Experiment : Friction in Pipes

EXPERIMENT 2: FRICTION IN PIPES
INTRODUCTION
In engineering practice, it is frequently necessary to estimate the head loss incurred by a fluid
as it flows along a pipeline. In this experiment scientific knowledge gained in classwork was
applied to investigate Frictional losses experienced by a fluid as it flows through a pipe.
Experiment set up
Figure 1: pipe friction apparatus
Image source: lab manual
METHOD
Pipe friction apparatus was set up as shown above and flowrate set using the dark blue gate
valve(D)
Readings were then taken using the Gravimetric Bench and Heights of piezometers 3 and 4
recorded as shown below.

RESULTS
Friction in Pipes Results:
Pipe length: 914mm =0.914m
Pipe Diameter: 13.6mm =0.0136m
Mass of water
collected (Kg)
Time
(s) H3(m
m)
H4(m
m)
hf(h3-
h4)
(m)
Friction
factor(f)
(Re)
6 23.61 400 200 0.200 0.0191 238
00
6 28.22 435 237 0.198 0.0274 198
56
6 30.29 438 265 0.173 0.0316 184
96
6 35.81 440 295 0.145 0.0441 156
40
6 39.08 440 315 0.125 0.0520 144
16
6 45.65 442 348 0.094 0.0714 122
94
Sample calculations
L= 0.914 m
Cross-sectional area, A= π ( d) 2
4
A= π (13.6 × 10−3) 2
4 =1.453 ×10-4 m2
flow rateQ= volume collected
elapsed time
(i); Q= 6.0 ×10−3 m3
23.61 sec =2.541×10-4 m3/sec

(ii); Q= 6.0 ×10−3 m3
28.22 sec =2.126×10-4 m3/sec
(iii) Q= 6.0 ×10−3 m3
30.29 sec =1.981×10-4 m3/s
(iv); Q= 6.0 ×10−3 m3
35.81 sec =1.676×10-4 m3/sec
(v); Q= 6.0 ×10−3 m3
39.08 sec =1.535×10-4 m3/sec
(vi) Q= 6.0 ×10−3 m3
45.65 sec =1.314×10-4 m3/s
velocity= flowrate Q
Area (equation 5)
v 1=
2.541 ×10−4 ( m3
sec )
1.453× 10−4 m2 =1.75 m/s
therefore f = 0.2 ( 13.6 × 10−3 ) 2 ×9.81
0.914 (1.75)2 =0.0191
v 2=
2.126 ×10−4 ( m3
sec )
1.453 ×10−4 m2 =1.46 m/s
therefore f = 0.2 ( 13.6 × 10−3 ) 2 ×9.81
0.914 (1.46)2 =0.0274
v 3=
1.981 ×10−4 ( m3
sec )
1.453× 10−4 m2 =1.36 m/ s
therefore f = 0.2 (13.6 × 10−3 ) 2 ×9.81
0.914 (1.36)2 =0.0316
v 4=
1.676 ×10−4 ( m3
sec )
1.453× 10−4 m2 =1.15 m/ s
therefore f = 0.2 ( 13.6 × 10−3 ) 2 ×9.81
0.914 (1.15) 2 =¿ 0.0441
v 5=
1.535 ×10−4 ( m3
sec )
1.453 ×10−4 m2 =1.06 m/ s

therefore f = 0.2 (13.6 × 10−3 ) 2 ×9.81
0.914 (1.06)2 =0.0520
v 6=
1.314 ×10−4 (m 3
sec )
1.453 ×10−4 m2 =0.904 m/ s
therefore f = 0.2 ( 13.6 × 10−3 ) 2 ×9.81
0.914(0.904)2 =0.0714
Average f= ( 0.0191+0.0274+ 0.0316+0.0441+0.0520+0.0714
6 )=0.0409
ℜ= ρVD
μ
( i ) ℜ= 1000 ×1.75 ×13.6 ×10−3
1 ×10−3 =23800
( ii ) ℜ=1000 ×1.46 × 13.6× 10−3
1×10−3 =19856
( iii ) ℜ= 1000× 1.36 ×13.6 ×10−3
1 ×10−3 =18496
( iv ) ℜ=1000 ×1.15 × 13.6× 10−3
1× 10−3 =15640
( v ) ℜ=1000 × 1.06× 13.6 ×10−3
1× 10−3 =14416
( vi ) ℜ=1000 × 0.904 ×13.6 ×10−3
1 ×10−3 =12294
Average Re= ( 23800+19856+18496+15640+14416+ 12294
6 ) =¿17417