Single Resistances in a Tube Flow

In addition to the friction factor, individual resistances in ducts have to be taken into account. These resistances are created by the velocity, through bends, branches, valves, and other obstructions in the duct.

The single resistance factor Ј is defined as

A p _ >-Xпь Pg 2g

 

Ap =

 

(4.52)

 

Or

 

The I values for different types of resistances are available in hydraulics text­books and also from the literature of pipe and valve manufacturers.

Single Resistances in a Tube Flow

^Ploss

подпись: ^ploss

2

P^m

2

подпись: 2
p^m
2

1.5

подпись: 1.5

0.9

подпись: 0.9

0.8

подпись: 0.8

103 1.5 2 3 4 5 6 8 104 1.5 2 3 4 J 6 8 10s 1.5 2 3 4 5 6 81061.5 2 3 4 5 6 107

Re

подпись: 103 1.5 2 3 4 5 6 8 104 1.5 2 3 4 j 6 8 10s 1.5 2 3 4 5 6 81061.5 2 3 4 5 6 107
re
The pressure loss and the corresponding resistance height in Eq. (4.53) and Eq. (4.54), respectively, are the sum of the friction losses and individual resistances:

(4.53)

When px = p2 and vOTl ance border 1:

= 0, we obtain for the system defined by bal-

= v

Ml

P _

Hf + (z-! — z2) ,

4vP8

2 / vml

2g

Hf+ tL-ll + (Z, _22) + Vl. Yjni

P = QvPg

2 g

Pg

And

 

V~

 

Fj = OSS =

F pg

 

(4.54)

 

2g-

 

Example I

Find the power P required by the pump in the system shown in Fig. 4.5. The energy balance of a continuing system, when qm = qvp, gives qm.

 

Single Resistances in a Tube Flow

(4.55)

 

Single Resistances in a Tube Flow
Single Resistances in a Tube Flow

(4.56)

 

Where

 

Single Resistances in a Tube Flow

Hf =

 

2#

 

Resistance factors Ј are taken from the Moody chart, when the Reynolds number and roughness are known.

Re = ^

 

And vm is determined by the volume flow qv = Avm, where A is the cross­sectional area of the tube.

Single resistances Ј can be found from the literature. One of the single re­sistances is the outflow loss at point 1. The outflow loss is the kinetic energy representing vm: and therefore Јoutflow = 1.

 

Single Resistances in a Tube Flow

The form of Eq. (4.56) depends on the chosen balance border. The border can be chosen arbitrarily. When the balance space is chosen according to the balance border 2, the energy equation is

= hf + -…. !… ‘ * + Zy — Zt +………… ……………………………………. . (4.57)

QvPg 1 Pg ~ 2g

Using the values corresponding to the states 1′ and 2′, /y’ is the resistance height between V and 2′, and pr and pT are determined by the pressure loss and initial pressure. Other quantities are determined correspondingly. It is worth noting that there are no outflow losses in this case. Generally, it is wise to use energy balances in calculations.

The above is valid for a liquid flow, when the effect of compressibility can be ignored when calculating gas flows with small pressure differences. For instance, in ventilating duct work, air is not compressed, so the density is considered as constant. In HVAC technology a unit of pressure fre­quently used for convenience is a water column millimeter, 1 mm H->0 =10 Pa.

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