Question

Lake Superior, the largest of the Great Lakes of North America, is also the world's largest freshwater lake by surface area (~ 82,100 km2), and the third largest freshwater lake by volume (~ 12,100 km3). The monthly average precipitation in 2018 in the lake area was 2.69 inches. According to the data provided by USGS water stations, the total flow rate in all the incoming streams was found to be 356,747 ft3/s (cubic feet per second). At the same time, the lake discharges 717,258 ft3/s to its surrounding water bodies. The monthly average evaporation was 18.7 mm. The groundwater inflow was 783.33 km3 less than the groundwater outflow in this year. (Hints: (1) The average monthly data can be used to calculate the annual data; (2) Volume = Area * Height; (3) Pay special attention to the units, convert all the units to be consistent first.)

(1) With the information above, please write the water budget for Lake Superior (10 points);
(2) Please estimate the change of storage (in m3) in the year of 2018 (20 points);
(3) Please estimate the increase or decrease of water level (in mm) in the year of 2018 (10 points).

Answer 1

1) V_win = 7.05 10¹⁰ m³ , 2)V_ lost = 1.8471 10¹⁰ m³, 3) h = 633.4 mm

Step-by-step explanation:

In this exercise we will start by reducing all units to the SI system

h1 = 2.69 inch (2.54 10-2 m / 1 inch) = 6.8326 10⁻² m

Ф1 = 356.747 ft3 / s (1 m3 / 35.3147 ft3) = 10.1019 m³ / s

Ф2 = 717,258 ft3 / s = 20,310 m³ / s

h2 = 18.7 mm (1 m / 1000 mm) = 18.7 10⁻³ m

V_subterránea = - 783.33 10³ m,

Now let's answer the questions

1) they ask us the amount of water that reaches the lake in 2018

volume of rainwater is

V₁ = A. h t

V₁ = 82 100 106 * 6.8326 10-2 * 12

V₁ = 6.7315 10 10 m3

stream water volume in a year

V₂ = Ф₁ t

t = 1 year (365 days / year) (24 h / 1 day) (3600 s / 1h) = 3.1536 10⁷ s

V₂ = 10.1019 3.1536 10⁷

v₂ = 31,857 10⁷ m³

The total water volume is

V_win = V₁ + V₂

V_win = 6.7315 10¹⁰ + 31.857 10⁷

V_win = 7.05 10¹⁰ m³

2) let's find the amount of water lost from the lake

volume of water to surrounding bodies

V₃ = Ф₂ t

V₃ = 20.310 3.1536 10⁷

V₃ = 6.40496 10 8 m3

Volume of evaporated water

V₄ = A h₂ t

V₄ = 82 100 10⁶ 18.1 10⁻³ 12

V⁴ = 1,783 10¹⁰ m³

groundwater volume

V⁵ = 7.83.33 10³ m³

The volume of water lost is

V_lost = V₃ + V₄ + V₅

V_ lost = 6.40496 10⁸ + 1.783 10¹⁰ + 783.33 10³

V_ lost = 1.8471 10¹⁰ m³

3) the change in the height of the lake

the change in volume is

ΔV = V_ won - V_ lost

ΔV = 7.05 10¹⁰ - 1.8471 10¹⁰

ΔV = 5.20 10¹⁰ m³

the volume is

v = A h

h = V / A

h = 5.20 10¹⁰/82100 10⁶

h = 6.334 10⁻¹ m

h = 633.4 mm