1.0
INTRODUCTION
1.1
Literature Review
Head loss is a measure of the reduction in the total head (sum of elevation head, velocity
head and pressure head) of the fluid as it moves through a fluid system. Head loss is unavoidable
in real fluids.
There are two categories of head loss in pipe. One of them is due to viscous resistance
extending throughout the total length of the circuit. Next is due to localized effects such as
valves, sudden changes in area of flow and bends. Many factors affect the head loss in pipes, the
viscosity of the fluid being handled, the sizes of the pipes, the roughness of the internal surface
of the pipes, the changes in elevation within the system and the length of travel of the fluid.
The resistance through various valves and fittings will also contribute to the overall head
loss. A method to model the resistances for valves and fittings will be of minor significance to
the overall head loss, many designers choose to ignore the head loss for valves and fittings at
least in the initial stages of a design.
Frictional loss is that part of the total head loss that occurs as the fluid flows through
straight pipes. The head loss for fluid flow is directly proportional to the length of pipe, the
square of the fluid velocity, and a term accounting for fluid friction called the friction factor. The
head loss is inversely proportional to the diameter of the pipe.
The friction factor has been determined to depend on the Reynolds number for the flow
and the degree of roughness of the pipe’s inner surface.
1
The overall head loss is a combination of both these categories.
The arrangement of the apparatus is shown in the table below:
Dark Blue Circuit
Light Blue Circuit
A) Straight pipe 13.7 mm diameter
E) Sudden expansion – 13.6 mm/26.2 mm
B) 90o sharp bend (mitre) 0 mm radius
F) Sudden contraction – 26.2 mm/13.6 mm
C) Proprietary 90o elbow 12.7 mm radius
G) Smooth 90o bend 50.8 mm radius
D) Gate valve
H) Smooth 90o bend 100 mm radius
J) Smooth 90o bend 152 radius
K) Globe valve
L) Straight pipe length 26.4 mm
2
1.2
Theory
Head Loss in Straight Pipe
Darcy-Weisback and Reynold Number equation:
HL=
Re=
f=
4 𝑓𝐿𝑉²
2𝑔𝑑
=
32 𝑓𝐿𝑄²
𝜋²𝑔𝑑⁵
= KQn
𝜌𝑉𝑑
𝜇
64
𝑅𝑒
f= 0.316×Re -0.25
Laminar flow
Turbulent flow
where:
hL= Head losses
f= Friction factor
L= Length
𝑄
V= Mean velocity ( )
𝐴
g= Gravity
d= Constant diameter
3
Head Losses due to Sudden Changes in Area of Flow
I. Sudden Expansion
Expanding pipe
hL=
𝐾 (𝑉 ₁−𝑉 ₂)
2𝑔
Where:
V= Mean velocity
K= Dimensionless coefficient
II. Sudden Contraction
Contracting pipe
hL=
𝐾𝑉 ₂²
2𝑔
Where:
K= Dimension coefficient which depends upon the area ratio as shown in Table 1
A1/A2 0
K
0.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.45
0.41
0.39
0.36
0.33
0.28
0.15
0.15
0.06
0
Table 1 : Loss coefficients for sudden contractions
4
Temperature (oC)
ρ (kg/m3)
μ (x 10-3 Ns/m2)
0
999.8
1.781
5
1000.0
1.518
10
999.7
1.307
15
999.1
1.139
20
998.2
1.002
25
997.0
0.890
30
995.7
0.798
40
992.2
0.653
50
988.0
0.547
60
983.2
0.466
70
977.8
0.404
80
971.8
0.354
90
965.3
0.315
100
953.4
0.282
Table 2 : Table of ρ and μ depend on temperature.
5
Head Loss due to Bends
hb=
𝐾в 𝑉²
2𝑔
Where:
KB= Dimensionless coefficient which depends upon the bend/ pipe radius ratio and the angle of
the bend.
Head Loss due to Valves
hv =
𝐾ѵ 𝑉²
2𝑔
Where:
KV= Dimensionless coefficient which depends upon type of valve and degrees of opening
Valve
K
Globe valve, fully open
10.0
Gate valve, fully open
0.2
Gate valve, half open
5.6
Table 3: Values of loss coefficient for gate valve and globe valve
1.3
Objective
1. To find Reynold Number, friction factor and compare with the theoretical value.
2. To find the head loss as a function of volume flow rate.
6
2.0
METHODOLOGY
2.1
Procedure
1. It is important for the globe valve; gate valve and water control valve on the hydraulic
bench are fully close. (open-anti-clockwise, close- clockwise)
2. The pump is switched on by pressing the black button on the hydraulic bench and the
pump is to let running for a few minutes.
3. The readings on the piezometer tubes (No 1-16) and the U-tube are recorded.
4. The piezometer tubes are identified by fill in Table 1 and refer to Figure 1.
5. The water control valve on the hydraulic bench is opened slowly to check there is no
leaking along the pipeline.
6. The water control valve and gate valve are fully-opened to obtain maximum flow through
the Dark Blue circuit.
7. Time to collect 15 L of water on the hydraulic bench is recorded. The piezometer (no 1-6)
and U-tube (Gate valve) are read. The temperature of water on the hydraulic bench is
recorded.
8. A half-turn is made on the gate valve and procedure no 6 is repeated.
9. The above procedures are repeated for a total of 8 different valve opening/radings.
10. The gate valve is fully-closed and the globe valve is fully-opened. The same procedures
are repeated for the Light Blue circuit.
11. For globe valve, piezometer is recorded as no. 7-16.
12. Before switching off the pump, the globe valve, gate valve and water control valve are
fully-closed.
7
3.0
RESULTS AND DISCUSSIONS
3.1
RESULTS
TABLE 1: IDENTIFYING PIPING SYSTEM
DARK BLUE CIRCUIT
Pipe description
Piezometer tube no.
Straight pipe 13.7mm diameter
3 and 4
90o sharp bend (mitre) 0mm radius
5 and 6
Proprietary 90o elbow 12.7mm radius
1 and 2
Table 1
LIGHT BLUE CIRCUIT
Pipe description
Piezometer tube no.
Sudden expansion – 13.7mm/26.4mm
7 and 8
Sudden contraction – 26.4mm/13.7mm
9 and 10
Smooth 90o bend 50.8 mm radius
15 and 16
Smooth 90o bend 100 mm radius
11 and 12
Smooth 90o bend 152 mm radius
13 and 14
Straight pipe length 26.4 mm
8 and 9
Table 2
8
FOR GATE VALVE
Test
Time
no.
(s)
-
U-tube (cm)
Piezometer tube readings (cm) water
Hg
T
(oC)
1
2
3
4
5
6
Gate Valve
69
690
375
230
15
1050
630
355
365
29
1
78
695
375
230
10
1048
630
355
368
29
2
76
694
373
230
13
1050
630
355
368
29
3
71
693
372
229
13
1050
630
355
368
29
4
68
693
375
229
18
1049
630
355
368
29
5
69
692
378
228
15
1049
633
350
368
29
Table 3
FOR GLOBE VALVE
Test
Time
no.
(s)
U-tube (cm)
Piezometer tube readings (cm) water
7
8
9
Hg
10
11
12
13
14
15
16
T (oC)
Globe Valve
Initial
72
395
444
432 240
385
135
440
200
385
150
340
330
30
1
65
395
445
433 240
385
135
440
195
385
150
340
330
30
2
63
394
443
430 239
385
130
440
195
385
145
342
326
30
3
64
395
445
434 240
385
130
440
195
386
149
340
326
30
4
65
395
444
433 244
385
140
440
200
385
150
340
332
30
5
65
395
444
433 245
390
140
440
200
385
155
335
333
30
Table 4
9
3.2 Discussions
Determination of the coefficient K consists of plotting the experimental local loss
coefficients versus the corresponding Reynolds numbers, Re=
𝑉𝐷
𝑛
, where D is the diameter of the
pipe and n s the kinematic viscosity for known discharges through the device. The head loss of
the devices included in a pipe systems must be determine in order to be used in design
calculations to find the amount of head loss in the system, and consequently, the needed source
pressure or energy to satisfy the design objective.
Determination of the friction factor first requires a value for the head loss to be obtained. The
head loss represents the conversion of mechanical energy to unwanted thermal energy. The
thermal energy is generated as the flow interacts with the surface of the pipe producing heat and
thus producing a pressure loss in the pipe. The pressure loss for the experiment is determined
through the use of a manometer. The manometer reading was a direct measurement of the head
loss that occurred between the two points of interest.
We try to find Reynolds Number and the Friction factor experimentally and compare the
values with the theoretical values in this experiment. We also determined the head loss as a
function of volume flow rate. Thus, our experimental values are not accurate compare to
theoretical values.
The factors that affecting the values of K includes the flow Reynolds number and proximity
to other things which is the tabulated values of K are components in isolation with long straight
runs of pipe upstream and downstream.
There are a few precaution steps in order to get the more accurate result in this experiment.
We must ensure that we read the manual before start the experiment. Before running anything,
we should know how to control the flow in the apparatus. We also have to make sure there are no
bubbles in the pipeline system to eliminate disturbance to the fluid flow. The head loss of the
devices included in a pipe systems must be determine in order to be used in design calculations
to find the amount of head loss in the system, and consequently, the needed source pressure or
energy to satisfy the design objective.
10
3.3 Questions
1. Plot graph of Log10 hL versus Log10 Q on graph paper to get the value of K and n.
Calculate the value of friction factor, f and Reynold number, Re
Head loss, hL in straight pipe, so we use Piezometer 3 and Piezometer 4.
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Q (m3/s)(x10-4)
Log10Q
hL (m)
Log10hL
1.923
-3.716
2.20
0.342
1.974
-3.705
2.17
0.336
2.113
-3.675
2.16
0.334
2.206
-3.656
2.11
0.324
2.174
-3.663
2.13
0.328
Table 5
Graph of Log10 hL vs Log10 Q
0.345
0.340
Log10 hL
0.335
0.330
0.325
0.320
0.315
-3.716
-3.705
-3.675
-3.656
-3.663
Log10 Q
Graph 1
11
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear.
To assume the gradient of graph, the first and last point is taken into calculation.
Gradient of graph, m =
=
y 2  y1
x 2  x1
0.342  0.328
 3.716  (3.663)
= -0.264
Y= mX + c
hL = KQn
log hL = log K + n log Q
n=m
thus, n = -0.264
log hL = log K + n log Q
0.342 = Log K + (-0.264) (-3.716)
Log K = -0.639
K = 2.3×10-1
12
Velocity, V 
=
Q
A
𝑄
𝜋𝑟²
Reynolds Number, Re 
Vd

For ρ,
From table 2 in module,
Temperature (°C)
ρ (kg/ m3)
25
997.0
29
ρ
30
995.7
Table 6
𝜌−997.0
995.7−997.0
=
29−25
30−25
ρ = 995.96 kg/m3
For μ,
From table 2 in module,
Temperature (°C)
μ(×10-3 Ns/ m2)
25
0.890
29
μ
30
0.798
Table 7
𝜇−0.890
0.798−0.890
=
29−25
30−25
μ= 0.8164 ×10-3 Ns/ m2
13
Friction factor,
f=
64
Re
(Laminar flow)
f = 0.316 x Re-0.25
(Turbulent flow)
Velocity, v (m/s)
Reynolds number, Re
Type of Flow
Fiction factor, f (x10-3)
1.304
21794
Turbulent
26.0
1.339
22379
Turbulent
25.8
1.433
23950
Turbulent
25.4
1.496
25003
Turbulent
25.1
1.475
24652
Turbulent
25.2
Table 8
14
2. Plot graph of hL versus (V1- V2)2/ 2g for Sudden Expansion Pipe to get the K value and
compare with the theoretical value, K=1.0. Calculate the error.
Head loss, hL for Sudden Expansion Pipe, so we use Piezometer 7 and Piezometer 8.
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Velocity, V 
Volume
Time
Q
A
Q (m3/s)(x10-4)
hL (m)
V1 (m/s)
V2 (m/s)
(V1-V2)2/2g (m)
2.308
0.50
1.566
0.422
0.067
2.381
0.49
1.615
0.435
0.071
2.344
0.50
1.590
0.428
0.069
2.308
0.49
1.566
0.422
0.067
2.308
0.49
1.566
0.422
0.067
Table 9
Graph of hL vs (V1- V2)2/ 2g
0.502
0.5
hL (m)
0.498
0.496
0.494
0.492
0.49
0.488
0.0665
0.067
0.0675
0.068
0.0685
0.069
(V1- V2)2/
0.0695
0.07
0.0705
0.071
0.0715
2g (m)
Graph 2
15
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
0.49−0.50
0.071−0.067
= -2.5
hL  K
(V1  V2 ) 2
2g
Y= mX+ c
Therefore, m= K
Hence, K= -2.5
Calculated K value is smaller than theoretical K value.
Error =
1.0  (2.5)
100%
1.0
= 350%
16
3. Plot graph of hL versus (V2)2/ 2g for Sudden Contraction pipe to get the K value and
compare with the theoretical value, K= 0.30. Calculate the error.
Head loss, hL for Sudden Expansion Pipe, so we use Piezometer 9 and Piezometer 10.
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Velocity, V 
Q
A
Q (m3/s)(x10-4)
hL (m)
V2 (m/s)
(V2)2/2g (m)
2.308
1.93
1.566
0.125
2.381
1.91
1.615
0.133
2.344
1.94
1.590
0.129
2.308
1.89
1.566
0.125
2.308
1.88
1.566
0.125
Table 10
Graph of hL vs (V2)2/ 2g
1.95
1.94
1.93
hL (m)
1.92
1.91
1.9
1.89
1.88
1.87
0.124
0.125
0.126
0.127
0.128
0.129
(V2)2/
0.13
0.131
0.132
0.133
0.134
2g (m)
Graph 3
17
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
1.94−1.93
0.129−0.125
= 2.5
2
hL  K
V2
2g
Y= mX+ c
Therefore, m= K
Hence, K= 2.5
Calculated K value is larger than theoretical K value.
Error =
2.5  0.30
100%
0.30
= 733.33%
18
4. Plot graph of hL versus V2/ 2g for Bending pipe to get the value of K and compare with
the theoretical value, K= 0.37. Calculate the error.
For proprietary 90o elbow 12.7mm radius, we use Piezometer 1 and Piezometer 2.
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Velocity, V 
Q
A
Q (m3/s)(x10-4)
hL (m)
V (m/s)
V2/2g (m)
1.923
3.20
1.518
0.117
1.984
3.21
1.566
0.125
2.101
3.21
1.659
0.140
2.212
3.18
1.746
0.155
2.174
3.14
1.716
0.150
Table 11
Graph of hL vs V2/ 2g
3.22
3.21
3.2
hL (m)
3.19
3.18
3.17
3.16
3.15
3.14
3.13
0
0.02
0.04
0.06
0.08
V2/
0.1
0.12
0.14
0.16
0.18
2g (m)
Graph 4
19
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
3.21−3.20
0.140−0.117
= 0.435
hL  K
V2
2g
Y= mX+ c
Therefore, m= K
Hence, K= 0.435
Calculated K value is larger than theoretical K value.
Error =
0.435  0.37
100%
0.37
= 17.57%
20
For smooth 90o bend 50.8 mm radius, we use Piezometer 15 and Piezometer 16.
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Velocity, V 
Q
A
Q (m3/s)(x10-4)
hL (m)
V (m/s)
V2/2g (m) (x10-4)
2.315
2.35
0.114
6.624
2.381
2.40
0.117
6.977
2.358
2.37
0.116
6.858
2.315
2.35
0.114
6.624
2.284
2.30
0.113
6.508
Table 12
Graph of hL vs V2/ 2g
2.42
2.4
hL (m)
2.38
2.36
2.34
2.32
2.3
2.28
6.4
6.5
6.6
6.7
6.8
6.9
7
7.1
V2/ 2g (×10-4) (m)
Graph 5
21
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
2.37−2.35
6.858−6.624
= 0.085
hL  K
V2
2g
Y= mX+ c
Therefore, m= K
Hence, K= 0.085
Calculated K value is smaller than theoretical K value.
Error =
0.37  0.085
100%
0.37
= 77.0%
22
For smooth 90o bend 100 mm radius, we use Piezometer 11 and Piezometer 12.
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Velocity, V 
Q
A
Q (m3/s)(x10-4)
hL (m)
V (m/s)
V2/2g (m) (x10-4)
2.315
2.50
0.029
4.286
2.381
2.55
0.030
4.587
2.358
2.55
0.030
4.587
2.315
2.45
0.029
4.286
2.284
2.50
0.029
4.286
Table 13
Graph of hL vs V2/ 2g
2.58
2.56
hL (m)
2.54
2.52
2.5
2.48
2.46
2.44
4.2
4.25
4.3
4.35
4.4
4.45
4.5
4.55
4.6
4.65
V2/ 2g (×10-4 ) (m)
Graph 6
23
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
2.55−2.50
4.587−4.286
= 0.166
hL  K
V2
2g
Y= mX+ c
Therefore, m= K
Hence, K= 0.166
Calculated K value is smaller than theoretical K value.
Error =
0.37  0.166
100%
0.37
= 55.0%
24
For smooth 90o bend 152 mm radius, we use Piezometer 13 and Piezometer 14.
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Velocity, V 
Q
A
Q (m3/s)(x10-4)
hL (m)
V (m/s)
V2/2g (m) (x10-6)
2.315
2.45
0.0128
8.35
2.381
2.45
0.0131
8.75
2.358
2.45
0.0130
8.61
2.315
2.40
0.0128
8.35
2.284
2.40
0.0126
8.09
Table 13
Graph of hL vs V2/2g
2.46
2.45
hL (m)
2.44
2.43
2.42
2.41
2.4
2.39
8
8.1
8.2
8.3
8.4
V2/
2g
(×10-6 )
8.5
8.6
8.7
8.8
(m)
Graph 7
25
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
2.40−2.45
8.09−8.35
= 0.19
hL  K
V2
2g
Y= mX+ c
Therefore, m= K
Hence, K= 0.19
Calculated K value is smaller than theoretical K value.
Error =
0.37  0.19
100%
0.37
= 48.6%
26
5. Plot graph of hL versus V2/ 2g for loss due to valves to get the value of K.
For gate valves,
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Velocity, V 
Q
A
Q (m3/s)(x10-4)
hL (m)
V (m/s)
V2/2g (m)
1.923
0.13
1.305
0.0868
1.984
0.13
1.346
0.0923
2.101
0.13
1.425
0.1035
2.212
0.13
1.501
0.1148
2.174
0.18
1.841
0.1727
Table 14
Graph of hL vs V2/ 2g
0.2
0.18
0.16
hL (m)
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
0.02
0.04
0.06
0.08
0.1
V2/
0.12
0.14
0.16
0.18
0.2
2g (m)
Graph 8
27
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
0.18−0.13
0.1727−0.0923
= 0.62
hL  K
V2
2g
Y= mX+ c
Therefore, m= K
Hence, K= 0.62
28
For globe valve,
Head loss, hL = Piezometer height 1 – Piezometer height 2
Volume flow rate, Q 
Volume
Time
Velocity, V 
Q
A
Q (m3/s)(x10-4)
hL (m)
V (m/s)
V2/2g (m) (×10-3)
2.315
0.10
0.423
9.120
2.381
0.16
0.435
9.644
2.358
0.14
0.431
9.468
2.315
0.08
0.423
9.120
2.294
0.02
0.419
8.948
Table 15
Graph of hL vs V2/ 2g
0.18
0.16
0.14
hL (m)
0.12
0.1
0.08
0.06
0.04
0.02
0
8.9
9
9.1
9.2
9.3
V2/
9.4
9.5
9.6
9.7
2g (×10-3) (m)
Graph 9
29
Due some unforeseen error, the result obtained is not accurate. Hence the graph is not linear. To
assume the gradient of graph, only two points are taken into calculation.
Gradient of graph, m =
=
𝑦₂−𝑦₁
𝑥₂−𝑥₁
0.16−0.10
9.644−9.120
= 0.115
hL  K
V2
2g
Y= mX+ c
Therefore, m= K
Hence, K= 0.115
30
4.0
Conclusion
The K values obtained from the experiment deviates greatly from the theoretical value of K. It is
shown that head loss is affected by volume flow rate and there is a specific relationship between
them. Therefore, precaution is essential to reduce error in the experiment.
31
5.0
References



http://www.engineeringtoolbox.com/total-pressure-loss-ducts-pipes-d_625.html
http://me.queensu.ca/courses/MECH441/losses.htm
http://www.kmisystemsinc.com/files/Technical%20References/PressureLoss.pdf

Gupta, R. S. (1989). Hydrology and Hydraulic Systems, Waveland Press, NY

Rouse, H. (1949). Engineering Hydraulics, John Wiley and Sons, NY

White, F.M. (1994). Fluid Mechanics, 3rd edition, McGraw-Hill, Inc., New York, NY.
32