Problem 4.204 - ⋅ M c v ⋅ dT dt ⋅ u m exit ⋅ p ρ m exit ⋅ = dT dt m exit p ⋅ M c v ⋅ ρ ⋅ − = But M ρ V ⋅ =(where V is volume

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Problem 4.204 [Difficulty: 3] Given: Compressed air bottle Find: Rate of temperature change Solution: Basic equations: Continuity; First Law of Thermodynamics for a CV Assumptions: 1) Adiabatic 2) No work 3) Neglect KE 4) Uniform properties at exit 5) Ideal gas Given data p 500 kPa = T2 0 ° C = T 293K = V 100 L = m exit 0.01 kg s = Also R air 286.9N m kg K = c v 717.4 Nm kg K = From continuity t M CV m exit + 0 = where m exit is the mass flow rate at the exit (Note: Software does not allow a dot!) t M CV m exit = From the 1st law 0 t M u d u p ρ + m exit + = u t M M t u + u p ρ + m exit + = Hence um exit
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Unformatted text preview: ⋅ M c v ⋅ dT dt ⋅ + u m exit ⋅ + p ρ m exit ⋅ + = dT dt m exit p ⋅ M c v ⋅ ρ ⋅ − = But M ρ V ⋅ = (where V is volume) so dT dt m exit p ⋅ V c v ⋅ ρ 2 ⋅ − = For air ρ p R air T ⋅ = ρ 500 10 3 × N m 2 ⋅ kg K ⋅ 286.9 N ⋅ m ⋅ × 1 20 273 + ( ) K ⋅ × = ρ 5.95 kg m 3 = Hence dT dt 0.01 − kg s ⋅ 500 × 10 3 × N m 2 ⋅ 1 100 L ⋅ × L 10 3 − m 3 ⋅ × kg K ⋅ 717.4 N ⋅ m ⋅ × m 3 5.95 kg ⋅ ⎛ ⎜ ⎝ ⎞ ⎠ 2 × = 1.97 − K s ⋅ = 1.97 − C s ⋅ =...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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