Hw 3 solutions

Hw 3 solutions - Solution of Homework 3 3-28 Complete the following table for Refrigerant-134a T-8 30 C P 320 kPa v m3 kg 0.0007569 0.015 0.11041

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution of Homework 3 3-28 Complete the following table for Refrigerant-134a : T, ° C P, kPa v , m 3 / kg Phase description -8 320 0.0007569 Compressed liquid 30 770.64 0.015 Saturated mixture -12.73 180 0.11041 Saturated vapor 80 600 0.044710 Superheated vapor 3-41 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states. Analysis This is a constant volume process. The specific volume is /kg m 0014 . 0 kg 10 m 014 . 0 3 3 2 1 = = = = m V v v The initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A-12 by interpolation C 0.61 ° = = kPa 300 @ sat 1 T T and kJ/kg 52 . 54 ) 13 . 198 )( 009321 . 0 ( 67 . 52 009321 . 0 /kg m ) 0007736 . 0 067978 . 0 ( /kg m ) 0007736 . 0 0014 . 0 ( 1 1 3 3 1 1 = + = + = = = = fg f fg f h x h h x v v v The total enthalpy is then kJ 545.2 = = = ) kJ/kg 52 . 54 )( kg 10 ( 1 1 mh H The final state is also saturated mixture. Repeating the calculations at this state, C 21.55 ° = = kPa 600 @ sat 2 T T kJ/kg 64 . 84 ) 90 . 180 )( 01733 . 0 ( 51 . 81 01733 . 0 /kg m ) 0008199 . 0 034295 . 0 ( /kg m ) 0008199 . 0 0014 . 0 ( 2 2 3 3 2 2 = + = + = = = = fg f fg f h x h h x v v v kJ 846.4 = = = ) kJ/kg 64 . 84 )( kg 10 ( 2 2 mh H R-134a 300 kPa 10 kg P 2 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
3-63 Heat is supplied to a piston-cylinder device that contains water at a specified state. The volume of the tank, the final temperature and pressure, and the internal energy change of water are to be determined. Properties The saturated liquid properties of water at 200 ° C are: v f = 0.001157 m 3 /kg and u f = 850.46 kJ/kg (Table A-4).
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/16/2008 for the course ME 2322 taught by Professor Oler during the Spring '06 term at Texas Tech.

Page1 / 4

Hw 3 solutions - Solution of Homework 3 3-28 Complete the following table for Refrigerant-134a T-8 30 C P 320 kPa v m3 kg 0.0007569 0.015 0.11041

This preview shows document pages 1 - 3. Sign up to view the full document.

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