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Gill's Chapter 4
1. What is a material element? [p.63]. [material element = infinitesimally small fluid sample that retains
its identity.]
2. Why is it that if molecular diffusion is =0, the material element will consist of the same
particles? What are "particles" in this case? [p.64]. [If diffusion=0, then molecules (=particles) cannot
spread, so material element consists of the same particles.]
3. Why is (4.2.1) valid (i.e. why can the LHS be broken up into the 3 terms on the RHS)?
[Use d(ab)/dk = adb/dk+bda/dk).]
4. Derive (4.2.3), (4.2.4) & (4.2.5).
5. What is the difference between advection and convection? [p.66]
6. Derive (4.1.8) from (4.3.2) and (4.2.4) (== (4.2.3)).
7. Derive (4.2.5) [Use Fig.4.2].
8. Explain the flux vector F in (4.3.3) - as advective + diffusive fluxes.
9. Note how (4.3.4) [or 4.3.7] is similar to the Salinity equation used in POM. Is it identical? Why (yes or
not)?
10. The entropy of a material element is fixed if the element is moved
(a) without exchanging heat with its surrounding and if
(b) it retains its fixed composition.
The motion is then called isentropic.
11. So, the salinity or 'salt' equation is fairly straight-forward, since the Qv on p.67 = rho*s = mass of salt
per unit volume; and the "mass of salt" is a 'measurable' scalar quantity - i.e. a "mass."
12. Temperature is somewhat different however. It is still a scalar of course, but it is closely tied to the
agitation of molecules and therefore also to the entropy or "heat content." Temperature is derived
from the "heat equation." To understand this, we need to explore a little thermodynamics.
13. The "heat" equation is obtained from the 2nd law of thermodynamics:
dE = Tdpdvs
E = internal energy per unit mass
Vs = specific volume (volume per unit mass)
T and p = temperature and pressure respectively
 = specific entropy (entropy per unit mass)
(3.2.1)
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Energy increases because of (i) agitation (molecules) d, and (ii) compression dvs < 0. We assume no
phase change and salinity is fixed – i.e. the fluid is of fixed composition; so  = (p, T).
Tdchange in heat content per unit mass
pdvs = work done by p in compressing fluid’s volume per unit mass.
14. Eqn.(3.2.6) is used often later – so we will derive it. From (3.2.1):
Td = dE + pdvs
(N4.1)
So, increase in heat content for a change in T while keeping p constant is then, from (N4.1):
Cp  T(/T)p = (E/T)p + p(vs/T)p
(N4.2)
The Cp is called the specific heat at constant pressure.
Similarly, we obtain the following for increase in heat content for a change in p while keeping T
constant:
T(/p)T = (E/p)T + p(vs/p)T
(N4.3)
Now, recall that  = (p, T), the Td (i.e. LHS of (3.2.6)) is:
Td = T(/T)p dT + T(/p)T dp
= CpdT + T(/p)T dp
(N4.4)
where the definition of Cp from (N4.2) has been used for the first term. For the second term, T(/p)T,
we may use (N4.3), but we can get rid of the “E” as follows. We can [T(N4.3)p(N4.2)]:
(/p)T +T(/pT)T(/Tp) = (E/pT)(E/Tp)+p(vs/pT)p(vs/Tp)(vs/T)p
so that,
(/p)T = (vs/T)p
(3.2.5)
which can be used in (N4.4) to give (3.2.6):
Td = CpdT  T(vs/T)pdp.
(3.2.6)
15. A process is reversible if it changes very, very slowly due to infinitesimal changes in some property of
the system, so that no dissipation occurs. For example, if a gas in an insulated cylinder is compressed by
pushing a piston (see Fig.N4.1) and there is no friction between the piston and the cylinder wall, then if
the piston is let go the compressed gas would expand back so the piston is reversed to its original
position. If there is friction, then heat is permanently loss and the process is irreversible.
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In practice, no process is reversible. But if the applied changes are slow and/or if the system’s response
is fast, then the process may be considered as being reversible. Gill assumes this [p.42, approx. line 6].
piston
gas
cylinder
Fig.N4.1
16. Since Td = increase in heat content (of the system), then if the process is isentropic and reversible,
then it is also called adiabatic. In our case, we assume “reversible processes,” so isentropic = adiabatic.
Thermodynamic Definition: An adiabatic process a process in which no heat is transferred to or from the
working fluid.
17a. When a parcel of fluid sinks adiabatically, say into the deep ocean, it experiences increased
pressure and therefore warms up. Similarly, when air parcel rises (to mountain top) it experiences less
pressure and cools down. In this case, “adiabatic” means that there is no turbulent or molecular
diffusion and no radiation across the parcel’s boundary, and also no change in composition or phase.
Thus, in the case of adiabatic sinking or rising fluid parcel, because of pressure compression (expansion),
DT/Dt  0; it would be nice to define a quantity such that the D(.)/Dt = 0.
Also, in both cases (sinking or rising parcel), the (vertical) column of fluid (air) is apparently unstable (i.e.
“warm” is under “cool”). But this is not true – i.e. the fluid is actually stable. To avoid this apparent
confusion, and to make D(.)/Dt = 0, the quantity potential temperature  is introduced.
The potential temperature “” is the temperature of the parcel at pressure p would have if the parcel
were brought adiabatically (i.e. isentropically; i.e. without exchange of heat with its surrounding) to a
reference (i.e. common) pressure pr.
By using , we eliminate the apparent warming or cooling due to changes in pressure. Since now the
various temperatures at different depths have been brought to a common pressure, we can compare
them without worrying that the ’s represent false temperatures. The variation of  with height can
therefore be used to judge if the vertical column is statically stable or not.
The situation is a bit similar to trying to judge if person (A) on Jupiter is heavier than person (B) on Pluto
(Fig.N4.2). It would appear that (A) is heavier than (B). But to more fairly judge them, we can move
them (without loosing even a single hair; constant entropy) all to a reference place, say the earth, then
re-weigh them. The weight measured on earth may then be called “potential weight.”
(A) I’m 2 tons
Jupiter
(B) I’m 2kg
Pluto
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Fig.N4.2
17b. An example (for air) and some exercises will make the concept of potential temperature clearer. It
is convenient to do this with air than with water because air behaves like an “ideal gas” and explicit
formulae may then be obtained.
Suppose we follow an air parcel from some height with (p,T) which sinks adiabatically (i.e. isentropically)
from a high mountain to the ground. For isentropic process, we have from (3.2.6) with d = 0:
dT = T(vs/T)p,s dp/Cp = T(/T)p,s dp/(2Cp).
(N4.5)
Assume that air obeys the ideal-gas law:
 = p/(RT)
(N4.6)
[note that high p gives rise to high T and also smaller volume (per unit mass) vs = -1], we have then,
(/T)p,s = p/(RT2)
(N4.7)
and therefore (using (N4.6) again):
(/T)p,s /(2Cp) = +(R/Cp)/p = (2/7)/p =  /p,
(N4.8)
where  = (R/Cp) = 2/7 for dry air from Gill’s (3.3.2). Equation (N4.6) then becomes:
dT/T =  dp/p
(N4.9)
Integrating (N4.9) from a height where pressure is p and temperature is T to the reference level where
the pressure is pr and temperature is, by definition, the potential temperature , we have
ln()ln(T) =  [ln(pr)ln(p)], or
ln(/T) =
/T
=
ln(pr/p)2/7, or
(pr/p)2/7.
(N4.10)
It is clear that potential temperature is only meaningful when we have prescribed a reference level; the
most common one is the surface, where the pressure is  1013 mb. It is convenient to simply let pr =
1000 mb.
Exercise 17b1: Air parcel “A” at a height where p = pA = 100 mb has an in situ temperature TA = 175 K.
Another parcel “B” closer to the ground where p = pB = 400 mb has a TB = 220 K. Is the air column
stable? [Hint: calculate the corresponding .]
Please do not read beyond this line before you attempt exercise 17b1.
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More Thoughts on Exercise 17B1: In the above exercise, TA < TB and according to (N4.6), the density A
seems to be greater than B, i.e. A > B, i.e. air column appears to be unstable. But (N4.6) cannot be
used this way to determine the densities because the pressures pA and pB are different. In fact, since pA
< pB, it may be possible for A to be less than B. This is why we introduce the potential temperature 
by “moving” both “A” and “B” adiabatically to the common level where p = pr. By doing this “moving”
adiabatically, we ensure that both parcels’ heat-gain (and rise in temperature) as they are moved closer
to the ground is “kept inside” the parcels. Therefore, from (N4.10), A  338 K > B  286 K, so that the
air column is in fact stable.
Exercise 17b2: If we increase TB, the air column would then tend to the state of “less stability.” (we will
learn the precise definition later). Calculate the value of TB such that the air column is neutrally stable.
Exercise 17b3: In Exercise 17b1, is the air column stable if TB = 300 K?
Exercise 17b4: Figure N4.3 plots T as a function of p for three values of ’s. Think about this in view of
the exercises you have just done. By choosing a few numbers, convince yourself that you have
understood the concept of .
Fig.N4.3. The drop in local temperature (T) with height (or pressure) for various potential temperatures
 = 240, 280 and 320 K, computed from (N4.10), rewritten as T =  (p/pr)2/7, pr = 1000 mb.
Answers to Exercises 17B2&3: TB_neutral = 260 K; unstable.
18. Static Stability (section 3.6).
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Suppose the fluid is at rest; to judge a parcel’s static stability, we move the parcel up or down
isentropically, calculate its density change, d, then compare this change with the change in the
surrounding (i.e. ambient) density, da. If the parcel is moved up, then for stability, d must > da so
that the parcel can sink back to its original depth when let go.
The general formula for the change in density d(p,T,s) of the fluid (ambient or parcel fluid) as a result of
a change in height “dz” is, by chain-rule:
d(p,T,s)  (d/dz).dz = [ (/p)T,s (dp/dz) + (/T)p,s (dT/dz) + (/s)p,T (ds/dz) ].dz
(3.6.8)
where s = salinity. For a parcel of fluid that is moved isentropically, with  = (p, T), and fixed s; see
“13”, (3.6.8) becomes:
d = (/p)T,s dp + (/T)p,s dT
(N4.11)
We can replace “dp” and “dT” with “dz” as follows. From (3.2.6), for isentropic process (d = 0):
dT = T(vs/T)p,s dp/Cp = T(/T)p,s dp/(2Cp) = Tdp/(Cp) = dz
(3.6.1)
where  = thermal expansion coefficient and  = adiabatic lapse rate:
 = -1 (/T)p,s > 0
(3.6.2)
 = gT/Cp,
(3.6.5)
and the last equality in (3.6.1) (involving ) is obtained by using the hydrostatic equation:
dp/dz = g
(N4.12)
Substituting “dT” from (3.6.1) and “dp” from (N4.12) into (N4.11), which becomes:
d =  [g(/p)T,s + ] dz
(3.6.7)
Reminder: equation (3.6.7) is the change in density, dpar, of the fluid parcel as it is moved. On the other
hand, equation (3.6.8) is general, and gives (for example) the change, damb, of the ambient (i.e.
surrounding) fluid density with z. Suppose that the parcel is moved upward (i.e. dz > 0), we must have,
for stability, that dpar > damb, so that the parcel can move back (i.e. downward) to its original position
when let go:
 [ + dT/dz]  s ds/dz > 0
(3.6.9)
Cp-1 2gT + dT/dz  s ds/dz > 0
(3.6.10)
or
where
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s = -1 (/s)p,T > 0
(3.6.11)
is the expansion coefficient for salinity (I use s to avoid possible confusion with the planetary beta ).
Exercise 18.1: Show that the same stability condition is obtained if similar arguments as above are
applied but the parcel is moved downward instead of upward.
Answers: If parcel is moved down, then for stability we must have dpar < damb. Then from (3.6.7) and (3.6.8):
[g(/p)T,s + g(/p)T,s +  + dT/dz  sds/dz] dz < 0.
But since dz < 0, the […] must be > 0, and (3.6.9) is obtained.
19. The buoyancy frequency (or BruntVaisala frequency; section 3.7) squared is:
N2 = g [( + dT/dz)  s ds/dz] = g[Cp-1 2gT + dT/dz  s ds/dz].
(3.7.1)
When N2 > 0 (< 0), then the medium is stable (unstable). In a stable medium, N is real and represents
the frequency of (vertical) oscillation of a fluid parcel – closely linked to internal waves. In the ocean N 
10-4 to about 10-2 s-1, or periods of about 10 minutes to a few hours.
20. Since  is the temperature that a parcel has if it is moved isentropically to a reference pressure (see
“17a”), then a particular  has a particular entropy . Therefore  = constant when  = constant. So if s
= fixed,  = (), i.e.  is a function of  only. The function can be obtained by fixing p = pr (reference
pressure) in (3.2.6) which then becomes, since T is then = :
d/d = Cp(pr, )/
(3.7.6)
21. Assume that s = constant (i.e. fixed composition). Instead of p and T, we can use p and , so that
 = (p, ).
Then for a parcel that is moved isentropically (i.e. d = 0), the change in its density is (c.f. N4.11)
dpar = (/p),s dp + (/)p,s d
(N4.13)
The change of the surrounding density is from (3.6.8) with T replaced by :
damb(p, ,s)  (d/dz).dz = [ (/p),s (dp/dz) + (/)p,s (d/dz) + (/s)p,T (ds/dz) ].dz (N4.14)
For stability, dpar  damb > 0 (if dz > 0) as in “18”, then:
g-1N2 = ’(d/dz)  ’s (ds/dz) > 0
(3.7.9)
where ’ = -1(/)p,s and ’s = -1(/s)p, . In the form written as (3.7.9), if s = fixed, then the
stability condition becomes very simple, i.e. the fluid medium is stable if the potential temperature
increases upwards. This fact was used in Exercise 17b.1-4.
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22. The Heat Equation (isentropic motion):
If fluid elements do not exchange heat with their surroundings, and further more if they retain a fixed
composition, then their motions are adiabatic or isentropic, and we have:
D/Dt = 0 = Cp(pr, )-1 D/Dt
(4.4.1)
the last equality is from (3.7.6). Thus for adiabatic motion, the heat equation takes on this simple form:
D/Dt = 0,
(N4.15)
i.e. the potential temperature is constant following the parcel’s movement. The heat equation looks a
little more complicated if written in terms of the in situ temperature. From (3.2.6), we have
TD/Dt  CpDT/Dt  (T/)Dp/Dt = 0.
(N4.16)
The heat equation can also be written in terms of the internal energy E (per unit mass; why? See the
pressure work term in eqn.3.2.1). From (3.2.1):
TD/Dt  DE/Dt + pDvs/Dt = 0.
(N4.17)
This last equation can be re-written by making use of the continuity equation (4.2.4) which is:
/t + .(u) = 0,
(4.2.4)
so that
DE/Dt + E*LHSof(4.2.4) = t + u.E + E*LHSof(4.2.4) = (E)/t + .(uE); or:
DE/Dt = (E)/t + .(uE).
(4.3.6)
Then,  times the second equation of (N4.17) gives:
DE/Dt  (E)/t + .(uE) = pvs-1Dvs/Dt = p.u,
(4.4.3)
in which (4.3.6) and (4.2.2) have been used. Equation (4.4.3),
(E)/t + .(uE) = p.u,
(N4.18)
says that the rate of increase of energy per unit volume following a fluid parcel, D(E)/Dt > 0, is equal to
the rate at which work is being done due to the parcel’s compression (i.e. convergence) .u < 0.
Similarly, there is a decrease in E if the parcel expands (i.e. divergence) .u > 0.
By the way, one can use Figure 4.2 and its derivation of the time rate of change, /t, of mass per unit
volume (= ), to apply to energy per unit volume = E.
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23. The Heat Equation (non-isentropic motion):
Then additional terms are added to account for various heat transfers and sources: Frad, kT and QH.
Similar to the “advective heat flux density” Eu, we can identify Frad and kT as the “radiative and
conductive heat flux densities” respectively. In other words, they are “fluxes” that ‘flow’ in and out of
the side faces of the rectangular volume element of Figure 4.2 (see the comment of the last sentence of
“22” above). By contrast, the “QH” (= rate of heating per unit volume) is a heat source (or sink) inside
the volume element and is therefore grouped together with the pressure work p.u. The result is
therefore equation (4.4.4):
(E)/t + .(Eu + Frad  kT) = QH p.u.
(4.4.4)
An alternate form using the in situ temperature T is obtained by equating  times (N4.16) to  times
(N4.17), and then using (4.3.6) and (4.4.4), to obtain:
CpDT/Dt  T Dp/Dt = .( kT  Frad) + QH.
(4.4.6)
Yet another alternative form using the potential temperature  is simpler, using (4.4.1), (N4.16) and
(4.4.6):
TCp(pr, )-1 D/Dt = .(kT  Frad) + QH.
(4.4.7)
The Princeton Ocean Model (POM) uses  (so do most other ocean models), hence it uses equation
(4.4.7). In fact an approximate form is used, in which QH = 0, the k is replaced by the Smagorinsky’s
horizontal diffusivity coefficient Asmag (why??), and Frad represents the radiative penetration of light from
the ocean’s surface and is significant only near the surface z > 30 m (where   T):
D/Dt = .(Asmag)  (Frad)/z.(Cp)-1,
(N4.19)
where .(Asmag) is supposed to represent mixing effects by small-scale eddies; we replace T by ,
and also lump the (Cp)-1 into Asmag (so that it has a unit of m2s-1) thus acknowledging that we don’t know
much about this small-scale eddy mixing process, and therefore the errors due to these “replacement”
and “lumping” of terms are inconsequential. In actual numerical simulation, we set Asmag to be very
small, in fact often zero.
24. Momentum Balance (or the Equation of Motion):
This is (4.5.20) [or pom_v13.pdf (or *.doc) equations (1-24a,b,c)]:
(Du/Dt + 2 × u) = p’  ’g + .(u),
(4.5.20)
u = (u, v, w);  = (1, 2, 3);
(N4.20)
where
g = (0, 0, g).
25. Mechanical Energy (Kinetic Energy) Equation:
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This is (4.6.1) [with (4.6.2)]; it is derived by taking the scalar product of “u” with eqn.(4.5.20)  i.e.
u.(4.5.20). Term by term, we have:
u.Du/Dt = u.(u/t + uu/x + vu/y + wu/z) = D(u.u/2)/Dt [decompose into components]
u.(2 × u) = 0 [the 2 vectors are perpendicular, so their dot product = 0]
u.p’ = .(u p’) + p’.u [by chain rule]
u. (’g) = wg’
u.[.(u)]   u.[.(u)] = .(u.u)   (u .u)
= .((u.u/2))   [(u/x)2+(u/y)2+(u/z)2]
Putting all these together, we have then for u.(4.5.20):
D(u2/2)/Dt = wg’ .(u p’) + .((u2/2))   [(u/x)2+(u/y)2+(u/z)2] + p’.u
(N4.21)
or,
D(u2/2)/Dt = wg’ +.(u p’ + (u2/2))   + p’.u
(4.6.1)
 =  [(u/x)2+(u/y)2+(u/z)2] = dissipation rate
(4.6.2)
where
26. It is also useful to think of the rate of change of K.E. for a fixed volume – Gill’s equations (4.6.34.6.4). Also understand (4.6.5) and interpretations based on Fig.4.6.
27. For a large volume, we have eqn.(4.6.7):
dK/dt + Fn’dS =   dxdydz,
(4.6.7)
where Fn’ = n.F’ for F’ = energy flux density given by (4.6.4), and n is the unit vector normal to the
bounding surface dS.
28. Importance of Viscosity!
We normally say (or hear the statement) that viscosity is not important on the large scales (say a few
kilometers or larger) typical of the energy input (say by wind). But let’s think about this more carefully.
Let’s take the simple case of a homogeneous ocean. Input of energy is due to the pressure and viscous
stress (see equation 4.6.4) at the free-surface only, since no energy can pass through the solid bottom and
side walls. Yet despite this constant energy input, we know that the ocean’s energy is not continually
increasing, i.e. dK/dt  0 in (4.6.7). Therefore, in that equation, the large-scale energy flux input Fn’dS
can only be balanced by the small-scale energy dissipation  dxdydz by molecular viscosity! How
small is small? Let’s call this small scale = l. The l can be assumed to depend only on  and :
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l = (3/)1/4  O(10-3 m)
(N4.22)
Therefore, because of the very disparate scales, the only way that the large-scale energy can be dissipated
by the small-scale (l ) viscous dissipation is that there will be a transfer of energy from large to small.
This transfer is accomplished by the nonlinear terms (e.g. uu/x etc; if u ~ eikx, then uu/x ~ ei2kx etc) –
“breaking” down large eddies into smaller ones, which are in turn broken down again into even smaller
ones etc until the smallest eddies have scales of O(l ) when molecular viscosity can act to dissipate the
(kinetic) energy into heat.
The fact that we have disparate scales creates big problem for numerical models which typically must use
some kind of eddy viscosity >> molecular viscosity  in order to dissipate input of large-scale (say wind)
energy. However, this “eddy viscosity” parameterization does not guarantee that we are removing the
energy in a realistic way – thus we have (for example) Mellor-Yamada scheme or the Smagorinsky
viscosity to try to model this “energy-removing” process. If we have a super-super computer that allow a
grid resolution of 10-3 m, then we will not need to use Mellor-Yamada or Smagorinsky!