# State Changes of Humid Air

Now we will consider a balance borderline of the system presented in Fig. 4.14. The system can be any part of the air surrounding the process device. If an air-handling application is considered, the balance can be calculated over the inner air of a room, an office, or an industrial hall, for example.

The energy balance of the system, consisting of the area inside the balance borderline, is in a stationary situation:

<j>- Wm = + (4.139)

+ ihvl) + (mh2bh2 — thhlbhi)]

Where 4> is the net heat power received by the system, Wm is the net work power to the environment preferred by the system, mv is the water flow?

(1 = inflow, 2 = outflow), thj is the ice flow, and mh is the separate clean

Steam flow not included in the air flows. The steam flows included in the

Thjоhjп m,hu mhhh Trnhki M, hk2 —t-

Mah,2 ntvihvi mh^hhi FIGURE 4.14 The energy-humidity borderline.

Humid air flow are mxj and rix2 and they have an effect on the energy bal­ance through terms hkl and hkl.

Correspondingly, the humidity or water balance is

M,(x2-X| ) = (thj] — thji) + (mvl — mvl) + (mh l — mh2) (4.140)

In many cases the incoming and outgoing air flows can consist of vari­

Ous air flows in different states (temperature, humidity), which must be treated separately. This means that the air enthalpy flow must be divided into corresponding parts.

Example 8

Mixing of two air flows. As illustrated in Fig 4.15a, the energy balance is mnhk +rni2hk2 = ml3hk3 (4.141)

And the humidity balance is

MnX] +mi2x 2 = mi3X) (4.142)

The dry air balance is

»7^ + riiji = (4.143)

From Eqs. (4,141)-(4.143) it follows that

Hk3-hk2 = %3-x2 = mil

Hk~hki *i~*3 mi2’

Which shows that the mixing point is on the line connecting points 1 and 2 in Fig 4.156.

(4.144)

Example 9

Heating of an air flow. From Eq. (4.139) it follows that

<t> = mti(hk2-hk i) (4.145) L ci

 Mn hkl

 Ma b *•



 (b) FIGURE 4.15 The mixing point in a Mollier diagram. If the supersaturated area (c) is considered, the state of the air is driven to point 4 and (x3 — x4) kg H20/kg d. a. of water is condensed in the mixing chamber.

Th,(x2-xi) = 0,

So x2 — x. When air is heated, the state point moves up along the constant humidity line.

Cooling of an airflow. From Eq. (4.139) it follows that

<b = fhj(hk2 — bk,) + riiv2hv2 , (4.146)

Where now <p < 0. When air is cooled, some water can be condensed. This depends on the surface temperature of the cooling coil, and therefore we have the term n,2hu2 in the equation above. From Eq. (4.140) we obtain the water balance FIGURE 4.16 Air state change in a cooling coil. If the surface temperature ft is under the dew point 0k, there will be condensation. If f)p > 9k cooling takes place along the constant humidity line x. Mj(x2 -*i) = — fhv2 , so the final humidity x2 — Xj. The air cooling process is illustrated in Fig. 4.16.

When the airflow meets a surface whose temperature is lower than the dewpoint, water vapor from the air condenses on the surface of the cooling coil. If all air comes into contact with the cold surface, the state of the air after the process will be at point 3. Some air always escapes the cold surface, and therefore the state of air after contact with the coil is a mixture of saturated air (3) and escaped air (1). The mixing point (2) lies on the line connecting points and 3, as shown in Example 8. The nearer point 2 is to point 3, the more effective is the cooling coil.

Adding steam to the air. From Eqs. (4.139) and (4.140) it follows that mhhh = fhi(hk2~hkx)

Mh = rhi(x2-x),

Where mh is the added steam flow and hk its enthalpy. From these equations it follows that

HMi~hx — hh. {4.147)

X2-Xi ‘

In differential form Eq. (4.147) is

D-ji = h. (4.148)

On the other hand, differentiating Eq. (4.90b), hk = cpi\$ + x(cph9 + lho), with respect to the variables 0 and x, we obtain

Dhk = (cpi + xc, ph)d0 + (0cph + lh0)dx,

And substituting this in Eq. (4.148) we get

D0 _ hj,, — (//;q + ,) Hb — hfj(9y 149)

Dx cpj + xcpf, cpi + xcp)j

Where hh(0) = lho + cpf,0 is the steam enthalpy at air temperature 0.

From Eq. (4.149) we notice that if the temperature of the steam added to

Air is below the temperature of the air, the air will cool down and

D0/dx < 0. If the steam temperature is higher than the air temperature, the temperature of air will rise ((d0)/dx > 0).

The room temperature is required to be 20 °C and the relative humidity (p — 50%. The net heat load developing in the room is 2.45 kW and the net steam flow 1.5 • 10~3 kg/s. What should the inlet air temperature and humid­ity be when the inlet air is (a) m,: = 0.3 kg/s and (b) m, = 0.6 kg/s

P’h (20 °C) = 0.02337 ph = 0.5 • 0.02337 bar = 0.01169 bar,

Room air humidity x — 0.6220 ■ = 0.00736 .

1.0 — 0.01169

When the inlet air is well mixed with the room air, the humidity of the

Outlet air (2) is the same as the room air humidity, so x2 — 0.00736, and its

Temperature is the same as the temperature of the room air, 0 = 20 °C.

The enthalpy of the outlet air k^x = 1.006 • 20 + 0,00736 • 2501 + 1.85 • 20 = 38.8 kj/kg. The enthalpy of inlet air hkl and its humidity xi are determined by the energy and humidity balances:

= 4> + mhhh

Mi(x2-xx) = mh

The net heat power also includes the enthalpy flow, mhh),. Then

4> + mhhf, = 2.45 kW

And therefore

(a) bk2 -hk} = ^- = 8.2 kj/kg —> hkl = 38.8 — 8.2 = 30.6 kj/kg

(b) bk2-hkl = 4.1 kj/kg -> bki = 38.8-4.1 = 34.7 kj/kg

The arising steam flow’ is mh = 1.53 • 10~3 kg/s, and thus

*,-*!= L510 ■■ = 0.0051

(a) — 1 0.3

X, = 0.00736 — 0.00255 = 0.00481

*■ r “ 1-53 ■ 10 _ a

(b) *2~Xl 06 0.00255

Xt = 0.00736 — 0.00255 = 0.00481

The corresponding air temperatures can be calculated with the equation

0 — ^k ~

Cpi xcph

 (a) X = 0.00226 bk = 30.6 kj/kg

A = 30.6-0.00226-2501 =?44 oC 1.006 + 0.00226- 1.85

(b) x = 0.00481 hk = 34.7 kj/kg 0 = 22.2 °C The result of Example 12 is illustrated in Fig. 4.17.