Thermo solution4 - Copy

Thermo solution4 - 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: doyleknight@gmail.com OFce hours: Tuesday and ±riday, 9:00 am to 11:30 am and by appointment

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Homework Assignment No. 4 Problem 5.24 A 1200 kg car accelerates from zero to 100 km/hr over a distance of 400 m. The road at the end of the 400 m is 10 m higher elevation. What is the total increase in the car kinetic and potential energy ? Solution The increase in kinetic energy is ΔKE = 1 2 mV 2 = 462 . 96 kJ The increase in potential energy is ΔPE = mgH = 117 . 6 kJ 650:351 Thermodynamics Page 1
Homework Assignment No. 4 Problem 5.28 Find the missing properties a. Water at 250 C, v = 0 . 02 m 3 /kg From Table B.1.1, at T = 250 C, v f = 0 . 001251 m 3 /kg and v fg = 0 . 04887 m 3 /kg. Thus, it is saturated liquid-vapor. From Table B.1.1, p = 3973 kPa. Also, 0 . 02 = 0 . 001251 + ( x )(0 . 04887) and thus x = 0 . 384. Answer: p = 3973 kPa, x = 0 . 384. b. N 2 at T 120 K and p = 0 . 8 MPa From Table B.6.1, at T = 120 K, p = 2 . 51 MPa. Thus, it is superheated vapor. From Table B.6.2, at p = 0 . 8 MPa and T = 120 K, h = 114 . 0 kJ/kg and x = 1. Answer: h = 114 . 0 kJ/kg, x = 1. 650:351 Thermodynamics Page 2

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Homework Assignment No. 4 Problem 5.28 c. Water at T = 2 C and p = 100 kPa This is ice at 1 atmosphere. From Table A.3, ρ = 917 kg/m 3 and thus v = ρ 1 = 0 . 00109 m 3 /kg. The internal energy is u = u ref + i T T ref c p dT Since u = 0 at T = 0 . 01 C, using c p = 2 . 04 kJ/kg, u = 4 . 1 kJ/kg. Answer: u = 4 . 1 kJ/kg, v = 0 . 00109 m 3 /kg d. R-134a at p = 200 kPa and v = 0 . 12 m 3 /kg From Table B.5.1, at p = 201 kPa, v f = 0 . 000755 m 3 /kg and v g = 0 . 09921 m 3 /kg. Thus, it is a superheated vapor. From Table B.5.2, at p = 200 kPa, interpolate between T = 30 C and T = 40 C to obtain T = 32 . 5 C at v = 0 . 12 m 3 /kg. Using the same interpolation, u = 405 . 1 kJ/kg. Answer: T = 32 . 5 C, u = 405 . 1 kJ/kg 650:351 Thermodynamics Page 3
Homework Assignment No. 4 Problem 5.30 Find the missing properties and give the phase of the substance a. NH 3 (ammonia) at T = 65 C and p = 600 kPa From Table B.2.1, at T = 65 C, p = 2947 kPa. Thus, it is a superheated vapor. From Table B.2.2, for

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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.

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Thermo solution4 - Copy - 650:351 Thermodynamics Doyle...

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