HW5_solution

HW5_solution - Solution to HW 5 July 27, 2007 1 Isothermal...

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Solution to HW 5 July 27, 2007 1 Isothermal Compression Isothermal compression of a mixture of gases with given initial and final conditions (pressure and temperature). Assumptions: Closed system Reversible process Non-reactive process Given Data: Species: CO identified as 1 , and O 2 identified as 2. Masses: m 1 = 2 kg, m 2 = 3 kg T 1 = 1200 K T 2 = 1200 K p 1 = 1 bar p 2 = 10 bar Find: 1
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final volume, V 2 . Heat transfer, Q . Work done by the system (gas), W . Solution The molar masses are M 1 = 28 . 01 kg/kmole and M 2 = 32 kg/kmole. The mole numbers are n 1 = m 1 /M 1 = 0 . 0714 and n 2 = m 2 2 = 0 . 0938 . The total number of moles is, n = n 1 + n 2 = 0 . 1652; finally the mole fractions are y 1 = n 1 /n = 0 . 4323 and y 2 = n 2 /n = 0 . 5677 . a) V 2 = n ¯ RT 2 /p 2 = 1 . 6478 m 3 b) for a reversible isothermal process Q = mT Δ s = nT Δ¯ s. Where, Δ¯ s = y 1 ( ¯ s 2 , CO - ¯ s 1 , CO ) + y 2 ( ¯ s 2 , O 2 - ¯ s 1 , O 2 ) - ¯ R ln( p 2 /p 1 ) , the ¯ s terms drop out of the equation because the temperature is constant, and the only term remaining is Δ¯ s = - ¯ R ln( p 2 /p 1 ) . Thus, Q = - 3794 . 2 kJ c) From the first law of thermodynamics Δ U = Q - W. For a mixture of ideal gases, the internal energy is a function of the temperature only, so its change is zero in a isothermal transformation. W = Q = - 3794 . 2 kJ 2
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2 Adiabatic compression Adiabatic compression of a mixture of ideal gases between the given initial condition and the final pressure.
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HW5_solution - Solution to HW 5 July 27, 2007 1 Isothermal...

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