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Modern Engineering Thermodynamics

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

Table 6.2 The Effect of Height on Potential Energy<br />

Height, Z<br />

Potential Energy, gZ/g c<br />

ft m Btu/lbm kJ/kg<br />

0 0 0 0<br />

1.0 0.30 1.3 × 10 −3 2.9 × 10 −3<br />

10. 3.0 1.3 × 10 −2 2.9 × 10 −2<br />

100. 30.5 1.3 × 10 −1 2.9 × 10 −1<br />

1000. 305 1.3 2.9<br />

There are, of course, exceptions to this rule of thumb. If h in − h out ≈ 0, then the enthalpy term loses its dominance.<br />

In this case, a small specific kinetic energy term may be quite significant to the analysis. A nozzle or a<br />

diffuser is an example of such an exception.<br />

Now consider the specific potential energy term, gZ/g c . Taking g =32.174ft/s 2 and, g c =32.174lbm· ft/(lbf · s 2 ), then<br />

0<br />

1<br />

<br />

gZ<br />

= Z g <br />

B 32:174 ft/s 2 C<br />

= ð½ZŠ ftÞ× @<br />

g c g c<br />

32:174 lbm A= ½ZŠ<br />

.ft<br />

ft .lbf<br />

lbm<br />

lbf .s 2<br />

where the symbol [Z] stands for the numerical value of Z in units of feet. Again, converting to Btu/lbm gives<br />

gZ<br />

<br />

= ½ZŠ ft <br />

<br />

.lbf<br />

×<br />

1Btu<br />

=<br />

g c lbm 778:16 ft.lbf<br />

In the SI units system, g c = 1.0 and is dimensionless, so<br />

<br />

gZ<br />

= Z g <br />

9:807 m/s2<br />

= ð½ZŠ mÞ ×<br />

g c g c<br />

1<br />

where 1 m 2 /s 2 =1J/kg=10 −3 kJ/kg.<br />

<br />

½ZŠ<br />

778:16<br />

Btu<br />

lbm<br />

= ½ΖŠ × 9:807 J 9:807<br />

= ½ΖŠ<br />

kg 1000<br />

Table 6.2 gives values of specific potential energy for various heights using these equations. Note that, for<br />

systems with normal engineering dimensions, say less than 1000 ft (305 m) high, the specific potential energy<br />

is very small. Consequently, it is common to neglect the effect of a flow stream’s specific potential energy when<br />

the flow stream enters the system less than 1000 ft (305 m) above or below the potential energy baseline of the<br />

system, or gZ/g c < ~1.0 Btu/lbm (2.3 kJ/kg).<br />

There are also exceptions to this rule of thumb. Again, if h in − h out ≈ 0, then flow stream height changes may be<br />

very important in the analysis. A hydroelectric power plant is an example of such an exception.<br />

Deciding whether to neglect factors such as specific kinetic and potential energies is not always an easy task for<br />

the beginner. Self-confidence comes only with experience. However, one more rule of thumb applies to most<br />

textbook thermodynamic problems: If values for velocity and height are not given in the problem statement and are not<br />

among the problem’s unknowns, then you are supposed to neglect the kinetic and potential energy terms in your analysis.<br />

This means that the person who wrote the problem knew that either V in ≈ V out and Z in ≈ Z out or that all the<br />

velocities and heights were relatively small. The only exception to this last rule of thumb is when you know the<br />

mass flow rate _m , the diameter D, or cross-sectional area A, and the fluid density ρ or specific volume v of a<br />

flow stream. With this information you can calculate the flow stream velocity using Eq. (6.1) as<br />

V = _m ρA = _mv<br />

A<br />

kJ<br />

kg<br />

= 4 _mv<br />

πD 2 (6.14)<br />

If you can make this calculation for V, then you might as well use it in your energy rate balance equation, unless<br />

it is so small that you are certain it will not affect the results of your analysis.<br />

6.5 NOZZLES AND DIFFUSERS<br />

Nozzle is the generic name of any device whose primary function is to convert the pressure energy _mpv of an<br />

inlet flow stream into the kinetic energy _mV 2 /2 of an outlet flow stream. Thus, a nozzle is a very simple energy<br />

conversion device. Similarly, diffuser is the generic name of any device whose primary function is to convert the<br />

kinetic energy of an inlet flow stream into the pressure energy of an outlet flow stream. Note that nozzles and

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