A hydroturbine operates with a head of 50 m of water. Inlet and outlet conduits are 2 m in diameter. Estimate the mechanical power developed by the turbine for an outlet velocity of 5 m? s?1.

Question

What will be an ideal response?

Answer 1

A general energy balance on the hydroturbine can be written as fs



The system can reasonably be assumed to operate at steady-state, so the first term is zero. The change in kinetic energy will also be negligible, because the inlet and outlet pipes have the same diameter, and therefore the water will have the same velocity at the inlet and outlet. A somewhat stronger assumption is that the change in enthalpy of the water is negligible. All real processes, including hydroturbines, have some degree of irreversibility. However, it turns out that a well-designed hydroturbine can be quite efficient, and can convert more than 90% of the potential energy of the water into mechanical energy (to be used to drive a generator, which will have its own inefficiencies, so that the final electrical energy output will be smaller). Thus, for the hydroturbine, it is reasonable to neglect the change in enthalpy of the water, as well as heat flow to or from the system. Doing so gives simply



This will be an upper limit for the mechanical power (rate of work) produced by the turbine. To compute it, we need to know the mass flow rate of the water through the turbine (which is the same at the inlet and outlet, by a trivial mass balance). The cross-sectional area of the 2 m diameter inlets and outlets is ?·(1 m)2= 3.14 m2. Multiplying this by the water velocity of 5 m s-1 gives a volumetric flow rate of 15.7 m3/s. Multiplying this by the density of water (1000 kg/m3) gives the mass flow rate as 15,700 kg/s. Thus, the mechanical work is estimated as



A crude estimate is that the average power consumption rate of a U.S. household is around 1 kW. Allowing for an overall efficiency (turbine plus generator) of 85%, this is still enough to power about 6500 households.