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TAMU CHEN 205 - extra-sol234

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ME 24-221 THERMODYNAMICS I Solutions to extra problems in Chapters 2, 3 and 4: December 5, 2000 J. Murthy 2.25 Two reservoirs, A and B, open to the atmosphere, are connected with a mercurymanometer. Reservoir A is moved up/down so the two top surfaces are level at h3as shown in Fig. P2.25. Assuming that you know ρA, ρHg and measure theheights h1, h2 , and h3, find the density ρB.Solution:Balance forces on each side:P0 + ρAg(h3 - h2) + ρHggh2 = P0 + ρBg(h3 - h1) + ρHggh1⇒ ρB = ρAh3 - h2h3 - h1 + ρHgh2 - h1h3 - h1 2.32 Two piston/cylinder arrangements, A and B, have their gas chambers connected bya pipe. Cross-sectional areas are AA = 75 cm2 and AB = 25 cm2 with the pistonmass in A being mA = 25 kg. Outside pressure is 100 kPa and standard gravitation.Find the mass mB so that none of the pistons have to rest on the bottom.Solution: A B P P 0 0 Force balance for both pistons: F↑ = F↓ A: mPAg + P0AA = PAA B: mPBg + P0AB = PABSame P in A and B gives no flow between them. mPAgAA + P0 = mPBgAB + P0 => mPB = mPA AA/ AB = 25 × 25/75 = 8.33 kg2.36 Two cylinders are connected by a piston as shown in Fig. P2.36. Cylinder A isused as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg andthere is standard gravity. What is the gas pressure in cylinder B?Solution: Force balance for the piston: PBAB + mpg + P0(AA - AB) = PAAAAA = (π/4)0.12 = 0.00785 m2; AB = (π/4)0.0252 = 0.000491 m2PBAB = PAAA - mpg - P0(AA - AB) = 500× 0.00785 - (25 × 9.807/1000) - 100 (0.00785 - 0.000491) = 2.944 kNPB = 2.944/0.000491 = 5996 kPa = 6.0 MPa 4.21 A cylinder having an initial volume of 3 m3 contains 0.1 kg of water at 40°C. Thewater is then compressed in an isothermal quasi-equilibrium process until it has aquality of 50%. Calculate the work done in the process. Assume the water vaporis an ideal gas.Solution: C.V. Water2 1 3 v T P G 1 P 40o C v1 = V1/m = 30.1 = 30 m3/kg ( > vG )Tbl B.1.1 => PG = 7.384 kPa very lowso H2O ~ ideal gas from 1-2P1 = PG vGv1 = 7.384 × 19.5230 = 4.8 kPaV2 = mv2 = 0.1 × 19.52 = 1.952 m3T = C: W12 =⌡⌠1 2 PdV = P1V1ln V2V1 = 4.8 × 3.0 × ln 1.9523 = -6.19 kJv3 = 0.001008 + 0.5 × 19.519 = 9.7605 => V3 = mv3 = 0.976 m3P = C = Pg: W23 = ⌡⌠2 3 PdV = Pg (V3-V2) = 7.384(0.976 - 1.952) = -7.21 kJTotal work: W13 = -6.19 - 7.21 = -13.4 kJ3.12 Air in a tank is at 1 MPa and room temperature of 20°C. It is used to fill aninitially empty balloon to a pressure of 200 kPa, at which point the diameter is 2m and the temperature is 20°C. Assume the pressure in the balloon is linearlyproportional to its diameter and that the air in the tank also remains at 20°Cthroughout the process. Find the mass of air in the balloon and the minimumrequired volume of the tank.Solution: Assume air is an ideal gas.Balloon final state: V2 = (4/3) π r3 = (4/3) π 23 = 33.51 m3 m2bal = P2 V2 / RT2 = 200× 33.51 / 0.287 × 293.15 = 79.66 kgTank must have P2 ≥ 200 kPa => m2 tank ≥ P2 VTANK /RT2Initial mass must be enough: m1 = m2bal + m2 tank = P1V1 / R T1P1VTANK / R T1 = m2bal + P2VTANK / RT2=>VTANK = RTm2bal / (P1 – P2) = 0.287 × 293.15 × 79.66/ (1000 – 200 ) = 8.377 m33.22 Is it reasonable to assume that at the given states the substance behaves as an idealgas?Solution:a) Oxygen, O2 at 30°C, 3 MPa Ideal Gas ( T » Tc = 155 K from A.2)b) Methane, CH4 at 30°C, 3 MPa Ideal Gas ( T » Tc = 190 K from A.2)c) Water, H2O at 30°C, 3 MPa NO compressed liquid P > Psat (B.1.1)d) R-134a at 30°C, 3 MPa NO compressed liquid P > Psat (B.5.1)e) R-134a at 30°C, 100 kPa Ideal Gas P is low < Psat (B.5.1)ln PTVaporLiq.Cr.P.a, bc, de3.35 Determine the mass of methane gas stored in a 2 m3 tank at −30°C, 3 MPa. Estimatethe percent error in the mass determination if the ideal gas model is used.Solution:The methane Table B.7.2 linear interpolation between 225 and 250 K.⇒ v ≅ 0.03333 + 243.15-225250-225×(0.03896-0.03333) = 0.03742 m3/kgm = V/v = 2/0.03742 = 53.45 kgIdeal gas assumptionv = RT/P = 0.51835 × 243.15/3000 = 0.042m = V/v = 2/0.042 = 47.62 kgError: 5.83 kg 10.9% too small 4.10 A piston/cylinder contains 1 kg of liquid water at 20°C and 300 kPa. Initially thepiston floats, similar to the setup in Problem 4.7, with a maximum enclosedvolume of 0.002 m3 if the piston touches the stops. Now heat is added so a finalpressure of 600 kPa is reached. Find the final volume and the work in the process.Solution:Take CV as the water which is a controlmass: m2 = m1 = m ;Table B.1.1: 20°C => Psat = 2.34 kPaState 1: Compressed liquid v = vf(20) = 0.001002 m3/kgState 1a: vstop = 0.002 m3/kg , 300 kPa1a21PVP1VstopState 2: Since P = 600 > Plift then v = vstop = 0.002 and V = 0.002 m3For the given P : vf < v < vg so 2-phase T = Tsat = 158.85 °CWork is done while piston moves at Plift = constant = 300 kPa so we get1W2 = ∫ P dV = m Plift(v2 -v1) = 1 × 300(0.002 - 0.001002) = 0.30 kJ4.13 Air in a spring loaded piston/cylinder has a pressure that is linear with volume, P= A + BV. With an initial state of P = 150 kPa, V = 1 L and a final state of 800kPa and volume 1.5 L it is similar to the setup in Problem 3.16. Find the workdone by the air.Solution:Knowing the process equation: P = A + BV giving a linear variation ofpressure versus volume the straight line in the P-V diagram is fixed by the twopoints as state 1 and state 2. The work as the integral of PdV equals the areaunder the process curve in the P-V diagram.0 PV21State 1: P1 = 150 kPa V1 = 1 L = 0.001 m3State 2: P2 = 800 kPa V2 = 1.5 L = 0.0015 m3Process: P = A + BV linear in V ⇒ 1W2 = ⌡⌠ 1 2 PdV = (P1 + P22)(V2 - V1) = 12 (150 + 800)(1.5 - 1)× 0.001 = 0.2375 kJ 4.31 A vertical cylinder (Fig. P4.31) has a 90-kg piston locked with a pin trapping 10 Lof R-22 at 10°C, 90% quality inside. Atmospheric pressure is 100 kPa, and thecylinder cross-sectional area is 0.006 m2. The pin is removed, allowing the pistonto move and come to rest with a final temperature of 10°C for the …


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