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Sonntag, Borgnakke, Van Wylen. Solution Manual Chapters 1-9 for Classical Mechanics and Thermodynamics(427s)_PG_.pdf |
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Size 1.3Mb Date Nov 14, 2006 |
Balance forces in the manometer: (H - h2) - (H - h1) = ∆hHg = h1 - h2 P1A + ρ H Oh1gA + ρ Hg(H - h1)gA 2 = P2A + ρ H Oh2gA + ρ Hg(H - h2)gA
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2-12 2.36 Two cylinders are connected by a piston as shown in Fig. P2.36. Cylinder A is used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and there is standard gravity. What is the gas pressure in cylinder B? Solution: Force balance for the piston: AA = (π/4)0.12 = 0.00785 m2;...
= [(10/12) × 62.3 × 32.174] / 32.174 ×144 = Pgauge = 0.36 lbf/in2 h = H × sin 30° ⇒ H = h/sin 30° = 2h = 20 in = 0.833 ft...
3-3 3.4 A substance is at 2 MPa, 17°C in a rigid tank. Using only the critical properties can the phase of the mass be determined if the substance is nitrogen, water or propane ? Solution: a) b) c) Find state relative to critical point properties which are: : 3.39 MPa 126.2 K Nitrogen N2 Water H2O : 22.12 MPa 647.3 K Propane C3H8 : 4.25 MPa 369.8 K...
3-5 3.9 A 1-m3 rigid tank with air at 1 MPa, 400 K is connected to an air line as shown in Fig. P3.9. The valve is opened and air flows into the tank until the pressure reaches 5 MPa, at which point the valve is closed and the temperature inside is 450K. a. What is the mass of air in the tank before and after the process? b. The tank eventually cools to room temperature, 300 K. What is the pressure inside the tank then? Solution: P, T known at both states and assume the air behaves as an ideal gas. P 1V 1000 × 1 = = 8.711 kg mair1 = RT1 0.287 × 400 mair2 = P 2V RT2 = 5000 × 1 = 38.715 kg 0.287 × 450...
∆H = (V2 -V3)/A = (1.96-1.54) × 0.001/0.00785 = 0.053 m = 5.3 cm 3.12 Air in a tank is at 1 MPa and room temperature of 20°C. It is used to fill an initially empty balloon to a pressure of 200 kPa, at which point the diameter is 2 m and the temperature is 20°C. Assume the pressure in the balloon is linearly proportional to its diameter and that the air in the tank also remains at 20°C throughout the process. Find the mass of air in the balloon and the minimum required volume of the tank. Solution: Assume air is an ideal gas. Balloon final state: Tank must have 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 kg P2 ≥ 200 kPa => m2 tank ≥ P2 VTANK /RT2 Initial mass must be enough: m1 = m2bal + m2 tank = P1V1 / R T1 P1VTANK / R T1 = m2bal + P2VTANK / RT2
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3-12 3.24 Determine whether refrigerant R-22 in each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor. Solution: All cases are seen in Table B.4.1 a. 50°C, 0.05 m3/kg b. 1.0 MPa, 20°C c. 0.1 MPa, 0.1 m3/kg d. 50°C, 0.3 m3/kg e − 20°C, 200 kPa superheated vapor, v > vg =0.01167 at 50°C compressed liquid, P > Pg = 909.9 kPa at 20°C mixture liq. & vapor, vf < v < v g at 0.1 MPa superheated vapor, v > vg = 0.01167 at 50°C superheated vapor, P < Pg = 244.8 kPa at -20°C
P C.P. T f b c a d e T v b c...
3-14 3.28 Give the phase and the specific volume. Solution: a. H2O b. H2O c. CO2 T = 275°C T = −2°C T = 267°C P = 5 MPa P = 100 kPa Table B.1.1 or B.1.2 Table B.1.5 Psat = 5.94 MPa => superheated vapor v = 0.04141 m3/kg Psat = 0.518 kPa =>compressed solid v ≅ vi = 0.0010904 m3/kg P = 0.5 MPa Table A.5 v= RT 0.18892 × 540 = = 0.204 m3/kg P 500 Table A.5 RT 0.287 × 293 = = 0.420 m3/kg P 200...
Determine the mass of methane gas stored in a 2 m3 tank at −30°C, 3 MPa. Estimate the 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. 243.15-225 ⇒ v ≅ 0.03333 + × (0.03896-0.03333) = 0.03742 m3/kg 250-225 m = V/v = 2/0.03742 = 53.45 kg Ideal gas assumption v = RT/P = 0.51835 × 243.15/3000 = 0.042 m = V/v = 2/0.042 = 47.62 kg Error: 5.83 kg 10.9% too small...
A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up by 5°C per hour due to a failure in the refrigeration system. How long time will it take before the methane becomes single phase and what is the pressure then? Solution: Use Table B.7.1 Assume rigid tank v = const = v1 = 0.002439 + 0.25×0.30367 = 0.078366 All single phase when v = vg => T ≅145 K P = Psat= 824 kPa ∆t = ∆Τ/5°C ≅ (145 – 120 ) / 5 = 5 hours...
m1 = V/v = 0.005/0.18513 = 0.027 kg State 2: P2 = 1.2 MPa, Flow in so: m2 = 2 m1 = 0.054 kg Process: Piston Fext = KV2 = PA => P = CV2 => P2 = P1 (V2/V1)2 From the process equation we then get: 1200 1/2 1/2 V2 = V1 (P2/P1) = 0.005 ( ) = 0.006984 m3 615 v2 = V/m = At P2, v2: 0.006984 = 0.12934 m3/kg 0.054 T2 = 70.9°C...
Check state 2 in Table B.1.1 vg(T2) = 0.12736; Pg(T2) = 1.5538 MPa If v2 = vg(T2) ⇒ P2 = 12.3 MPa > Pg not OK v2 = 0.0161 m3kg < vg sat. OK, x2 = (0.0161 - 0.001156) / 0.1262 = 0.118...
Tr = ( 460 – 30 ) / 271.4 = 1.58, P r = 450/706 = 0.64 ⇒ Ideal gas Z = 1 ⇒ m = PV/RT = 390 lbm...
3-34 3.76E Saturated water vapor at 200 F has its pressure decreased to increase the volume by 10%, keeping the temperature constant. To what pressure should it be expanded? Solution: v = 1.1 × vg = 1.1 × 33.63 = 36.993 ft3/lbm Interpolate between sat. at 200 F and sup. vapor in Table C.8.2 at 200 F, 10 lbf/in2 P ≅ 10.54 lbf/in2...
State 1: v1 = 1.7524 ft3/lbm P = Cv ⇒ C = P1/v1 = 256.79 State 2: sat. vap. x2 = 1 Trial & error on T2 or P2 At 350 lbf/in2: Pg/vg = 263.8 > C
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Solution : C.V. Balloon and the tank. Control mass. Balloon State 1: (T, x) and size m1 = 5 kg D...
W23 = P2=3 × m(v3 - v2) = 0.95 × 0.2 ×(175 - 350) ×144 / 778 = -6.16 Btu W13 = - 6.16 - 4.33 = -10.49 Btu...
1π -W12 = ( ) 30×106 × 1 × (10-3)2 = 2.94 ft•lbf 2 16 4.77E The sun shines on a 1500 ft2 road surface so it is at 115 F. Below the 2 inch thick asphalt, average conductivity of 0.035 Btu/h ft F, is a layer of compacted rubbles at a temperature of 60 F. Find the rate of heat transfer to the rubbles....
5-17 5.23 Find the heat transfer in Problem 4.10. Solution: Take CV as the water. Properties from table B.1 m2 = m1 = m ; State 1: Compressed liq. m(u2 - u1) = 1Q2 - 1W2 v = vf (20) = 0.001002, u = uf = 83.94...
5-19 5.27 Two kilograms of nitrogen at 100 K, x = 0.5 is heated in a constant pressure process to 300 K in a piston/cylinder arrangement. Find the initial and final volumes and the total heat transfer required. Solution: Take CV as the nitrogen. m2 = m 1 = m ; State 1: Table B.6.1 v1 = 0.001452 + 0.5 × 0.02975 = 0.01633 m3/kg, h1 = -73.20 + 0.5 × 160.68 = 7.14 kJ/kg State 2: P = 779.2 kPa , 300 K => sup. vapor interpolate in Table B.6.2 v2 = 0.14824 + (0.11115-0.14824)× 179.2/200 = 0.115 m3/kg, V2 = 0.23 m3 h2 = 310.06 + (309.62-310.06) × 179.2/200 = 309.66 kJ/kg Process: P = const. ⇒ 1W2 = ⌠PdV = Pm(v2 - v1) ⌡
1Q 2 = m(u2 - u1) + 1W2 = m(h2 - h1) = 2 x (309.66 - 7.14) = 605 kJ...
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