Equations of Continuity – Conservation and Box Models
Relation between Distribution of Properties and Currents in the Sea (From SFJ)
Consider any scalar quantity, call it s, and it could be salinity, temperature, pressure, oxygen, calcium,
etc. The distribution of s is continuous in space and time so it can be represented by
𝑠 = 𝑓(𝑡, 𝑥, 𝑦, 𝑧)
Assume that this scalar, s, can be considered a property of the individual particles of the fluid.
A particle in motion after a time dt will be in a new locality
𝑥 + 𝑑𝑥, 𝑦 + 𝑑𝑦, 𝑧 + 𝑑𝑧
where the scalar quantity, s, under consideration has the value
𝑠 + 𝑑𝑠 = 𝑓(𝑡 + 𝑑𝑡, 𝑥 + 𝑑𝑧, , 𝑦 + 𝑑𝑦, 𝑧 + 𝑑𝑧)
The property, s, of the individual particle has therefore been changed by the amount ds in the time dt;
that is the time rate of change is ds/dt
This time rate can alsw be expressed by the characteristics of the field because, by means of the Taylor’s
expansion (remember that?) we have
𝑓(𝑡 + 𝑑𝑡, 𝑥 + 𝑑𝑥, 𝑦 + 𝑑𝑦, 𝑧 + 𝑑𝑧) = 𝑓(𝑡, 𝑥, 𝑦, 𝑧) +
𝜕𝑓
𝜕𝑓
𝜕𝑓
𝜕𝑓
𝑑𝑡 +
𝑑𝑥 +
𝑑𝑦 +
𝑑𝑧
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
or, since 𝑠 = 𝑓(𝑡, 𝑥, 𝑦, 𝑧)
𝑑𝑠 =
𝜕𝑠
𝜕𝑠
𝜕𝑠
𝜕𝑠
𝑑𝑡 + 𝑑𝑥 +
𝑑𝑦 + 𝑑𝑧
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
Dividing by dt we get
𝑑𝑠 𝜕𝑠 𝑑𝑡 𝜕𝑠 𝑑𝑥 𝜕𝑠 𝑑𝑦 𝜕𝑠 𝑑𝑧
=
+
+
+
𝑑𝑡 𝜕𝑡 𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 𝜕𝑧 𝑑𝑡
and since dx/dt, dy/dt and dz/dt are the components of velocity
𝑑𝑠 𝜕𝑠 𝜕𝑠
𝜕𝑠
𝜕𝑠
=
+ 𝑣𝑥 +
𝑣𝑦 + 𝑣𝑧
𝑑𝑡 𝜕𝑡 𝜕𝑥
𝜕𝑦
𝜕𝑧
Individual time change = local time change plus advection (currents)
1. The distribution of any scalar quantity (salinity for example) is stationary, that is, independent of
𝜕𝑠
time fi the local change is zero ( 𝜕𝑡 = 𝑜)
2. The advection terms disappear if there is no motion or if the field is uniform
3. When the individual time change is zero, ds/dt =0, the local change equals the advection but of
the opposite sign
4. If the property is stationary and the individual time change is zero the equation is reduced to
𝜕𝑠
𝜕𝑠
𝜕𝑠
𝑣𝑥 +
𝑣𝑦 + 𝑣𝑧 = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧
This only happens when flow is along equiscalar surfaces.
Exercise –
Assume no vertical motion, speed in x direction is 0.1 m/s, in y is 0.01 m/s, s gradient in x
direction is 0.0005 s/m. What is the s gradient in the y direction?
Concentration
Now let’s talk about not just a scalar but concentration – the measurable amount of anything in a
volume of seawater: heat content, salinity, oxygen, nitrate, even phytoplankton.
So concentration is just like the scalar s we have been looking at
Two kinds of processes can alter concentration:
External – processes at the boundaries. For example at the sea surface heat concentration can change or
oxygen for example thru interaction with the atmosphere.
Internal processes – that happen within the ocean away from the boundaries.
Conservative Concentrations – Only changed by advection or diffusion – no reactions, no biology, no
radioactivity.
Heat and salt content good examples of conservative concentrations.
Now add diffusion
The main equation from SVJ page 159 V5
𝜕𝑠
𝜕 𝐴𝑥 𝜕𝑠
𝜕 𝐴𝑦 𝜕𝑠
𝜕 𝐴𝑧 𝜕𝑠
𝜕𝑠
𝜕𝑠
𝜕𝑠
=
(
)+
(
)+ (
) − ((𝑣𝑥 ) + (𝑣𝑦 ) + (𝑣𝑧 ))
𝜕𝑡 𝜕𝑥 𝜌 𝜕𝑥
𝜕𝑦 𝜌 𝜕𝑦
𝜕𝑧 𝜌 𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
local change = diffusion in x and y and z – advection in x, y and z
Local time change of concentration, s, equals the effect of diffusion minus effect of advection
Az usually = 10-4 to 10-3 m2/s
Ax and Ay often referred to as Ah is 10-1 to 103 m2/s
Exercise
What is the equation if velocity is only along the X axis, only vertical diffusion is acting, A/ρ is
constant, and there is on change in concentration S. (v7)
What is the equation if there is no gradient in the x, y or z direction and only diffusion in the z
direction? V8
If the gradient of S in the X direction is 0.01/1000 m and Vx = 1m/s and there is no diffusion
what is the local change in S?
Box Models, Residence and Flushing Time
Straits of Gibraltar example from Knauss page 76.
Exercise (box 4.3)
Redo the calculation 4.23 confirming the results
How does the inflow and outflow change if freshwater comes in so Sin = 36.0? This might happen
if there is more rain in the North Atlantic for example.
How long is a particle in the Med? Page 77
Exercise – back of the envelope calculation for Chesapeake Bay
Flow into Bay (T in Knauss) - The Bay mouth is 40 km wide and 10 m deep and the average flow
in is 0.1 m/s. How much water is flowing into the Bay every second (cubic meters)?
Assume the salt concentration is 30 units per cubic meter (n in Box 4.3)
Volume of Bay (V in box .3)- What is the volume of the Bay? Assume is it 200 km long, 50 km
wide and 10 m deep? In cubic meters.
So now you have T, n and V
N is the amount of new salinity in the volume.
The flushing rate is (N/V)T. What is it for the above numbers? Are the units right.
Exercise for Lynnhaven River – guess at
the volume of Lynnhaven River,
the cross section area of the inlet at the Bridge
the speed of the current during a tide.
How much water flows into LR during one tide?
Divide the volume of LR by the flow thru the inlet. How days will it take to ‘flush’ the River?