Question

A source of heat at 1000 K transfers 1000 kW of power to a power generation device, while producing 400 kW of useful work. Determine the second law efficiency of the system, if the environment is at 300 K.

Answer 1

Answer:

The value is $\eta _2 = 0.57$

Explanation:

From the question we are told that

    The temperature of the heat source is $T_s = 1000 \ K$

    The amount of power transferred is  $P_s = 1000 \ kW = 1000 *10^{3} \ W$

      The work produced is  $W = 400 \ kW$

       The temperature of the environment $T_e = 300 \ K$

Gnerally the Carnot efficiency of the system is mathematically represented as

        $\eta_c = 1 -\frac{T_e}{T_s}$

=>    $\eta_c = 1 -\frac{300}{1000}$

=>      $\eta_c = 0.7$

Generally the first  law efficiency of the system is mathematically represented as

            $\eta _1 = \frac{W}{P_s}$

=>         $\eta _1 = \frac{400}{1000}$  

=>         $\eta _1 = 0.40$

Generally the second  law efficiency of the system is mathematically represented as

                $\eta _2 = \frac{\eta_1}{\eta_c}$

=>               $\eta _2 = \frac{0.4}{0.7}$

=>               $\eta _2 = 0.57$