Lagrangian description from the perspective of a parcel moving within the flow. Streamline Eulerian, tangent line to instantaneous velocity field.


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1 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 Eulerian, tangent line to instantaneous velocity field. Pathline Lagrangian, path of a fluid parcel. Streakline Locus of parcels that have passed through a given point. 2.2 Hydrostatics and Pressure In many fluid problems the velocity is zero or the velocity is constant τ = 0. If the tangential stresses are zero this leaves only the normal stresses, which in the absence of acceleration arise only from the pressure. Then our question is one of statics, meaning we have no motion (or relative motion), and how does the pressure vary with 25
2 CEE 3310 Hydrostatics, Sept. 5, direction? The answer is the pressure at a point is independent of direction!  see section 2.1 in your textbook for the derivation. Hence we speak of pressure as a scalar quantity as it is a quantity with no dependence on direction (as opposed to velocity, which is a vector). This result is known as Pascal s law and says that if τ = 0 the pressure at a point is the same in all directions. 2.3 Equation of Motion in absence of Shear Stresses Let s investigate the force balance on a small volume of fluid to determine the equation of motion in the absence of shear stresses
3 CEE 3310 Hydrostatics, Sept. 5, Surface Forces d F s d = P = Force Volume Body Forces What body forces might act on the fluid element? Gravity, electromagnetic,... F g = ρ g x y z Taking the limit as x, y, z 0 we have d F g d = γ k
4 CEE 3310 Hydrostatics, Sept. 5, Now, putting it all together we have F = m a df s d + d F g d = ρ a P γ k = ρ a 2.4 Hydrostatic Pressure We found the equation of motion for a fluid with τ = 0: P γ k = ρ a Now, if the fluid is at rest (or at least moving at a constant velocity): Hence we can write: a = 0 P = γ k P x = 0, P y = 0 P z = γ What does this tell us? Well, we see that P = P (z) only, hence the pressure at a given elevation (z position) is constant. In the vertical we have Hence as z P dp dz = γ where we have replaced with d now Incompressible Fluids dp = γ dz P2 P 1 z2 z2 dp = γ dz P 2 P 1 = γ dz z 1 z 1
5 CEE 3310 Hydrostatics, Sept. 5, Now let s consider particular solutions: Constant γ It is usually reasonable to assume that all liquids essentially have a constant γ, certainly true for all fluids at constant temperature and pressure. Under this condition we have Therefore P 2 P 1 = γ z2 z 1 dz = γ(z 2 z 1 ) = γh P 1 = γh + P 2 or h = P 1 P 2 γ where we refer to h as the pressure head as it is a pressure measured in units of length. Liquids with a Free Surface Examples include lakes, rivers, oceans, reservoirs, your pint glass of water,... The surface is the datum (at this point the pressure is the pressure of the atmosphere or surrounding gas). Therefore we have z = 0, P = P atm. P 1 = γh + P atm but h = z 2 z 1 = z 1 P = P atm γz where z < 0 in liquid (water) and z > 0 in gas (atmosphere) Compressible Gases How do we solve this? dp dz = γ = ρg = P RΘ g separation of variables! P2 P 1 dp z2 P = g dz z 1 RΘ
6 CEE 3310 Hydrostatics, Sept. 5, we assume g is a constant but that Θ may vary (e.g., the atmosphere). Then we can write the above ln P 2 = g z2 dz P 1 R z 1 Θ If we assume that the temperature is constant within the elevation range of interest, an assumption known as isothermal conditions (i.e., Θ = Θ 0 = a constant), we can write ln P 2 = g ( (z 2 z 1 ) or P 2 = P 1 exp g(z ) 2 z 1 ) P 1 RΘ 0 Rθ The Standard Atmosphere In the troposphere (first 11.0 km of atmosphere) the temperature drops linearly with elevation. The rate of temperature drop is known as the lapse rate and is given by β = K/m (6.5 C/km or 3.6 F/1000 ). Hence in the troposphere we write Θ = Θ a βz where Θ a is the temperature at sea level which for the standard atmosphere is taken to be 15 C. We can use the expression we found for compressible gasses (recall we allowed for temperature variation!) and integrate to find ( P = P a 1 βz ) g Rβ Θ a This is the expression for the pressure in the troposphere.
7 CEE 3310 Hydrostatics, Sept. 5, Altitude (km) Altitude (km) Temperature ( C) Pressure (kpa) Figure 2.1: U.S. standard atmsophere to 20.1 km 2.5 Measurement of Pressure Absolute pressure is the pressure relative to a perfect vacuum, hence it is always a positive (or zero) value. Gage pressure is the pressure relative to the local atmospheric value P > P a P gage = P P a > 0 P < P a P gage = P P a < 0 P vacuum = P suction = P a P > 0 Units of pressure are force/area hence pounds per square inch (psi), pounds per square foot (psf), newtons per square meter (N/m 2 =pascal (Pa) ).
8 CEE 3310 Hydrostatics, Sept. 5, We will take all pressures, unless otherwise noted, to be gage. Example Liquids with a Free Surface Barometer P atm P vapor = γh P atm = γh + P vapor For mercury P vapor 0 P atm = γh. Typical values of pressure (from the U.S. standard atmosphere at sea level) are: 14.7 psia, 101 kpa, or, 760. mm Hg, 29.9 in Hg, 33.9 ft water (recall γ water = 62.4 lbs/ft 3 ) and S.G. mercury = 13.6).
9 CEE 3310 Hydrostatics, Sept. 5, Manometry A manometer is a vertical or inclined tube used to measure pressure. There are three fundamental types of manometers: Piezometer tube, Utube, and inclined tube. Manometer analysis is straight forward hydrostatics  the key is to keep track of the signs of the pressure terms! Consider the following Utube manometer: P 2 P 1 = γ(z 2 z 1 ) Now we track the pressure from A to 1 and from 1 to 2. A simple rule, based on our understanding that as we move down in a static fluid the pressure increases, is: P down = P up + γ z This obviates the need to keep track of signs of directions in the vertical. Hence for our Utube problem we have P A + γ 1 z A z 1 γ 2 z 2 z 1 = P atm and we solve for P A
10 CEE 3310 Hydrostatics, Sept. 5, Example  Inclined Manometer
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