Lecture -3

1. Lecture -3 Class Problems– Chapter 1

2. Problem # 1.22 A spherical balloon holding 35 lbmof air has a diameter of 10 ft. For the air, determine the specific volume in ft3/lbm and ft3/lbmol, and b)the weight in lbf. Let g = 31.0 ft/s2. m = 35lbm; volume of sphere = 4/3 p R3 = 4/3 (3.1417)(5)3 = 523.6 ft3 v = specific volume = 523.6 / 35 = 14.96 ft3/lbm M air = 28.97 lbm / lbmol. Hence molar specific volume = M v = 28.97 x 14.96 = 433.4 ft3/lbmol Weight = m g = 35 x 31 lbm. ft/s2 = 1085 lbm ft/s2 = 1085 / 32.2 lbf = 33.696 lbf.

3. 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 p1 = 1 bar, and v1 = 0.2 m3 / kg. and ends with p2 = 0.25 bar. Determine the final volume in m3, and plot the process on a graph with pressure and volume as axes. m = 5 kg. p1 v11.3 = p2 v21.3 ; (1) (0.2)1.3 = (0.25) (v2) 1.3 (v2) 1.3 = (1/0.25)(0.2) 1.3; v2 = (4)1/1.3 (0.2) = 2.9039 (0.2) = 0.5808 m3 /kg V = m v = (5)(0.5808) = 2.9039m3

5. Problem 1.31 A gas contained within a piston-cylinder assembly undergoes three processes in series. Process 1-2: compression with pV = constant from p1 = 1 bar, V1 = 1.0m3 to V2 = 0.2m3 Process 2-3: Constant pressure expansion to V3 = 1.0m3 Process 3-1 Constant Volume Sketch the processes on a p-V diagram and label the pressures and volumes at each point. p2 = p1 V1 / V2 = (1) (1) / 0.2 = 5 bars p3 = p2 Back to state 1.

7. Problem 1.32 A manometer is attached to a tank of gas in which the pressure is 104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3. If g = 9.81 m/s2, and the atmospheric pressure is 101.35 kPa. Calculate the difference in mercury level in the manometer, in cm. b) the gage pressure of gas in kPa. p gage = (104.0 – 101.35) kPa = 2.65 kPa = (r) (g) (L) = (13.59)(1000) (9.81) (L) = 2650 Pa = (kg / m3) (m/s2) (m) = kg . m/s2 / m2 L = 2650 / (13.59 x 1000 x 9.81) = 1.9877 x 10-2 m = 1.9877 cm

8. Problem 1.39 A vacuum gage indicates that the pressure of carbon dioxide in a closed tank is -10 kPa. A mercury barometer gives the local atmospheric pressure as 750 mm Hg. Determine the absolute pressure of the carbon dioxide in kPa. The density of mercury is 13.59 g/cm3, and g = 9.81 m/s2 750 mm Hg = (r) (g) (L) = (13.59) (1000 kg/m3) (9.81) m/s2 (750) x 10-3 (m) = 99988.425 kg. m. m/s2 / m3 = 99988.425 N/m2 = 99988.425Pa Absolute Pressure = gage pressure + atmospheric pressure = = - 10000 + 99988.425 = 89988.425 Pa

9. Problem 1.52 Natural gas is burned with air to produce gaseous products at 1985 oC. Express this temperature in K, oR, oF 1985 oC = 1985 + 273.15 K = 2258.15 K 2258.15 K = (1.8) x (2258.15) oR = 4064.67 oR 4064.67 oR = (4064.67 - 459.67) oF = 3605 oF

10. Problem 1.57 What is the maximum increase and maximum decrease in body tmperature each in oC from a normal body temperature of 37 oC that humans can experience before serious medical complications result? Maximum body temperature =104 oF =(104-32)/1.8 oC = 40 oC Minimum temperature = 96oF = (96–32) /1.8=35.55 oC. Maximum increase from 37 oC = 3 oC Maximum decrease from 37 oC = 1.45 oC

11. Homework problems # 1 Problems from Moran & Shapiro 6th edition Problem # 1.24, 1.28, 1.33, 1.38, 1.53