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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/kgK at p = 25 kPa For saturated liquidvapor water at p = 25 kPa, s f = 0 . 893 kJ/kgK and s fg = 6 . 9383 kJ/kgK from Table B.1.2. Assuming that the substance is saturated liquidvapor, x = ( s s f ) s fg = (7 . 70 . 893) 6 . 9383 = 0 . 981 Since 0 < x < 1, it is a saturated liquidvapor. 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/kgK c. R12 at T = 0 C and p = 200 kPa From Table B.3.2, the temperature for saturated liquidvapor 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/kgK may be read directly from the Table 650:351 Thermodynamics Page 2 Homework Assignment No. 7 Problem 8.27 d. R134a 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/kgK e. NH 3 at T = 20 C and s = 5 . 50 kJ/kgK Assume a saturated liquidvapor. From Table B.2.1, s g = 5 . 086 kJ/kgK 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/kgK at 400 kPa s = 5 . 4244 kJ/kgK at 500 kPa Interpolating to s = 5 . 5 kJ/kgK, p = 441 . 0 kPa and u = 1356 . 8 kJ/kg....
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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.
 Spring '08
 Knight

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