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Hydrology Basics

Hydrology Basics. W e need to review fundamental information about physical properties and their units. http://www.engineeringtoolbox.com/average-velocity-d_1392.html. Vectors 1: Velocity. A scalar is a quantity with a size, for example mass or length

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Hydrology Basics

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  1. Hydrology Basics • We need to review fundamental information about physical properties and their units.

  2. http://www.engineeringtoolbox.com/average-velocity-d_1392.htmlhttp://www.engineeringtoolbox.com/average-velocity-d_1392.html Vectors 1: Velocity • A scalar is a quantity with a size, for example mass or length • A vector has a size (magnitude) and a direction. 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, is called Velocity velocity is the rate and direction of change in position of an object.

  3. 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

  4. Example Vector Components • A folded sandstone layer is exposed along the coast of Lake Michigan. In some places it is vertical, on others gently dipping. Which surface is struck by a greater wave force?

  5. 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

  6. SI Units • 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. http://en.wikipedia.org/wiki/International_System_of_Units

  7. What’s in it for me? • Hydrologists will take 1/5th of GS jobs. • Petroleum Geologists make more money, 127K vs. 80K, but have much less job security during economic downturns. • Hydrologists have much greater responsibility. • When a geologist makes a mistake, the bottom line suffers. When a hydrologist makes a mistake, people suffer. http://www.bls.gov/oco/ocos312.htm

  8. Issaquah Creek Flood, WA http://www.issaquahpress.com/tag/howard-hanson-dam/

  9. What does a Hydrologist do? • Hydrologists provide numbers to engineers and civil authorities. They ask, for example: • “When will the crest of the flood arrive, and how high will it be?” • “When will the contaminant plume arrive at our municipal water supply? http://www.weitzlux.com/dupont-plume_1961330.html

  10. 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. To convert from one unit to another by using conversion factors. • Conversion Factors involve multiplication by one, nothing changes • 1 foot = 12 inches so 1 foot = 1 12 “ http://waterdata.usgs.gov/nj/nwis/current/?type=flow http://climate.rutgers.edu/njwxnet/dataviewer-netpt.php?yr=2010&mo=12&dy=1&qc=&hr=10&element_id%5B%5D=24&states=NJ&newdc=1

  11. Example • Water is flowing at a velocity of 30 meters per second from a spillway outlet. 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. • (1) (2) • 30 meters x 3.281 feet = 98.61 feet second meter second

  12. Flow Rate • The product of velocity and area is a flow rate • Vel [m/sec] x Area [meters2] = Flow Rate [m3/sec] • Notice that flow rates have units of Volume/ second • Discussion: recognizing units

  13. Example Problem • Water is flowing at a velocity of 30 meters per second from a spillway outlet that has a diameter of 10 meters. What is the flow rate?

  14. 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 feet per hour? • 3000 m3 x 60 sec x 60 min = sec min hour

  15. 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

  16. 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 • 1kg . m is called a Newton (N) sec2

  17. Statics and Dynamics • If all forces and Torques are balanced, an object doesn’t move, and is said to be static • Discussion Torques, See-saw • Reference frames • Discussion Dynamics F=2 F=1 -1 0 +2 F=3

  18. 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

  19. Density • Density is the mass contained in a unit volume • Thus density must have units kg/m3 • The symbol for density is r

  20. Chaining Conversion Factors • Suppose you need the density of water in kg/m3. You find that 1 cubic centimeter (cm3) of water has a mass of 1 gram. • 1 gram water x (100 cm)3x 1 kilogram = 1000 kg / m3 • (centimeter)3 (1 meter)3 1000 grams • rwater=1000 kg / m3

  21. Mass Flow Rate • Mass Flow Rate is the product of the Density and the Flow Rate • i.e. Mass Flow Rate = rAV • Thus the units are kg m2 m = kg/sec m3 sec

  22. 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)

  23. Conservation of Mass for a horizontal Nozzle Consider liquid water flowing in a horizontal pipe where the cross-sectional area changes. r1A1V1(mass inflow rate) = r2A2V2( mass outflow rate) Liquid water is incompressible, so the density does not change and r1= r2. The density cancels out, r1A1V1 = r2A2V2 so A1V1 =A2V2 Notice If A2 < A1 then V2 >V1 In a nozzle, A2 < A1 .Thus, water exiting a nozzle has a higher velocity than at inflow The water is called a JET V1 -> A1 A2 V2 -> Q2 = A2V2 A1V1 = A2V2 Q1 = A1V1

  24. Example Problem Water enters the inflow of a horizontal nozzle at a velocity of V1 = 10 m/sec, through an area of A1 = 100 m2 The exit area is A2 = 10 m2. Calculate the exit velocity V2. V1 -> A1 A2 V2 -> Q2 = A2V2 Q1 = A1V1 A1V1 = A2V2

  25. 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 • 1kg . m2 is called a N.m or Joule (J) sec2 • Energy is conserved • KE + PE + P/v + Heat = constant

  26. 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

  27. 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

  28. 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. • 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

  29. Conservation of Energy • We said earlier “Energy is Conserved” • This means KE + PE + P/v + Heat = constant • For simple systems involving liquid water without friction heat, at two places 1 and 2 1/2 mV12 + mgh1 + P1/v = 1/2 mV22 + mgh2 + P2/v If both places are at the same pressure (say both touch the atmosphere) the pressure terms are identical • 1/2 mV12 + mgh1 + P1/v = 1/2 mV22 + mgh2 + P2/v

  30. Example Problem • A tank has an opening h = 1 m below the water level. The opening has area A2 = 0.003 m2 , small compared to the tank with area A1 = 3 m2. Therefore assume V1 ~ 0. • Calculate V2. Method: Assume no friction, then mgh1=1/2mV22 V2 = 2gh 1/2mV12 + mgh1 = 1/2mV22 + mgh2

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