Example of Cooling Tower Dimensioning

Cooling towers are commonly used for water cooling, but they can also be used for heat recovery from outlet air. If the water temperature is higher than the dew — point of the air, water will cool in the tower. Cooling is caused by vaporization on the surface of the water drops. The vaporization energy comes from the inner en­ergy of the water and in a certain phase, when the water temperature is lower than the dry bulb temperature of the air, also from the airflow. When the water temperature drops to near the air wet bulb temperature at the observation point.

SYMBOLS

H = specific enthalpy, kj/kg dry air

X = absolute humidity, kg water/kg dry air

0 = dry bulb temperature of air, °C

Tf., = wet bulb temperature of air, °C

P* = quantity ot dry air.

Kg dry air/m moist air

Ah/Ax = proccss direction

(p= relative humidity, %

= partial pressure ot water vapor, mbar

The diagram applies to atmospheric

Pressure of 1000 mbar

S500
»000	7000 6500 6000
« 50000 20000	10000
0.020
0.010
0.005	0.0 5
0.010 0.015 —► a: 0.0201 ^ ^i x_ — i v _ ^ a
A/;/Ax
-5000 -2000 -1000

"о О

FIGURE 4.1 8 Water drop cooling in the air flow. Point (I) represents the air state surrounding the water drop (2).

Water will not cool further even though there is still water vaporization from the drop surface. This is due to the fact that the temperature difference between the air dry bulb temperature and drop surface is so large that the energy needed for vaporization comes convectively from the air. This is illustrated in Fig. 4.18.

If the air dewpoint is higher than the water temperature (or more accu­rately, the surface temperature of the drops), water vapor condenses from the air on the surface of the water drops. Now the water warms up and the air cools down and at the same time dries up; in other words, the cooling tower recovers heat from the outlet air. We will now consider the operation of a cooling tower more closely with the notations of Fig. 4.19.

The energy balance for a distance dL is

-mvcpv dOv = m’h l(6v) dA — a(6- 0V) dA.

From Eqs. (4.112), (4.113), and (4.121), we have, when Le = 1, for vaporization ri% = (a/cp)(x'(8v) — x), and substituting this in the equation above, we get

-m„cpv ^ d0t, = -[cp(6-0v) + {x-x'(0v))H0v)] dA.

Droplet _ separator

Dl

R -….. r T……….. — ‘ r 1 …….. t….. 1….. -..

Nij mv Tiki °v)

9„,hk Bv + dd^ hfr + dhk

Guiding Plates

Uuuuu

External Process

Rnh, em

On the other hand, while [Eq. (4.119)] cp — (pi/p){cpi + xcpj,) s cpi + xcph and /0 = l(6v), we obtain, according to the definition of bk [Eq. (4.90b)], an approximate value of

Cp(fl-0„) + (x-x'(0„))/(e„) = hk-h’k (0,) ,

And thus we have

(4.150)

D<}> = — mvcpv dev = f-(h’k (0V) — hk) dA.

H

-P

Alternatively, we can write the energy balance with the help of the en­thalpy flows as

-mt dhk = rhvcpv dtv-dmvcpvt,

Where dri = rhj dx, where dx is the humidity change of the airflow. Then we can write

-mvcpv dBv = — m,(dbk — cpvBv dx).

While dbk = {cpi+ cph) d6 + l0 dx and cpvQv«l0, to a good accuracy dhk — Cpv6v dx = dbk and therefore

Dip = — mlrcpv dOv = —mi dh Considering Eq. (4.150), we have

(4.151)

Tit dhk = jr(hk-h’k (6V)) dA.

Lti

P

From Eq. (4.151) we see that in a dampening process the state of air tends to change toward the saturated air state corresponding to the surface temperature of the water. We will now solve the Eq. (4.151) approximately. First we state that

Dhk = d(bk-b’k (6v)) + dh’k (flj = d{hk-h’k + —dhk

Using the approximation

Db’k (6„) _b’t (e, A)-H (0v2) dhk bkl-bkl

D{bk-b’k(Qv)) _ {bki-bkl)d{hk-h’k{Qv))

And substituting this in Eq. (4.151), we have

Nti(hki-hk2)= — A bkln, CP

We get

Where the definition Ahk ln is

(bki ~~ b’k (Ovi)) ~ (hki ~ h’k (0V2))

Mkln =

In^i — b’k (Ai)

Bki~h’k (^1/2)

And

A = AkApL

The quantity aAp is defined separately for each type of cooling tower. It depends on many variables: jet pressure, jet division, airflow velocity, and oth­ers. The total energy balance for a cooling tower is (see Fig. 4.19)

™i(hk2 — hkl)= mhCf>v0vo + ™hCt, v(Gv2 — <W) 5

Where 0„o is the temperature of the excess feed water. The need for excess feed water represents the rate of vaporization n%. Usually the term mhcpvOvo has minor significance (vaporization rate rh/, corresponds usually to just a small percentage of the water flow n, ), so on the basis of Eq. (4.152) we get an equation for the cooling power <pv:

4>v = rhvCpv(Qvi ~ ~ ln (4.153)

P

We can dimension a cooling tower according to the equation above.

Example 13

A paper industry’s cooling tower recovers heat from the outlet air. This

Situation is represented by the following values:

• Inlet air enthalpy hki = 293 kj/kg

• Outlet air enthalpy hk2 = 208 kj/kg (saturated 44.3 °C air)

• Outlet water temperature 6vl = 40 °C

• Inlet water temperature 0v2 — 5.0 °C

• Water flow mv = 44 kg/s

• Air flow riti = 44 • 4.186 • (40 — 5)/(293 -208) = 75.8 kg/s

• Cross-sectional area of the cooling tower Ak =31 m2 and height L=3m

It is discovered that in the cooling tower the water moving down­ward from the jets changes its direction to upward after drop formation. There is an effective heat transfer process when the drops move upward: heat transfers from the outlet air to the drops through convection and condensation.

Drops collide with the drop separator and drain down to the lower part of the tower. These drops are large, so their total surface area is small and insig­nificant. The effective heat transfer process takes place when the drops move with the air flow, so this arrangement has to be treated as a parallel flow heat transfer.

(a) Calculate aAp. According to the parallel flow principle, the situation is as shown in Fig. 4.20.

Bk2 = 208kJ/kg 0vl = +40 °C

Hkl^ 293 kj/kg

0„, = +5°C

FIGURE 4.20 Heat transfer according to the parallel flow principle.

First we calculate the logarithmic enthalpy difference:

And

B’k (8ul) = b’k (5°C) = 18 kj/kg b’k (0,2) = h’k (40°C) = 168 kj/kg

A^kin = ^^1^(201^168) = 12L9 kJ/kg ln208- 168

The specific heat capacity of humid air calculated per kilogram of dry air is cpK = cp, + xcph = 1.006 4- 08 ■ 1.85 = 1.154 kj/kg °C cp = *cpK = 1.154( kj/kg °C)

From Eq. (4.153),

01A = ~&h^^cpv^tvl~tvl’) = ~rTn>’ 44-0 ’4-186 ’ (5~40)

= 61.0 kW/°C On the other hand, AAp Ak L, and therefore

AAp =………… = °-656 kW/°C m3

(b) The same cooling tower is to function as a water cooler. Let the out­door air be 24 °C, <p = 50%, and the air flow 100 kg/s (dry air). The water in­let temperature is 24 °C and the water flow 30 kg/s. What is the cooling capacity if we assume that a ~ Jv; and Ap ~ mv, or aAp — k Jv^ and also that the active cooling process is parallel-flow heat transfer?

For case (a) was 44 kg/s and the air flow velocity = (75.5/1.0)/31 = 2.44 m/s. Therefore the heating capacity (or cooling capacity, depending on the sign) can be presented as

AAf. AuL k [Ujrit.,

4> = — Jr-— Abkin = ~~r AkLAkkXn,

Lp CP

And solving factor k out of this equation,

K = ^ = 1-154 (44.0-4.186 (40-5)) = 9 «.«. 10_3

JU^AkLAh^ JIM • 44.0 • 31.0 • 3 • 121.9 ‘ ‘

Then the capacity tf> is

(j) — — ■ AkL ■ Jv, mvAban~ ^ 154 31.3 • Jv^th^Ah^n

And after calculations,

</> = 0-770 ■ Jvt ■ mv ■ AbUn

Where Vj is in m/s, mv in kg/s, Abkln in kj/kg, and tfi is in kW. It was found that mv — 30 kg/s and w^=100 kg/s, and now vt = (100/1.2)/31 = 2.69 m/s. Thus

$ = 0.770 ■ JT69 ■ 30.0 ■ Abkln = 37.9 • Abkin

The cooling tower functioning in an air cooling situation is illustrated in Fig. 4.21. Because of the logarithmic enthalpy difference the solution must be iterated;

1. Guess: 0v2 = 20°C; h’k (20°C) = 57.9 kj/kg

4>= 30 ■ 4.186 (24-20) = 502 kW hk,-hkl = 502/100 = 5.02 kj/kg; bkl = 53.3 kf/kg ^ = 211,^31.1,8^

Ln 57.9-53.3 4> = 37.9 11.8 = 446 kW bk2 = 24 — 446/30 • 4.186 = 20.4°C

2. We choose 6v2 = (20.0 + 20.4)/2 = 20.2°C; h’k (20.2 °C) = 58.6 kj/kg.

4> = 30 • 4.186 • (24 — 20.2) = 477 kW hk2 = 48.3 + 477/100 = 53.1 kj/kg

A*tln~244-(5M zm12.6 kj/kg

‘"SsJHn

4> = 37.9 • 12.6 = 477 kW

Now 0u2 = 20.2°C and cooling capacity <t> — 477 kW.

When the cooling tower was operating as a heat recovery device, its ca­pacity was considerably higher because of the high temperatures and humidi­ties. In case (a) we had

</> = 44 • (40 — 5) ■ 4.186 = 6450 kW.

FIGURE 4.21 Water cooling in a cooling tower, which operates according to the parallel flow principle; in other words, the water spray turns in the direction of the air flow.

If we assume that the outlet air is saturated, the air state change process is as presented in Fig. 4.22. The exact determination of the air humidity at the end of the process would demand separate mass and heat transfer examinations.

Posted in INDUSTRIAL VENTILATION DESIGN GUIDEBOOK