Explanation Express partial pressure of water vapor at inlet. P ν , 1 = ϕ 1 P g , 1 = ϕ 1 P sat@20°C (I) Here, the relative humidity at inlet is ϕ 1 , initial vapor pressure is P g , 1 and saturation pressure at temperature of 20°C is P sat@20°C . Express the final and initial specific humidity. ω 2 = ω 1 = 0.622 P ν , 1 P − P ν , 1 (II) Here, initial and final specific humidity is ω 1 and ω 2 respectively, and pressure is P . Express the final temperature. T 2 = T 1 [ P 2 P 1 ] k − 1 k (III) Here, inlet or initial temperature is T 1 , specific heat ratio is k , initial and final pressure is P 1 and P 2 respectively. Express gas pressure at outlet. P g , 2 = P sat@258°C (IV) Here, saturation pressure at temperature of 258°C is P sat@258°C . Express vapor pressure at the exit. P ν , 2 = ω 2 P 2 ω 2 + 0.622 (V) Express the relative humidity of the air at the compressor outlet. ϕ 2 = P ν , 2 P g , 2 (VI) Conclusion: Refer Table A-4, “saturated water-temperature table”, and write saturation pressure at temperature of 20°C . P sat@20°C = 2.3392 kPa Substitute 2.3392 kPa for P sat@20°C and 0.90 for ϕ 1 in Equation (I). P ν , 1 = ( 0.90 ) ( 2.3392 kPa ) = 2.105 kPa Substitute 2.105 kPa for P ν , 1 and 100 kPa for P in Equation (II). ω 2 = 0.622 ( 2.105 kPa ) 100 kPa − 2.105 kPa = 0.0134 kg H 2 O / kg dry air Refer Table A-2 (a), “ideal gas specific heats of various common gases”, and write specific heat ratio of dry air. k = 1.4 Substitute 20°C for T 1 , 1.4 for k , 800 kPa for P 2 and 100 kPa for P 1 in Equation (III). T 2 = ( 20°C ) [ 800 kPa 100 kPa ] 1.4 − 1 1.4 = ( 20 + 273 ) K ( 8 ) 0

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Chapter 14.7, Problem 19P

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Chapter 14 Solutions

THERMODYNAMICS-SI ED. EBOOK >I<

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The relative humidity of the air at the compressor outlet.

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