Modern Engineering Thermodynamics

11.9 Gas Tables 383
where p r is the relative pressure, defined as
p r = p/p o = expðϕ/RÞ
consequently,
p 2 /p 1 = p r2 /p r1 (11.36)
Equation (7.34) expresses the specific entropy in terms of temperature and specific volume as
Again, for an isentropic process, s 2 = s 1 and
− 1 R
Z T2
where v r is the relative volume, defined as
consequently,
s 2 − s 1 =
Z T2
ðc v /TÞdT + R ln ðv 2 /v 1 Þ (7.34)
T 1
ðc v /TÞdT = lnðv 2 /v 1 Þ = ln v 2v o
= ln ðv r2 /v r1 Þ
T 1
v 1 v o
v r = v/v 0 = exp − 1 R
ðc v /TÞdT
T 0
Z T
v 2 /v 1 = v r2 /v r1 (11.37)
The p r and v r columns of the gas tables are to be used only for isentropic processes, and their values are to be used
only in Eqs. (11.36) and (11.37).
EXAMPLE 11.12
A diesel engine has a compression ratio of 19.2 to 1. Air at 60.0°F and 14.7 psia is drawn into the engine during the intake
stroke and compressed isentropically during the compression stroke. Using the gas tables, determine the final temperature
and pressure of the air at the end of the compression stroke and the work required per lbm of air present.
Solution
First, draw a sketch of the system (Figure 11.2).
Air at 60.0°F
and 14.7 psia
v 2 = v 1 /19.2
T 2 = ?
State 1 State 2
FIGURE 11.2
Example 11.12.
Theunknownsherearethefinaltemperatureandpressureoftheairattheendofthecompressionstrokeandthework
required per lbm of air present.
The piston-cylinder arrangement of a diesel engine forms a closed system for the air being compressed. The unknowns are T 2 ,
p 2 , and 1 W 2 /m. The energy balance for this system (neglecting any changes in the potential and kinetic energies of the air) is
1Q 2 − 1 W 2 = mu ð 2 − u 1 Þ
(Continued )