Ch.6 Hydropower

ME – 481 (2, 0)
ENERGY RESOURCES &
UTILIZATION (ERU)
Arranged By
PROF. DR.ASAD NAEEM SHAH
anaeems@uet.edu.pk

HYDRO POWER
“The harnessing power from
stored water by mini and micro-
hydro power is efficient,
renewable, sustainable and
environment friendly.”

INTRODUCTION
• The term hydro-power is usually restricted to the generation of
shaft power from falling water. The power is then used for
direct mechanical purposes or, more frequently, for generating
electricity.
• Hydro-power is by far the most established and widely used
renewable resource for electricity generation and commercial
investment.
• The electricity generation from hydro-turbines started in 1880.
• Hydro-power now accounts for about 20% of world’s electric
generation. In about one-third of the world countries, hydro-
power produces more than half the total electricity.
• Table 1 reviews the importance of hydroelectric generation for
various countries and regions, while Fig. 1 indicates the global
increase.
Arranged by Prof. Dr. Asad Naeem Shah

Table 1: Hydro-power capacity by country/region

Fig. 1: World hydroelectricity generation (in 𝑇𝑊ℎ𝑦−1
); mainly from large hydro.
Extrapolated from the year of publication.

INTRODUCTION Cont.
• By the 1940s, most of the best sites in industrialized countries
had already been exploited. Almost all the increase in Fig. 1 is in
developing countries, notably India, China and Brazil, as reflected
in the ‘under construction’ column in Table 1.
• However, global estimates can be misleading for local hydro-
power planning, as many small scale applications have been
ignored. Thus the potential for hydro generation from run-of-
river schemes is often underestimated.
• Social and environmental factors are also important, and these
too cannot be judged by global surveys but only by evaluating
local conditions.
• Hydro installations and plants are long-lasting with routine
maintenance, e.g. turbines for about fifty years and longer with
minor refurbishment, dams and waterways for hundred years.

INTRODUCTION Cont.
• Hydro turbines have a rapid response for power generation
and so the power may be used to supply both base load and
peak demand requirements on a grid supply. Power generation
efficiencies may be as high as 90%.
• Turbines are of two types:
1) Reaction turbines − The turbine is totally embedded in the
fluid and powered from the pressure drop across the device.
2) Impulse turbines − The flow hits the turbine as a jet in an
open environment, with the power deriving from the kinetic
energy of the flow.
• The disadvantages of hydro-power are associated with effects
other than the generating equipment, particularly for large
systems.

DISADVANTAGES OF HYDRO-POWER
The major disadvantages include:
1. The possible adverse environmental impact.
2. The effect on fish.
3. The silting of dams.
4. The corrosion of turbines in certain water conditions.
5. The social impact of displacement of people from the
reservoir site.
6. The loss of potentially productive land.
7. The relatively large capital costs compared with those of
fossil power stations.

PRINCIPLES OF HYDRO-POWER
• A volume per second, 𝑄, of water falls down a slope. The density of
the fluid is 𝜌 . Thus the mass falling per unit time is 𝜌𝑄, and the rate of
potential energy lost by the falling fluid is:
𝑷 𝟎 = 𝝆𝑸𝒈𝑯 →→ (𝟏)
where 𝑔 is the acceleration due to gravity, 𝑃0 is the energy
change per second (power in Watts) and 𝐻 is the vertical
component of the water path.
• The purpose of a hydro-power system is to convert this power to shaft
power with very small frictional losses−an obvious advantage.
• For a given site, 𝐻 is fixed and 𝑄 can usually be held fairly constant by
ensuring that the supply pipe is kept full. Therefore, the actual output
is close to the design output, and it is not necessary to install a
machine of capacity greater than normally required.

PRINCIPLES OF HYDRO-POWER Cont.
• However, the site must have sufficiently high 𝑄 and 𝐻. In
general this requires a rainfall > ~40𝑐𝑚𝑦−1
dispersed
through the year, a suitable catchment and, if possible, a
water storage site.
• Moreover, civil works (in the form of dams, pipework, etc.)
often cost more than the mechanical and electrical
components.
• It is pertinent to note that the cost per unit power of turbines
tends to increase with 𝑄. Thus for the same power output,
systems with higher 𝐻 will be cheaper unless penstock costs
become excessive.

ASSESSING THE RESOURCE FOR
SMALL INSTALLATIONS
It is clear from the equation 𝑷 𝟎 = 𝝆𝑸𝒈𝑯 that to estimate the
input power 𝑃0 we have to measure the flow rate 𝑄 and the
available vertical fall 𝐻 (called head).
 MEASUREMENT OF HEAD 𝑯:
• The power input to the turbine depends not on the geometric
(or total) head 𝐻𝑡, but on the available head 𝐻 𝑎:
𝐻 𝑎 = 𝐻𝑡 − 𝐻𝑓 →→ (1)
where 𝐻𝑓 denotes the friction losses in the pipe and
channels leading from the source to the turbine.
• By a suitable choice of pipework it is possible to keep
𝐻𝑓 < ~ 𝐻𝑡 3

SMALL INSTALLATIONS Cont.
MEASUREMENT OF FLOW RATE 𝑸:
• The flow through the turbine produces the power, and this
flow will usually be less than the flow in the stream. However,
the flow in the stream varies with time.
• For power generation it is usually required to know the
minimum (dry season) flow to avoid the overcapacity of
machinery, and maximum flow & flood levels to avoid the
damage to installations.
• The measurement of 𝑄 is more difficult than the
measurement of 𝐻. The method chosen will depend on the
size and speed of the stream concerned.

• The Flow rate or discharge is given as:
𝑄 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒 →→ (2)
𝑄 = (𝑚𝑒𝑎𝑛 𝑠𝑝𝑒𝑒𝑑 𝑢) × (𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝐴) →→ (3)
𝑄 = 𝑢 𝑑𝐴 →→ (4)
where 𝑢 is the streamwise velocity (normal to the
elemental area 𝑑𝐴).
• The methods based on the above three Equations (2−4) are
known as basic, refined and sophisticated methods,
respectively.

o BASIC METHOD: It is appropriate to divert the whole stream
into a containing volume (Fig. 1). So it is possible to measure
the flow rate from the volume trapped using Eqn. (2). This
method makes no assumptions about the flow, is accurate and
ideal for small flows, such as those at a very small waterfall.
Fig. 1: Measuring water flow using
Basic Method.

o REFINED METHOD: Equation (3) defines the mean speed 𝑢 of
the flow. Since the flow speed is zero on the bottom of the stream,
the mean speed will be slightly less than the speed 𝑢 𝑠 on the top
surface. For a rectangular cross-section, for example, it has been
found that:
𝑢 ≈ 0.8𝑢 𝑠
where 𝑢 𝑠 can be measured by simply placing a float, e.g. a
leaf, on the surface and measuring the time it takes to go
a certain distance along the stream.
• For best results the stream should be straight and of uniform cross-
section. The cross-sectional area 𝐴 can be estimated by measuring
the depth at several points across the stream (Fig. 2), and integrating
across the stream in the usual way:

𝐴 ≈
1
2
𝑦1 𝑧1 +
1
2
𝑦2 − 𝑦1 𝑧1 + 𝑧2 +
1
2
𝑦3 − 𝑦2 𝑧2 + 𝑧3
+
1
2
𝑦4 − 𝑦3 𝑧3
Fig. 2: Measuring water flow using Refined method.

o SOPHISTICATED METHOD: This is the most accurate
method for large streams and is used by professional
hydrologists. The forward speed 𝑢 is measured with a small
flow meter at the points of a two-dimensional grid extending
across the stream (Fig. 3), and thus Eqn. (4) is used.
Fig. 3: Measuring water flow using Sophisticated Method.

IMPULSE TURBINE
Impulse turbines are easier to understand than reaction turbines,
so a particular impulse turbine – the Pelton wheel is considered.
 FORCES:
• The potential energy of the water in the reservoir is changed into
kinetic energy of one or more jets. Each jet then hits a series of
buckets or ‘cups’ placed on the perimeter of a vertical wheel (Fig. 1)
leading to a change in momentum of the fluid. Thus a tangential
force is applied to the wheel which causes it (wheel) to rotate.
Fig. 1: Schematic diagram of a
Pelton wheel impulse turbine.

IMPULSE TURBINE Cont.
• Figure 2(a) shows a jet, of density 𝜌 and volume flow rate 𝑄𝑗,
hitting a cup. The cup moves to the right with steady speed 𝑢 𝑐 and
the input jet speed is 𝑢𝑗.
• Figure 2(b) shows the frame of the cup with relative jet speed
𝑢𝑗 − 𝑢 𝑐 ; since the polished cup is smooth, friction is negligible,
and so the jet is deflected smoothly through almost 180° with no
loss in speed; 𝑢2 = 𝑢𝑗 [as per Fig. 2(a)].
Fig. 2: Speed of cup and fluid, in (a) the laboratory frame (b) the frame of the cup.

• Thus in the frame of the cup, the rate of change of
momentum, and hence the force 𝐹 experienced by the cup is:
𝐹 = 2𝜌𝑄𝑗 𝑢𝑗 − 𝑢 𝑐 →→ (1)
(This force is in the direction of the jet)
• The power 𝑃𝑗 transferred to the single cup is:
𝑃𝑗 = 𝐹𝑢 𝑐 = 2𝜌𝑄𝑗 𝑢𝑗 − 𝑢 𝑐 𝑢 𝑐 →→ (2)
where 𝑄𝑗 is the flow through the jet.
• Differentiating Eqn. (2) with respect to 𝑢 𝑐 leads to:
𝑢 𝑐/𝑢𝑗 = 0.5 →→ (3)
• So putting the value of 𝑢 𝑐 in Eqn. (2) to get the maximum 𝑃𝑗 :
𝑃𝑗 𝑚𝑎𝑥
=
1
2
𝜌𝑄𝑗 𝑢𝑗
2
→→ (4)
i.e. the max. output power equals the input power (rate of
change of K.E), and this ideal turbine has 100% efficiency.

• For this ideal case, in the laboratory frame, the absolute velocity of
the water leaving the cup is zero i.e. 𝑢2 = 0. Therefore the water
from the horizontal jet falls vertically from the cup.
• Although the ideal turbine efficiency is 100%, in practice values range
from 50% for small units to 90% for accurately machined large
commercial systems. The design of a practical Pelton wheel (Fig.3)
aims for the ideal performance described.
• The ideal cannot be achieved in practice, because an incoming jet
would be disturbed both by the reflected jet and by the next cup
revolving into place.
• Pelton made several improvements in the turbines of his time (1860)
to overcome these difficulties. Notches in the top of the cups gave the
jets better access to the turbine cups. The shape of the cups
incorporated a central splitter section so that the water jets were
reflected away from the incoming water.

Fig. 3: Impulse turbine runner (Pelton type) with buckets cast integrally with the hub.

JET VELOCITY AND NOZZLE SIZE:
• As indicated in Fig. 1, the pressure is atmospheric both at the
top of the supply pipe and at the jet. Therefore Bernoulli’s
theorem implies that, in the absence of friction in the pipe:
𝑢𝑗
2
= 2𝑔𝐻𝑡
• In case of Pipe friction, however, the total head 𝐻𝑡 can be
replaced by the available head 𝐻 𝑎, so:
𝑢𝑗
2
= 2𝑔𝐻 𝑎 →→ (5)
• In practice the size of the pipes is chosen so that 𝑢𝑗 is
independent of the nozzle area. If there are 𝒏 nozzles, each of
area 𝒂, then the total flow (discharge) from all jets is:
𝑄 = 𝑛𝑎𝑢𝑗 = 𝑛𝑄𝑗 →→ (6)

• If the 𝜂 𝑚 is the mechanical efficiency, then the mechanical power
output 𝑃𝑚 from the turbine with 𝒏 jets is [from Eqns. (4)−(6)]:
𝑃𝑚 = 𝜂 𝑚 𝑛 𝑃𝑗 𝑚𝑎𝑥
= 𝜂 𝑚 𝑛
1
2
𝜌𝑄𝑗 𝑢𝑗
2
= 𝜂 𝑚 𝑛
1
2
𝜌 𝑎𝑢𝑗 𝑢𝑗
2
⇒ 𝑷 𝒎 =
𝟏
𝟐
𝜼 𝒎 𝒏𝒂𝝆 𝟐𝒈𝑯 𝒂
𝟑 𝟐 →→ (𝟕)
• Eqn. (7) shows the importance of 𝐻 𝑎 between turbine and
reservoir. Also, the output power is proportional to the total jet
cross-sectional area 𝐴 𝐴 = 𝑛𝑎 . However, 𝒂 is limited by the size
of cup, so if 𝒂 is to be increased, a larger turbine is needed.
• It is usually easier to increase the number of nozzles 𝒏 than to
increase the overall size of the turbine, but the arrangement
becomes unworkably complicated for n ≥ 4. For small wheels, n = 2
is the most common.

• Moreover, as the total flow 𝑄 through the turbine cannot be more
than the flow in the stream 𝑄𝑠𝑡𝑟𝑒𝑎𝑚, thus from Eqns. (5) & (6):
𝑛𝑎𝑗 ≤ 𝑄𝑠𝑡𝑟𝑒𝑎𝑚 / 2𝑔𝐻 𝑎
1 2
→→ (8)
 ANGULAR VELOCITY AND TURBINE SIZE:
• If the nozzle size and number have been selected in accordance with
Eqns. (6) and (7) to give the maximum power available. It means the
nozzle size has fixed the size of the cups, but not the overall size of
the wheel.
• The overall size of the wheel is determined by geometric constraints,
and also by the required rotational speed.
• For electrical generation, the output variables, e.g. voltage,
frequency and efficiency, depend on the angular speed of the
generator. Most electric generators have greatest efficiency at large
rotational frequency, commonly at ~1500 rpm.

• Thus the turbine should operate at large frequency i.e.,
rotational speed.
• If the wheel has radius 𝑅 and turns at angular velocity 𝜔, then:
𝑃 = 𝐹𝑅𝜔 →→ (9)
• As 𝑢 𝑐 = 𝑅𝜔; but 𝑢 𝑐 = 0.5𝑢𝑗; & 𝑢𝑗 = 2𝑔𝐻 𝑎 , so:
⇒ 𝑅 =
𝑢 𝑐
𝜔
=
0.5 2𝑔𝐻 𝑎
1 2
𝜔
→→ (10)
• The nozzles usually give circular cross-section jets of area 𝒂
and radius 𝒓 so that 𝑎 = 𝜋𝑟2
and:
𝑟2 =
𝑎
𝜋
=
𝑃 𝑚
2 𝜋 𝜂 𝑚 𝜌 𝑛 𝑔𝐻 𝑎
3
2
→→ (11) (using Eqn. (7))

• Combining Eqns. (10) and (11) yields:
𝑟
𝑅
=
𝜔
0.5 2𝑔𝐻 𝑎
1 2
𝑃𝑚
𝜂 𝑚 𝜌𝑛𝜋 𝑔𝐻 𝑎
3 2 2
1
2
= 0.68 𝜂 𝑚 𝑛 −1 2 𝒮 →→ 11
where
𝒮 =
𝑃𝑚
1 2
𝜔
𝜌1 2 𝑔𝐻 𝑎
5 4
→→ (12)
• 𝒮 is a non-dimensional measure of the operating conditions,
called the shape number of the turbine.
• From Eqn. (11), it is clear that 𝜼 𝒎 at any instant is a function of:
1. The fixed geometry of a particular Pelton wheel (measured by
the non-dimensional parameters 𝑟𝑗/𝑅 and 𝑛) &
2. The shape number 𝒮 (characterizing the operating conditions).

• However, engineering texts usually use a dimensioned characteristic
called specific speed 𝑁𝑠 instead of the dimensionless shape number
𝒮. Thus Eqn. (12) leads to the specific speed, defined in terms of
variables 𝑃, 𝑣 = 𝜔 2𝜋 , and 𝐻 𝑎 as:
𝑁𝑠 =
𝑃1/2 𝑣
𝐻 𝑎
5/4
→→ (13)
• The 𝑁𝑠 has dimensions and units, and the units vary between USA
(rpm, shaft Horsepower, ft) and Europe (rpm, metric Horsepower, m)
with a standard version for SI units yet to become common.
• Moreover, for a particular shape of Pelton wheel (specified here by
𝑟𝑗 𝑅 and n), there is a particular combination of operating conditions
(specified by 𝒮) for maximum efficiency.

EXAMPLE
Determine the dimensions of a single jet Pelton wheel to
develop 160 kW under a head of (1) 81 m and (2) 5.0 m. What is
the angular speed at which these wheels will perform best?
Assume that 𝑟𝑗 = 𝑅 12; 𝜂 𝑚 = 0.9 and 𝒮 = 0.11.
HINTS:
(a)
𝒮 =
𝑃𝑚
1 2
𝜔
5 4
⇒ 𝜔 = 36 𝑟𝑎𝑑 𝑠−1
𝑢𝑗
2
= 2𝑔𝐻 𝑎 ⇒ 𝑢𝑗 = 40 𝑚𝑠−1
𝑅 =
1
2
𝑢𝑗
𝜔 ⇒ 0.55 𝑚
⇒ 𝑟𝑗 is also known now

REACTION TURBINES
• It is clear from the expression 𝑃0 = 𝜌𝑄𝑔𝐻 that inorder to have the
same power from a lower head, we have to maintain a greater flow
𝑄 through the turbine.
• As the shape number 𝒮 is 𝒮 =
𝑃 𝑚
1 2
𝜔
5 4 , for the same 𝜔 and 𝑃
with a lower 𝐻, turbine with larger 𝒮 is required.
One way of doing this is to increase the
number of nozzles on a Pelton wheel, as
(
𝑟
𝑅
= 0.68 𝜂 𝑚 𝑛 −1 2 𝒮 and Fig. 1(a)).
However, the pipework becomes unduly
complicated if 𝑛 > 4 , and thus the
efficiency decreases.
Fig. 1: Methods of increasing the power from a given size of
machine, working at the same water pressure. (a) A four-jet Pelton
wheel, the power of which is four times greater than that from a
one-jet wheel of the same size and speed.

REACTION TURBINES Cont.
• To have a larger flow through the turbine it is, therefore, necessary
to make a significant change in the design. The entire periphery of
the wheel is made into one large ‘slot’ jet which flows into the
rotating wheel, as shown in Fig. 1(b). Such turbines are called
reaction machines.
This turbine contrasts with impulse
machines (Pelton wheels), where the blades
(cups) receive a series of impulses. For a
reaction turbine, the wheel (called runner) is
adapted so that the fluid enters radially
perpendicular to the turbine axis, but leaves
parallel to this axis (e.g., Francis reaction
turbine) as shown in Fig. 1(b). The fluid has a
radial component of velocity in addition to
the tangential velocity.
Fig. 1: Methods of increasing the power from a given size of machine, working at the same water
pressure. (b) a reaction or radial flow turbine: e.g. Francis turbine.

REACTION TURBINES Cont.
A larger water flow can be obtained by
making the incoming water ‘jet’ almost
as large in cross section as the wheel
itself. This concept leads to a turbine in
the form of a propeller, with the flow
mainly along the axis of rotation as
shown in Kaplan turbine in Fig. 1(c).
However, the flow is not exactly axial.
Guide vanes on entry provide the fluid a
whirl (rotary) component of velocity,
thus the tangential momentum produced
from this whirl component is transferred
to the propeller, and making it rotate.
Fig. 1(c): A propeller turbine: e.g. Kaplan turbine.

Ch.6 Hydropower

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