1.4 Review. 1.5 Thermodynamic Properties. CEE 3310 Thermodynamic Properties, Aug. 26,

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1 CEE 3310 Thermodynamic Properties, Aug. 26, 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 will form a free surface in a gravitational field (and mix minimally with any atmosphere) while a gas will form an atmosphere (and eventually mix with an existing atmosphere). While buoyancy forces are important in each, gravity is generally an important forcing term in free surface liquid flows and not in atmospheric (gas) flows. We consider a fluid to be a continuum i.e., it is continuously differentiable. Dimensional consistency F = ma [MLT 2 ] = [M][LT 2 ]. A dimensionally consistent equation may be based on physics, an inconsistent equation is certainly not physically based. 1.5 Thermodynamic Properties Temperature Measure of internal energy level. Pressure Measure of compressive (normal) stress at a point. P = 1 3 (σ xx +σ yy +σ zz ) = F A It is created by the bombardment of the surface by molecules of fluid.

2 Density (kg/m 3 ) T emperature ( o C) Figure 1.1: Density of water Density ρ = Mass Volume There is less than a 1% change in the density of water over the standard range of temperatures seen in the environment yet this difference can be very important! Specific Weight γ 60 o F=62.4lbs/ft 3 γ = ρg = weight volume γ 20 o C = 9790N/m Specific Gravity The specific gravity is the density of a substance normalized by the density of water at a certain temperature, often 4 C, the temperature of maximum density at normal

3 CEE 3310 Thermodynamic Properties, Aug. 26, pressures. Hence we write S.G. = ρ ρ 4 C = ρ in S.I. units 1000 kg/m3 S.G. of sands and gravels is about Perfect Gas Law P = ρrθ where R is the specific gas constant which can be expressed as R = C P C V where C P is the specific heat at constant pressure and C V is the specific heat at constant volume. We also can write: R = Λ MW gas whereλis the universal gas constant (8314 kg m 2 s 2 K 1 kmol 1 ) andmw gas (kg kmol 1 ) is the molecular weight of the gas. Let s check the units P = [M] [L 3 ] [L 2 ] [M] Θ = [T 2 Θ] [LT 2 ] = [ML] [L 2 T 2 ] = Force Area Example Find ρ for CO 2 at 20 C and 1atm. ρ=1.44 kg/m 3 ( S.G.=14.1 N/m 3 )

4 Viscosity d 1 = u 1 dt d 2 = u 2 dt = (u 1 +du)dt Strain = d 2 d 1 = (u 1 +du u 1 )dt = du dt For solids we know that stress is proportional to strain. In fluids we find that stress is proportional to strain rate. Strain rate = Therefore since du dt dt = du the velocity gradient is the strain rate! Therefore µ = τ du stress strain rate (1.1) τ du (1.2) τ = µ du (1.3) = [MLT 2 L 2 ] [LT 1 ] [L] = [M] [LT] what is this? Momentum has the units of mass times velocity hence we can interpret µ as having the dimensions of momentum per area. Thus we can think of µ, known as the viscosity, as the amount of momentum transported by molecular activity across a given area. Thus highly viscous fluids (honey) transport lots of momentum and tend to be harder to move (be more sticky) as you have to move the whole fluid while low viscosity

5 CEE 3310 Thermodynamic Properties, Aug. 26, fluids (water) tend to be easier to move as only a small parcel of fluid is affected by trying to move a thin slab of fluid Kinematic Viscosity If we normalize the viscosity by the density we have the kinematic viscosity. ν = µ ρ = [L2 ] [T] At 20 C water has an absolute or dynamic viscosity of Nsm 2 (or Pas) and a kinematic viscosity of m 2 s 1. Now, if we have a thin gap filled with a fluid but the solid surfaces on either side of the gap have some relative velocity (e.g., one surface is fixed but the other is moving) then there will be stress on either solid surface transmitted by the fluid. The molecules on either solid boundary must be moving at the speed of the boundary, this is known as the no-slip boundary condition. If the fluid filled gap is long compared to its width then we can ignore what happens at the end of the gap and, if the system is at steady state (meaning the velocity profile of the fluid in the gap is no longer changing in time) then we would find that the velocity profile just varies linearly, going from the velocity of the one boundary to the velocity of the other boundary. We will actually solve for the exact solution from the equations of motion later in the semester! A linear velocity variation is a constant velocity gradient hence the fluid stress is constant, just equal to the fluid viscosity times the constant velocity gradient. This is perhaps best illustrated by an example Example - A block sliding down an inclined plane If the block has a mass of 1 kg: 1. Determine the viscosity, µ, of the lubricant fluid in the gap.

6 16 Figure 1.2: Sliding block 2. What speed will the block travel if the angle, θ, is adjusted to 10 and the gap, δ, is decreased to 0.5 mm 1) µ = kg m s 2) V = m s (= N s m 2 = Pa s);

7 CEE 3310 Thermodynamic Properties, Aug. 29, Review System of units B.G., S.I. Be careful and be comfortable in both! Thermodynamic properties Θ, P, ρ(θ,p) Perfect Gas Law Viscosity stress strain rate τ = µ du 1.9 Vapor Pressure If initially we start with a vacuum, over time a pressure will form as the result of molecular action. Particles leave the surface. Eventually an equilibrium pressure is achieved as the same number of particles leave the surface as return to it. This pressure is known as the vapor pressure of the fluid and is denoted p v. As we will see in a few weeks, fluid motions can lead to very low pressures. If p p v the fluid will boil. This process is known as cavitation Surface Tension The water molecule is polar. The O attracts the H +. Within the fluid this attraction is in balance, i.e., the net force due to all of the polar pairs is zero. However, at the surface half of this force is missing and the surface is pulled toward the fluid interior with

8 18 a certain energy. surface energy = J m 2 = Nm m 2 = N m = force length = tension hence we refer to this energy as the surface tension (Υ) Example the pressure in a bubble Tension force = 2πRΥ Pressure force = (P I P E )πr 2 P = P I P E = 2πRΥ πr 2 = 2Υ R The Contact Angle In the case of a bubble we only had to concern ourselves with a liquid gas interface but often we find we have three phases present (a liquid-gas-solid interface) for example, when you fill your glass with water and you get a contact line around the circumference of the glass at the air-water-glass interface. You ve all likely noticed that the contact line rises locally, appearing to adhere to and be lifted by the glass boundary forming what is known as a meniscus, the region local to the solid boundary where the gas-liquid

9 CEE 3310 Thermodynamic Properties, Aug. 29, interface is curved. The angle that is formed at this three-phase interface is known as the contact angle and is defined as the angle between the line originating from the three-phase contact point tangent to the liquid-gas interface and the tangent to the solid boundary as measured through the liquid, e.g., When the liquid seems to spread easily over the boundary, the contact angle is θ c < 90 and we refer to the liquid as wetting, as in the case of water on glass, which is totally wetting giving θ c 0. When the liquid resists spreading over the boundary, instead trying to form a droplet, the contact angle is θ c > 90 and we refer to the liquid as non-wetting, as in the case of water on teflon (θ c 110) nd Example A water barometer You are planning on constructing your own water barometer. This will be constructed by filling a long cylindrical glass tube sealed at one end with water and then carefully inverting it so that the mouth of the tube stays wet. The free surface of the water drops to a given elevation and you can measure the height of the water above the reservoir below to calculate the atmospheric pressure. There are at least two important fluid properties that affect the accuracy of your water barometer, what are they?

10 20 What minimum diameter must the tube be if you want the capillary induced rise in the tube to be less than 1mm (assume 20 o C water)? If the atmospheric pressure is 30in Hg, what is the correction you need to apply to the barometer reading to account for the effect of vapor pressure on your reading?

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