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Chapter 3 Moist Air Properties and Conditioning Processes

Chapter 3 Moist Air Properties and Conditioning Processes

Chapter 3 Moist Air Properties and Conditioning Processes

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..M.. = i = 1150.4 BtuJJbm<br />

t.w \I'<br />

3-5 ClaSSIc Moi ,;, <strong>Air</strong> <strong>Processes</strong> 63<br />

where iw is read from Table A-la. A parallel line is drawn from state 2 as shown in<br />

Fig. 3-7. State X is determi ned by the intersection on lines 1 - X <strong>and</strong> X - 2. The heal<br />

Iransfer rate is then given by<br />

where<br />

In = 0(60) = 1600 60 = 7296Ibma/hr<br />

" vI J3.16<br />

<strong>and</strong> i I <strong>and</strong> it' read from Chart 1 tI, are L6.8 <strong>and</strong> 29. 2 BtuJlbma, respectivel y. Then<br />

if = 7296(29.2 -16.8) -::::: 90,500Btu/hr<br />

The mass flow rate of the waler vapor is given by<br />

,ill' = ,j1a(W2 - W I )<br />

where W 2 <strong>and</strong> W I are read from Chart La as 0.0193 <strong>and</strong> O.00221bmvllbma. respectively.<br />

Then<br />

,hI. = 7296(0.0 193 - 0.0022) = 1251bmv I hr<br />

Adiabatic Mixing of Two Streams of <strong>Moist</strong> <strong>Air</strong><br />

The mixing of airstreams is quite common in air-conditioning systems. The mixing<br />

usually occurs under steady. adiabatic flow conditions. Figure 3-9 illustrates the mixing<br />

of two airstreams. An energy balance gi ves<br />

The milss balance on lhe dry air is<br />

<strong>and</strong> the mass balance 01) the water vapor is<br />

Iha l l{) + nZa2 W 2 = li1a3 W]<br />

Combining Eqs. 3-41. 3-42. <strong>and</strong> 3-43 <strong>and</strong> eliminating m ll3 yields<br />

i ) - i,<br />

----- = W2 - IV3 = m al<br />

",::, - IVI III a2<br />

Figure 3-9 Schematic of the adiabatic lllixing of lWO airstreams.<br />

(3-4 1 )<br />

(3-42)<br />

(3-43)<br />

(3-44)


3-6 Space <strong>Air</strong> <strong>Conditioning</strong>- Design Condition, 71<br />

In Example 3-9 the outdoor air was hot <strong>and</strong> humid. This is not always the caf>C.<br />

<strong>and</strong> statc 0 (outdoor ai r) can be almost anywhere on Chart la , For example, the sou(h­<br />

\vcstern part of the United Stales is hot <strong>and</strong> dry during the summer, <strong>and</strong> evaporative<br />

cooling can ofte n be used to advantage under these conditions. A simple system of<br />

this type is shown in Fig. 3-15. The dry outdoor air flows through an adiabatic spray<br />

chamber <strong>and</strong> is cooled <strong>and</strong> humidified. An energy balance on the spray chamber will<br />

show that the enthalpies io <strong>and</strong> i j are equal: therefore. the process is as shown in Fig.<br />

3-16. Ideally the cooling process terminates at the space condition line. The air then<br />

flows through the space <strong>and</strong> is exhausted. Large quantities of ai r are required , <strong>and</strong> this<br />

system is nO! satisfactory \vhere the outdoor relative humidity is high. If Wo is too<br />

high. the process 0-1 cannot intersect the condition line.<br />

Evaporat ive cool ing can be combined with a conventional system as shown in Fig ,<br />

3- 17 when outdoor conditions are suitable. There arc a l1umber of possibilities. First,<br />

Cond itioned<br />

space<br />

Figure 3-1 5 A simple evaporativc cooling system,<br />

Figurt> 3·16 Psychromctric diagram for the cvaporative cooling system of Fig , 3-1 S,<br />

Exhaust<br />

®<br />

Evaporative qc<br />

cooler<br />

Figure 3-17 Combination evaporatlve <strong>and</strong> regular cooli ng system<br />

@


3-6 Space <strong>Air</strong> Condllloning-Desig1J Conditions 73<br />

225.000<br />

SHF = ----- = 0.80<br />

225,000 + 56,250<br />

The state of the supply air lies 01) a line drawn through state point 3 parallel 10 (he<br />

SHF :::: 0.8 line on the protractor of Char! la. Figure 3-20 shows th is construction.<br />

State 2 is located at 120 F dry bulb <strong>and</strong> the intersection of this line. An energy balance<br />

on the space gives<br />

or<br />

'i = Jil a2 (i Z - i3)<br />

From Chart la, i2 :::: 42 Btu/lbma. i3 = 28. 2 Btu/lbma, <strong>and</strong><br />

. q 281,250 20400 1b /1<br />

m aZ = -.--. = ::::, nla If<br />

12 - l3 42 - 28.2<br />

From Chart la, v 2 = 14.89 ft 3 /lbma, <strong>and</strong><br />

01 :::: 20.400 x 14.89 = 5060 cfrn<br />

- 60<br />

To find the conditions at state 1. the mixing process must be considered. A mass balance<br />

on the mixing section yields<br />

or<br />

m (l4 = m a2 - lIloO<br />

r;l :::: 0 0 <strong>and</strong> Vo :::: 12.54 ft 3 1lbma<br />

(I Q v o<br />

Figure 3·20 Psychrornelric processes for Example 3- 10.

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