Sphw6 - HOMEWORK SET 6 8.17 Find the missing properties and...

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

View Full Document Right Arrow Icon
HOMEWORK SET 6 8.17 Find the missing properties and give the phase of the ammonia, NH 3 . a. T = 65 ° C, P = 600 kPa s = ? v = ? b. T = 20 ° C, P = 100 kPa v = ? s = ? x = ? c. T = 50 ° C, v = 0.1185 m 3 /kg s = ? x = ? P = ? a) B.2.2 average between 60 ° C and 70 ° C v = (0.25981 + 0.26999)/2 = 0.26435 m 3 /kg s = (5.6383 + 5.7094)/2 = 5.6739 kJ/kgK b) B.2.1: P < P sat = 857.5 kPa => B.2.2 superheated vapor so x is undefined v = 1.4153 m 3 /kg, s = 6.2826 kJ/kgK c) B.2.1: v > v g = 0.06337 m 3 /kg => B.2.2 superheated vapor so x is undefined very close to 1200 kPa, s = 5.1497 kJ/kgK 8.28 Consider a Carnot-cycle heat pump with R-22 as the working fluid. Heat is rejected from the R-22 at 40 ° C, during which process the R-22 changes from saturated vapor to saturated liquid. The heat is transferred to the R-22 at 0 ° C. a. Show the cycle on a T–s diagram. b. Find the quality of the R-22 at the beginning and end of the isothermal heat addition process at 0 ° C. c. Determine the coefficient of performance for the cycle. Solution: a) 1 2 3 4 40 0 T s b) From Table B.4.1, state 3 is saturated liquid s 4 = s 3 = 0.3417 kJ/kg K = 0.1751 + x 4 (0.7518) => x 4 = 0.2216 State 2 is saturated vapor so from Table B.4.1 s 1 = s 2 = 0.8746 kJ/kg K = 0.1751 + x 1 (0.7518) => x 1 = 0.9304
Background image of page 1

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

View Full DocumentRight Arrow Icon
c) β′ = q H w IN = T H T H – T L = 313.2 40 = 7.83 8.45 A piston cylinder contains 0.25 kg of R-134a at –20 o C, 100 kPa. It is compressed in a reversible adiabatic process to 400 kPa. How much work is needed? Solution: C.V. R-134a, Control mass, adiabatic 1 q 2 = 0 Energy Eq.5.11: u 2 u 1 = 1 q 2 1 w 2 = - 1 w 2 Entropy Eq.8.3: s 2 s 1 = dq/T Process: Adiabatic and reversible => s 2 = s 1 State 1: (T, P) B.5.2 u 1 = 367.36 kJ/kg, s 1 = 1.7665 kJ/kg K State 2: (P, s) B.5.2 P 2 = 400 kPa, s 2 = s 1 = 1.7665 kJ/kg K interpolate to get T 2 = 22.6 o C, u 2 = 394.36 kJ/kg 1 w 2 = u 1 - u 2 = 367.36 – 394.36 =
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.

Page1 / 7

Sphw6 - HOMEWORK SET 6 8.17 Find the missing properties and...

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