(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.

8th Edition

ISBN: 8220100257056

Author: CENGEL

Publisher: YUZU

See similar textbooks

Concept explainers

Properties of Pure Substances

A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substan…

Videos

Question

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.7, Problem 67P

Chapter 14.7, Problem 69P

Students have asked these similar questions

Saturated water at constant pressure is entering a boiler and producing wet steam with dryness fraction 0.9. if the enthalpy of saturated water is 500 kJ/kg and enthalpy of Saturated steam is 2700 kJ/kg, the heat rate per kg of steam is

steam at a pressure of 1.4MN/m2 with a dryness fraction of 0.7 is throttled to a pressure of 0.11MN/m2. Determine the dryness fraction of the steam after throttling.

Steam at 3000 kPa and a temp of 630.0 C is expanded adiabatically until its pressure becomes 1.5 kPa. Determine the final dryness fraction of the steam. The adiabatic expansion may be assumed to be isentropic. XX = Enter the number as a decimal (e.g. 0.7216)

Chapter 14 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 14.7 - What is the difference between dry air and...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 - Is the relative humidity of saturated air...Ch. 14.7 - Is it possible to obtain saturated air from...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. 8P Ch. 14.7 - Consider a tank that contains moist air at 3 atm...Ch. 14.7 - Why are the chilled water lines always wrapped...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The exit temperature.

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

Concept explainers

Videos

The exit temperature.

The exit relative humidity of the air.

The exit velocity.

Want to see the full answer?

Chapter 14 Solutions