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Modern Engineering Thermodynamics

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374 CHAPTER 11: More Thermodynamic Relations<br />

EXAMPLE 11.8<br />

In Chapter 3, an equation of state developed in 1903 by Pierre Berthelot (1827–1907) was briefly discussed. Using this<br />

equation of state, develop equations based on measurable properties for the changes in (a) specific internal energy, (b) specific<br />

enthalpy, and (c) specific entropy for an isothermal process.<br />

Solution<br />

The Berthelot equation is given in Eq. (3.46) as<br />

pv− ð b<br />

where a and b are constants. Solving this equation for p gives<br />

Þ = RT − av− ð bÞ/ Tv 2<br />

p = RT/ ðv − bÞ− a/ Tv 2<br />

a. The change in specific internal energy is given by Eq. (11.20), for which we need<br />

<br />

∂p<br />

= R/ ðv − bÞ+ a/ T 2 v 2<br />

∂T<br />

Then, for an isothermal process (T 1 = T 2 ), Eq. (11.20) gives<br />

ðu 2 − u 1 Þ T =<br />

Z v2<br />

v1<br />

v<br />

<br />

RT/ ðv − bÞ+ a/ Tv 2 − RT/ðv − bÞ + a/ Tv 2 dv<br />

= −ð2a/TÞð1/v 2 − 1/v 1 Þ = 2aðv 2 − v 1 Þ/ ðTv 1 v 2 Þ<br />

b. To find the change in specific enthalpy, we could use Eq. (11.23). However, to evaluate this equation, we need to be<br />

able to determine the relation ð∂v/∂TÞ p<br />

: Since the Berthelot equation is not readily solvable for v = v (T, p), we choose<br />

instead to use the simpler approach, utilizing only the definition of specific enthalpy, h = u + pv. Then,<br />

ðh 2 − h 1 Þ T<br />

= ðu 2 − u 1 Þ T<br />

+ p 2 v 2 − p 1 v 1 = 2aðv 2 − v 1 Þ<br />

+ p 2 v 2 − p 1 v 1<br />

Tv 1<br />

v 2<br />

= 3aðv <br />

2 − v 1 Þ v<br />

+ RT 2<br />

Tv 1 v 2 v 2 − b − v <br />

1<br />

v 1 − b<br />

c. Finally, since we already evaluated the relation ð∂p/∂TÞ v , we choose to use Eq. (11.25) for the isothermal specific entropy relation:<br />

Z v2<br />

<br />

∂p<br />

ðs 2 − s 1 Þ T = dv<br />

∂T<br />

v<br />

=<br />

v1<br />

Z v2<br />

v1<br />

<br />

R/ðv − bÞ + a/ðT 2 v 2 <br />

Þ dv<br />

= R ln ½ðv 2 − bÞ/ðv 1 − bÞŠ+ aðv 2 − v 1 Þ/ðT 2 v 1 v 2 Þ<br />

Exercises<br />

22. Setting a = 0 in the Berthelot equation of state used in Example 11.8 produces the Clausius equation of state, p(v – b) = RT<br />

(see Eq. (3.43)). Determine equations for the change in specific internal energy, specific enthalpy, and specific entropy for a<br />

Clausius gas undergoing an isothermal process. Answer:(u 2 – u 1 ) T = 0, (h 2 – h 1 ) T = RT [(v 2 /(v 2 – b) – v 1 /(v 1 – b)], and (s 2 – s 1 ) T = R<br />

ln[(v 2 – b)/(v 1 – b)].<br />

23. Evaluate the change in specific internal energy of water vapor when it is modeled as a Berthelot gas, with a = 4.30 MN·m 4·K/kg 2<br />

and b = 4.50 × 10 –3 m 3 /kg, and undergoes an isothermal compression from a specific volume 40.0 m 3 /kg to a specific volume<br />

of 5.00 m 3 /kg at a constant temperature of 100.°C. Answer: (u 2 – u 1 ) T = –4.03 kN·m/kg = –4.03 kJ/kg.<br />

24. Evaluate the change in specific enthalpy of water vapor when it is modeled as a Berthelot gas, with a = 4.30 MN·m 4·K/kg 2 and<br />

b = 4.50 × 10 –3 m 3 /kg, and undergoes an isothermal expansion from a specific volume of 10.0 × 10 –3 m 3 /kg to a specific volume<br />

of 1.00 m 3 /kg at a constant temperature of 500.°C. Use R water = 461 N·m/(kg·K). Answer:(h 2 – h 1 ) T = 261 kN·m/kg = 261 kJ/kg.<br />

Note that, for an ideal gas undergoing an isothermal process,<br />

ðu 2 − u 1 Þ T = ðh 2 − h 1 Þ T = 0 and ðs 2 − s 1 Þ T = R lnðv 2 /v 1 Þ<br />

Therefore, the equations developed in Example 11.8 can be considered to be Berthelot corrections to ideal gas<br />

behavior.<br />

Equation (11.24) has the same Mdx + Ndy form as Eq. (11.11), so that we can utilize Eq. (11.12) to produce<br />

the property relation<br />

<br />

∂ðc v /TÞ<br />

∂v<br />

T<br />

<br />

= ∂<br />

∂T<br />

<br />

∂p<br />

∂T<br />

<br />

v<br />

v

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