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Sagot :
Argon gas enters an insulated diffuser at 300 k and 100 kpa and flows in at a speed of 250 m/s. the gas's exit temperature is 359.5 K when the exit velocity is 25 m/s.
m₁ = me & assume no heat transfe
Energy Eq.
he+1/2 Ve² = hi + 1/2 Vi²
he = hi - 1/2 Ve²+ 1/2 Vi²
he-hi≈Cp (Te-Ti)
= 1/2 (Vi² - Ve² ) = ( 250² – 25² )
=30937.5 J/kg = 30.938 kJ/kg
Specific heats for ideal gases are
Argon
Cp = 0.52 kJ/kg K;
ΔT = 30.938/0.52 = 59.5
Te=359.5 K
Chemically speaking, argon is an element with the symbol Ar and atomic number 18. It belongs to the group 18 of the periodic table and is a noble gas. At 0.934%, argon is the third-most common gas in the atmosphere of the Earth (9340 ppmv). More than twice as much of it exists as water vapour.
Learn more about Argon here:
https://brainly.com/question/133359
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