Explanation As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process. m ˙ a 1 = m ˙ a 2 = m ˙ a Here, the mass flow rate of air at inlet is m ˙ a 1 , mass flow rate of dry air at exit is m ˙ a 2 and mass flow rate of dry air is m ˙ a . Express the water mass balance: ∑ m ˙ w , i = ∑ m ˙ w , e m ˙ 3 + m ˙ a 1 ω 1 = m ˙ 4 + m ˙ 2 ω 2 m ˙ 3 − m ˙ 4 = m ˙ a ( ω 2 − ω 1 ) Here, mass flow rate of water at inlet and exit is m ˙ w , i and m ˙ w , e respectively, specific humidity at state 1 and 2 is ω 1 and ω 2 respectively; mass flow rate at state 2, 3 and 4 is m ˙ 2 , m ˙ 3 and m ˙ 4 respectively. Express the energy balance. E ˙ in − E ˙ out = Δ E ˙ system E ˙ in − E ˙ out = 0 E ˙ in = E ˙ out ∑ m ˙ i h i = ∑ m ˙ e h e 0 = ∑ m ˙ e h e − ∑ m ˙ i h i 0 = m ˙ a 2 h 2 + m ˙ 4 h 4 − m ˙ a 1 h 1 − m ˙ 3 h 3 h 4 = m ˙ 3 h 3 − m ˙ a ( h 2 − h 1 ) m ˙ 3 − m ˙ a ( ω 2 − ω 1 ) (I) Here, the rate of total energy entering the system is E ˙ in , the rate of total energy leaving the system is E ˙ out , the rate of change in the total energy of the system is Δ E ˙ system , initial and exit mass flow rate is m ˙ i and m ˙ e respectively, enthalpy at inlet and exit is h i and h e respectively and enthalpy at state 1, 2, 3 and 4 is h 1 , h 2 , h 3 and h 4 respectively. Express the water’s exit temperature. T 4 = T sat @ h 4 = T sat @ h f (II) Here, saturation temperature corresponding to enthalpy at state 4 and enthalpy of saturation liquid is T sat @ h 4 and T sat @ h f respectively. Conclusion: Refer Figure A-31E, “psychometric chart at 1 atm total pressure”, and write the properties corresponding to dry bulb temperature of 60°F and relative humidity of 20% . h 1 = 16.8 Btu / lbm dry air ω 1 = 0.00219 lbm H 2 O / lbm dry air ν 1 = 13.15 ft 3 / lbm dry air Here, specific volume at inlet is v 1 . Refer Figure A-31E, “psychometric chart at 1 atm total pressure”, and write the properties corresponding to dry bulb temperature of 75°F and specific humidity of 0.018 lbm H 2 O / lbm dry air . h 2 = 37.7 Btu / lbm dry air ϕ 2 = 0.957 = 95.7 % ω 2 = 0.018 lbm H 2 O / lbm dry air Here, the relative humidity at which the air leaves the tower is ϕ 2

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Chapter 14.7, Problem 110P

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Chapter 14.7, Problem 109P

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Chapter 14 Solutions

THERMODYNAMICS LLF W/ CONNECT ACCESS

Ch. 14.7 - What is the difference between dry air and...Ch. 14.7 - What is vapor pressure?Ch. 14.7 - What is the difference between the specific...Ch. 14.7 - Can the water vapor in air be treated as an ideal...Ch. 14.7 - Explain how vapor pressure of the ambient air is...Ch. 14.7 - Is the relative humidity of saturated air...Ch. 14.7 - Moist air is passed through a cooling section...Ch. 14.7 - How will (a) the specific humidity and (b) the...Ch. 14.7 - Prob. 9P Ch. 14.7 - Consider a tank that contains moist air at 3 atm...

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The relative humidity at which the air leaves the tower and the water’s exit temperature.

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