Modern Engineering Thermodynamics

178 CHAPTER 6: First Law Open System Applications
EXAMPLE 6.2 (Continued )
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
water shows us that the water is in a compressed liquid state and therefore can be considered to be an incompressible fluid.
Also, since the amount of excess pressure here is relatively small (94.7 psia vs. the saturation pressure of 0.2563 psia at
60.0°F), we need not use the compressed water tables but can get sufficient accuracy using the saturated liquid tables at
60.0°F for all the inlet properties we may need.
The first law formulation that applies to this problem is the modified energy rate balance, Eq. (6.12). The auxiliary
equations needed include the equation of state for an incompressible fluid (Eq. (6.19)) and the mass flow rate formula
(Eq. (6.1)). For the case of the system shown in Figure 6.4, we have the standard nozzle configuration, except that we can
calculate V in in this problem from Eq. (6.14) as
V in = 4 _mv
πD 2
where v ≈ v f ð60:0°FÞ = 0:01603 ft 3 /lbm: Then, for system a,
ðV in Þ a = 4ð0:800 lbm/sÞð0:01603 ft3 /lbmÞ
πð1:00 in:Þ 2 ð1 ft/12 in:Þ 2
= 2:35 ft/s
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
solution for the time being.
The solution to part a is obtained by solving Eq. (6.12), the modified energy rate balance, for (V out ) a
, with the enthalpy values
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
about any heat transfer to or from the nozzle, we assume that there is none. This is justifiable on the following basis:
Since the nozzle is very small (hand size), the water is not inside of it long enough for any significant heat transfer of energy
to occur. This is a common circumstance that often occurs in obviously uninsulated small systems with no significant heat
transfer. If the residence time of the fluid in the system is very small, then in the absence of an extraordinarily large temperature
difference between the environment and the system, the time is simply insufficient for any significant heat transfer to
occur, regardless of whether the system is insulated or not. The modified energy rate balance for system a reduces to a form of
Eq. (6.21):
ðV out Þ a
= fVin 2 + 2g 1/2
c½cðT in − T out Þ + vðp in − p out ÞŠg a
The problem statement told us to assume the water flow through the nozzle is isothermal, so we set T in = T out . Actually, the
water flow is not exactly isothermal, due to an increase in internal energy of the water from viscous effects, turbulence, and
so forth. However, for a small nozzle, these effects are negligible. For an isothermal flow, we obtain
The data for system a are as follows:
Then
ðV out Þ a
= ½ðV 2 in Þ a + 2g cvðp in − p out Þ a
Š 1/2
ðV in Þ a
= 2:35 ft/s
v = v f ð60:0°FÞ = 0:01603 ft 3 /lbm
ðp in Þ a
= 80:0 psig = 94:7 psia
ðp out Þ a
= 0:00 psig = 14:7 psia
ðV out Þ a
= ð2:35 ft/sÞ 2 + 2 32:174 lbm .ft
ð0:01603
lbf .s 2 ft 3 /lbmÞ½ð80:0 − 0:00Þ lbf/in 2 Š × 144 in 2 /ft 2
1/2
= 109 ft/s
Notice that (V in ) a
was only about 2% of (V out ) a
and therefore could have been neglected in this case.
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
modified energy rate balance. The following assumptions are now made for system b:
_Q = _W = 0 ðZ out Þ b = ?
ðV out Þ a = ðV in Þ b = 109 ft/s ðp in Þ b = ðp out Þ b = 14:7 psia
ðV out Þ b
≈ 0ft/s ðT in Þ b
= ðT out Þ b
= 60:0°F
ðZ in Þ b
= 0
(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
through a large surface area. The conservation of mass law for an incompressible fluid requires that _m =constantandso
(V out ) b =(V in ) b (A in /A out ) b ≪ (V in ) a here.