can some one help me on this math problem. the book is Hydrology and floodplain analysis, third edition.?

Question

this is the question exactly the way it is printed in the book

for a given month, a 300- acre lake has 15 cfs of inflow, 13 cfs of outflow, and a total storage increase of 16 ac-ft. a USGS guage next to the lake recorded a total of 1.3 in of precipitation for the lake for the month. assuming that infiltration loss is insignificant for the lake, determine the evaporation loss, in inches, over the lake for the month.
SOLUTION- solving the water balance for inflow I and outflow O in a lake gives, for evaporation
E = I - O + P - change in S
E= evaporation
I= inflow
O= outflow
P= precipitation
change in S= change in storage

so can anyone help me solve the problem, and/or tell me the best way to set up and work problems similar to this

Answer 1

This is a mass balance problem ... calculate the volumes.
Assume a 30 day month.
I am also assuming that you know about mass balance.

E = I - O + P - change in S

.........^-Evap
13cfs.|................. 15cfs
<-----[300 Ac lake]<---
..1.3" Precip -^

Start with I - O = 15cfs - 13cfs = 2cfs
For the month:

... 2 ft^3/s * (3600s/hr)(24hr/day)(30days)
= 5,184,000 ft^3 net inflow.

Now, Precipitation volume (depth * area):
1.3in. (1ft/12in) * (300Ac)(43,560 ft^2/Ac) = 16,988,400 ft^3 Precipitation.

Storage increase : 16 ac-ft. * (43,560 ft^2/Ac) = 696,960 ft^3

So, E = 5,184,000 ft^3 + 16,988,400 ft^3 - 696,960 ft^3

E = 21,475,440 ft^3

Depth over the area of the lake = Vol / Area

= 21,475,440 ft^3 / (300Ac * (43,560 ft^2/Ac))
= 21,475,440 ft^3 / 13,068,600 ft^2
= 1.64 ft = 19.7 in.

You should check my arithmetic, but you get the idea...

Answer 2

It's actually pretty straightforward, but the units are not what you are accostomed to. The equation is a water balance- what comes in = what goes out + increases in the lake + evaporation. What goes in-what goes out = 2 cfs * 86400 sec/day * 30 days/mo. That is I-O in ft^3. Now, the acre-foot is one acre of area covered by 1 ft of water or about 43500 ft^3. So the change in storage is 300*16*43500 in ft^3. The input from precip is
(1.3/12)*300*43500 in ft^3. Now you can sub and crank out the answer.