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Thermo solution7 - Copy

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

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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: [email protected] Office hours: Tuesday and Friday, 9:00 am to 11:30 am and by appointment

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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 0 . 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

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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. 650:351 Thermodynamics Page 3
Homework Assignment No. 7 Problem 8.32 In a Carnot engine with ammonia as the working fluid, the high temperature is 60 C and as Q H is received, the ammonia changes from saturated liquid to saturated vapor. The ammonia pressure at the low temperature is 190 kPa. Find

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

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