## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The flow rate of the water through the turbines is $3.9\times 10^5~kg/s$
Each second, 80% of the kinetic energy is used to produce 50 MW of electric power. Let $M_w$ be the mass of water which flows through the turbines each second. We can find the flow rate of the water as; $\frac{(0.80)~KE}{1~s} = 50\times 10^6~W$ $(0.80)~\frac{1}{2}M_w~v^2 = (50\times 10^6~W)(1~s)$ $M_w = \frac{(2)(50\times 10^6~J)}{(0.80)~v^2}$ $M_w = \frac{(2)(50\times 10^6~J)}{(0.80)~(18~m/s)^2}$ $M_w = 3.9\times 10^5~kg$ The flow rate of the water through the turbines is $3.9\times 10^5~kg/s$.