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A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces upstream and downstream of the dam is 131 m. The water is supplied to the turbine at a rate of 201 kg/s. If the shaft power output from the turbine is 234 kW, the efficiency of the turbine is_________.

Respuesta :

MechEngineer

Answer:

0.906

Explanation:

Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second

[tex]\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW[/tex]

So the efficiency of the water turbine is the ratio of output power over input power:

[tex]\frac{234}{258.307} = 0.906[/tex]

hamzaahmeds

This question involves the concepts of efficiency and potential energy.

The efficiency of the turbine is "90.6%".

The efficiency of the turbine can be given by the following formula:

[tex]Efficiency = \frac{Turbine\ shaft\ power}{Potential\ Energy\ power\ of\ Water}\\\\[/tex]

where,

Turbine shaft power = 234 KW

Potential energy power of water = mgh/t = (201 kg/s)(9.81 m/s²)(131 m)

Potential energy power of water = 258307.11 W = 258.31 KW

Therefore,

[tex]Efficiency=\frac{234\ KW}{258.31\ KW}[/tex]

Efficiency = 0.906 = 90.6%

Learn more about efficiency here:

https://brainly.com/question/5035443?referrer=searchResults

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