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

Replacing dhby cpdTand duby cvdT and dividing the resulting expression
bydT,we obtain

(4 –29)

This is an important relationship for ideal gases since it enables us to deter-
mine cvfrom a knowledge of cpand the gas constant R.
When the specific heats are given on a molar basis,Rin the above equa-
tion should be replaced by the universal gas constant Ru(Fig. 4 –29).

(4 –30)

At this point, we introduce another ideal-gas property called the specific
heat ratiok, defined as

(4 –31)

The specific ratio also varies with temperature, but this variation is very
mild. For monatomic gases, its value is essentially constant at 1.667. Many
diatomic gases, including air, have a specific heat ratio of about 1.4 at room
temperature.

k

cp

cv

cpcvRu¬¬ 1 kJ>kmol#K 2

cpcvR¬¬ 1 kJ>kg#K 2

Chapter 4 | 183

∆u = cv ∆T

T 1 = 20 °C

P = constant

AIR

T 2 = 30 °C

Q 1 Q 2

T 1 = 20 °C

V = constant

AIR

T 2 = 30 °C

= 7.18 kJ/kg

∆u = cv ∆T

= 7.18 kJ/kg

FIGURE 4 –27

The relation ucvTis valid for

anykind of process, constant-volume

or not.

∆u = u 2 – u 1 (table)

∆u =

2

1

cv(T) dT

∆u ≅cv,avg ∆T

FIGURE 4 –28

Three ways of calculating u.

EXAMPLE 4 –7 Evaluation of the uof an Ideal Gas

Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine

the change in internal energy of air per unit mass, using (a) data from the air

table (Table A–17), (b) the functional form of the specific heat (Table A–2c),

and (c) the average specific heat value (Table A–2b).

Solution The internal energy change of air is to be determined in three differ-

ent ways.

Assumptions At specified conditions, air can be considered to be an ideal

gas since it is at a high temperature and low pressure relative to its critical-

point values.

Analysis The internal energy change uof ideal gases depends on the ini-

tial and final temperatures only, and not on the type of process. Thus, the

following solution is valid for any kind of process.

(a) One way of determining the change in internal energy of air is to read the

uvalues at T 1 and T 2 from Table A–17 and take the difference:

Thus,

(b) The c–p(T) of air is given in Table A–2cin the form of a third-degree poly-

nomial expressed as

cp 1 T 2 abTcT^2 dT^3

¢uu 2 
u 1  1 434.78
214.07 2 kJ>kg220.71 kJ>kg

u 2 u @ 600 K434.78 kJ>kg

u 1 u @ 300 K214.07 kJ>kg