Question

The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30°C enters it with a velocity of 350 m/s and the exit state is 200 kPa and 90°C.

Answer 1

Explanation:

Expression for energy balance is as follows.

        $\Delta E_{system} = E_{in} - E_{out}$

or,          $E_{in} = E_{out}$

Therefore,  

         $m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})$

          $h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}$

Hence, expression for exit velocity will be as follows.

           $V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}$

                      = $V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}$

As $C_{p}$ for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.

       $V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}$                    

                  = $[(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}$

                 = 40.7 m/s

Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.