A jet engine a flow of air at 1000 K,

Betsy Rhone

Betsy Rhone

Answered question

2021-12-24

A jet engine a flow of air at 1000 K, 200 kPa, and 40 m/s enters a nozzle, where the air exits at 500 m/s, 90 kPa. What is the exit temperature, assuming no heat loss?

Answer & Explanation

godsrvnt0706

godsrvnt0706

Beginner2021-12-25Added 31 answers

Step 1 
The information on the jet engine is as follows:
Ti=1000K 
Pi=200kPa 
Pe=90kPa 
ve=500ms 
vi=40ms 
Mass flow: 
m=mi=me 
Both heat transfer and work are absent. The energy equation is as follows:
he+12ve2=hi+12vi2 
he=hi+12(vi2ve2) 
From Air's ideal gas characteristics table A.7.1, which corresponds to Ti=1000K we can find inlet specific enthalpy: 
hi}=1046.22kJkg 
Calculating exit specific enthalpy: 
he=hi+12(vi2ve2)=1046.22+12(4025002)11000=922.02kJkg 
Step 2 
Table A.7.1, Ideal Gas Properties of Air, yields values that are h=877.4kJkg we find temperature 
T=850K 
Table A.7.1, Ideal Gas Properties of Air, yields values that are h=933.15kJkg we find temperature 
T=900K 
Now, using the linear interpolation method, we can get the exit temperature:
Te=850+922.02877.4933.15877.4(900850)=890K 
Te=890K

Gerald Lopez

Gerald Lopez

Beginner2021-12-26Added 29 answers

Continuity m˙i=m˙e=m
Energy: m˙i(hi+12Vi2)=m˙(he+12Ve2)
Due to the high T we take the h value
he=hi+12Vi212Ve2
=1046.22+0.5×(4025002)(11000)
=1046.22124.2=922.02kJkg
Interpolation
Te=850+50922.02877.4933.15877.4=890K

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