ch02_eng

# ch02_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 2...

• Homework Help
• 18

This preview shows pages 1–5. Sign up to view the full content.

SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 2 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN

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

Sonntag, Borgnakke and Wylen CHAPTER 2 SUBSECTION PROB NO. Concept-Study Guide Problems 87-91 Properties and Units 92 Force, Energy and Specific Volume 93-96 Pressure, Manometers and Barometers 97-103 Temperature 104-105 Correspondence table The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and the SI number refers to the corresponding 6 th edition SI unit problem. New 5 th Ed. SI New 5 th Ed. SI 87 new - 97 43E 43 88 new 11 98 new 50 89 new 12 99 new 53 90 new 19 100 45E 70 91 new 20 101 46E 45 92 new 24 102 new 82 93 39E 33 103 48E 55 94 40E - 104 new 80 95 new 47 105 47E 77 96 42E 42
Sonntag, Borgnakke and Wylen Concept Problems 2.87 E A mass of 2 lbm has acceleration of 5 ft/s 2 , what is the needed force in lbf? Solution: Newtons 2 nd law: F = ma F = ma = 2 lbm × 5 ft/s 2 = 10 lbm ft/s 2 = 10 32.174 lbf = 0.31 lbf 2.88 E How much mass is in 0.25 gallon of liquid mercury (Hg)? Atmospheric air? Solution: A volume of 1 gal equals 231 in 3 , see Table A.1. From Figure 2.7 the density is in the range of 10 000 kg/m 3 = 624.28 lbm/ft 3 , so we get m = ρ V = 624.3 lbm/ft 3 × 0.25 × (231/12 3 ) ft 3 = 20.86 lbm A more accurate value from Table F.3 is ρ = 848 lbm/ft 3 . For the air we see in Figure 2.7 that density is about 1 kg/m 3 = 0.06243 lbm/ft 3 so we get m = ρ V = 0.06243 lbm/ft 3 × 0.25 × (231/12 3 ) ft 3 = 0.00209 lbm A more accurate value from Table F.4 is ρ = 0.073 lbm/ft 3 at 77 F, 1 atm.

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

Sonntag, Borgnakke and Wylen 2.89 E Can you easily carry a one gallon bar of solid gold? Solution: The density of solid gold is about 1205 lbm/ft 3 from Table F.2, we could also have read Figure 2.7 and converted the units. V = 1 gal = 231 in 3 = 231 × 12 -3 ft 3 = 0.13368 ft 3 Therefore the mass in one gallon is m = ρ V = 1205 lbm/ft 3 × 0.13368 ft 3 = 161 lbm and some people can just about carry that in the standard gravitational field. 2.90 E What is the temperature of –5F in degrees Rankine?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern