15.5 SUMMER AIR CONDITIONING-APPARATUS DEW POINT

In summer, the outside air temperature and humidity are both high. The room, therefore, gains heat as well as moisture. It is thus required to cool and dehumidify the recirculated room air in the air-conditioning apparatus either by the use of a cooling coil or by an air washer in which chilled water is sprayed. The process follows the room sensible heat factor (RSHF) line. The room sensible heat factor is the ratio of the room sensible heat to the room total heat

RSHF = RSH

RSH +RLH = RSHRTH

In a cooling and dehumidification process, the temperature at which the RSHF or condition line intersects the saturation curve is called the room apparatus dew point (Room ADP). Thus t AD P in Fig. 15.22 denotes the effective surface temperature t S. The condition line i-S represents the locus of all possible supply air states. One extremity of the condition line is i which would be the supply air state with an infinite quantity of supply air. The other extremity is S which is the supply state with the minimum supply air requirement corresponding to the given condition line. It is not possible to have any other supply air state with a DBT lower or higher than the ADP on the saturation curve which would satisfy the given condition line.

Fig. 15.22 Locus of supply air states for cooling and apparatus dew point

The minimum quantity of supply air will then be given by either of the following three equations:

(cmm)s ,min = RSH

0.0204(t i−t ADP) (15.27a)

= RLH

50(ω i−ωADP)

(15.27b)

= RTH

0.02(hi−hADP)

(15.27c)

In the case of an actual coil with a bypass factor of X, the leaving air state will be at s. It is seen that the effect of the bypass factor is to decrease the difference in temperature between the room air and supply air, and hence to increase the supply air quantity over its minimum value. For any supply air state, the temperature difference (t i−t s) available to counteract the room sensible heat load is called the dehumidified rise and the corresponding dehumidified air quantity (cmm)d which is equal to the quantity of the supply air, is obtained by the equation for sensible heat balance, and considering the effect of bypass factor

(cmm)d = (cmm)s = RSH

0.0204(t i−t ADP) =

RSH0.0204 (t i−t ADP) (1−X ) (15.28)

It can also be found from the equations of latent heat or total heat balances.

15.5.1 Summer Air-Conditioning System with Ventilation Air-Zero Bypass Factor

The introduction of fresh outside air for the ventilation of conditioned space is necessary to dilute the carbon dioxide and odours and other air contaminants for maintaining the purity of room air. Accordingly, the simple air-conditioning system of Fig. 15.20 is modified, so that the supply air to the room comprises fresh air and recirculated room air. An amount equivalent to the fresh air is ejected from the room. The schematic diagram of the system is shown in Fig. 15.23, and the processes for the cases of cooling and dehumidification are shown in Fig. 15.24.

Fig. 15.23 Schematic diagram of system with ventilation air

In Fig. 15.24, 0 and I represent the outside and inside air states and 1 is the state of air after the mixing of recirculated room air with ventilation air. The mixture entering the conditioning apparatus comprises recirculated room air mai and ventilation air mao. The room sensible heat factor (RSHF) line is drawn from the inside condition i to intersect the saturation curve at room ADP at 2. Point 2 represents the supply air state for a minimum rate of supply air. The line 1-2, therefore, represents the condition line for the apparatus and is called the grand sensible heat

factor (GSHF) line. It is noted that the line i-2 is the condition line for the room or the RSHF line and line 1-2 is the condition line for the apparatus or the GSHF line intersecting the saturation curve at coil apparatus dew point (Coil ADP). Note that, in this case, the coil ADP and the room ADP are the same. In the absence of ventilation air the load on the air-conditioning apparatus is that due to the room sensible heat and room latent heat. When ventilation air is used, there is an additional load on the apparatus named the ventilation load equivalent to the change of state of the ventilation air from the outside condition to inside condition. This becomes evident when we write for the total load on the air-conditioning apparatus in terms of the change of state of the mixture air mas from 1 to 2. Thus

Fig. 15.24 Summer air-conditioning processes with ventilation air and zero bypass factor

Q = mas(h1−h2) = ¿ + maOhO) −mas

h2

= (mas −maO)hi + maOhO−mas

h2

= mas(hi−h2) + maO(hO−hi) (15.29)

The first term on the right-hand side in Eq. (15.29) represents the room load and the second term, the load due to the ventilation air as explained earlier. Accordingly, if(cmm)O is the outside ventilation air volume flow rate, then the outside air sensible heat (OASH) and outside air latent heat (OALH) loads are

OASH = QS O= 0.0204(cmm)O(tO−t i) (15.30) OALH = QLO = 50(cmm)O(ωO−ωi) (15.31)

Also for the outside air total heat (OATH), we have OATH = QO = OASH + OALH = 0.02(cmm)O (hO−hi) (15.32)

Note: These equations apply to winter air conditioning as well.

The break-up of the load on the air conditioning apparatus is now as follows:Room Load Sensible RSH Latent RLH Total RTH = RSH + RLHVentilation Load Sensible OASH Latent OALH Total OATH = OASH + OALHAir-conditioning Equipment Load Total sensible TSH = RSH + OASH Total latent TLH = RLH + OALH Grand total GTH = TSH + TLH

In Fig. 15.24, the process line 1-2 represents the grand sensible heat factor line for the process in an air-conditioning apparatus. The grand sensible heat factor is the ratio of the total sensible heat to the grand total heat. Thus

GSHF = TSH

TSH+TLH = TSHGTH

Example 15.6 The air-handling units of an air-conditioning plant supplies a total of 4500 cmm of dry air which comprises by weight 20 percent fresh air at 40° C DBT and 27°C WBT, and 80 percent recirculated air at 25°C DBT and 50 per cent RH. The air leaves the cooling coil at 13°C saturated state. Calculate the total cooling load, and room heat gain.

Solution Refer to Fig. 15.24. From the psychrometric chart the following conditions are noted:

Condition DBT°C

WBT°C

RH%

Sp. Hu.g w.v/kg g.a.

EnthalpykJ/kg d.a.

Outside 40 27 17.2 85Inside 25 50 10.0 50.8ADP 13 100 9.4 37.0

Condition of air entering the cooling coilω1 = 0.2 (17.2) + 0.8 (10) = 11.44 g w.v/kg d.a.

h1 = 0.2 (85) + 0.8 (50.8) = 57.64 kJ/kg d.a.t 1 = 0.2 (40) + 0.8 (25) = 28°CSpecific volume of air entering the cooling coil v1 = 0.869 m3/kg d.a.

Mass flow rate of air entering the cooling coil

ma1 = 4500

(60 )(0.869) = 86.31 kg d.a./s

Total cooling load Q = ma1(h1−h2) = 86.31 (57.64 – 37) = 1781.4 kWFresh air load QO = maO(hO−hi) = 0.2 (86.31) (85 – 50.8) = 590.4 kWRoom heat gain RTH = Q−QO = 1781.4 – 590.4 = 1191 kW

Note A point to be noted here is that the fresh air load is a significant part of the cooling load. In the present case, it is 33 per cent. In some applications such as operation theatres in hospitals where 100 per cent fresh air is taken, it is of overriding importance. In cinema halls and theatres also, the predominant load is due to occupancy and fresh air. In such cases, the peak load occurs when the outside wet bulb temperature is maximum. This usually occurs between 2 and 5 p.m.

Example 15.7 An air-conditioned space is maintained at 27°C DBT and 50 per cent RH. The ambient conditions are 40°C DBT and 27°C WBT. The space has a sensible heat gain of 14 kW. Air is supplied to the space at 7°C saturated.Calculate:(i) Mass of moist air supplies to the space in kg/h.(ii) Latent heat gain of space in kW.(iii) Cooling load of the air washer in kW if 30 per cent of the air supplied to the space is fresh, remainder being recirculated.

Solution Refer to Fig. 15.24. From the psychrometric chart.

Condition DBT°C

WBT°C

RH%

SP. Hu.g w.v/kg d.a.

EnthalpykJ/kg d.a.

Outside 40 27 17.2 85Inside 27 50 11.2 56.1Supply 7 100 6.2 23.0

(i) Mass of dry air supplied to space

ma = Q s

Cp Δt = 14(3600)

1.0216(27.7) = 2467 kg d.a./h

Ratio of moist air to dry air in supply air = (1 + ω2) = 1.0062 Mass of moist air supplied to space m = (1 + ω2) ma = 1.0062 (2467) = 2482 kg/h(ii) Latent heat gain of space

QL = ma (ω1−ω2¿h f gO =

2467(11.2−6.2)(3600 )(1000)

(2500) = 8.57 kW

(iii) For point 1 t 1 = 0.7 (27) + 0.3 (40) = 30.9°C On the line joining o to I, locate point 1 at 30.9°C. Then, from the psychrometric chart h1 = 64.9 kJ /kg d.a. Cooling load of the air washer

Q=ma(h1−h2) = 24673600 (64.9 – 23.0) = 28.71 kW

15.5.2 Summer Air-conditioning System with Ventilation Air-Bypass Factor X

In the case when the bypass factor of cooling and dehumidifying apparatus is not zero, it is evident from Fig. 15.24 that if the surface temperature is t S, viz., equal to room ADP, the leaving air state 2 will not be at S; it will be on the line joining 1 to S. Hence the supply air state will not lie on the room sensible heat factor line which is essential to satisfy the room sensible and latent heat load requirements. In such a case, which usually occurs in actual practice, it will be necessary to lower the apparatus dew point, i.e., the effective surface temperature of the air-conditioning apparatus, as shown in Fig. 15.25, in such a way that the leaving air state 2 lies on the RSHF line i-S, and also the new surface temperature t S

' or coil ADP is such that the following condition for the bypass factor X is satisfied.

X = t2− tS

'

t1− tS' =

ω2−ωS

ω1−ωS =

h2−hS'

h1−hS'

Fig. 15.25 Summer air-conditioning process with ventilation air and finite bypass factor

It will be seen that the effect of the bypass factor is to lower the ADP of the surface, and hence to decrease the coefficient of performance of the refrigerating machine. It is also seen that the position of the grand sensible heat factor line is changed. This is explained in greater detail in

Sec. 19.7. The supply air temperature t i is now increased to leaving air temperature t 2. Further, the dehumidified rise now is t i−t 2. The dehumidified air quantity may be calculated accordingly, which will be found to be more than that of the apparatus with a zero bypass factor. The recirculated air quantity can then be calculated by the difference of the dehumidified (supply) air and ventilation air quantities. Since ventilation air quantity is fixed according to requirement, this leads to a variation in the recirculated air quantity. The greater the BPF, the more the recirculated air quantity. As a result, point 1 after mixing, shifts closer towards i.

Example 15.8 A building has the following calculated cooling loads: RSH gain = 310 kW RLH gain = 100 kW The space is maintained at the following conditions: Room DBT = 25°C Room RH = 50% Outdoor air is at 28°C and 50% RH. And 10% by mass of air supplied to the building is outdoor air. If the air supplied to the space is not to be at a temperature lower than 18°C, find: (a) minimum amount of air supplied to space in m3/s.(b) Volume flow rates of return (recirculated room) air, exhaust air, and outdoor air.(c) State and volume flow rate of air entering the cooling coil.(d) Capacity, ADP, BPF and SHF of the cooling coil.

Solution Refer to Fig. 15.25. Room SHF is 0.756. Draw room SHF line. Its intersection with t = 18°C vertical gives supply air state point s which is the same as coil leaving air state point 2. From psychrometric chart hi = 50.5 kJ/kg d.a., h2 = hs = 41.2 kJ/kg d.a., vs = 0.836 m3/kg d.a. h0 = 92.0 kJ/kg d.a.(a) Supply air quantity and volume flow rate (minimum)

mas = RTHhi−hs

= 410

505−41.2 = 44.09 kg/s

Qvs = mas vs = (44.09) (0.836) = 36.86 m3/s(b) Quantity and volume flow rate of outdoor/exhaust air. ma0 = 0.1 mas = 0.1 (44.09) = 4.41 kg/s Qv0 = ma0 v0 = 4.41 (0.91) = 4.01 m3/s Quantity and volume flow rate of return air mai = mas−ma0 = 44.09 – 4.41 = 39.68 kg/s Qvi = mai vi = 39.68 (0.86) = 34.05 m3/s Here, v0 = 0.91 and vi = 0.86 m3/¿kg d.a. are the specific volumes of outdoor and indoor air respectively.(c) State of air entering cooling coil

t 1 = 0.9 t i + 0.1 t 0 = 0.9 (25) + 0.1 (38) = 26.3°C t 1

' = 19.2°C at 26.3°C DBT on the line joining i to o v1 = 0.865 from psychrometric chart h1 = 54.6 kJ/kg d.a. Volume flow rate of air entering the cooling coil Qv1 = mas v1 = 44.09 (0.865) = 38.14 m3/s(d) Refrigerating capacity of the coil Qcoil = GTH = mas (h1−h2) = 44.09 (54.6 – 41.2) = 591 kW Coil ADP is obtained by the intersection of the line joining 1 to 2 with the saturation curve. Thus t ADP = 9°C BPF and SHF of the coil

BPF = t2−t ADP

t1−t ADP =

18−926.3−9 = 0.52

GSHF = TSHGTH =

masC p(t 1−t 2)591

= 44.09 (1.0216 )(25−18)

591 = 0.533

Note Room SHF of 0.756 is normal. But GSHF of 0.533 is very low. This is due to high OALH; because of the outside air being very humid and hot, ω0 = 21 g/kg d.a. and h0 = 92.0 kJ/kg d.a. Low GSHF and high BPF of the coil chosen, viz., 0.52 has resulted in low ADP of 9°C. This would result in low evaporator temperature and high power consumption of the refrigerating machine. It would be better to choose a coil of low BPF. That would require a higher coil ADP and evaporator temperature, and hence would give better performance. But, then the supply air temperature would be lower than 18°C, and quantity of supply air would be reduced.

Example 15.9 The conditioning plant of a room consists of a fresh air intake, a cooling coil-followed by a mixing chamber or the cooled fresh air and recirculated room air, and a supply fan as shown in Fig. 15.26. The cooling coil handles all fresh air and has a BPF of 0.1. (See also Sec. 23.7). The ratio of fresh air to recirculated air is determined by modulating dampers. The other data is as follows: Inside conditions : DBT = 24°C, RH = 50% Outside conditions : DBT = 30°C, WBT = 23.3°C Heat gains : RSH = 14.7 kW, RLH = 3 kW Supply air quantity : 191 cmm Neglecting temperature changes in the fan and duct, determine:(i) DBT and moisture content of supply air.(ii) Mass flow rate of moist air supplied to room.(iii) DBT and moisture content of air leaving cooling coil.(iv) Load on the cooling coil.

Fig. 15.26 Conditioning plant for Example 15.9Solution (iii) The processes are shown in Fig. 15.27

RSHF = 14.7

14.7+3 = 14.717.7 = 0.831

This is the slope of line i-s-2. However, it is not possible to fix points s and 2 at this stage. But, we know that

BPF = Line2−SLine0−S = 0.1

Fig. 15.27 Processes in Example 15.9 Accordingly, point 2 can be fixed by trial and error on the RSHF line. However, the following construction is simpler and more accurate. Draw line o-i and extend it to A such that:

Line i−ALineo−A = 0.1

And then draw AS parallel to RSHF line i – s. Intersection with saturation curve gives coil ADP of 11.1°C at S. Join 0 – S. It cuts i – s extended at 2. Thus, we get condition of air leaving coil as: t 2 = 12.2°C, ω2 = 8.5 g/kg d.a., h2 = 33.8 kJ/kg d.a.(i) DBT and moisture content of supply air

mas = Qvs

vs =

19160v s

RTH = mas (hi−hs) = 19160v s

(48.8−hs) = 17.7

This equation can be solved by trial and error. Thus, point s can be located on RSHF line. It is found that:

vs = 0.856 m3/kg d.a., t s = 20°C, hs = 43.8 kJ/kg d.a., ωs = 9.1 g/kg d.a.(ii) Mass flow rate of moist air supplies to room

ms = 19160

0.856 (1 + 0.0091) = 3.753 kg/s

(iv) Fresh air through coil ma2 t 2 + (mas−ma2) t i = mas t 1

ma2 (12.2) + (191

0.856−ma2) (24) =

1910.856 (20)

ma2 = 76.1 kg/min = 1.268 kg/s = ma0

Load on cooling coil GTH = ma2 (h0−h2) = 1.268 (69.8 – 33.8) = 45.7 kW

Example 15.10 Given for a conditioned space: Room sensible heat gain = 20 kW Room latent heat gain = 5 kW Inside design conditions: 25°C DBT, 50% RH Bypass factor of the cooling coil = 0.1 The return air from the space is mixed with the outside air before entering the cooling coil in the ratio of 4 : 1 by weight. Determine:(i) Apparatus dew point.(ii) Condition of air leaving cooling coil.(iii) Dehumidified air quantity.(iv) Ventilation air mass and volume flow rates.(v) Total refrigeration load on the air conditioning plant.

Solution Refer to Fig. 15.25. From psychrometric chart

Condition DBT°C

WBT°C

RH%

Sp. Hu.g w.v/kg

d.a.

EnthalpykJ/kg d.a.

Sp. Vol.m3/kg d.a.

Outside 43 27.5 17.0 87.5 0.922Inside 25 50 10.0 50.8

Condition of air entering the cooling coil ω1 = 0.8 ωs + 0.2 ω0 = 0.8 (10) + 0.2 (17) = 11.4 g w.v./d.a. h1 = 0.8 (50.8) + 0.2 (87.5) = 598.1 kJ/kg d.a. t 1 = 0.8 (25) + 0.2 (43) = 28.6°C (i) and (ii). Both parts have to be worked out together. There are two methods. One method is to draw the RSHF line and then draw a line from 1 to S on the saturation curve so that

(S – 2)/(1 – 2) are given in the ratio of 1 : 9. Another method is to do the same thing using calculations as given below. Ratio of room sensible and latent heats

RSHRLH =

0.0204(25−t 2)50(0.01−ω2)

= 205

(I) Relations for bypass factor

t2−t ADP

t1−t ADP =

t2−t ADP

28.6−t ADP = 0.1

(II)

ω2−ωADP

ω1−ωADP =

ω2−ωADP

0.0114−ω ADP = 0.1

(III) Solving Eqs (I), (II) and (III) by iteration for t ADP, t 2 and ω2, we obtain t ADP = 11.8°C (Corresponding ω ADP = 8.6 g w.v./kg d.a.) t 2 = 13.5°C ω2 = 0.0089 kg w.v./kg d.a.

(iii) Dehumidified air quantity

(cmm)d = RSH

0.0204(t i−t 2) =

200.0204(25−13.5) = 85.25

(iv) Specific volume of supply air v2 = 0.822 m3/kg d.a. Mass flow rate of supply air

mas = (cmm)d

(60 ) v2 =

85.25(60 )(0.822) = 1.729 kg d.a./s

Mass flow rate of fresh air ma0 = 0.2 mas = 0.2 (1.729) = 0.346 kg d.a./s Volume flow rate of fresh air Qv0 = ma0 v0 (60) = (0.346) (0.922) (60) = 19.12 cmm(v) Outside air total heat OATH = ma0 (hi−h0) = 0.346 (87.5 – 50.8) = 12.7 kW Total refrigeration load on the air conditioning plant GTH = RTH + OATH = (20 +5) + 12.7 = 37.7 kW

15.6 WINTER AIR CONDITIONING

In winter, the building sensible heat losses are partially compensated by the solar heat gains and the internal heat gains such as those from occupancy, lighting, etc. Similarly, the latent heat loss

due to low outside air humidity is more or less offset by the latent heat gains from occupancy. Thus in winter, the heating load is likely to be less than the cooling load in summer. However, the actual situation both in summer and winter depends on the swing of the outside temperature and humidity with respect to the inside conditions. Further, certain sensible heat gains (negative loads) such as the solar heat may not be present at the time of peak load, and hence they are not counted. On the other hand, latent heat gains from occupancy, etc., are always present and should be taken into account. As a result, the design heating load for winter air conditioning is present dominantly sensible. In general, the process in the conditioning apparatus for winter air conditioning for comfort involves heating and humidifying. Two of the typical process combinations are: (i) Preheating the air with steam or hot water in a coil followed by adiabatic saturation and reheat. (ii) Heating and humidifying air in an air washer with pumped recirculation and external heating of water followed by reheat. The processes for the two systems are shown in Fig. 15.28. The first system with preheating and adiabatic saturation follows processes 1 – 2 and 2 – 3 respectively. The second system replaces the two processes with heated water spray in the air washer and the process line is 1 – 3. The leaving air state 3 from the air may be affected by the saturation efficiency. The reheating process 3 – s is common to both. The supply air states should lie on the room sensible heat factor line. It is, therefore, determined by the RSHF and by the choice of supply air rate which is usually known from summer air-conditioning calculations.

Fig. 15.28 Winter air-conditioning processes

Example 15.11 In an industrial application for winter air conditioning, an air washer is used with heated water spray followed by a reheater. The room sensible heat factor may be taken as unity. The design conditions are: Outside: 0°C DBT and dry Inside: 22°C DBT and 50% RH Room heat loss: 703 kW The following quantities are known from the summer design. Ventilation air 1600 cmm Supply air 2800 cmm Spray water quantity 500 kg/min

The air washer saturation efficiency is 90 per cent. The make-up water is available at 20°C. Calculate:(i) The supply air condition to space.(ii) The entering and leaving air conditions at the spray chamber.(iii) The entering and leaving spray water temperatures.(iv) The heat added to the spray water.(v) The reheat, if necessary.

Solution Refer to Fig. 15.29(i) Supply air temperature

t s = RSH

0.0204(cmm)s + t i =

7030.0204(2800) + 22 = 34.2°C

A RSHF = 1 line from the room condition i can be drawn as shown in Fig. 15.29. Intersection with the 34.2°C DBT line locates the supply air state. The supply air WBT is found to be 19.6°C, and its specific humidity ωs to be 0.0082 kg/kg d.a.

Fig. 15.29 Figure for Example 15.11

(ii) Entering air conditions (calculate from the volume flow rates).

t 1 = (cmm)0 t 0+(cmm)i ti

(cmm)i

= (1600 ) ( 0 )+(2800−1600 )(22)

2800 = 9.43°C

Similarly, ω1 = (1600 ) ( 0 )+ (1200 )(0.0082)

2800 = 0.0035 kg w.v./kg d.a.

Wet bulb temperature of entering air (from psychrometric chart) t 1

' = 4.8°C Specific humidity of leaving air ω2 = ωs = ωi = 0.0082 kg w.v./kg d.a. Expression for saturation or humidifying efficiency

ηH = ω2−ω1

ωS−ω1

0.9 =0.0083−0.0035

ωS−0.0035which gives the specific humidity at the wetted-surface temperature t S as ωS = 0.00873 kg w.v./kg d.a. From the psychrometric chart t S = 11.8°C Dry bulb temperature of leaving air t 2 = ηH (t S−t1) + t 1 = 0.9 (11.8 – 9.43) + 9.43 = 11.6°C Wet bulb temperature of leaving air (from the psychrometric chart) t 2

' = 11.5°C(iii) The temperature of the leaving spray water and the wet bulb temperature of the leaving air may be taken to be the same. Hence, the leaving spray water temperaturetw2 = 11.5°C. Energy balance of the air washer

(cmm)s

vs (h2−h1) = mw C pw (tw1−tw2)

28000.88 (33.0 – 18.2) = (500) (4.187) (tw1−11.5)

tw1 = 34°(iv) Make-up water

∆ mw=¿ (cmm)s

vs (ω2−ω1) =

28000.88 (0.0082 – 0.0035) = 14.95 kg/min

Heat added to make-up water

Q1 = 14.95

60 (4.187) (34 – 20) = 14.61 kW

Heat added to raise temperature of spray water

Q2 = 50060 (4.187) (34 – 11.5) = 785.1 kW

Heat added to spray water Q = Q1+Q2 = 14.61 + 785.1 = 799.71 kW(v) Reheat = 0.0204 (cmm)s (t s−t 2) = 0.0204 (2800) (34.2 – 11.6) = 1290.9 kW