# Dimensioning of a Duct with Liquid Flow

This example demonstrates the dimensioning of a duct with a frictional incompressible fluid flow. Now the Bernoulli equation can be written as

| 7 1 2 /?, + pgz 1 + 2P^T = Pi + PgZz + 2Pw/2 + Ap, (4.65)

Where

( l sr ) W2

 Ap =

+ 2-! ^ ^"2” = PreSSUre ^OSS term!

L w2

^qP~2” = t^le Pressure l°ss due to the friction,

W2 • ■

/ ^p~2~ =the pressure loss due to the single resistances.

To determine the pressure losses, we have to find out whether the flow is lam­inar or turbulent, because Ј = /(Re, k/d). In practical dimensioning, Eq. (4.66) and the Moody chart are used.

 2.52 +.

 (4.66)

-L =-2ln

Re^ 3.71 d

Empirical tables are used to determine the value of the single resistances.

The determination of the desired volume flow V and the pressure differ­ence p2 — p leads to an iterative procedure in a turbulent case, due to terms Ј and XC • a laminar flow

T _ 64

5 Re’

So no iteration is needed. This is due to the fact that the pipe diameter d is re­duced from its value in the formula for the pressure loss caused by friction, be­cause Re = 4qv/tt dv.

Example 3

A pipe of diameter d0 conveys water at a volume flow rate qv as shown in Fig. 4.6. The lateral branches 1-2 and 3-4 are each required to have volume flows of q„/6. The pressures at points 0,2, and 4 are known; the water leaves the system at points 2 and 4. Determine diameters d4 and d2.

 Si’ 6

 №

U

FIGURE 4.6 The system to be dimensioned.

The following values are known:

Volume flow = 10 m3 s_1 Diameter d0 = 79.8 mm Pressure p0 = 600 kPa Pi = P4 = 250 kPa Temperature T = 283 K Length L = 50 m Roughness k = 0.03 mm

Water properties can be found from tables, such as the VDl-Wasser- dampftafeln; The NBS/NRC Steam Tables; the ASHRAE and CIBSE Guides; or Thermodynamic and Transport Properties of Fluids by Mayhew and Rodgers. First we dimension d4, and now Eq. (4.65) is

 (4.67)

PQ + pgZ + Jpwl = p4 + pgZ4 + ^_pw + A/?4

Taking z0 = 0, when T = 283 K, p = 999.7 kg/m-3, v = 1.3 x 10-6 m2s and qv = qv = 10 m3 s_1. Thus the flow velocity wQ = 2 ms-1. From Eq. (4.67) we re­ceive

 (4.68)

106 823 = 499.85«4 + Ap4.

The formula for the pressure loss is now

 50
 2 25′
 —— 1- —————— ; 79.8 xlO-3 999.7^4 999.7^4
 999.7 • 2l
 50

 Ap4 =

 79.8×10

 (4.69)

 X999_7. MM_ +

The Reynolds number is

Re01 = = 1.227 x 105.

Trdnv

Hence the flow is turbulent.

 1/4
 9.854 x 10’9 + 8-554 xl° — З34
 D 4 =
 999.710%
 Api = 0.0193-50^-999.7.2- + |-0>88 + ^25 + 1
 1/4
 6.809 x10"
 D7 =
 7.844 x 10 +
 12
 ^12
 = 24 178.2 + The iteration equation is
 939.7+ 12 496.25
 989.7 + ^4i^^-
 ■y W4 .
 Ap4 — 41 080 +
 When k = 0.03 mm, the ratio k/d = 0.000 375, and therefore the friction factor Ј01 = 0.0193. Correspondingly, Re1? = 10 xlO5 and = 0.0196. The formula for pressure loss is

 I’4.70)

 The dimensioning of pipe 3-4 happens by means of Eqs. (4.68) and (4.70). The iteration equation is now

 (4.71)

 Calculations are dealt with in the following stages 1. d4 is chosen. 2. Re and k/d4 are calculated. 3. |!4 is read from the Moody chart as a function of Re and k/d4 4. A new d4 is calculated with Eq. (4.71) until the d4 value is suitable. Calculations are best carried out in a tabular form. The value of d4 is 29.7 mm. In practice the next larger standard pipe size would be selected. The diameter d2 is solved analogously. The Bernoulli equation at the in­terval 1-2 is

 106 823 = 499.S5w2 + &p2 ■

 (4.72)

 The pressure loss is

 79.8×10-2

 (4.73)

 (4.74)

 The diameter = 28.8 mm. Example 4 A pump lifts water from a lake. At the pump suction entry a foot valve is fitted. Determine the maximum static delivery height the water can be raised without cavitation taking place. The saturation pressure of water is 1.23 kPa at 10 °C and the dynamic viscosity is 1.3 x 10-3 kg m-1 s-1. The suction pipe water velocity is 2.0 m s-1, the internal pipe diameter is 100 mm, and the pipe roughness is 0.03 mm. The resistance of the foot valve is 4.5.

 Note: Cavitation occurs when the pressure at a point in a liquid flow field is equal or less than the vapor pressure of the liquid. At this point bubbles of vapor are formed, this is cavitation. It has serious effects such

 63

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As loss of duty, loss in efficiency, and serious erosion on the suction side of the pump impeller.

Pressure at the suction inlet pt:

 (4.751

Pt — p, ~ PShs — P8hf — 1 -23 kPa

W 2 g

Friction factor:

Re = t^Ј =………………… :„1.0QQ = 153 8.50

V 1.3 xlO“3

K = 0.03 mm, d = 100 mm => Ј, = 0.0185

 Pi ~ Pghs ~ Pg

By substituting the resistance head loss equation in Eq. (4.75) the suction head loss is determined from

= 1.23 kPa

2 g

4 L

 1+4

 Pg

I. w d’ 2g

 4.0 2-9.81

100-1.23-1000-9.81-4,5-

1000-9.811 1 +0.0185 !.. 4.0

0.1 2-9.81

8.82 m.