What is Dimensional Analysis in Atmospheric Science?

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SUMMARY

This discussion focuses on dimensional analysis in atmospheric science, specifically calculating energy fluxes using the equations for sensible heat flux (Q(sub)s) and latent heat flux (Q(sub)e). The final units for both Q(sub)s and Q(sub)e are established as W/m², indicating power per square meter. Key variables include wind speed (U), temperature (T), and water vapor mixing ratio (w), with specific constants such as ρ (1.023 kg/m³) and L (2.5 x 10^6 J/kg) being utilized in the calculations. The analysis confirms that the derived units represent power rather than work, clarifying the relationship between energy transfer and time.

PREREQUISITES
  • Understanding of dimensional analysis in physics.
  • Familiarity with atmospheric science concepts, particularly energy fluxes.
  • Knowledge of thermodynamic properties such as specific heat (C(sub)p) and latent heat (L).
  • Basic proficiency in manipulating units and equations in scientific contexts.
NEXT STEPS
  • Explore the principles of dimensional analysis in fluid dynamics.
  • Study the impact of wind speed on energy transfer in atmospheric models.
  • Learn about the role of specific heat and latent heat in climate systems.
  • Investigate the applications of energy flux calculations in meteorology and oceanography.
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Atmospheric scientists, meteorologists, and students studying environmental science who seek to deepen their understanding of energy transfer processes in the atmosphere.

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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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