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Chapter Two Psychrometrics of Air Condition Processes

2.1- Psychrometric Chart It is a graphical representation of various thermodynamic properties of moist air. The psychrometric chart is very useful for finding out the properties of air (which are required in the field of air condition) and eliminate lot of calculations. There is a slight variation in the charts prepared by different air-conditioning manufactures but basically they are all alike. The psychrometric chart is normally drawn for standard atmospheric pressure of 760 mm of Hg (or 1.01325 bar).

Fig

.(2.

1) P

sych

rom

etric

cha

rt


In a psychrometric chart, dry bulb temperature is taken as abscissa and specific heat i.e. moisture content as ordinate, as shown in Fig.(2.1). Now the saturation cure is drawn by plotting the various saturation points at corresponding dry bulb temperatures. The saturation curve represent 100% relative humidity at various dry bulb temperatures. It also represents the wet bulb and dew point temperatures. Though the psychrometric chart has a number of details, yet the following lines are important from subject point of view: 1. Dry bulb temperature lines: The dry bulb temperature lines are vertical i.e. parallel to the

ordinate and uniformly spaced as shown in Fig.(2.2). Generally, the temperature range of these lines on psychrometric chart is from -6ºC to 45ºC. The dry bulb temperature lines are drawn with difference of every 5ºC and up to the saturation curve as shown in the figure. The values of dry bulb temperatures are shown on the saturation curve. 2. Specific humidity or moisture content lines: The specific humidity (moisture content) lines are

horizontal i.e. parallel to the abscissa and are also uniformly spaced as shown in Fig.(2.3). Generally, moisture content range of these lines on psychrometric chart is from 0 to 30 g / kg of dry air (or from 0 to 0.030 kg / kg of dry air). The moisture content lines are drawn with a difference of every 1 g (0.001 kg) and up to the saturation curve as shown in the figure. 3. Dew point temperature lines: The dew point temperature lines are horizontal i.e. parallel to the

abscissa and non-uniformly spaced as shown in Fig.(2.4). At any point on the saturation curve, the dry bulb and dew point temperatures are equal. The values of dew point temperatures are generally given along the saturation curve of the chart as shown in the figure.

Fig.(2.2) Dry bulb temperature lines Fig.(2.3) Specific humidity lines

Fig.(2.4) Dew point temperature lines


4. Wet bulb temperature lines: The wet bulb temperature lines are inclined straight lines and non-uniformly spaced as shown in Fig.(2.5). At any point on the saturation curve, the dry bulb and wet bulb temperatures are equal. The values of wet bulb temperatures are generally given along the saturation curve of the chart as shown in the figure. 5. Enthalpy (total heat) lines: The enthalpy (or total heat) lines are inclined straight lines and

uniformly spaced as shown in Fig.(2.6). These lines are parallel to the wet bulb temperature lines, and are drawn up to the saturation curve. Some of these lines coincide with the wet bulb temperature lines also. The values of total enthalpy are given on a scale above the saturation curve as shown in the figure. 6. Specific volume lines: The specific volume lines are obliquely inclined straight lines and

uniformly spaced as shown in Fig.(2.7). These lines are drawn up to the saturation curve. The values of volume lines are generally given at the base of the chart. 7. Relative humidity lines: The relative humidity lines are curved and follow the saturation curve.

Generally, these lines are drawn with values of relative humidity 10%, 20%, 30% etc. and up to 100%. The saturation curve presents 100% relative humidity. The values of relative humidity lines are generally given along the lines themselves as shown in Fig.(2.8).

Fig.(2.5) Wet bulb temperature lines Fig.(2.6) Enthalpy lines

Fig.(2.7) Specific volume lines Fig.(2.8) Relative humidity lines


Example 2.1. For a sample of air having 22ºC DBT, relative humidity 30 percent at barometric pressure of 760 mm of Hg, calculate: 1.Vapour pressure, 2. Humidity ratio, 3. Vapour density, and 4.Enthalpy. Verify your result by psychrometric chart. Solution: Given :td =22ºC ; φ =30%=0.3 ; pb=760 mm of Hg=760×133.4= 101384N/m2=1.01384 bar

1.Vapour pressure Let pv=Vapour pressure From steam tables (table 1.4), we find that the saturation pressure of vapour corresponding to dry bulb temperature of 22ºC is ps=0.026448 bar We know that relative humidity (φ ),

026448.0

3.0 v

s

v p

p

p ==

∴ pv=0.3×0.02642=0.007934 bar 2.Humidity ratio

We know that humidity ratio,

007934.001384.1

007934.0622.0622.0

−×=

−×=

vb

v

pp

pW

= 0.0049 kg / kg of dry air 3.Vapour density

We know that vapour density, ( ) ( )

( )22273287

10007934.001384.10049.0 5

+−=−=

da

vbv TR

ppWρ

= 0.00582 kg / m3 of dry air 4.Enthalpy

We know that enthalpy, ( ) ( )dd tWth 84.12501026.0007.1 ++−=

= (1.007×22-0.026) + 0.0049× (2501+1.84×22) =34.58 kJ/kg dry air Verification from psychrometric chart The initial condition of air i.e. 22ºC dry bulb temperature and 30% relative humidity is marked on the psychrometric chart at point A as shown in Fig.(2.9) From point A, draw a horizontal line meeting the humidity ratio line at C. From the psychrometric chart, we find that the humidity ratio at point C, W≈5 g/kg of dry air ≈ 0.005 kg/kg of dry air

Fig.(2.9)


We also find from psychrometric chart that the specific volume at point A is 0.843 m3/kg of dry air. ∴ Vapour density, ρv=W/ρa=0.005/0.843=0.0059 kg/m3 of dry air Now from point A, draw a line parallel to the wet bulb temperature line meeting the enthalpy line at point E. Now the enthalpy of air as read from the chart is 34.8 kJ/kg of dry air. 2.2 Psychrometric Processes The various psychrometric processes involved in air conditioning to vary psychrometric properties of air according to the requirement are as follows:

1. Adiabatic mixing of air streams 2. Sensible heating 3. Sensible cooling 4. Humidification and dehumidification 5. Cooling and adiabatic humidification 6. Cooling and humidification by water injection 7. Heating and humidification 8. Humidification by steam injection 9. Adiabatic chemical dehumidification

2.2.1 Adiabatic Mixing of Two Air Streams When two quantities of air having different enthalpies and different specific humidities are mixed, the final condition of the air mixture depends upon the masses involved, and on the enthalpy and specific humidity of each the constituent masses which enter the mixture. Now consider two air streams 1 and 2 mixing adiabatically as shown in Fig.(2.10)(a) Let m1= Mass of air entering at 1, h1= Enthalpy of air entering at 1, W1= Specific humidity of air entering at 1, m2, h2, W2=Corresponding values of air entering at 2, and m3, h3, W3=Corresponding values of air mixture at 3.

Assuming no loss of enthalpy and specific humidity during the air mixture process, we have for the mass balance of dry air, 321 mmm =+ ………………… (2.1)

Fig(2.10)Adiabatic mixing of two air streams


For the energy balance, 332211 hmhmhm =+ ………………… (2.2)

and for the mass balance of water vapour, 332211 WmWmWm =+ ………………… (2.3)

Substituting the value of m3 from equation (2.1) in equation (2.2), ( ) 32313212211 hmhmhmmhmhm +=+=+

or 22323111 hmhmhmhm −=−

( ) ( )232311 hhmhhm −=−

23

31

1

2

hh

hh

m

m

−−=∴ ………………… (2.4)

Or 21

22113 mm

hmhmh

++= ………………… (2.5)

Similarly, substituting the value of m3 from equation (2.1) in equation (2.3), we have

23

31

1

2

WW

WW

m

m

−−= ………………… (2.6)

Or 21

22113 mm

WmWmW

++= ……………………… (2.7)

Now from equations (2.4) and (2.6)

23

31

23

31

1

2

WW

WW

hh

hh

m

m

−−=

−−=∴ ………………… (2.8)

The adiabatic mixing process is represented on the psychrometric chart as shown in Fig.(2.10) (b). The final condition of the mixture (point 3) lies on the straight line 1-2. The point 3 divides the line 1-2 in the inverse ratio of the mixing masses. Example 2.2. 300 m3/min of fresh air at 30 ºC (DBT) dry bulb temperature and 50% RH is to be mixed with 800 m3/min of recirculated air at 22 ºC (DBT) dry bulb temperature and 10 ºC dew point temperature. Determine the enthalpy, specific volume, humidity ratio, and dew point temperature of the mixture. Solution. Given: v1=300 m3/min; td1=30 ºC; 1φ =50%; v2=800 m3/min; td2=22 ºC; tdp2=10 ºC Enthalpy of the mixture Let h3= Enthalpy of the mixture The condition of recirculated air at 22 ºC DBT and 10 ºC dew point temperatures is marked on the psychrometric chart at point 2 as shown in Fig.(2.11). Now mark the condition of fresh air at 30 ºC dry bulb temperature and 50% relative humidity at point 1 as shown in the figure. Join 1 and 2.


From the psychrometric chart, we find that enthalpy of air at point 1, h1=64.6 kJ/kg of dry air Enthalpy of air at point 2, h2=41.8 kJ/kg of dry air Specific humidity of air at point 1, W1= 0.0134 kg/kg of dry air Specific humidity of air at point 2, W2= 0.0076 kg/kg of dry air Specific volume at point 1, vs1= 0.877 m3/kg of dry air and Specific volume at point 2, vs2= 0.846 m3/kg of dry air We know that mass of fresh air at point 1,

min/07.342877.0

300

1

11 kg

v

vm

s

===

and mass of recirculated air at point 2,

min/6.945846.0

800

2

22 kg

v

vm

s

===

We know that

3

3

31

23

2

1

6.64

8.41

6.945

07.342

h

hor

hh

hh

m

m

−−=

−−=

∴ h3=47.86 kJ/kg of dry air Ans. Specific volume, humidity ratio, and dew point temperature of the mixture Plot point 3 on line joining the points 1 and 2 corresponding to enthalpy h3=47.86 kJ/kg of dry air, as shown in Fig.(2.11). From point 3 on psychrometric chart, we find that specific volume of the mixture at point 3, vs3= 0.855 m3/kg of dry air Ans. Humidity ratio of the mixture at point 3, W3= 0.0092 kg/kg of dry air Ans. And dew point temperature of the mixture at point 3, tdp3≈12.7 ºC Ans.

W1

W2

W3

2

1

3

10 ºC

tdp3

h1

h3

h2

50%

vs3

0.846 m3/kg

22

0.876 m3/kg

30

Fig(2.11)


2.2.2 Sensible Heat Factor Actually, the heat added during a psychrometric process may be split up into sensible heat and latent heat. The ratio of the sensible heat to the total heat is known as sensible heat factor (briefly written as SHF) or sensible heat ratio (briefly written as SHR). Mathematically,

ls

s

QQ

Q

LHSH

SH

heatTotal

heatSensibleSHF

+=

+== ………………(2.9)

where SH=Sensible heat, and LH=Latent heat

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2.2.3 Sensible Heating The heating of air, without any change in its specific humidity, is known as sensible heating. Let air at temperature td1 passes over a heating coil of temperature td3, as shown in Fig.(2.13)a. It may be noted that the temperature of air leaving the heating coil (td2) will be less than td3. The process of sensible heating, on the psychrometric chart, is shown by a horizontal line 1-2 extending from left to right as shown in Fig.(2.13) b. The point 3 represents the surface temperature of the heating coil. The heat absorbed by the air during sensible heating may be obtained from the psychrometric chart by the enthalpy difference (h2-h1) as shown in Fig.(2.13) b. It may be noted that the specific humidity during the sensible heating remains constant (i.e.W1=W2). The dry bulb temperature increases from td1 to td2 and relative humidity reduces from 1φ to 2φ as shown in Fig.(2.13) b. The amount of heat added during sensible heating may also be obtained from the relation: Heat added, 12 hhq −=

( ) ( )

( )( ) ( )1212

1212

ddpmddpspa

ddpsddpa

ttcttcWc

ttcWttc

−=−+=

−+−=

2

1

SHR

∆hl

∆hs

∆ht

State point

Parallel

Sensible heat ratio line

Fig(2.12) Sensible heat ratio line


The term ( )pspa cWc + is called humid specific heat (cpm) and its value is taken as 1.022 kJ/kg K.

∴ Heat added ( ) kgkJttq dd /022.1 12 −= ………….(2.10)

Notes: 1- For sensible heating, steam or hot water is passed through the heating coil. The heating coil may be electric resistance coil. 2- The sensible heating of moist air can be done to any desired temperature. 2.2.4 Sensible Cooling The cooling of air, without any change in its specific humidity, is known as sensible cooling. Let air at temperature td1 passes over a cooling coil of temperature td3, as shown in Fig.(2.14)a. It may be noted that the temperature of air leaving the heating coil (td2) will be more than td3. The process of sensible cooling, on the psychrometric chart, is shown by a horizontal line 1-2 extending from right to left as shown in Fig.(2.14) b. The point 3 represents the surface temperature of the cooling coil. The heat rejected by the air during sensible cooling may be obtained from the psychrometric chart by the enthalpy difference (h1-h2) as shown in Fig.(2.14) b. It may be noted that the specific humidity during the sensible cooling remains constant (i.e. W1=W2). The dry bulb temperature reduces from td1 to td2 and relative humidity increases from 1φ to 2φ as shown in Fig.(2.14) b. The amount of heat rejected during sensible cooling may also be obtained from the relation: Heat rejected, 21 hhq −=

( ) ( )

( )( ) ( )2121

2121

ddpmddpspa

ddpsddpa

ttcttcWc

ttcWttc

−=−+=

−+−=

Fig(2.14) Sensible Cooling

Fig(2.13) Sensible Heating


The term ( )pspa cWc + is called humid specific heat (cpm) and its value is taken as 1.022 kJ/kg K.

∴ Heat rejected, ( ) kgkJttq dd /022.1 21 −=

Generally, Heat rejected, ( )21 hhmq −×= ………………(2.11) Notes: 1- For sensible cooling, the cooling coil may have refrigerant, cooling water or cool gas flowing through it. 2- The sensible cooling can be done only upto the dew point temperature (tdp) as shown in Fig.(2.14) b. The cooling below this temperature will result in the condensation of moisture. 2.2.5 By-pass Factor of Heating and Cooling Coil We already discussed that the temperature of the air coming out of the apparatus (td2) will be less than *td3 in case the coil is a heating coil and more than td3 in case the coil is a cooling coil. Let 1 kg of air at temperature td1 is passed over the coil having its temperature (i.e. coil temperature surface) td3 as shown in Fig. (2.15). A little consideration will show that when air passes over a coil, some of it (say x kg) just by-passes unaffected while the remaining (1-x) kg comes in direct contact with the coil. This by-pass process of air is measured in terms of a by-pass factor. The amount of air that by-passes or by-pass factor depends upon the following factors:

1. The number of fins provided in a unit length i.e. the pitch of the cooling coil fins; 2. The number of rows in a coil in the direction of flow ; and 3. The velocity of flow air.

It may be noted that by-pass factor of a cooling coil decreases with decrease in fin spacing and increase in number of rows. Balancing the enthalpies, we get

( ) ( )

( ) 2313

231 ...11

dddd

pmdpmdpmdpm

ttttx

heathumidSpecificcwheretctcxtcx

−=−

=×=−+

13

23

dd

dd

tt

ttx

−−=∴ …………………..(2.12)

where x is called by-pass factor of the coil and is generally written as BPF. Therefore, by-pass factor for heating coil,

13

23

dd

dd

tt

ttBPF

−−= …………………..(2.13)

Fig(2.15) By-pass factor


* Under ideal condition, the dry bulb temperature of the air leaving the apparatus (2dt ) should be equal to that of the

coil ( 3dt ). But it is not so, because of the inefficiency of the coil. This phenomenon is known as by-pass factor.

Similarly, by-pass factor for cooling coil,

31

32

dd

dd

tt

ttBPF

−−= …………………..(2.14)

The value of x (BPF) also is obtained by balancing the enthalpies as follows: ( ) 312 1 hxhxh −+=

for heating coil

13

23

hh

hhx

−−= ……………………….. (2.15)

for cooling coil

31

32

hh

hhx

−−= ………………….. (2.16)

Note: The performance of a heating or cooling coil is measured in terms of a by-pass factor. A coil with low by-pass factor has better performance. 2.2.6 Efficiency of Heating and Cooling Coils The term (1-BPF) is known as efficiency of coil or contact factor. ∴ Efficiency of the heating coil,

13

12

13

2311dd

dd

dd

ddH tt

tt

tt

ttBPF

−−=

−−−=−=η

Similarly, efficiency of the cooling coil,

31

21

31

321dd

dd

dd

ddC tt

tt

tt

tt

−−=

−−−=η …………………..(2.17)

Example 2.3. In a heating application, moist air enters a steam heating coil at 10º C, 50% RH and leaves at 30º C. Determine the sensible heat transfer, if mass flow rate of air is 100kg of dry air per second. Also determine the steam mass flow rate if steam enters saturated at 100º C and condensate leaves at 80º C. Solution. Given: td1=10º C ; 1φ =50% ; td2=30º C; ma=100kg/s ; ts= 100º C; tC= 80º C Sensible heat transfer First, mark the initial condition of air, i.e. 10 ºC dry bulb temperature and 50% relative humidity on the psychrometric chart at point 1, as shown in Fig.(2.16). Draw a constant specific humidity line from point 1 to intersect the vertical line drawn through 30º C dry bulb temperature at point 2. the line 1-2 represents sensible heating of air. From the psychrometric chart, we find that enthalpy at point 1, h1=19.5 kJ/kg of dry air and enthalpy at point 2, h2=40 kJ/kg of dry air We know that sensible heat transfer, Q=ma(h2-h1)=100(40-19.5)=2050 kJ/s Ans.

Fig(2.16)


Steam mass flow rate From steam tables (table (1.4)), corresponding to temperature of 100º C, we find that enthalpy of saturated steam, hg=2675.44 kJ/kg and enthalpy of condensate, corresponding to 80 ºC, hf=335 kJ/kg ∴ Steam mass flow rate

skghh

Q

fg

/876.033544.2675

2050 =−

=−

=

hkg /31533600876.0 =×= Ans. Example 2.3. The air enters a duct at 10 ºC and 80% RH at the rate of 150 m3/min and is heated to 30 ºC without adding or removing any moisture. The pressure remains constant at 1 atmosphere. Determine the relative humidity of air at exit from the duct and the rate of heat transfer. Solution. Given: td1=10º C; 1φ =80%; v1=150 m3/min; td2=30º C; p=pb= 1 atm=101.325 kPa Relative humidity of air at exit First, mark the initial condition of air, i.e. 10 ºC dry bulb temperature and 80% relative humidity on the psychrometric chart at point 1, as shown in Fig.(2.17). Since air is heated to 30 ºC without adding or removing any moisture, therefore it is a case of sensible heating. Draw a constant specific humidity line from point 1 to intersect the vertical line drawn through 30º C dry bulb temperature at point 2. The line 1-2 represents sensible heating of air. From the psychrometric chart, we find that the relative humidity of air exit i.e. at point 2,

2φ ≈23.5% Rate of heat transfer From the psychrometric chart, we also find that the specific volume of air at point 1, υ1=0.81 m3/kg of dry air enthalpy of air at point 1, h1=26 kJ/kg of dry air and enthalpy of air at point 2, h2=45.5 kJ/kg of dry air We know that amount of air supplied,

Fig(2.17)


min/2.18581.0

150 3

1

1 mv

ma ===υ

The amount of air supplied may also be obtained as discussed below: From steam tables (table (1.4)), pressure of vapour corresponding to dry bulb temperature of 10 ºC, we find that saturation pressure of vapour, ps1=0.001228 bar We know that partial pressure of vapour, pv1= 1φ × ps1=0.8×0.001228=0.00982 bar ….( )111 / sv pp=φQ

( ) ( )( ) min/2.185

10273287

1501000982.0013.1 35

1

11 mTR

vppm

da

vba =

+×−=−=

∴ Rate of heat transfer =ma(h2-h1)=185.2(45.5-26)=3611.4 kJ/min Ans. 2.2.7 Humidification and Dehumidification The addition of moisture to the air, without change in its dry bulb temperature, is known as humidification. Similarly, removal of moisture from the air, without change in its dry bulb temperature is known as dehumidification. The heat added during humidification process and heat removed during dehumidification process is shown on the psychrometric chart in Fig.(2.18) a and b respectively. It may be noted that in humidification, the relative humidity increases from 1φ to 2φ and specific humidity also increase from W1 to W2 as shown in Fig.(2.18) a. Similarly, in dehumidification, the relative humidity decrease from 1φ to 2φ and specific humidity also increase from W1 to W2 as shown in Fig.(2.18) b. It may be noted that in humidification, change in enthalpy is shown by the intercept (h2-h1) on the psychrometric chart. Since the dry bulb temperature of air during the humidification remains constant, therefore its sensible heat also remains constant. It is thus obvious, that the change in enthalpy per kg of dry air due to the increased moisture content equal to (W2 -W1) per kg of dry air is considered to cause a latent heat transfer (LH). Mathematically, ( ) ( )1212 WWhhhLH fg −=−= …………………..(2.18)

where hfg is the latent heat transfer of vaporization at dry bulb temperature(1dt ).

Notes: 1. For dehumidification, the above equation may be written as:

Fig.(2.18) Humidification and Dehumidification


( ) ( )2121 WWhhhLH fg −=−= …………………..(2.19)

2. Absolute humidification and dehumidification processes are rarely found in practice. These

are always accompanied by heating and cooling processes. 3. In air conditioning, the latent heat load per minute is given as:

fgfgaa hWhmhmLH ρν=∆=∆= …………………..(2.20)

where ν = Rate of dry air flowing in m3/s. ρ = Density of moist air =1.2 kg/m3 of dry air. hfg= Latent heat of vaporization = 2450 kJ/kg, and ∆W = Difference of specific humidity between the entering and leaving conditions of air=( 12 WW − ) for humidification and (W1-W2) for dehumidification. Substituting these values in the above expression, we get kWorskJWWLH /294025002.1 ∆×=∆×××= νν …………………..(2.21) 2.2.8 Cooling and Dehumidification This process is generally used in summer air conditioning to cool and dehumidify the air. The air passed over a cooling coil or through a cold water spray. In this process, the dry bulb temperature as well as the specific humidity of air decreases. The final relative humidity of the air is generally higher than of the entering air. The dehumidification of air is only possible when the effective surface temperature of the cooling coil (i.e. td4) is less than the dew point temperature of the air entering the coil (i.e. tdp1). The effective surface temperature of the coil is known as apparatus dew point (briefly written as ADP). The cooling and dehumidification process is shown in Fig.(2.19). Let td1= Dry bulb temperature of air entering the coil, tdp1= Dew point temperature of air entering = td3, and td4 = Effective surface temperature or ADP of the coil. Under ideal conditions, the dry bulb temperature of the air leaving the cooling coil (i.e. td4) should be equal to the surface temperature of the cooling coil (i.e. ADP), but it is never possible due to inefficiency of the cooling coil. Therefore, the resulting condition of air coming out of the coil is

Fig.(2.19) Cooling and dehumidification


shown by a point 2 on the straight line joining the points 1 and 4. The by-pass factor in this case is given by

ADPt

ADPt

tt

ttBPF

d

d

dd

dd

−−=

−−=

1

2

41

42 …………………..(2.22)

Also 21

24

41

42

hh

hh

WW

WWBPF

−−=

−−= …………………..(2.23)

Actually, the cooling and dehumidification process follows the path as shown by a dotted curve in Fig.(2.19) a, but for calculation of psychrometric properties, only end points are important. Thus the cooling and dehumidification process shown by a line 1-2 may be assumed to have followed a path 1-A (i.e. dehumidification) and (i.e. cooling) as shown in Fig.(2.19)a. We see that the total heat removed from the air during the cooling and dehumidification process is ( ) ( ) SHLHhhhhhhq AA +=−+−=−= 2121

where AhhLH −= 1 = Latent heat removed due to condensation of vapour of reduced moisture content (W1-W2), and 2hhSH A −= =Sensible heat removed. We know that sensible heat factor,

21

2

hh

hh

LHSH

SH

heatTotal

heatSensibleSHF A

−−=

+== …………………..(2.24)

Note: The line 1-4 (i.e. the line joining the point of entering air and the apparatus dew point) in Fig. (2.19) b is known as sensible heat factor line. Example 2.4. The atmospheric air at 30 ºC dry bulb temperature and 75% relative humidity enters a cooling coil at the rate 200 m3/min. the coil dew point temperature is 14 ºC and the by-pass factor of the coil is 0.1. Determine: 1. the temperature of air leaving the cooling coil; 2. the capacity of the cooling coil in tones of refrigeration and in kilowatt; 3. the amount of water vapour removed per minute; and 4. the sensible heat factor for the process. Solution. Given: td1=30 ºC ; 1φ =75% ; v1=200 m3/min ; ADP= td4=14 ºC; BPF=0.1 1. Temperature of air leaving the cooling coil Let td2= Temperature of air leaving the cooling coil First, mark the initial condition of the air, i.e. 30 ºC dry bulb temperature and 75% relative humidity on the psychrometric chart at point 1, as shown in Fig.(2.20). From the psychrometric chart, the dew point temperature of the entering air at point 1. tdp1= 25.2 ºC Since the coil dew point temperature (or ADP) is less than the dew point temperature of entering air, therefore it a process of cooling and dehumidification. We know that the by-pass factor,


1430

141.0 2

1

2

41

42

−−=

−−=

−−=

d

d

d

dd

dd

t

ADPt

ADPt

tt

ttBPF

∴ td2= 15.6 Ans. 2. Capacity of the cooling coil The resulting condition of air coming out of the coil is shown by point 2, on the line joining the points 1 and 4, as shown in Fig.(2.20). The line 1-2 represent the cooling and dehumidification process which may be assumed to have followed the path 1-A (i.e. dehumidification) and A-2 (i.e. cooling). Now from psychrometric chart, we find that Water vapour in the entering air or the specific humidity of entering air at point 1, W1=0.0202 kg/kg of dry air Water vapour in the leaving air or the specific humidity of leaving air at point 2, W2=0.011 kg/kg of dry air Specific volume of entering air at point 1, υ1=0.886 m3/kg of dry air Enthalpy of entering air at point 1, h1=82 kJ/kg of dry air Enthalpy air at point A, hA=58 kJ/kg of dry air and enthalpy of leaving air at point 2 h2=43.5 kJ/kg of dry air We know that mass of air flowing through the cooling coil,

min/7.225886.0

200

1

1 kgv

ma ===υ

∴ Capacity of cooling coil in tones of refrigeration =ma(h1-h2)=225.7(82-43.5)=8690.7 kJ/min =8690.7/210=41.38 TR Ans. … (Q 1 TR=210 kJ/min) and capacity of the cooling coil in kilowatt =8690.7/60=144.845 kW Ans. 3. Amount of water vapour removed per minute We know that amount of water vapour removed =ma(W1-W2)=225.7(0.0202-0.011)=2.076 kg/min Ans. 4. Sensible heat factor for the process We know that sensible heat factor,

377.05.4382

5.4358

21

2 =−−=

−−=

hh

hhASHF Ans.

W1

W2

Fig(2.20)


2.2.9 Methods of Obtaining Humidification and Dehumidification The humidification is achieved either by supplying or spraying steam or hot water or cold water into the air. The humidification may be obtained by the following two methods:

1. Direct method: In this method, the water is sprayed in a highly atomized state into the room to be air-conditioned. This method of obtaining humidification is not very effective.

2. Indirect method: In this method, the water is introduced into the air in the air-conditioning plant, with the help of an air washer, as shown in Fig.(2.21). This conditioned air is then supplied to the room to be air-conditioned. The air washer humidification may be accomplished in the following three ways:

a. By using re-circulated spray water without prior heating of air. b. By pre-heating the air and then washing it with re-circulated water, and c. By using heated spray water.

The dehumidification may be accomplished with the help of an air-washer or by using

chemicals. In the air washer system, the outside or entering air is cooled below its dew point temperature so that it looses moisture by condensation. The moisture removal is also accomplished when the spry water is chilled water and its temperature is low than the dew point temperature of the entering air. Since the air leaving the air washer has its dry bulb temperature much below the desired temperature in the room, therefore a heating coil is placed after the air-washer. The dehumidification may also be achieved by using chemicals which have the capacity to absorb moisture in them. Two types of chemicals known as absorbents (such as calcium chloride) and adsorbents (such as silica gel and activated alumina) are commonly used for this purpose. 2.2.10 Air Washer Figure (2.21) shows the schematic representation of an air washer. It involves the flow of air through a spray of water. During the course of flow, the air may be cooled or heated, humidified or dehumidified, or simply adiabatically saturated, depending on the mean surface temperature of water. The water is, accordingly, externally cooled or heated or simply recirculated by pump. Make up water is added for any loss in the case of humidification of air. Eliminator plates are provided to minimize the loss of water droplets. Figure (2.22) shows the thermodynamic changes of state of air along paths 1-2 in air washer, depending on the mean surface temperature of water droplets tsurf. which is equal to the actual temperature of water tw.

Fig(2.21) Air washer


Thus, the droplets of water act as wetted surface, and both sensible and latent heat transfer take place. Their directions depend on the temperature and vapour pressure potentials. The following processes are possible: Process 1-2A: Heating and humidification (tsurf. > td) The mean surface temperature of water is greater than the dry bulb temperature of air. The water is externally heated. Process 1-2B: Humidification (tsurf. =td) The mean surface temperature of water is equal to the dry bulb temperature of air. The enthalpy of air increases. Hence the water is required to be externally heated. Process 1-2C: Cooling and humidification (tw<tsurf. <td) The mean surface temperature of water is less than the dry bulb temperature of air but greater than the wet bulb temperature of air. Though the air is cooled, its enthalpy increases as a result of humidification. The water is, therefore, required to be externally heated. Process 1-2D: Adiabatic saturation (tsurf. =tw) This is the case of pumped recirculation of water without any external heating or cooling as discussed in Sec. 1.5. The recirculated water reaches the equilibrium temperature which is equal to the thermodynamic wet bulb temperature of air. Process 1-2E: Cooling and humidification (tdp<tsurf. <tw) The process is similar to 1-2C with the difference that the enthalpy of air decreases in this case. Accordingly, water is required to be externally cooled. Process 1-2F: Cooling (tsurf. = tdp) The temperature of water is equal to the dew point temperature of air. water is required to be externally cooled. Process 1-2G: Cooling and dehumidification (tsurf. < tdp) The mean surface temperature of water is lower than the dew point temperature of air. Ai is simultaneously cooled and dehumidified. The process is exactly similar to that of a cooling and dehumidifying coil. Again, the limiting process is along the condition line tangent to the saturation line drawn from initial state 1. It is thus seen that the air washer affords means for a year-round air-conditioning system.

W1

td1 tw1 tdp1

Fig.(2.22) Range of psychrometric processes with air washer


The effectiveness or humidifying efficiency of the air washer or spray chamber is given by

humidityspindropIdeal

humidityspindropActual

inDBTdropIdeal

inDBTdropActualH .

.==η

1

12

1

12

WW

WW

tt

tt

sds

dd

−−

=−−

= …………………..(2.25)

2.2.11 Humidification by injection of water or steam Let water, or steam, at certain temperature be injected and sprayed into a flowing air stream with the help of nozzles as shown in Fig.(2.23). When use steam or water and by assume that all water is evaporating or drift the vapour with air and not condensate steam on the walls of the channel and assume adiabatic process, we get on the following equation of energy balance:

22311 hmhmhm w =+ …………………..(2.26) For mass balance for dry air m1=m2=m …………………..(2.27) for water vapour

2211 WmmWm w =+ …………………..(2.28) Or

( )12 WWmmw −= ……………………(2.29) also

( )123 hhmhmw −= ……………………(2.30) By dividing Equ.(2.30)on Equ.(2.29), we get:

W

h

WW

hhh

∆∆=

−−=

12

123 …………………..(2.31)

as shown in Fig.(2.24) Humidification by water injection (1-a) constant enthalpy humidification (1-b) adiabatic humidification (1-c) humidification with water at 100 ºC Humidification by steam injection (1-d) isothermal humidification with steam at 100 ºC (1-e) humidification with steam at 234 ºC and 30 bar and hs=2803 kJ/kg as shown in Fig.(2.25)

m1 t1 W1 h1

m2 t2 W2 h2

h3

Fig.(2.23)Steam or water injection

Fig.(2.24) Gradation of enthalpy of water or steam injection on the protractor


2.2.12 Cooling and Humidification by Water Injection (Evaporative Cooling) Let water at a temperature tl is injected into the flowing stream of dry air as shown in Fig.(2.26). The final condition of air depends upon the amount of water evaporation. When the water is injected at the temperature equal to the wet bulb temperature of the entering air (tw1), then the process follows the path of constant wet bulb temperature line, as shown by the line 1-2 in Fig.(2.26 )b.

Example 2.5. 200 m3 of air per minute is passed through the adiabatic humidifier. The condition of air at inlet is 40 ºC dry bulb temperature and 15% relative humidity and the outlet condition is 25 ºC dry bulb temperature and 20 ºC wet bulb temperature. Find the dew point temperature and the amount of water vapour added to the air per minute. Solution. Given: v1=200 m3/min; td1=40 ºC ; 1φ =15% ; td2=25 ºC; tw2=20 ºC First, mark the inlet condition of air at 40 ºC dry bulb temperature and 15% relative humidity on the psychrometric chart at point 1, as shown in Fig.(2.27). Now mark the outlet condition of air at 25ºC dry bulb temperature and 20 ºC wet bulb temperature, as point 2. The line 1-2 represents the adiabatic humidification. Dew point temperature On the psychrometric chart, draw a horizontal line through point 2 up to the saturation curve. From the chart, we find that dew point temperature, tdp2=17.6 ºC Ans. Amount of water vapour added to the air per minute From the psychrometric chart, we find that specific volume of air at point 1, υ1=0.896 m3/kg of dry air Specific humidity at point 1, W1=0.007 kg/kg of dry air and specific humidity at point 2, W2=0.1026 kg/kg of dry air We know that mass of air supplied,

min/2.223896.0

200

1

1 kgv

ma ===υ

∴ Amount of water vapour added to the air =ma(W2- W1) =223.2(0.0126-0.007) =1.25 kg/min Ans. Fig.(2.27)

Fig.(2.26) Cooling and Humidification by water injection


Example 2.6. Determine the final dry bulb temperature and relative humidity of air washer with recirculated spray water if the air is initially at dry bulb temperature 35 ºC and 50% relative humidity as it enters an air washer which has humidifying efficiency of 85 percent. Solution. Given: td1=35 ºC ; 1φ =50% ; ηH=85%=0.85 First, mark the initial condition of air at 35 ºC dry bulb temperature and 50% relative humidity on the psychrometric chart at point 1, as shown in Fig.(2.28). The wet bulb temperature of the entering air as read from the psychrometric chart is tw1=26.1 ºC Final dry bulb temperature Let td2 = final dry bulb temperature of the air leaving the air washer We know that humidifying efficiency of an air washer (ηH),

1.2635

3585.0 2

31

21

−−=

−−= d

dd

dd t

tt

tt

∴ td2 = 35-0.85×8.9=27.435 ºC Ans. Final relative humidity On the constant wet bulb temperature line 1-3, mark point 2 such that td2 = 27.435 ºC. Now the relative humidity of the air leaving the air washer (corresponding to point 2) as read from the psychrometric chart is

2φ =90% Ans. 2.2.13 Heating and Humidification by Steam Injection The steam is normally injected in to the air in order to increase its specific humidity as shown in Fig.(2.29) a). This process is used for the air conditioning of textile mills where high humidity is to be maintained. The dry bulb temperature of air changes very little during this process, as shown on the psychrometric chart in Fig.(2.29) b).

Fig.(2.28)

Fig.(2.29)Heating and Humidification by steam injection


Let ms= Mass of steam supplied, ma= Mass of dry air entering, W1= Specific humidity of entering air, W2= Specific humidity of leaving air, h1= Enthalpy of entering air, h2= Enthalpy of leaving air, and hs= Enthalpy of steam injected into the air, Now for the mass balance,

a

s

m

mWW += 12 …………………(2.23)

and for heat balance,

( ) ssa

s hWWhhm

mhh ×−+=×+= 12112 …. [From equation (2.23)]

Example 2.7. The atmospheric air at 25 ºC dry bulb temperature and 12 ºC wet bulb temperatures is flowing at the rate of 100 m3/min through the duct. The dry saturated steam at 100 ºC is injected into the air steam at the rate of 72 kg per hour. Calculate the specific humidity and enthalpy of the leaving air. Also determine the dry bulb temperature, wet bulb temperature and relative humidity of the leaving air. Solution. Given: td1=25 ºC ; tw1=12 ºC v1=100 m3/min; ts=100 ºC; ms=72 kg/h = 1.2 kg/min Specific humidity of the leaving air Let W2= Specific humidity of the leaving air. First, mark the initial condition of air at 25 ºC dry bulb temperature and 12 ºC wet bulb temperature on the psychrometric chart at point 1, as shown in Fig.(2.30). Now from the psychrometric chart, we find that the specific volume of air at point 1, υ 1= 0.849 m3/kg of dry air Specific humidity of air at point 1, W1= 0.0034 kg/kg of dry air Enthalpy of air at point 1, h1= 34.2 kJ/kg of dry air We know that mass of air flowing,

min/78.117849.0

100

1

1 kgv

ma ===υ

We know that

airdryofkgkgm

mWW

a

s /0135.078.117

2.10034.012 =+=+= Ans.

Enthalpy of the leaving air Let h2= enthalpy of the leaving air. From steam table (1.4), we find that enthalpy of dry-saturated steam corresponding to 100 ºC hs= 2675.44 kJ/kg

We know that airdryofkgkJhm

mhh s

a

s /45.6144.267578.117

2.12.3412 =×+=×+= Ans.

Dry bulb temperature, wet bulb temperature and relative humidity of the leaving air.

Fig.(2.30)


Mark the condition of leaving air on the psychrometric chart as point 2 corresponding to W2= 0.0135 kg/kg of dry air and h2= 61.45 kJ/kg of dry air. Now from the psychrometric chart corresponding to point 2, Dry bulb temperature of the leaving air, td2=26.1 ºC Wet bulb temperature of the leaving air, tw2=21.1 ºC and relative humidity of the leaving air, 2φ =64% Ans. Note: Solve this example by using the protractor 2.2.14 Heating and Dehumidification - Adiabatic Chemical Dehumidification This process is mainly used in industrial air conditioning and can also be used for some comfort air conditioning installations requiring either a low relative humidity or low dew point temperature in the room. In this process, the air is passed over chemicals which have an affinity for moisture. As the air comes in contact with these chemicals, the moisture gets condensed out of the air and gives up its latent heat. Due to the condensation, the specific humidity decreases and the heat condensation supplies sensible heat for heating air and thus increasing its dry bulb temperature. The processes, which is the reverse of adiabatic saturation process, is shown by the line 1-2 on the psychrometric chart as shown in Fig.(2.31). The path followed during the process is along the constant wet bulb temperature line or constant enthalpy line. The effectiveness or efficiency of the dehumidifier is given as

12

13

dd

ddH tt

tt

etemperaturbulbdryinincreaseIdeal

etemperaturbulbdryinincreaseActual

−−==η

Notes: 1. In actual practice, the process is accompanied with a release of heat called heat of adsorption, which is very large. Thus the sensible heat gain of air exceeds the loss of latent heat and the process is shown above the constant wet bulb temperature line in Fig.(2.31). 2. Two types of chemicals used for dehumidification are absorbents and absorbents. The absorbents are substances which can take up moisture from air and during this process change it chemically, physically or in both respects. This includes water solutions or brines of calcium

W1

W2 W3

Actual adsorption

dehumidification

Fig.(2.31)Heating and Dehumidification


chloride, lithium chloride, lithium bromide and ethylene glycol. These are used as air dehydrators by spraying or otherwise exposing a large surface of the solution in the air stream. The adsorbents are substances in the solid state which can take up moisture from the air and during this process do not change it chemically or physically. These include silica gel (which is a form of silicon dioxide prepared by mixing fused sodium silicate and sulphuric acid) and activated alumina (which is a porous amorphous form of aluminium oxide). Example 2.8. Saturated air at 21 ºC is passed through a drier so that its final relative humidity is 20%. The drier uses silica gel adsorbent. The air is then passed through a cooler until its final temperature is 21 ºC without a change in specific humidity. Determine: 1. the temperature of air at the end of the drying process; 2. The heat rejected during the cooling process; 3. the relative humidity at the end of the cooling process; 4. the dew point temperature at the end of the drying process; and 5.the moisture removed during the drying process. Solution. Given: td1= td3=21 ºC; 2φ =20% 1. Temperature of air at the end of the drying process First, mark the initial condition of air at 21 ºC dry bulb temperature upto the saturation curve (because the air is saturated) on the psychrometric chart at point 1, as shown in Fig.(2.32). Since the drying process is a chemical dehumidification process, therefore, it follows a path along the constant wet bulb temperature or the constant enthalpy line as shown by the line 1-2 in Fig.(2.32). Now mark the point 2 at relative humidity of 20%. From the psychrometric chart, the temperature at the end of drying process at point 2, td2=38.5 ºC 2. Heat rejected during the cooling process The cooling process is shown by the line 2-3 on the psychrometric chart as shown in Fig.(2.32). From the psychrometric chart, we find that enthalpy of air at point 2, h2=61 kJ/kg of dry air and enthalpy of air at point 3, h3=43 kJ/kg of dry air ∴ Heat rejection during the cooling process = h2- h3=61-43 =18 kJ/kg of dry air Ans. 3. Relative humidity at the end of the cooling process From the psychrometric chart, we find that relative humidity at the end of the cooling process (i.e. at point 3), 3φ =55% Ans. 4. Dew point temperature at the end of the drying process From the psychrometric chart, we find that the dew point temperature at the end of the drying process, tdp2=11.6 ºC Ans. 5. Moisture removed during the drying process.

Fig.(2.32)


From the psychrometric chart, we find that the moisture in air before the drying process at point 1, W1=0.0157 kg/kg of dry air And moisture in air after the drying process at point 2, W2=0.0084 kg/kg of dry air ∴ Moisture removed during the drying process = W1- W2=0.0157-0.0084=0.0073 kg/kg of dry air Ans. 2.3 Cycles of air conditioning To design the air conditioning system for any space and for any season of the year, the internal and external condition of the place must be taken in the account. All the air conditioning cycles may be used to obtain the internal condition of the space. The most important air conditioning cycles are the summer air conditioning cycle and the winter air conditioning cycle. 2.3.1. The summer typical air conditioning cycle Figure (2.33) represent a typical air conditioning cycle in summer for comfort purpose. Point (r) represents internal room condition while point (o) represents the external design condition. The mixing of these two streams occur due to ventilation purpose at point (m), after that the mixture enters a cooling coil to cooling it and reduce the moisture content to be at point (a). The point (a) depends on the by-pass-factor and the apparatus dew point (c). If the air is supplied directly to the room then the process is (ra) (the sensible heat ratio curve), or the air is heated sensibly to a point (s) the supplying point. The air after that supplied to the room. The sensible heating from (a) to (s) is caused by the heat from the fan or distributing ducts. 2.3.2. The winter typical air conditioning cycle Figure (2.34) represent a typical air conditioning cycle in winter for comfort purpose. Point (r) represents internal room condition while point (o) represents the external design condition, and point (m) represents the mixture condition, from point (m) the mixture is heated sensibly to the supplied point (s), then to the room by the path (rs), or from point (s) to point (a) by humidification process and then sensible cooling from (a) to (r) due to the heat losses through the ducts and other sources.

Fig.(2.33) Typical summer air conditioning cycle


Example 2.9.An air conditioned auditorium is to be maintained at 27°C dry bulb temperature and 60% relative humidity. The ambient condition is 40°C dry bulb temperature and 30°C wet bulb temperature. The total sensible heat load is 100 000 kJ/h and the total latent heat load is 40 000 kJ/h. 60% of the return air is recirculated and mixed with 40% of make-up air after the cooling coil. The condition of air leaving the cooling coil is at 18°C. Determine 1.Room sensible heat factor; 2.The condition of air entering the auditorium; 3.The amount of make-up air; 4.Apparatus dew point; and 5.By-pass factor of the cooling coil. Show the processes on the psychometric chart. Solution. Given: td4=27 ºC; 4φ =60% ; td1=40 °C; tw1=30 °C; RSH=100000 kJ/h; RLH=40000kJ/h; td2=18 ºC 1.Room sensible heat factor We know that room sensible heat factor,

714.040000100000

100000 =+

=+

=RLHRSH

RSHRSHF

2. Condition of air entering the auditorium The line diagram for processes involved in the air conditioning of an auditorium is shown in Fig.( ). These processes are shown on the psychometric chart as discussed below:

First, mark the ambient condition of air (outside air) at 40 ºC dry bulb temperature and 30 °C wet bulb temperature on the psychrometric chart at point 1, as shown in Fig.(2.36). Now mark the condition of air in the auditorium at 27°C dry bulb temperature and 60% relative humidity, as point 4. Mark the calculated value of SHR=0.714 on the sensible heat factor scale on the protractor and draw line from the center of the protractor to value of SHR as shown in Fig.(36). Now from point 4,

Wr=Wa

Wm=Ws

Fig.(2.34) Typical winter air conditioning cycle

Fig.(2.35)


draw a line 4-5(known as RSHF line) parallel to the line drawn in the protractor. Since the condition of air leaving the cooling coil is at 18°C, therefore, mark point 2 such that td2=18 ºC. Join points 1 and 2 and produce upto point 6 on the saturation curve. The line 1-2-6 is given that 60% of the air from the auditorium is recirculated and mixed with 40% of the make-up air after the cooling. The condition of air entering the auditorium is given by point 3. From the psychrometric chart, we find that at point 3,

Ctm

mt

m

mt

a

a

a

a oQ 4.23276.0184.04

3,

4,2

3,

2,3 =×+×=+=

Wet bulb temperature tw3=19.7 °C and relative humidity 3φ =71% 3. Amount of make-up air From the psychrometric chart, we find that enthalpy of air at point 4, h4=61.5 kJ/kg of dry air and enthalpy of air at point 3, h3=56.5 kJ/kg of dry air

3434 hh

RLHRSH

hh

heattotalRoomms −

+=−

=

hkg /280005.565.61

40000100000 =−+=

Since the make-up air is 40% of supply air, therefore mass of make-up air =0.4×28000=11200 kg/h Ans. 4.Apparatus dew point From the psychrometric chart, we find that the apparatus dew point of the cooling coil at point 6 is ADP= td6=13.3 °C Ans. 5. By pass factor of the cooling coil We know that by-pass factor of the cooling coil,

176.03.1340

3.1318

1

2 =−−=

−−=

ADPt

ADPtBPF

d

d Ans.

RSHF

Fig.(2.36)


Example 2.10.Air at 10°C dry bulb temperature and 90% relative humidity is to be heated and humidified to 35°C dry bulb temperature and 22.5°C wet bulb temperature. The air is pre-heated sensibly before passing to the air washer in whish water is recalculated. The relative humidity of the air coming out of the air washer is 90%. This air is again reheated sensibly to obtain the final desired condition. Find : 1.The temperature to which the air should be preheated. 2. The total heating required; 3.The make up water required in he air washer; and 4.the humidifying efficiency of the air washer. Solution. Given: td1=10 ºC; 1φ =90% ; td2=35°C; tw2=22.5°C First, mark the initial condition of air at 10 ºC dry bulb temperature and 90% relative humidity, on the psychrometric chart at point 1, as shown in Fig.(2.37). Now mark the final condition of air at 35°C dry bulb temperature and 22.5°C wet bulb temperature at point 2. From point 1, draw a horizontal line to represent sensible heating and from point 2 draw horizontal line to intersect 90% relative humidity curve at point B. Now from point B, draw a constant wet bulb temperature line, which intersects the horizontal line drawn through point 1 at point A. The line 1-A represents preheating of air, line AB represents humidification and line B-2 represents reheating to final condition. 1.Temperature to which the air should be preheated From the psychrometric chart, the temperature to which the air should be preheated (corresponding to point A) is tdA=32.6 ºC Ans. 2.Total heating required From the psychrometric chart, we find that enthalpy of air at point 1, h1=27.2 kJ/kg of dry air Enthalpy of air at point A, hA=51 kJ/kg of dry air and enthalpy of air at point 2, h2=68 kJ/kg of dry air we know that heat required for preheating of air = hA- h1=51-27.2=23.8 kJ/kg of dry air and heat required for reheating of air = h2- hB=68-51=17 kJ/kg of dry air … (QhB = hA) ∴ Total heat required =23.8+17=40.8 kJ/kg of dry air Ans. 3. Make up water required in the air washer From the psychrometric chart, we find that specific humidity of entering air, W1=0.00068 kg/kg of dry air and specific humidity of leaving air, W2=0.0122 kg/kg of dry air ∴Make up water required in the are washer = WB- WA= W2- W1 =0.0122-0.00068=0.0054 kg/kg of dry air 4.Humidifying efficiency of the air washer From the psychrometric chart, we find that tdB=19.1 ºC and tdB' =18 ºC We know that humidifying efficiency of the air washer

'dBdA

dBdAH tt

tt

DBTindropIdeal

DBTindropActual

−−==η

%4.92924.0186.32

1.196.32or=

−−

Fig.(2.37)


Example 2.11. In order to compare the space conditions produced and the relative energy requirements of bypass and reheat systems, consider a space that has a sensible load of ∑Qs=205kW and a latent load of ∑QL=88kW when the space is maintained at a dry bulb temperature of 25°C and the outdoor condition are 35°C dry bulb temperature and 40% relative humidity. The space dry bulb condition is to be met by using either of the systems shown in Figs.(2.38) or (2.40). Supply air is to be introduced to the space at a flow rate of 30kga/s.The flow rate of exhaust air is 4.5 kga/s. The exit condition of the systems cooling coil are a dry bulb temperature of 10°C and relative humidity of 95%. For by-pass system shown in fig. (2.38)determine:(a) the relative humidity in the space and (b) the required system cooling capacity. For the reheat system shown in fig. (2.40)determine (c) the relative humidity in the space , (d) the rate of heat required for the reheat coil, and (e) the required system cooling capacity. Solution. The nomenclature of Fig.(2.38) will be used for parts (a) and (b). Standard atmospheric pressure is assumed and the solution carried out using psychrometric chart. States 4 and 6 can be located from given information. States 6,7, and 2b fall on the mixing line connecting 6 and 2b. The slope of this line is determined by observing that it is also the space-condition line from state 1 to state 2. The sensible-heat ratio for the space-condition line is given by:

7.088205

205 =+

=+

=ls

s

QQ

QSHR

(a) State 2 is located at the intersection of the line with an SHR=0.7 drawn through state 6 and td2=25 ºC. The resulting point falls at 2φ =50%. Thus using the system shown in Fig.(2.35), the

space conditions will be td2=25 ºC and 2φ =50%. (b) State 5 is on the mixing line connecting state 4 and 2a.(Note that states 2, 2a, 2b, and 3 are the same) In order to determine the division of the return air between mº

a,2a and mºa,2b, locate state 7,1.

( )srps ttcmQ −= Where tr=td2 , ts=td1, and cp=1.02 kJ/kg .K

205=30×1.02×(25- td1) ⇒ td1=18.3 ºC

12

61

12

61

6

2

tt

tt

hh

hh

m

m b

−−=

−−=Q

From Fig.(2.38) we observing that mºa,2b= mº

a,1- mºa,6 and that mº

a,5 = mºa,6

45.01025

3.1825

62

12

62

12

1

5 =−

−=−−=

−−=

tt

tt

hh

hh

m

mQ

Thus mº

a,5 =0.45×30=13.5 kg/s mº

a,2b=30-13.5=16.5 kg/s Also, from Fig.(2.38) it is seen that mº

a,2a= mºa,1- m

ºa,3- m

ºa,2b=30-4.5-16.5=9 kg/s

Cm

tmtmt

a

aaaa oQ 3.28

5.13259355.4

5,

22,44,5 =×−×=

−=

Thus, state 5 is located at the intersection of the line connecting 2a and 4 and a dry bulb temperature 28.3 ºC. from chart airdrykgkJh /8.575 =⇒ The required system cooling capacity


( )65565 hhmQ −×= o

=13.5×(57.8-28.3)= 398.25kW The nomenclature of Fig.(2.40) will be used to analyze the reheat system. (c) The dry bulb temperature of the supply air is found in the same manner as it was in part (b) of this example. Therefore, td1=18.3 ºC. The state 1,7 is located by following a sensible-heating process line (constant W) beginning at state 6 and ending at td1= td7=18.3 ºC. State 2 is located by drawing the space condition line beginning at state 1 and ending at the prescribed space dry bulb temperature of td2=25 ºC. In part (a) of this example the SHR for the space-condition line was calculated to be SHR =0.70. Locating state 2 on the psychrometric chart, we read 2φ =45%. Thus it is seen that use of the reheat system results in a lower relative humidity in the space than does the use of the bypass system (45% vs. 50%). From psychrometric chart h1,7=37.1 kJ/kg of dry air (d) The reheat energy rate is

( )67576 hhmQ −×= o =30×(37.1-28.3)=264 kW (e) State 5 is the intersection of td5 with the adiabatic mixing process line connecting states 2 and 4. Notice that in the reheat system the entire supply-air flow rate passes through the cooling coil. Therefore, mº

a,5 = mºa,1=30 kg/s and

mºa,2= mº

a,1- mºa,3=30-4.5=25.5 kg/s

from chart h2=46.5 kJ/kg of dry air

Cm

tmtmt

a

aa oQ 5.26

30255.25355.4

5,

22,44,5 =×−×=

−=

After locating state 5, we read h5=50.2 kJ/kg of dry air The system cooling capacity required is

Fig.(2.38)

4

5

2,2a,2b,3

1,7 6

SHR

SHR

Fig.(2.39)


( )65565 hhmQ −×= o

=30×(50.2-28.3)=657 kW The result of example are summarized in table

Result or Set point Bypass System Reheat System

Space dry-bulb temperature 25 ºC 25 ºC Space relative humidity 50% 45%

Required cooling capacity 398.25kW 657 kW Required reheat ---- 264 kW

Total energy rate required 398.25kW 921 kW

Fig.(2.40)

4

5

2,3

1,7 6 SHR

SHR

Fig.(2.41)


Example 2.13. The below figure show layout of winter system. Sensible load for the space 30 kW and the space lose humidity at rate 8 kg/hr. Indoor design condition DBT=22 ºC and 30% Rh and outdoor design condition DBT=5 ºC and WBT=2 ºC. The air supplied to the conditioned space at rate 10200 m3/hr and the air humidified after supplied to the conditioned space by injection of saturated steam at temperature 100 ºC. Rate of the outdoor ventilation air (fresh air) 2000 m3/hr. Determine: 1. Condition of supply air to the conditioned space (DBT & RH); 2.Consumption rate of steam in humidifier; 3. Load of heating load; 4. Draw the cycle on the psychrometric chart. Solution:

fgwL hmQ o= , hfg=2450 kJ/kg

QL=8/3600×2450=5.444 kW

85.0846.044.530

30 ≈=+

=+

=Ls

s

QQ

QSHR

( )rsps ttcmQ −= Where tr=td1 , ts=td5, and cp=1.02 kJ/kg .K

( )rss ttvQ −= o22.1 30 =1.22×(10200/3600)×( td5-22) td5 =30.67 ºC from the psychrometric chart, υ1=0.843 m3/kg dry air υ2=0.792 m3/kg dry air m1=m5-m2 v1=v5-v2

hkgv

ma /16.9727843.0

200010200

1

11 =−==

υ

hkgv

ma /25.2525792.0

2000

2

22 ===

υ

m5= m1+m2=9727.16+2525.25= 12252.41kg/h=3.4 kg/s

CCmm

tmtmt oo

Q 5.18494.1825.252516.9727

525.25252216.9727

21

22113 ≈=

+×−×=

+−

=

From psychrometric chart td4 =30.4 ºC, h4=43.5 kJ/kg d.a. , h5=47.1 kJ/kg d.a. hs from table(1.4)(at t=100 ºC) = 2675.44 kJ/kg

Outdoor air

Exhaust air

Supply air

Return air

Heating coil Steam

at 100 ºC

Humidifier Conditioned space

Fig.(2.42)


By using value of hs at the protractor, from psychrometric chart, as shown in Fig.(2.43) RH5=19% , h3=32 kJ/kg d.a. By using energy balance on the humidifier m4 h4+ms hs=m5 h5 , 3.4×43.5+ms×2675.44=3.4×47.1 ms=4.575×10-3=16.47 kg/h

( )343 hhmQ coilheating −×= o

=3.4×(43.5-32)=39.1 kW

4

5

2

1

3

Humidifier

SHR

hs SHR

Fig.(2.43)

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