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).

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Complete Question

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Answer:

The value is [tex]P = 294594.3 \ W[/tex]

Explanation:

From the question we are told that

   The height is  [tex]h  =  41 \  m[/tex]

   The efficiency of the turbine is [tex]\eta =  0.77[/tex]

   The flow rate is [tex]\r  V  = 1 m^3 / s[/tex]

    The head loss is  g =  2 m

Generally the head gain by the turbine is mathematically represented as      

        [tex]H  =  h -  d[/tex]

=>     [tex]H  =  41 - 2[/tex]

=>     [tex]H  =  39 \  m [/tex]

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

          [tex]P =  \eta  *\gamma *    \r V * H[/tex]

Here [tex]\gamma[/tex] is the specific density of water with value

        [tex]\gamma   =  9810 N/m^3 [/tex]

So

       [tex]P =0.77  *9810 *   1 *  39[/tex]

       [tex]P = 294594.3 \ W[/tex]

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