Transcript Lecture 1 units v10

Hydrology 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
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?
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
What’s in it for me?
• Hydrologists will take 1/5th of Geol. 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
http://www.issaquahpress.com/tag/howard-hanson-dam/
Issaquah Creek Flood, WA
What does a Hydrologist do?
• Hydrologists provide numbers to
engineers and civil authorities. Clients
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 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 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. Then calculate.
• (1)
(2)
• 30 meters x 3.281 feet
= 98.61 feet
second
meter
second
Flow Rate
• 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
• Discussion: recognizing units
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?
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
sec2
is called a Newton (N)
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
F=2
• Reference frames
F=1
-1
0
F=3
• Discussion Dynamics
+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 units kg/m3
• The symbol for density is r
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
• 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 compressable
air), horizontal (no Pressure differences) incompressible fluid.
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
V1 ->
A2 V2 ->
A1
Notice If A2 < A1 then V2 > V1
In a nozzle, A2 < A1 .Thus, water
exiting a nozzle has a higher
velocity than at inflow
Q2 = A2V2
Q =A V
A1V1 = A2V2
The water exiting is called a JET
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 ->
A2 V2 ->
A1
Q2 = A2V2
Q1 = A1V1
A1V1 = A2V2
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 + P/v + Heat = constant
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
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
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
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
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.
1/2mV12 + mgh1 = 1/2mV22 + mgh2
• Calculate V2.
Method: only PE at 1, KE at 2
mgh1=1/2mV22 V2 = 2gh