Muhammed Fuad Rashid Petroleum Engineering Department at Koya Universiy Fluid Mechanics Laboratory 2020

1. Erbil Polytechnic University
Koya Technical Institute
Petroleum Technology
Operation and Control
Report
Fluid Mechanic Lab.
Test no: (9)
Test name:
(Minor Losses)
Supervised by:
Karwan A. Ali
Date of Test: 05/04/2018
Date of Submit: 03/05/2018
Prepared by: Muhammed Shwan Ali

2. Title Page No.
Introduction 3
Aim of the
experiment
3
Unit description 3-5
Calculation 5-9
Discussion 9-10
Table of Reading 11

3. -Introduction:
Pipe system which include valves, elbows (Bends), enlargements,
contractions, inlets, outlets, and other fittings that cause additional
losses, each of these devices causes a change in the magnitude or the
direction of the velocity vectors and hence results 1n a loss. A minor
loss is expressed in terms of a loss coefficient (K), defined by:
𝑉2
2𝑔
=eH
In this experiment we will calculate the minor losses due special pipe
component such as pipe bends or elbows, pipe branches, changes in
cross-section, valves, and flaps
-Aim of experiment:
Calculating the loss coefficient (K) for pipe bends or elbows, pipe
branches, changes in cross-section, valves, and flaps.
-Unit description:
The unit as shown in the figure consists of a square tubular steel
frame with a Powder-‘coated back wall, on which a, pipe system is
mounted with sections which can be. Individually shut off. The back
wall also features two level-tube pressure gauges attached using star-
type nuts. The gauges can be fitted in two positions on the back Wall.
Various Measurement objects can be accommodated in an adjustable
Measurement system“
Water is supplied either-by way of the HM 150 Hydraulics Bench
or via the laboratory mains. The HM 150 permits construction of
closed water Circuit.

4. -gauge:PressureDouble
-The double pressure gauge is suitable for measuring both differential
pressures and gauge pressures in mm w.g.; these can then be
converted into absolute pressures with allowance for the atmospheric
pressure. -The measuring range is O-1000mm w.g.
-The gauge consists of two glass level tubes backed by a metal mm
scale. -The two level tubes are interconnected at the top and have a
joint vent valve. -Differential pressure is measured with the vent
valve closed and gauge pressure with the valve open.
-The measurement points are connected to the lower end of the Level
tubes using rapid-action hose couplings with automatic shut-off.
-A drain valve 13 provided at the bottom of each level tube.
-of experiment:Performance
*The following instructions for experimentation and the performance
of the experiments A, B are based on the HM 150 Hydraulics Bench.
Position test set-up on the HM 150 Hydraulics Bench with drainage
via volumetric tank.
* Loosen star-type nuts for pressure gauge attachment on back of unit
and move gauges down a hole. Then retighten nuts
* Make hose connection between HM 150 and unit.
*Open drain of HM 150.
* Switch on pump and slowly open main coke of HM 150. * Connect
pressure gauges to desired measurement points.
* Slowly Open ball cock of appropriate measurement system and
vent pressure gauges; see Section of double pressure gauge.

5. * By simultaneously adjusting vent and drain valve on pressure
gauge, set water level such that both water columns are in the
measuring range.
*Determine volumetric flow. To do so, establish time t required to
raise the level in the volumetric tank of the HM 150 from 10 to 20 or
30 litters. The drain cock beneath the tank is to be closed for this
purpose.
-Pipe elbow experiment:
For pipe elbows, the loss coefficient (K) depends on the angle of
deviation of the flow and the ratio of the elbow radius to the pipe
diameter. In addition, the coefficient of resistance is influenced by the
shape of the elbow. For this special case of a pipe elbow with 90°
deviation, the following diagram is applicable for smooth and rough
pipes.
For pipe angles, i. e Elbow radii less than the pipe diameter (R/d<1)
the losses coefficients for knee pieces are approximately applicable.

6. For example, for a 90° knee piece / kink, with a smooth pipe, the K is
1.13 and for rough pipes the K 18 1. 68, while for a 45 piece
= 0.36.rough=.0.24 and KsmoothK
-Table of reading:
Time
second
Volume
liter
Qrotameter
Valve
Pipe elbow
No.
Elbow
single
45 ̊
Double
elbow
90 ̊
Single
elbow
90 ̊
Δh (cm)Δh (cm)Δh (cm)
19.84511901.87.74.21
16.2514001.510.452
14.86515203.61263
-Calculation:
Volume=5 L → 5000cm3
g=981cm/s2
d=20mm→0.2cm
A=3.14cm2
A/single elbow:
t=19.84sΔh=4.2cmNo.1:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
19.84
= 252.016=1Q
V1=
𝑄
𝐴
=
252.016
3.14
= 80.26cms
K1=
2𝑔∆ℎ
𝑉2 =
2×981×4.2
80.262 =1.217
cm4.2=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e1h

7. t=16.2sΔh=5cmNo.2:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
16.2
= 308.641=2Q
V2=
𝑄
𝐴
=
308.641
3.14
= 98.293cms
K2=
2𝑔∆ℎ
𝑉2 =
2×981×5
98.2932 = 1.015
= 5cm𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e2h
s4.86t=1Δh=6cmNo.3:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
16.2
= 336.473=3Q
V3=
𝑄
𝐴
=
336.473
3.14
= 107.157cms
K3=
2𝑔∆ℎ
𝑉2 =
2×981×5
107.1572 = 1.025
cm6=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e3h
B/Double elbow:
t=19.84sΔh=7.7cmNo.1:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
19.84
= 252.016=1Q
V1=
𝑄
𝐴
=
252.016
3.14
= 80.26cms
K1=
2𝑔∆ℎ
𝑉2 =
2×981×7.7
80.262 =2.345
cm7.7=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e1h

8. t=16.2sΔh=10.4cmNo.2:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
16.2
= 308.641=2Q
V2=
𝑄
𝐴
=
308.641
3.14
= 98.293cms
K2=
2𝑔∆ℎ
𝑉2 =
2×981×10.4
98.2932 = 2.112
cm10.4=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e2h
t=14.86sΔh=12cmNo.3:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
16.2
= 336.473=3Q
V3=
𝑄
𝐴
=
336.473
3.14
= 107.157cms
K3=
2𝑔∆ℎ
𝑉2 =
2×981×12
107.1572 = 2.050
cm12=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e3h
C/Elbow Angle 45 ̊ :
t=19.84scmΔh=1.8No.1:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
19.84
= 252.016=1Q
V1=
𝑄
𝐴
=
252.016
3.14
= 80.26cms
K1=
2𝑔∆ℎ
𝑉2 =
2×981×1.8
80.262 =0.548
cm1.8=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e1h

9. t=16.2scm2.5Δh=No.2:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
16.2
= 308.641=2Q
V2=
𝑄
𝐴
=
308.641
3.14
= 98.293cms
K2=
2𝑔∆ℎ
𝑉2 =
2×981×2.5
98.2932 = 0.507
5cm2.=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e2h
t=14.86s6cm3.Δh=No.3:
s3
cm
𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑖𝑚𝑒
=
5000
16.2
= 336.473=3Q
V3=
𝑄
𝐴
=
336.473
3.14
= 107.157cms
K3=
2𝑔∆ℎ
𝑉2 =
2×981×3.6
107.1572 = 0.615
6cm3.=𝐾
𝑉2
2𝑔
→
2𝑔∆ℎ
𝑉2 ×
𝑉2
2𝑔
= ∆ℎ=e3h
Table of Calculation:
V
(cm/s)
Pipe elbow
No.
Elbow Angle 45 ̊Double elbow 90 ̊Single elbow 90 ̊
he
(cm)
K
Δh
(cm)
he
(cm)
K
Δh
(cm)
he
(cm)
K
Δh
(cm)
80.261.80.5481.87.72.3457.74.21.2174.21
98.2932.50.5072.510.42.11210.451.01552
107.1573.60.6153.6122.0501261.02563
-Discussion:

10. * Which elbow has minimum loss coefficient and why?
A/Elbow Angl
./2g2
* Draw the relating between Ah & velocity head V
3159638
3344638
3529638
3714638
3899638
4084638
4269638
4454638
4639638
4824638
5009638
5194638
5379638
5564638
5749638
5934638
4.2 4.4 4.6 4.8 5 5.2 5.4 5.6 5.8 6 6.2
V2/2g
Δh
Single Elbow 90̊
3159638
3344638
3529638
3714638
3899638
4084638
4269638
4454638
4639638
4824638
5009638
5194638
5379638
5564638
5749638
5934638
7.6 8 8.4 8.8 9.2 9.6 10 10.4 10.8 11.2 11.6 12
V2/2g
Δh
Double Elbow 90 ̊

11. 3159638
3344638
3529638
3714638
3899638
4084638
4269638
4454638
4639638
4824638
5009638
5194638
5379638
5564638
5749638
5934638
1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 3.1 3.2 3.3 3.4 3.5 3.6
V2/2g
Δh
Elbow Angle 45̊