help please!!! Problem 7.14: The definition (7.55) of equivalent potential tempe

Question

help please!!!

Problem 7.14: The definition (7.55) of equivalent potential temper- ature assumes that all condensation results in the appearance of liq- uid water, regardless of the temperature. If freezing occurs, addi- tional latent heat is released, which further raises the temperature of the parcel. Assume that a parcel is at saturation at a temperature of freezing and a pressure of 900 hPa. a) What is the equivalent potential temperature 8, according to b) By how much would increase if all condensate were as- the traditional definition? sumed to freeze? Answer: 1.5 K

Explanation / Answer

Part-A

we know that

e = exp(Lw/(cpTLcL))

Now, = T(Po/P)^(R/Cp)

T = Freezing temperature = 0 C = 273 K

Po = 1000 hpa

P = 900 hpa

R/Cp = 0.286 for air

therefore = 273(1000/900)^(0.286) = 281.35 K

now, it is at saturation level therefore, Lw = 0

therefore, e = exp(0) = = 281.35 K

Part-B

The value of rvs at 0 C and 900 mb is about 4.3 g kg-1 = 0.0043 kg kg-1

L = latent heat of freezing = 3.34*10^5 J/kg

dq = -Ld(rvs) = cp dT

integrating we get,

change in T = -L * rvs / cp

substituting value in above equation we get,

change in T = 3.34*10^5*0.0043 / 1005 = 1.429 K

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