AC Air Cycles

7/27/2019 AC Air Cycles

1/18

Website: http://guider55.4t.com Refrigeration and Air Conditioning

Air Conditioning Systems

Summer airconditioningWinter air conditioning

.))Summer Air Conditioning Systems

225 = CT oR5%50 =R

Room heat load SQ LQ,Room sensible heat factor

RSHF.
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(Summer air conditioning system without by pass)

)-(.)-(

Fig. 8-1 Summer air conditioning system

Fig. 8-2 Summer air conditioning processes

Return airFresh air Airhandling unit


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S5 ~ 10 oC90 %.

Ro m

S %90=SmS%100=a

,Apparatus dew Point AdpS.

Room heat load SR,)( SRat hhmQ = & (8-1)

Cooling coil capacity and efficiency mSand a,

)( Smacoil hhmQ = & (8-2)

am

Smcoil

hh

hh

= (8-3)

Condensate water through dehumidification process mS,)(

Smawmm = && (8-4)

Example 8-1

An air conditioning room is maintained at 25oC db and RH 50 %. The ambient

conditions are 38oC db and 27

oC wb. The room has sensible heat gain of 40 kW

and latent heat gain of 10 kW. The re-circulated air contains 30 % by weight fresh

air. The re-circulated air is supplied to the room at 14oC. Calculate, the sensible

heat factor, the total amount of conditioned air, the cooling coil dew point, the

cooling coil capacity in TR, the cooling coil efficiency, and the rate of water

removed from the air.

Data: aoo

ooR

o

R mmCtCtRHCt && 3.0,wb27,db38,%50,25o

=====

CtkWQkWQ SLSo14,10,40 ===

Required: wcoila mRCAdpmRSHF && ,,,,,


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Solution

As shown in Fig. 8-2 of Psychrometric chart and the air properties are,kgkJhkgkJh

oR/55.84,/34.50 ==

aRao mmmm &&&& 7.0,3.0 ==

kWQQQ LSt 501040 =+=+=

8.01040

40=

+=

+=

LS

S

QQ

QRSHF

Heat balance of mixing point m,

maRRoo hmhmhm &&& =+

maaa hmmm &&& 34.507.055.843.0 =+

damm kggkgkJh /5.12,/60.6 ==

Processes mSR, cooling and dehumidification from m to Sat 90 % RH, from

StoR, room total heat released parallel toRSHF.CAdptkgkJhkggkgkJh o

aadaSS9,/27,/7.8,/5.35 =====

Re-circulated air,

skgmmhhmQ aaSRat /369.3),5.3534.50(50),( === &&&

Cooling coil capacity,

TRhhm

RC Sma 162.245.3

)5.356.60(369.3

5.3

)(=

=

=&

Cooling coil efficiency,

%7.74747.0276.60

5.356.60==

=

=

am

Smcoil

hh

hh

Condensate water,

min/768.06010)7.85.12(369.3),( 3 kgmmmwSmaw

===

&&&

Example 8-2

In a summer air conditioning system shown the fresh air is mixed with return air

by equal mass before entering the cooling coil. After pre-cooled the air

adiabatically humidified in air washer to 90 % RH. The outside air conditions

are 35 C db and 17 C wb. The inside air conditions are 27 C db and 21 C wb.

Internal sensible heat gain is 35 kW and latent heat gain is 15 kW. Estimate; the


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temperature to which the air is pre-cooled, the temperature to which the sprayed

water must be held, the amount of supplied air, the capacity of the pre-cooler, and

the make up water.

Data: wb17,db35,,wb21,db27 o CtCtCtCt oooo

Ro

R ====

%90,15,35 === SLS RHkWQkWQ

Required: waS mRCmtt && ,,,,1

Solution

From Psychrometric chart and the air properties are,kgkJhkgkJh oR /3.47,/68.60 ==

aRommm &&& 5.0==

kWQQQ LSt 501535 =+=+=

7.01535

35=

+=

+=

LS

S

QQ

QRSHF



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maRRoo hmhmhm &&& =+

maaa hmmm &&& 68.605.03.475.0 =+

damm kggkgkJh /9,/54 ==

Processes m1SR, sensible cooling from m to 1, but Sat cross ofRSHFwith

90 %RH, from Sto 1 adiabatic line at cross with sensible cooling m1.

m

oCt == 11 dbt,6.23

CAdptCtkggkgkJh oao

SdaSS 6.16db,9.17,/5.11,/47 =====

Re-circulated air,

skgmmhhmQ aaSRat /655.3),4768.60(50),( === &&&

Cooling capacity of pr-cooler,

TRhhm

RC ma 31.75.3

)4754(655.3

5.3

)( 1=

=

=&

Air washer efficiency,

%43.18143.06.166.23

9.176.23

1

1==

=

=

a

Swasher

tt

tt

Make up water in air washer,

min/5483.06010)95.11(655.3),( 31 kgmmm wSaw ===

&&&

(Summer air conditioning system with by pass)

Fig. 8-3 Summer air conditioning system with by pass


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)-(

)-(S

.

Fig. 8-4 Summer air conditioning processes with by pass

Example 8-3

A summer air conditioning systems consists of water chiller and air re-heater as

shown below. The return air is mixed partially before the water chiller and by

passed after it with equal masses. The inside conditions are 25C db and 50 % RH

and outside conditions are 38 C db and 26C wb. Fresh air for ventilation is 0.5

m3/s. The internal sensible heat gain is 21 kW and internal latent heat gain is 7

kW. The air leaving water chiller saturated at 10oC and the temperature difference

between inside and supply air is 8 C db. Determine the refrigeration capacity of

the water chiller and the heating capacity of the air reheated.

Data: smmCtCtRHCt oo

ooR

o

R/5.0,wb26,db38,%50,25 3o ===== &

CtRHCtkWQkWQ oSLS 10%,100,17825,7,21 11o

======

Required: reha PowerRCmRSHF ,,, & Solution


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As shown in Fig. 8-4 of Psychrometric chart and the air properties are,

kgkJhmkgkJhkgkJh ooR /5.29/kg,9047.0v,/05.80,/34.50 13

====

skgV

mmmo

ooRR /5527.0

9047.0

5.0

v,21 ====

&&&&

kWQQQ LSt 28721 =+=+=

75.028

21===

t

S

Q

QRSHF

State Sis at cross ofRSHFwith ts= 17oC

/5.37 kgkJhS =

skgmmhhmQaaSRat

/181.2),5.3734.50(28),( === &&&

skgmmmmmm

RR

oaRR

/8142.02/)5527.0181.2(21

21

===

=+

&&

&&&&


mRoRRoohmmhmhm )(

11&&&& +=+

kgkJhh mm /35.62,)8142.0(0.552734.508142.005.805527.0 =+=+

Heat balance of mixing point 1 withR and 2Rm& ,

2211 )( hmhmhmm aRRRo &&&& =++

kgkJhh /28.37,2.18134.508142.05.29)8142.05527.0( 22 ==++

Cooling capacity water chiller,

TRhhm

RC ma 47.205.3

)5.2935.62(181.2

5.3

)( 1=

=

=&

Capacity of re-heater,

kWhhmPowerSa

479.0)28.375.37(181.2)(2 === &

.Example 8-4

In an air conditioning unit, air is supplied at a rate of 0.65 m3/s with 25C db and

60 % RH. The unit consists of a cooling coil where air is cooled and dehumidified


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to 5C db and 4 C wb. Then, the air passes through an electric heater and leaves it

at 20C db. After that, air passes through an air washer, where it is humidified to

90 % RH. Part of the air is by passed across the air washer to have final air relative

humidity of 60 %. Calculate, the capacity of cooling coil, the heater power, the by

pass factor for air washer, and the bypassed low rate.

Data: smmCtCtRHCt oo

o

o

o /65.0,wb4,db5,%60,253

1

o

1 ===== &

%60%,90,25 3o

2 === SRHRHCt

Required: ,,, bwasher mBFPowerRC &

Solution

From Psychrometric chart and the air properties are,kgkJhhhkgkJhkgkJh So /33,/6.16,/55 321 =====

kgmkggkggodada

/8613.0v,/2.8,/8.4 332 ===

daadaS kggkgg /5.8,/8.6 ==

skgV

mo

oo /755.0

8613.0

65.0

v===

&&

TRkWRChhmRCoo

283.8992.28)6.1655(755.0),(1

==== &

kWPowerhhmPower o 61.16)3355(755.0),( 12 === &


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Air washer efficiency and bypass factor,

%89.919189.08.45.8

8.42.8

2

23==

=

=

a

washer

%1.8081.09189.011 ==== washerwashweBF

Mass balance of air after bypassed and air washer,

Sobbommmm &&&& =+ 23)(

8.6755.08.42.8)755.0(,)( 23 =+=+ bbSobbo mmmmmm &&&&&&

skgmb /311.0=&

))Air Conditioning SystemsWinter

Pre-

heaterAir washer Re-heater)-(.

Fig. 8-5 Winter air conditioning system

)-(CTTS RS o10~5+=

RRSS


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SRRRSHFRRS\S \>

S\RRRSHF .Humidification

Fig. 8-6 Winter air conditioning processes

Example 8-5

Air flow of 11 m3/min is supplied to air conditioning room at 24

oC db and 50 %

RH. The outside air conditions at 10oC db and 90 % RH. The return air is mixed

with the fresh air before entering the primary heater in ratio of 2 to 1 by weight.

The air is first preheated until its wet bulb temperature is equal to the room wet

bulb temperature. Then, the air is adiabatic saturated in air washer to 90 % RH.Finally the air is reheated to 35

oC before being supplied to the room. Determine

the heat added in both two heaters in kW, the mass of water evaporated in the air

washer, and the air washer efficiency.

Data: min/11,%90,db10,%50,db24 3o mmRHCtRHCt aooRo

R ===== &

%90,db35,2/ 2o

=== RHCtmm SoR &&

Required: washerPowerPower ,, 21

SolutionFrom Psychrometric chart as Fig. 8-6 and the air properties are,


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kgmkgkJhkgmkgkJh ooRR /8111.0v,/35.27,/8546.0v,/83.4733

====

skgv

Vm

R

aa /215.0

8546.060

11=

==

&&

Heat balance of mixed point m,

ooRaoRmaRRoo mmmmmmhmhmhm &&&&&&&&& 3,2/, =+===+

moooo hmmmm )2(83.47235.27 &&&& +=+

skghhhskgh Rm /83.47,/41 21 ====

Process m12S, sensible heating from m to 1at h1 = hR , adiabatic air washer

from 1 to 2 at 90 %RH, sensible heating from 2 to Sat 35oC db.

daadadaS kggkggkggkgkJh /3.12,/8.11,/5.8,/3.65 21 ====

kWhhmPower ma 468.1)4183.47(215.0)( 11 === &

kWhhmPower Sa 756.3)83.473.65(215.0)( 22 === &

%84.8684.86.05.83.12

5.88.11

1

12==

=

=

a

washer

Example 8-6

A winter air conditioning unit, consists of preheating coil, adiabatic air washer and

reheating coil is used to maintain the condition inside a room at 25oC db and 50 %

RH. The re-circulated air is mixed with fresh air at equal parts by weight before

the preheating coil. An amount of 56.6 m3/min fresh air is supplied to the unit at 5

oC db and 90 % RH. The air leaves the humidifier at 85 % RH and leaves the

reheating coil at 33oC db and 35 % RH. Representation the processes on the

Psychrometeric chart and calculate the heating capacity of each heating coil, the

rate of water to be consumed in the humidifier, and the humidifying efficiency.

Data: oRooRoR mmRHCtRHCt && ===== ,%90,db5,%50,db25 o %35,db33%85/min,6.56 o2

3==== SSo RHCtRHmV

&

Required: washerwmPowerPower ,,, 21 &

Solution

From Psychrometric chart as Fig. 8-5 and the air properties are,

kgmkgkJhkgkJh ooR /7943.0v,/23.17,/34.503

===

skgv

Vm

o

oo /188.1

7943.060

6.56=

==

&&


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Heat balance of mixing process stateR with state o by equal mass,

skgmmmmmmhmhmhm ooRaoRmaRRoo /376.2188.122,, ===+===+ &&&&&&&&&

skgmmmmmm ooRaoR /376.2188.122, ===+== &&&&&&

mooo hmmm &&& 223.1734.50 =+

skghm /79.33=

State Sat, %35,db33 o == SS RHCt , daSS kggkgkJh /01.11,/42.61 ==

Process S2, sensible heating from 2 to S, state 2 atRH = 85 %

daS kggkgkJh /01.11,/1.46 22 ===

Process 12, adiabatic air washer from 1 to 2 at 85 %RH,daadam kggkggkgkJh /5.11,/4.7,/1.46 11 ====

kWhhmPower ma 249.29)79.331.46(376.2)( 11 === &

kWhhmPower Sa 4.36)1.4642.61(376.2)( 22 === &

skgmm aw /515.06010)4.701.11(376.2)(3

12 ===&&

%05.888805.04.750.11

4.701.11

1

12==

=

=

a

washer

Example 8-7

A winter air conditioning unit consists of a preheating coil a spray system and a

reheating coil. The unit is designed to handle 3200 kg of air per hour and 100 % of

which is fresh air. The heat loss from the space is 9 kW. The outside air enters the

unit at 7 C db and 4 C wb. The air leaves the spray system at 90 % RH. The unit

delivers the air at 33C db and 30 % RH. Note that SHF=1, determine the

condition maintained inside the space, and the heating capacities of preheating and


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reheating coil.

Data: skghrkgmRHCtCt aoo

o /889.0/3200,%90,wb4,db7 2o

===== &

%30,db33,1,9 o ==== SSt RHCtRSHFkWQ

Required: 21 ,,conditionsRoom PowerPower Solution

From Psychrometric chart, and the air properties are,kgkJhkgkJhkgkJh oS /6.16,/5.38,/5.57 2 ===

Process 2S, sensible heating and state Sat 33o

Cdb and 30 %RH. The state 2 isat the horizontal line atRH= 90 %.

dakggkgkJh /6.9,/5.38 22 ==

Process 12, adiabatic air washer and state 1 at cross with sensible heating line

from state o which at 7oCdb and 4

oCwb.

dakggkgkJh /8.3,/5.38 11 ==

Room heat load from state Sto stateR horizontal line, SHF = 1kgkJhhhhmQ RRRSat /4.47),5.57(889.09),( === &

%55db,231 == R

o RHCt

kWhhmPower oa 469.19)6.165.38(889.0)( 11 === &

kWhhmPower Sa 891.16)5.385.57(889.0)( 22 === &

Example 8-8

In air conditioning system shown below, all dehumidification occurs in the

adsorbent dehumidifier (Adiabatic dehumidification). Sketch the cycle on the

Psychrometric chart and find the condition of air leaving the adsorbent


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dehumidifier, and the amount of water required for cooling the air in each cooler.

Data: CttmVCtCt oo 27min,/8.70,wb24,db35 523

1

o

11 =====&

min/127,wb5.17,db27 38

o

88mVCtCt o === &

min/42min,/85

33

mVmV ba== &&

%30,db33,1,9 o ==== SSt RHCtRSHFkWQ

Required: 21 ,4,stateofConditions ww mm &&

Solution

From Psychrometric chart, and the air properties are,dakggkgmkgkJh /24.14,/8931.0v,/75.71 1

3

11 ===


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dakggkgmkgkJh /58.8,/8622.0v,/06.49 83

88 ===

skgv

Vm a /321.1

8931.060

8.70

1

11 =

==

&&

skgv

V

m

b

b /643.18622.060

85

8

1

1=

==

&&

skgv

Vm bb /812.0

8622.060

42

8

22 =

==

&&

Process 12, sensible cooling and state 2 at 27oCdb, kgkJh /49.632 =

Heat balance at mixing point b1, to locate state 3 at 27oC.

3118111)( hmmhmhm

bb&&&& +=+

3)643.1321.1(06.49643.175.71321.1 h+=+

dakggkgkJh /5.12,/17.59 33 ==

Supplied air and room total heat load,skgmmmm bba /776.3)812.0643.1321.1()( 211 =++=++= &&&&

kgkJhhhhmQ at /32.38),06.49(776.363),( 7778 === &

Room latent heat load, sensible heat, and room sensible heat factor,kgkJhmQ fgwL /62.278.2437)60/68.0( === &

kgkJQQQ LtS /37.3562.2763 ===

56.06337.35/ === tS QQRSHF

State 7 at cross ofRSHFwith h7,

Process 67 sensible cooling and state 6 at 27oCdb, kgkJh /5.41

6

=

Heat balance at mixing point b2, to locate state 5 at at 27oCdb.

6258255 )( hmmhmhm bb &&&& +=+

5.41)812.0643.1321.1(06.49812.0)643.1321.1( 5 ++=++ h

kgkJh /43.395 =

Process 45 sensible cooling and state 4 at the intersect of adiabatic line from

state 3 and horizontal line from state 5.

da

o kggCtkgkJh /8.4db,5.45,/17.59 444 ===

Heat balance of cooler 1,

)( 2111 hhmtCm wPw w = && kg/s5801.0),49.6375.71(321.1)135.17(18.4 11 == ww mm &&

Heat balance of cooler 2,)( 7662 hhmtCm wPw w = &&

kg/s8308.1),38.325.41(776.3)135.17(18.4 22 == ww mm &&

Problems

1.

A certain building is to be air conditioned in summer. The following data aregiven. The outside air is 38 C db, 25 C wb, and inside air 24 C db, 50 % RH.


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Estimated heat from the building is 17.6 kW, electric lighting and machines are

40 kW. The total number of persons is 100, and sensible heat per person is 80

W and latent heat per person is 103 W. The ratio of return air and fresh air is

2:1 by mass. The air leaves the cooling coil at 90 % R.H. Determine the

sensible heat factor, the total amount of conditioned air, the cooling coil dew

point, the cooling coil capacity in TR, the cooling coil efficiency, and the rate

of water removed from the air.

2. In a summer air conditioning system, employing evaporative cooling equalmasses of fresh air and return air, after mixing air first pre-cooled then

adiabatically humidified to 100 % R.H by passing through a spray water

system. The outside air conditions are 35 C db and 17 C wb. The inside air

conditions are 27 C db, and 21 C wb. Internal sensible heat gain is 30 kW and

latent heat gain is 15 kW. Find the temperature to which the air is pre-cooled,

the temperature to which the sprayed water must be held, the amount of

supplied air, the capacity of the pre-cooler, and the make up water.

3. The following information was obtained in designing the summer airconditioning plant. The outside design conditions are 38 C db, 28 C wb and

the inside design conditions are 21 C db, 70 % RH. Internal sensible heat gain

is 175 kW and internal latent heat gain is 65 kW. Outside fresh air to total air is

50 % by mass. Temperature difference between supply air and return air is 8 C

db. Determine the conditions of the supply air, the quantity of supply air, the

cooling capacity of the plant, the cooling coil efficiency and by pass factor, and

the rate of water removed from the air.

4. A summer air conditioning systems consists of water chiller and by pass. Usesreturn air before the water chiller and by pass air after it. The given data are,

inside conditions 26C db & 50 % R.H and outside conditions 35C db & 26C

wb. Fresh air for ventilation is 2 m3/s. The internal sensible heat gain is 40 kW

and internal latent heat gain is 17.5 kW. The relative humidity of air leaving

water chiller is 90 % and the temperature difference between inside and supply


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air is 9 C db. Equal masses for return and by pass air. Determine the

refrigerating capacity of the water chiller and the condensate water by lit/min.

5. A winter air conditioning unit, consists of preheating coil, adiabatic air washerand reheating coil is used to maintain the condition inside a room at 25 oC db

and 50 % RH. The re-circulated air is mixed with fresh air at equal parts by

weight before the preheating coil. An amount of 56.6 m3/min fresh air is

supplied to the unit at 5oC db and 90 % RH. The air leaves the humidifier at 85

% RH and leaves the reheating coil at 30oC db and 45 % RH. Draw a sketch

for this unit and its representation on the Psychrometeric chart. Then calculate

the heating capacity of each heating coil, the rate of water to be consumed in

the humidifier, and the humidifying efficiency.

6. A winter air conditioning unit consists of air pre-heater, air washer and airreheater. The re-circulated air is admitted partially as return air before the air

preheated and the rest is by passed after the air washer. Inside condition is 25

C db and 50 % RH. Outside condition is 5 C db and 70 % RH. Fresh air

enters the systems at the rate of 28 m3

/min. Temperature of air leaving pre-heater equal 21 C db and the relative humidity of air leaving air washer is 95

%. The temperature difference of supply air is high than room air by 7 C db.

Equal masses of fresh air, return air and by-pass air are used, draw the system

with its all details and represent it in the Psychrometric chart, then calculate the

capacity of both pre-heater and re-heater, internal heating load, and make up

water.

Date post:	14-Apr-2018
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