Question

Water flows from one reservoir to another a height, 41 m below. A turbine (η=0.77) generates power from this flow. 1 m3/s passes through the turbine. If 12 m head loss occurs between the two reservoirs, determine the actual (i.e. useable) power generated by the turbine (in kW).

Answer 1

Complete Question

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Answer:

The value is
`P = 294594.3 \ W`

Step-by-step explanation:

From the question we are told that

The height is
`h  =  41 \  m`

The efficiency of the turbine is
`\eta =  0.77`

The flow rate is
`\r  V  = 1 m^3 / s`

The head loss is g = 2 m

Generally the head gain by the turbine is mathematically represented as

`H  =  h -  d`

=>
`H  =  41 - 2`

=>
`H  =  39 \  m`

Generally the actual power generated by the turbine is mathematically represented as

`P =  \eta  *\gamma *    \r V * H`

Here
`\gamma` is the specific density of water with value

`\gamma   =  9810 N/m^3`

So

`P =0.77  *9810 *   1 *  39`

`P = 294594.3 \ W`