Airflow through transition pieces

It is customary to express the total pressure loss, Apt, through a duct fitting by the product of a coefficient, Ј and a velocity pressure, pv. The velocity used is the actual mean velocity, V, determined by the ratio Q/A, where Q is the volumetric airflow rate and A is the cross sectional area of the fitting. For the case of transition pieces in which the section is expanding or contracting, the smaller section, where the velocity is greater, is generally used for A. In the following examples, Ј values from the CIBSE (1986a) have been used but coefficients from other sources yield approximately similar pressure drops.

(a) Expanders

It is possible to derive an expression from first principles according to Prandtl (1953) and Lewitt (1948) for the total pressure loss when an airstream flows through an abruptly expanding duct section (as in Figure 15.8):

APt = VapM — V2)2 (15.28)

Airflow through transition pieces

Fig. 15.8 Airflow through an abrupt expansion.

Since V2 = Vt(ai/a2) this can be re-written as

Apt = 1/2pV12(l -axla2)2

The term in the brackets is the coefficient, Ј, used to evaluate the total pressure loss, since V2 pVj2 is the upstream velocity pressure. Thus, for abruptly expanding duct sections:

Ј=(l-a,/a2)2 (15.29)

More recent work by Idelchik et al. (1986) and ASHRAE (1993) expresses the loss coefficient in general terms, referred to the included angle of the transition piece, 0, from 0° to 90° to cover the case of expanding sections and from 90° to 0° for reducing sections. It is to be noted that the coefficients for a 90° abrupt expander are somewhat greater than those

Predicted by equation (15.29). At the time of writing, the CIBSE (1986a) use coefficients

For abrupt expanders in agreement with equation (15.29) but, for gradual expansion pieces, this is multiplied by a further coefficient to express the total pressure loss as:

APt = CaCgPv (15.30)

Where i^a is the coefficient for an abrupt expander, Јg the coefficient for a gradual expander

And pv is the velocity pressure in the upstream, smaller duct section. ASHRAE (1997a) use

A single coefficient.

If air flows through an expansion piece (Figure 15.9), without friction or turbulence, there is no loss and pt{ equals pt2. Hence, by equation (15.18):

Psi + Pvi = Ps2 + Pv2 and so Pvl _ Pv2 = Ps2 _ Psl

This states that there is full conversion of kinetic energy (velocity pressure) to potential energy (static pressure). In other words, 100 per cent static regain occurs.

In the actual case (Figure 15.9) pt2 is less than /?tl and the loss is:

APt = Pti — Pt2

Airflow through transition pieces

Fig. 15.9 Airflow through a gradual expander, with and without loss.

A full conversion of velocity pressure to static pressure does not take place and the static regain is given by the fall in velocity pressure minus the total pressure loss:

Static regain = ps2 — psi

Ps2-Psi =Pvi — Ps2- Apt (15.31)

EXAMPLE 15.3

Calculate the total pressure loss and the static regain through a symmetrical, gradual expansion piece having an initial velocity of 20 m s-1 and a final velocity of 10 m s_1, using

(a) CIBSE (1986a) data and (b) ASHRAE (1993) data.

Answer

A{/a2 = V2/V i = 10/20 = 0.5

(a) From CIBSE (1986a) data, Јa = 0.25 and Јg = 0.8.

Pv = 0.6 x 202 = 240 Pa by equation (15.26)

Hence Apt = 0.25 x 0.8 x 240 = 48 Pa by equation (15.30) and, by equation (15.31), static regain is given by

Ps2 — Psi = 0.6 x 202 — 0.6 X 102 — 48 = 132 Pa

One hundred per cent static regain would be 0.6 x 202 — 0.6 x 102, or 180 Pa. Hence the percentage static regain is only 132 x 100/180, namely, 73 per cent.

(b) From ASHRAE (1993) data, = 0.24.

Hence

Apt = 0.24 x 240 = 58 Pa

And the static regain is

Ps2 ~ Psi = 0.6 x 202 — 0.6 x 102 — 58 = 122 Pa

Which is 68 per cent of the maximum possible.

Static regain can also occur in a main duct when its section remains constant as air is fed from it through a branch, or a supply grille. This is because the velocity in the main reduces when the volumetric airflow rate in it falls, after a branch or a grille.

(b) Static regain duct sizing

It is possible to use static regain methods according to Carrier (1960), Shataloff (1966) and Chun Lun (1983) for sizing but it is impossible to be precise about the loss incurred as air flows through an expander, because of variations in duct construction and the lack of exact data regarding loss coefficients. However the following offers an approximate solution.

Refer to Figure 15.10. The exact form of the expanding section, 0—1, is not defined except by assuming that the fraction of static regain occurring is denoted by r. The further

TOC o "1-5" h z 0 1 y 2 3

Vo v, = v2 V3

I I

Fig. 15.10 Diagram of a duct system for the derivation of an approximate equation for duct sizing by

Static regain. Also refers to Example 15.4.

Assumption is made that the rate of pressure drop in the duct section to be sized, 1—2, is proportional to the square of the mean velocity, Vt(= V2). The static regain across the section of duct, 0—1, is to be equated to the loss of total pressure in the downstream main duct from 1 to 2, in order to determine the mean velocity (Vj = V2) in the duct and so size it.

Static regain from 0 to 1 = r(pv0 — pv])

Total pressure loss in the main from 1 to 2 is given by Aptl2 = yApt0(V,2/V02), where y is the duct length in m and Apt0 is the pressure drop rate in the duct approaching section 0. Equating the static regain to the downstream total pressure loss:

K0.6Vo — 0.6V2) = yApt0(V]lV20) r(0.6V40 — 0.6V2Vq) = y&PtoV] V(yApto + 0.6 rV) = 0.6rVo

V2 = (0.6rV40)/(yApt0 + 0.6 rV)

Vi = Vl ————- ———— r (15.32)

I (y&Pto + 0.6rVl)

EXAMPLE 15.4

Refer to Figure 15.10. The volumetric flow rate in the main duct before section 0 is 1.0 m3 s-1 and 0.1 m3 s~* is fed through the first branch. The mean velocity in the main duct leading to the branch is 12 m s_1, the diameter is 326 mm and the pressure drop rate CIBSE (1986a) is 5.5 Pa m-1. Size the main duct, between sections 1 and 2, assuming that 70 per cent static regain occurs from 0 to 1. Consider two cases: (a) when the length of the duct 1-2 is 10 m and, (b), when it is 3 m.

Answer

(a) By equation (15.32)

Vi = 122 ———— 06 X-°———— =- = 8.68 m s"1

V 10 x 5.5 + 0.6 x 0.7 x 12

By equation (15.12)

A = 0.9/8.68 = 0.1037 m2

Whence

D = 363 mm

Check

Static regain = 0.7(0.6 x 122 — 0.6 x 8.682) = 29 Pa

By CIBSE (1986a) the pressure drop rate for a duct of 363 mm diameter handling

0. 9 m3 s-1 is about 2.3 Pa m-1. Hence the pressure drop is approximately 23 Pa over a length of 10 m.

(.b) By equation (15.32)

Vi = 122 —————— 0 6 x °’7—————- =- = 10.64 ms-1

Y (3 x 5.5 + 0.6 x 0.7 x 122)

By equation (15.12)

A = 0.9/10.64 = 0.084 59

Whence

D — 328 mm

Check

Static regain = 0.7(0.6 x 122 — 0.6 x 10.642) = 13 Pa

By CIBSE (1986a) the pressure drop rate for a 328 mm diameter duct handling 0.9 m3 s_1 is about 3.3 Pa m_1. Hence for a length of 3 m the total pressure drop is about 10 Pa.

Reading a pressure drop rate from the duct sizing chart in CIBSE (1986a) is not very accurate and therefore the above checks can be regarded as not unreasonable.

The difficulty when using equation (15.32) is in deciding on the value of the static regain factor r. It is suggested that the designer uses ASHRAE (1993) (which contains a large selection of duct transition pieces and branches) and chooses the type of branch that most closely resembles the actual fitting likely to be used. The coefficient given will then enable the loss to be calculated and a value for the static regain fraction, r, determined.

(c) Reducers

Airflow through an abrupt reducer is illustrated in Figure 15.11. The major source of loss is the pocket of turbulence that forms downstream is the narrower duct section, where the higher velocity prevails. Equation (15.28) is fundamental and can be used here to refer to the losses incurred by expansion from the vena-contracta at section c—c, to section 2—2, in Figure 15.11:

Apt =tP(K — V2)2

1

Airflow through transition pieces

Since Vc = V2/CA this can be rewritten as

APl = l/2pV22[(/CA)-l]2 (15.33)

There is an insignificant error because the upstream loss has been ignored.

According to Lewitt (1942), experiment shows that, for waterflow, CA is about 0.62 which yields 0.376 for the value of [(1/CA) — l]2, the loss coefficient that multiplies the velocity pressure to give the total pressure loss. The CIBSE (1986a) quotes loss factors in terms of the ratio of the downstream to upstream areas, A2!AX, and these are used to

Multiply the downstream velocity pressure, pv2. The same reference gives loss factors in

Terms of the included angle only, for the case of gradually reducing fittings and, unlike the procedure for expansion pieces, this is not multiplied by the coefficient for the corresponding abrupt fitting. Thus

Apt = ЈaPv2 (15.34)

And

EXAMPLE 15.5

Calculate the total pressure loss through a symmetrical, gradual reducer, of circular section, having an included angle of 30° if the initial velocity is 10 m s-1 and the final velocity is 20 m s“1 using CIBSE (1986a) data.

Answer

For an included angle of 30° and a symmetrical section

= 0.02. Hence, by equation (15.35)

Apt = 0.02 x 0.6 x 202 = 5 Pa

Figure 15.12 illustrates the case of a gradual reducer. If there is no loss, the fall in static pressure is used to accelerate the air from V to V2 and exactly equals the increase in velocity pressure. In the real case, when there is a loss, the static pressure must provide the energy to do two things: to accelerate the air and to make good the loss.

Airflow through transition pieces

Fig. 15.12 Airflow through a gradual reducer, with and without loss.

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