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Handbook of air conditioning and refrigeration / Shan K

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22.34 CHAPTER TWENTY-TWO<br />

�a � <strong>air</strong> density, lb/ft3 (kg/m 3 )<br />

� <strong>air</strong> supply volume flow rate, cfm (m3 /s)<br />

Ao � cross-sectional area <strong>of</strong> free <strong>air</strong> passage in st<strong>air</strong>well, ft2 (m2 )<br />

<strong>and</strong> the ratio R <strong>of</strong> the area <strong>of</strong> free flow passage (orifice) to the interior cross-sectional area <strong>of</strong> the<br />

st<strong>air</strong>well As, in ft2 (m2 V˙<br />

), can be calculated as<br />

R � Ao/As (22.13)<br />

In Eq. (22.12), De, in ft (m), indicates the circular equivalent <strong>of</strong> the st<strong>air</strong>well, <strong>and</strong> it can be calculated as<br />

De � (22.14)<br />

where Ps � perimeter <strong>of</strong> the cross-sectional area <strong>of</strong> the st<strong>air</strong>well, ft (m).<br />

The pressure drop coefficient is K. It is mainly a function <strong>of</strong> the configuration <strong>of</strong> the st<strong>air</strong>well<br />

<strong>and</strong> the occupant density on the st<strong>air</strong>s. According to the experimental results <strong>of</strong> Achakji <strong>and</strong> Tamura<br />

(1988), for a st<strong>air</strong>well tested under the following conditions<br />

4As Ps ● With a cross-sectional area <strong>of</strong> 134 ft2 (12.5 m2 )<br />

● A floor height <strong>of</strong> 8.5 ft (2.6 m)<br />

● An occupant density <strong>of</strong> 0.18 person/ft 2 (2.0 persons/m 2 )<br />

the frictional loss <strong>of</strong> <strong>air</strong>flow per floor �p f,s has the following values:<br />

vo, fpm (m/s) �pf,s, in. WC (Pa)<br />

Open-tread st<strong>air</strong>, bottom injection 879 (4.47) 0.128 (32)<br />

659 (3.35) 0.070 (17.5)<br />

439 (2.23) 0.032 (8)<br />

Closed-tread st<strong>air</strong>, bottom injection 1055 (5.36) 0.200 (50)<br />

848 (4.31) 0.116 (29)<br />

527 (2.68) 0.05 (12.5)<br />

Here v o represents the <strong>air</strong> velocity calculated based on the free flow area, in fpm (m/s). The difference<br />

in �p f,s between top <strong>and</strong> bottom injection is small if other conditions remain the same.<br />

When calculating the system pressure loss <strong>of</strong> a st<strong>air</strong>well pressurization system with multiple injection<br />

�p sy, in in. WC (Pa), as shown in Fig. 22.6, is calculated as<br />

�p sy ��p a � b ��p b � o<br />

(22.15)<br />

where �p a � b, �p b � o � pressure loss between points a <strong>and</strong> b <strong>and</strong> points b <strong>and</strong> o, respectively, in.<br />

WC (Pa) Pressure loss �p b � o usually varies between 0.10 <strong>and</strong> 0.40 in. WC (25 <strong>and</strong> 100 Pa) <strong>and</strong> is<br />

only a small portion <strong>of</strong> the system pressure. However, it is difficult to calculate accurately because<br />

<strong>of</strong> complicated <strong>air</strong>flow paths at various operating conditions <strong>and</strong> the influence <strong>of</strong> outdoor conditions.<br />

Pressure loss �p a � b should be calculated between points a <strong>and</strong> b according to the procedure<br />

described in Chap. 17, including the velocity pressure at the supply inlet, <strong>and</strong> �p b � o should be estimated<br />

as 0.4 in. WC (100 Pa). The calculated system pressure �p sy should be multiplied by a<br />

safety factor 1.2. The actual �p b � o can be adjusted during acceptance testing.<br />

In a st<strong>air</strong>well pressurization system with a bottom single injection, as shown in Fig. 22.8a, the<br />

maximum pressure difference between the st<strong>air</strong>well <strong>and</strong> outdoors <strong>of</strong>ten occurs at the top <strong>of</strong> the<br />

st<strong>air</strong>well �p c � o because the cross section <strong>of</strong> the st<strong>air</strong>well is constant. The system pressure can<br />

therefore be calculated as<br />

�p sy ��p a � b ��p b � c ��p c � o<br />

(22.16)

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