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