The turbines in a hydroelectric plant are fed by water falling from a 50 m height. Assuming 91% efficiency for conversion of potential to electrical endrgy, and 8% loss of the resulting power in transmission, what is the mass flow rate of water required to power a 200 W light bulb?

From  the information given;

the height of the water stream = 50 mthe efficiency of conversion from potential energy to electrical energy is  91%loss of power transmission = 8%

To determine the mass flow rate, let's start by understanding some concepts and parameters.

The power is known to be the energy per unit of time. Mathematically, it can be written as:

where;

is the potential energy of the streamm = mass flow rateg = acceleration under gravityz = height

Thus;

= m × 9.81 m/s² × 50 m

= m × 490.5 (m²/s²)

Recall that:

The power P = 200 W, and;the conversion of the P.E = 91% = 0.91

∴

= 0.91 × 490.5m (m²/s²)

= 446.355m (m²/s²)

Since the resulting power transmission is said to be 8%

Then;

the loss in the power transmission (P) = 100% - 8% ×  446.355m (m²/s²)

the loss in the power transmission (P) = 92%  ×  446.355m (m²/s²)

the loss in the power transmission (P) = 0.92  ×  446.355m (m²/s²)

the loss in the power transmission (P) =  410.65m (m²/s²)

Finally;

P = 410.65m (m²/s²)

replacing the values, we have:

since 1 J/s = 1 kgm²/s²)

Then:

mass flow rate of the water (m) = 0.487 kg/s

Therefore, we can conclude that the mass flow rate of the water required to power a 200 W bulb light is 0.487 kg/s

Learn more about the hydroelectric plant here:

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