Answer:
[tex]Q_{down}=8 * 10^4 \:\:m^3/year[/tex]
[tex]C_{down}=5.4\:\:mg/L[/tex]
Step-by-step explanation:
[tex]Q_{down}=Q_{in}-Q_{evap}=9 * 10^4-1 * 10^4=8 * 10^4 \:\:m^3/year[/tex]
[tex]V\frac{dC}{dt} =Q_{up}C_{up}-Q_{evap}C_{evap}-Q_{down}C_{down}-kVC\\\\where \:\:\:C_{down}=?,\:C_{up}=C=6\:\:mg/L\:\:\:and\:\:\:\frac{dC}{dt}=0,\:C_{evap}=0[/tex]
[tex]0=Q_{up}C-(Q_{down}+kV)C_{down}\\where\:\:\:V=2*10^5\:\:m^3 \:\:\:and\:\:\:k=0.10/year[/tex]
[tex]C_{down}=\frac{Q_{up}C}{Q_{down}+kV}=\frac{9*10^4*6}{8*10^4+0.10*2*10^5}=5.4\:\:mg/L[/tex]
Answer:
Qdown = 0 m³/year
C = 27 mg/L
Step-by-step explanation:
Given
V = 2*10⁵ m³
Qin = 9*10⁴ m³/year
Cin = 6 mg/L
Qevap = 10⁴ m³/year
K = 0.10/year
Qdown = ?
C = ?
Take volume V of the entire lake as the control volume. If we assume that evaporation is the only means of water loss in the lake and it is at steady state (= situation unchanging over time) we have
Qdown = 0 m³/year
Budget reduces to:
0 = Qin*Cin - K*V*C
⇒ C = (Qin*Cin)/(K*V)
⇒ C = (9*10⁴ m³/year*6 mg/L)/(0.10/year*2*10⁵ m³) = 27 mg/L
