Homework No. 6 - Solution

1. Consider a homogeneous snowpack of 0.6 m depth and density 0.5 g/cm3 and which is in thermodynamic equilibrium at a temperature of 271.15 ēK.

a) Compute the water equivalent depth of the snowpack, the cold content in units of energy per unit area, and the thermal quality of the snowpack.

Assume that rain having a temperature of 275.15 ēK begins to fall at a constant rate of 0.3 cm/hr.

b) How long will it be before snow melt begins?
c) What is the water equivalent of the snowpack at the beginning of snowmelt?
d) How long will it be before runoff from snowmelt begins?
e) How long will it be before runoff from rainfall begins?

The specific heat of snow, c ps , is 0.5 cal/g/ēK, that of liquid water c p , is 1.0 cal/g/ēK and the latent heat of fusion of ice is L f = 80 cal/g. The maximum water holding capacity, Wmax, is 0.035 for the conditions given. Assume a seepage velocity, v s , of 18 cm/h. Also, assume that the only important heat fluxes are those associated with advection and latent heats.

> restart;with(student);

Define cold content function, Hcc :

> H[cc]:=Ts->rho[s]*C[ps]*d[s]*(273.15-Ts);

Define cold content in units of equivalent depth of melt, dcc

> d[cc]:=Ts->H[cc](Ts)/L[f]/rho[w];

Define latent heat of fusion of snow, Lfs

> L[fs]:=L[f]*(1-(W[max]-W[def]));

Define the total heat deficit of the snowpack, Hdef

> H[def]:=Ts->H[cc](Ts)+rho[s]*d[s]*L[fs];

Define Thermal Quality function

> theta:=Ts->L[fs]/L[f]+C[ps]*(273.15-Ts)/L[f];

Define Advection Heat (i.e. heat added by the precipitation), Hp . The temperature of precipitation is Tr and the total depth of precipitation is dr .

where dr is the total precipitation depth over the given time interval.

> H[p]:=(Tr,Ts)->rho[w]*C[pw]*d[r]*(Tr-Ts);

Define Latent Heat (resulting from the freezing of the rain water):

where dr is the total precipitation depth over the given time interval.

> H[e]:=rho[w]*L[f]*d[r];

Define depth of water required to supply the Cold Content. In this case, energy will be supplied from the advection of energy by rain and from the release of latent heat by freezing rain. That is, Hcc = He + Hp , where and . Use the resulting equation to solve for dp .

dr will be equal to time delay multiplied by the rain rate.

> d[r]:=(Tr,Ts)->(rho[s]*C[ps]*d[s]*(273.15-Ts))/(rho[w]*C[pw]*(Tr-Ts)+rho[w]*L[f]);

Define time delay before snowmelt begins assuming a rainfall intensity of ir .

> t[c]:=(Tr,Ts)->d[r](Tr,Ts)/ir;

Define water equivalent at the time when cold content is satisfied. In this case, because additional heat is being supplied by advection, the total depth of rain water is not equal to dcc but to dr , which is smaller that dcc .

> d[eq]:=(Tr,Ts)->(rho[w]*d[r](Tr,Ts)+rho[s]*d[s])/rho[w];

Define water deficit at the beginning of snowmelt

> d[def]:=(Tr,Ts)->W[max]*d[eq](Tr,Ts);

Defien time delay between the beginning of snowmelt and fully ripe conditions:

where m is the melt rate. Thus, we need to compute the melt rate. After the pack is at 0 C, the only heat source is the advection heat. Thus, the melt rate is:

Since the cold content has already been satisfied, the temperature of the snowpack is 273.15 K.

> m:=Tr->rho[w]*C[pw]*ir*(Tr-273.15)/rho[w]/L[f];

> t[s]:=(Tr,Ts)->d[def](Tr,Ts)/(ir+m(Tr));

Define input data

> L[f]:=80;C[ps]:=0.5;d[s]:=60;rho[s]:=0.5;W[max]:=0.05;W[def]:=0.05;rho[w]:=1;C[pw]:=1;v[s]:=18;ir:=.25;Tss:=271.15;Tpp:=275.15;

> H[cc](Tss);

> d[cc](Tss);

> d[eq](273.15,273.15);

> theta(Tss);

> L[fs];

> H[def](Tss);

> H[p](Tpp,Tss);

> H[e];

> d[r](Tpp,Tss);

> t[c](Tpp,Tss);

> d[eq](Tpp,Tss);

> d[def](Tpp,Tss);

> t[s](Tpp,Tss);

> t[t]:=(Tr,Ts)->(d[s]-m(Tr)*t[s](Tr,Ts))/v[s];

> t[t](Tpp,Tss);