Modern Engineering Thermodynamics

4.3 The First Law of Thermodynamics 103
alternatively,
and
E G = U 2 − U 1 + KE 2 − KE 1 + PE 2 − PE 1 (4.7)
E G = mðu 2 − u 1 + ke 2 − ke 1 + pe 2 − pe 1 Þ (4.8)
In most of the engineering situations we encounter, either the system is not moving at all or it is moving without
any change in velocity or height. In these cases,
E G = U 2 − U 1 = mðu 2 − u 1 Þ = E T
EXAMPLE 4.1
Figure 4.2 shows that 3.00 lbm of saturated water vapor at 10.0 psia is sealed in a rigid container aboard a spaceship traveling
at 25,000. mph at an altitude of 200. mi. What energy transport is required to decelerate the water to zero velocity and bring
it down to the surface of the Earth such that its final specific internal energy is 950.0 Btu/lbm? Neglect any change in the
acceleration of gravity over this distance.
p 1 = 10.0 psia, x 1 = 1.00
u 2 = 950.0 Btu/1bm
Z 2 = V 2 =0
V 1 = 2500. mph
Sealed rigid
container
Z 1 = 200. miles
State 1 State 2
FIGURE 4.2
Example 4.1.
Solution
Let the system in this example be just the water in the container, then the process followed by the water is a constant
volume process (the water is in a “rigid, sealed container”). Therefore, the problem statement can be outlined as follows:
State 1
m = 3:0 lbm, V = constant State 2
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p 1 = 10:0 psia
u 2 = 950:0 Btu/lbm
x 1 = 1:00ðsaturated vaporÞ v 2 = v 1 = 38:42 ft 3 /lbm
v 1 = v g ðat 10:0 psiaÞ = 38:42 ft 3 /lbm
Notice how the process path gives us the value of a property (v 2 ) in the final state. To determine the required energy
transport, we use the energy balance Eq. (4.1), along with the definition of the energy gain term E G from Eq. (4.5):
EB: E G = E T + E P0 ðas required by the first lawÞ
and, assuming g is constant during this process,
E G = E T = U 2 − U 1 + m V2 2 2g − mg
V2 1 +
c
g
ðZ 2 − Z 1 Þ
c
Here, V 2 = Z 2 = 0, so
E T = U 2 − U 1 − m V1 2 2g − mg
c
g
Z 1 c
Table C.2a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics gives
u 1 = u g ð10:0 psiaÞ = 1072:2 Btu/lbm
and the problem statement requires that u 2 = 950.0 Btu/lbm. Therefore,
U 1 = mu 1 = ð3:00 lbmÞð1072:2 Btu/lbmÞ = 3216:6 Btu
(Continued )