EXPERIMENT 1: HYDROSTATICS
I. OBJECTIVE
To understand the basic equation of the hydrostatics and apply it to some problems associated
with uncompressed fluid at static state.
II. SUMMARY OF THEORY
Figure 1.1
The state of the static, uncompressed fluid is described by the basic equation:
𝑝
(1.1)
𝑧 + = 𝑐𝑜𝑛𝑠𝑡
𝛾
With z being the elevation of any point in the static fluid mass with constant specific weight
γ and p is the hydrostatic pressure at that point, the value of 𝑧 + 𝑝 is the water pressure gauge,
𝛾
it is in meters. We can have several uses from this equation:
1.
Isobaric surface
The isobaric surface is the surface on which the pressure is the same. When the pressure at
the points on the opened sides is equal, the elevation of the points must be equal. That means
the surface is horizontal.
2.
Fluid manometer
Apply the basic equation (1.1) to calculate the pressure at any point in a static fluid mass. If
at a point with elevation z0, the pressure is p0, then at the point of elevation z (see Figure 1.1),
the pressure will be:
(1.2)
𝑝 = 𝑝0 + 𝛾 (𝑧0 − 𝑧) = 𝑝0 + 𝛾ℎ
In which the 0-0 plane is the standard plane for comparison.
By measuring height h, we will determine the pressure p.
3. Calculate the specific weight of the liquid
If knowing both p0, p, z0 and z, we can calculate the specific weight of the liquid from the
formula:
p  p0
(1.3)
 
z0  z
Basic equation (1.1) is established when the influence of capillary phenomenon is neglected.
In case the liquid inside a small tube which has a diameter of d ≤ 3mm, the effect of the
capillary phenomenon is significant, equation (1.1) is no accurate. If the liquid is water,
alcohol, oil ... the liquid level in the small diameter tube will rises higher than in the large
diameter tube, and if the liquid is mercury, the liquid level in the small diameter tube will
decrease lower than in the large diameter tube.
1
Ruler (mm)
I.
III.
EQUIPMENT
Hydrostatics experiment Set
The pressure gauge consists of a group of tubes numbered 1 to 10 mounted on a board with
rulers to measure elevation, and accuracy to the millimetre shown in Figure 1.2. Diameter of
tubes di = 5mm (except for group of tubes 2 with diameter ≤ 3mm).
In these tubes:
−
Pipes 1-3 are used as fluid manometer to measure the pressure in the tank T
Group of tubes 2, consisting of 3 tubes: 21, 22, 23 with diameters of 1mm, 2mm, 3mm,
−
respectively, are used to observe capillary phenomena.
−
Pairs of tubes: 4-5, 6-7 and 8-9 contain liquid to determine their specific weight.
Tubes 10 and 3 are used to observe the isobaric surface. The pressure of the tubes 1,
−
4, 6 or 8 is the gas pressure in the tank T. The liquid level in the tubes is denoted by Li
The pressure generating part consists of the tank T in the back of the board and the tank D
hanging on the pulley 12. By rotary handle 11 we can change the height of the tank D, making
gas pressure in the tank T change.
II.
IV.
PROCEDURE
1- Check if the standard surfaces of the measured rulers (on the millimetre table) is horizontal
or not by reading the water levels in tubes 3 and 10. These water levels must be horizontal.
2- Use the rotary handle 11 to put the tank D up to a high position (the opened surface of the
tank D is higher than the opened surface of the tank T about zĐ - zT = 15 ÷ 20cm).
− Measure the liquid level in tubes from 1 to 10 and write down in the table in the report.
(including group of tubes 2).
2
−The tank D is lowered to the average position (zĐ - zT = 5 ÷ 7cm) and low position (zĐ - zT
= -15 ÷ -20cm). Measured similarly to the first time and write down last time and record the
results in the table (of the report).
V. ANALYSIS
1. Calculate hydrostatic pressure of gas in the tank T
From equation (1.2), the absolute pressure in the tank T is calculated:
p T  p a   H2 O L 3  L1 
(1.4a)
The specific weight of the water in the tank is determined by the temperature of surrounding
environment. If we take pa = 0, we can calculate the gauge pressure
(1.4b)
pT   H O  L3  L1 
2
With three positions of the tank D, we can calculate three values of the gas pressure in the
tank T.
2. Calculate to the specific weight of the fluid in the U-tubes
Consider to fluid has the specific weight γ, in static state in both branch i and i + 1 of the Utube. Pressure in the tank T can also be calculated:
p T  p a   L i 1  L i 
(1.5)
Comparing this formula with formula (1.4a), we derive the formula for the specific weight of
this liquid
L 3  L1 
(1.6)

Li1  Li 
H2 O
3. Calculate the error of the method of pressure measurement method
Applying the theory of error, from the formula (1.4b) we determine the relative error δp of the
gauge pressure measurement:
p 


 H2 O L 3  L 1
 p T    H2 O L 3  L 1 



pT
 H2 O L 3  L 1
 H2 O
L 3  L1
(1.7)
Thus, the relative error of the gauge pressure measurement is the sum of the two relative error
values
The relative error due to the determination of the specific weight of the water:
 H2 O   H2 O  H2 O
The relative error due to reading elevation on the scale
 L 3 1 
Thus:
L 3  L 1
L 3  L1
 p   H O   L
31
2
(1.8)
(1.9)
Usually we take:
 H O = 0,12% and ΔL1 = ΔL3 = 0,5mm
4. Calculate the error when calculating γ of the liquids in the U-tubes
From formula (1.7), we determine the relative error of specific weight measurement of the
liquid in the U-tubes
(1.10)
  H 2 O L 3  L1 L i  1  L i
2
 


 H2 O


L 3  L1
L i 1  L i
    H O   L   L
or:
2
3 1
i 1 i
We also take  H O = 0,12% và ΔL1 = ΔL3 = ΔLi = ΔLi+1 = 0,5 mm
2
3
(1.11)
DATE:
NAME OF INSTRUCTOR:
EXPERIMENT 1: HYDROSTATICS
PREPARATION:
(Students must complete this section before coming to the lab. Failure to do so will result
in the student not being allowed to participate in the experiment.)
1. What should be done to check the reference surface of the measuring tools?
I.
2. What data do we need to collect when conducting the hydrostatic experiment?
3. How many scenarios are tested in the hydrostatic experiment, and what are these scenarios?
4. How can the air pressure in the vessel be altered between measurements?
5. Why do we need to measure the air pressure and temperature in the room?
4
DATE:
NAME OF INSTRUCTOR:
II. MEASUREMENT DATA:
The atmospheric pressure and temperature during the experiment are:
Pa = …………………mmHg; t0 …………………………0C
The water specific weight:
γH2O = ………………………………………. N/m3
For the three positions of container Đ compared to container T, record the measured values
from the 9 pressure tubes and group 2 tubes into Table 1.
Table 1a. Measurement (Unit: cm)
No
L1
L3
L4
L5
L6
L7
L8
L9
L10
NOTE
1
2
3
Table 1b. Measurement in group of tubes 2 (Unit: cm)
No
L21
L22
L23
NOTE
1
2
3
III. RESULTS CALCULATION AND PRESENTATION OF EXPERIMENTAL
RESULTS
1. Calculate the absolute pressure and the excess pressure of the gas in vessel T, along with
the relative error of the pressure for each measurement. Enter the results into Table 2.
2. Calculate the specific weights of three liquids (from pairs of tubes 4-5, 6-7, and 8-9), along
with the relative error for these specific weights. Enter the results into Table 2
Table 2. Result
pt
No
103N/m2
pd
δρ
γ4-5
γ6-7
103N/m3
%
5
γ8-9
δγ4-5
δγ6-7
%
δγ8-9
1
2
3
3. In the hydrostatic system, which tubes or tanks have equal water levels? Why?
4. In the hydrostatic system, which tubes do not follow the hydrostatic law? Why?
5. How do the pressure in tank T change in experiments? Why?
6. In the same measurement case, how does the measurement error 𝛿𝛾 change in relation to
𝛾, and why?
6
7
1
EXPERIMENT 3A: ENERGY EQUATION
I.
FUNDAMENTAL THEORY
a.
Objectives
Investigating the energy equation and calculate related parameters in flow when changing
the area of cross sections.
b.
Theory
The energy of a unit weight of fluid at a section is
determined by 3 components:
-
Potential energy: zi
-
Flow energy (pressure head): pi/g
-
Kinetic energy (velocity head):
Figure 2.1
In which:
-
zi; pi, Vi are height, pressure and velocity at section i-i, respectively.
 is kinetic energy correction factor. The occurrence of  is due to the
inhomogeneous distribution of velocity in the flow section, given by formula:
3

1 u
   dA
A AV
(3.1)
In which:
-
V: average velocity
-
u: point velocity
To simplify calculation,
for turbulent flows
The energy equation from section 1-1 to section 2-2:
z1 
p 1  1V12
p
 V2

 z 2  2  2 2  h f12

2g

2g
(3.2)
In which:
h f12 : The head loss due to friction from section 1 to section 2
Neglecting total head loss, the energy equation becomes:
(3.3)
From the equation above (3.3), if neglecting total head loss, the total head of the considered
interval is a constant and there is transformation of energy between potential energy and
kinetic energy.
2
II. EQUIPMENT:
Figure 2.2
The experimental equipment (see Figure 2.2) consists of the following components:
Water is pumped from the reservoir (6) to the holding tank (1), flowing into the glass channel
via valve (2), which adjusts the flow rate.
●
●
The glass channel (3) is a rectangular, horizontal channel with a bottom width of B = 78 mm.
A broad crested weir (4) with a trapezoidal cross-section, where the sides are inclined at a
45° angle and the step height is a = 33.1 mm.
●
The water level downstream of the step is adjusted using valve (5) located at the end of the
channel. The water then flows back into the reservoir (6) through a rectangular weir.
●
A coordinate measuring probe (7) is mounted on the glass channel (3) to measure the elevation
of the channel bed and the water level inside the channel.
●
.
Figure 2.3
III.
III. PROCEDURE
Determine the location of the sections on the glass channel corresponding to points 1 to 6 in
the order from the upstream to downstream levels as shown in Figure 3.4. The distance
between sections is as follows:
L1-2 = L2-3 = L4-5 = L5-6 = 20cm; L3-4 =3.7cm
Determine bottom of the channel from 1 to 6 by point gauge and vernier. Write down the
results in Table 1 (in the report).
Use the valve ⑤ to adjust the flow and water level in the channel so that the downstream
water level is higher than the water level on the broad crested weir.
3
Figure 3.4
The flow in the channel is stabilized, using point gauge and vernier to determine Zi of water
level from sections 1 to 6. Note down the results in Table 1 (in the report).
Keeping the discharge constant, adjust the water level at downstream by the valve ⑤ (this
time the downstream water level is lower than the water level on the broad crested weir). Wait
for the water level in the glass channel to be stable, repeat the procedure in step 4. Write down
the results in Table 1 (in the report).
IV. CALCULATING
1.
Calculations of velocity:
The average velocity at section i:
Vi = Q/Ai
(3.4)
In which:𝑄 = 𝐶𝑑 2 𝐵0 ℎ0 √2𝑔ℎ0
3
Ai is the area of the cross-section i: Ai = Bhi;
Channel width is 𝐵0 = 1,15m;
𝑃1 = 0,42𝑚 (height of weir)
ℎ0 = 𝑍𝑑𝑏 − 𝑍0 ; 𝑍𝑑𝑏 is elevation of weir, 𝑍0 𝑖𝑠 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑖𝑛 𝑤𝑒𝑙𝑙
ℎ0
𝐶𝑑 = 0,605 + 0,08
𝑃1
Water level from the bottom of the channel: hi = |Zđi - Zi|; (see figure 3.3)
Notice:
2.
Check if the whole system is safe to operate
Switch ON switchboard.
Observe the water level in the glass channel to prevent the overflow of water.
Calculations of velocity head:
The average velocity head at section i:
hVi  Vi2 / 2g
(3.5)
3.
Investigate the energy loss:
The energy loss between sections i and sections j was determined by applying the Bernoulli
equation (3.1) for these sections:
𝑝
𝛼𝑉𝑖2
𝛾
2𝑔
ℎ𝑓𝑖𝑗 = (𝑧𝑖 + 𝑖 +
𝑝
𝛼𝑉𝑗2
𝛾
2𝑔
) − (𝑧𝑗 + 𝑗 +
4
) = (𝑧𝑖 + ℎ𝑉𝑖 ) − (𝑧𝑗 + ℎ𝑉𝑗 )
(3.6)
In which zi and zj are the water level at i section and j section; pi


pj

0
. where α = 1 and
average velocity head is given by (3.5).
Apply (3.6) to calculate the energy loss form section 1 to section 2 (hf1-2), from section 2 to
section 3 (hf2-3), from section 3 to section 4 (hf3-4), from section 4 to section 5 (hf4-5), from
section 5 to section 6 (hf5-6). Calculate for cases of water level.
4.
Draw water level lines in the channel:
From the measured values hi and the calculated values hvi, plot the energy variation along the channel
from cross-section 1 to cross-section 6 in two cases:
Case 1: Ignore energy losses between cross-section 1 and cross-section 6 for the case water
level downstream is lower than the water level on the broad crested weir.
Case 2: Include energy losses between cross-section 1 and cross-section 6 for the case water
●
level downstream is lower than the water level on the broad crested weir
●
5
DATE:
NAME OF INSTRUCTOR:
EXPERIMENT 3A: ENERGY EQUATION
I.
PREPARATION:
(Students must complete this section before coming to the lab. Failure to do so will result
in the student not being allowed to participate in the experiment.)
1.
How can the
water level and the channel bed elevation be measured?
2.
How can the
water level in the glass channel be adjusted? How many downstream water levels are
tested?
How many types of energy losses are encountered in this experiment?
6
DATE:
NAME OF INSTRUCTOR:
II.
MEASUREMENT DATA:
Zdb =
cm.
Z0r in the well=
cm.
3
Q=
m /s
Measurement of the bottom of the glass channel Zđ, the water surface Zi in the glass channel
at the cross section corresponding to different water level, the results recorded in Table 1
Table 1 Measurement data
Section
No
1
Bottom height
Zđ, cm
1
2
3
4
0
90
90
18
0
90
180
198
5
6
Water level Zi,
cm
Distance from section i
to section i+1, cm
Accrual distance from
section 1 to section i+1,
cm
III.
2
CALCULATIONS AND RESULTS
1. Calculate the velocity of the flow and the average velocity head at the cross-sections of
Equations (3.4), (3.5). Calculating for two measured cases. The results are recorded in Table
2.
2. For two measured cases, calculating the energy loss (hf1-2) between the cross sections 1 - 2,
hf2- 3 between the cross sections 2 - 3, hf3-4 between the cross sections 3-4, hf4-5 between the
cross sections 4 - 5, hf5-6 between the cross sections 5 - 6 according to the formula (3.6). The
results are recorded in Table 2.
3. In figure 1, draw the bottom of the channel:
a) Based on the results in table 1, drawing on Figure 1 a "measuring waterline" (drawing for
the first water level).
b) Based on the results in table 1, drawing on Figure 1 a " ideal " waterline (drawing for the
first water level).
c) Discussing two "measuring" waterlines and "ideal" waterlines?
7
4. Discussing the water level between section 5 and section 6?
5.
Please comment, compare, explain the energy loss of calculations power between the
sections in table 2.
8
Table 2 Results
Energy losses
hf5 -6, cm
Energy losses
hf4 - 5, cm
Energy losses
cm
hf3–4,
Energy losses
hf2 -3, cm
Energy losses
hfl -2, cm
sec
6
Average
Velocity
head hvi,
cm
sec
5
sec
4
sec
3
sec
2
sec
1
sec
6
sec
5
sec
4
Average
Velocity
Vi, cm/s
sec
3
sec
2
sec
1
No
1
9
2
EXPERIMENT 3D MEASUREMENT OF VOLUMETRIC FLOW RATE
I.
OBJECTIVES OF EXPERIMENT
Calculate air flow from pressure difference.
Comparison of flow measurement devices in a duct:
Orifice plate.
−
Venturi nozzle.
−
II. EQUIPMENT SET - UP
The fan inlet is a duct 149 mm diameter provided with pressure tapings whereby the static
pressure may be measured simultaneously at each of 4 sections. All four pressure tapings are
connected to a bank of pressurized manometer tubes (1,2,3,4). Two flow measurement
devices are:
65mm orifice plate (1)
149mm – 65mm diameter venturi nozzle (2).
Figure1: Experimental flowchart
In which:
1. Orifice plate
2. Venturi nozzle
3. Fans and electric motors
4. Inverter
5. Measuring tubes
6. Pressure gauges
7. Silicon tubes
1,2,3,4: Order number of the measuring tubes.
III. SUMMARY OF THEORY
The volume flow rate at the orifice plat and venturi nozzle in the pipe is determined by
formula as follows
(3.1)
Where:
Q: volumetric flow rate.
C: discharge coefficient.
10
Δp: pressure difference from inlet to throat. The manometer containing liquid of density ρ1 is
used to indicate Δp, the pressure difference may be expressed in terms of the manometric
head differential Δh by:
Δp = (ρ1 –ρ).g.Δh
(3.2)
ρ : flow density ρ = 1.226 kg/m3
ρ1: water density
ρ1 = 1000kg/m3
β : diameter ratio = d/D.
ε: expandability factor. The expandability factory is also detailed in the code and allows for
the effects of density change in gas flows where a high pressure reduction occurs. For liquid
flows and gas flows with moderate variation in pressure at the meter, ε = 1.00.
The discharge coefficients of the orifice plat and the venturi nozzle can be determined by
empirical formula. For the orifice plate:
(3.3)
𝑅𝑒 =
𝑉𝐷
(3.4)
𝜈
Where:
Re: Reynolds number
U: upstream pipe velocity.
Q: discharge in pipe
D: diameter of pipe
μ: dynamic viscosity
When determining Q from Δp, it is necessary to estimate a value C initially as Re cannot be
calculated until Q is known. From an initial estimate of C (example C = 1), Q can be
calculated and thus Re found. The value of C can then be corrected and new values of Q and
Re cure calculated.
For the venturi nozzle:
(3.5)
IV. PROCEDURE
i. Check that there are no obstructions at the air intake and outlet of the gas pipe.
ii. Turn on the fan switch.
iii. Adjust the inverter speed to around 400 - 450 revolutions per minute (RPM).
iv. Read the water level in pressure tubes 1 and 2, and record the values in Table 1, in the
first row under the column "pressure tubes 1, 2." Read the pressure gauge on the left and
record the value in Table 1, in the first row under the column "pressure gauge."
v. Read the water level in pressure tubes 3 and 4, and record the values in Table 2, in the
first row under the column "pressure tubes 3, 4." Read the pressure gauge on the right and
record the value in Table 2, in the first row under the column "pressure gauge."
Repeat steps iii to v for three additional inverter speed values: 650-700 RPM, 900-950
RPM, and 1150-1200 RPM. Record the corresponding values in Tables 1 and 2.
11
DATE:
NAME OF INSTRUCTOR:
EXPERIMENT 3D MEASUREMENT OF VOLUMETRIC FLOW RATE
PREPARATION:
(Students must complete this section before coming to the lab. Failure to do so will result
in the student not being allowed to participate in the experiment.)
1.
What
equipment is used to measure gas flow in the pipe for this experiment??
I.
2.
How can the
gas flow rate in the pipe be adjusted between measurements?
3. For a gas flow meter, how many measurements need to be taken, and what data should be
collected during each measurement??
12
DATE:
NAME OF INSTRUCTOR:
MEASUREMENT DATA:
The air temperature: t0 =……… oC ;
The air density:
γair = ………. kg/m3
The air kinematic viscosity: νair = ………. m2/s
The water density: γwater =………….kg/m3
IV.
I. MEASUREMENTS AND CALCULATES
Table 1 Orifice plat
Frequency tube 1,
h1 (m)
tube 2,
h2 (m)
h2 – h1
(m)
p1 – p2
(Pa)
Difference
Pressure (Pressure
gauge
gauge
(Pa)
- measuring
tube) (%)
C
Q
(m3/s)
C
Q
(m3/s)
Table 2 Venturi nozzle
Frequency tube 3, tube 4,
h3 (m) h4 (m)
h4 – h3
(m)
P3 – p4
(Pa)
Difference
Pressure (Pressure
gauge
gauge
(Pa) - measuring
tube) (%)
II. REPORT
1. Determine the discharge coefficients, the volumetric flow rate in 4 experiments by using
venturi nozzle and venturi nozzle (table 1, 2)
2. Explain the difference from the results of U tubes and pressure gauge?
13
3. Compare the flow rates measured using the thin-walled orifice and the nozzle. Explain the
results.?
4. Which method (thin-walled orifice or nozzle) provides more accurate results? Why?
EXPERIMENT 5A FRICTION LOSS IN PIPE
14
I. OBJECTIVES OF EXPERIMENT
- To investigate the variation of friction head along a circular pipe with the mean flow velocity
in the pipe.
- To investigate the friction factor against Reynolds number and roughness.
II. EQUIPMENT SET – UP
A pipe of 10.64cm diameter is supplied by water a centrifugal pump. Four test sections with
interval of 3 m are connected to a bank of pressurized manometer tubes. Water from the pipe
flow into the concrete channel, and at the end of channel a vee-notch, is installed to measure
the flow rate in the channel, this flow rate is equal to the flow rate in the pipe. Water level
over the vee-notch is measured by a point gauge vernier mounted on a small tank which is
opened to the channel.
The flow rate over the vee-notch is calculated by formula as follow:
Q
8  
tg  C D 2 g .h05 / 2
15  2 
(5.1)
α = 90o
CD = 0.58
g = 9.81 m/s2
ho = ZCR – z
Where z is water level in channel, ZCR is the elevation of the crest of Vee-notch, ZCR = 27.8
cm
Where:
15
The flow rate over the Vee-notch is regulated by the inverter of the pump on the electric box
corresponding to the flow rate in the pipe. Inverter is adjusted at 8 different speeds (1000rpm,
900rpm, 800rpm, 700rpm, 650 rpm, 600 rpm, 550 rpm, 500 rpm). The difference of pressure
between the test sections in the pipe are measured by reading the water level in the tubes of
the manometer. Write the reading values into Table 1.
THEORY
Considering flow at two sections i, j in a pipe, Bernoulli’s equation may be written as:
V.
Vi 2
2g
Where:

Pi

 Zi 
V j2
2g

Pj

 Z j  hij
(5.2)
Vi, Vj : velocity at section i, j respectively
Pi, Pj : pressure at section i, j respectively
Zi, Zj : elevation at water surface section i, j
hi,j
: friction loss from section i to section j
For this apparatus, Zi = Zj, Vi = Vj, hence
hi , j 
Pi  Pj

 hi , j
(5.3)
Δhi,j : the difference of the manometer reading at section i and section j.
On the other hand, the friction factor can be determined by Darcy’s formula:
L V2
D 2g
hi , j  f
Where:
(5.4)
f : friction factor
L : the distance between section i and j
D : diameter of pipe
V : velocity of pipe
The friction coefficient depends upon the Reynolds number of flow and upon the ratio D/D,
the relative roughness of the pipe.


f  F  Re , 
D

Where:
Re : Reynolds’ number
Re = VD/ν (ν: viscosity coefficient)
D: size of roughness
 : relative roughness
D
For a given pipe,  is a constant
D
IV. PROCEDURE:
16
(5.5)
1.
Before testing, check valves (2), (4) and locks (9). Make sure they are closed and
check the spindle of the pump and the motor by turning lightly to see whether it is hard or
not, if it moves as well.
2.
Fully open the valve (4) by turning counterclockwise until it is no longer rotatable.
Note: If valve 4 does not open fully, when opening the valve 2 will break the manometer tubes
because of high pressure. Fully open the valve (2). On the electric box, you press: POWER
ON button and adjust the inverter at 8 different speeds, and then open clocks (9).
3. Measurements are made eight times.
4. Important note when you want to shut down. Adjust the inverter to 0 rpm, then press the
OFF button, Close valve (2), then close the valve (4) and clocks (9)
a- The first measurement:
. Open the locks (9) of the manometer tubes at the section (1) and (2)
. Adjust the inverter to change the three flow rate levels corresponding to the frequency of
1000 rpm – 800rpm (the first - the flow rate corresponding to 1000 rpm, the second - the flow
rate corresponding to 900rm; The third - flow rate corresponds to 800rpm)
. Wait for the water level in the channel to stabilize (the water level in point gauge (5) is
constant), read the following values:
Water level in manometer tube (1)
Water level in manometer tube (2)
Elevation Z at point gauge and vernier (5).
The measurement results are recorded in Table 1 of the report.
b- The second measurement:
• Adjust the inverter to change the five discharge levels corresponding to the frequency 700
rpm – 500 rpm
• Continue to open the locks of manometer tubes (3), (4)
• For each discharge level, wait for the water level in the channel to stabilize, taking the
following measurements:
- Reading water level from the manometer tub (1) to the manometer tube (4).
- Elevation Z at the point gauge (5).
The results are recorded in Table 1 of the report
17
DATE:
NAME OF INSTRUCTOR:
EXPERIMENT 5A. FRICTION LOSS IN PIPE
PREPARATION:
(Students must complete this section before coming to the lab. Failure to do so will result
in the student not being allowed to participate in the experiment.)
1.
What types of
energy losses are studied in this experiment? What causes these losses?
IV.
2.
How
is
the
flow rate in the pipe adjusted?
3. Why are the pressure measurement tubes connected to each other but not exposed to
atmospheric pressure?
4.
What data needs to be measured during the experiment?
18
DATE:
NAME OF INSTRUCTOR:
MEASUREMENT DATA:
The water temperature t0H2O = ……………………………………0C
The water kinematic viscosity: ν =…………………….m2/s
The elevation of the crest of Vee-notch: ZCR=…………cm
VI.
Table 1. Measurement
No
Frequency
1
2
3
4
5
6
7
8
(rpm)
1000
900
800
750
700
650
600
550
Manometer reading (cm)
(1)
(2)
(3)
(4)
Water level
(Z)
(cm)
II. Calculation and discussion
1. Select any three levels of flow rate at the second measurement (Frequency <800rpm),
calculate the flow rate (Q) in the pipe and friction loss between the sections: 4 - 3, 4 - 2, 4 –
1, the results are recorded in Table 2. And then Plot the head loss (hd) against length (L) for
each flow rate level
Table 2: Relationship between hd and L
Q=
Q=
Q=
Cross Section Length (L) (m)
hi-j (cm)
4 -3
3
4-2
6
4-1
9
19
2. With eight measured discharge levels, calculate the flow rate (Q), friction losses along the
pipe, hi-j , between two sections 1 and 2. The results are recorded in Table 3. And then plot
the friction loss (hi-j) against the flow rate (Q)
No.
frequency
1
500
2
550
3
600
4
650
5
700
6
800
7
900
8
1000
Q, l/s
Table 3
hi-j, cm
f
Re
3. With eight measured discharge levels, Calculating the Reynolds number (Re) and friction
coefficients (f). The results are recorded in Table 3.
4. Use the Moody’s chart and mark the calculated points (value pairs f and Re) on the
Moody’s chart to conclude about the flowing state in the pipe. Based on Moody's chart,
approximate the absolute roughness value D.
Moody Chart ( 𝛥 = 𝑒 )
𝐷
20
𝐷
Answer:
The flowing state: .............................................................................................
Δ/D = ......................................................Δ = ...................................... mm
Comment:
a) What is the relationship between hd and L? Is that reasonable? Why?
b) What is the relationship between hd and Q? Is that reasonable? Why?
c) Is the flowing state, relative roughness and absolute roughness reasonable? Why?
21
Plot graph
22
Appendix
POINT GAUGE AND VERNIER
I- STRUCTURE
Point Gauge and Vernier are used in laboratories to
measure elevation (or depth) as well as coordinates
of points (see Figure 1).
1- POINT GAUGE is a steel bar which is carved
in mm. One end of the ruler has a small needle
attached to the point of measurement.
2- VERNIER Used to determine the measure of
the ruler. On the vernier is carved, the number of
lines is the divider of 1mm of the ruler. These
measurements in the laboratory are accurate to 1 /
20mm (corresponding to a 20 lines). The
measurement of the ruler is determined by the
location of the remaining lines on the vernier.
3- WHEEL: used to move up and down to pointer
in contact to the surface
4PLATFORM: used to keep the vernier
stable.
II- Using:
1- Uses the wheel to move the ruler up and down
so that the Pointer contact area of the water surface
to be measured. (Note the surface at this time must
be stable).
2- Read water depth values on the ruler at the "0"
of the vernier. Normally, the "0" line is between
two lines on the ruler, the actual value is in this middle.
3- Determine the odd part of the 1mm: Find a line on the vernier that coincides with a certain
line on the ruler. The odd part of the 1mm is the number of lines on the chain multiplied by 1
/ 20mm.
EX: In step 2, the line "0" is found between the bars of 7.8 and 7.9 on the ruler. In Step 3, we
see that "15" on vernier coincides with a certain line on the Ruler.
The measured value is:
7,8 + 15 × (1/20)/10 = 7,8 + 0,075 = 7, 875 cm.
23
Properties of water
Temperature
o
C
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
specific
weight Density ρ
γ
(Kg/m3)
(KN/m3)
9.806
999.9
9.807
1000.0
9.804
999.7
9.798
999.1
9.789
998.2
9.778
997.1
9.764
995.7
9.749
994.1
9.730
992.2
9.710
990.2
9.690
988.1
9.666
985.7
9.642
983.2
9.616
980.6
9.589
977.8
9.560
974.9
9.530
971.8
9.499
968.6
9.466
965.3
9.433
961.9
9.399
958.4
μ
(Pa.s)
ν
(m2/s)
1.792E-3
1.519E-3
1.308E-3
1.140E-3
1.002E-3
0.890E-3
0.801E-3
0.723E-3
0.656E-3
0.599E-3
0.549E-3
0.506E-3
0.469E-3
0.436E-3
0.406E-3
0.380E-3
0.357E-3
0.336E-3
0.317E-3
0.299E-3
0.284E-3
1.792E-6
1.519E-6
1.308E-6
1.141E-6
1.007E-6
0.897E-6
0.804E-6
0.727E-6
0.661E-6
0.605E-6
0.556E-6
0.513E-6
0.477E-6
0.444E-6
0.415E-6
0.390E-6
0.367E-6
0.347E-6
0.328E-6
0.311E-6
0.296E-6
Properties of Air
Temperature (oC)
20
25
30
35
40
45
Density ρ
(kg/m3)
1.2047
1.1845
1.1649
1.1459
1.1275
1.1098
24
ν
(m2/s)
1.5111E-5
1.5571E-5
1.6036E-5
1.6507E-5
1.6982E-5
1.7462E-5