OC210: Ocean Circulation:
Answers to questions at the end of Topic 1B
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?
ΔV/Δt = Inflow – Outflow = 10 m3/s + 5 m3/s – 20 m3/s = -5 m3/s
-You would need the surface area of the lake (As), which for Lake Washington is 90 km2.
Then you could calculate the time rate of change in the lake level (depth).
ΔZ/Δt = ΔV/Δt / As = [-5 m3/s * 3.14x107 sec/year] / 90x106 m2 = -1.7m/yr. Is this a rapid or
slow rate? (Hint: the mean depth of L. Washington is about 25m)
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.
Res. time = Volume/Outflow = (surface area*mean depth)/Outflow
= (1000 km2 * (106m2/km2) * 140m)/ 20000 m3/s
= 7x106 sec * (1 year/3.15x107 sec)
= 0.22 years or 2.7 months
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?
Generally, you would want to discharge wastewater into an estuary where the water
residence time is short rather than long, assuming the less time the wastewater spends in the
estuary, the better. For example, the discharge of nutrients from septic tanks and agricultural
runoff into Hood Canal is thought to be one factor in the recent increase in fish kills due to
low oxygen levels. Thus in Puget Sound it would be better to have the wastewater discharge
plants located in the northern portion of the Sound versus the southern portion of the Sound,
because of the water residence times are shorter as you get closer to the site of the
inflow/outflow at Admiralty Inlet.
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?
The melting of sea ice and snow on land reduces the reflectivity (albedo) of the earth’s
surface and causes the earth to adsorb more solar insolation which enhances the warming
effect. This is an example of a positive feedback. For example, the earth is warming due to
greenhouse gases and the melting of ice and snow accelerates this warming. In contrast, a
negative feedback would be a situation where a response of the system slowed or decelerated
the warming.
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.
Use a heat budget to solve this problem.
Δ(V*ρ*c*T)/Δt = I*ρi*c*Ti – O*ρo*c*To + Solar Heating
You know the following:
Z=100m, As=100 km2, Solar Heating = 120 Watts/m2, Inflow=Outflow= 1000 m3/sec,
density=1024 kg/m3 , heat capacity = 3985 Joules/(kg ºK) and T of inflow = 20ºC. The
system is at steady-state.
You want to determine the temperature of the Outflow (To).
To start, the time rate of change = 0.
Δ(V*ρ*c*T)/Δt = I*ρi*c*Ti – O*ρo*c*To + As*Solar Heating = 0
Rearrange the equation to solve for To:
To = (As*Solar Heating + I*ρi*c*Ti) / (O*ρo*c)
First, let’s check the units:
To=(m2*Joules/(m2 sec)+m3/sec*kg/m3*Joules/(kg ºK)*ºK)/(m3/sec*kg/m3*Joule/kg ºK)
The units look good since To is in ºK.
Let’s calculate the value for To
To = (100 km2 *(106 m2/km2)* 120 J/(m2 s)+ 1000 m3/s*1024 kg/m3*3985 J/(kg
ºK)* (20+273 ºK) / (1000 m3/s *1024 kg/m3 * 3985 J/(kg ºK)
To = 295.94 ºK = 22.94 ºC