Heat Flux Calculation: August to February

In summary: In this case, the denominator is the volume of 10 meters of seawater, and the numerator is the temperature change over the six month period.In summary, the question is asking for the net heat flux lost from the ocean to the atmosphere from August to February, assuming a steady state condition. The given variables and equations allow for the calculation of individual fluxes, but not for the partitioning of radiation, conduction, convection, and evaporation heat losses. However, using the bulk equations, it can be estimated that the net heat loss is around 50W/m2.
  • #1
geojon
19
0
If mean seawater temperatures are 20C in August and 3C in February, for a patch of water with a depth of 10 meters, what is the net heat flux (in W m^-2) to the atmosphere from August to February?
 
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  • #2
Looks like homework to me; what have you done so far with it?
 
  • #3
Bystander said:
Looks like homework to me; what have you done so far with it?

I have attached a document with my work so far. And here are the variables I am using in my work.

Q = total heat flux = zero
Qs = shortwave solar radiation (150 W m^-2)
Qb = net longwave infrared radiation (50 W m^-e) and is derived from Qb,out - Qb,in
Qh = sensible heat flux (10 W m^-2)
Qe = latent heat flux (90 W m^-2)

Direction of fluxes are: Qs into the ocean ; Qb,h,e are out of the ocean

Pa = density of air (1.5 kg/m^3)
Ce = latent heat transfer coefficient (1.35 x 10^-3)
Ch = sensible heat transfer coefficient (0.9 x 10^-3
Tsst = sea surface temperature
T10 = temperature of air 10 meters above surface
U10 = wind speed at 10 meters above the sea surface
L = latent heat of evaporation (2.5 x 10^6 J/kg)
Qsst = specific humidity of air (kg water vapor/kg air)
Q10 = specific humidity of air 10 meters above surface (kg water vapor/kg air)

I should also rewrite the question to state it exactly as it is...

"Mean seawater temperatures vary greatly throughout the year in the Baltic Sea with mean water temperatures as high as 20C in August and as low as 3C in February. For a patch of water with a depth of 10 meters, what is the net heat flux lost (in W m^-2) to the atmosphere from August to February? (Ignore thermal expansion)".

Also, I am working under the assumption that to ignore thermal expansion means that the water parcel can be brought adiabatically to the surface.

The gist of it is that I am trying to move from working with a heat budget for the ocean, which involves fluxes in and out, to evaluating just those parameters which have a direction of out of the ocean and into the atmosphere. I think that I have a decent start, so far..

Again, attached is where I am at at this point
 

Attachments

  • Heat Flux.pdf
    63 KB · Views: 212
  • #4
My IE doesn't want to open pdfs, and I don't feel like fighting it tonight. From your variable list, you're prepared to attack the problem in detail; from the problem statement (unless there were verbal instructions given in class), "20 C in August and 3 C in February for ten cubic meters of sea water, find the net heat flux," you've not been given sufficient details to partition radiation, conduction, convection, and evaporation heat losses. If you've been asked to write the general expression for energy transfer, the variable list looks reasonably complete, and I'll take it on faith that you've used them correctly in the pdf; if you've been asked to integrate your expression for a meteorological data set for some past late summer, fall, and winter, I ain't here.

If my suspicion that the exercise is aimed at generating a "mean net heat loss rate to atmosphere plus CMB over the August to February time period" is correct, you should come up with something in the neighborhood of ~ 50W/m2.
 
  • #5
Assessing the conductive (sensible heat) loss, without direct measurement of parameters like the temperature of air 10 meters above the sea surface, is impossible. I do have some lecture notes that I can look through to see if there is a value somewhere for this, though.

Yes, your suspicion is correct; I am trying to calculate the net heat loss from the ocean to the atmosphere from August to February. CMB is not needed. And yes, I am only given the above variables ans bulk flux formulas for (a) net surface flux (b) sensible heat flux and (c) latent heat flux.
 
  • #6
Whoops! Looking back at your list of variables, "Q = total heat flux = zero" might be worth re-examination, given the temperatures listed for August and February.
 
  • #7
Q = Qs - Qb - Qh - Qe = zero is to say that solar radiation (Qs) which is a flux into the ocean is equal to infrared longwave radiation + sensible heat + latent heat (Qb + Qh + Qe) which are fluxes out of the ocean.

There are also bulk equations for Qh and Qe, however certain components of them, such as air temperature 10 meters above the surface, and, dew point 10 meters above the surface I do not have values for, and do not know how to find from the given list of variables.

So, at this point I am shooting for partial credit. However, I would love to solve it, as it's a very interesting question, and it would be neat to find an acceptable value to quantify the measure.
 
  • #8
The water column has cooled by 17 degrees over a time period of six months. How much heat has been lost from the water column? Sorry I didn't catch on earlier that you were using "Q = zero" for a "steady state" condition (no change in temperature of the water column) rather than Q = del T x CP/(6 mo.).
 
  • #9
geojon said:
So, at this point I am shooting for partial credit.
Why are you shooting for partial credit? I think you have enough information to answer the question as stated in #3...
 
  • #10
Bystander said:
The water column has cooled by 17 degrees over a time period of six months. How much heat has been lost from the water column? Sorry I didn't catch on earlier that you were using "Q = zero" for a "steady state" condition (no change in temperature of the water column) rather than Q = del T x CP/(6 mo.).
I have been working from the bulk equations. They exist for Q, Qs and QL.

delQ = delT / [(density of seawater) * (height of water column)] = 17C / [(1,030 kg/m3 * 10 meters)] = 0.00165 (units here are difficult to figure out).

Q = delT * CP / 6 months = 17C * 1.01 J/gC = 17.17 J/g in 6 months.

These (above) are not bulk equations, but was given to me by a classmate to use. The latter (when looking at the value you get) seems really only to say how much the water changed in temperature over 6 months, as the number is very close to 17.
 
  • #11
olivermsun said:
Why are you shooting for partial credit? I think you have enough information to answer the question as stated in #3...
delQ = delT / [(density of seawater) * (height of water column)] = 17C / [(1,030 kg/m3 * 10 meters)] = 0.00165 (units here are difficult to figure out).

Q = delT * CP / 6 months = 17C * 1.01 J/gC = 17.17 J/g in 6 months.

These (above) are not bulk equations, but was given to me by a classmate to use. The latter (when looking at the value you get) seems really only to say how much the water changed in temperature over 6 months, as the number is very close to 17.
 
  • #12
Where did you get the Cp of 1.01 J/g °C from?

Also, can you explain your units confusion?
 
  • #13
I got the CP value from my lecture notes.

My confusion with units is that the calculation leaves me with C -kg-m2
 
  • #14
I think the specific heat for seawater should be closer to 4 J/g °C.

But anyway it should not be a problem, units-wise. The question is asking for the net surface heat flux, which is in W/m2 = J/s/m2. You have a specific heat in J/g °C (or kJ/kg °C,) you have density in kg/m3, and a thickness of the water in m. Should work out.
 
  • #15
What is the mass of water contained in a column of water 1 m2 in cross section and 10 m high? How much heat has to be removed from this column of water to cool it 17 C? If it takes 6 months to remove this amount of heat, what is the average rate of heat loss? What is the average rate of heat loss per unit area (at the top of the column)?

Chet
 
  • #16
geojon said:
delQ = delT / [(density of seawater) * (height of water column)] = 17C / [(1,030 kg/m3 * 10 meters)] = 0.00165 (units here are difficult to figure out).

Q = delT * CP / 6 months = 17C * 1.01 J/gC = 17.17 J/g in 6 months.

These (above) are not bulk equations, but was given to me by a classmate to use. The latter (when looking at the value you get) seems really only to say how much the water changed in temperature over 6 months, as the number is very close to 17.

I took it on faith --- my bad.

So, one (mis)step at a time problem solving proceeds as follows: 1) read the problem; 2) identify the quantity/information to be determined; 3) sort given information for what is and what ain't relevant; 4) understand (units, equations, physical principles, etc.) the given information; 5) translate units as necessary (months to seconds, power to W to J/s or energy per unit time, energy to CPdelT, CP to mass x sp. ht., calories to joules) so that a uniform set is present throughout whatever equations or expressions are used to find an answer to the original question; 6) solve; 7) check steps 1 through 6.

Someone moved this to Elem. Phys. Homework --- should be in other sciences; it is, however, useful to illustrate differences in nomenclature between physical and Earth sciences. For instance, "Q" in the given list of variables is more correctly written "dQ/dt" or "Q dot," the time rate of energy transfer, or, "power." When I write "CP," I am indicating the constant pressure heat capacity for the system (ten cubic meters of Baltic surface water), units of J/K, not specific heat (cal/gK, or J/gK) or molal/molar heat capacity.

The problem statement gives a time, a mass, and an observed temperature difference for a ten cubic meter mass of sea water; one proceeds by calculating time in seconds and energy in joules to produce the temperature change, and gets a mean value for total power applied to the system. That total power is the sum of radiation input and radiation loss from and to the sky beyond the atmosphere, and losses to the atmosphere through conduction, convection and evaporation
(the answer being sought).

My apologies for failing to hold your feet to the fire for a specific question from you at the beginning of this thread, for failing to quiz you on your fluency with units, and for letting you get away with quoting a textbook and lecture notes as evidence of a higher level of understanding of the material than actually exists.

"Apology:" noun---a promise that the same infraction/insult/offense WILL be repeated.
 

Related to Heat Flux Calculation: August to February

What is heat flux?

Heat flux is the rate of heat transfer per unit area. It is a measure of how much heat is moving through a given area in a certain amount of time.

How is heat flux calculated?

Heat flux is calculated using the formula q = -k * (∆T/∆x), where q is the heat flux, k is the thermal conductivity of the material, ∆T is the temperature difference, and ∆x is the distance over which the heat is being transferred.

What is the significance of calculating heat flux from August to February?

Calculating heat flux from August to February provides information on the seasonal variation in heat transfer. This is important in understanding the energy balance in a given region and can be used to predict future climate patterns.

What factors affect heat flux?

The factors that affect heat flux include the thermal conductivity of the material, the temperature difference across the material, the distance over which the heat is being transferred, and the presence of any insulating materials.

How is heat flux used in practical applications?

Heat flux calculations are used in a variety of practical applications, such as designing energy-efficient buildings, predicting crop growth and yield, and understanding the effects of climate change. They are also used in industries such as aerospace and automotive to design and optimize thermal management systems.

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