Dear Ignacio,

One way is to do it in the following different approach

Normally Hcold = Rn – G – 1.05λ ETr

But using your cold pixel you can do the following

dTcold = Hcold × rah / (r × cp)

Hcold = dTcold X (r × cp)/ rah

where: r is air density (kg/m3), cp is air specific heat (1004 J/kg/K), dT (K) is the

temperature difference (T1 – T2) between two heights (z1 and z2), and rah is the

aerodynamic resistance to heat transport (s/m).

r =1000 P/ 1.01(TS – dT)

r is air density (kg/m), P is mean atmospheric pressure, R is the specific gas

constant (287 [J kg/K]), Ts is the surface temperature (K) for a pixel, and dT is the near surface air temperature difference (K)

 rah = ln(z2/z1)/u* x k

where: z1 and z2 are heights in meters above the zero plane displacement (d) of the

vegetation, u* is the friction velocity (m/s) which quantifies the turbulent velocity fluctuations

in the air, and k is von Karman’s constant (0.41).

u* =kux/ln(zx/zom)

k is von Karman’s constant, ux is the wind speed (m/s) at height zx, and zom is the

momentum roughness length (m).

zom = 0.12h

where: h is the vegetation height (m).

Thanks for the answer!

However, I'm still having a doubt about this. In the approach that you mention (dTcold = Hcold × rah / (r × cp) and Hcold = dTcold X (r × cp)/ rah), I don't know the air temperature at my cold pixel. In fact, one of the premises of the model is that the temperature difference between the two heights is unknown (dT). Actually, I should know this difference in my meteorological station, but not in my cold pixel, and that is the reason for the use of the equation that I mention, in order to estimate sensible heat flux and then, estimate dT.

Maybe I'm wrong...

However, I still can't figure out how to obtain a proper H value or a LE value in this pixel. In this case I am computing the values before to continue with the iterative correction process (Monin-Obukhov) and if I think the issue, those initial values are always going to be way too high because of the λ (latent heat of vaporization) value, which is:

λ=(2.501−0.00236(Ts−273.15))10^6 (j/kg)

As an example of the calculation:

for Ts = 10°C; λ=2,477,400 j/kg

for Ts = 20°C; λ=2,453,800 j/kg

Therefore, the LE = 1.05λ*ETr, at a very small ETr of 0.2 mm/h is going to be between 515,298  and 520,254 W/m2, and H should retrieve something around -514,000 to - -519,000 W/m2, considering that the range of Rn is between 100-700 W/m2, and G is between 0.05Rn to 0.15Rn in agricultural areas.

I'm not sure if this make any sense, but in most of the literature the equation is mentioned without detailing very much about this, and the final values that they obtain for H or LE are not higher than 1000 W/m2.

Regards

I think that LE ( 1.05λ ETr ) in cold pixel must be divided by 3600 because the ETr value is calculated in hourly time step and the energy fluxes are instantaneous values (W/m2/s).

Regards.

1 Watt /m2 = 0.0864 MJ /m2/day

1 MJ /m2/day  =0.408 mm /day

For a ETr of 0.2mm/h, LEcold in METRIC should be around ~ 143 W/m2 (1.05*0.2*681.6, where 681.6 is the scale factor to convert mm/h to W/m2). In SEBAL, LEcold is typically taken as Rn-G (i.e. H = 0).