1. The document discusses various fluid flow measurement devices and their coefficients, including the coefficient of discharge, coefficient of velocity, and coefficient of contraction. 2. It also covers different flow measurement structures like orifices, weirs, and pitot tubes. It provides the theoretical equations to calculate flow based on these structures. 3. Sample problems at the end demonstrate using the equations to calculate things like flow rate, time to empty a tank, and other flow properties based on given device dimensions and flow conditions.

1. MODULE 5:
FLUID FLOW MEASUREMENT

2. DEVICE COEFFICIENTS
Coefficient of Discharge (C or 𝑪𝒅)
• The ratio of the actual discharge through the device to the
ideal or theoretical discharge which would occur with out
losses.
C or 𝑪𝒅 =
𝐀𝐜𝐭𝐮𝐚𝐥 𝐃𝐢𝐬𝐜𝐡𝐚𝐫𝐠𝐞,𝐐
𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐃𝐢𝐬𝐜𝐡𝐚𝐫𝐠𝐞,𝑸𝑻
• Actual discharge may be accomplished by series of
observation, usually by measuring the total amount of fluid
passing through the device of a known period.
• Theoretical value can be accomplished using the Bernoulli’s
theorem neglecting losses.

3. Coefficient of Velocity, 𝑪𝒗
• The ratio of the actual velocity to the ideal or theoretical
velocity which would occur without any losses.
𝑪𝒗 =
𝐀𝐜𝐭𝐮𝐚𝐥 𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲,𝒗
𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 ,𝒗𝑻
Coefficient of Contraction, 𝑪𝒗
The ratio of the actual area of the contracted section of the
stream or jet to the area of the opening through which the
fluid flows.
𝑪𝒄 =
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝑺𝒕𝒓𝒆𝒂𝒎 𝒐𝒓 𝑱𝒆𝒕, 𝒂
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐎𝐩𝐞𝐧𝐢𝐧𝐠 ,𝑨

4. Relationship Between the Three Coefficient
C= 𝑪𝒄 𝑪𝒗

5. Head Lost
• The head lost through Venturi Meters, Orifices, Tubes and Nozzles may be
expressed as:
HL = (
𝟏
𝒄𝒗
𝟐 − 𝟏)(𝟏 − (
𝑨𝟐
𝑨𝟏
)
𝟐
)
𝒗𝟐
𝟐𝒈
• If the orifice or nozzle takes off directly from a tank where 𝑨𝟏 is very much
greater than the 𝑨𝟐 , then the velocity of approach is negligible and
equation reduces to:
HL = (
𝟏
𝒄𝒗
𝟐 − 𝟏)
𝒗𝟐
𝟐𝒈
Where v = actual velocity

6. ORIFICE
• An orifice is an opening (usually circular) with a closed
perimeter through which fluids flows.

7. Theoretical Velocity (𝒗𝑻) = 𝟐𝒈𝑯
Actual Velocity (v) = 𝑪𝒗 𝟐𝒈𝑯
Theoretical Discharge (𝑸𝑻) = 𝐀 𝟐𝒈𝑯
Actual Discharge = 𝐂𝐀 𝟐𝒈𝑯
H = h +
𝒗𝑨
𝟐
𝟐𝒈
+
𝑷𝑨
𝜸
-
𝑷𝑩
𝜸
= Head upstream – Head
Downstream

10. Contraction of the Jet
• The section on the jet where the contraction ceases is
called vena contracta which is approximately located one
half of the orifice diameter (D/2) from the upstream
surface.

11. Pitot Tube
• Named after French Physicist and Engr. Henry Pitot.
• It is a bent (L-shaped or U-shaped) tube with both ends open and is
used to measure the velocity of fluid or velocity of airflow as used in
airplane speedometer.
v = 𝟐𝒈𝒉

13. Sample Problem
1) A volumetric tank 1.20m in diameter and 1.50m high was filled with oil in
16 minutes and 32.4 seconds. What is the average discharge.
(Ans. 0.1025 cu.m/min)
2) Calculate the discharge in liters per second through a 100mm diameter
orifice under a head of 5.5m of water. Assume 𝑪𝒄 = 0.61 and 𝑪𝒗= 0.98.
(Ans. 48.77 L/s)
3) An open cylindrical tank, 2.4m in
diameter and 6m tall has 1 m of glycerin
(sg=1.5) . 2.5 m of water, and 1.5 m of oil
(sg = 0.82). Determine the discharge
through the 125 mm diameter located at
the bottom of the tank. Assume C = 0.65
(Ans. 0.65 cu.m/sec)

14. 4) A pitot tube having a coefficient of 0.98 is used to measure the
velocity of water at the center of the pipe, as shown in the figure. What is
the velocity.
(Ans. 13.8 ft/s)

15. 5) Water flows from a large tank through an orifice and discharges to the
atmosphere, as shown in the figure. The coefficient of velocity and
contraction are 0.96 and 0.62, respectively. Find the diameter and actual
velocity in the jet and the discharge from the orifice.
(Ans. D jet = 0.1313 ft or 1.58 in ; v jet = 24.3 ft/s; Q jet = 0.329

16. 6) A 75mm-diameter orifice discharges 23.41 L per second
of liquid under a head of 2.85m . The diameter of the jet at
the vena contracta is found by callipering to be 66.25mm.
Calculate the three orifice coefficient.
(Ans. 𝑪𝒄 = 0.78; 𝑪𝒗 = 0.909; C = 0.709)
7) The actual velocity in the contracted section of a jet of
liquid flowing from a 2 in diameter orifice is 28.0 ft/sec
under a head of 15 ft.
• Find the coefficient of velocity (Ans. 0.901)
• If the measured discharge is 0.403 cfs, determine the
coefficients of contraction and discharge.
(Ans. Cc=0.659 ; C = 0.594)

17. UNSTEADY FLOW
• The flow through an orifice, weirs or tubes is said to be steady only if
the total head producing flow, H, is constant. The amount of fluid being
discharged for a time t can be computed using the formula Vol = Qt
where Q is constant.
• In some condition, however, the head over an orifice, tube or weir may
vary as the fluid flows out and thus causing the flow to be unsteady.

18. t = 𝒉𝟏
𝒉𝟐 𝑨𝒔𝒅𝒉
𝑸𝒊𝒏−𝑸𝒐𝒖𝒕
When there is no inflow (𝑸𝒊𝒏 = 0), the formula becomes:
t = 𝒉𝟐
𝒉𝟏 𝑨𝒔𝒅𝒉
𝑸𝒐𝒖𝒕

19. • For tanks with constant cross-sectional area and outflow is through an
orifice or tube (with no inflow), the time for the head to change from
𝑯𝟏 to 𝑯𝟐 is:
• If liquid flows through a submerged orifice or tube connecting two tanks
as shown, the time for the head to change from 𝑯𝟏 to 𝑯𝟐 is:
Where:
𝐴𝑠1 𝑎𝑛𝑑 𝐴𝑠2 = water surface areas in the tanks at any time.
H = the difference in water surfaces in the two tanks at any time
t =
𝟐𝑨𝒔
𝑪𝑨𝒐 𝟐𝒈
( 𝑯𝟏 − 𝑯𝟐 )
t =
𝑨𝒔𝟏𝑨𝒔𝟐
𝑨𝒔𝟏+ 𝑨𝒔𝟐
𝟐
𝑪𝑨𝒐 𝟐𝒈
( 𝑯𝟏 − 𝑯𝟐 )

21. Weir
• Weir are overflow structures which are built across an open channel for
purpose of measuring or controlling the flow of liquids.
Classification of Weirs
Nape – the overflowing stream in a weir
Crest of Weir – the edge or top surface of a weir with which the following
liquid comes in contact.
Contracted Weir – weirs having sides sharp-edged, so that the nappe is
contracted in width or having end contractions, either one end or two
According to Shape According to form of the crest
Rectangular
Triangular
Circular
Parabolic
Trapezoidal
Sharp-crested
Broad-crested

22. Suppressed weir or full-width weir – weirs having its length L being equal
to the width of the channel so that the nappe suffers no end contractions.
Drop-down Curve – the downward curvature of the liquid surface before
the weir.
Head, H –the distance between the liquid surface and the crest of the weir,
measured before the drop-down curve.

23. Rectangular Weir
Q =
𝟐
𝟑
C 𝟐 𝒈 L ((H+ 𝒉𝒗)
𝟑
𝟐 - 𝒉𝒗
𝟑
𝟐 )
Or
Q = 𝑪𝒘L ((H+ 𝒉𝒗)
𝟑
𝟐 - 𝒉𝒗
𝟑
𝟐 )
Where:
ℎ𝑣 = velocity head of approach =
𝑣𝑎
2
2𝑔
L = length
𝐶𝑤 =
2
3
C 2 𝑔
• If the ratio of H/P is sufficiently small, the velocity of approach becomes
very small and the term 𝒉𝒗
𝟑
𝟐 may be neglected. The discharge formula
become:
Q = 𝑪𝒘L H
𝟑
𝟐

24. In situations where the discharge is required considering the velocity of
approach. The following simplified equations maybe used:
Q = 𝑪𝒘L H
𝟑
𝟐 ( 1+ 𝑪𝟏(H/𝒅)
𝟑
𝟐 )
Where
𝐶1 = (3/2)(C2
/2𝑔)
d = depth of water upstream = H + P
STANDARD WEIR FACTOR (𝑪𝒘) FORMULAS
1. Francis Formula:
𝑪𝒘 = 1.84 (1+9.26 (H/𝒅)𝟐
) SI Units
For H/P < 0.4, the ff value of 𝑪𝒘 maybe used.
S.I. Units 𝑪𝒘 = 1.84
English 𝑪𝒘 = 3.3

25. 2. Rehbock and Chow Formula
English Unit 𝑪𝒘 = 3.27 + 0.40 H/P
SI Unit 𝑪𝒘 = 1.8 + 0.22 H/P
3. Bazin Formula
𝑪𝒘 = 0.5518 (3.248 + 0.02161/H) (1 + 0.55 (H/𝒅)𝟐
)

26. Contracted Rectangular Weirs
The effective length of L of a contracted weir is given by:
L = L
′
– 0.1 NH
N= number of end contraction (1 or 2)
L
′
= measure length of crest
H = measured head

27. Triangular Weir (V-Notch)
For Triangular V-Notch Weir:
Q =
𝟖
𝟏𝟓
𝑪 𝟐𝒈 𝒕𝒂𝒏
Ɵ
𝟐
H
𝟓
𝟐 = 𝑪𝒘tan
Ɵ
𝟐
H
𝟓
𝟐
For 90
𝟎
Weir:
Q= 1.4H
𝟓
𝟐 (SI Units)
Q = 2.5H
𝟓
𝟐 (English)

28. Trapezoidal Sharp Crested Weir
the discharge from a trapezoidal weir is assumed the same as that
from a rectangular weir and a triangular weir in combination.
Q = 𝑪𝒘𝟏L H
𝟑
𝟐 + 𝑪𝒘𝟐𝐙H
𝟓
𝟐
Z = b/H, substituted for tan
Ɵ
2

29. Cipolletti Weir
Cipolletti weirs are trapezoidal weirs with side slope of 1 horizontal
to 4 vertical. The additional area at the sides adds approximately enough
effective width of the stream to offset the side contractions.
Q = 1.859 LH
𝟑
𝟐 (SI Units)
Q = 3.37 LH
𝟑
𝟐 (English)

30. Sample Problem
8) A 1.5m diameter vertical cylinder tank 3m high contains 2.5m of water. A
100mm diameter circular sharp-edged orifice is located at the bottom.
Assume C = 0.60
• How long will it take to lower the water level to 1m deep after opening
the orifice? (Ans. 98.4 sec)
• How long will it take to empty the tank?
9) An open cylindrical tank 4m in diameter and 10m high contains
of water and 4m of oil (sg = 0.8). Find the time to empty the tank
through a 100mm dimeter orifice at the bottom. Assume C= 0.9 and
Cv = 0.98
(Ans. T= 44.3 min)

31. 10) Two vertical cylindrical tanks 1 and 2 having diameters 2m and 3m,
respectively, are connected with a 200mm tube at its lower portion, and
having C=0.60. When the tube is closed, the water surface in tank 1 is 5m
above tank 2. How long will it take after opening the tube, for the water
surface in the tank 2 to rise by 1 m.
(Ans. 47.57 sec)
11) A rectangular, sharp crested weir with end contractions is 1.4m long.
How high should it be placed in a channel to maintain an upstream depth
of 2.35m for a flow of 400 L/second.
(Ans. P = 2.05m)

32. 12) The discharge from a 150mm diameter orifice under a head of 3.05m
and coefficient of discharge C = 0.60 flows into a rectangular suppressed
weir. The channel is 1.83m wide and the weir has height P=1.50m ad length
L=0.31m. Determine the depth of water in the channel. Use Francis Formula
and neglect velocity of approach.
(Ans. d= 1.74m)
13) The discharge over a trapezoidal weir is 1.315 cu.m/sec . The crest length
is 2m and the sides are inclined at 750 57′ 49" with the horizontal. Find the
head on the weir in meters.
(Ans H=0.50m)

33. 14) A trapezoidal weir having a side slope of 1H to 2V discharges 50
cu.m/sec under a constant head of 2m. Find the length of the weir
assuming C = 0.60.
(Ans. L=9.18m)