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Basics • We need to review fundamental information about physical properties and their units. http://www.engineeringtoolbox.com/average-velocity-d_1392.html Scalars and Vectors • A scalar is a quantity with a size, for example mass or length • A vector has a size (magnitude) and a direction. http://www.engineeringtoolbox.com/average-velocity-d_1392.html Velocity • Velocity is the rate and direction of change in position of an object. • For example, at the beginning of the Winter Break, our car had an average speed of 61.39 miles per hour, and a direction, South. The combination of these two properties, speed and direction, forms the vector quantity Velocity Vector Components • Vectors can be broken down into components • For example in two dimensions, we can define two mutually perpendicular axes in convenient directions, and then calculate the magnitude in each direction • Vectors can be added • The brown vector plus the blue vector equals the green vector Trig & • Geometry• • • • cos q = adj / hyp = abs /hyp sine q = opp / hyp = ord / hyp tan q = ord / abs Usually angle known, solve eqn. to find size of an unknown The sum of angles inside a triangle = 180o Vectors 2: Acceleration. • Acceleration is the change in Velocity during some small time interval. Notice that either speed or direction, or both, may change. • For example, falling objects are accelerated by gravitational attraction, g. In English units, the speed of falling objects increases by about g = 32.2 feet/second every second, written g = 32.2 ft/sec2 SI Units: Kilogram, meter, second • Most scientists and engineers try to avoid English units, preferring instead SI units. For example, in SI units, the speed of falling objects increases by about 9.81 meters/second every second, written g = 9.81 m/sec2 • Unfortunately, in Hydrology our clients are mostly civilians, who expect answers in English units. We must learn to use both. Système international d'unités pron dooneetay http://en.wikipedia.org/wiki/International_System_of_Units Data and Conversion Factors • In your work as a hydrologist, you will be scrounging for data from many sources. It won’t always be in the units you want. We convert from one unit to another by using conversion factors. http://waterdata.usgs.gov/nj/nwis/current/?type=flow http://climate.rutgers.edu/njwxnet/dataviewernetpt.php?yr=2010&mo=12&dy=1&qc=&hr=10&element_id%5B%5D=24&states=NJ&newdc=1 • Conversion Factors involve multiplication by one, nothing changes • 1 foot = 12 inches so 1 foot = 1 12 “ Example • Water is flowing at a velocity of 30 meters per second through a canyon. What is this speed in feet per second? • Steps: (1) write down the value you have, then (2) select a conversion factor and write it as a fraction so the unit you want to get rid of is on the opposite side, and cancel. Then calculate. • (1) (2) • 30 meters x 3.281 feet = 98.61 feet second meter second Flow Rate Q = V . A • The product of velocity and area is a flow rate • V [meters/sec] x A [meters2] = Flow Rate [m3/sec] • Notice that flow rates have units of Volume/ second • It is very important that you learn to recognize which units are correct for each measurement or property. Example Problem • Water is flowing at a velocity of 30 meters per second though a sea arch that has a diameter of 10 meters. What is the flow rate? Radius r = D/2 = 5 m A = p x 52 = 78.54m2 Q = VA = 30 m/s x 78.54 m2 Q = 2356.2 m3/s Chaining Conversion Factors • Water is flowing at a rate of 3000 meters cubed per second from a spillway outlet. What is this flow rate in feet3 per hour? • • 3000 m3 x 60 sec x 60 min sec min hour 10800000 m3 hour x (3.281 feet)3 ( 1 meter) 3 3/hour 10800000 m = = 381454240. ft3/hr Momentum (plural: momenta) • Momentum (p) is the product of velocity and mass, p = mv • In a collision between two particles, for example, the total momentum is conserved. • Ex: two particles collide and m1 = m2, one with initial speed v1 , the other at rest v2 = 0, • m1v1 + m2v2 = constant Force • Force is the change in momentum with respect to time. • A normal speeds, Force is the product of Mass (kilograms) and Acceleration (meters/sec2), so Force F = ma • So Force must have SI units of kg . m sec2 • 1 kg . m is called a Newton (N) sec2 An example of Force is weight, F = mg Statics and Dynamics • If all forces and torques are balanced, an object doesn’t move, and is said to be static. We will use force balances shortly. • Torque is force at some distance • Demo Torques, ruler, See-saw F=2 F=1 Both forces and torques balanced -1 0 F=3 +2 Pressure • Pressure is Force per unit Area • So Pressure must have units of kg . m sec2 m2 • 1 kg . m is called a Pascal (Pa) sec2 m2 Density • Density is the mass contained in a unit volume • Thus density must have SI units kg/m3 • The symbol for density is r, pronounced “rho” • Very important r is not a p, it is an r • It is NOT the same as pressure Chaining Conversion Factors Suppose you need the density of water in kg/m3. You may recall that 1 cubic centimeter (cm3) of water has a mass of 1 gram. 1 gram water x (100 cm)3 x 1 kilogram = 1000 kg / m3 (centimeter)3 (1 meter)3 1000 grams r water = 1000 kg / m3 Don’t forget to cube the 100 Mass Flow Rate • Mass Flow Rate is the product of the Density and the Flow Rate • i.e. Mass Flow Rate = rAV elocity • Thus the units are kg m2 m m3 sec = kg/sec Conservation of Mass – No Storage Conservation of Mass : In a confined system “running full” and filled with an incompressible fluid, all of the mass that enters the system must also exit the system at the same time. r1A1V1(mass inflow rate) = r2A2V2( mass outflow rate) What goes in, must come out. Notice all of the conditions/assumptions confined (pipe), running full (no compressible air), horizontal (no Pressure differences) incompressible fluid. Energy • Energy is the ability to do work, and work and energy have the same units • Work is the product of Force times distance, • W = Fd • 1 kg . m2 is called a N.m or Joule (J) sec2 • • Energy in an isolated system is conserved KE + PE + Pv + Heat = constant N.m is pronounced Newton meter, Joule sounds like Jewel. KE is Kinetic Energy, PE is Potential Energy, P-v is Pressure energy, v is unit volume An isolated system, as contrasted with an open system, is a physical system that does not interact with its surroundings. Kinetic Energy • Kinetic Energy (KE) is the energy of motion • KE = 1/2 mass . Velocity 2 = 1/2 mV2 • SI units for KE are 1/2 . kg . m . m • sec2 Note the use of m both for meters and for mass. The context will tell you which. That’s the reason we study units. Note that the first two units make a Newton (force) and the remaining unit is meters, so the units of KE are indeed Energy s Potential Energy • Potential energy (PE) is the energy possible if an object is released within an acceleration field, for example above a solid surface in a gravitational field. • The PE of an object at height h is PE = mgh Units are kg . m . m sec2 Note that the first two units make a Newton (force) and the remaining unit is meters, so the units of PE are indeed Energy Note also, these are the same units as for KE KE and PE exchange • An object falling under gravity loses Potential Energy and gains Kinetic Energy. • A pendulum in a vacuum has potential energy PE = mgh at the highest points, and no kinetic energy because it stops • A pendulum in a vacuum has kinetic energy KE = 1/2 mass.V2 at the lowest point h = 0, and no potential energy. • The two energy extremes are equal Stops v=0 at high point, fastest but h = 0 at low point. Without friction, the kinetic energy at the lowest spot (1) equals the potential energy at the highest spot, and the pendulum will Conservation of Energy • We said earlier “Energy is Conserved” • This means KE + PE + Pv + Heat = constant • For simple systems involving liquid water without friction heat, at two places 1 and 2 1/2 mV12 + mgh1 + P1v = 1/2 mV22 + mgh2 + P2v If both places are at the same pressure (say both touch the atmosphere) the pressure terms are identical • 1/2 mV12 + mgh1 + P1v = 1/2 mV22 + mgh2 + P2v Example Problem • A Watchung Lava flow dammed a proglacial lake, Lake Passaic, south of the melting Wisconsinan glacier. A leaky area had an opening h = 100 m below the water level. The opening had an area A2 = 10 m2 , small compared to the lake surface with area A1 = 3,000,000 m2. Therefore assume V1 ~ 0. • Calculate V2. note m1 = m2 Method: only PE at 1, KE at 2 mgh1=1/2mV22 V2 = 2gh 1/2mV12 + mgh1 = 1/2mV22 + mgh2 44.29m/sec