline 720 Glycerin 1260 Mercury 13600 Water (at 4°C) 1000 Gravitational acceleration SPECIFIC WEIGHT/UNIT WEIGHT/WEIGHT DENSITY(γ) – weight per unit volume weight π m × g N lb γ= = = = 3= 3 Volume π π m ft g = 9.81 m/s2 = 32.2 ft/s2 SPECIFIC VOLUME( Vs) – volume occupied γ= by a unit mass of fluid 1 1 m3 Vs = = = density ρ kg weight π m × g = = Volume π π γ = ρg γ = 1000 kg/m3 x 9.81 m/s2 = 9810 γ = 9810 N/m3 SPECIFIC GRAVITY(s) – is the ratio of specific weight of liquid to water s= πliquid πwater = Densities of Common Fluid πΎliquid πΎwater Fluid = 9810 N/m3 (SI) = 62.4 lb/ft3 (English) ππ β π m3 β s2 π in kg/m3 Air (STP) 1.29 Alcohol 790 Ammonia 602 Gasoline 720 Glycerin 1260 Mercury 13600 Oil 800 Water (at 4°C) 1000 πΉ = ππ Length 1 ft = 0.3048 m 1 mi = 5280 ft. = 1609.344 m 1 nautical mile = 6076 ft = 1852 m 1 yd = 3 ft = 0.9144 m Volume 1 ft3 = 0.028317 m3 1 L = 0.001 m3 = 0.0353515 ft3 Mass 1 slug = 14.594 kg 1 tonne = 1000 kg Velocity 1 ft/s= 0.3048 m/s 1 mi/h = 1.46666 ft/s = 0.44704 m/s Area 1 ft2 = 0.028317 m3 1 mi2 = 2.78784 x 107 ft2 = 2.59 x 106 ft3 Acceleration 1 ft/s2 = 0.3048 m/s2 Mass Flow 1 slug/s = 14.594 m/s 1 lbm/s = 0.4536 kg/s SPECIFIC VOLUME( Vs) – volume occupied γ= by a unit mass of fluid 1 1 m3 Vs = = = density ρ kg weight π m × g = = Volume π π γ = ρg γ = 1000 kg/m3 x 9.81 m/s2 = 9810 γ = 9810 N/m3 SPECIFIC GRAVITY(s) – is the ratio of specific weight of liquid s= πliquid πwater = Densities of Common Fluid πΎliquid πΎwater Fluid = 9810 N/m3 (SI) = 62.4 lb/ft3 (English) ππ β π m3 β s2 π in kg/m3 Air (STP) 1.29 Alcohol 790 Ammonia 602 Gasoline 720 Glycerin 1260 Mercury 13600 Oil 800 Water (at 4°C) 1000 πΉ = ππ SPECIFIC VOLUME( Vs) – volume occupied by a unit mass of fluid 1 1 m3 Vs = = = density ρ kg γ= weight π m × g = = Volume π π γ = πg γ = 1000 kg/m3 x 9.81 m/s2 = 9810 SPECIFIC GRAVITY(s) – is the ratio of ππ β π m3 β s2 γ = 9810 N/m3 specific weight of liquid π πΎ s = πliquid = πΎliquid water π = 1000 kg/m3 water = 9810 N/m3 (SI) = 62.4 lb/ft3 (English) πΉ = ππ 1 slug x x 14.594 kg π = 1.940 slug/ft3 γ = πg γ = 1.940 slug/ft3 x 32.2 ft/s2 γ = 62.468 lb/ft3 0.3048 m 1 ft 3 ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 1 A glycerin has a mass of 1200 kg and a volume of 0.952 cu.m. (a) Find its weight , (b) unit weight, (c) mass density, (d) and specific gravty. Solution a π = πg = 1200 kg (9.81 m/s2) = 11772 N = 11.772 kN π π 11.772 kN = 12.366 kN/m3 = 3 0.952 m b πΎ= π π 1200 kg = 1260.504 kg/m3 = 3 0.952 m πππππ’ππ d π = ππ€ππ‘ππ 1260.504 kg/m3 = 1.261 = 1000 kg/m3 c π= ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 2 If an object has a mass of 22 kg at sea level, (a) what will be its weight at a point where acceleration due to gravity = 9.75 m/s2 (b) what will be its mass at that point? Solution a π = πg = 22 kg (9.75 m/s2) = 214.5 N b π = 22 kg ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 3 Specific gravity of a certain oil is 0.82 (a) Find its unit weight in kN/m3, lb/ft3 (b) Find its density in kg/m3, slug/ft3 Solution πΎππππ’ππ a π = πΎπ€ππ‘ππ 0.82 = πΎπππ 9.81 kN/m3 πΎπππ = 8.044 kN/m3 b 0.82 = πΎπππ 62.4 lb/ft3 πΎππππ’ππ = 51.168 lb/ft3 πΎπππ = ππππ g 8.044 × 103 N/m3 = ππππ (9.81 m/s2) 51.168 lb/ft3 = ππππ (32.2 ft/s2) πππππ’ππ = 819.980 kg/m3 πππππ’ππ = 1.589 slug/ft3 ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 4 If a specific volume of a gas is 0.7848 m3/kg. What is its unit weight? Solution ππ = 1 π 0.7848 m3 /kg = 1 π π= 1.274 kg/m3 πΎπππ = ππππ g = 1.274 kg/m3(9.81 m/s2) πΎπππ = 12.498 N/m3 ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 5 What is the specific weight if the pressure is 480 kPa absolute at 21°C? Solution π= π π π (480 × 103 )Pa = (287)(J/kg − K)(21° + 273)(K) π = 5.689 kg/m3 πΎ = πg = 5.689 kg/m3(9.81 m/s2) πΎ = 55.809 N/m3 ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 6 Find the mass density of helium at temperature 4°C, and a pressure of 184 kPa gage, if atmospheric pressure is 101.92 kPa. (R = 2079 J/kg-K) Solution π= π π π 184 + 101.92 103 Pa = (2079)(J/kg − K)(4° + 273)(K) π = 0.496 kg/m3 πππππππππ (μ) mu The property of fluid which determines the amount of its resistance to shearing forces. Area=A y U F viscous fluid F ∝ AU y πΉ π ∝ π΄ π¦ πΉ = π (shearing stress) π΄ π π π∝ π=π π¦ π¦ Where the k is called the dynamic absolute viscosity denoted as μ π τ=μ π¦ π μ= π/π² N m2 Where: π = shear stress in lb/ft2or Pa μ = absolute viscosity in lb sec/ft2 or Pa-sec y= distance between the plates in ft or m U = velocity in ft/s or m/s πππππππππ πππππππππ (ν) nu Kinematic viscosity is the ratio of the dynamic viscosity of the fluid, μ, to its mass density, ρ π π= π Common Units of Viscosity System Absolute, μ Kinematic, ν English lb-sec/ft2 (slug/ft-sec) ft2/sec Metric Dyne-s/cm2 (poise) cm2/s (stoke) Pa-s (N-s/m2) m2/s S.I Note: 1 poise = 1dyne-s/cm2 = 0.1 Pa-sec 1 stoke = 0.0001 m2/s (1 dyne = 10-5N) ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 7 Two large plane surfaces are 25 mm apart and the space between them is filled with a liquid of viscosity of μ = 0.958 Pa-s. Assuming the velocity gradient to be a straight line, what force is required to pull a very thin plane of 0.37 m2 area at a constant speed of 0.3 m/s if the plate is 8.4 from one of the surfaces Solution 8.4 mm F1 F2 25 mm A = 0.37 m2 16.6 mm μ = 0.958 Pa-s F ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 7 Two large plane surfaces are 25 mm apart and the space between them is filled with a liquid of viscosity of μ = 0.958 Pa-s. Assuming the velocity gradient to be a straight line, what force is required to pull a very thin plane of 0.37 m2 area at a constant speed of 0.3 m/s if the plate is 8.4 mm from one of the surfaces Solution 8.4 mm F1 πΉ = πΉ1 + πΉ2 F2 25 mm A = 0.37 m2 π πΉ/π΄ π= 16.6 mm π= π/π¦ π/π¦ μ = 0.958 Pa-s πΉ1 πΉ2 2 0.37 m 0.37 m2 0.958 Pa β s = 0.958 Pa β s = 0.3 m/s 0.3 m/s −3 8.4 × 10 m 16.6 × 10−3 m 0.3m/s πΉ1 = 0.958 Pa β s (0.37 m2) −3 m 8.4 × 10 N m2 πΉ1 = 12.659 N πΉ2 = 6.406 N πΉ = πΉ1 + πΉ2 πΉ = 12.659 N + 6.406 N πΉ = 19.065 N F ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 8 A cylinder of 125 mm radius rotates concentrically inside a fixed cylinder of 130 mm radius. Both cylinders are 300 mm long. Determine the viscosity of the liquid which fills the space between the cylinders if a torque of 0.88 N-m is required to maintain an angular velocity of 2π radians/sec. Assume the velocity gradient to be straight line. Fixed Solution 29.879 Pa π π = 0.005 m π= m π/π¦ 0.785 s /0.005m F U = 0.785 π = ππ π = 0.190 Pa β s F = 0.125(2π) π = 0.785 m/s Fixed cylinder T Torque = Force(radius) Rotating cylinder 0.88 N β m = πΉ(0.125 m) πΉ = 7.04 N πΉ 7.04 N = π΄ 2π πrotating cylinder (πΏ) 7.04 N = 2π 0.125m (0.3m) π= π = 29.879 N/m2 π = 29.879Pa 0.3 m 0.125 m 0.13 m ππππ πππ πππππππ (σ) sigma Is a force per unit arc length created on an interface between two immiscible fluids as a result of molecular attraction/cohesion. Surface tension allows a needle to be floated on a free surface of water. Surface tension allows insect to land on water surface without getting wet Surface tension also causes droplets to take on a spherical shape ππππ πππ πππππππ (σ) sigma Is a force per unit arc length created on an interface between two immiscible fluids as a result of molecular attraction/cohesion. Surface tension allows a needle to be floated on a free surface of water. πΉresisting σ = Surface tension allows insect to land on water surface without getting wet πΉ πΉresisting = π΄ 2ππ Pressure(p) = πΉresisting = σ2ππ πΉπππ‘πππ ππ’π π‘π ππππ π π’ππ Surface tension also causes droplets to take on a spherical shape πΉ πΉacting = π΄ ππ 2 πΉacting = pππ 2 By equilibrium σ πΉ = 0 ; πΉresisting = πΉacting σ2ππ = P ππ 2 σ= pπ 2 N/m Where: σ=surface tension in N/m r = radius of the droplet p = gage pressure in Pa ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 9 What is the value of the surface tension of a small drop of water 0.3 mm in diameter which is in contact with air if the pressure within the droplet is 561 Pa? Solution σ= pπ 2 N/m 561 Pa(0.15 × 10−3 m) σ= 2 π = 0.042 N/m πππππππππππ/CAPILLARY ACTION Is the name given to the behavior of the liquid in a thin-bore tube. The rise or fall of a fluid in a capillary tube is caused by surface tension and depends on the relative magnitudes of the cohesion of the liquid and the adhesion of the liqud to the walls of the containing vessel. ADHESION COHESION The attraction force between different molecules. The attraction force between molecules of the same substance H2 0 Hg Capillary Tube πππππππππππ/CAPILLARY ACTION Is the name given to the behavior of the liquid in a thin-bore tube. The rise or fall of a fluid in a capillary tube is caused by surface tension and depends on the relative magnitudes of the cohesion of the liquid and the adhesion of the liqud to the walls of the containing vessel. ADHESION COHESION The attraction force between different molecules. The attraction force between molecules of the same substance rise Densities of Common Fluid Fluid h fall h π in kg/m3 Mercury 13600 Water (at 4°C) 1000 H2 0 Adhesion > Cohesion Hg Cohesion > Adhesion Capillary Tube πππππππππππ/CAPILLARY ACTION Is the name given to the behavior of the liquid in a thin-bore tube. The rise or fall of a fluid in a capillary tube is caused by surface tension and depends on the relative magnitudes of the cohesion of the liquid and the adhesion of the liqud to the walls of the containing vessel. By equilibrium: σ πΉπ£ = 0 F Ο΄ F F W F πΉ cos Ο΄ = W πΉ σ = π΄ πΉ = σ (ππ) F π = γπ h π2 π = π (β) 4 π2 π = γπ (β) 4 d Contact Angles π2 σ(πd) cos Ο΄ = πΎπ (β) 4 β= 4σ cos Ο΄ πΎπ Where: h=capillary rise or depression in m. γ = unit weight of the fluid in N/m3 d = diameter of the tube in m σ= surface tension in Pa Materials Angle, Ο΄ Mercury-glass 140° Water-paraffin 107° Water-silver 90° Kerosene-glass 26° Glycerin-glass 19° Water-glass 0° Ethyl alcohol-glass 0° πππππππππππ/CAPILLARY ACTION Is the name given to the behavior of the liquid in a thin-bore tube. The rise or fall of a fluid in a capillary tube is caused by surface tension and depends on the relative magnitudes of the cohesion of the liquid and the adhesion of the liqud to the walls of the containing vessel. By equilibrium: σ πΉπ£ = 0 F Ο΄ F F W F πΉ cos Ο΄ = W πΉ σ = π΄ πΉ = σ (ππ) F π = γπ h π2 π = π (β) 4 π2 π = γπ (β) 4 d Contact Angles π2 σ(πd) cos Ο΄ = πΎπ (β) 4 β= 4σ cos Ο΄ πΎπ Where: h=capillary rise or depression in m. γ = unit weight of the fluid in N/m3 d = diameter of the tube in m σ= surface tension in Pa Materials Angle, Ο΄ Mercury-glass 140° Water-paraffin 107° Water-silver 90° Kerosene-glass 26° Glycerin-glass 19° Water-glass 0° Ethyl alcohol-glass 0° ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 10 Estimate the capillary depression for mercury in a glass capillary tube 2 mm in diameter. Use σ = 0.514 N/m and Ο΄ = 140°. Densities of Common Fluid Solution Fluid 4σ πππ π 4(0.514)(πππ 140°) β= = γπ (13600 × 9.81)(0.002) β = −0.0059 m (the negative sign indicates capillary depression) Capillary depression, β = 5.9 mm π in kg/m3 Air (STP) 1.29 Alcohol 790 Ammonia 602 Gasoline 720 Glycerin 1260 Mercury 13600 Oil 800 Water (at 4°C) 1000 ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 11 Estimate the height to which water will rise in a capillary tube of a diameter 3 mm. Use σ = 0.0728 N/m an γ = 9810 N/m3 for water. Solution Note: θ= 0° for water in clean tube Capillary rise, β = 4σ γπ 4(0.0728) Capillary rise, β = 9810(0.003) Capillary rise, β = 0.0099 π = 9.9 ππ COMPRESSIBILITY Compressibility, β (beta), (also known as the coefficient of compressibility) is the fractional change in the volume of a fluid per unit change in pressure in a constant temperature process. βπ − π 1 π½= = βπ πΈπ΅ π½=− ππ/π ππ Where: ΔV π Δp dV/V = change in volume, m3 = original volume, m3 = change in pressure, Pa = change in volume (usually in percent) heat BULK MODULUS OF ELASTICITY The bulk modulus of elasticity, EB, of the fluid expresses the compressibility of the fluid. It is the ratio of the change in unit pressure to the corresponding volume change per unit of volume. βπ stress = πΈπ΅ = βπ strain π ππ πΈπ΅ = − ππ/π PRESSURE DISTURBANCES Pressure disturbances imposed on a fluid move in waves. The velocity or celerity (c) of pressure wave (also known as acoustical or sonic velocity) is expressed as: π= πΈπ΅ = π 1 π½π PROPERTY CHANGES IN IDEAL GAS For any ideal gas experiencing any process, the equation of state is given by π1π1 π2π2 = π1 π2 When temperature is held constant, the above equation is reduced to (Boyle’s Law) π1π1 = π2π2 When pressures is held constant (isothermal constant), the ideal gas equation is reduced to (Charle’s Law) π1 π2 = π1 π2 FOR ADIABATIC OR ISENTROPIC CONDITIONS (no heat exchanged) π1π1π = π2π2π π1 or π2 and π = π2 = Constant π1 π2 π2 = π1 π1 π−1 π Where: p1 π2 V1 π2 T1 π2 k = initial absolute pressure of gas, Pa = final absolute pressure of gas, Pa = initial volume of gas, m3 = final volume of gas, m3 = initial absolute temperature of gas in °K (°K = °C + 273) = final absolute temperature of gas in °K = ratio of the specific heat at constant pressure to the specific heat at constant volume. Also known as adiabatic exponent. ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 12 A liquid compressed in a container has a volume of 1 liter at a pressure of 1 MPa and a volume of 0.995 liter at a pressure of 2 MPa. The bulk modulus elasticity (πΈπ΅ ) of the liquid is: Solution ππ ππ/π 1−2 =− (1 − 0.995)/1 πΈπ΅ = − πΈπ΅= 200 MPa ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 13 What pressure is required to reduce the volume of water by 0.6 percent? Modulus of elasticity of water, EB = 2.2 GPa. Solution πΈπ΅ = − ππ ππ/π ππ = π2 − π1 π1 = 0 ππ = π2 ππ = π2 − π1 ; π1 = π; = (π − 0.6% V) −π ππ = −0.6%π = −0.006V π2 πΈπ΅ = − = 2.2 −0.006π/π π2 = 0.0132 GPa π2 = 13.2 MPa π2 = π − 0.6% V; ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 14 Water in a hydraulic press, initially at 137 kPa absolute, is subjected to a pressure of 116,280 kPa absolute. Using πΈπ΅ = 2.5 πΊππ, determine the percentage decrease in the volume of water. Solution πΈπ΅ = − 2.5 × ππ ππ/π 109 116,280 − 137 × 103 =− ππ/π ππ = −0.0465 π ππ = 4.65% πππππππ π π ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 15 If 9 m3 of an ideal gas at 24 °C and 150 kPa abs is compressed to 2 m3, (a) what is the resulting pressure assuming isothermal conditions. (b) What would have been the pressure and temperature if the process is isentropic. Use k=1.3. Solution π πΉππ ππ ππ‘βπππππ ππππππ‘πππ: π2 π2 = π1 π1 π1π1 = π2π2 150(9) = π2(2) π2 = 675 πππ πππ π πΉππ ππ πππ‘πππππ ππππππ π : 1.3 π2 1,060 = 24 + 273 150 (1.3−1)/1.3 π2 = 466.368°πΎ ππ 193.4°πΆ π1π1π = π2π2π 150 9 (π−1)/π . = π2 2 1 3 π2 = 1,059.906 πππ πππ ππ΄πππΏπΈ ππ ππ΅πΏπΈπ 16 A sonar transmitter operates at 2 impulses per second. If the device is held to the surface of fresh water (EB = 2.04 x 109 Pa), what is the velocity of the pressure wave? Solution Sonar transmitter The velocity of the pressure wave (sound wave) is: π= πΈπ΅ π π= 2.04 × 109 = 1,428 π/π 1000 Sound Wave h Bottom Echo
