Question

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occurs at an average outer surface temperature of 315 K at the rate of 30 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. For air as an ideal gas with Cp = 1.1 Kj/kg * K, determine
(a) the rate power is developed, in kJ per kg of air flowing, and
(b) the rate of entropy production within the turbine, in kJ/kg per kg of air flowing.

Answer 1

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is   $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

     From  the question we are told

          The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

          The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

          The inlet temperature is  $T_1 = 970K$

          The outlet temperature is $T_2 = 670K$

           The pressure at the inlet of the turbine is $p_1 = 400 kPa$

          The pressure at the exist of the turbine is $p_2 = 100kPa$

           The temperature at outer surface is $T_s = 315K$

         The individual gas constant of air  R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

               $\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as          

        $h = c_p T$

The above equation becomes

             $\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

              $\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

           $\r W$ is the work output from the turbine

            $\r m$ is the mass flow rate of air

             $\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

              $\frac{\r W}{\r m} = (-30)+1.1(970-670)$

                   $\frac{\r W}{\r m}= 300kJ/kg$

The general balance  equation for an entropy rate is represented mathematically as

                       $\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

          =>          $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

    generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

                      $\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

                      Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

 substituting values

                $\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

                    $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$