Exam (elaborations) TEST BANK FOR Thermodynamics An Engineering Approach 7th Edition By Yunus A. Cengel and Michael A. Boles (Solution Manual)

1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle. 1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill. 1-3C There is no truth to his claim. It violates the second law of thermodynamics. preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-3 Mass, Force, and Units 1-4C The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English system. You should get into the habit of never writing the unit “lb”, but always use either “lbm” or “lbf” as appropriate since the two units have different dimensions. 1-5C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit. 1-6C There is no acceleration, thus the net force is zero in both cases. 1-7E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives 210.5 lbm 1 lbf 32.174 lbm ft/s 32.10 ft/s 210 lbf 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = ⎯⎯→ = = g W mg m W Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be lbf 35.8 = ⎟⎠ ⎞ ⎜⎝ ⎛ ⋅ = = 2 2 32.174 lbm ft/s W mg (210.5 lbm)(5.47 ft/s ) 1 lbf 1-8 The interior dimensions of a room are given. The mass and weight of the air in the room are to be determined. Assumptions The density of air is constant throughout the room. Properties The density of air is given to be ρ = 1.16 kg/m3. ROOM AIR 6X6X8 m3 Analysis The mass of the air in the room is m = ρV = (1.16 kg/m3 )(6×6×8 m3 ) = 334.1kg Thus, N 3277 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = 2 2 1 kg m/s 1 N W mg (334.1 kg)(9.81 m/s ) preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-4 1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 0.5% is to be determined. 0 z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 ×10−6 z) In our case, W = 0.995Ws = 0.995mgs = 0.995(m)(9.81) Substituting, 0.995(9.81) = (9.81− 3.32×10−6 z) ⎯⎯→ z = 14,774 m ≅ 14,770 m Sea level 1-10 The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be W = mg = (200 kg)(9.6 m/s2 ) = 1920N 1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units. Analysis Applying Newton's second law, the weight is determined in various units to be 0.240 Btu/lbm F 0.240 kcal/kg C 1.005 J/g C 1.005 kJ/kg K ° ⋅ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⋅ ° ⋅ ° = ⋅ ° ° ⋅ = ⎟⎠ ⎞ ⎜⎝ = ⋅ ° ⎛ ° ⋅ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ = ⋅ ° ⎛ ⋅ = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⋅ ° ⋅ = ⋅ ° 4.1868 kJ/kg C (1.005 kJ/kg C) 1 Btu/lbm F 4.1868 kJ (1.005 kJ/kg C) 1 kcal 1000 g 1 kg 1 kJ (1.005 kJ/kg C) 1000 J 1 kJ/kg C (1.005 kJ/kg C) 1 kJ/kg K p p p p c c c c preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-5 1-12 A rock is thrown upward with a specified force. The acceleration of the rock is to be determined. Analysis The weight of the rock is 29.37 N 1 kg m/s 1 N (3 kg)(9.79 m/s ) 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ W = mg = Then the net force that acts on the rock is Fnet = Fup − Fdown = 200 − 29.37 =170.6 N Stone From the Newton's second law, the acceleration of the rock becomes 2 m/s 56.9 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = = 1 N 1 kg m/s 3 kg 170.6 N 2 m a F preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-6 1-13 Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. "The weight of the rock is" W=m*g m=3 [kg] g=9.79 [m/s2] "The force balance on the rock yields the net force acting on the rock as" F_up=200 [N] F_net = F_up - F_down F_down=W "The acceleration of the rock is determined from Newton's second law." F_net=m*a "To Run the program, press F2 or select Solve from the Calculate menu." SOLUTION a=56.88 [m/s^2] F_down=29.37 [N] F_net=170.6 [N] F_up=200 [N] g=9.79 [m/s2] m=3 [kg] W=29.37 [N] m [kg] a [m/s2] 1 2 3 4 5 6 7 8 9 10 190.2 90.21 56.88 40.21 30.21 23.54 18.78 15.21 12.43 10.21 1 2 3 4 5 6 7 8 9 10 0 40 80 120 160 200 m [kg] a [m/s2] preparation. If you are a student using this Manual, you are using it without permission. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 1-7 1-14 During an analysis, a relation with inconsistent units is obtained. A correction is to be found, and the probable cause of the error is to be determined. Analysis The two terms on the right-hand side of the equation E = 25 kJ + 7 kJ/kg do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally homogeneous; that is, every term in the equation will have the same unit. Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage. 1-15 A resistance heater is used to heat water to desired temperature. The amount of electric energy used in kWh and kJ are to be determined. Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of electric energy used in 2 hours becomes Total energy = (Energy per unit time)(Time interval) = (4 kW)(2 h) = 8 kWh Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ, Total energy = (8 kWh)(3600 kJ/kWh) = 28,800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy. 1-16 A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is to be obtained for the filling time. Assumptions Gasoline is an incompressible substance and the flow rate is constant. Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unit of time is ‘seconds’. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have t [s] ↔ V [L], and V& [L/s} It is obvious that the only way to end up with the unit “s” for time