Nozzle and Diffuser (Exercise1)

Download the exercise with solution: Nozzle and Diffuser

Air flows through an adiabatic and horizontal placed pipe with changeable cross section. At the inlet side air flows with velocity cin=10 m/s under the pressure pin=1bar and the temperature ϑin=25°C. At the outlet side air flows out of the pipe with velocity cout=150 m/s under the pressure pin=0.8bar.

1)      Determine whether this pipe is nozzle or diffuser without calculating the ratio of cross section of outlet and inlet Aout/Ain

2)      Determine now the ratio of cross section of outlet and inlet Aout/Ain

Air can be considered as perfect gas with cp=1 J/K∙g and R=0.29 J/K∙g

Solution:

ex1

1)

The 1.Law of thermodynamics:1 hauptsatzStationary → d/dτ=0

Adiabatic → ΣQ=0

Without input and output (work) → ΣW=0

Horizontal placed → no change in potential energy → Δ(g·z)=0

Then we can obtain the simplified 1.Law: ex2wobei hin=cp·Tin und hout=cp·Tout

Therefore the temperature of air at the outlet side is: ex3

We now calculate the entropy change during this process: ex4

Hence it is Nozzle.

2)

In order to determine Aout/Ain, we need additional the law of conservation of mass and the ideal gas law: ex5

Ain>Aout , it is also proved that this pipe is a nozzle.

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