Modern Engineering Thermodynamics

6.4 Flow Stream Specific Kinetic and Potential Energies 173
Integrating Eq. (6.12) over time gives the open system modified energy balance (MEB) as
1Q 2 − 1 W 2 +
The open system modified energy balance (MEB)
Z 2
1
_m½h in − h out + ðV 2 in − V2 out Þ/ð2g cÞ + ðZ in − Z out Þðg/g c ÞŠ dt = 0 (6.13)
However, the vast majority of open system problems are set up on a rate basis, so the modified energy rate
balance is the equation most often used in open system analysis.
When the conditions of steady state (steady flow), single inlet, or single outlet do not exist in any particular problem,
we must return to the more general energy rate balance of Eq. (6.4) as a starting point for the analysis.
This is illustrated with the unsteady state examples presented later in this chapter.
6.4 FLOW STREAM SPECIFIC KINETIC AND POTENTIAL ENERGIES
Before we can begin analyzing thermodynamic problems, we must establish a criterion for when the specific
kinetic and potential energy flow stream terms of Eqs. (6.4) and (6.12) are important and when they are not.
To get some feeling for the importance of these terms, we look at how their magnitude varies over a wide range
of velocities and heights. First, consider the specific kinetic energy term V 2 /2g c : If we work in the Engineering
English units system, then V is normally in feet per second, and g c = 32:174 lbm.ft/ðlbf .s 2 Þ: Hence,
V 2
2g c
=
½VŠ 2 ft 2 /s 2
2 × 32:174 lbm = ½VŠ2
.ft 64:348
lbf .s 2
ft.lbf
lbm
where the symbol [V] stands for the numerical value of V in units of ft/s. The remaining term in the flow stream
energy transport equation to which the specific kinetic and potential energy terms are to be added is the specific
enthalpy h. In the Engineering English units system, the specific enthalpy has units of Btu/lbm. If we convert the
specific kinetic energy into these units, we get
V 2
2g c
=
½VŠ 2
64:348
ft.lbf
lbm
In the SI units system, g c = 1.0 and is dimensionless, so
where 1 m 2 /s 2 = 1 J/kg = 10 −3 kJ/kg.
l Btu
778:16 ft.lbf
V 2
= ½VŠ2 m 2 /s 2
= ½VŠ2
2g c 2ð1Þ 2
J
kg = ½VŠ2
2000
= ½VŠ2
50,070
Table 6.1 gives values of the specific kinetic energy for various velocities using these equations. Since the specific
enthalpy values for most substances fall roughly between 100 and 1000 Btu/lbm, Table 6.1 shows that the specific
kinetic energy is very small when compared to these h values for velocities less than about 250 ft/s (76 m/s),
or V 2 /2g c is less than about 1:0 Btu/lbm ð2:3 kJ/kgÞ. Consequently, it is common to neglect the effect of a flow
stream’s specific kinetic energy when the flow stream velocityislessthanabout250ft/s(76m/s).Thisisa
relatively high velocity (~170 mi/h or 270 km/h), and most engineering applications do not have such rapid
flow streams.
kJ
kg
Btu
lbm
Table 6.1 The Effect of Velocity on Kinetic Energy
Velocity, V
Kinetic Energy, V 2 /2g c
ft/s m/s Btu/lbm kJ/kg
0 0 0 0
1 0.3 2 × 10 −5 4.7 × 10 −5
10. 3.0 2.0 × 10 −3 4.7 × 10 −3
100. 30.5 2.0 × 10 −1 4.70 × 10 −1
1000. 305 20.0 470.