# Design of an Air Curtain Device

An air curtain is to be used in the doorway of an industrial building to prevent the penetration of cold air into the building. The task is to dimension this air curtain device for the following climatic conditions:

Case I :

• Outdoor air temperature 0out = -12 °C

• Wind velocity vw = 0 m/s

Case 2:

• Outdoor air temperature 0out = -5 °C

• Wind velocity vw = 5 m/s

The wind velocities are local values at the height of the doorway, so no addi­tional wind sheltering coefficients or height corrections are necessary. A solu­tion is sought for which both situations above can be handled with the same air curtain device by just changing the fan speed and the heating power of the air curtain.

7.7.5.2 Data

The building (see Fig. 7.95) is 12 m high and has three large openings. The doorway for which the air curtain is designed is on the ground level, has dimensions 4 m (height) x 6 m (width), and is intended primarily for vehicular traffic.

 A

 N  Gate Opening 1 Opening 2 Width L[m) 6 4 2 Height H[m] 4 2 1 Area A[m*] 24 8 2 Center height above ground Bg [m] 2 9 5.5 Pressure coefficient Cv H 0.7 -0.5 -0.3 Discharge coefficient ^ [-] 0.25 0.8 0.8

There are two large openings in the upper part of the building: L701, which is 4 m wide x 2 m high, and U02, which is 2 m wide x 1 m high. The lower — level of 5 m. Otherwise the building envelope is assumed to be airtight. This information is summarized in Table 7.29.

The indoor air temperature and humidity are assumed to be uniform:

• Temperature: 0^ = 20 °C

• Humidity: 4> = 60%

The general ventilation airflows are in balance.

The unbalanced local exhausts take their makeup air through the building openings:

4v, m = 4 m3/s> <7m, m = 4-8 kg/s.

Calculation is made in accordance with Section 7.8.2, Infiltration and Ex­filtration.

Mass balance of the airflows through the building envelope:

Tfm, g 1 4m. 2 4m. m ^ > i ‘ -223)

Where

Flow through opening i: qmj = (jL,;A;j~2Ap(~|1/2 [kg/s] (7.224)

Pressure difference across opening i: Apt = p{ out — p, in (7.225)

Pressure inside building: in = px = const (7.226)

Air densities:

Outdoor air density: pout, i = 1-35 kg/m3

Indoor air density: pin = 1.2 kg/m3

Reference level 0 is located in the center of the gate opening, i. e., 2 m above ground level.

Cose I: Temperature -12 °C, Calm Conditions

Design Pressures

1, out — ®

APl, out = 0 + {bx — /?2)g(Pout, 1 “ Pm, l)

= (9 m-5.5 m) • 9.81 m/s2(1.35 kg/m3- 1.2 kg/m3)

= 5.15 Pa

APg, out = 0 + (hx-hg)g(pout> i-p1I1;i)

= (9 m-2 m) • 9.81 m/s2(1.35kg/m3 — 1.2kg/m3)

= 10.3 Pa

The design airflows are found from Eq. (7.224):

Qm g = 0.25 • 24 m2[2(10.3 Pa — px) ■ 1.35 kg/m3]1/2 ‘ = 9.86(10.3 Pa — px.)1/2[kg/s]

QmA = 0.8 ■ %[2{px — 0.675) ■ 1.2 kg/m3F2 = 9.91 (px-5.15)i/2[kg/s]

Qm>2 = 0.8 ■ 22{px — 0.0) • 1.2 kg/m3]1/2 = 2.48(pjr)^2

Qm, m = 4.8 kg/s

The mass balance equation (7.223) now gives 9.86(10.3 — px)wl = 9.91(px — 5.15)1/2 + 2.48(px)l/1 + 4.8px = 6.05 Pa.

The design differential pressure is

Apgl = 10.3 Pa-6.05 Pa = 4.25 Pa.

Case 2: Temperature -5 °C, Wind 5 mis

Design pressures

&Pl, out = 0

APl, out = 0 + (b1- b2)g(poutii — Pin, l) + (Cf,’2 ~ cp,)Pw

= (9-5.5)-9.81 -(1.32-1.2)+ [-0.3-(-0.5)]- 16.5 = 7.42 Pa

APg, out ^ (^1 “ bg)g(pomA — Pin? i) + (Cp, g ~~ Cp^ypiv

= (9 — 2) • 9.81 — (1.32 — 1.2) + [0.7-(-0.5)] • 16.5 = 28.04 Pa

The mass balance equation (7.223) is

9.86(28.04 — px)1/2 = 9.91 (p* — 7.42)1/2 + 2.48(px)1/2 + 4.8 px~ 13.2 Pa.

The design differential pressure is

Apg = 28.04 Pa — 13.2 Pa = 14.84 Pa.

7.7.5.4 Calculation of the Parameters of the Air Curtain

We choose the ratio between the gate area and the discharge opening area,

Which implies an air curtain discharge aperture of

As = 1.2 m2.

The aperture height equals the height of the gate, 4 meters, which yields the width of the air outlet slot:

Bs = AJ4 m = 0.2 m.

The discharge angle of the air curtain jet is

A = 30°(i. e., sina = 0.5).

The efficiency factor E, according to Table 7.27 is

E = 0.3.

Based on the given value f the values of the factors mi and m2 are deter­mined from Table 7.28:

My = 2.9 m1 = -1.3 .

To determine the supply temperature of the curtain, 0O, and the thermal capacity P, we use

^mix ^in ‘

The parameters of the air curtain can now be determined.

The initial discharge velocity is calculated according to formula (7.217):

Version a

Pq = [(4.25 • 20)/(2 ■ 1.05 • 1.2 • 0.3)]0.5 = 10.6 m/s

Version b

VQ = [(14.84 • 20)/(2 • 1.05 • 1.2 ■ 0.3)]0.5 = 19.8 m/s

The mass flow of the air supplied by the air curtain is determined accord­ing to formula (7.218):

Version a qmfi = 1.2 • 10.6- 1.2 = 15.26 kg/s Version b qmfi = 1.2-19.8 ■ 1.2 = 28.51 kg/s

The discharge temperature of the air curtain is determined from formula (7.221):

Version a

0O = — 12 + 2.9(20 + 12) — 1.3(20 + 12) = 39.2 °C Version b

The heat capacity of the air curtain is determined according to formula (7.222): ‘

Version a

P = 1.01 • 103 • 15.26(39.2 — 20) = 295.9 ■ 103 W Version b P = 1.01 • 103 • 28.51(35-20) = 431.9 ■ 103 W

Thus, version b is more unfavorable, as the differential pressures AP are the greatest. The selection of air curtain equipment is made for version b.

The fan of the air curtain is to be provided with a device regulating the supply airflow, for example, by frequency transformer, multispeed motor, etc. The supply airflow is to change according to the dependence qv\$ ~ ApQ-s, which enables, in particular, provision of the necessary efficiency for version a.

As the temperature of the supplied air increases in case of decreasing outdoor temperature, the heater of the air curtain selected for version b is to be tested ac­cording to the calculation designed for the lowest outdoor temperature (version a). The heater is to be provided with devices for automatic regulation of the tem­perature of the supplied air according to the dependence G0 ~ 0mit.