## 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)

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)

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

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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