M FLUID ECHANICS FLUID PROPERTIES Module 1 FLUID PROPERTIES Module 1 Density and Specific Weight Viscosity Compressibility Surface Tension Capillary Action Vapor Pressure M FLUID ECHANICS Density and Specific Weight Density is defined as mass per unit volume. π= π π½ The density for air at standard atmospheric conditions, that is, at a pressure of 101.3 kPa and a temperature of 15°C, is 1.23 kg/m3. For water, the nominal value of density is 1000 kg/m3. Specific gravity is the ratio of the density of a substance to the density of water at a reference temperature of 4°C. πΊ= ππππππ πΎ ππ = = ππ π½ π½ ππππππ = πΈ πΈπππππ The density and specific weight of water do vary slightly with temperature, as described below: Specific weight is defined as weight per unit volume. πΈ= π π»−π = ππππ − πππ π where g is the local gravity. The units of specific weight are N/m3. For water, the nominal value is 9800 N/m3. M FLUID ECHANICS Density and Specific Weight (Sample Problem 1) A reservoir of glycerin has a mass of 1200 kg and a volume of 0.952 m3. Find the glycerin’s weight (π), mass density (π), specific weight (πΎ), and specific gravity (π . π.). Solution π = ππ = 1200 ππ 9.81 ππ 2 = πππππ π΅ ππ ππ. ππ ππ΅ π= π 1200 ππ = = ππππ ππ/ππ 3 π 0.952 π πΎ= π 11.77 ππ = = ππ. ππ ππ΅/ππ 3 π 0.952 π π . π. = πΎ πΎπ»2π = 12.36 = π. ππ 9.81 M FLUID ECHANICS Density and Specific Weight (Sample Problem 2) Estimate the density of water at 80°C. Solution The variation of density of water with temperature is given by: π ππππππ π»−π = ππππ − πππ 2 ππ€ππ‘ππ 80 − 4 = 1000 − 180 ππ€ππ‘ππ = 968 ππ/π3 M FLUID ECHANICS Viscosity Lesson 1.2 M FLUID ECHANICS Viscosity Viscosity is a measure of a fluids resistance to flow, or the stickiness of a fluid. Consider Fig. 1 in which the fluid particles move in the x-direction at different speeds, so that particle velocities π vary with the y-coordinate π = π π . π΅ ππ π΅−π ππ where π − shear stress π − viscosity Figure 1. Shear stress in fluid. We can define the viscosity π of the fluid by the relationship: π=π π π π π π π − velocity gradient or strain rate π π The concept of viscosity can also be illustrated by considering a fluid within the small gap between two concentric cylinders. M FLUID ECHANICS Viscosity a) the two cylinders; b) rotating inner cylinder; c) velocity distribution; d) the inner cylinder. The outer cylinder is fixed and the inner cylinder is rotating. The only stress that exists to resist the applied torque for this simple flow is a shear stress, which is observed to depend directly on the velocity gradient; that is; π=π π π π π where: π − tangential velocity component which depends only on π π π − velocity gradient π π For a small gap (β βͺ π ) this gradient can be approximated by using a linear velocity distribution in the gap. Figure 2. Shear stress in fluid between two concentric cylinders. M FLUID ECHANICS Viscosity Thus, π π ππΉ = π π π where: π − is the gap width. Thus, we can relate the applied torque π to the viscosity and other parameters by the equation π» = π π‘πππ π π₯ ππππ π₯ ππππππ‘ πππ π» = π π ππ πΉπ³ π πΉ π»=π ππΉ π ππ πΉπ³ π πΉ π ππ πΉπ ππ³π π»= π π³ − represents the length of the rotating cylinder. Figure 2. Shear stress in fluid between two concentric cylinders. M FLUID ECHANICS Viscosity Newtonian Fluids – fluids with constant viscosity no matter the amount of shear applied for a constant temperature. The shear rate is directly proportional to shear stress. Figure 3. Velocity profile of a Newtonian fluid. M FLUID ECHANICS Viscosity Non-newtonian fluid has viscosity which does not vary linearly with shear stress; viscosity can change when under stress to either more liquid or more solid. Pseudoplastics become less resistant to motion with increased strain rate. (b) (c) (a) Figure 4. Pseudoplastics: (a) Honey, (b) Paint, (c) Catsup. M FLUID ECHANICS Viscosity Dilatants (quicksand, slurries) become more resistant to motion as the strain rate increases. (a) (b) Figure 5. Dilatants: (a) Liquid armor, (b) Walking on top of a dilatant. M FLUID ECHANICS Viscosity Ideal plastics (or Bingham fluids) require a minimum shear stress to cause motion. (a) (b) Figure 6. Bingham fluids: (a) Toothpaste, (b) Mayonnaise. M FLUID ECHANICS Viscosity Figure 7. Classification of fluids according to viscosity. M FLUID ECHANICS Viscosity Since the viscosity is often divided by the density in the derivation of equations, it has become useful and customary to define kinematic viscosity to be π= π π where: π − is in ππ /π M FLUID ECHANICS Viscosity (Sample Problem 1) Rearranging Eq. 3 A viscometer is constructed with two 30-cm-long concentric cylinders, one 20.0 cm in diameter and the other 20.2 cm in diameter. A torque of 0.13 N-m is required to rotate the inner cylinder at 400 rpm (revolutions per minute). Calculate the viscosity. Solution π= = π»π ππ πΉπ ππ³ 0.13(0.001) 2π 0.1 3 (41.89)(0.3) = π. ππππππ π΅ − π/ππ The applied torque is just balanced by a resisting torque due to the shear stresses. Given πΏ = 30 ππ = 0.3 π π1 = 20.0 ππ = 0.2 π π2 = 20.2 ππ = 0.202 π π = 0.13 π − π π = 10 ππ 2 2π π = 400 π₯ = 41.89 πππ/π 60 π = β = (π2 −π1 )/2 = 0.1 ππ π = 400 πππ M FLUID ECHANICS Viscosity (Sample Problem 2) The velocity distribution in a 4-cm-diameter pipe transporting 20°C of water is given by π’ π = 10(1 − 2500π 2 ) m/s. Calculate the shearing stress at the wall. Viscosity of water at 20°C is 10-3 N.s/m2. Solution The shear stress is given by: π=π ππ’ ππ We determine du/dr from the given expression for u as: ππ’ π = 10(1 − 2500π 2 ) = −50,000π ππ ππ At the wall r = 2 cm = 0.02 m. Substituting r in the above equation: ππ’ = 50,000π = 1000/π ππ Substitute in the equation for shear stress to get π=π ππ’ π. π 1000 π = 10−3 2 π₯ = 1 2 = π π·π (π¨ππ. ) ππ π π π M FLUID ECHANICS Viscosity (Sample Problem 3) Consider a fluid flow between two parallel fixed plates 5 cm apart as shown. The velocity distribution for the flow is given by π’ π¦ = 120 0.05π¦ − π¦ 2 m/s where π¦ is in meters. The fluid is water at 10°C. Calculate the magnitude of the shear stress acting on each of the plates. At 10°C, π = 1.308 π₯ 10−3 π. π /π2 . Solution π=π The shear stress is given by: ππ’ π = 120 0.05π¦ − π¦ 2 ππ¦ ππ¦ ππ’ ππ¦ = 120 0.05 − 2π¦ At the lower plate where y = 0: π=π ππ’ = 1.308 π₯ 10−3 π₯ 120(0.05 − 2 0 ) ππ¦ ππ=π = π. πππ π ππ−π π΅ ππ (π¨ππ. ) At the upper plate where π¦ = 0.05 π: π=π ππ’ = 1.308 π₯ 10−3 π₯ 120(0.05 − 2 0.05 ) ππ¦ ππ=π = π. πππ π ππ−π π΅ ππ (π¨ππ. ) M FLUID ECHANICS Compressibility All fluids compress if the pressure increases, resulting in a decrease in volume or an increase in density. A common way to describe the compressibility of a fluid is by bulk modulus of elasticity B. Bulk modulus of elasticity, also called coefficient of compressibility is defined as the ratio of the change in pressure to the relative change in density or volume while the temperature remains constant. It has the same units as pressure. π©= βπ βπ = βπ/π βπ½/π½ The bulk modulus for water at standard conditions is approximately 2100 MPa. To cause 1% change in the density of water a pressure of 21 MPa (210 atm) is required. The bulk modulus can also be used to calculate the speed of sound in a liquid; π= π· π This yields approximately 1450 m/s for the speed of sound in water at standard conditions. M FLUID ECHANICS Compressibility (Sample Problem 1) Two engineers wish to estimate the distance across a lake. One pounds two rocks together under water on one side of the lake and the other submerges his head and hears a small sound 0.62 s later, as indicated by a very accurate stopwatch. What is the distance between the two engineers? Solving for the speed of sound c: π= 211 π₯ 10−7 = 1453 π/π 999.7 The distance across the lake is, Solution The sound will travel across the lake at the speed of sound in water. The speed of sound in water is calculated using: π= πΏ = πβπ‘ = 1453 π₯ 0.62 = πππ π (π¨ππ. ) π½ π Assuming the temperature of the lake is 10°C Bulk modulus of elasticity π½ = 211 π₯ 10−7 ππ Density ρ = 999.7 ππ/π3 M FLUID ECHANICS
