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6.107

A modern jet engine has a temperature after combustion of about 1500 K at 3200

kPa as it enters the turbine setion, see state 3 Fig. P.6.107. The compressor inlet is

80 kPa, 260 K state 1 and outlet state 2 is 3300 kPa, 780 K; the turbine outlet state

4 into the nozzle is 400 kPa, 900 K and nozzle exit state 5 at 80 kPa, 640 K.

Neglect any heat transfer and neglect kinetic energy except out of the nozzle. Find

the compressor and turbine specific work terms and the nozzle exit velocity.

Solution:

The compressor, turbine and nozzle are all steady state single flow devices

and they are adiabatic.

We will use air properties from table A.7.1:

h1 = 260.32, h2 = 800.28, h3 = 1635.80, h4 = 933.15, h5 = 649.53 kJ/kg

Energy equation for the compressor gives

wc in = h2 – h1 = 800.28 – 260.32 = 539.36 kJ/kg

Energy equation for the turbine gives

wT

= h3 – h4 = 1635.80 – 933.15 = 702.65 kJ/kg

Energy equation for the nozzle gives

h4

= h5 + ½ V

2

5

½ V

2

5 = h4 - h5 = 933.15 – 649.53 = 283.62 kJ/kg

V

5 = [2( h4 – h5) ] 1/2 = ( 2􀁵 283.62 􀁵1000 ) 1/2 = 753 m/s

+ نوشته شده در  یکشنبه هجدهم اردیبهشت 1390ساعت 23:11  توسط Reza  |