Finding Depth of Water Loss Using Surface Area

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Kova Nova

Homework Statement


A city with 30,000 residents lies next a lake which is their only source of water. The average family of four in this city uses 1100 liters of water per day. Neglecting rain and evaporation, how much depth would the lake lose per year if it covered 40 square kilometers?

Homework Equations


unknown

The Attempt at a Solution


To begin with, I of course divided the 30,000 residents by 4 to gain 7,500 families living in the city. I then multiplied the number of families by 1,100 liters and got 8,250,000 L/day and then multiplied that product by 365 days to get 3,011,250,000 L/year being used.
I know that 1 decimeter^3 is equal to 1 liter and that there are 1,000,000,000,000 liters in a kilometer^3. So I attempted to raise (1x10^12 dm^3) to the power of 2/3 which in turn gave me 1x10^8 dm^2. I then converted 1x10^8 dm^2 to 1 km^2 as they are one in the same and then I divided 1 km^2 by 40 km^2 and then raised that fraction to the power of 1/2 thus giving me 0.15811 km or 158.11 m of water being lost per year.
I know at least half of this is wrong, (I just realized I didn't even implement the amount of L/year being used), I tried to assume a certain depth as well but that led me nowhere either. If someone could just push me in the right direction I would be very thankful.
 
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First, it asks how much depth will it lose... But it does not specify a time period. So either you missed typing that into the problem statement, or just specify a time period.

Next, try this. Figure how many square meters the lake is. This is area or Length2. How can you get from volume (Length3) to depth (Length1), by using area (Length2)? Hint: you do not have to raise to powers of 2/3 or 1/2.
 
scottdave said:
First, it asks how much depth will it lose... But it does not specify a time period. So either you missed typing that into the problem statement, or just specify a time period.

Next, try this. Figure how many square meters the lake is. This is area or Length2. How can you get from volume (Length3) to depth (Length1), by using area (Length2)? Hint: you do not have to raise to powers of 2/3 or 1/2.
Thank you very much, I got 7.53 cm lost/year. I'm not sure if it's correct but it makes a lot more sense than what I had before.
 
Kova Nova said:

Homework Statement


A city with 30,000 residents lies next a lake which is their only source of water. The average family of four in this city uses 1100 liters of water per day. Neglecting rain and evaporation, how much depth would the lake lose per year if it covered 40 square kilometers?
Maybe I missed the per year, the first time I read it.
I also calculated 7.53 cm per year.

Kova Nova said:
... then I divided 1 km^2 by 40 km^2 and then raised that fraction to the power of 1/2 thus giving me 0.15811 km or 158.11 m
Notice what you did here: taking (1 km2) / (40 km2) causes the km2 to cancel, giving you a dimensionless number. So that should tell you that is not the right approach.
 
Kova Nova said:

Homework Statement


A city with 30,000 residents lies next a lake which is their only source of water. The average family of four in this city uses 1100 liters of water per day. Neglecting rain and evaporation, how much depth would the lake lose per year if it covered 40 square kilometers?

Homework Equations


unknown

The Attempt at a Solution


To begin with, I of course divided the 30,000 residents by 4 to gain 7,500 families living in the city. I then multiplied the number of families by 1,100 liters and got 8,250,000 L/day and then multiplied that product by 365 days to get 3,011,250,000 L/year being used.
I know that 1 decimeter^3 is equal to 1 liter and that there are 1,000,000,000,000 liters in a kilometer^3. So I attempted to raise (1x10^12 dm^3) to the power of 2/3 which in turn gave me 1x10^8 dm^2. I then converted 1x10^8 dm^2 to 1 km^2 as they are one in the same and then I divided 1 km^2 by 40 km^2 and then raised that fraction to the power of 1/2 thus giving me 0.15811 km or 158.11 m of water being lost per year.
I know at least half of this is wrong, (I just realized I didn't even implement the amount of L/year being used), I tried to assume a certain depth as well but that led me nowhere either. If someone could just push me in the right direction I would be very thankful.

Are the sides of the lake vertical, or are they sloped? The volume vs. depth computation is different if the lake is a cylindrical hole in the ground, than if it is a cone-shaped depression, for example.
 
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Ray Vickson said:
Are the sides of the lake vertical, or are they sloped? The volume vs. depth computation is different if the lake is a cylindrical hole in the ground, than if it is a cone-shaped depression, for example.
I was thinking on similar track, but since the OP did not specify, I tried vertical walls first. Ballpark calculations in my head, I arrived at 0.2 mm per day, which is close to what the OP came up with.
 
Ray Vickson said:
Are the sides of the lake vertical, or are they sloped? The volume vs. depth computation is different if the lake is a cylindrical hole in the ground, than if it is a cone-shaped depression, for example.
Isn't that second order? I.e. it just means the rate of change would diminish year on year.
 
haruspex said:
Isn't that second order? I.e. it just means the rate of change would diminish year on year.
If it does slope as most lakes would (at least for some of the shoreline), then as level drops, the surface area will decrease. Then for the same amount of volume extracted, the water level will drop more.
 
scottdave said:
If it does slope as most lakes would (at least for some of the shoreline), then as level drops, the surface area will decrease. Then for the same amount of volume extracted, the water level will drop more.
Yes, I should have gone back to read the question exactly. I was thinking of evaporation. But my point is that the the effect of the slope is second order, so can be ignored.
 
haruspex said:
Yes, I should have gone back to read the question exactly. I was thinking of evaporation. But my point is that the the effect of the slope is second order, so can be ignored.

Yes, for one day's withdrawal that is true. But the question asked for an annual figure, so we have 365 small effects.