answersthermo - Answers to Selected Problems 3.1 (a) 268.8...

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Updated: 1/14/04 Answers to Selected Problems 3.1 (a) 268.8 K and 4.625 × 10 5 Pa; (b) at 90 o (horizontal) 282.6 K and 4.863 × 10 5 Pa; (c) at 180 o (vertical downward) 296.4 K and 5.1 × 10 5 Pa. 3.2 (a) δ Q / δ t = 3.51 × 10 8 J/hr [contractor pays] (b) δ Q / δ t = 3.23 × 10 8 J/hr [contractor collects fee] 3.5 (a) L A = 0.4352 m, L B = 0.0218 m, T A = T B = 311 K, P A = P B = 7 bar (b) Same as (a) (c) There is no real solution given the constraints, we must specify more about the expansion (or compression) of A or B. For example, if A expands adiabatically, P A = P B = 6.68 bar, T A = 263.6 K, T B = 881.3 K, L A = 0.3917 m, and L B = 0.0655 m. 3.7 Many variations are possible. If we assume gas in the cylinder is ideal and that it expands adiabatically, then (a) total time t = 0.9 s, (b) height h ( max ) = 4.78 m, by reducing tube length to 1.64 m. 3.12 (a) After 6 s, T = 349.5 K, P = 1.139 × 10 5 Pa in the bulge. (b) After 3 s, T = 454.6 K, P = 6.26 × 10 3 Pa in the large tank. 4.1 For the minimum condition T A (final) = 132 K and for the maximum condition T A (final) = 617 K, these extrema require using the object with the lowest mass x heat capacity product. 4.2 W = 9.807 y in J/kg evaporated, for typical values of ambient dry bulb and wet bulb temperatures of 300 K and 280 K respectively, y = 16.8 km! 4.3 (a) n ± B = 0.166 mol/s, n ± C = 0.833 mol/s (b) T D = 299.1K (c) W = ! 26.4 J/s (d) S = 13.3 J/K (e) Same as (d). (f) max W = ! 4000 J/s 4.4 W = 1.11 × 10 4 J, T f = 288.7 K, P f = 0.54 bar 4.6 (a) 21.6 s (b) 391.8 K and 2.2 bar at 10 s (c) S gas = 4.93 × 10 3 J/K = S universe (d) S (gas in tank) = ! 813.6 J/K S (gas vented) = +2020 J/K S (surroundings) = +1203 J/K (e) W min = +2.83 × 10 5 J
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Updated: 1/14/04 4.7 (a) Q ± = 2400 W , T = 327.7 K, and Q = 22,200 J after 10 s (b) T = 1750 K at t = 57.7 s with all stored work consumed. 4.9 W net = 3.17 × 10 6 J 4.12 (a) T 2 = 227.9 K, T 1 = 392.9 K (2) after venting (1) before venting (b) T 2 = 211.1 K, T 1 = 363.9 K (c) W = 2547 J/mol (d) W max = 1.38 × 10 5 J/kg 4.15 (a) S = 1.38 × 10 3 J/K hr; (b) S = 2 J/K hr 4.16 (a) Case (1) 1.98 × 10 2 kWhr; (2) 2.78 × 10 2 kWhr; (3) 2.16 × 10 2 kWhr; (4) 2.78 × 10 2 kWhr (b) Case (1) 420 K, 42 J/K; (2) 300 K, 0 J/K; (3) 300 K, 73.9 J/K; (4) 300 K, 0 J/K 4.18 W = 5.23 × 10 3 J = ! Q 4.22 (a) P = 6.35 × 10 4 Pa (b) T = 272.8 K (c) P = 3.68 × 10 4 Pa with 2 pumps (d) P = 1.91 × 10 4 Pa and T (e) Power (minimum) = 20.9 W 4.24 (a) max W = ! 198.76 kJ/kg (Carnot + expansion work) W p = ideal pump work (estimated to include PE for lifting water and for the compression of gas space in storage tanks) 4.30 (a) max W ± = 6.38 × 10 6 J/s (b) u η = 0.90 = (actual power)/(maximum power) (c) no, the maximum power outputs are the same 5.2 (a) () ( ) () V Pd d N dT S dy N S T y y j n j j n j j µ = µ = = + 1 1 0 1 + n (b) ( ) = µ + = = n j j j dN T V d T P T d U dy U T S y 1 1 1 / / / 1 / 1 5.4 (a) i N , V T P / (b) ( ) ( ) N P PT P T V T V G T G T P = N T, , / / = + (c) U (d) ∂∂ H V T,N =V G TG G PP PT PP // / bg
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Updated: 1/14/04 5.7 (a) () ( ) T C V T V S T S P P N G, / / / / + = (b) ( ) T T N T, P V = V T V P G A ln / ln / / / = 5.10 T f = 1087 K 5.12 T 2 1 22 / / / κ α + = = p p v V T NC T NC y 5.21 V c = κ RT = ( C p / C v ) RT 330 m/s (use mass units for R with molecular weight) distance at 2 s 660 m 5.22 (a) T 2 / κ α = p v p V C C 5.23 (b) No, the ratio as shown in Problem 5.17 is expressed in terms of PVT properties and an isentropic derivative which in turn requires non- PVT property information
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This note was uploaded on 09/29/2011 for the course CHEM 101 taught by Professor Smith during the Spring '11 term at Portland State.

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answersthermo - Answers to Selected Problems 3.1 (a) 268.8...

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