Modern Engineering Thermodynamics

22 CHAPTER 1: The Beginning
EXAMPLE 1.7 (Continued )
Now,
Therefore,
1m
Z = ð8:00 ftÞ = 2:438 m
3:281 ft
Weight = 2000. lbf
PE = mgZ
g c
= ð907 kgÞð9:81 m=s2 Þð2:438 mÞ
= 21,700 kg.m 2 /s 2
1
= 21,700 N.m = 21,700 J = 21:7kJ
In the Engineering English units system, we have
m = Fg ð2000: lbfÞ 32:174 lbm
.ft
c
=
lbf .s 2
g
32:174 ft=s 2 = 2000: lbm
Here, Z = 8.00 ft, and from Table 1.2, we find that in the Engineering
English units system g c = 32.174 lbm·ft/(lbf·s 2 ). Therefore,
8.00 ft
PE = mgZ
g c
= ð2000: lbmÞð32:174 ft=s2 Þð8:00 ftÞ
32:174 lbm .ft
lbf .s 2
= ð16,000 ftl ≅ bfÞ
= 16,000 ft .lbf
!
1 Btu
= 20:6 Btu
778:17 ft .lbf
FIGURE 1.14
Example 1.7.
Exercises
19. If the automobile in Example 1.7 weighs 4000. lbf instead of 2000. lbf, determine its potential energy in SI and Engineering
English units. Assume all other variables (i.e., the height) remain unchanged). Answer: PE= 43.4 kJ = 41.1 Btu.
20. Determine the potential energy in both SI and Engineering English units of a 3.00 lbf textbook sitting on a table
28.0 inches above the floor. Answer: PE= 9.49 J = 7.00 ft·lbf.
21. Determine the potential energy in SI and Engineering English units of a 4000. kg moose standing on the top of a house
5.00 m above the ground. Answer: PE= 196 kJ = 186 Btu.
EXAMPLE 1.8
Determine the translational kinetic energy of a bullet having a mass of 10.0 grams traveling at a velocity of 3000. ft/s in both
SI and Engineering English units (Figure 1.15).
V = 3000. ft/s
Mass = 10.0 grams
FIGURE 1.15
Example 1.8.
Solution
The formula for the kinetic energy (KE) of an object with mass m traveling at velocity V is given by Eq. (1.11) as
KE = 1 2 k 1mV 2 = mV2
2g c
In the SI units system, m = 10.0 g = 0.0100 kg, and
V = 3000: ft
1m
= 914:4 m/s
s 3:281 ft
From Table 1.2, we find that g c = 1 (dimensionless). Therefore,
KE =
ð0:0100 kgÞð914:4 m/sÞ2
2ð1Þ
= 4180 kg .m 2
s 2
= 4180 N.m = 4180 J = 4:18 kJ