Thermo solution7 - Copy

Thermo solution7 - Copy - 650:351 Thermodynamics Doyle...

Info iconThis preview shows pages 1–5. 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

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: 650:351 Thermodynamics Doyle Knight Department of Mechanical and Aerospace Engineering Rutgers - The State University of New Jersey New Brunswick, NJ USA Tel: 732 445 4464 Email: doyleknight@gmail.com Office hours: Tuesday and Friday, 9:00 am to 11:30 am and by appointment Homework Assignment No. 7 Problem 8.27 Find the missing properties and give the phase of the substance Solution a. H 2 O for s = 7 . 70 kJ/kg-K at p = 25 kPa For saturated liquid-vapor water at p = 25 kPa, s f = 0 . 893 kJ/kg-K and s fg = 6 . 9383 kJ/kg-K from Table B.1.2. Assuming that the substance is saturated liquid-vapor, x = ( s s f ) s fg = (7 . 70 . 893) 6 . 9383 = 0 . 981 Since 0 < x < 1, it is a saturated liquid-vapor. From Table B.1.2, T = 64 . 97 C. The specific enthalpy is h = h f + xh fg = 271 . 9 + (0 . 981)(2446 . 3) = 2573 . 8 kJ/kg 650:351 Thermodynamics Page 1 Homework Assignment No. 7 Problem 8.27 b. H 2 O for u = 3400 kJ/kg at p 10 MPa From Table B.1.2, u g = 2544 . 41 kJ/kg at p = 10 MPa. Since u > u g , it is a superheated vapor. Therefore, x = 1 (or, in the terminology of your text, x is undefined). The values for T and s can be interpolated from Table B.1.3, to yield T = 682 C and s = 7 . 1223 kJ/kg-K c. R-12 at T = 0 C and p = 200 kPa From Table B.3.2, the temperature for saturated liquid-vapor at p = 200 kPa is T sat = 12 . 53 C. Thus, it is a superheated vapor. Hence, x is undefined and s = 0 . 7325 kJ/kg-K may be read directly from the Table 650:351 Thermodynamics Page 2 Homework Assignment No. 7 Problem 8.27 d. R-134a at T = 10 C and x = 0 . 45 From Table B.5.1, at T = 10 C the values of v f , v fg , s f and s fg may be read directly to yield v = v f + xv fg = 0 . 000755 + (0 . 45)(0 . 09845) = 0 . 04506 m 3 /kg s = s f + xs fg = 0 . 9507 + (0 . 45)(0 . 7812) = 1 . 3022 kJ/kg-K e. NH 3 at T = 20 C and s = 5 . 50 kJ/kg-K Assume a saturated liquid-vapor. From Table B.2.1, s g = 5 . 086 kJ/kg-K at T = 20 C. Since s > s g , it is a superheated vapor. Thus, x is undefined. From Table B.2.2, s = 5 . 5525 kJ/kg-K at 400 kPa s = 5 . 4244 kJ/kg-K at 500 kPa Interpolating to s = 5 . 5 kJ/kg-K, p = 441 . 0 kPa and u = 1356 . 8 kJ/kg....
View Full Document

This homework help was uploaded on 04/07/2008 for the course MECHANICAL 351 taught by Professor Knight during the Spring '08 term at Rutgers.

Page1 / 18

Thermo solution7 - Copy - 650:351 Thermodynamics Doyle...

This preview shows document pages 1 - 5. Sign up to view the full document.

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