1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Dimensional Analyis Review

  1. Feb 21, 2006 #1
    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.
     
  2. jcsd
  3. Feb 21, 2006 #2

    HallsofIvy

    User Avatar
    Science Advisor

    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.
     
    Last edited by a moderator: Feb 21, 2006
  4. Feb 21, 2006 #3
    Thank you!

    It makes more sense know... just an ice breaker.:redface:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...
Similar Threads for Dimensional Analyis Review Date
1-Dimensional Distance with Drag Mar 5, 2018
Dimension of length using h,G,c Dec 26, 2017
Dimensional analysis problem Nov 17, 2017
Physical Chem review Jan 22, 2017