Midterm I Review - ME 300 -D Midterm I Review Problem #...

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ME 300 -D Midterm –I Review
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Problem # 1.22 A spherical balloon holding 35 lbm of air has a diameter of 10 ft. For the air, determine the specific volume in ft 3 /lbm and ft 3 /lbmol, and b)the weight in lbf. Let g = 31.0 ft/s 2. m = 35lbm; volume of sphere = 4/3 π R 3 = 4/3 (3.1417)(5) 3 = 523.6 ft 3 v = specific volume = 523.6 / 35 = 14.96 ft 3 /lbm M air = 28.97 lbm / lbmol. Hence molar specific volume = M v = 28.97 x 14.96 = 433.4 ft 3 /lbmol 2 2
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Problem 1.27 A closed system consisting of 5 kg. of gas undergoes a process during which the relationship between pressure and specific volume is p v 1.3 = constant. The process begins with p 1 = 1 bar, and v 1 = 0.2 m 3 / kg. and ends with p 2 = 0.25 bar. Determine the final volume in m 3 , and plot the process on a graph with pressure and volume as axes. m = 5 kg. p 1 v 1 1.3 = p 2 v 2 1.3 ; (1) (0.2) 1.3 = (0.25) (v 2 ) 1.3 (v 2 ) 1.3 = (1/0.25)(0.2) 1.3 ; v 2 = (4) 1/1.3 (0.2) = 2.9039
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Problem 1.31 A gas contained within a piston-cylinder assembly undergoes three processes in series. Process 1-2: compression with pV = constant from p 1 = 1 bar, V 1 = 1.0m 3 to V 2 = 0.2m 3 Process 2-3: Constant pressure expansion to V 3 = 1.0m 3 Process 3-1 Constant Volume Sketch the processes on a p-V diagram and label the pressures and volumes at each point. p 2 = p 1 V 1 / V 2 = (1) (1) / 0.2 = 5 bars
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Problem 1.57 What is the maximum increase and maximum decrease in body tmperature each in o C from a normal body temperature of 37 o C that humans can experience before serious medical complications result? Maximum body temperature =104 o F =(104-32)/1.8 o C = 40 o C Minimum temperature = 96 o F = (96–32) /1.8=35.55 o C.
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Problem 2.8 An object whose mass is 1 lb has a velocity of 100ft/s. Determine the final velocity in ft/s. if the kinetic energy of the object decreases by 100 ft. lbf. Determine the change in elevation in ft. associated with a 100 ft. lbf change in potential energy. Let g = 32.0 ft/s 2 . m = 1 lb. V 1 = 100 ft/s. (KE) 1 = ½ m V 1 2 = ½ (1)(100) (100) = 5000 lbm ft 2 /s 2 . 1 ft. lbf = (1) (ft) (32.2) lbm ft/s 2 KE 2 = 5000 – 32.2 x 100 = 1780 lbm. ft 2 /s 2 V 2 2 = 2 x (1780) / (1); V 2 = 59.66 ft/s m g z = 32.2 x 100; z = 32.2 x 100 / (32.0 x 1) =
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Problem 2.19 An object of mass 80 lb. initially at rest experiences a constant acceleration of 12 ft/s2 due to the action of a resultant force applied for 6.5 s. Determine the work of the resultant force in ft. lbf and in Btu. m = 80 lb.
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Midterm I Review - ME 300 -D Midterm I Review Problem #...

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