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Modern Engineering Thermodynamics

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178 CHAPTER 6: First Law Open System Applications<br />

EXAMPLE 6.2 (Continued )<br />

The material flowing through this system is water at 80.0 psig (94.7 psia) and 60.0°F. A check of the saturation tables for<br />

water shows us that the water is in a compressed liquid state and therefore can be considered to be an incompressible fluid.<br />

Also, since the amount of excess pressure here is relatively small (94.7 psia vs. the saturation pressure of 0.2563 psia at<br />

60.0°F), we need not use the compressed water tables but can get sufficient accuracy using the saturated liquid tables at<br />

60.0°F for all the inlet properties we may need.<br />

The first law formulation that applies to this problem is the modified energy rate balance, Eq. (6.12). The auxiliary<br />

equations needed include the equation of state for an incompressible fluid (Eq. (6.19)) and the mass flow rate formula<br />

(Eq. (6.1)). For the case of the system shown in Figure 6.4, we have the standard nozzle configuration, except that we can<br />

calculate V in in this problem from Eq. (6.14) as<br />

V in = 4 _mv<br />

πD 2<br />

where v ≈ v f ð60:0°FÞ = 0:01603 ft 3 /lbm: Then, for system a,<br />

ðV in Þ a = 4ð0:800 lbm/sÞð0:01603 ft3 /lbmÞ<br />

πð1:00 in:Þ 2 ð1 ft/12 in:Þ 2<br />

= 2:35 ft/s<br />

Now, this ðV in Þ a probably produces a negligible inlet kinetic energy, but since we have its value, we carry it along in the<br />

solution for the time being.<br />

The solution to part a is obtained by solving Eq. (6.12), the modified energy rate balance, for (V out ) a<br />

, with the enthalpy values<br />

given by Eq. (6.19). This produces an equation similar to Eq. (6.20) except with a V in term included. Since we have no information<br />

about any heat transfer to or from the nozzle, we assume that there is none. This is justifiable on the following basis:<br />

Since the nozzle is very small (hand size), the water is not inside of it long enough for any significant heat transfer of energy<br />

to occur. This is a common circumstance that often occurs in obviously uninsulated small systems with no significant heat<br />

transfer. If the residence time of the fluid in the system is very small, then in the absence of an extraordinarily large temperature<br />

difference between the environment and the system, the time is simply insufficient for any significant heat transfer to<br />

occur, regardless of whether the system is insulated or not. The modified energy rate balance for system a reduces to a form of<br />

Eq. (6.21):<br />

ðV out Þ a<br />

= fVin 2 + 2g 1/2<br />

c½cðT in − T out Þ + vðp in − p out ÞŠg a<br />

The problem statement told us to assume the water flow through the nozzle is isothermal, so we set T in = T out . Actually, the<br />

water flow is not exactly isothermal, due to an increase in internal energy of the water from viscous effects, turbulence, and<br />

so forth. However, for a small nozzle, these effects are negligible. For an isothermal flow, we obtain<br />

The data for system a are as follows:<br />

Then<br />

ðV out Þ a<br />

= ½ðV 2 in Þ a + 2g cvðp in − p out Þ a<br />

Š 1/2<br />

ðV in Þ a<br />

= 2:35 ft/s<br />

v = v f ð60:0°FÞ = 0:01603 ft 3 /lbm<br />

ðp in Þ a<br />

= 80:0 psig = 94:7 psia<br />

ðp out Þ a<br />

= 0:00 psig = 14:7 psia<br />

<br />

ðV out Þ a<br />

= ð2:35 ft/sÞ 2 + 2 32:174 lbm .ft<br />

ð0:01603<br />

lbf .s 2 ft 3 /lbmÞ½ð80:0 − 0:00Þ lbf/in 2 Š × 144 in 2 /ft 2<br />

1/2<br />

= 109 ft/s<br />

Notice that (V in ) a<br />

was only about 2% of (V out ) a<br />

and therefore could have been neglected in this case.<br />

Part b of this example is basically a mechanics problem, but it can be easily solved using system b in Figure 6.4 and the<br />

modified energy rate balance. The following assumptions are now made for system b:<br />

_Q = _W = 0 ðZ out Þ b = ?<br />

ðV out Þ a = ðV in Þ b = 109 ft/s ðp in Þ b = ðp out Þ b = 14:7 psia<br />

ðV out Þ b<br />

≈ 0ft/s ðT in Þ b<br />

= ðT out Þ b<br />

= 60:0°F<br />

ðZ in Þ b<br />

= 0<br />

(V out ) b is ≪ (V in ) b here because the water stream spreads into a large fan at the top of its trajectory and therefore exits system b<br />

through a large surface area. The conservation of mass law for an incompressible fluid requires that _m =constantandso<br />

(V out ) b =(V in ) b (A in /A out ) b ≪ (V in ) a here.

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