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