# Pipe

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CHEE 4500 Summer 2002

Centrifugal Pumps in a Piping Network
Objective The objective of this experiment is to illustrate the practical application of pumping and piping theory. Additionally, this experiment will exercise your ability to deal with open-ended problems. Introduction Consider a centrifugal pump used to deliver a fluid through a pipeline that contains various fittings and an elevation change. A typical performance curve for the pump is given in Figure 1. Figure 1 indicates that the head increase, ∆h, across the pump declines as the flow rate, Q, increases. However the head loss increases with increasing flow (why?) for the piping system and is typically of the form ∆h = α + βQ 2 for turbulent flow. The constant, α, is a hydrostatic term and if positive would imply that the fluid is being pumped to a higher elevation. The constant, β, represents the frictional losses through the system due to the resistance of the pipe and associated fittings. In this particular example, there is s stable point of operation, P, where the two curves cross. Clearly the same flow passes through the pump and piping system, and the head increase provided by the pump must equal the head loss in the pipeline.

Pump Head Loss or Increase ∆h P

Piping System

Q Figure 1. Centrifugal Pump Performance in a Piping Network

Flow Rate

In addition to exploring the behavior of a centrifugal pump in a piping system, you will also study frictional losses through pipes, valves and fittings in this experiment. Equipment The equipment is a pumping-piping system located in the Unit Operations Laboratory. It allows flow to be directed through any of six flow paths with different restrictions. Flow is measured using rotameters, paddle wheel flow meters and/or timed weighings. The pressure drop across a

CHEE 4500 Summer 2002

CHEE 4500 Summer 2002

Figure 2. Physical Schematic of Pressure Measurement System

Figure 3. Manometer Schematic Equations for the Manometer Force balance equations:
PA − (hH − hT )( g / g c ) ρ H 2O + (hH − (hx + hA ))( g / g c ) ρ H 2O = PM A PB − (hH − hT )( g / g c ) ρ H 2O + (hH − (hx + hB ))( g / g c ) ρ H 2O = PM B

Subtracting the second equation from the first we get:

CHEE 4500 Summer 2002

PA − PB + (hB − hA )( g / g c ) ρ H 2O = PM A − PM B

Thus:
∆P = ( PA − PB ) = ( PM A − PM B ) + (hA − hB )( g / g c ) ρ H 2O
(pressure drop across device) = (difference in meter readings) + (correction for difference in fluid height in clear tube)

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