Practice Problems

Answers to Practice Problems, Thermo 8th Edition (Moran et al.)

Ch. 1

[1.4] (a). 61 in³, (b). 0.616 BTU, (c). 99.596 ft lb/s, (d). 50 lb/min, (e). 44.09 lb_f/in² or psi, (f). 0.54 ft³/s, (g). 45.57 ft/s, (h). 1 ton

[1.25] Hint: Perform a force balance for the piston

[1.31] Sketch

Ch. 2

[2.26] (a). 0.033 m³, (b). -20 kJ (in, or "work done on system")

[2.33] Graph; Process 1-2:0, Process 2-3:-554.5 kJ, Process 3-4: -400 kJ

[2.67] (a). 0.25 m³/kg (b). -50.4 kJ (in, or "work done on system") (c). -10.4 kJ (out, or "heat loss")

[2.74] W₁₂=1,200, ΔE₂₃=800, W₂₃=0, ΔE₃₄=0, Q₄₁=1000

[2.80] (a). 0.103 m³ (b). W₁₂ = Q₁₂ = -18.8 kJ (c). refrigeration or heat pump cycle (must show justification!)

[2.90] Q_dot_in = 0.45 kW, COP (β) = 3

[2.92] W_cycle = 10.42 kW*h, Q_dot_in = 2.70 kW, Total cost = $1.04 (for 24 hours)

Ch. 3

[3.6] T-v and p-v diagrams: (a). Entire two-phase (saturated) region (b). Subcooled (c). Superheated (d). Subcooled (e). Solid

[3.7] (a). 0.1879 m³/kg(b). 260 ^oC (c). 0.15567 m³/kg

[3.10] (a). p = 3.613 bar (b). ν = 0.001029 m³/kg (c). ν ≈ 0.0319 m³/kg (d). ν = 2.556 m³/kg

[3.14] T-v and p-v diagrams: (a). Subcooled (b). Saturated (c). Superheated

[3.19] V = 0.9056 m³, vapor volume fraction = 0.977 (i.e., 97.7% by volume)

[3.23] p₁ = 15.54 bar, p₂ = 1.014 bar (also: x₁ = 100%, x₂ = 7.6%); sketch process on T-v and p-v diagrams

[3.38] (a). T = -14.66 ^oC, u = 153.67 kJ/kg (b). p = 5 bar, h = 2855.4 kJ/kg (c). T = 44.46 ^oC, v = 0.2971 m³/kg

[3.46] Q/m (heat transfer per mass) = 407.6 kJ/kg

[3.78] γ = 20

[3.92] (a). 0.0534 m³ where Z=0.86  (b). 0.05282 m³ (more accurate)

[3.112] m = 4.65 kg, Q = W = 277.3 kJ

[3.142] η = 0.204 = 20.4%

[3.144] w₁₂=q₁₂=38.42 kJ/kg, w₂₃=33.72 kJ/kg, q₂₃=0, w₃₄=-55.39 kJ/kg, q₃₄=-193.77 kJ/kg, w₄₁=0, q₄₁=172.08 kJ/kg; η = 0.08 = 8%

Ch. 4

[4.16] (a). m_dot₁= 0.351 kg/s (b). A₂= 5.1 cm²

[4.24] (a). m_dot₁= 0.621 kg/s (b). v₂= 52.23 m/s (c). Q_dot = 6.82 kW

[4.31] (a). 600.8 m/s (b). A₁= 22.5  cm², A₂= 7.1  cm²

[4.42] Turbine power = W_dot_T = 6927 kW = 6.927 MW

[4.75] (a). m_dot = 3.775 kg/min = 0.063 kg/s (b). Q_dot = -574 kJ/min = -9.6 kW

Ch. 5

[5.17] (a). η_max = 0.7 Irreversibly (b). η_max = 0.7 Reversibly (c).W_cycle=200kJ Impossible (d). η_max = 0.75 Impossible

[5.43] (a). Impossible (b). Irreversibly (c). Reversibly (d). Irreversibly

[5.68] (a). W_dot_cycle = 0.8kW (b). The minimum theoretical temperature is 246.7K

[5.74] (a). W_dot_elec = Q_dot_H = 12.31 kW (b). 3.52 kW (c). W_dot_{elec_min} = 0.84 kW

Ch. 6

[6.3] (a). h = 1430.55 kJ/kg (b). s = 7.7622 kJ/kgK (c). s = 0.7649 kJ/kgK (d). u = 1331.94 kJ/kg

[6.7] (a). T₂ = 368.8^oC, Δh = 638.8 kJ/kg (b). T₂ = 204.1^oC, Δh = 826.8 kJ/kg

[6.15] (a). Q = -2392 kJ (heat loss) (b). ΔS = -5.5238 kJ/K

[6.41] (a). p₂ = 3.33 bar (b). A-22: W = -676.2 kJ, A-20: W = -675.2 kJ (c). A-22: σ = 1.725 kJ/K, A-20: σ = 1.724 kJ/K

[6.49] Cannot occur adiabatically (must show proof!); heat loss must exist

[6.119] (a). 774 K (b). -334.3 kJ/kg (work done on system)

[6.143] (a). η_t = 0.947 (94.7%) (b). σ_dot_cv = 0.465 ^kW⁄_K

[6.162] (a). m_dot_steam/m_dot_air = 0.303 (b). m_dot_steam/m_dot_air = 0.232