# HW12s - Spring 2006 Thermodynamics Homework #12 Solution 1....

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Spring 2006 1 ME 201 Thermodynamics Homework #12 Solution 1. A 2 ft 3 scuba diver’s air tank is to be filled with air from a compressed air line at 120 psia, 100 ° F. Initially, the air in the tanks is at 20 psia and 70 ° F. Assuming that the tank is well insulated, determine the temperature and mass in the tanks when it is filled to 120 psia. Solution: System Type: Transient Working Fluid: Air (ideal gas) Process: Constant total volume, adiabatic State 1 (initial) State 2 (final) State 3 (inlet) T 1 = 70 ° F = 530 R T 2 = 724 R T 3 = 100 ° F = 560 R P 1 = 20 psia P 2 = 120 psia P 3 = 120 psia u 1 = 90.33 Btu/lb m u 2 = 123.95 Btu/lb m h 3 = 133.86 Btu/lb m m 1 = 0.20377 lb m m 2 = 0.895 lb m m 3 = 0.689 lb m Italicized values are from air table. Bold values are calculated. Initial State: Fixed Inlet State: Fixed Final State: UNKNOWN Q = 0 W sh = 0 W bnd = 0 KE and PE are negligible 1st Law: m 2 u 2 - m 1 u 1 = m 3 h 3 mass: m 2 - m 1 = m 3 Approach: To fix the final state we will have to simultaneously solve the 1 st law, conservation mass and ideal gas law. First we go to the air tables and find u 1 = 90.33 Btu/lb m and h 3 = 133.86 Btu/lb m Next we use the ideal gas law to calculate the initial mass or m = P V RT = (20)(2) (10.73/ 28.97)(530) = 0.2038 lb 1 1 1 1 m

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ME 201 Thermodynamics Spring 2006 2 We begin by taking the 1 st law and eliminating m 3 with the use of the conservation of mass equation or ( ) m u - m u = m - m h 2 2 1 1 2 1 3 Rearranging ( ) 2 3 1 2 1 1 2 m h m - m u m = u + Next we use our ideal gas law to eliminate m 2 m = P V RT 2 2 2 2 Through the air tables we have a relationship between T
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## This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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HW12s - Spring 2006 Thermodynamics Homework #12 Solution 1....

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