ME320 solution4 - ME320 Thermodynamics Homework set #4 Due:...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ME320 Thermodynamics Homework set #4 Due: Wednesday 10/1/03 at the start of class 1. 265C (Textbook) Solution: A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to the critical temperature and pressure. 2. 270 (Textbook) Solution: Initially, the absolute pressure in the tire is P 1 = P g + P atm = 210 kPa + 100 kPa = 310 kPa Treating the air as an ideal gas, it follows that P 1 V 1 T 1 = P 2 V 2 T 2 . Thus, P 2 = P 1 V 1 T 2 V 2 T 1 and, assuming that the volume of the tire remains constant, the final pressure in the tire is given by P 2 = P 1 T 2 T 1 = (310 kPa)(323 K) 298 K = 336 kPa Thus, the pressure rise is P = P 2- P 1 = (336- 310) kPa = 26 kPa . Finally, using the gas constant of air given in Table A-1, we find that m 1 = P 1 V 1 RT 1 = (310 kPa)(0 . 025 m 3 ) (0 . 287 kPa m 3 / kg K)(298 K) = 0 . 0906 kg and m 2 = P 2 V 2 RT 2 = (310 kPa)(0 . 025 m 3 ) (0 . 287 kPa m 3 / kg K)(323 K) = 0 . 0906 kg , whereby the mass of air that needs to be bled off to restore pressure...
View Full Document

This note was uploaded on 04/11/2008 for the course MASE 320 taught by Professor Shen during the Spring '08 term at Washington University in St. Louis.

Page1 / 3

ME320 solution4 - ME320 Thermodynamics Homework set #4 Due:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online