p0702 - first law of thermodynamics ∆ E = Q − W = ⇒ W...

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7.2. CHAPTER 7, PROBLEM 2 749 7.2 Chapter 7, Problem 2 We know that, for an isentropic process, p = A ρ γ ,wh e r e A is a constant. From the perfect-gas law, p = ρ RT = ρ = p RT so that p = A w p RT W γ = A R γ p γ T γ = p 1 γ = A R γ T γ Consequently, temperature and pressure are related according to T =constant · p ( γ 1) / γ = w T 2 T 1 W = X p 2 p 1 ~ ( γ 1) / γ Thus, noting that the tables in the text tell us γ =1 . 4 for hydrogen, the final temperature is T 2 =(273 . 16 K) w 0 . 4MPa 0 . 2MPa W (1 . 4 1) / 1 . 4 = 332 . 98K = 59 . 8
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Unformatted text preview: first law of thermodynamics, ∆ E = Q − W = ⇒ W = − ∆ E But, ∆ E = m ∆ e = mc v ∆ T , so that W = − mc v ( T 2 − T 1 ) where W is the work done by the system on the surroundings. Hence, W c = mc v ( T 2 − T 1 ) So, since c v = c p − R , for the given values W c = m ( c p − R ) ( T 2 − T 1 ) = (5 kg)(14180 J / kg / K − 4124 J / kg / K)(332 . 98K − 273 . 16 K) = 3 . 01 · 10 6 J = 3 . 01 MJ...
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