Explanation The following figure shows the schematic diagram of the humidification device with control volume. Figure-(1) The temperature at state 1 is T 1 , temperature at state 2 is T 2 , temperature at state 3 is T 3 , the pressure at state 1 is p 1 , the pressure at state 2 is p 2 and the pressure at state 3 is p 3 . Write the expression for ideal gas constant of air. R a = R ¯ M a (I) Here, the universal gas constant is R ¯ and the molecular weight of air is M a . Write the expression for ideal gas constant of vapour. R v = R ¯ M v (II) Here, the molecular weight of air is M v . Write the expression for the mass flow rate of vapour. m ˙ v 1 = ω 1 m ˙ a 1 (III) Here, humidity ratio at state 1 is ω 1 and the mass flow rate of dry air at state 1 is m ˙ a 1 . Write the expression for number of moles for dry air at state 1 . n ˙ a 1 = m ˙ a 1 M a (IV) Write the expression for number of moles for vapour at state 1 . n ˙ v 1 = m ˙ v 1 M v (V) Write the expression for the mole fraction of dry air at state 1 . y a 1 = n ˙ a 1 n ˙ a 1 + n ˙ v 1 (VI) Write the expression for the mole fraction of vapour at state 1 . y v 1 = n ˙ v 1 n ˙ a 1 + n ˙ v 1 (VII) Write the expresson for the mass flow rate of vapour at state 3 . m ˙ v 3 = m ˙ v 1 + m ˙ v (VIII) Write the expression for number of moles for dry air at state 3 . n ˙ a 3 = m ˙ a 3 M a (IX) Write the expression for number of moles for vapour at state 3 . n ˙ v 3 = m ˙ v 3 M v (X) Write the expression for the mole fraction of dry air at state 3 . y a 3 = n ˙ a 3 n ˙ a 3 + n ˙ v 3 (XI) Write the expression for the mole fraction of vapour at state 3 . y v 3 = n ˙ v 3 n ˙ v 3 + n ˙ v 3 (XII) Write the expression for the rate of entropy production for control volume. σ ˙ c v = ( m ˙ a ( c p a ln T 3 T 1 − R a ln y a 3 y a 1 ) + m ˙ v 3 ( s g ( T 3 ) − R ln y v 3 p p g ( T 3 ) ) − m ˙ v 1 ( s g ( T 1 ) − R ln y v 1 p p g ( T 1 ) ) − m ˙ v s g ( T 2 ) ) (XIII) Here, the specific heat at constant pressure of dry air is c p a . Write the expression for the rate of energy destruction. E ˙ d = T 0 σ ˙ c v (XIV) Here, the ambient temperature is T 0 and the rate of entropy production at control volume is σ ˙ c v . Conclusion: Substitute 1545 lbf ⋅ ft / lbmol ⋅ ° R for R ¯ and 28.97 lb / lbmol for M a in Equation (I). R a = 1545 lbf ⋅ ft / lbmol ⋅ ° R 28.97 lb / lbmol = 1545 lbf ⋅ ft / lbmol ⋅ ° R 28.97 lb / lbmol × 1 Btu 778.169 lbf ⋅ ft = 0.0685 Btu / lb ⋅ ° R Refer to Table A-1E, “Atomic or molecular weights and critical properties of selected elements and compounds” to obtain the value of M v for water as 18.02 lb / lbmol . Substitute 1545 lbf ⋅ ft / lbmol ⋅ ° R for R ¯ and 18.02 lb / lbmol for M v in Equation (II). R v = 1545 lbf ⋅ ft / lbmol ⋅ ° R 18.02 lb / lbmol = 1545 lbf ⋅ ft / lbmol ⋅ ° R 18.02 lb / lbmol × 1 Btu 778.169 lbf ⋅ ft = 0.1102 Btu / lb ⋅ ° R Substitute 0.01406 for ω 1 and 400 lb / min for m ˙ a 1 in Equation (III). m ˙ v 1 = ( 0.01406 ) ( 400 lb / min ) = 5.624 lb / min Substitute 400 lb / min for m ˙ a 1 and 28.97 lb / kmol for M a in Equation (IV). n ˙ a 1 = 400 lb / min 28.97 lb / kmol = 13.8074 kmol / min Substitute 5.624 lb / min for m ˙ v 1 and 18.02 lb / kmol for M v in Equation (V). n ˙ v 1 = 5.624 lb / min 18.02 lb / kmol = 0.3121 kmol / min Substitute 13.8074 kmol / min for n ˙ a 1 and 0.3121 kmol / min for n ˙ v 1 in Equation (VI). y a 1 = 13.8074 kmol / min 13.8074 kmol / min + 0.3121 kmol / min = 13.8074 kmol / min 14.1195 kmol / min = 0.9779 Substitute 13.8074 kmol / min for n ˙ a 1 and 0.3121 kmol / min for n ˙ v 1 in Equation (VII). y v 1 = 0.3121 kmol / min 13.8074 kmol / min + 0.3121 kmol / min = 0.3121 kmol / min 14.1195 kmol / min = 0.0221 Substitute 5.624 lb / min for m ˙ v 1 and 1.404 lb / min for m ˙ v in Equation (VIII). m ˙ v 3 = 5.264 lb / min + 1.404 lb / min = 7.028 lb / min Substitute 400 lb / min for m ˙ a 1 and 28.97 lb / kmol for M a in Equation (IX). n ˙ a 3 = 400 lb / min 28.97 lb / kmol = 13.8074 kmol / min Substitute 7.028 lb / min for m ˙ v 1 and 18.02 lb / kmol for M v in Equation (X). n ˙ v 3 = 7.028 lb / min 18.02 lb / kmol = 0.39 kmol / min Substitute 13.8074 kmol / min for n ˙ a 3 and 0.39 kmol / min for n ˙ v 3 in Equation (XI)

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Fundamentals of Engineering Thermodynamics

Ch. 12.9 - Prob. 1E Ch. 12.9 - Prob. 2E Ch. 12.9 - Prob. 3E Ch. 12.9 - Prob. 4E Ch. 12.9 - Prob. 5E Ch. 12.9 - Prob. 6E Ch. 12.9 - Prob. 7E Ch. 12.9 - Prob. 8E Ch. 12.9 - Prob. 9E Ch. 12.9 - Prob. 10E

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