This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 MAE 301: Engineering Thermodynamics Spring 2010 HOMEWORK #5 ( Solutions ) Problem 1. Problem 4.61 from the book. Solution: The gas constant of neon is R = 0.4119 kJ/kg ⋅ K and the constantpressure specific heat of neon is 1.0299 kJ/kg ⋅ K (Table A2 a ). At the compressor inlet, the specific volume is /kg m 207 . 1 kPa 100 K) 273 K)(20 /kg m kPa (0.4119 3 3 1 1 = + ⋅ ⋅ = = P RT v Similarly, at the compressor exit, /kg m 2414 . kPa 500 K) 273 K)(20 /kg m kPa (0.4119 3 3 2 2 = + ⋅ ⋅ = = P RT v The change in the specific volume caused by the compressor is /kg m 0.966 3 − = − = = Δ 207 . 1 2414 . 1 2 v v v Since the process is isothermal, kJ/kg = Δ = Δ T c h p Problem 2. Problem 4.68E from the book. Solution: Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of − 221.5 ° F and 547 psia. The kinetic and potential energy changes are negligible, pe ke ≅ Δ ≅ Δ . Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and airconditioning applications. The tank is insulated and thus heat transfer is negligible. We take the entire tank as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as ) ( 1 2 energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net T T mc U E E E out in − = Δ = Δ = − v 3 2 1 43 42 1 Since the internal energy does not change, the temperature of the air will also not change. Applying the ideal gas equation gives psia 50 = = = = = ⎯→ ⎯ = 2 psia 100 2 2 / 1 2 2 1 2 1 1 2 2 2 1 1 P P P P P P V V V V V V 2 Problem 3. Problem 4.74 from the book. Problem 4....
View
Full Document
 Spring '08
 Hassan
 Thermodynamics, Energy, Heat, potential energy changes

Click to edit the document details