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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 |
2-4 2.10 On the moon the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 2.1), what is the reading? Solution: Moon gravitation is: g = gearth/6...
P = ρ g∆l = 13550 × 9.80665 × 0.725 × 10-3 = 96.34 kPa 2.16 A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gunpowder is burned a pressure of 7 MPa is created in the gas behind the ball. What is the acceleration of the ball if the cylinder (cannon) is pointing horizontally? Solution: The cannon ball has 101 kPa on the side facing the atmosphere. ma = F = P 1 × A - P 0 × A a = (P1 - P0 ) × A / m = ( 7000 - 101 ) π [ ( 0.152 /4 )/5 ] = 24.38 m/s2 2.17 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative to the horizontal direction. Solution: ma = F = ( P1 - P0 ) A - mg sin 40 = 121.9 - 31.52 = 90.4 N a = 90.4 / 5 = 18.08 m/s2
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Process 2 → 3 is constant V, constant mass cooling to T3 P3 = P2 × (T3/T2) = 5000 × (300/450) = 3.33 MPa 3.10 A hollow metal sphere of 150-mm inside diameter is weighed on a precision beam balance when evacuated and again after being filled to 875 kPa with an unknown gas. The difference in mass is 0.0025 kg, and the temperature is 25°C. What is the gas, assuming it is a pure substance listed in Table A.5 ? Solution: π Assume an ideal gas with total volume: V = (0.15)3 = 0.001767 m3 6 _ 0.0025 × 8.3145 × 298.2 mRT M= = = 4.009 ≈ MHe PV 875 × 0.001767 => Helium Gas...
3-30 3.64E A cylinder is fitted with a 4-in.-diameter piston that is restrained by a linear spring (force proportional to distance) as shown in Fig. P3.16. The spring force constant is 400 lbf/in. and the piston initially rests on the stops, with a cylinder volume of 60 in.3. The valve to the air line is opened and the piston begins to rise when the cylinder pressure is 22 lbf/in.2. When the valve is closed, the cylinder volume is 90 in.3 and the temperature is 180 F. What mass of air is inside the cylinder? π Solution: V1 = V2 = 60 in3; Ap = × 42 = 12.566 in2 4 2 ; V = 90 in3 , T = 180°F = 639.7 R P P2 = 22 lbf/in 3 3
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b) Ideal gas T = constant ⇒ PV = mRT = constant W = ⌠PdV = P1V1 ln ⌡ P1 P2 = 100 × 0.02 × ln 100 = -2.2 kJ 300...
Solution: Take CV as the R-134a which is a control mass. Process: Pv1.5 = constant until P = 500 kPa P = Psat = 201.7 kPa from Table B.5.1 m 2 = m1 = m...
4-26 4.47 The sun shines on a 150 m2 road surface so it is at 45°C. Below the 5 cm thick asphalt, average conductivity of 0.06 W/m K, is a layer of compacted rubbles at a temperature of 15°C. Find the rate of heat transfer to the rubbles....
Solution : Steady conduction through the 4 mm steel wall. . ∆T Q=k A ⇒ ∆x . Α = Q ∆x/k∆Τ...
I. C.V. Food. a) short term.: -Q from warm food to cold refrigerator air. Food cools. b) Long term: -Q goes to zero after food has reached refrigerator T. II. C.V. refrigerator space, not food, not refrigerator system a) short term: +Q from the warm food, +Q from heat leak from room into cold space. -Q (sum of both) to refrigeration system. If not equal the refrigerator space initially warms slightly and then cools down to preset T. b) long term: small -Q heat leak balanced by -Q to refrigeration system. Note: For refrigeration system CV any Q in from refrigerator space plus electrical W input to operate system, sum of which is Q rejected to the room. 4.61 A room is heated with an electric space heater on a winter day. Examine the following control volumes, regarding heat transfer and work , including sign. a) The space heater. Electrical work (power) input, and equal (after system warm up) Q out to the room. b) Room Q input from the heater balances Q loss to the outside, for steady (no temperature change) operation. c) The space heater and the room together Electrical work input balances Q loss to the outside, for steady operation....
uA1 = 227.03 + 0.15 x 162.16 = 251.35 mA1 = VA/vA1 = 163.854 kg Process: Constant temperature and total volume. m2 = mA1 ; V2 = VA + VB = 2 m3 ; v2 = V2/m2 = 0.012206 m3/kg State 2: x2 = 100%, v2 = 0.012206 ⇒ T2 = 55 + 5 x (0.012206 – 0.01316)/(0.01146 – 0.01316) = 57.8°C u2 = 406.01 + 0.56 x (407.85 – 406.01) = 407.04 kJ/kg
1Q 2 = m2(u2 - uA1) = 163.854 x (407.04 - 251.35) = 25510 kJ...
5-34 5.51 A 1-L capsule of water at 700 kPa, 150°C is placed in a larger insulated and otherwise evacuated vessel. The capsule breaks and its contents fill the entire volume. If the final pressure should not exceed 125 kPa, what should the vessel volume be? C.V. Larger vessel. Continuity: m2 = m1 = m = V/v1 = 0.916 kg Process: expansion with 1Q2 = 0 , 1W2 = 0 / / Energy: m(u2 - u1) = 1Q2 - 1W2 = 0 ⇒ u2 = u1 / u1 ≅ uf = 631.66 kJ/kg State 1: v1 ≅ vf = 0.001091 m3/kg;...
State 2: Highest T in Table B.1.3 is 1300°C Using a ∆u from the ideal gas tables, A.8, we get h(1500°C) - h(1300°C) = 61367.7 - 51629.5 = 9738.2 kJ/kmol u1500 - u1300 = ∆h/M - R(1500 - 1300) = 540.56 - 92.3 = 448.26 kJ/kg Since the ideal gas change is at low P we use 1300°C, lowest P available 10 kPa from steam tables, B.1.3, ux = 4683.7 kJ/kg as the reference. u2 - u1 = (u2 - ux)ID.G. + (ux - u1) = 448.26 + 4683.7 - 83.95 = 5048 kJ/kg 5.68 For an application the change in enthalpy of carbon dioxide from 30 to 1500°C at 100 kPa is needed. Consider the following methods and indicate the most accurate one. a. Constant specific heat, value from Table A.5. b. Constant specific heat, value at average temperature from the equation in Table A.6. c. Variable specific heat, integrating the equation in Table A.6. d. Enthalpy from ideal gas tables in Table A.8. Solution: a) b) ∆h = Cp∆T = 0.842 (1500 - 30) = 1237.7 kJ/kg Tave = 1038.2 K ; θ = T/100 = 10.382 Table A.6 Cp = 54.64 ⇒ Cp = Cp/M = 1.2415 ∆h = Cp,ave∆T = 1.2415 x 1470 = 1825 kJ/kg c) For the entry to Table A.6: ∆h = ⌠CpdT = ⌡ = 100 ⌠ C dθ M⌡p θ2 = 17.7315 ; θ1 = 3.0315...
5-50 A spherical elastic balloon contains nitrogen (N2) at 20oC, 500 kPa. The initial volume is 0.5 m3. The balloon material is such that the pressure inside is proportional to the balloon diameter. Heat is now transferred to the balloon until its volume reaches 1.0 m3 , at which point the process stops. a) Can the nitrogen be assumed to behave as an ideal gas throughout this process? b) Calculate the heat transferred to the nitrogen. Solution: C.V. Nitrogen, which is a control mass. Continuty: m 2 = m1 Energy: m(u2 − u1) = 1Q2 − 1W2...
5-57 5.88 Consider the 100-L Dewar (a rigid double-walled vessel for storing cryogenic liquids) shown in Fig. P5.88. The Dewar contains nitrogen at 1 atm, 90% liquid and 10% vapor by volume. The insulation holds heat transfer into the Dewar from the ambient to a very low rate, 5 J/s. The vent valve is accidentally closed so that the pressure inside slowly rises. How long time will it take to reach a pressure of 500 kPa? State 1: T1 = 77.3 K, Vliq1 = 0.9 V, Vvap1 = 0.1 V, Table B.6.1: mliq1 = v1f = 0.00124 m3/kg, v1g = 0.21639 m3/kg,...
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