Modern Engineering Thermodynamics

3.9 Thermodynamic Equations of State 79
EXAMPLE 3.6
Determine the change in specific internal energy and specific
enthalpy of an incompressible hardwood as it is heated from a
temperature and pressure of 20.0°C, 0.100 MPa to a temperature
and pressure of 100.°C and 1.00 MPa (Figure 3.20). Assume the
wood has a constant density of 515 kg/m 3 and a constant specific
heat over this temperature and pressure range.
Solution
The changes in specific internal energy and specific enthalpy of a
constant specific heat incompressible material are given
by Eqs. (3.33) and (3.34), and the specific heat of
wood is found in Table 3.6 to be c =1.76kJ/kg· K.
Then, Eq. (3.33) gives
u 2 − u 1 = cðT 2 − T 1 Þ
20.0°C
0.100 MPa
State 1
FIGURE 3.20
Example 3.6.
100. °C
1.00 MPa
State 2
= ð1:76 kJ=kg . KÞ½ð100 + 273:15Þ − ð20:0 + 273:15ÞKŠ = 141 kJ=kg
Notice that we could have used either °C or K for the temperature difference T 2 − T 1 . This is because the Celsius and Kelvin
degree sizes are exactly the same, only their zero points differ. From Eq. (3.34), we have v =1/ρ = 1/515 = 0.00194 m 3 /kg
and
h 2 − h 1 = cðT 2 − T 1 Þ + vðp 2 − p 1 Þ = u 2 − u 1 + vðp 2 − p 1 Þ
= 141 + 0:00194 m 3 =kg 1:00 × 10 3 − 100: kN=m 2 = 143 kJ=kg
where we have converted the units of pressure into kN/m 2 so that the units of u and of the pv product match exactly.
Exercises
9. Determine the changes in specific internal energy and specific enthalpy of liquid propane as it is heated from −90.0°C,
0.100 MPa to −50.0°C, 1.00 MPa. Assume that liquid propane has a constant density of 615 kg/m 3 and a constant
specific heat over this temperature and pressure range. Answer: u 2 − u 1 = 96.5 kJ/kg and h 2 − h 1 = 97.9 kJ/kg.
10. Determine the changes in specific internal energy and specific enthalpy of a block of iron as it is heated in an oven at
atmospheric pressure from 70.0°F to 250.°F. Assume that iron has a density of 490. lbm/ft 3 and a constant specific heat
over this temperature and pressure range. Answer: u 2 − u 1 = 19.3 Btu/lbm and h 2 − h 1 = 19.3 Btu/lbm.
3.9.2 Ideal Gases
The next simplest equation of state is that of an ideal gas. It is important because all gases approach ideal gas
behavior at low pressure. Like an incompressible substance, an ideal gas is also defined by two state equations,
both of which must be obeyed if a gas is to be called ideal. The first equation of state is the common ideal gas
law, which has the following four equivalent forms:
pV = mRT
pv = RT
pV = nRT
(3.35a)
(3.35b)
(3.35c)
pv = RT
(3.35d)
where n = m/M is the number of moles, v = V/n is the molar specific volume, and R is the universal gas constant
whose value is
R = 1545:35 ft ⋅ lbf/ ðlbmole ⋅ RÞ = 1:986Btu/ ðlbmole ⋅ RÞ
= 8314 joule/ ðkgmole ⋅ KÞ = 8:314 kJ/ ðkgmole ⋅ KÞ
The second state equation used to define an ideal gas is that its specific internal energy is only a function of
temperature, or
u = uðTÞ (3.36)