Lecture Guidelines for GEOF110 Chapter 2 + Thermodynamics (2 hours) Chapters 3+ 4 (2 hours) Ilker...

Lecture Guidelines for GEOF110

Chapter 2 + Thermodynamics (2 hours)Chapters 3+ 4 (2 hours)

Ilker Fer

Guiding for blackboard presentation.Following Pond & Pickard, Introductory Dynamical Oceanography

GEOF110 Guidelines / 2 2

Pond & Pickard, Ch.2 Density: Fresh water (P,T)

Seawater (P,T, S)Air (specific humidity, T, P)

Salinity measure of mass of dissolved salt in 1 kg of seawater (about 35 gr/kg; titration not practical to measure modern: practical salinity scale)

Extreme properties of seawater:- high specific heat (Def.: heat energy required to increase T of a unit quantity of

substance by a certain T interval, unit: J/(kgK)) ocean currents carry much thermal energy- high latent heat of fusion (Def.: amount of thermal energy absorbed or

released to change state from solidliquid, vice-versa) T is maintained close to freezing point (Tf 0.054xS) in polar regions- high latent heat of evaporation heat transfer sea to air- high molecular heat conductivity


Density

as T or S t = - 1000 kg/m3

(P,T,S) is non-linear ; more so for Trule-of-thumb: t 1 by T = -5K S = 1

P = 2000 kPa (= 200 dbar 200 m)(note kPa is SI; dbar not SI)

Narrow range of ocean T-S properties (show volumetric T/S plot). How do we measure in the ocean CTD (in situ C,T, P; then work out

S, t and Practical salinity scale– make use of electrical conductivity easier

to measure (K15 = ratio of cond of seawater at 15degC, 1 atm to cond of KCl with

same T and P and a certain mass fraction. Salinity = f(K15). K15=1 S=35)


Thermodynamics

1st law: dq – p d = de (1)

dq : change of heat per unit massd : change in specific volume ( = 1/)p d : work done by pressure (expansion)de : change in internal energy

If the fluid particle receives heat (dq > 0) and work done by p is positive (i.e., compressing particles, d < 0) internal energy will increase: de > 0. Note e is proportional to Tabs hence Tabs increases. (Tabs = 273.15K)

2nd law: dq = Td (2)

d : entropy per unit mass

Entropy is a measure of how close a thermodynamic system is to equilibrium. It is unavailability of a system's energy to do work!

From (1) & (2): Td – p d = de (3)


For a gas or fluid, eq. of state gives the relation between the state variables:f(p,T,S,)=0, where S is composition (e.g. salinity for seawater, specific humidity for air). For simplicity assume composition does not change, dS = 0.Density then changes with p and T only. Specific entropy change:d dp dT

p T

heat capacity (at constant p) :

1 1Thermal expansion coeff. :

Gibbs function (or energy) per unit mass:

P

T

i

C TT

T T

g e p T

Change in gi: dgi = de + pd + dp - Td - dTUsing (3): dgi = Td - pd + pd + dp - Td - dT dgi = dp - dT (4)

Because ( , ) : (5)

Because (4)=(5): / ; /

(take / of and / of ) / / (6)

i ii

i i

g gg f p T dg dp dT

p T

g p g T

T p p T

Define T: Tendency to change in volume in response to change in T

Define gi: Max. amount of non-expansion work which can be extracted from a closed system, or the max can be attained in a completely reversible process. It is the thermodynamic potential obtainable form an isothermal, isobaric thermodynamic system.


(7)T

P

P

P

Td T dp T dTp T

Td T dp C dTp

Td T dp C dT

Td T dp C dTT

1st law ofthermodynamics


Adiabatic Lapse Rate

Adiabatic means to alter the state of gas or fluid without adding/removing heat.So, in Eq.(1), dq = 0. Following from Eq. (2), entropy must be constant.

From Eq. 7, using d = 0 and =1/:T

P

TdT dp

C

From hydrostatic pressure (fluid at rest), we have dp = -gdz, so

T

P

T

ad P

TdT gdz

C

gdTT

dz C

Adiabatic Lapse Rate, orAdiabatic Temperature Gradient

For fluids with simple Eq. of state (NOT SEAWATER), easy to compute . For dry, ideal gas = p/RT (R is universal gas constant). So,

2

1 1 , hence, T

TP P

g TRT p g

T p RT T C C

This is, DRY ad. lapse rate, for dry air about 1K / 100 m. Smaller if there’s moist.

0.1-0.2K / 1000 m in Ocean


Potential Temperature

Adiabatic: 0

Integrate:

ln ln ln ;

exp

r

r

r

T P

T

P

p

T

PT p

p

T

Pp

p

T

Pp

Td T dp C dT

d

TdT dp

C

dTdp

T C

T dpT C

T dpC

is the T a parcel of fluid would acquire if moved adiabatically to a reference pressure level. Derivation: Start with enthalpy form of 1st law of thermodynamics:

For seawater, final eq. can be integrated using Eq. of state or using tables. For the atmosphere ideal gas behavior is typically assumed, i.e. = p/RT and T=1/T(1/ )

( / )

Integrate:

ln ln ;

[Poisson's Eq.]

r

P

P

p

PT p

r

P

R

Cr

T TdT dp

p RT C

dT Rdp

T pC

pR

T C p

pT

p

For dry air; R = 287.05 J/kgK CP = 1004 J/kgK


Mindanao Trench


Density: (for water 1000-1028) Relative density: d Specific volume: = 1/ t = (S,T,P=0)-1000 kg/m3

= (S, ,P=0)-1000 kg/m3

Also 1; 2; 4;Sigma introduced simply because the variability is 10XX.

Pressure for density calculations refers to hydrostatic pressure. Atmospheric pressure P = 0

(show plots of typical S, T, theta, sigma_t, sigma_theta)


90

% o

f O

cean


Left in situ and potential temperature and Right sigma-t and sigma-theta in the Kermadec Trench in the Pacific at 175.825°E and 28.258°S. Warren (1973).


Specific volume: = 1/For perfect gas Eq. of State: = RTa/P(R = Gas constant; Ta = Absolute temperature)

Seawater = (S, T, P) : complicated Atmosphere = (water vapor, T, P)

IES80: International Eq. of state for Seawaterbetter than Knudsen-Ekman tables.Systematic difference between tables and IES80

IES80 slightly denser than tables:about 0.01 t at P=0about 0.03 t at z=5000 mabout 0.09 t at z=10,000 m


Upper: σθ, showing an apparent density inversion below 3,000 m. Lower: σ4 showing continuous increase in density with depth. From Lynn and Reid (1968).


Chapter 3: Basic Laws

1. Conservation of Mass– leads to continuity equation2. Conservation of Energy (split: heat (leads to heat budgets) and

mechanical energy (leads to wave eq.))

3. Newton #1 law of motion (Fnet = 0 no change of motion)

4. Newton #2 law of motion (a = rate of change of motion Fnet)

5. Newton #3 law of motion (for any force acting on a body, there’s equal and opposite force acting on some other body)

6. Conservation of angular momentum (leads to conservation of vorticity)

7. Newton’s law for gravitation

Laws 3-5 are aspects of conservation of linear momentum (leads to Navier-Stokes).

Law 7 relates to astronomical tides; pressure distribution; gravitational instability


Forces

Primary Forces (cause motion) Gravitation ( pressure gradient, buoyancy, tides) Wind stress (parallel to surface: friction force) Atmospheric pressure (inverted barometer effect: adjustment of sea-

level to changes in barometric pressure: +1mbar -1 cm in sea level) Seismic (driven by earthquakes, e.g. Tsunami)

Secondary Forces (result from motion) Coriolis force (apparent force on a moving body observed relative to

the rotating Earth) Friction (turbulence)

[Force is a vector magnitude & direction]

Solid Fluid

From J. Price.


Motion

Thermohaline Wind driven (major upper circulation, surface waves,

upwelling) Tidal currents (driven by tidal potential) Tsunami (note: surface wave) Turbulent motions Waves in the interior

Internal gravity waves Planetary waves (e.g. Rossby, Kelvin, Equatorial waves)


Chapter 4: Eq. of Continuity (of Volume)

A1A2

u1 u2

In a pipe there is no loss of flow. Incoming flow at cross-section A1 equals outgoing flow at cross-section A2.

Volume flow (flux) is mean speed x cross-section area:

u1A1 = u2A2

If A2 = 2A1; then u2 =u1/2 to satisfy continuity.


Example

Constant channel width; no rainfall; no evaporation; layer 1 thickness h1 is constant;Cross-section area at A and B is equal. But uB>uA in layer 1.Inlet is NOT emptying conservation of volumeOutflow in layer 1 must be balanced by inflow in layer 2.

AA uA = AB uB is not valid (AA = AB but uA < uB) vertical motion

AA uA + AA-B w= AB uB

RIVER

INLET SEA

h1LAYER 1

LAYER 2

A B


Derivation of Eq. of Continuity

x

y

z

ax

z

y

u, u+u+

direction:

mass flow in =

mass flow out = + y z

+ y z

x

I u y z

O u u

uO x u x

x x

0

2

small in lim

Net flow

+ y z -

x y z

x

O I

u uu x u x x u y z

x x x x

uuu V

x x x

Control volume, V= x y z


All 3-D:

Short-hand:

u v wNet Flow V

x y z

Net Flow v V

The mass in changes by per unit time

To conserve mass: ( ) 0

0

0

Remember:

10

0 d u v w

d

V C Vt

C O I

vt

v vt

dv

dt t

d

t x y zv

dt

If fluid is incompressible ( =0):

(Bossinesq approxim

0

0

ation)

10

or

u v w

x

t

y z

u

d

d


Alternative: Consider the rate of change of volume of a fluid element

with mass

V x y z

m V

x

y

z

ax

y

u /u u x x

x-dir: In time δt, the change in volume is:

/ /

Using all 3D:

u u x x t y z u t y z u x t V

d V u v wV

dt x y z

Mass is conserved, using

0

0

m V

d dm V

dt dtd Vd

Vdt dt

d Vd u v w

dt V dt x y z

d V

dt

10

du

dt


Infer vertical velocity from horizontal currents

ShipCourse steered

Position fix t=0

Ship

Position fix t=24h

Act

ual

Cou

rse

Position usingdead reckoning(from constant speed and course)

Mean surface current during t=24 hours

Mean surface current from ship drift

u = 0.3 m/sv = 0.03 m/s

x = 500 km

y=

500 k

m

Bu = 0.25 m/sv = 0.05 m/s

u =-0.25 m/sv =-0.01 m/s

u = -0.25 m/sv = 0.0 m/s

DC

A

E

8 15 5

8 1 8 1

8 1

0.25 ( 0.25) 0.25 (0.3)0; 10 10

5 10 5 10

: 5 10 ; : 8.3 10

13.3 10

A B

at E

u us

x xu v

Average s Average sx y

w u vs

z x y


8 1

50

0

0 at surface; 13.3 10 is downward below surface

Vertical velocity at - is

e.g. at 50 0.58 /h

h

ww s ve w

zz h

ww dz m w m day

z

2 2

1 1

61 2 2 18

Recall, 0 at surface less near surface

Total time par a particle to sink m is t = /

1 17.5 10 ln /

13.3 10

to 50 this takes close to 1 year.

z z

z z

w w

z z w

t dz dz z zw z

m

w << uTake horizontal scale for whole ocean: L = 4000 km (Earth’s radius 6371 km)Vertical scale H = 4 km (mean ocean depth)Aspect ratio: H/L = O(10-3) : Like a sheet of paperScaling: w Hu/L u/1000 !!!


Example: Extension to conservation of mass and salt

Observations give the salinity of the inflow and outflow across the Gibraltar Strait, and the outflow volume flux in Sv (1 Sv = 106 m3/s) What is R+P-E ?What is flushing time?(Assume in/outflow density is equal.)

Volume=4x106 km3

Conservation of mass: Total In Total Out i o i o

i oV R P V E

Source: Robert Stewart

0.836

i i i o o o i o

o oi

i

V S V S

V SV Sv

S

Conservation of salt:

4 34.6 10 m /so iR P E V V

Volume of SeaMinimum Flushing Time = about 151 years

Incoming Water

Lecture Guidelines for GEOF110 Chapter 2 + Thermodynamics (2 hours) Chapters 3+ 4 (2 hours) Ilker...

Documents

Transcript of Lecture Guidelines for GEOF110 Chapter 2 + Thermodynamics (2 hours) Chapters 3+ 4 (2 hours) Ilker...