Civil Engineering Handbook

37 Laminar flow of inelastic fluids in noncircular ducts

Analytical solutions for the laminar flow of time-independent fluids in non-axisymmetric conduits are not possible. Numerous workers have obtained approximate numerical solutions for specific flow geometries including rectangular and triangular pipes [Schechter, 1961; Wheeler and Wissler, 1965; Miller, 1972; Mitsuishi and Aoyagi, 1973]. On the other hand, semi-empirical attempts have also been made to develop methods for predicting pressure drop for time-independent fluids in ducts of non-circular cross-section. Perhaps the most systematic and successful friction factor analysis is that provided by Kozicki et al. [1966, 1967]. By noting the similarity between the form of the Rabinowitsch-Mooney equation for the flow of time-independent fluids in circular pipes (equation 3.25) and that in between two plates, they suggested that it could be extended to ducts having a constant cross-section of arbitrary shape as follows:

dr w d rw where a and b are two geometric parameters characterising the cross-section of the duct, is the hydraulic diameter (= 4 times flow area/wetted perimeter), and rw is the mean value of shear stress at the wall, and is related to the pressure gradient as:

For constant values of a and b, equation (3.86) can be integrated to obtain:

It should be noted that for a circular pipe of diameter D, = D; a = 1/4 and b = 3/4; equation (3.88) then reduces to equation (3.21). For the flow of a power-law fluid, /(r) = (r/m)1/n and integration of (3.88) yields:

which can be re-written in terms of the friction factor, f = 2rw/pV2 as:

Table 3.3 Values of a and b to be used in equations (3.90) and (3.91)

1. Concentric annuli

a = R/Ro	a	b
0.00	0.2500	0.7500
0.01	0.3768	0.8751
0.03	0.4056	0.9085
0.05	0.4217	0.9263
0.07	0.4331	0.9383
0.10	0.4455	0.9510
0.20	0.4693	0.9737
0.30	0.4817	0.9847
0.40	0.4890	0.9911
0.50	0.4935	0.9946
0.60	0.4965	0.9972
0.70	0.4983	0.9987
0.80	0.4992	0.9994
0.90	0.4997	1.0000
1.00	0.5000	1.0000
2. Elliptical ducts
=	a	b
0.00	0.3084	0.9253
0.10	0.3018	0.9053
0.20	0.2907	0.8720
0.30	0.2796	0.8389
0.40	0.2702	0.8107
0.50	0.2629	0.7886
0.60	0.2575	0.7725
0.70	0.2538	0.7614
0.80	0.2515	0.7546
0.90	0.2504	0.7510
1.00	0.2500	0.7500
3. Rectangular ducts
E = H/W	a	b
0.00	0.5000	1.0000
0.25	0.3212	0.8182
0.50	0.2440	0.7276
0.75	0.2178	0.6866
1.00	0.2121	0.6766

Table 3.3 (continued)

4. Isosceles triangular ducts

2a	a	b
10°	0.1S47	0.6278
20°	0.1693	0.6332
40°	0.1840	0.6422
60°	0.187S	0.6462
80°	0.1849	0.6438
90°	0.1830	0.639S
S. Regular polygonal ducts
N	a	b
4	0.2121	0.6771
S	0.224S	0.6966
6	0.2316	0.7092
8	0.2391	0.7241

N sides where the generalised Reynolds number

oV Dn

The main virtue of this approach lies in its simplicity and the fact that the geometric parameters a and b can be deduced from the behaviour of Newtonian fluids in the same flow geometry. Table 3.3 lists values of a and b for a range of flow geometries commonly encountered in process applications. A typical comparison between predicted and experimental values of friction factor for rectangular ducts is shown in Figure 3.22. Similar agreement has been reported by, among others, Mitsuishi et al. [1972] and, more recently, by Xie and Hartnett [1992] for visco-elastic fluids in rectangular ducts. Kozicki et al. [1966] argued that equation (3.37) can be generalised to include turbulent flow in non-circular ducts by re-casting it in the form:

4= = -075 log10(Reg/(2-n)/2) - n4 C 4n025 log (3.91)

Note that since for a circular tube, a = 1/4 and b = 3/4, equation (3.91) is consistent with that for circular pipes, equation (3.37). The limited data available on turbulent flow in triangular [Irvine, Jr., 1988] and rectangular ducts [Kostic and Hartnett, 1984] conforms to equation (3.91). In the absence

Figure 3.22 Experimental friction factor values for power-law fluids in laminar regime in rectangular channels. o, Wheeler and Wissler (1965); a, Hartnett et al. (1986); a, Hartnett and Kostic (1985)

of any definite information, Kozicki and Tiu [1988] suggested that the Dodge-Metzner criterion, Regen < 2100, can be used for predicting the limit of laminar flow in non-circular ducts.

Scant analytical and experimental results suggest that visco-elasticity in a fluid may induce secondary motion in non-circular conduits, even under laminar conditions. However, measurements reported to date indicate that the friction factor - Reynolds number behaviour is little influenced by such secondary flows [Hartnett and Kostic, 1989].

Example 3.12

A power-law fluid (m = 0.3Pa-s" and n = 0.72) of density 1000 kg/m3 is flowing in a series of ducts of the same flow area but different cross-sections as listed below:

(ii) circular pipe

(v) isosceles triangular with half-apex angle, a = 20°.

Estimate the pressure gradient required to maintain an average velocity of 1.25 m/s in each of these channels. Use the geometric parameter method. Also, calculate the value of the generalised Reynolds number as a guide to the nature of the flow.

Solution

(i) For a concentric annulus, for a = 0.4, and from Table 3.3: a = 0.489; b = 0.991 The hydraulic diameter, Dh = 2R(1 — a)

Thus, the flow is laminar.

and —^ = f! = 2 x 0.0276 x 1000 x 1.252 = 1963 Pa/m L Dh 0.044

For the sake of comparison, equation (3.82) yields a value of 1928 Pa/m which is remarkably close.

(ii) For a circular tube, the area of flow

= nR2(1 — a2) = 3.14 x 0.0372(1 — 0.42) = 0.003613m2

which corresponds to the pipe radius of 0.0339 m or diameter of 0.0678 m.

For a circular pipe, a = 0.25, b = 0.75, Dh = D = 0.0678 m.

Thus, the flow is laminar and f = 16/1070 = 0.01495 and pressure gradient,

-Ap _ 2fpV2 _ 2 x 0.01495 x 1000 x 1.252 L = Dh = 0.0678

(iii) For a rectangular duct with H/W = 0.5, H = 0.0425m and W = 0.085m (for the same area of flow), and from Table 3.3: a = 0.244, b = 0.728

4 x Flow area ihe hydraulic diameter, Dh =

wetted perimeter

4x A 4 x 0.003613

_ pV2-nDl _ (1000) x (1.25)2-°'72 ( 0.0 5 6 7)0'72

Again, the flow is laminar and f = 16/960 = 0.0167 -Ap 2fpV2

The pressure gradient,

L Dh

0 0567

(iv) The cross-sectional area of an elliptic pipe with semi-axes a' and b' is na'b', while a = 0 . 2629 and b = 0 . 7886 from Table 3.3 corresponding to b'/a' = 0 . 5

Solving with b'/a' = 0.5, a' = 0.04795 m and b' = 0.024 m. The hydraulic diameter Dh is calculated next as:

4 x Flow area

Wetted perimeter

No analytical expression is available for the perimeter of an ellipse; however, it can be approximated by 2^((a'2 + b'2)/2)1/2

4 x 0.003613

0.047952 + 0.0242

0.2629 0.72

and f = 16/Reg = 16/953 = 0.0168 -Ap 2fpV2 2 x 0.0168 x 1000 x 1.252

0.0607

(v) For the pipe of triangular cross-section, with a = 20° a = 0.184 and b = 0.6422

Let the base of the triangle be x

\ height of the triangle =

2 tan 20

The hydraulic diameter is calculated next:

sin 20

The Reynolds number of flow,

1006

0.184 0.72

980 Pa/m

0.0508

0.72

0.72

0.28

This example clearly shows that the pressure drop is a minimum in the case of circular pipes, followed by the elliptic, rectangular and triangular cross-sections, and the concentric annulus for flow. On the other hand, if one were to maintain the same hydraulic diameter in each case, the corresponding pressure gradients range from 2500 Pa/m to 4000 Pa/m.

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