0 Problem Ce (10 points) Po=6ookPa D=0.02m To=32K Use: Y=1,4 R=0.287 KJ Kokg ER = 6.0045 Y-1 Kokg = + 200 172 = bo kJ/kg

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answerhappygod
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0 Problem Ce (10 points) Po=6ookPa D=0.02m To=32K Use: Y=1,4 R=0.287 KJ Kokg ER = 6.0045 Y-1 Kokg = + 200 172 = bo kJ/kg

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0 Problem Ce 10 Points Po 6ookpa D 0 02m To 32k Use Y 1 4 R 0 287 Kj Kokg Er 6 0045 Y 1 Kokg 200 172 Bo Kj Kg 1
0 Problem Ce 10 Points Po 6ookpa D 0 02m To 32k Use Y 1 4 R 0 287 Kj Kokg Er 6 0045 Y 1 Kokg 200 172 Bo Kj Kg 1 (202.38 KiB) Viewed 37 times
0 Problem Ce (10 points) Po=6ookPa D=0.02m To=32K Use: Y=1,4 R=0.287 KJ Kokg ER = 6.0045 Y-1 Kokg = + 200 172 = bo kJ/kg Go- KT Air at stagnation temperature 323 K and stagnation pressure 600 kPa flows isentropically through a converging nozzle, as shown in the figure. The nozzle feeds a frictionless, constant- area circular duct having diameter D = 0.02 meters. Heat transfer to the duct occurs at a rate 192 = + 200 kJ/kg Determine the maximum mass-flow-rate through the system, and the range of back-pressures which will allow this maximum mass-flow-rate. Determine the stagnation temperature and stagnation pressure at the exit of the duct (for maximum mass-flow-rate).
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