MECH 310 Problems Chapter 3 Solutions, Fall 2011

1 MECH 310 Thermodynamics I October 10, 2011 American University of Beirut, Fall 2011 Handout # 1 Chapter 03 Solved Problems Problem 3.23 a) H2O T = 260°C P = 5 MPa Table B.1.1 or B.1.2 B.1.1 For given T: Psat = 4.689 MPa P > Psat => compressed liquid B.1.2 For given P: Tsat = 264°C T < Tsat => compressed liquid b) H2O T = −2°C P = 100 kPa Table B.1.1 T < Ttriple point Table B.1.5 at −2°C: Psat = 0.518 kPa since P > Psat => compressed solid Problem 3.24 a) Water 100°C, 500 kPa From Table B.1.1 Psat(100°C) = 101.3 kPa 500 kPa > Psat then it is compressed liquid OR from Table B.1.2 Tsat(500 kPa) = 152°C 100°C < Tsat then it is subcooled liquid = compressed liquid b) Ammonia (NH3) -10°C, 150 kPa Table B.2.1: P < Psat(-10 °C) = 291 kPa Superheated vapor c) R-410a 0°C, 350 kPa Table B.4.1: P < Psat(0 °C) = 799 kPa Superheated vapor. Problem 3.25 a) P = 10 MPa, v = 0.003 m3/kg B.1.2 at 10 MPa vf = 0.001452; vg = 0.01803 m3/kg, => vf < v < vg => so mixture of liquid and vapor. 2 b) 1 MPa, 190°C : Only one of the two look-ups is needed P < Psat = 1254.4 kPa so it is superheated vapor T > Tsat = 179.91°C so it is superheated vapor c) 200°C, 0.1 m3/kg: B.1.1 vf = 0.001156 m3/kg; vg = 0.12736 m3/kg, => vf < v < vg => so mixture of liquid and vapor. d) 10 kPa, 10°C : Only one of the two look-ups is needed B.1.1: P > Pg = 1.2276 kPa so compressed liquid B.1.2: T < Tsat = 45.8°C so compressed liquid Problem 3.33 P [kPa] T [°C] v [m3/kg] x a) 500 20 0.001002 Undefined b) 500 151.86 0.20 0.532 c) 1400 200 0.14302 Undefined d) 8581 300 0.01762 0.8 a) Table B.1.1 P > Psat so it is compressed liquid => Table B.1.4 b) Table B.1.2 vf < v < vg so two phase L + V x = (v – vf)/vfg = (0.2 – 0.001093) / 0.3738 = 0.532 T = Tsat = 151.86°C c) Only one of the two look-up is needed Table B.1.1 200°C P < Psat = => superheated vapor Table B.1.2 1400 kPa T > Tsat = 195°C Table B.1.3 sub-table for 1400 kPa gives the state properties d) Table B.1.1 since quality is given it is two-phase v = vf + x × vfg = 0.001404 + 0.8 × 0.02027 = 0.01762 m3/kg Problem 3.35 a) –15°C, 500 kPa Table B.4.1: P > Psat = 480.4 kPa, so compressed liquid. v ≈ vf = 0.000815 m3/kg 3 b) 20°C, 1000 kPa Table B.4.1: P < Psat = 1444 kPa, so superheated vapor Table B.4.2: v = 0.02838 m3/kg c) 20°C, quality 25% Table B.4.1: vf = 0.000815 m3/kg, vfg = 0.01666 m3/kg so v = vf + x vfg = 0.000815 + 0.25 × 0.01666 = 0.00498 m3/kg Problem 3.40 You want a pot of water to boil at 105°C. How heavy a lid should you put on the 15 cm diameter pot when Patm = 101 kPa? Table B.1.1 at 105°C : Psat = 120.8 kPa = 0.01767 m2 Fnet = (Psat –Patm) A = (120.8 - 101) kPa × 0.01767 m2= 0.3498 kN = 350 N Fnet = mLID g mLID = Fnet/g = 350/9.807 = 35.7 kg Problem 3.41 State 1 from Table B.1.1 at 120°C v = vf + x vfg = 0.001060 + 0.25 × 0.8908 = 0.22376 m3/kg State 2 has same v at 140°C also from Table B.1.1 x =(v – vf)/vfg = (0.22376 - 0.00108)/0.50777 = 0.4385 P = Psat = 361.3 kPa Problem 3.43 Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. State 1: B 1.2 v1 = vg (200 kPa) = 0.8857 m3/kg, T1 = 120.2°C Process: P = constant = 200 kPa State 2: P, v2 = v1/2 = 0.44285 m3/kg Table B.1.2 v2 < vg so two phase T2 = Tsat = 120.2°C Height is proportional to volume 4 h2 = h1 × v2/v1 = 0.1 × 0.5 = 0.05m Problem 3.50 Control volume: both tanks. Constant total volume and mass process. State A1: (P, v) mA = VA/vA = 1/0.5 = 2 kg State B1: (P, T) Table B.1.3 vB = 0.6173 m3/kg ⇒ VB = mBvB = 3.5 × 0.6173 = 2.1606 m3 Final state: mtot = mA + mB = 5.5 kg Vtot = VA + VB = 3.1606 m3 v2 = Vtot/mtot = 0.5746 m3/kg Problem 3.53 Initial state: v = 7.6707 m3/kg from table B.1.1 Final state: v = 1.10 × vg = 1.1 × 7.6707 = 8.4378 m3/kg Interpolate at 60°C between saturated (P = 19.94 kPa) and superheated vapor P = 10 kPa in Tables B.1.1 and B.1.3 P ≅ 19.941 + (10 − 19.941) x (8.4378 − 7.67071)/(5.3345 − 7.6707) = 18.9 kPa 5 Problem 3.70 The mass comes from knowledge of state 1 and ideal gas law m = P1V1 /RT1 = 1000 kPa × 0.1 m3/(0.287 kJ/kgK × (227 + 273) K )= 0.697 kg The final pressure is found from the ideal gas law written for state 1 and state 2 and then eliminate the mass, gas constant and volume (V2 = V1) between the equations P1 V1 = m RT1 and P2 V2 = m RT2 P2 = P1 × T2/T1 = 1000 × 1500500 = 3000 kPa Problem 3.72 m = PV/RT = 600 × 1/(0.2968 × 400) = 5.054 kg m2 = m - 0.5 = 4.554 kg P2 = m2RT2/V = 4.554 × 0.2968 × 375/1 = 506.9 kPa Problem 3.111 Real gas behavior: P = 1000 kPa from Table B.4.2 Ideal gas constant: R = /M = 8.31451/72.585 = 0.1146 kJ/kg K P = RT/v = 0.1146 × (273.15 + 60) / 0.0347 = 1100 kPa which is 10% too high Generalized chart Fig D.1 and critical properties from A.2: Tr = 333.2/(273.15 + 71.3) = 0.967; Pc = 4901 kPa Assume P = 1000 kPa => Pr = 0.204 => Z ≅ 0.92 v = ZRT/P = 0.92 × 0.1146 × 333.15 /1000 = 0.03512 too high Assume P = 1050 kPa => Pr = 0.214 => Z ≅ 0.915 v = ZRT/P = 0.915 × 0.1146 × 333.15 / 1050 = 0.03327 too low P = 1000 + ( 1050 − 1000 ) × (0.03470 − 0.03512)/(0.03327 − 0.03512) P = 1011 kPa 1.1 % high