Question

In an adiabatic mixing device (two inlets and one exit) operating at steady state, a 10 kg/s flow of fluid-A (with a specific entropy of 5 kJ/kg.K) is mixed with a 5 kg/s flow of fluid-B (with a specific entropy of 10 kJ/kg.K). At the exit (the flow rate is 15 kg/s), the specific entropy is measured as 7 kJ/kg.K. Determine the rate of entropy generation in the device.

Answer 1

Explanation:

Formula for entropy balance for the given situation is as follows.

   $\frac{Q}{T} + m_{A}S_{A} + m_{B}S_{B} - m_{AB}S_{AB} + S_{gen}$ = 0

As the given system is adiabatic, hence Q = 0.

So,     $S_{gen} = m_{AB}S_{AB} - (m_{A}S_{A} + m_{B}S_{B})$

As the given data is as follows.

         $m_{A}$ = 10 kg/s,    $S_{A}$ = 5 kJ/kg K

         $m_{B}$ = 5 kg/sec,    $S_{B}$ = 10 kJ/kg K

        $S_{AB}$ = 7 kJ/kg K,    

             $m_{AB} = m_{A} + m_{B}$

                         = 10 + 5

                         = 15 kg/sec

  $S_{gen} = 15 \times 7 - (10 \times 5 + 5 \times 10)$

               = 5 kW/K

Thus, we can conclude that the rate of entropy generation in the device is 5 kW/K.