Solution Manual for Thermodynamics An pdf

[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 faces in kN. SOLUTION (a) From the free body diagram of the book at the top, a vertical force balance produces: F  mg  kg m kN  kN 1 1000   s 2 N   F1  19.81  0.00981 kN 1000 (b) Using the free body diagram of the book at the bottom, F  F  mg kN 2 1 1000  F2  0.00981 19.81  0.01962 kN F mg 1 F1 F mg 2 0-1-1 [UA] Determine (a) the pressure 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 cm2 . (b) What-if Scenario: How would a change in atmospheric pressure affect your answer (0: No change; 1: increase; -1 decrease)? SOLUTION (a) p  F ; kN =kPa  A  m 2    p  mg ; 1000A  p  3.924 kPa  p  19.81 ; 10000.0025 0-1-2 [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 Fthrust  Fgravity  ma; 1.1mg  mg  ma;  a  0.1g;  a  0.19.81;  a  0.981 m s 2   0-1-3 [UF] A body weighs 0.05 kN on earth where g = 9.81 m/s2 . Determine its weight on (a) the moon, and (b) on mars with g = 1.67 m/s2 and g = 3.92 m/s2 , respectively. SOLUTION m  w ; g  m  0.051000 ; 9.81  m  5.097 kg; (a) On the moon: 5.0971.67 w ; 1000  w  0.008512 kN (b) On Mars: 5.0973.92 w ; 1000  w  0.01998 kN  0-1-4 [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/s2 and (b) g = 9.78 m/s2 respectively. SOLUTION (a) w  mg ;  w  509.8 ;  w  0.49 kN (b) 509.78 w ; 1000  w  0.489 kN 0 2  2  0-1-5 [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 earth. 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 w  GmM 0 R 2  g0m; (b)  w0  9.811;  w0  9.81 N At h  45,000 ft0.304810 3 km 13.716 km ft GmM GmM R 2 R 2 w  ; R  h  w  R 2 ; R  h  w  g m; R  h  w  gm;  9.81 m  (6,378 km)2 s m  g    ; (6,378 km 13.716 km)2  g  9.76 ; s 2  w  9.7671;  w 9.76 N