1. KV
FLUID DISCHARGE
{for energy conversion}

2. KV
Identify the unique vocabulary used in the description and
analysis of fluid flow with an emphasis on fluid discharge
Describe and discuss how fluid flow discharge devices
affects fluid flow.
Derive and apply the governing equations associated with
fluid discharge
Determine the power of flow in a channel or a stream and
how this can be affected by height, pressure, and/or
geometrical properties.
OBJECTIVE
S

3. KV
FLOW THROUGH
ORIFICES and
MOUTHPIECES
An orifice
Is an opening having a closed perimeter
A mouthpiece
Is a short tube of length not more than two to
three times its diameter

4. KV
FLOW THROUGH
ORIFICES and
MOUTHPIECES
u1
z1
z2 ②
H
u = u2
①
Orifice
area A
Applying Bernoulli’s equation to
stations ① and ②
putting
z1 - z2 = H, u1 = 0, u2 = u and p1 = p2

5. KV
TORRICELLI’s
THEOREM
Torricelli’s theorem states that the velocity of the
discharging jet is proportional to the square root of the
head producing flow. This is support by the preceding
derivation;
The discharging flow rate can be determined theoretically
if A is the cross-sectional area of the orifice
Actual discharge

6. KV
The velocity of the jet is less than that determined by the velocity of jet equ because
there is a loss of energy between stations ① and ② i.e
actual velocity, u = Cu √(2gH)
where Cu is a coefficient of velocity which is determined experimentally and is of the
order 0.97 - 0.98
The paths of the particles of the fluid converge on
the orifice and the area of the discharging jet at B
is less than the area of the orifice A at C
actual area of jet at B = Cc A
where Cc is the coefficient of contraction -
determined experimentally - typically 0.64
Vena contracta
C B
pB
u
uC
pC

7. KV
Actual discharge = Actual area at B × Actual velocity at B
we see that the relationship between the coefficients is Cd = Cc × Cu
To determine the coefficient of discharge measure the actual discharged volume
from the orifice in a given time and compare with the theoretical discharge.

8. KV
(a) A jet of water discharges horizontally into the atmosphere from an orifice in
the side of large open topped tank. Derive an expression for the actual
velocity, u of a jet at the vena contracta if the jet falls a distance y vertically for
a horizontal distance x, measured from the vena contracta.
(b) If the head of water above the orifice is H, calculate the coefficient of velocity.
(c) If the orifice has an area of 650 mm2 and the jet falls a distance y of 0.5 m in a
horizontal distance x of 1.5 m from the vena contracta, calculate the values of
the coefficients of velocity, discharge and contraction, given that the
volumetric flow is 0.117 m3/min and the head H above the orifice is 1.2 m
EXAMPLE 1

9. KV
Let t be the time taken for a particle of fluid to travel from the vena contracta
A to the point B.

10. KV
putting x = 1.5m, H = 1.2m and Area, A = 650×10-6m2

11. KV
THEORY OF
LARGE
ORIFICES
H1
H2
h
δh
D
B

12. KV
(a) A reservoir discharges through a sluice gate of width B and height D. The
top and bottom openings are a depths of H1 and H2 respectively below
the free surface. Derive a formula for the theoretical discharge through
the opening
(b) If the top of the opening is 0.4 m below the water level and the opening is
0.7 m wide and 1.5 m in height, calculate the theoretical discharge (in
meters per second) assuming that the bottom of the opening is above the
downstream water level.
(c) What would be the percentage error if the opening were to be treated as
a small orifice?
EXAMPLE 2

13. KV
Given that the velocity of flow will be greater at the bottom than at
the top of the opening, consider a horizontal strip across the
opening of height δh at a depth h below the free surface
For the whole opening, integrating from h = H1 to h = H2

14. KV
putting B = 0.7 m, H1 = 0.4 m and H2 = 1.9 m

15. KV
NOTCHES
& WEIRS
H
h
δh
H
b

16. KV
Consider a horizontal strip of width b and height δh at a depth h below the
free surface.
Integrating from h = 0 at the free surface to h = H at the bottom of the notch
b must be expressed in terms of h before integrating

17. KV
For a rectangular notch, put b = constant = B
b = constant
B
H
For a vee notch with an included angle θ, put b
= 2(H - h)tan(θ⁄2)
b = 2(H - h)tan(θ⁄2)
H
h
θ

18. KV
In the foregoing analysis it has been assumed that
• the velocity of the liquid approaching the notch is very small so that its
kinetic energy can be neglected
• the velocity through any horizontal element across the notch will depend
only on the depth below the free surface
These assumptions are appropriate for flow over a notch or a weir in the side of
a large reservoir
If the notch or weir is located at the end of a narrow channel, the velocity of
approach will be substantial and the head h producing flow will be increased by
the kinetic energy;
where ū is the mean velocity and α is the kinetic energy correction factor to
allow for the non-uniform velocity over the cross section of the channel

19. KV
Therefore
at the free surface, h = 0 and x = αū2/2g, while at the sill , h = H and
x = H + αū2/2g. Integrating between these two limits
For a rectangular notch, putting b = B = constant

20. KV
Pressure, p, velocity, u, and elevation, z, can cause a stream of fluid to do
work. The total energy per unit weight H of a fluid is given by
If the weight per unit time of fluid is known, the power of the stream can be
calculated;
THE POWER OF
A STREAM OF
FLUID

21. KV

22. KV

23. KV
In a hydroelectric power plant, 100 m3
/s of water flows from an elevation of
12 m to a turbine, where electric power is generated. The total irreversible
heat loss is in the piping system from point 1 to point 2 (excluding the
turbine unit) is determined to be 35 m. If the overall efficiency of the turbine-
generator is 80%, estimate the electric power output.
EXAMPLE 3

24. KV
Assumptions
1.The flow is steady and incompressible
2.Water levels at the reservoir and the discharge
site remain constant
Properties
We take the density of water to be 1000 kg/m3
The mass flow rate of water through the turbine is
We take point ➁ as the reference level, and thus z2 = 0. Therefore the
energy equation is
(Çengel, et al 2008)

25. KV
Also, both points ➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the
flow velocities are negligible at both points (V1 = V2 = 0). Then the energy
equation for steady, incompressible flow reduces to
Substituting, the extracted turbine head and the corresponding turbine power
are
Therefore, a perfect turbine-generator would generate 83,400 kW of electricity
from this resource. The electric power generated by the actual unit is
Note that the power generation would increase by almost 1 MW for each
percentage point improvement in the efficiency of the turbine-generator unit.

26. KV
Flow through orifices and mouthpieces
Theory of small orifice discharge
Torricelli’s theorem
Theory of large orifices
Notches and weirs
The power of a stream of fluid

27. KV
Andrews, J., Jelley, N., (2007) Energy science: principles, technologies
and impacts, Oxford University Press
Bacon, D., Stephens, R. (1990) Mechanical Technology, second edition,
Butterworth Heinemann
Boyle, G. (2004) Renewable Energy: Power for a sustainable future,
second edition, Oxford University Press
Çengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid
sciences, Third edition, McGraw Hill
Douglas, J.F., Gasoriek, J.M., Swaffield, J., Jack, L. (2011), Fluid
Mechanics, sisth edition, Prentice Hall
Turns, S. (2006) Thermal fluid sciences: An integrated approach,
Cambridge University Press
Young, D., Munson, B., Okiishi, T., Huebsch, W., 2011Introduction to
Fluid Mechanics, Fifth edition, John Wiley & Sons, Inc.
Some illustrations taken from Fundamentals of thermal fluid sciences