Hydrology BasicsHydrology Basics
• We need to review fundamental We need to review fundamental information about physical properties information about physical properties and their units.and their units.
Vectors 1: VelocityVectors 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
http://www.engineeringtoolbox.com/average-velocity-d_1392.html
velocity is the rate and direction of change in position of an object.
Vector ComponentsVector 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
Example Vector ComponentsExample 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?
Vectors 2: Acceleration.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 UnitsSI 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
What’s in it for me?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
Issaquah Creek Flood, WAIssaquah Creek Flood, WAhttp://www.issaquahpress.com/tag/howard-hanson-dam/
What does a Hydrologist do?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
Data and Conversion FactorsData 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
ExampleExample
• 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
Flow RateFlow 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
Example ProblemExample 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?
Chaining Conversion FactorsChaining 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
Momentum (plural: momenta)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
ForceForce• 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
Statics and DynamicsStatics 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
PressurePressure
• 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
DensityDensity
• Density is the mass contained in a unit volume
• Thus density must have units kg/m3
• The symbol for density is
Chaining Conversion FactorsChaining 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)3 x 1 kilogram = 1000 kg / m3
(centimeter)3 (1 meter)3 1000 grams
water = 1000 kg / m3
Mass Flow RateMass Flow Rate
• Mass Flow Rate is the product of the Density and the Flow Rate
• i.e. Mass Flow Rate = AV
• Thus the units are kg m2 m = kg/sec
m3 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.
1A1V1(mass inflow rate) = 2A2V2( mass outflow rate)
Conservation of Mass for a horizontal Nozzle
Liquid water is incompressible, so the density does not change
and 1= 2. The density cancels
out, 1A1V1 = 2A2V2
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 JETQ1 = A1V1
A1
V1 ->
Q2 = A2V2
A1V1 = A2V2
A2 V2 ->
1A1V1(mass inflow rate) = 2A2V2( mass outflow rate)
Consider liquid water flowing in a horizontal pipe where the cross-sectional area changes.
Example Problem
Q1 = A1V1
A1
V1 ->
Q2 = A2V2
A1V1 = A2V2
A2 V2 ->
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.
EnergyEnergy
• 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 is conserved • KE + PE + P/v + Heat = constant
Kinetic EnergyKinetic 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
Potential EnergyPotential 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
KE and PE exchangeKE 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
Conservation of EnergyConservation 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 mV2
2 + 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 mV2
2 + mgh2 + P2/v
Example ProblemExample 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=1/2mV2
2 V2 = 2gh
1/2mV12 + mgh1 = 1/2mV2
2 + mgh2
