Question

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each second. The water is released 220 m below the top of the reservoir.If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?Express your answer with the appropriate units.

Answer 1

Answer:

The maximum electric power output is $P_{max} =1.339*10^{9} \ W$

Explanation:

From the question we are told that

        The capacity of the hydroelectric plant is $\frac{V}{t} = 690 \ m^3 /s$

         The level at which water is been released is $h = 220 \ m$

        The efficiency is  $\eta =$0.90

       

The electric power output is mathematically represented as

       $P = \frac{PE_l - PE _o}{t}$

Where  $PE_l$ is the potential energy at  level h which is mathematically evaluated as  

          $PE_l = mgh$

and  $PE_o$  is  the potential energy at ground level which is mathematically evaluated as  

          $PE_o = mg(0)$

         $PE_o = 0$

So  

         $P = \frac{mgh}{t}$

here  $m = V * \rho$

where V is volume  and  $\rho$ is density of water whose value is  $\rho = 1000 kg/m^3$

 So  

         $P = \frac{(\rho * V) * gh}{t}$

        $P = \frac{V}{t} * gh \rho$

substituting values  

       $P =690 * 9.8 * 220 * 1000$

      $P =1.488*10^{9} \ W$

The maximum possible electric power output is

           $P_{max} = P * \eta$

substituting values  

         $P_{max} =1.488*10^{9} * 0.90$

         $P_{max} =1.339*10^{9} \ W$