water budget problem

W

willowthewisp

New member
Joined
Aug 25, 2013
Messages
5
Hi folks, this is my first post and I'm not sure I picked the right forum but I have a surface water budget word problem that I am having trouble with solving. Any help would be greatly appreciated.

"a lake has a surface area of 200 km2. Annual precipitation rates are 250 mm (assume no infiltration) and evapotranspiration rates are 95 mm annually. Stream inflow is 1690 ft3/sec and stream outflow is 921 ft3/sec. groundwater seepage into the lake is 1 cm/day. Determine how much water annually can be diverted from the lake for agriculture to prevent long-lasting water level changes in the lake."

the equation to be used is: S = P + Qin +Qg - Qout - Es - Ts - I.

S is change in storage, P is precipitation, Qin is stream inflow, Qg is groundwater to stream, Qout is stream outlfow, Es is evaporation, Ts is transpiration and I is infiltration.

I need my answer to be in cm/year.

thanks for looking.
 
willowthewisp said:
Hi folks, this is my first post and I'm not sure I picked the right forum but I have a surface water budget word problem that I am having trouble with solving. Any help would be greatly appreciated.

"a lake has a surface area of 200 km2. Annual precipitation rates are 250 mm (assume no infiltration) and evapotranspiration rates are 95 mm annually. Stream inflow is 1690 ft3/sec and stream outflow is 921 ft3/sec. groundwater seepage into the lake is 1 cm/day. Determine how much water annually can be diverted from the lake for agriculture to prevent long-lasting water level changes in the lake."

the equation to be used is: S = P + Qin +Qg - Qout - Es - Ts - I.

S is change in storage, P is precipitation, Qin is stream inflow, Qg is groundwater to stream, Qout is stream outlfow, Es is evaporation, Ts is transpiration and I is infiltration.

I need my answer to be in cm/year.

thanks for looking.
Click to expand...
Most of the terms are in depth per time, which can be converted to cm/year. The two exceptions are stream inflow and outflow, netting +769 ft^3/s. Convert ft^3 to metric volume units, perhaps m^3. To convert to depth units, divide by the area, 200 km^2 (or 2.00×10^8 m^2). You also have to calculate how many seconds in a year. Once you get all the terms converted to cm/year, add 'em up.
 
Okay, thanks for the help.

this is what I got and I'd be thankful if someone can confirm this for me.

My conversions
250 mm/yr = 25 cm/yr
95 mm/yr = 9.5 cm/yr
1 cm/day = 365 cm/yr
(this is the conversion I am not too sure about) 1690 ft3/sec - 921 ft3/sec = 769 ft3/sec. 769 ft3 = 21.8 m3. 21.8 m3/200 km2 = 109 nanometers. 109 nanometers = .0000109 cm. .0000109 cm * 31536000 seconds per year = 343.7 cm/yr.

So then my equation in all cm/yr is: 25 cm/yr + 343 cm/yr + 365 cm/yr - 9.5 cm/yr. And for a final answer I get 724.2 cm/yr.

Does this final answer make sense and look correct?
 
willowthewisp said:
Okay, thanks for the help.

this is what I got and I'd be thankful if someone can confirm this for me.

My conversions
250 mm/yr = 25 cm/yr
95 mm/yr = 9.5 cm/yr
1 cm/day = 365 cm/yr
(this is the conversion I am not too sure about) 1690 ft3/sec - 921 ft3/sec = 769 ft3/sec. 769 ft3 = 21.8 m3. 21.8 m3/200 km2 = 109 nanometers. 109 nanometers = .0000109 cm. .0000109 cm * 31536000 seconds per year = 343.7 cm/yr.

So then my equation in all cm/yr is: 25 cm/yr + 343 cm/yr + 365 cm/yr - 9.5 cm/yr. And for a final answer I get 724.2 cm/yr.

Does this final answer make sense and look correct?
Click to expand...
Looks good. Don't include the fraction .. just 724 cm/yr ("significant digits").
 
You must log in or register to reply here.
Top