Water enters a turbine at v_1 = 4 m/s and leaves at v_2 = 1…

Water enters a turbine at $v_1 = 4\ \mathrm{m/s}$ and leaves at $v_2 = 1\ \mathrm{m/s}$; the inlet is $h = 1.5\ \mathrm{m}$ above the outlet; the mass flow is $Q' = 300\ \mathrm{kg/s}$; the turbine's efficiency is $\eta = 0.85$. Find the power.

Explanation of Method

Power of a water turbine.

Step 1: Energy given up per kilogram

\[ \frac{v_1^2 - v_2^2}{2} + gh = 7.5 + 14.72 = 22.2\ \mathrm{J/kg}. \]

Step 2: Power

\[ P = Q'\left(\frac{v_1^2 - v_2^2}{2} + gh\right) = 300 \cdot 22.2. \]

\[P \approx 6.66\ \mathrm{kW}, \quad \eta P \approx 5.67\ \mathrm{kW}\]

$\tfrac{v_1^2}{2} = 8$ $gh = 14.7$ $\text{per kg in}$ $\text{to the turbine: } 22.2\ \mathrm{J/kg}$ $\text{per kg out}$ $P = Q'(22.2\ \mathrm{J/kg}) = 6.66\ \mathrm{kW}$ $P_{\text{useful}} = 0.85\,P \approx 5.67\ \mathrm{kW}$ $\tfrac{v_2^2}{2}$

Energy per kilogram of water (J/kg). Each kilogram brings its kinetic energy and its height; it leaves with 0.5 J/kg of kinetic energy, and the rest is given to the turbine.

Result

\[\boxed{\,P \approx 6.66\ \mathrm{kW}, \quad \eta P \approx 5.67\ \mathrm{kW}\,}\]

Remark

Check. The original results (6.7 kW and 5.7 kW) are correct.