(a) Explanation Express the mass flow rate of dry air through the cooling section. m ˙ a = 1 v 1 V 1 A 1 = 1 v 1 V 1 [ π D 2 4 ] (I) Here, initial specific volume is v 1 , initial velocity is V 1 , initial area is A 1 and diameter is D . Express final specific enthalpy using energy balance on air in the cooling section. − Q ˙ out = m ˙ a ( h 2 − h 1 ) (II) Here, rate of heat removed from the air is Q ˙ out , initial specific enthalpy is h 1 . Express final specific humidity of air. ω 2 = ω 1 (III) Here, initial specific humidity of air is ω 1 . Conclusion: Refer Figure A-31, “psychometric chart at 1 atm total pressure”, and write the properties corresponding to dry bulb temperature of 35°C and relative humidity of 45 % . h 1 = 76.14 kJ / kg dry air ω 1 = 0.01594 kg H 2 O / kg dry air v 1 = 0.8953 m 3 / kg dry air Here, initial specific volume of air is v 1 . Substitute 0.01594 kg H 2 O / kg dry air for ω 1 in Equation (III). ω 2 = 0.01594 kg H 2 O / kg dry air Perform unit conversion of diameter from cm to m . D = 30 cm = 30 cm [ m 100 cm ] = 0.3 m Substitute 0.8953 m 3 / kg dry air for v 1 , 18 m / s for V 1 and 0 (b) To determine The exit relative humidity of the air. (c) To determine The exit velocity.

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ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Chapter 14.7, Problem 68P

(a)

To determine

The exit temperature.

(b)

To determine

The exit relative humidity of the air.

(c)

To determine

The exit velocity.

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Chapter 14.7, Problem 67P

Chapter 14.7, Problem 69P

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The exit temperature.

Solution Summary: The author explains the exit temperature and mass flow rate of dry air through the cooling section.

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The exit temperature.

The exit relative humidity of the air.

The exit velocity.

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Chapter 14 Solutions