minnor losses

Friction Losses...

Pipe Fittings...

Invariably a system containing piping will have connections which change th size and / or direction of the conduit. These fittings andfriction, called " minor losses ", to the system head. Fitting losses are generally the result of changes in velocity and / or direction.

A decreasing velocity results in more loss in head than an increasing velocity as the former causes energy - dissipating eddies.Experimental results have indicated that minor losses vary approximately as the square of the velocity through the fittings.

Valves and Standard Fittings...

The resistance to flow through valves and fittings may be estimated by any of the following methods. The " Hydraulic Institute -Pipe Friction Manual ", lists losses through valves and fittings in terms of the average velocity head in a pipe of corresponding

diameter and a " resistance coefficient " as shown in figures given below.

Source : www.gouldspumps.com

Toprak Home Page http://web.deu.edu.tr/atiksu/ana58/friction.html

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The frictional resistance in feet is found from the equation ;

h= ( K ) ( V 2 / 2 x g )

where ; K : resistance coefficient which depends on design and size of valve or fitting, V : average velocity in pipe of corresponding

diameter ( ft / s ) and g : acceleration of gravity ( = 32.17 ft / s 2 ).

Table given below indicates the wide variation in published values of K.


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A comparison of the " Darcy - Weisbach " equation and the above equation suggests that K = f ( L / D ) to produce the same headloss in a straight pipe as in the valve or fitting. The ratio ( L / D ) or " equivalent length in pipe diameters " of straight pipe may

then be used as another method to estimate valve and fitting losses. Tests have shown that while K decreases with size ofdifferent lines of valves and fittings, ( L / D ) is almost constant. In the zone of complete turbulence, as shown in figure below, K for

a given size and ( L / D ) for all sizes of valves and fittings are constant. In the transition zone K increases as does the frictionfactor f with decreasing.


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" Reynolds " number Re while L / D remains approximately constant. Table given below lists suggested values of L / D for variousvalves and fittings.


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Multiplying L / D by the inside diameter of pipe corresponding to the schdule number shown in table given below gives theequivalent length of straight pipe.

When using the equivalent length method, the friction head loss is determined by employing the " Darcy - Weisbach " equation. Thismethod therefore takes into consideration the viscosity of the liquid, which in turn determines the " Reynolds " number and the

friction factor.

The loss of head through valves, particularly control valves, is sometimes expressed in terms of the " flow coefficient, C V ". The

flow in gpm at 60 O F to produce a pressure drop of 1 lb / in 2 is defined as the flow coefficient for a particular valve opening. Valuesof C V may be obtained from various manufacturers for their different lines of valves. The pressure loss for liquids with viscosity

close to water at 60 O F may be found for different flows from ;

lb / in 2 = sp gr at 60 O F ( gpm / C V ) 2

The following examples illustrate the use of the " resistance coefficient, K " and the " equivalent length in pipe diameters, L / D "methods for estimating losses in valves and fittings.


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Example - 1...

A pumping system consists of 20 ft of 2 - in suction pipe and 300 ft of 1 1/2 - in discharge pipe, both " Schedule " 40 new steel.

Also included are a bellmouth inlet, a 90 O LR suction elbow, a suction - gate valve, a discharged - gate valve and a swing - checkvalve. The valves and fittings are screw - connected and the same size as the connecting pipe. Determine the pipe, valve, and fitting

losses when pumping 60 gpm of 60 O F oil having a sp gr of 0.855. Use the resistance coefficient method.

Solution...

- Suction pipe :

* ID = 2.067 in* EPSILON / D = 0.00087

* NU = 0.0009 ft 2 / s

* V = ( 60 / 2.067 2 ) ( 0.408 ) = 5.73 ft / s* ( V ) ( D" ) = ( 5.73 ) ( 2.067 ) = 11.8 ft / sec . in

* Re = 1 x 10 4

* f = 0.032

h FS = ( 0.032 ) [ ( 20 ) ( 12 ) / ( 2.067 ) ] [ ( 5.73 2 ) / ( 2 ) ( 32.17 ) ] = 1.9 ft

- Discharge pipe :

* ID = 1.610 in* EPSILON / D = 0.0011

* NU = 0.00009 ft 2 / s

* V = ( 60 / 1.610 2 ) ( 0.408 ) = 9.44 ft / s* ( V ) ( D" ) = ( 9.44 ) ( 1.610 ) = 15.2 ft / sec . in

* Re = 1.5 x 10 4

* f = 0.030

h FD = ( 0.030 ) [ ( 300 ) ( 12 ) / ( 1.610 ) ] [ ( 9.44 2 ) / ( 2 ) ( 32.17 ) ] = 92.9 ft

- Valve and fitting losses :

* 2 - in bellmouth, K = 0.05

h F - 1 = ( 0.05 ) [ ( 5.73 2 ) / ( 2 ) ( 32.17 ) ] = 0.026 ft

* 2 - in LR, 90 O elbow, K = 0.4 +- 25 %

h F - 2 = ( 0.4 ) [ ( 5.73 2 ) / ( 2 ) ( 32.17 ) ] = 0.202 +- 0.051 ft

* 2 - in gate valve, K = 0.16 +- 25 %

h F - 3 = ( 0.16 ) [ ( 5.73 2 ) / ( 2 ) ( 32.17 ) ] = 0.082 +- 0.021 ft

* 1 1/2 - in gate valve, K = 0.19 +- 25 %

h F - 4 = ( 0.19 ) [ ( 9.44 2 ) / ( 2 ) ( 32.17 ) ] = 0.263 +- 0.066 ft

* 1 1/2 - in swing - check valve, K = 2.4 +- 30 %

h F - 5 = ( 2.4 ) [ ( 9.44 2 ) / ( 2 ) ( 32.17 ) ] = 3.32 +- 1.0 ft

- Total pipe, valve and fitting losses :

TOTAL h F = 1.9 + 92.9 + 0.026 + 0.202 + 0.082 + 0.263 + 3.32 = 98.69 ft

TOTAL variation = +- ( 0.202 + 0.021 + 0.066 + 1.0 ) = +- 1.29 ft

Example - 2...

Determine the value and fittings losses in examle given above by the equivalent length pipe diameters method and compareresults.

Solution...

* 2 - in bellmouth, h F - 1 = 0.026 ft

* 2 - in LR, 90 O elbow, L / D = 20L 2 = ( 20 ) ( 2.067 / 12 ) = 3.44 ft

* 2 - in gate valve, L / D = 13L 3 = ( 13 ) ( 2.067 / 12 ) = 2.24 ft* 1 1/2 - in gate valve, L / D = 13

L 4 = ( 13 ) ( 1.610 / 12 ) = 1.74 ft* 1 1/2 swing - check valve, L / D = 135

L 5 = ( 135 ) ( 1.610 / 12 ) = 18.1 ft

Using f from example given above, total valve and fitting losses are ;


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Total valve and fitting losses from example given above ;

TOTAL h F = ( 0.026 + 0.202 + 0.082 + 0.263 + 3.32 ) +- 1.29 = 3.89 +- 1.29 ft

Increasers :

The loss of head for a sudden increase in diameter with velocity changing from V 1 to V 2 in the direction of the flow can becalculated analytically. Computed results have been confirmed experimentally to be true to within +- 3 percent. The head loss is

expressed as shown below with K computed to be equal to unity ;

The value of K is also approximately equal to unity if a pipe discharges into a relatively large reservoir. This indicates that all the

kinetic energy V 1 2 / 2g is lost, and V 2 equals zero. The loss of head for a gradual increase in diameter through a diffuser can be

found from figure given below.

The diffuser converts some of the kinetic energy into pressure. Values for the coefficient used with the above equations for

calculating head loss are shown in the figure. The optimum total angle appears to be 7.5 O . Angles greater than this result in

shorter diffusers and less friction, but separation and turbulence occur. For angles greater than 50 O it is preferable to use a suddenenlargement.

Reducers :

Figure shown below gives values of the resistance coefficient to be used for sudden reducers.


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Bends :

Figure given below may be used to determine the resistance coefficient for 90 O pipe bends of uniform diameter.

Figure shown below gives resistance coefficients for less than 90 O pipe bends and can be used for surfaces having moderateroughness such as clean steel and cast iron.


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These two figures are not recommended for elbows with R / D below 1.

Table shown below gives values of resistance coefficients for miter bends.


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Pump Suction Elbows :

Figures given below illustrate two typical rectangular to round reducing suction elbows. Elbows of this configuration are sometimesused under dry pit vertical volute pumps. These elbows are formed in concrete and are designed to require a minimum height, thuspermitting a higher pump setting with reduced excavation. Figure (first) shows a long - radius elbow and figure (second) shows ashort - radius elbow. The resulting velocity distribution into the impeller eye and the loss of head are shown for these selected two

designs.


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Meters :

Orifices, nozzles, and venturi meters are used to measure rate of flow. These metering devices, however, introduce additional lossof head into the pumping system. Each of these meters is designed to create a pressure differential through the primary element.The magnitude of the pressure differential depends on the velocity and the density of the liquid and the design of the element. Theprimary element restricts the area of flow, increases the velocity, and decreases the pressure. An expanding section following the

primary element provides pressure head recovery and determines the meter efficiency. The pressure differential between inlet andthroat taps measures rate of flow ; the pressure differential between inlet and outlet taps measures the meter head loss (an outlettap is not usually provided). The meters offering the least resistance to flow are in the following decreasing order ; venturi, nozzle,

and orifice. Figures given below illustrate these different meter designs.


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When meters are designed and pressure taps are located as recommended, figures given below may be used to estimate theoverall pressure loss. The loss of pressure is expressed as a percentage of the differential pressure measured at the appropriate

taps and values are given for various size meters.


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minnor losses

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ft discharge pipe

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connecting pipe

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