OC210 TOPIC 1:
BASIC CONCEPTS
PART 2: TRANSPORT AND BUDGETS OF MASS, SALT, HEAT AND
NUTRIENTS
I. Transport of mass, salt, heat, and nutrients by ocean currents
1. Currents move mass, heat and salt and thus affect the observed distributions of temperature,
salinity and density in the ocean.
2. The amount of heat carried by ocean currents plays a major role in controlling climate on earth.
-Currents (and winds) are a major mechanism that moves heat poleward on earth
3. Nutrients carried by ocean currents fuel photosynthesis by phytoplankton. The spatial and
temporal variations in currents and mixing have a major impact on the distribution of plankton and
spatial pattern of photosynthesis in the ocean.
4. Quantifying the amount of water, mass, salt, heat, nutrients, etc. transported by ocean currents is a
fundamental calculation in oceanography.
A. How do we calculate the amount of water and mass an ocean current carries or transports?
1. First, let’s think of a river
-we often refer to the amount (volume) of water flowing in a river as its discharge (Q)
-the units for the rate of water flow (transport) or river discharge are volume/time and,
typically, is expressed in m3/sec
-water transport or discharge represents volume of water per unit time passing a certain point
-there is an animation of the ‘discharge’ concept in a stream at this Web site
http://www.sciencecourseware.org/VirtualRiver/Files/page04.html (Fig. 1)
2. How do we calculate the volume transport or discharge of a river?
-River has an average velocity (V, m/sec or cm/sec) and cross sectional area (Ax, m2)
-then the discharge Q (m3/s) = Ax (m2) * Vel (m/s), where s=sec
3. Calculate the discharge of the Amazon River
-On average, the river is 3 miles wide, 20m deep and has an average velocity of 2 m/s
Ax= 3miles * 1km/0.6mile * 1000m/km * 20 m = 100,000 m2
V= 2 m/s
Q (m3/s) = Ax (m2) * V (m/s) = 200,000 m3/s
-In contrast, the Columbia R. has discharge of about 5,000 m3/s
4. The flow rate or discharge of an ocean current is usually referred to as its volume transport.
-Volume Transport (m3/s) = Ax (m2) * V (m/sec)
5. Determining the cross sectional area of an ocean current often is more difficult than for a river
-how do we estimate width and depth?
-sometimes we can use temperature and salinity to define boundaries of currents
-In certain oceanographic settings, current flow is clearly constrained by topographic features
and we can treat them as rivers.
-an example is the Florida Current
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6. The Florida Current is a fast northward flowing current off the east coast of Florida, which
eventually turns into the Gulf Stream when is passed the coast of North Carolina (Fig. 2)
7. Determine the volume transport of the Florida Current, a deep fast current off the east coast of
Florida (Fig. 3)
a. For the Florida Current, we’ll calculate transport using the computed current velocities
across a channel between Miami, FL and Bimini Is.
-average width = ~ 70kms: average depth = ~600m: average velocity = ~0.6 m/s
-Florida Current volume transport = 70,000m * 600m * 0.6 m/s = ~ 25x106 m3/s
-Florida Current volume transport is >100x discharge of Amazon R.
b. Oceanographers use the term Sverdrups (Sv) as a unit of water volume transport in the
ocean
-where 1 Sv = 106 m3/s (1,000,000 m3/sec)
-thus the calculated Florida Current transport is 25 Sv
7. The large volume of water transported by ocean currents results from their large cross sectional
areas not their high velocities
8. How do we calculate the mass of water transported by the Florida Current?
-Mass transport equals volume transport times density
-Units for mass transport is kg/sec = Volume transport * density = m3/s* kg/m3
-For the Florida Current, assume the density is 1026 kg/m3
-thus the mass transport = 25x106 m3/s * 1026 kg/m3 = 25.65x109 kg/s
B. How much salt is carried by the Florida Current?
a.Assume the average salinity = ~36 (Fig 3)
-Remember: salinity = gms of salt per kg of seawater
-thus S = 36 gms salt/ kg seawater
b. Salt transport = water mass transport * salt concentration (salinity)
= 25.7x109 kg seawater/sec * 36 gms salt /kg seawater
= 923 x109 gms salt /sec
= 923x106 kg salt /sec
-(this is the equivalent of 50,000 dump trucks of salt per second)
C. How much heat is carried by the Florida Current?
1. We estimate that the temperature of the water is, on average, ~15°C (Fig. 3)
-but temperature is not equivalent to heat
2. How do we convert temperature to heat?
-heat typically measured in joules (or calories) whereas temperature is measured in degrees
3. Heat capacity of a substance (or specific heat) relates heat (in joules or calories) to temperature (in
°K).
- °K refers to temperature above absolute zero
-that is °K = 273 + °C
-for example, 0°C = 273 °K
4. The heat capacity (c) of freshwater is 4186 Joules/(kg°K) or 1000 calories/(kg°K)
-we define 1 calorie as the heat needed to raise 1 gm of freshwater by 1°K (when water at 4°C)
-the heat capacity of seawater is on average 3985 Joules/(kg°K) or ~950 cal/(kg°K)
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5. Calculate heat transported by the Florida Current assuming its average temperature is 15°C or
288°K (Fig. 3)
a. Heat transport = Mass transport* temperature* heat capacity
= 25.7x109 kg water /s * 288°K * 3985 Joules/(kg°K)
(= 25.7x109 kg water /s * 288°K * 950 cal/(kgºK) = 7.03x1015 cal /s)
= 29.5 x1015 Joules/s
-since 1 Joule/s = 1 Watt,
-then the heat transport = 29.5x1015 Watts (Note: 1 Watt = 0.24 calorie/sec)
-Remember: use absolute temperatures (ºK) and not ºC for heat transport
-seawater at a temperature of 0ºC still contains lots of heat
D. How important is the heat transported by the Florida Current to the sun’s heating rate on
earth?
1. Let’s compare the heat transported by the Florida Current to the solar heating rate poleward
of ~30°N
-the net annual solar insolation heat flux (north of 30°N) is about 130 Watts/ m2 (Fig. 4)
2. Let’s distribute the heat carried by the Florida Current evenly over the earth surface area
north of 30°N
-earth’s area = ~500x1012m2
-earth’s area Northern Hemisphere = 250x1012 m2
-poleward of 30°N, the earth’s area = 125x1012 m2
-thus the average heating rate resulting from Florida Current transport, if evenly distributed
over this area, = 238 Watts/m2 (= 29.5x1015 Watts / 125x1012 m2 )
3. Although the heat transported northward by the Florida Current is almost double the solar
heating rate north of 30ºN, we have to remember that there are southward flowing currents in
the ocean that carry heat equatorward
a. Thus some of the northward heat transported by the Florida Current is offset by heat
carried by currents flowing southward
4. If the southward flowing currents had the same temperature as the Florida Current, then there
would be no net northward heat flux
a. The heat transported northward by the Florida Current would be exactly matched by
the heat transported southward by other currents
5. If the southward flowing water was 5°C cooler than the northward flowing Florida Current (closer
to reality), then the net northward heat transport would = 5.1x1014 Watts
-that is, the net heat transport = 25.7x109 kg water /s * 288°K * 3985 J / kg°K –
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25.7x10 kg water /s * 283°K * 3985 Joules/(kg°K)
= 5.1x1014 Joules/s
= 5.1x1014 Watts
6.Thus our estimate of net northward heat flux by only the Florida Current represents ~3% of the
solar heating rate (i.e., 5.12x1014 Watts /125x1012m2 = 4.1 Watt/m2 / 130 Watts/m2 = 0.032)
7. If we look at the poleward heat transport of ~3x1015 W at 30ºN by all ocean currents (Fig. 5), then
dividing this heat by area north of 30ºN yields 24 W/m2 or ~20% of the solar insolation north of
30ºN.
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8. Just for perspective, the amount of global warming resulting from the increase of atmospheric
carbon dioxide levels over the last century due to human activity is ~1.5 W/m2.
9. Thus heat carried by ocean currents is an important component of the earth’s heat budget.
10. The classic view is that the heat lost from the warm waters of the Gulf Stream (which is a
northward extension of the Florida Current) to the atmosphere as its moves northward helps warm
Europe during winter
-at the same latitudes, SSTs in eastern N. Atlantic are higher than in western N. Atlantic
(Fig. 6)
-at the same latitudes, Europe is significantly warmer (~10-20ºC) during the winter than the
eastern US (Fig. 7)
-thus heat carried by Gulf Stream has substantial impact on air temperatures in Europe
11. Climatic implications:
a. If ocean circulation changes significantly, then temperature distribution on earth would
change as a result
-i.e., some regions would get warmer and some colder
b. if the Florida Current/Gulf Stream currents slowed down, the temperatures north of 40°N
would likely decrease and Europe, for example, would be colder than it is presently
c. a slow down of the northward flowing warm currents in the N. Atlantic is thought to have
occurred during the last Ice Age (~20,000 years ago).
E. Nutrient cycling in the ocean
a. What are the important nutrients required for plankton growth?
a. Macronutrients: Carbon, nitrogen, phosphorous, silica (diatoms)
b. Micronutrients: iron, other metals, vitamins, etc.
b. These nutrients appear as dissolved chemical species in the ocean
a. Carbon as bicarbonate (HCO3-), carbonate (CO3=) and dissolved carbon dioxide gas
(CO2)
b. Nitrogen as nitrate (NO3-) and ammonia (NH4+)
c. Phosphorous as phosphate (PO4-3)
d. Silica as silicate (SiO3)
c. Plankton covert these inorganic nutrients species into organic compounds (carbohydrates,
amino acids, lipids, etc.)
a. If one adds up the organic compounds in the plankton, typically, there is a molar
ratio of carbon:nitrogen:phosphorous (C:N:P) of 106:16:1. This is often referred to
as the Redfield Ratio after the scientist who first measured plankton composition.
d. The equation that describes the photosynthetic conversion of inorganic nutrients into organic
compounds by plankton can be expressed as the following: (Fig. 8)
106CO2 + 16HNO3 + 1H3PO4 + 122H2O + energy → (CH2O)106(NH3)16(H3PO4) +138O2
-where [(CH2O)106(NH3)16(H3PO4)] represents an ‘average’ organic compound in plankton
-this equation describes the photosynthetic production of biological organic matter
-the energy required for this reaction to occur is supplied by sunlight
-during photosynthesis, inorganic nutrients are converted to organic compounds (represented
by [(CH2O)106(NH3)16(H3PO4)]) and oxygen (O2) is produced
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e. Phytoplankton must have sunlight to photosynthesize
-Sunlight is completely adsorbed by seawater within the top 150m or so of the ocean
-Thus photosynthesis is restricted to a very shallow portion of the ocean
-The surface layer where photosynthesis occurs is called the photic or euphotic layer
f.
Respiration is the reverse of photosynthesis (Fig. 9)
- during respiration organic compounds are converted (oxidized) to inorganic nutrients
(CO2, NO3, PO4) and oxygen is consumed
- the respiration reaction is the reverse of the photosynthesis reaction
(CH2O)106(NH3)16(H3PO4) +138O2 → 106CO2 + 16HNO3 + 1H3PO4 + 122H2O + energy
-
respiration is a source of energy for plants, animals, bacteria, etc.
in contrast to photosynthesis, which requires light energy
g. Respiration occurs almost everywhere in the ocean
a. Respiration occurs mainly by bacteria but also by plants and animals.
b. At depths greater than 150m in the ocean, respiration dominates since it does not
require sunlight
c. Respiration consumes oxygen and produces nutrients and CO2
h. Typical depth profile of nutrients and oxygen in the ocean (Fig. 10)
a. CO2, NO3 and PO4 concentrations typically increase with depth in the ocean
b. Oxygen concentrations typically decrease with depth
c. Phytoplankton mass typically decreases with depth
d. Explain these observations.
i.
Since nutrient concentrations typically increase with depth in the ocean (Fig. 10), mixing (or
upwelling) of thermocline water parcels into the surface layer provides nutrients for plankton
growth.
-in certain regions of the surface ocean, like the subtropical gyres, the photosynthesis
rate is thought to be limited by nutrient supply
-in other regions of the surface ocean, where there are abundant nutrients (like in
polar latitudes (>50º), other factors (light, iron, temperature) limit photosynthesis
rates
F. Transport of nutrients by currents
1. Nutrient (and oxygen) concentrations are typically measured in µmoles/kg
- where 1 µmole = 10-6 moles = one millionth of a mole
2. Transport of nutrients by currents equals mass transport times the nutrient concentration
3. Mass transport of Florida Current is 25.7x109 kg /s
a. Assume concentration of nitrate is 5 µmole/kg
4. Nitrate transport = 25.7x109 kg /sec * 5x10-6 moles/kg of nitrate = 1.29x105 moles nitrate/sec
a. Calculation is similar for phosphate and CO2
II. Budgets (Volume, Mass, Heat, Salt, nutrients)
A. Water Budget in a Bath Tub
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1. A budget adds up the inputs (sources) and subtracts the outputs (sinks)
-if there is a balance between sources and sinks, then the total amount doesn’t change
-if there isn’t a balance, then the total amount either increases (when sources > sinks) or
decreases (when sinks > sources)
2. We use the term steady-state to describe a situation where sources (inputs) equals sinks (losses)
- Steady-state means that there is no change in the amount of material (e.g., water, salt, heat,
nutrients, etc.) over the time period of interest
-steady-state does not mean that there is no input or output, only that the total input (sources)
equals the total output or loss (sinks)
3. Let’s look at the water budget for a bathtub (Fig. 11)
a. Initially filling tub---amount of water in the tub increases because there is no water loss
(assuming the drain in closed)
-at this time, the water input rate exceeds the loss rate and the tub fills
b. At some later time, the water level reaches bathtub rim and starts to over flow
-under this condition, the total amount of water in the tub remains the same,
-we’ve reached a steady-state, where the filling rate equals the overflow rate,
-i.e, water input = water loss
c. Are the same water molecules always in the tub at steady-state?
-No!….even though total amount or volume of water remains constant there is
constant renewing of the water itself
4. To construct a water budget you have to add up all the inputs and subtract away all the losses
-the larger the difference between inputs and losses, the larger the time rate of change of
water volume
-the time rate of change of volume is expressed mathematically as V/t
-remember the symbol Δ is often used to express change
-if water sources are greater than sinks, then the volume of water increases (V/t >0)
-if water sources are less than sinks, then the volume of water decreases (V/t <0)
-if water sources equals sinks, then the volume of water doesn’t change (V/t =0)
-at steady-state, ΔV/Δt = 0
5. Class Exercise: Draw a graph of the volume of water in tub versus time
-indicate the filling stage and overflow stage on the graph
-indicate when the system reaches steady-state
-How would the graph change if the water input rate increased with time rather than stayed
constant?
Mass Budget
1. Usually oceanographers refer to the ‘amount’ of water in terms of its mass (kg)
- Note that two parcels of water with the same volume can have different masses
-under what conditions would this occur?
2. To construct a mass budget you have to add up all the mass inputs and subtract away all the mass
outputs or losses
-the larger the difference between inputs and losses, the larger the time rate of change of
mass (ΔM/Δt)
-remember: Mass (kg) = volume (m3) * density (kg/m3)
3. The time rate of change of mass is expressed mathematically as (V*ρ)/t
- where V = volume, ρ = density and t = time
- remember, the symbol Δ is often used to express change
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- if mass sources are greater than sinks, then (V*ρ)/t > 0
- if water sources are less than sinks, then (V*ρ)/t < 0
- if water sources equals sinks, then (V*ρ)/t = 0
- at steady-state, Δ(V*ρ)/Δt = 0
4. Mathematically, the mass budget for a volume of water is expressed as follows:
Δ(V*ρ)/Δt = I * ρi – O * ρo, where I = volume input and O= volume output
- Units: Δ(V*ρ)/Δt = (volume* density)/time = (m3 * kg/m3)/sec = kg/sec
- I and O have units of volume/time (m3/sec) and ρ is mass/volume (kg/m3)
- this means that the terms I*ρi and O*ρo are in units of kg/sec (same as for Δ(V*ρ)/Δt)
- At steady-state, Δ(V*ρ)/Δt = 0
-thus at steady-state: 0 = I*ρi – O*ρo
-thus at steady-state: I*ρi = O*ρo
5. If there is a balanced mass budget (steady-state) but the densities of the inflow and outflow differ,
what does this imply about the magnitude of I and O?
- how would the magnitude of I and O compare at a balanced mass budget (steady-state),
if ρi = ρo?
Salt Budget
1. What if saltwater, rather than freshwater, is flowing from the faucet into the tub?
2. The salt budget for the tub is expressed as follows:
Δ(V*ρ*S)/Δt = I*ρi*Si – O*ρo*So
- where S = salinity of the tub
- Note: we have to include the density of the water because salinity is expressed as gms Salt
per kg seawater
- Units: Δ(V*ρ*S)/Δt is in (m3*kg/m3*gms Salt/kg)/sec = gms Salt/sec
- this means that I*ρi*Si and O*ρo*So also have units of gms salt/sec
- I and O are in m3/sec, ρ is in kg/m3 and S is in gms Salt/kg seawater
- Important: the units for each term in any equation have to be the same
3. At steady-state in the tub
- thus Δ(V*ρ*S)/Δt = 0
- thus at steady-state: I*ρi*Si = O*ρo*So
- also at steady-state, the mass balance implies that I*ρi = O*ρo
- thus at steady-state in the tub, then Si =So
- that is, the salinity of the water in the outflow (S) equals the salinity of the water inflow
from the faucet
4. How does the salinity of the outflow (So) compare to the salinity of the tub (S)?
- thus So = S, because the water in the outflow is coming from the tub
5. What if there were two faucets (inflows) into the tub?
a. The mass budget would be expressed as follows:
a. Δ(V*ρ)/Δt = I1*ρ1 + I2*ρ2 – O*ρo (units: kg/s)
b. The salt budget would be expressed as follows:
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a. Δ(V*ρ*S)/Δt = I1*ρ1*S1 + I2*ρ2 *S2 – O*ρo*So (units: gm Salt/s)
c. At steady-state:
a. Mass: I1*ρ1 + I2*ρ2 = O*ρo
b. Salt: I1*ρ1*S1+ I2*ρ2*S2 = O*ρo*S
c. This means that at steady-state the salinity of the tub (S) will be intermediate
between S1 and S2 and depend on the mass discharge of each faucet (I1*ρ1 and I2*ρ2)
- mathematically: S = I1*ρ1*S1+ I2*ρ2*S2 at steady-state
O * ρo
- remember that at steady-state I1*ρ1 + I2*ρ2 = O*ρo
-thus: S = I1*ρ1*S1+ I2*ρ2*S2 at steady-state
I1*ρ1 + I2*ρ2
d. if you know the inflow rates, densities and salinities of the two inflows (faucets) you can
calculate the salinity of the tub at steady-state (don’t have to measure it)
Heat Budget
1. What if the two faucets were adding water with different temperatures?
a. the Heat budget:
a. Δ(V*ρ*c*T)/Δt = I1*ρ1*c*T1 + I2*ρ2 *c*T2 – O*ρo*c*To
i. Where T is temperature in ºK and c = heat capacity
b. Units: Δ(V*ρ*c*T)/Δt = Joules/sec = [m3*kg/m3*joules/(kg ºK) * ºK / sec]
-all terms in heat budget have the same units in J/s
1. At steady-state, the total heat input equals the total heat loss
Δ(V*ρ*c*T)/Δt = I1*ρ1*c*T1 + I2*ρ2 *c*T2 – O*ρo*c*To = 0
thus I1*ρ1*c*T1 + I2*ρ2 *c*T2 = O*ρo*c*To
3. This means that at steady-state the temperature of the tub (T) will be intermediate between T 1 and
T2 depending on the mass discharge of each faucet (I1*ρ1 and I2*ρ2)
- mathematically: T = I1*ρ1*c*T1+ I2*ρ2*c*T2 at steady-state
O*ρo*c
- the “c” cancels in all the terms,
thus: T = I1*ρ1*T1+ I2*ρ2*T2 at steady-state
O*ρo
- since at steady-state I1*ρ1 + I2*ρ2 = O*ρo
-thus: T = I1*ρ1*T1+ I2*ρ2*T2 at steady-state
I1*ρ1 + I2*ρ2
Problems focused on the concept of steady-state.
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1. What would be the steady-state temperature of the outflow under the following conditions?
I1 = 5 m3/s, I2 = 10 m3/s, T1 = 20ºC, T2 = 10ºC, ρ1 = 1024 kg/m3 and ρ2 = 1027 kg/m3 and c
(freshwater) = 4186 Joules/(kg°K)
The answer is: 13.33ºC. See if you can obtain this result.
Hints: ºK or ºC? What is the ρ of the outflow? Did we need to specify the value of c?
2. What is the steady-state salinity of the second input (S2) for the following conditions?
I1 = 5 m3/s, I2 = 10 m3/s, S1 = 35, So = 36.5, ρ1 = 1024 kg/m3 and ρo = 1026 kg/m3
The answer is: 37.25. See if you can obtain this result.
Hint: What is density of second inflow (ρ2)?
3. What are the inflow rates of the two faucets (I1 and I2) at steady-state based on the following
measurements?
- You measure the outflow rate at 15 m3/s. The measured temperatures of the two inflows
and outflow are: T1 = 20ºC, T2 = 10ºC and To = 14ºC
Hint: Assume all densities are equal. Set up two budgets for water and heat. Solve for I1 and I2.
Answer: I2 = 9 m3/s and I1 = 6 m3/s. See if you can get this result.
4. We put the tub outside in the sunlight and we observe that the temperature of the outflowing
water is increasing from its initial temperature of 13.33ºC. After a short while, the
temperature of the outflow reaches a constant temperature of 13.45 ºC. We know that I1 = 5
m3/s, I2 = 10 m3/s, T1 = 20 º C, T2 = 10 ºC, ρ1 = 1024 kg/m3, ρ2 = 1027 kg/m3 and ρo = 1026 kg/m3,
and c (seawater) = 3985 Joules/(kg°K).
Why does the temperature initially rise in the outflow after we put the tub outside?
How do we use this information to calculate the solar heating rate of the water in the tub?
Under these conditions, the heat budget is expressed as follows:
Δ(V*ρ*c*T)/Δt = I1*ρ1*c*T1 + I2*ρ2 *c*T2 – O*ρo*c*To + S, where S = solar heating rate
-Note: S has the same units as all the other terms (joules/sec)
Calculate the value of solar heating rate at steady-state.
Answer: 7.55x106 joules/sec. See if you can obtain this result (which, by the way, is much too high
for Earth. Maybe the tub is on the planet Mercury).
Nutrient Budget
1. Let’s say there is no life (no plankton, bacteria, fish, etc.) in the bathtub
-oceanographers use the term abiotic to mean ‘without life’
2. Under these abiotic conditions the nutrient (N) budget would be expressed as follows:
- the equation: Δ(V*ρ*N)/Δt = I*ρi*Ni – O*ρo*No (looks like the salt budget)
- Units: Δ(V*ρ*N)/Δt = (m3 * kg/m3 * moles/kg) / sec = moles/sec
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- At steady-state: I*ρi*Ni = O*ρo*No
- Since I = O and ρi = ρo at steady-state, then Ni = No.
3. Now let’s say the tub is put out in the sunlight and plankton begin to grow.
-Do you think the nutrient concentration in the outflow will be less than (<), equal to (=),
or greater than (>) the nutrient concentration in the inflow? Explain your reasoning.
4. How is the nutrient budget expressed including a phytoplankton growth term?
a. Is the plankton growth term a source or sink for nutrients?
b. Δ(V*ρ*N)/Δt = I*ρi*Ni – O*ρo*No – B, where B= biological uptake rate
-the units for B must be the same as for the other terms (moles/sec)
5. Let’s measure the inflow and outflow rates and their nutrient concentrations and see what
this implies about biological utilization rate of nutrients.
a. I = 5 m3/sec, Ni = 20 µmole/kg, No = 10 µmole/kg and ρi = ρo = 1024 kg/m3.
1 µmole = 10-6 mole or one millionth of a mole
b. At steady-state:
i. Δ(V*ρ*N)/Δt = 0 and I = O and ρi =ρo.
ii. 0 = I*ρi*Ni – O*ρo*No – B
iii. Rearranging yields: B = I*ρi*Ni – O*ρo*No
iv. B = 5 m3/s*1024 kg/m3 *20 µmole/kg – 5 m3/s*1024 kg/m3 *10 µmole/kg
v. B = 51200 µmole/sec = 0.0512 moles/sec
6. Thus we calculated the biological uptake rate of a nutrient indirectly using a steady-state
budget approach.
a. In reality, it isn’t easy to measure directly the nutrient uptake rate by plankton, so
this budget approach is useful, especially in laboratory experiments.
7. Next, let’s cover the tub with black plastic to eliminate sunlight.
a. After one day, we measure a nutrient concentration of 25 µmole/kg in the outflow
which is not changing over time.
b. What is the biological uptake rate under these conditions with the same inflow
conditions as in the previous example?
i. B = I*ρi*Ni – O*ρo*No
ii. B = 5 m3/s*1024 kg/m3 *20 µmole/kg – 5 m3/s*1024 kg/m3 *25 µmole/kg
iii. B = -25600 µmole/sec = -0.0256 moles/sec
iv. Why is the biological uptake rate (B) negative?
v. What process is going on in the tub?
vi. Why can you assume a steady-state condition?
B. Surface Ocean Heat Budget under Non Steady-State Conditions
Non steady-state conditions implies that there is a time rate of change of either mass, salt,
heat, nutrients, etc.
-this implies that the inputs and losses are not equal
Seasonal Changes in Sea Surface Temperature
1. In this application, it means there is a change in heat content (H) of the seawater over some
time period
-that is, dH/dt is either > or < 0 (but not = 0)
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-this implies that the temperature of the water (T) is changing over time
-this is a result of the input of heat being either greater than (dH/dt>0) or less than
(dH/dt<0) the losses of heat for this water parcel
2. Calculating the effect of the adsorption of solar radiation (sunlight) by seawater on the seasonal
change of the temperature in the surface ocean.
a. Let’s assume:
- A solar heating rate of 100 Watts/m2 or 100 Joule/m2/s. (see Fig. 4)
- This means that for every 1 m2 of surface area 100 Joules of solar heating occurs
every second.
- Solar insolation is adsorbed over a 100m deep column of water
- We will assume there are no currents transporting heat into and out of this column
of water, i.e. I = 0 and O = 0.
c.
Calculate the increase in temperature over six months resulting from this solar heating rate
(SH) for a 100m deep surface layer.
i.
What does the heat budget look like?
Δ(V*ρ*c*T)/Δt = I*ρi*c*Ti – O*ρo*c*To + SH
ii.
Since I = 0 and O = 0, then the budget is simplified.
Δ(V*ρ*c*T)/Δt = SH
iii.
We will assume the volume (V), heat capacity [c] and density ρ of the water
column do not change over time. This means those terms can move outside
the parenthesis.
- thus, V*ρ*c* (ΔT/Δt) = SH
- only the temperature of the water column changes with time
iv.
We want to calculate the magnitude of ΔT/Δt because it represents the time rate
of change of temperature. (Remember: When there is a change in conditions
over time, then the system is not at steady-state.)
-thus ΔT/Δt = SH/(V*ρ*c)
-Units: ºK/sec
v.
However, SH is in units of 100 Joules/m2/sec, rather than Joules/sec which is the
unit of V*ρ*c*ΔT/Δt. The thing to realize is that the solar heating is 100
Joules/sec per 1 m2 of surface area. If our column of water is 1m2 in area,
then solar heating is 100 Joules/sec for a column of water with a 1 m2
surface area.
vi.
So let’s assume the column of water has a surface area of 1 m2
-then ΔT/Δt = SH/(V*ρ*c) = 100 Joule/sec / (1 m2 * 100m * 1024
kg/m3*3985 joules/(kg ºK)
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-thus ΔT/Δt = 2.4x10 ºK/sec
vii.
The amount of temperature increase over 6 months (6 month = 1.57x107 sec)
equals 3.8 ºK or ºC.
ΔT = ΔT/Δt * time interval
ΔT = 2.4x10-7 ºK/sec * 1.57x107 sec = 3.8 ºK or ºC
Note: the change in temperature is equal in ºK or ºC. (Why?)
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viii.
Question: Would the calculated temperature change be any different if we chose
a 100 km2 surface area for our water column (rather than 1 m2)? Explain.
ix.
Question: How does this estimated seasonal (6 month) SST change compare to
observed SST cycle in the surface ocean (Fig. 12)
C. Residence Time
a. Residence Time represents the average length of time material spends in a reservoir.
- residence time is usually represented by τ (tau)
b. We’ll use the Bath Tub as a helpful example.
c. Let’s calculate the residence time of the water in the tub.
d. Residence time of water equals the volume of the reservoir (tub) divided by the total inflow
or outflow rate.
a. Let’s assume the tub holds 10,000,000 (or 107) liters (huge tub).
i. Since 1 m3 = 1000 liters, then the tub volume = 10,000 m3
b. Let’s assume the inflow rate is 10 m3/sec (huge faucet)
c. Residence Time (sec) = volume/inflow
= 10,000m3 / 10 m3/sec
= 1000 sec or 16.7 mins.
e. This means that under these conditions, on average, the water spends only ~17 mins in the
tub before it overflows.
a. some water parcels will spend a shorter time and others a longer time, but on average
a water parcel will spend about 16.7 minutes in the tub
b. picture a histogram of water residence times in the tub as a Gaussian curve with the
center of the Gaussian distribution at 16.7 minutes
f.
Let’s calculate the residence time of nitrate (N) in the bathtub
a. The residence time of nitrate equals the total amount of nitrate in the reservoir
divided by the total input (or total output) rate of nitrate.
b. Scenario 1: No photosynthesis. Assume V = 10000 m3 and O= 10 m3/sec (as
above) and Ni = 20 µmole/kg (as on pg. 10) and steady-state.
i. τ = (V*ρ*N) / O*ρo*No
ii. since No = N = Ni and ρo = ρ
1. then τ = V/O = 10000 m3 / 10 m3/sec
2. τ = 1000 secs or 16.7 mins
iii. under these conditions, the residence time of nitrate equals the residence
time of water.
c. Scenario 2: With a photosynthetic uptake rate of nitrate (B) = 51200 µmole/sec (as
in example on p.10), we measure a nitrate concentration in the tub of 15 µmole/kg
i. Then τ = (V*ρ*N)/ (O*ρo*No + B)
ii. τ = (10000m3*1024 kg/m3*15 µmole/kg)/ (10 m3/sec*1024 kg/m3*15
µmole/kg + 51200 µmole/sec)
iii. τ = 750 secs or 12.5 mins.
iv. Under these conditions, the residence time of nitrate is shorter than in
Scenario 1 (the abiotic case) because the nitrate concentration in the tub is
lower than it was without biological uptake
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g. Calculate the residence time of water in the Deep Sea.
a. Inflow rate = 35 Sv (1 Sv = 106 m3 /sec)
b. Area of ocean = 360x1012 m2. Depth of the Deep Sea = 2500m (i.e., 4000m-1500m)
The answer: 816 years. See if you can obtain this result. (1 Year = 3.15x107 sec)
-If this estimate of τ is correct, then this means that on average the water in the Deep Sea is
renewed about every millennium (~1000 years). Fairly slow, huh?
-however, not every parcel of water will have an age of 816 years
- the long residence time of the Deep Sea means that the content of the Deep Sea doesn’t
change quickly even if the input (heat, nutrients, etc.) changes year to year
D. Anthropogenic Effects on Earth’s Temperature (non steady-state situation)
1. The addition of CO2 (and other greenhouse gases) to the atmosphere, due to fossil fuel combustion
and the burning of forests due to land cultivation practices, has caused an increased retention of heat
on the earth’s surface.
-this is an example of a non steady-state situation where the loss of heat from the earth’s
surface no longer matches the input of heat from solar radiation
-since the heat loss is now less than the heat input, the temperature of the earth is increasing
2. Greenhouse gases, like CO2, trap long wave radiation from the earth’s surface and reradiate back
towards the surface (Fig. 13)
-thus there is less heat loss and the temperature of the earth’s surface increases
-Venus, with an atmosphere composed of 98% CO2, has a surface temperature of 425ºC
3. Surface temperatures on earth have increased sharply during the last century and especially the
last few decades (Fig. 14)
4. Since the surface ocean is constantly exchanging heat with the atmosphere, the increase in surface
air and surface ocean temperatures are linked
5. Climatologists have developed mathematical models that predict the surface temperature as a
function of the greenhouse gas concentrations in the atmosphere (Fig. 15)
-the predicted temperature increase when the CO2 concentration reaches about twice (~700
ppm) today’s level (380 ppm) is around 3°C (globally averaged)
-the greatest temperature increases are predicted for the polar regions in the northern
hemisphere where temps could increase by as much as 10°C (Fig. 16)
6. Presently it is not well known how the projected increase in ocean temperatures over the next
century will affect the ocean.
- Impact on circulation
-will circulation rates slow down in a more stratified ocean?
-will increase ice melting change decrease the density of surface waters in key deep
water formation regions?
- Impact on biological
-how will ecosystems respond to a warmer ocean that may circulate more sluggishly?
-how will plankton and corals respond to a warmer ocean?
-plankton can migrate but corals can’t
7. One result of the ocean’s uptake of anthropogenic CO2 is the increased acidification of the ocean
(CO2 reacts with water to form a weak acid called carbonic acid H2CO3)
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- some plankton (forams and coccolithophores) and corals, which make calcareous shells, will
have a tougher time in a more acidic ocean
III. KEY POINTS
1. One can determine the amount of water, mass, salt, heat and nutrients carried by ocean currents if
the velocity, cross sectional area, density, salinity, temperature and nutrient concentration are known.
2. If the amount of water, salt, heat or nutrients isn’t changing with time (e.g., V/t=0, S/t=0,
T/t=0, N/t=0) then the system is at steady-state with regard to that property.
3. Heat transported by ocean currents is a very important process to redistribute the heating caused
by the solar radiation distribution on earth.
4. To convert heat fluxes to oceanic temperature change (non-steady state) you need to know the
density, heat capacity of water and the volume over which the heating occurs.
5. The residence time of material and heat (water, salt, heat, nutrients, etc.) equals the total amount of
material divided by the output rate (or input rate).
6. The production of greenhouse gases, like CO2, are perturbing the surface ocean heat balance and
causing temperatures to increase. Models predict average surface ocean warming of ~3°C over the
next century, with the greatest changes in the high (polar) latitudes on the N. Hemisphere.
IV. Questions/Problems
1. Determine the time rate of change of the volume of water (ΔV/Δt) in Lake Washington if the
Cedar River inflow is 10 m3/s and the Sammamish River inflow is 5 m3/s to the lake and the outflow
from the lake (at the Ballard Locks) is 20 m3/s? Assume other water inputs are negligible. (This is a
non-steady state situation.)
-What additional information would you need to calculate the time rate of change of lake
level or depth in this problem?
2. Calculate the residence time of water in Puget Sound given a surface area of 1000 km2, mean
depth of 140m and outflow rate of 20000 m3/sec at Admiralty Inlet.
3. How is the location of a waste water treatment plant in Puget Sound affected by the water
residence time? Explain why you think it is better to have either long or short water residence times
in estuaries for ameliorating pollution problems?
4.What positive climate feedback mechanism would likely contribute to the largest temperature
increases (~10°C) being predicted for the northern polar regions over the next 100 years? What is
meant by the terms positive and negative feedbacks?
5. A 100m deep column of water with a surface area of 100 km2 in the subtropical North Pacific
adsorbs solar insolation at a rate of 120 Watts/m2, yet the temperature of the water column does not
change over time (steady-state). The currents are flowing into and out of this water column at a rate
of 1000 m3/sec. Assume ρi = ρo = 1024 kg/m3 and heat capacity = 3985 joules/(kg ºK). What is the
temperature of the outflowing water, if the temperature of the inflowing water is 20.0ºC? The
answer is 23ºC. See if you can obtain this answer.
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