What is Dimensional Analysis in Atmospheric Science?

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In summary, the subscript notices come before the variables in the equations for the dimensional analysis problem. The equations are: a)Q(sub)s= ρC(sub)PC(sub) H U(T(sub)sfc - T(sub) air) b)Q(sub)e= ρLC(sub)E U(w(sub)sfc-w(sub)air)
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It's been three semesters since I was last required to do a dimensional analysis problem, and I could use a little help here. Just know that the subscript notices come before the variables. In this case, the final set of units must work out to be W/m^2. The equations are:

a)Q(sub)s= ρC(sub)PC(sub) H U(T(sub)sfc - T(sub) air)

b)Q(sub)e= ρLC(sub)E U(w(sub)sfc-w(sub)air)

Note:
C(sub)H =C(sub)E =1.5*10^-3 (dimensionless)
U=windspeed (m/s) at 10 m height
T= temperature in Kelvin (K)
w= water vapor mixing ratio (g/kg)
L= 2.5*10^6 J/kg
ρ= 1.023 kg/m^3
C (sub) p= specific heat of air pressure = 1004 J/(kgK).
Note: Final energy fluxes Q(sub)s and Q(sub)e have units of W/m^2 and are a measure of the amount of energy being transferred across the sea surface per unit time. Recall that W= J/s.
Note: Values T(sub) sfc refer to sea-surface air layer and T(sub) air assumes a height of 10 m in the boundary layer.
 
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I think you are saying
[tex]Q_s= \rho C_p C_H U(T_{sfc}- T_{air})[/tex]
with:
CH =CE (dimensionless)
U=windspeed (m/s) at 10 m height
T= temperature in Kelvin (K)
w= water vapor mixing ratio (g/kg) ?? If that's "grams/kg" then it is dimensionless.
L= J/kg
ρ= kg/m^3
Cp= J/(kgK).
Okay, taking a deep breath and jumping right in replacing each quantity by its units:
Qs= (kg/m3)(J/(kgK))(m/s)(K)
Well, that's not so bad! I see that "K"s and "kg"s will cancel and the "m" in the numerator cancels one of the "m3" in the denominator leaving "m2" in the denominator:
Qs= J/(m2s).

You say you want units of "Work/ m2" and J (Joule's) is work, but we have that "s" still in the denominator. I don't see how to avoid that- windspeed is the only thing you have there that depends on time. What you wind up with is POWER per square meter (Watts per square meter) rather than WORK per square meter.

For the second one,
[tex]Q_e= \rho L C_E U(w_{sfc}- w_{air})[/tex]
we have (kg/m3)(J/kg)(m/s) and again, "kg" cancels and "m" cancels with "m3" to leave "m2 in the denominator. This is again J/(m2s), a power per square meter unit.
 
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Thank you!

It makes more sense know... just an ice breaker.:redface:
 

Related to What is Dimensional Analysis in Atmospheric Science?

1. What is Dimensional Analysis?

Dimensional Analysis is a mathematical technique used to convert one unit of measurement to another by using conversion factors and canceling out unwanted units. It is commonly used in scientific and engineering fields.

2. Why is Dimensional Analysis important in science?

Dimensional Analysis is important because it helps ensure the accuracy and consistency of experimental data and calculations. It also allows for easy conversion between different units of measurement, making it easier to compare and analyze data.

3. How do you perform Dimensional Analysis?

To perform Dimensional Analysis, you must first identify the units you are converting from and to. Then, you multiply the given value by conversion factors until the unwanted units cancel out, leaving you with the desired unit. Finally, you check to make sure the units are correct and the conversion is accurate.

4. Can Dimensional Analysis be used for any type of unit conversion?

Yes, Dimensional Analysis can be used for any type of unit conversion as long as the units are related by a conversion factor. It is a universal technique that can be applied to various types of measurements, such as length, time, volume, and more.

5. How do you know if your Dimensional Analysis calculation is correct?

You can check the accuracy of your Dimensional Analysis calculation by making sure the units cancel out correctly and that the final unit matches what you were trying to convert to. Additionally, you can double-check your work by using a calculator or consulting a conversion chart.

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