Question 10.1.2: Water Balance on a City Reservoir The water level in a munic......
Water Balance on a City Reservoir
The water level in a municipal reservoir has been decreasing steadily during a dry spell, and local meteorologists predict that the drought could continue for another 60 days. The water company estimates that the consumption rate in the city is approximately 10^{7} L/day. The State Conservation Service estimates that rainfall and stream drainage into the reservoir coupled with evaporation from the reservoir should yield a net water input rate of 10^{6} exp(- t/100) L/day, where t is the time in days from the beginning of the drought, at which time the reservoir contained an estimated 10^{9} liters of water.
(a) Write a differential balance on the water in the reservoir.
(b) Integrate the balance to calculate the reservoir volume at the end of the 60 days of continued drought.
(a) We will write a balance on the mass of water in the reservoir, M(kg), but express the equation in terms of volumes to make use of the given data, using the relationship M(kg) = ρ(kg/L)V(L). Since water is neither generated nor consumed by chemical reactions in the reservoir (more precisely, we assume that its rates of generation and consumption are negligible), the differential balance equation is
accumulation = input – output
\frac{dM}{dt} = \dot{m}_{in} +\dot{r}_{gen} – \dot{m}_{out} – \dot{r}_{cons} (each term in kg/day) \\ \left. \large{\Downarrow } \right. \begin{matrix}\frac{dM}{dt} = \frac{d}{dt} (\rho V) = \rho(kg/L)\frac{dV}{dt} (L/day) (since \rho is constant) \\ \dot{m}_{in} = \rho(kg/L) [10^{6} e^{-t/100}(L/day)] \\ \dot{m}_{out} = \rho(kg/L)(10^{7} L/day) \\ \dot{r} _{gen} = \dot{r} _{cons} = 0 (water is not produced or consumed in the reservoir) \\ Cancel ρ \end{matrix} \\ \boxed{\begin{matrix} \frac{dV(t)}{dt} =10^{6} \exp (-t/100) – 10^{7}\\ t=0 , V=10^{9} L \end{matrix} }
(b) We now separate variables and integrate the differential balance equation from t = 0 to t = 60 days:
\int_{V(0)}^{V(60)}{dV} = \int_{0}^{60 d}{[10^{6} \exp (-t/100) – 10^{7}]dt} \\ \left. \Large{\Downarrow} \right.\\V(60 days) – V(0) = \int_{0}^{60 d}{10^{6} e^{-t/100} dt – \int_{0}^{60 d} 10^{7} dt} \\ \left. \Large{\Downarrow} \right. V(0) = 10^{9} liters\\ V(60 days) = 10^{9} – 10^{6} (10^{2})e^{-t/100} \left. \Large{]} \right.^{60 d}_{0} – 10^{7} t\left. \Large{]} \right.^{60 d}_{0} \\ = \boxed{4.45 × 10^{8} L} (verify)