Answer
$Income_{net}=66839\frac{dollars}{year}$
Work Step by Step
Here we are working with potential energy:
The potential work is: $W_{p}=mgh$
And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$
The potential power is:
$P_{p}=1000\frac{kg}{m^3}*(2\frac{m^3}{s})*9.81\frac{m}{s^2}*40m=784.8kW$
The pump electric power needed and its cost are:
$P_{pump,e}=\frac{784.8kW}{0.75}=1046.4kW$
$Cost=1046.4kW*10\frac{h}{day}*365\frac{day}{year}*0.05\frac{dollars}{kWh}=190,968\frac{dollars}{year}$
The turbine-generator electric power produced and its incomes are:
$P_{turbine-generator,e}=784.8kW*0.75=588.6kW$
$Income=588.6kW*10\frac{h}{day}*365\frac{day}{year}*0.12\frac{dollars}{kWh}=257807\frac{dollars}{year}$
Then the net income is:
$Income_{net}=257807\frac{dollars}{year}-190,968\frac{dollars}{year}=66839\frac{dollars}{year}$
