What is Dimensional Analysis in Atmospheric Science?

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Dimensional analysis in atmospheric science involves converting complex equations into understandable units. The discussion focuses on calculating energy fluxes, specifically Q(sub)s and Q(sub)e, which represent energy transfer across the sea surface in units of W/m^2. The equations provided demonstrate how to replace variables with their respective units, leading to the conclusion that both Q(sub)s and Q(sub)e yield units of power per square meter (J/(m^2s)). The analysis highlights that the presence of windspeed introduces a time component, resulting in power rather than work per square meter. Ultimately, the discussion clarifies the dimensional relationships in atmospheric energy transfer calculations.
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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
Q_s= \rho C_p C_H U(T_{sfc}- T_{air})
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,
Q_e= \rho L C_E U(w_{sfc}- w_{air})
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:
 
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