Midterm I Review

# Midterm I Review - ME 300-D Midterm I Review Problem 1.22 A...

This preview shows pages 1–10. Sign up to view the full content.

ME 300 -D Midterm –I Review

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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) =
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 37

Midterm I Review - ME 300-D Midterm I Review Problem 1.22 A...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online