Water flows from a reservoir at the rate of 3000 kg / min, to a turbine 120 m below. If the ef

step 1 In this example, I'm going to be looking at the relationship between power and energy. Okay, so

Water flows from a reservoir at the rate of 3000 kg / min,...

Water flows from a reservoir at the rate of $3000 \mathrm{~kg} / \mathrm{min}$, to a turbine $120 \mathrm{~m}$ below. If the efficiency of the turbine is 80 percent, compute the power output of the turbine. Neglect friction in the pipe and the small KE of the water leaving the turbine. Don't forget that it's only 80 percent efficient.

Key Concepts

Power Calculation

Power quantifies the rate at which work is done or energy is transferred. In the context of a turbine, it is calculated by multiplying the mass flow rate by the gravitational potential energy per unit mass and then adjusting for the system’s efficiency. This approach allows one to compute the actual useful power output derived from the conversion of gravitational energy.

Unit Conversion

Unit conversion is vital in solving engineering problems, ensuring that all quantities are expressed in compatible units. For example, mass flow rates given in kilograms per minute need to be converted into kilograms per second when calculating power (which is energy per unit time in joules per second). This step is fundamental for obtaining correct numerical results in power calculations.

Gravitational Potential Energy

This concept describes the energy possessed by an object due to its position in a gravitational field. In problems involving fluids like water in a reservoir, gravitational potential energy is converted into other forms of energy (such as mechanical energy) as the water falls. The potential energy available per unit mass is given by the product of the gravitational acceleration and the height difference, which is key to determining the energy input to systems like turbines.

Energy Conversion Efficiency

Efficiency in energy conversion quantifies how effectively a system transforms input energy into useful output energy. For turbines, not all gravitational potential energy is converted into mechanical or electrical energy due to various losses. The efficiency factor helps to account for these losses by scaling the theoretical energy available, thereby providing a more realistic estimation of the power output.

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