# Thread: A problem i'm having trouble with

1. Hey guys i am currently in Uni studying civil engineering and i'm having trouble with one of the question and would really appreciate some help.

This is the question;

If the net radiation in a day is 20 MJ/m^2 and the latent heat lost with evaporation is 15MJ/m^2, a well mixed 2m deep lake will heat up in one day by (choose the closest value):
a) 0.6 degrees Celsius
b) 2.39 degrees Celsius
c) 1.20 degrees Celsius
d) 0.006 degrees Celsius
e) 0.0005 degrees Celsius
(assume conductive heat fulx (Q)=0; Cp = 4.18 KJ/kg/degrees Celsius and pw=1000kg/m^3)

These are the formula i know;

A.Rn =dmv/dt*(hlg) +dT/dt*Cp m ± Q

and

dmv/dt = B ρw (Pvap,sat − Pvap )A

I can't seem to figure it out since they haven't given you the Temperature.

Any help is appreciated. Cheers  2.

3. Originally Posted by rushil01
These are the formula i know;

A.Rn =dmv/dt*(hlg) +dT/dt*Cp m ± Q
This is the one you need.

Q=0 (given)
dmv/dt*(hlg)=0 (I suppose, I'm not entirely sure what this is, but you don't need it)
Rn= incoming power per area - power going to evaporation per area (again, assuming this is what the symbol means)

rewrite: divide both parts by surface A with d being the depth of 2 m
you know all variable except , just fill them in.  4. Cheers for the reply. I was making such a small error however it was totally throwing me off. I got the answer to be B which i'm hoping is correct. Also just so you know dmv/dt is the evaporation rate and Hlg is the latent heat lost with evaporation.

I'm guessing that dmv/dt is zero because i'm assuming that all the energy goes into increasing the temperature of the water.  5. Ah, in that case, my assumptions about dmv/dt*(hlg) and Rn where wrong. dmv/dt*(hlg) isn't zero at all, since question clearly states that the evaporation energy of one day is 15 MJ/m^2. I put all of it in Rn, but that's not the case.    6. hmmm in that case the answer would be a) which is what my mate got. Cheers for your help Bender  Bookmarks
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