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Nezavi [6.7K]
3 years ago
8

The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any wo

rk 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.
Physics
1 answer:
wariber [46]3 years ago
5 0

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.

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