MECH 310 Problems Chapter 4 Solutions, Fall 2011

1 MECH 310 Thermodynamics I October 21, 2011 American University of Beirut, Fall 2011 Handout # 2 Chapter 04 Solved Problems Problem 4.20 W = ∫ F dx = F Δx = 1500 N × 100 m =150 000 J = 150 kJ W = ∫ F dz = ∫ mg dz = mg ΔZ = 500 kg × 9.807 m/s2 × 3 m = 14 710 J = 14.7 kJ Wtotal = 150 kJ+14.7 kJ= 164.7 kJ Problem 4.22 W = ∫ F dx = ∫ P dV = ∫ PA dx = PAx H = PΔV ΔV = W/P = 1 kJ/1200 kPa = 0.000 833 m3 Both cases the height is H = ΔV/A H1 = 0.000833/0.01 = 0.0833 m H2 = 0.000833/0.03 = 0.0278 m Problem 4.27 Control volume radiator. After the valve is closed no more flow, constant volume and mass. 1: x1 = 1, P1 = 110 kPa ⇒ v1 = vg = 1.566 m3/kg from Table B.1.2 2: T2 = 25°C Process: v2 = v1 = 1.566 m3/kg = [0.001003 + x2 × 43.359] m3/kg x2 = 1.566 – 0.001003/43.359 = 0.0361 State 2 : T2 , x2 From Table B.1.1 P2 = Psat = 3.169 kPa Work is 0 since constant volume process Problem 4.28 Table B.1.2 v1= 0.4625 m3/kg V1 = mv1 = 0.0925 m3 v2 = v1/ 2 = 0.23125 m3/kg V2 = V1 / 2 = 0.04625 m3 Process: P = Cst so the work term integral is W = ∫ PdV = P(V2-V1) = 400 kPa × (0.04625 – 0.0925) m3 = -18.5 kJ 2 Problem 4.30 Taking a control volume enclosing all the argon in both A and B. The boundary movement work done in cylinder B against constant external pressure of 150 kPa. Argon is an ideal gas, so writing out that the mass and temperature at state 1 and 2 are the same: PA1VA = mARTA1 = mART2 = P2( VA + VB2) => VB2 = 250 × 0.4/150 - 0.4 = 0.2667 m3 W12 = ∫ PextdV = Pext(VB2 - VB1) = 150 kPa (0.2667 - 0) m3 = 40 kJ Problem 4.33 P1 = P0 + ρgH = 101.32 + 997 × 9.807 × 5 / 1000 = 150.2 kPa ΔV = H × A = 5 × 0.1 = 0.5 m3 W12 = AREA = ∫ P dV = ½ (P1 + P0)/(Vmax -V1)= ½ (150.2 + 101.32) kPa × 0.5 m3= 62.88 kJ 3 Problem 4.35 a. State 1: Compressed liquid, saturated liquid at same temperature. B.1.1: v1 = vf(20°C) = 0.001002 m3/kg, State 2: v2 = V2/m = 0.1/1 = 0.1 m3/kg and P = 3000 kPa from B.1.3 => Superheated vapor close to T = 400°C so by interpolating: T2 = 404°C c. Work is done while piston moves at linearly varying pressure, so we get: W12 = ∫P dV= Pave(V2 - V1)=1/2(P1+P2)(V2-V1)= 0.5 (300+3000)(0.1-0.001) = 163.35 kJ Problem 4.47 Process: Pvn = Const = P1vn1 = P2 vn2 Ideal gas so : Pv = RT v1 = RT/P = 0.287 × 325/125 = 0.7462 m3/kg v2 = RT/P = 0.287 × 500/300 = 0.47833 m3/kg From the process equation (P2/ P1) = (v1/ v2)n => ln(P2/ P1) = n ln(v1/ v2) n = ln(P2/ P1) / ln(v1/ v2) = ln 2.4/ln 1.56 = 1.969 The work per unit mass (specific work) is: Problem 4.50 Process: Pvn = Const = P1vn1 = P2 vn2 To evaluate the work: 4 Problem 4.59 a) From Table B.4.2: v1 = 0.17732 m3/kg, m = V/v = 0.010/0.17732 = 0.056395 kg For the final state, if the piston were at the final stop, then V1a would have been of 11 L Then v1a= V/m= 0.011/0.056395=0.19505 m3/kg => T1a = -9.4°C but T2 = 15°C so T2> T1a => P2 >P1 If the piston didn’t reach the stops yet, the pressure would have remained constant at 150 kPa and by using 150 kPa and 15°C, we would have obtained the right specific volume Problem 4.86 Process: P = constant : Boundary work: δW = P dV => W= PΔV Assuming ideal gas behavior for air V2 = V1× (T2/T1) = 0.1 × (600/300) = 0.2 m3 = P ΔV/Δt = 300 × (0.2 - 0.1)/30 kPa m3/s = 1 kW Problem 4.90 Steady conduction through the bottom of the steel pot. Assume the inside surface is at the liquid water temperature. = k A ΔT/Δx ⇒ ΔΤ = Δx / kΑ ΔT = 250 W × 0.005 m /(50 W/m-K × π/4 × 0.22 m2) = 0.796 K T = 15 + 0.796 ≅ 15.8°C Problem 4.103 State 1: v1 = V/m = 0.2 / 2 = 0.1 m3/kg Process: 1 → 2 → 3 or 1 → 3’ State at stops: 2 or 2’ v2 = Vstop/m = 0.4 m3/kg & T2 = 600°C Table B.1.3 ⇒ Pstop = 1 MPa < P3 since Pstop < P then the process is as 1 → 2 → 3 5 State 3: P3 = 1.2 MPa, v3 = v2 = 0.4 m3/kg ⇒ T3 ≅ 770°C W13 = W12 + W23 = W12 = 0.5(P1 + P2)(V2 - V1) =0.5(100 + 1000) kPa × (0.8 - 0.2) m3 (this is only possible because the spring is linear and the force it applies, induces a pressure that varies linearly with the displacement and hence the swept volume W23 = 0 (constant volume) Hence W13= 330 kJ Problem 4.109 The process equation and T determines state 2. Use ideal gas law to say = 396 kPa = 0.2 ( 100/396 )1/1.1 = 0.0572 m3 For propane Table A.2: Tc = 370 K, Pc = 4260 kPa, Figure D.1 gives Z. Tr1 = 0.81, Pr1 = 0.023 => Z1 = 0.98 Tr2 = 0.92, Pr2 = 0.093 => Z2 = 0.95 Ideal gas model OK for both states, work is integrated to give: W12= -26.7 kJ