Explanation The following figure represents the dehumidifier and different state. Figure-(1) Write the expression for mass balance equation for the dry air. m ˙ a 1 = m ˙ a 3 = m ˙ a Here, mass flow rate of the air at the state-1 is m ˙ a 1 and mass flow rate of the air at the state-3 is m ˙ a 3 . Write the expression for mass balance equation for the water vapor. m ˙ v 1 = m ˙ v 2 + m ˙ w (I) Here, mass flow rate of the water vapor at the state-1 is m ˙ v 1 , mass flow rate of the water vapor at the state-2 is m ˙ v 2 and mass flow rate of the water is m ˙ w . Write the expression for the mass flow rate of water vapor. at the state-1. m ˙ v 1 = m ˙ a ω 1 Here, specific humidity at the state-1 is ω 1 . Write the expression for the mass flow rate of water vapor. at the state-2. m ˙ v 2 = m ˙ a ω 2 Here, specific humidity at the state-2 is ω 2 . Substitute m ˙ a ω 1 for m ˙ v 1 and m ˙ a ω 2 for m ˙ v 2 in the Equation (I). m ˙ v 1 = m ˙ a ω 1 + m ˙ a ω 2 = m ˙ a ( ω 1 + ω 2 ) (II) Write the expression for the vapor pressure at the state-1. p v 1 = ϕ 1 p g 1 (III) Here, relative humidity of the air at the state-1 is ϕ 1 , saturates pressure of the water pressure at state-1 is p g 1 . Write the expression for the specific humidity at the state-1. ω 1 = 0.622 ( p v 1 p 1 − p v 1 ) (IV) Write the expression for the vapor pressure at the state-2. p v 2 = ϕ 1 p g 2 (V) Here, relative humidity of the air at the state-2 is ϕ 2 , saturates pressure of the water pressure at state-2 is p g 2 . Write the expression for the specific humidity at the state-2. ω 2 = 0.622 ( p v 2 p 2 − p v 2 ) (VI) Write the expression for the energy rate balance equation at steady state to obtain the refrigerant flow rate. m ˙ R ( h i − h e ) = m ˙ a ( c p a T 2 − c p a T 1 + ω 2 h g 2 − ω 1 h g 1 + ω 1 h f 3 − ω 2 h f 3 ) m ˙ R m ˙ a = ( c p a T 2 − c p a T 1 + ω 2 h g 2 − ω 1 h g 1 + ω 1 h f 3 − ω 2 h f 3 ) ( h i − h e ) (VII) Here, refrigerant flow rate per kg of dry air is m ˙ R m ˙ a , specific enthalpy of the inlet refrigerant is h i , specific enthalpy of the exit refrigerant is h e , specific heat of the air is c p a , temperature at the state-2 is T 2 , temperature at the state-1 is T 1 , specific enthalpy saturated vapor at state-2 is h g 2 , T 1 , specific enthalpy saturated vapor at state-1 is h g 1 and specific enthalpy saturated liquid at state-3 is h f 3 . Conclusion: Refer Table A-2 “Properties of Saturated Water (Liquid–Vapor)” to obtain p g 1 as 0

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ISBN: 2818440116926

Author: MORAN

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Chapter 12.9, Problem 60P

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The refrigerant flow rate.

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Chapter 12.9, Problem 59P

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FUNDAMENTALS OF ENGINEERING THERMODYNAM

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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