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Fluid Flow in a Smooth Pipe Essay

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Experiment 1
Fluid Flow In A Smooth Pipe
Abstract
In this experiment, three variable flow meters are used to alter the flowrate. Changes in pressure drop due to the change in flowrate are then observed from the three pressure gauges that can measure pressure at different range and recorded. The shift from laminar flow to turbulent flow is seen from the results recorded, but it is observed more clearly from the water-soluble dye experiment that was carried out by the demonstrator. Laminar flow turns to be turbulent when the Reynolds Number goes above a certain value, around 2000.

Aims
To look at how the pressure drop changes when the average velocity is altered in a circular pipe and to plot a graph of Friction Factor versus Reynolds …show more content…

Fanning friction, f: f=∆PLd2ρV2 f=191001.50.01262999.44(3.56)2 f=0.0063 Head loss, hf:
∆P=ρghf
19100=999.449.81(hf) hf=1.948 m

Using the definition gz1+V122+P1ρ=gz2+V222+P2ρ , the condition z1≠z2 is true if the two pressure taps are not horizontal (at different height).
While the condition V1≠V2 is true if the cross sectional area of the pipe is not the same from the first pressure tap to the second pressure tap.

Considering a viscous liquid that is being pumped through a smooth pipe with the parameters: ρ=1460 kg/m3 μ=5.2×10-1Ns/m2 D=0.1 m Q=5×10-2 m3/s

To determine the velocity,
V=QA
V=5×10-2π0.124 m/s
V=6.37 m/s
Then find the Reynolds Number,
Re=ρVDμ
Re=14606.370.15.2×10-1
Re=1788.5
According to Figure 2, the Fanning friction factor is 0.007.
The Bernoulli equation:
∆Pρ+g∆z+∆12V2+2fLV2D+Ws=0
Horizontal pipe, so ∆z=0
Constant pipe cross sectional area, so ∆12V2=0
Also, work done by pump, WP=-Ws
So the Bernoulli equation is reduced to
∆Pρ+2fLV2D-WP=0
WP=∆P1460+20.007L6.3720.1
F=WPL=6.85×10-4∆PL+5.68 N
Conclusion
A

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