Gravity Niagara falls this time

Here, depth, h= 50m.

Average power generated = 5000 horse power = 5000×746 watt = $3.73×10^{5} Js^{(-1)}$

Efficiency = 85%

Mass of water, m = ?, t = 1 min = 60sec

Power generated = $\frac{100}{85} × avererage \quad power = \frac{100}{85}×37.3×10^{5}$

Therefore the total work done by falling water in 1min.

= $\frac{100}{85}×37.3×10^{5} × 60 joules$

$W = 26.32×10^{7} joules$

We know that $W=mgh$

Hence, $m= \frac{W}{gh}=\frac{26.32×10^{7}}{10×50} = 5.27×10^{5}kg = 527000kg$

Let M be the mass of water passing through the turbine in one second. It follows that 60M is the total mass of water passing through in one minute, which is the desired quantity.

The force of the water acting downwards is gM, since F =mg

Therefore the work done in moving the water 50m is 50gM (work = force * distance) The turbine is 85% efficient, so 0.85(50gM) J of energy are produced per second.

The output power is 746*5000 W .

Therefore, 0.85(50gM) = 5000*746

This can be solved for M = 8746.5 kg

Multiply this by 60 for the total mass of water in one minute, returning

526 588 kg

Rounding this:

527 000 kg