Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

184 | Thermodynamics

where a28.11, b0.1967 10 
^2 , c0.4802 10 
^5 , and

d
1.966 10 
^9. From Eq. 4–30,

From Eq. 4–25,

Performing the integration and substituting the values, we obtain

The change in the internal energy on a unit-mass basis is determined by

dividing this value by the molar mass of air (Table A–1):

which differs from the tabulated value by 0.8 percent.

(c) The average value of the constant-volume specific heat cv,avgis determined

from Table A–2bat the average temperature of (T 1 T 2 )/2 450 K to be

Thus,

Discussion This answer differs from the tabulated value (220.71 kJ/kg) by

only 0.4 percent. This close agreement is not surprising since the assump-

tion that cvvaries linearly with temperature is a reasonable one at tempera-

ture intervals of only a few hundred degrees. If we had used the cvvalue at

T 1 300 K instead of at Tavg, the result would be 215.4 kJ/kg, which is in

error by about 2 percent. Errors of this magnitude are acceptable for most

engineering purposes.

220 kJ>kg

¢ucv,avg 1 T 2
T 12  1 0.733 kJ>kg#K 231600
3002 K 4

cv,avgcv @ 450 K0.733 kJ>kg#K

¢u

¢u

M



6447 kJ>kmol

28.97 kg>kmol

222.5 kJ>kg

¢u6447 kJ>kmol

¢u

2

1

¬cv 1 T 2 ¬dT

T 2

T 1

¬^31 a
Ru^2 bTcT

(^2) dT 34 ¬ (^) dT
cv 1 T 2 cp
Ru 1 a
Ru 2 bTcT^2 dT^3
EXAMPLE 4 –8 Heating of a Gas in a Tank by Stirring
An insulated rigid tank initially contains 1.5 lbm of helium at 80°F and
50 psia. A paddle wheel with a power rating of 0.02 hp is operated within
the tank for 30 min. Determine (a) the final temperature and (b) the final
pressure of the helium gas.
Solution Helium gas in an insulated rigid tank is stirred by a paddle wheel.
The final temperature and pressure of helium are to be determined.
Assumptions 1 Helium is an ideal gas since it is at a very high temperature
relative to its critical-point value of
451°F. 2 Constant specific heats can be
used for helium. 3 The system is stationary and thus the kinetic and potential
energy changes are zero, KE PE 0 and EU. 4 The volume of
the tank is constant, and thus there is no boundary work. 5 The system is adi-
abatic and thus there is no heat transfer.
AIR at 300 K
cv = 0.718 kJ/kg · K
R = 0.287 kJ/kg. K {cp = 1.005 kJ/kg. K
or
cv = 20.80 kJ/kmol. K
Ru = 8.314 kJ/kmol. K cp = 29.114 kJ/kmol

. K
{

FIGURE 4 –29

The cpof an ideal gas can be
determined from a knowledge of
cvand R.