~ Chapter 13 Gas- Vapor Mixtures and Air-Conditioning 13-14 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition, n1v 0.3 kg OO 4 . (J)=-=-= .1 3 kgH2O/kgdryaIr ma 21 kg (b) The saturation pressure of water at 30°C is Pg = Pgli) :I1C = 4246kPa Then the relative humidity can be determined from fj) = wP = (O.OI43)(100 kPa) = 52.9% (0.622+w)Pg (0.622+0.0143)4.246 kPa (c) The volume of the tank can be determined from the ideal gas relation for the dry air , P" = fj)P 9 = (0529)(4.246 kPa) = 2.246 kPa Pa = P-P" = 100-2.246=97.754 kPa V = ~= (21 kg)(0.287 kJ / kg. K)(303 K) = 18.7 mJ Pa 97.754 kPa ~~ --~ ---
8
Embed
Chapter 13 Gas- Vapor Mixtures and Air-Conditioning O O (J ... · Chapter 13 Gas- Vapor Mixtures and Air-Conditioning 13-77 Air is first cooled, then dehumidified, and finally heated.
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
~
Chapter 13 Gas- Vapor Mixtures and Air-Conditioning
13-14 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative
humidity, and the volume of the tank are to be determined.
Assumptions The air and the water vapor are ideal gases.
Analysis (a) The specific humidity can be determined form its definition,
n1v 0.3 kg O O 4 .(J)=-=-= .1 3 kgH2O/kgdryaIr
ma 21 kg
(b) The saturation pressure of water at 30°C is
Pg = Pgli) :I1C = 4246kPa
Then the relative humidity can be determined from
fj) = wP = (O.OI43)(100 kPa) = 52.9%
(0.622+w)Pg (0.622+0.0143)4.246 kPa
(c) The volume of the tank can be determined from the ideal gas relation for the dry air ,
P" = fj)P 9 = (0529)(4.246 kPa) = 2.246 kPa
Pa = P-P" = 100-2.246=97.754 kPa
V = ~= (21 kg)(0.287 kJ / kg. K)(303 K) = 18.7 mJ
Pa 97.754 kPa ~~ --~
---
13-15 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative
humidity, and the volume of the tank are to be determined.
Assumptions The air and the water vapor are ideal gases.
Analysis (a) The specific humidity can be determined form its definition,
Chapter 13 Gas- Vapor Mixtures and Air-Conditioning
1200 kJ/min
13.68 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The
exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined.
Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant duringthe entire process (mal = ma2 = ma) .2 Dry air and water vapor are ideal gases. 3 The kinetic and potential
energy changes are negligible.
Analysis (a) The amount of moisture in the air remains constant ({J) I = {J)2) as it flows through the coolingsection since the process involves no humidification or dehumidification. The inlet state of the air is
completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state aredetermined from the psychometric chart (Figure A-33) to be
hl = 55.0 kJ / kg dry air
{J)J = 0.0089 kg H2O/kg dry air (= {J)2)
vi = 0.877 m3 / kg dry air f-:\ 32°C
\.!/ 30%18 m/s 0
AIRThe mass flow rate of dry air through the cooling section is
ma =~VIAI =- 1-(18m/s)(1l"xO.42/4m2)=258kg/s(0.877 m3 / kg)VI
From the energy balance on air in the cooling section,
-~\t. = irb(~- '1)
-l~ /00 kJ /s = (2.59 kg /s) ( ~ -550) kJ /kg
~ = 47.2 kJ /kg dry air
The exit state of the air is fixed now since we know both h2 and CV2. From the psychometric chart at this
state we read
(b)
T2 = 24.4°C
(12 = 46.6 %
V2 = 0.856 m3 / kg dry air
(c) The exit velocity is determined from the conservation of mass of dry air,
~ V2-=-
~ ~
~=~Ut ~
irbI=iT!t2 --,+ --+
~~~
13-19,
~
Chapter 13 Gas- Vapor Mixtures and Air-Conditioning
~
Heatingcoils
13-73 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at
the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be
determined.
Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant duringthe entire process (mal = ma2 = ma) .2 Dry air and water vapor are ideal gases. 3 The kinetic and potential
energy changes are negligible.
Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm.The properties of the:air at various states are determined from the psychometric chart (Figure A-33) to be
(c) The amount of water added to the air in the humidifying section is determined from the conservation of
mass equation of water in the humidifying section,
1'r6= irh((J)3-(J)2) = (433kg/~(Q~- QCX6.1 = 0.15 kg /mill
13-23
Chapter 13 Gas- Vapor Mixtures and Air-Conditioning
13-77 Air is first cooled, then dehumidified, and finally heated. The temperature of air before it enters the
heating section, the amount of heat removed in the cooling section, and the amount of heat supplied in theheating section are to be determined.
Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant duringthe entire process (mal = ma2 = ma) .2 Dry air and water vapor are ideal gases. 3 The kinetic and potential
energy changes are negligible.
Analysis (a) The amount of moisture in the air decreases due to dehumidification (fI) 3 < fI) 1), and remains
constant during heating (fI) 3 = fI) 2). The inlet and the exit states of the air are completely specified, and the
total pressure is 1 atm. The intermediate state (state 2) is also known since ~ = 100% and fI) 2 = fI) 3.
Therefore, we can determined the properties of the air at all three states from the psychometric chart (TableA-33) to be
hl = 95.2 kJ / kg dry air
{J)l = 0.0238 kg H2O / kg dry air
and Ti = 34°C
(>I = 70%h3 = 43.1 kJ / kg dry air
(t}3 = 0.0082 kg H2O / kg dry air ( = (t}2 )
Also,
h", = hf@IO°C = 42.0 kJ / kg
~ = 31.8 kJ / kg dry air
T2 = 11.1° C
(b) The amount of heat removed in the cooling section is determined from the energy balance equation
Chapter 13 Gas- Vapor Mixtures and Air-Conditioning
AIR ~ 34°CEXIT \2:; 90%
I~IWARM
WATER
40°C
60 kg/s
CD
AAAAAA'
13-106 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the
required makeup water are to be determined.
Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant duringthe entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are
negligible. 4 The cooling tower is adiabatic.
Analysis (a) The mass flow rate of dry air through the tower remains constant (mal =ma2 = ma) , but the
mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the
tower during the cooling process. The:water lost through evaporation must be made up later in the cycle to
maintain steady operation. Applying the mass and energy balances yields
Dry Air Mass Balance:
Lma.i = Lma.e ~ mal = ma2 = ma
Water Mass Balance:
L 1'f1ui = L 1'f1ue ~ ~ + iTh1{Ol = 1'14 + fT!O;{02