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Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition offers a comprehensive guide to solving every question in your textbook, helping you master the material.

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    0 - 1 - 1 [US] Two thermodynamics books, each with a mass of 1 kg, are stacked one on top of another. Neglecting the presence of atmosphere, draw the free body diagram of the book at the bottom to determine the vertical force on its (a) top and (b) bottom face s in kN. SOLUTION (a) From the free body diagram of the book at the top, a vertical force balance produces:    1 2 1 m kN kg k 0.00981 kN N 1000 s N 1 9.81 1000 mg F F            (b) Using the free body diagram of the book at the bottom,      2 1 2 kN 1000 1 9.81 0.00981 1000 0.01962 kN mg F F F       1 F mg 2 F mg 1 F

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    0 - 1 - 2 [UA] Determine (a) the pres sure felt on your palm to hold a textbook of mass 1 kg in equilibrium. Assume the distribution of pressure over the palm to be uniform and the area of contact to be 25 cm 2 . (b) What - if Scenario: How would a change in atmospheric pressure affect your answer (0: No change; 1: increase; - 1 decrease)? SOLUTION (a) 2 kN ; =kPa m F p A              1 9.81 ; ; 1000 1000 0.0025 mg p p A     3.924 kPa p   (b) A change in atmospheric pressure would not affect the answer because the net force created by a uniform pressure around any object is zero. Answer: 0

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    0 - 1 - 3 [UH] The lift - off mass of a Space Shuttle is 2 million kg. If the lift off thrust (the net force upward) is 10% greater than the minimum amount required for a lift - off, determine the acceleration. SOLUTION thrust gravity ; 1.1 ; 0.1 ; F F ma mg mg ma a g          0.1 9.81 ; a   2 m 0.981 s a  

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    0 - 1 - 4 [UF] A body weighs 0.05 kN on earth where g = 9.81 m/s 2 . Determine its weight on (a) the moon, and (b) on mars with g = 1.67 m/s 2 and g = 3.92 m/s 2 , respectively. SOLUTION 0.05 1000 ; ; 5.097 kg; 9.81 w m m m g       (a) On the moon:    5.097 1.67 ; 10 0.00 0 8 12 k 0 5 N w w    (b) On Mars:    5.097 3.92 ; 0.01998 N 00 k 1 0 w w   

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    0 - 1 - 5 [UD] Calculate the weight of an object of mass 50 kg at the bottom and top of a mountain with (a) g = 9.8 m/s 2 and (b) g = 9.78 m/s 2 respectiv ely. SOLUTION (a)    50 9.8 ; ; 1000 100 0.49 k 0 N mg w w w      (b)    50 9.78 ; 0.489 k 1000 N w w   

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    0 - 1 - 6 [UM] According to Newton's law of gravity, the value of g at a given location is inversely proportional to the square of the distance of the location from the center of the eart h. Determine the weight of a textbook of mass 1 kg at (a) sea level and (b) in an airplane cruising at an altitude of 45,000 ft. Assume earth to be a sphere of diameter 12,756 km. SOLUTION (a) At sea level 0 0 2 ; GmM w g m R      0 9.81 1 ; w   0 9.81 N w   (b)   3 km At 45,000 ft 0.3048 10 13.716 km ft h           2 2 0 2 2 2 2 ; ; ; ; GmM GmM R R w w w g m w gm R R h R h R h              2 2 2 2 m 9.81 (6,378 km) m s ; 9.76 ; (6,378 km 13.716 km) s 9.767 1 ; g g w              9.76 N w  

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    0 - 1 - 7 [UJ] The frictional force on a block of mass m A resting on a table (see accompanying figure) is given as F = μN , where N is the normal reaction force from the table. Determine the maximum value for m B that can be supported by friction. Assume the pulley to be frictionless. SOLUTION The maximum value of B m that can be supported by the friction:     ; kN 1000 ; kN 1000 ; A A B B A B m g F m g F F F     ; A B m g m g    B A m m   

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    0 - 1 - 8 [UW] If the block A in problem 0 - 1 - 7 [UJ] sits on a wedge with an angle θ with the horizontal, how would the answer change? SOLUTION The normal force on the block is:   cos ; kN 1000 A m g N   A balance between the tension in the string and friction produces: The normal force on t cos ; 1000 1000 A B m g m g    cos ; A B m g m g     cos A B m m    

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    0 - 1 - 9 [XR] A block with a mass of 10 kg is at rest on a plane inclined at 25 o to the horizontal. If μ s = 0.6, determine the range of the horizontal push force F if the block is (a) about to slide down, and (b) about to slide up. SOLUTION (a) As the block is about to slide down, friction acts upward and a force balance along the slope yields:   sin cos cos sin ; 1000 s mg F mg F                   (sin cos ) cos sin ; 1000 (sin cos ) ; 1000 cos sin 10 9.81 (sin 25 0.6 cos 25 ) ; 1000 cos 25 0.6sin 25 s s s s mg F mg F F                             0.010248 kN F    The negative sign indicates that a slight pull force is necessary to overcome friction for the block to slide down. (b) As the block is about to slide up, friction acts downward and a for ce balance along the slope yields:   cos sin sin cos ; 1000 s mg F mg F                    ( cos sin ) ( cos sin ) cos sin ; 1000 1000 cos sin 10 9.81 (0.6cos 25 sin 25 ) ; 1000 cos 25 0.6sin 25 s s s s mg mg F F F                             0.1452 kN F  

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    0 - 1 - 10 [XO] A vertical piston cylinder device contains a gas at an unknown pressure. If the outside pressure is 100 kPa, determine (a) the pressure of the gas if the piston has an area of 0.2 m 2 and a mass of 20 kg. Assume g = 9.81 m/s 2 . (b) What - if Scenario: What would the pressure be if the orientation of the device were changed and it were now upside down? SOLUTION (a) piston 2 piston 0 piston 2 m kN ; kPa m kg =kN 1000 s N i m g p A p A                   piston 0 2 2 2 piston kN m kN 1 ; kPa= kg 1000 m s N m 20 9.81 100 ; 1000 0.2 i i m g p p A p              100.981 kPa i p   (b) piston 2 piston 0 piston 2 m kN ; kPa m kg =kN 1000 s N i m g p A p A                   piston 0 piston ; 1000 20 9.81 100 ; 1000 0.2 i i m g p p A p       99.019 kPa i p   p o p i mg

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    0 - 1 - 11 [XB] Determine the mass of the weight necessary to increase the pressure of the liquid trapped in side a piston - cylinder device to 120 kPa. Assume the piston to be weightless with an area of 0.1 m 2 , the outside pressure to be 100 kPa and g = 9.81 m/s 2 . SOLUTION 2 piston 0 piston 2 m kN ; kPa m kg =kN 1000 s N w i m g p A p A                          0 piston 0 piston ; kPa 1000 1000 ; kg 120 100 0.1 1000 ; 9.81 w i i w w m g p p A p p A m g m          203.873 kg w m  

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    0 - 1 - 12 [XS] A mass of 100 kg is placed on the piston of a vertical piston - cylinder device containing nitrogen. The piston is weightless and has an area of 1 m 2 . The outside pressure is 100 kPa. Determine (a) the pressure inside the cylinder. The mass placed on t he piston is now doubled to 200 kg. Also, additional nitrogen is injected into the cylinder to double the mass of nitrogen. (b) Determine the pressure inside under these changed conditions. Assume temperature to remain unchanged at 300 K and R for nitrogen to be 0.296 kJ/kg - K. SOLUTION (a) A vertical force balance on the piston yields: 2 piston 0 piston 2 m kN ; kPa m kg =kN 1000 s N w i m g p A p A                     0 piston ; kPa 1000 100 9.81 10 100.98 k 0 ; 1000 a 1 P w i i i m g p p A p p         (b) As the weight is doubled, the force balance on the piston yields:       200 9.81 = 100; 1 1000 i p  101.96 kPa = i p  Note the amount of nitrogen (or the identity of the gas) is not relevant.

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    0 - 1 - 13 [XN] A piston with a diameter of 50 cm and a thickness of 5 cm is made of a composite material with a density ( ρ ) of 4000 kg/m 3 . (a) If the outside pressure is 101 kPa, determine the pressure inside the piston - cylinder assembly if the cylinder contains air. (b) What - if Scenario: What would the inside pressure be if the piston diameter were 100 cm instead? (c) Would the answers change if the cylinder contains liquid w ater instead? (1: yes; 2: no) SOLUTION (a)     2 2 2 0.5 ; ; 0.1963 m ; 4 4 d A A A          piston ; kg m At        2 piston piston 0.5 4000 0.05; 39.25 kg; 4 m m          piston 0 piston ; kPa 1000 i m g p p A         39.25 9.81 101 ; 1000 0.1 102.96 6 kPa 9 3 i i p p      (b)        2 piston piston 1 4000 0.05 ; 157 kg; 4 m m       piston 0 piston ; 1000 i m g p p A         157 9.81 101 ; 1000 0.785 102.96 kPa 4 i i p p      (c) No, the force balance on the piston does not depend on the identity of the fluids exerting pressures. Answer: 2 p o p i mg

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    0 - 1 - 14 [XA] A piston - cylinder device contains 0.02 m 3 of hydrogen at 300 K. It has a diameter of 10 cm. The piston (assumed weightless) is pulled by a connecting rod perpendicular to the piston surface. If the outside conditions are 100 kPa and 300 K, (a) determine the pull force necessary in kN to create a pressure of 50 kPa inside. (b) The piston is now released; as it oscill ates back and forth and finally comes to equilibrium, the temperature inside is measured as 600 K (reasons unknown). What is the pressure of hydrogen at equilibrium? (c) What - if Scenario: What would be the answer in part (a) if the gas were oxygen instead? SOLUTION (a) A vertical force balance on the piston yields:   0 ; kN i F p A p A         0 2 ; 100 50 0.393 kN 0.1 ; 4 i F p p A F F                (b) With the external force absent, a new force balance on the piston after it comes to equilibrium yields:   0 ; kN i p A p A  0 ; 100 kPa i i p p p     (c) The identity of the gas has nothing to do with the force balance. Therefore, the answer does not change. F = 0.393 kN.

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    0 - 1 - 15 [XH] Air in the accompanying piston - cylinder device is in equilibrium at 200 o C. If the mass of the hanging weight is 10 kg, atmospheric pressure is 100 kPa, and the piston diameter is 10 cm, (a) determine the pressure of air inside. Assume g = 9.81 m/s 2 . (b) What - if Scenario: What would the pressure be if the gas were hydrogen inste ad? Molar mass of air is 29 kg/kmol and that of hydrogen is 2 kg/kmol. Neglect piston mass and friction. SOLUTION (a) A vertical force balance on the piston yields:   0 ; kN 1000 i p p mg p A p A              0 2 ; kPa 1000 10 9.81 4 100 ; (100 12.49); 1000 (0.1) i p i i mg p p A p p           87.51 kPa in p   (b) The force balance does not depend on the identity of the gas. Hence, 87.51 kPa i p 

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