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Modern Engineering Thermodynamics

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4.3 The First Law of <strong>Thermodynamics</strong> 103<br />

alternatively,<br />

and<br />

E G = U 2 − U 1 + KE 2 − KE 1 + PE 2 − PE 1 (4.7)<br />

E G = mðu 2 − u 1 + ke 2 − ke 1 + pe 2 − pe 1 Þ (4.8)<br />

In most of the engineering situations we encounter, either the system is not moving at all or it is moving without<br />

any change in velocity or height. In these cases,<br />

E G = U 2 − U 1 = mðu 2 − u 1 Þ = E T<br />

EXAMPLE 4.1<br />

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<br />

at 25,000. mph at an altitude of 200. mi. What energy transport is required to decelerate the water to zero velocity and bring<br />

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<br />

acceleration of gravity over this distance.<br />

p 1 = 10.0 psia, x 1 = 1.00<br />

u 2 = 950.0 Btu/1bm<br />

Z 2 = V 2 =0<br />

V 1 = 2500. mph<br />

Sealed rigid<br />

container<br />

Z 1 = 200. miles<br />

State 1 State 2<br />

FIGURE 4.2<br />

Example 4.1.<br />

Solution<br />

Let the system in this example be just the water in the container, then the process followed by the water is a constant<br />

volume process (the water is in a “rigid, sealed container”). Therefore, the problem statement can be outlined as follows:<br />

State 1<br />

m = 3:0 lbm, V = constant State 2<br />

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ ƒ!<br />

p 1 = 10:0 psia<br />

u 2 = 950:0 Btu/lbm<br />

x 1 = 1:00ðsaturated vaporÞ v 2 = v 1 = 38:42 ft 3 /lbm<br />

v 1 = v g ðat 10:0 psiaÞ = 38:42 ft 3 /lbm<br />

Notice how the process path gives us the value of a property (v 2 ) in the final state. To determine the required energy<br />

transport, we use the energy balance Eq. (4.1), along with the definition of the energy gain term E G from Eq. (4.5):<br />

EB: E G = E T + E P0 ðas required by the first lawÞ<br />

and, assuming g is constant during this process,<br />

E G = E T = U 2 − U 1 + m V2 2 2g − mg<br />

V2 1 +<br />

c<br />

g<br />

ðZ 2 − Z 1 Þ<br />

c<br />

Here, V 2 = Z 2 = 0, so<br />

E T = U 2 − U 1 − m V1 2 2g − mg<br />

c<br />

g<br />

Z 1 c<br />

Table C.2a in Thermodynamic Tables to accompany <strong>Modern</strong> <strong>Engineering</strong> <strong>Thermodynamics</strong> gives<br />

u 1 = u g ð10:0 psiaÞ = 1072:2 Btu/lbm<br />

and the problem statement requires that u 2 = 950.0 Btu/lbm. Therefore,<br />

U 1 = mu 1 = ð3:00 lbmÞð1072:2 Btu/lbmÞ = 3216:6 Btu<br />

(Continued )

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