T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx

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Keith Vaugh BEng (AERO) MEng
CENTRIFUGAL
PUMPS

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PUMP
A pump is a hydraulic machine that
converts mechanical energy into
hydraulic energy in the form of
Pressure Energy
When the conversion of
mechanical energy into hydraulic
energy is done by means of a
centrifugal force, then the pump is
referred to as a Centrifugal Pump

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MAIN PARTS
①
②
③
④
1. Impeller
2. Volute or Casing
3. Suction Section
4. Delivery Section
At the eye of the
impeller there is a
negative pressure
causing the suction
from the sump

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PRINCIPLE
Rise of pressure head is directly
proportional to the tangential velocity

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therefore
AREA ⬆︎
increases
VELOCITY ⬆︎
increases
VELOCITY ⬇︎
decreases
AREA ⬇︎
decreases
1︎⃣
2︎⃣
1︎⃣ decreasing tangential velocity
2︎⃣ increasing area
The continuity equation states

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So how does this decrease in Kinetic
Energy increase the Pressure Head?
According to Bernoulli’s equation:
Pressure
Energy
Kinetic
Energy
Potential
Energy
+ + = Constant
In the case of a centrifugal pump, we can assume that the suction inlet
and discharge outlet are on the same level, therefore the potential
energy in Bernoulli’s equation can be ignored
Pressure
Energy
Kinetic
Energy
+ = Constant
So as the Pressure Energy increases, the Kinetic Energy decreases
or if the Kinetic Energy increases, the Pressure Energy decreases

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During a laboratory test on a centrifugal pump water at 20 ºC is pumped from a
sump to a reservoir. The flow rate of water passing through the pump was
measured at 15 m3/h and the torque applied to the pump shaft was measured to
be 6.4 N-m. Other measurements taken during the test are tabulated in Table 2.
Assume the pump operates at 3000rpm.
EXAMPLE 32
Parameter Inlet Section Outlet Section
Gauge pressure, p (kPa) 86 ?
Elevation above datum, z (m) 2.0 10
Velocity, U (ms-1) 2.0 4.6
Construct a detailed diagram and label the components in such a system, and
identify the key assumptions that can be made. If the pump and electric motor
efficiencies are 75% and 85% respectively, determine; (a) the electric power required,
(b) the gauge pressure at the outlet.

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ASSUMPTIONS:
• Flow is steady
• Irreversible losses are negligible
GOVERNING EQUATIONS
𝑃𝑜𝑤𝑒𝑟 = 𝜌 × 𝑔 × 𝑉
·
× 𝐻
𝑘𝑔
𝑚3
×
𝑚
𝑠2
×
𝑚3
𝑠
× 𝑚
𝑘𝑔
𝑚3
×
𝑚5
𝑠3
𝑘𝑔 ×
𝑚2
𝑠3
𝑘𝑔 × 𝑚
𝑠2
×
𝑚
𝑠
𝑁 ×
𝑚
𝑠
=
𝐽
𝑠
= 𝑊𝑎𝑡𝑡𝑠

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Manipulating to obtain the discharge pressure (pressure at the outlet)
Solving for the the electric power required to develop the head
From the governing equations:

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The flow system used to test a centrifugal pump at a nominal speed of 1750
rpm is shown in the figure. The liquid water at 27 °C and the suction and
discharge pipe diameters are 150 mm. Data measured during the test are given
in the table. The electric motor is supplied at 460 V, 3-phase, and has a power
factor of 0.875 and a constant efficiency of 90%.
EXAMPLE 33
Zd

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Calculate the net head delivered and the pump efficiency at a volumetric flow
rate of 227m3/s. Plot the pump head, power input and efficiency as functions of
the volumetric flow rate.
EXAMPLE 33
Table 1: Test results - Centrifugal pump
Volumetric Flow
Rate (m3/s)
Suction Pressure
(kPa-gauge)
Discharge Pressure
(kPa-gauge)
Motor Current (amp)
0 -25 377 18.0
114 -29 324 25.1
182 -32 277 30.0
227 -39 230 32.6
250 -43 207 34.1
273 -46 179 35.4
318 -53 114 39.0
341 -58 69 40.9

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GOVERNING EQUATIONS:
SOLUTION:
GIVEN: Pump test flow system and data shown
ASSUMPTIONS:
• Steady and incompressible flow
• Uniform flow at each section
• U1 = U2
• Correct all heads to same elevation
FIND:
• Pump head and efficiency at Q = 227 m3/s
• Pump head, power input, and efficiency as a function of volumetric flow rate
• Plot the results

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ANALYSIS:
Since U1 = U2, the pump head is
where the discharge and suction pressures, corrected to the same elevation,
are designated P2 and P1, respectively
Correct measured static pressures to the pump centreline

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Calculate the pump head
Compute the hydraulic power delivered to the fluid

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Calculate the motor power output (the mechanical power input to the pump)
from the electrical information
The corresponding pump efficiency is
Perform these calculations for each of the other points, and plot the results
against the Volumetric Flow Rate:

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IMPELLER CALCULATIONS
BEng Mechanical Engineering

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A fan has a bladed rotor of 0.3 m outside diameter, 0.13 m inside
diameter and runs at 1725 rpm. The width of each rotor blade is
0.025 m from blade inlet to outlet. The volumetric flow rate is steady
at 0.11m3/s and the absolute velocity of the air at the blade inlet U1
is purely radical. The blade discharge angle is 30 deg measured
with respect to the tangential direction at the outside diameter of the
rotor. Determine
• A reasonable blade inlet angle (measured with respect to the
tangential direction at the inside diameter of the rotor)?
• The power required to run the fan.
EXAMPLE 1 - FAN

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SOLUTION:
A fan turbine is driven at a specified angular speed by fluid entering along the axis of
rotation. The available power at the rotor shaft and an appropriate inlet angles at the
blades is to be determined.
ASSUMPTIONS:
The flow is steady
Irreversible losses are negligible
PROPERTIES:
Assume the density of the fluid to be 1.23 kg/m3
ANALYSIS:
The stationary and non-deforming control volume shown in the sketch is used. To
determine a reasonable blade inlet angle we assume that the blade should be tangent
to the relative velocity at the inlet. The inlet velocity triangle can be sketched as

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ω
Runner
blade
u1→ Absolute velocity vector. Velocity of the
flow at the inlet.
v1→ Velocity of blade at inlet i.e. 𝑣1 = 𝜔𝑟1
𝝑1→ Angle of flow at inlet; the angle between
the velocity vector of the blade and the
direction of the flow at inlet
𝝑1
v1

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β1
ω
Runner
blade
ur1
𝝑1
ur1→ The relative velocity vector of the flow at
the inlet
uf1→ The velocity of flow in the radial
direction
uw1→ The component of the velocity of the flow
in horizontal/tangential direction
β1→ Blade angle at inlet. Angle between the
relative velocity vector and the direction of
motion of the blade at the inlet
𝝑1
flow at the inlet.
𝝑1→ Angle of flow at inlet; the angle between
direction of the flow at inlet
v1

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β1
ω
Runner
blade
ur1
𝝑1
𝝑1
v1

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β1
ω
Runner
blade
ur1
𝝑1
𝝑1
v1
uf2
flow at the outlet.
ur2→ The relative velocity vector of the flow at
the outlet
direction
uw2→ The component of the velocity of the flow
in horizontal/tangential direction
𝝑2→ Angle of flow at outlet; the angle between
direction of the flow at outlet
β2→ Blade angle at outlet. Angle between the
relative velocity vector and the direction of
motion of the blade at the outlet

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ϑ
Opposite
Adjacent
𝑠𝑖𝑛𝜃 =
𝑂𝑃𝑃
𝐻𝑌𝑃
𝑐𝑜𝑠𝜃 =
𝐴𝐷𝐽
𝐻𝑌𝑃
𝑡𝑎𝑛𝜃 =
𝑂𝑃𝑃
𝐴𝐷𝐽
Also remember that
𝑡𝑎𝑛𝜃 =
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
𝑐𝑜𝑡𝜃 =
𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃
𝑠𝑒𝑐𝜃 =
1
𝑐𝑜𝑠𝜃
𝑐𝑠𝑐𝜃 =
1
𝑠𝑖𝑛𝜃
Recall from Trigonometry in Mathematics

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uf1
uf1
u1
uw1
ur1
uf1
u1
uf1 ur1
Velocity triangles Use trigonometry to solve for each vectors
uw1
Two triangles within
uw1
uw1
uw1 - v1
v1 - uw1
NOTE - These triangles use different notation - from Renewable module

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(a) With the velocity triangle, we conclude that
Now
𝛽1 = 𝑡𝑎𝑛−1 𝑢1
𝑣1
eq 1
and
𝑢1 =
𝑉
·
𝐴1
=
𝑉
·
2𝜋𝑟1ℎ1
𝑢1 =
0.11𝑚3
/𝑠
2𝜋 × 0.065𝑚 × 0.025𝑚
= 10.8𝑚/𝑠
𝑉1 = 𝑟1𝜔
𝑉1 =
2𝜋𝑟𝑎𝑑/𝑟𝑒𝑣 × 0.065𝑚 × 1725𝑟𝑒𝑣/𝑚𝑖𝑛
60𝑠/𝑚𝑖𝑛
= 11.7𝑚/𝑠
u1
v1
β1 𝝑1
r1
ur1

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Therefore from equation 1
𝛽1 = 𝑡𝑎𝑛−1
10.8𝑚/𝑠
11.7𝑚/𝑠
= 43∘
(b) The power required can be obtained from;
𝑊
·
𝑠ℎ𝑎𝑓𝑡 = 𝑚
·
2𝑣2𝑢𝑤2 eq 2
The mass flow rate maybe obtained as follows
𝑚
·
2 = 𝜌𝑉
·
= 1.23𝑘𝑔/𝑚3
× 0.11𝑚3
/𝑠 = 0.1353𝑘𝑔/𝑠

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and
𝑣2 = 𝑟2𝜔
𝑣2 =
2𝜋𝑟𝑎𝑑/𝑟𝑒𝑣 × 0.15𝑚 × 1725𝑟𝑒𝑣/𝑚𝑖𝑛
60𝑠/𝑚𝑖𝑛
= 27.1𝑚/𝑠
The value of Uθ2︎ maybe obtained by considering the velocity
triangle for the flow leaving the rotor at section 2. The
relative velocity at the rotor exit is considered to be tangent
to the blade there. The rotor exit flow velocity triangle is
sketched as
r2

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therefore
𝑢𝛽2 =
𝑢𝑓2
𝑡𝑎𝑛30∘
=
𝑉
·
2𝜋𝑟2ℎ2
𝑡𝑎𝑛30∘
=
0.11𝑚2/𝑠
2𝜋 × 0.15𝑚 × 0.025𝑚
𝑡𝑎𝑛30∘
= 8.1𝑚/𝑠
Now 𝑢𝑤2 = 𝑣2 − 𝑢𝛽2 and 𝑢𝑤2 = 27.1𝑚/𝑠 − 8.1𝑚/𝑠 = 19𝑚/𝑠
and from equation 2
𝑊
·
𝑠ℎ𝑎𝑓𝑡 = 0.1353𝑘𝑔/𝑠 × 27.1𝑚/𝑠 × 19𝑚/𝑠 = 69.7𝐽/𝑠

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TURBINE CALCULATIONS
BEng Mechanical Engineering & Renewable Energy

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ω
Runner
blade
β1
flow at the inlet. It’s direction is governed
by the guide vane angle α
v1→ Velocity of blade at inlet
ur1→ The relative velocity vector of the jet at
the inlet
direction
uw1→The component of the velocity of the jet
u1 in horizontal/tangential direction
α1→ Guide vane angle, or the angle between
direction of the jet
β1→ Blade angle at inlet. Angle between the
relative velocity vector and the direction
of motion of the blade at the inlet
u0
u0

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ω
Runner
blade
u0
flow at the inlet. It’s direction is governed
by the guide vane angle α
v1→ Velocity of blade at inlet
ur1→ The relative velocity vector of the jet at
the inlet
direction
uw1→The component of the velocity of the jet
α1→ Guide vane angle, or the angle between
direction of the jet
β1→ Blade angle at outlet. Angle between the
of motion of the blade at the inlet

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ω
Runner
blade
u0
u2→ Absolute velocity vector water leaving
the runner. It’s direction is governed
by the guide vane angle α and if it is in
the radial direction, will be equal to uf2
v2→ Velocity of blade at exit
ur2→ The relative velocity vector of the water
leaving the impeller
direction
uw2→The component of the velocity for exit
α2→ The angle between the velocity vector of
the blade and the direction of the u2
exiting
β2→ Blade angle at outlet, Angle between the
of motion of the blade at the outlet
uf2
β2
𝜶2
u0

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uf1
uf1
u1
uw1
ur1
uf1
u1
uf1 ur1
Velocity triangles Use trigonometry to solve for each vectors
uw1
Two triangles within
uw1
uw1
uw1 - v1
v1 - uw1

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A water turbine wheel rotates at the rate of 50 rpm in the
direction shown in figure 1. The inner radius r2 of the blade
row is 0.6 m and the outer radius r1 is 1.2 m. The absolute
velocity vector at the turbine rotor entrance makes an angle
of 20° with the tangential direction. The inlet blade angle is
60° relative to the tangential direction. The blade output
angle is 120°. The flow rate is 0.57m3/s. For the flow tangent
to the rotor blade surface at inlet and outlet, determine an
appropriate constant blade height, b, and the corresponding
power available at the rotor shaft.
EXAMPLE 1 - TURBINE

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SOLUTION:
A water turbine is driven at a specified angular speed by water entering it at
a specified rate and angle. The available power at the rotor shaft and an
appropriate height for the blades is to be determined.
ASSUMPTIONS:
The flow is steady
Irreversible losses are negligible
PROPERTIES:
Assume the density of water to be 1 kg/L = 1000 kg/m3

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The use of the “+” or “-” with V2Uθ,2 depends on whether Uθ,2 is opposite to or in
the same direction as V2 respectively. To determine the value of UR,1 use the
velocity triangle at section (1), thus:
ANALYSIS:
equation (1)
The shaft power is obtained with the moment of momentum power equation
equation (2)

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However
equation (3)
With the velocity triangle we get

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Therefore
and b can be calculated

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For Uθ,1 we use the velocity triangle at section (1) to obtain
For the blade velocities in equation (2)

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For Uθ,2 we construct the section (2)
velocity triangle sketch below and it is
noted that
equ (4)
From conservation of mass

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So with equation 4 we obtain
Finally from equation 2 we obtain

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A Francis radial flow hydroturbine has the following dimensions, where locations 2
is the inlet and 1 is the outlet; r1 = 2.0 m, r2 = 1.3 m, b1 = 0.85 m and b2 = 2.1 m.
The runner blade angles are β1 = 66° and β2 = 18.5° at the turbine inlet and outlet
respectively. The runner rotates a ṅ = 100 rpm. The volume flow rate at the design
conditions is 80.0 m3/s. Irreversible losses are neglected in this preliminary
analysis. Calculate the angle α1 through which the wicket gates should turn the
flow, where, α1 is measured from the radial direction at the runner inlet. Calculate
the swirl angle α2, where α2 is measured from the radial direction at the runner
outlet. Does this turbine have a forward or reverse swirl? Predict the power output
(MW) and the required net head (m).
EXAMPLE 2 - TURBINE

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SOLUTION
A Francis turbine is driven at a specified angular speed by water entering it
at a specified rate and angle.We are to calculate runner blade angles,
required net head, and power output for a Francis turbine.
ASSUMPTIONS:
• The flow is steady.
• The fluid is water at 20°C
• The blades are infinitesimally thin.
• The flow is everywhere tangent to the runner blades.
• We neglect irreversible losses through the turbine.
PROPERTIES:
Assume the density of water to be 998 kg/m3

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Analysis The angular velocity is 𝜔 =
2𝜋𝑛
60
= 10.4719𝑟𝑎𝑑/𝑠.
Solve for the normal component of velocity at the inlet,
𝑢𝑓1 =
𝑉
·
2𝜋𝑟1𝑏1
=
80𝑚3/𝑠
2𝜋 2.0𝑚 0.85𝑚
= 7.4896𝑚/𝑠 (1)
The tangential velocity component of the absolute velocity at the inlet is obtained from trigonometry to be
𝑢𝑤1 = 𝜔𝑟1 −
𝑢𝑓1
𝑡𝑎𝑛𝛽1
= 10.4719𝑟𝑎𝑑/𝑠 × 2𝑚 −
7.4896𝑚/𝑠
𝑡𝑎𝑛66∘ = 17.609𝑚/𝑠 (2)
From these two components of u1 in the absolute coordinate system, we calculate the angel α1 through
which the wicker gates should turn the flow,
𝛼1 = 𝑡𝑎𝑛−1
𝑢𝑓1
𝑢𝑤1
= 𝑡𝑎𝑛−1
7.4896𝑚/𝑠
17.609𝑚/𝑠
= 23.04∘
When measured from the radial direction - 90∘
− 23.04∘
= 66.96∘
(3)

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ω
Runner
blade
u0
u2→ Absolute velocity vector water leaving
the runner. If it is in the radial direction,
will be equal to uf2
v2→ Velocity of blade at exit
ur2→ The relative velocity vector of the water
leaving the impeller
direction
uw2→The component of the velocity for exit
α2→ The angle between the velocity vector of
the blade and the direction of the u2
exiting
β2→ Blade angle at outlet, Angle between the
of motion of the blade at the outlet
uf2
β2
𝜶2
u0

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Analysis The angular velocity is 𝜔 =
2𝜋𝑛
60
= 10.4719𝑟𝑎𝑑/𝑠.
Solve for the normal component of velocity at the inlet,
𝑢𝑓2 =
𝑉
·
2𝜋𝑟2𝑏2
=
80𝑚3/𝑠
2𝜋 1.3𝑚 2.1𝑚
= 4.6639𝑚/𝑠 (1)
The tangential velocity component of the absolute velocity at the inlet is obtained from trigonometry to be
𝑢𝑤2 = 𝜔𝑟2 −
𝑢𝑓2
𝑡𝑎𝑛𝛽2
= 10.4719𝑟𝑎𝑑/𝑠 × 1.3𝑚 −
4.6639𝑚/𝑠
𝑡𝑎𝑛18.5∘ = −0.325𝑚/𝑠 (2)
From these two components of u1 in the absolute coordinate system, we calculate the angel α1 through
which the wicker gates should turn the flow,
𝛼2 = 𝑡𝑎𝑛−1
𝑢𝑓2
𝑢𝑤2
= 𝑡𝑎𝑛−1
−0.325𝑚/𝑠
4.6639𝑚/𝑠
= −3.986∘
Since α2 is negative, this turbine operates with a small amount of reverse swirl.

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Using equations 2 and 4, the shaft output power is estimated from the euler turbo machinery equation
𝜌𝑉
·
𝑣1𝑢𝑤1 − 𝑣2𝑢𝑤2 = 𝜌𝜔𝑉
·
𝑟1𝑢𝑤1 − 𝑟2𝑢𝑤2
𝜌𝜔𝑉
·
𝑟1𝑢𝑤1 − 𝑟2𝑢𝑤2
= 998𝑘𝑔/𝑚3
× 10.472𝑟𝑎𝑑/𝑠 × 80𝑚3
/𝑠 2𝑚 × 17.609𝑚/𝑠 − 1.3𝑚 × −0.325𝑚/𝑠
𝜌𝜔𝑉
·
𝑟1𝑢𝑤1 − 𝑟2𝑢𝑤2 = 297.9𝑀𝑊
the required net head assuming that the turbine efficiency is 100% since we have irreversibility’s
𝐻 =
𝑊
·
𝜌𝑔𝑉
· =
297.9𝑀𝑊
998𝑘𝑔/𝑚3 × 9.81𝑚/𝑠2 × 80𝑚3/𝑠
= 38.05𝑚
Since the required net head is less than the gross net heat, the design is feasible

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𝑃𝑜𝑤𝑒𝑟 = 𝜌 × 𝑔 × 𝑉
·
× 𝐻
𝑘𝑔
𝑚3
×
𝑚
𝑠2
×
𝑚3
𝑠
× 𝑚
𝑘𝑔
𝑚3
×
𝑚5
𝑠3
𝑘𝑔 ×
𝑚2
𝑠3
𝑘𝑔 × 𝑚
𝑠2
×
𝑚
𝑠
𝑁 ×
𝑚
𝑠
=
𝐽
𝑠
= 𝑊𝑎𝑡𝑡𝑠
Examine the Units

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Water is pumped from a lower reservoir to a higher reservoir
by a pump that provides 20 kW of useful mechanical power
to the water. The free surface of the upper reservoir is 45 m
higher than the surface of the lower reservoir. If the flow rate
of water is measured to be 0.03 m3/s, determine the
irreversible head loss of the system and the lost mechanical
power during this process.
EXAMPLE 1

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ANALYSIS:
The mass flow rate of water through the system is
SOLUTION:
Water is pumped from a lower reservoir to a higher one. The head loss and
power loss associated with this process are to be determined.
ASSUMPTIONS
The flow is steady and incompressible.
The elevation difference between the reservoirs is constant.
PROPERTIES:
We take the density of water to be 1000 kg/m3 .

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We choose points 1 and 2 at the free surfaces of the lower and upper
reservoirs, respectively, and take the surface of the lower reservoir as the
reference level (z1 = 0). Both points are open to the atmosphere (P1 = P2 =
Patm) and the velocities at both locations are negligible (V1 = V2 = 0). Then
the energy equation for steady incompressible flow for a control volume
between 1 and 2 reduces to

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Substituting the lost mechanical power and head loss are calculated as
Noting that the entire mechanical losses are due to frictional losses in piping and thus,
the irreversible head loss is determined to be

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DISCUSSION:
The 6.76 kW of power is used to overcome the friction in the piping system.
Note that the pump could raise the water an additional 23 m if there were no
irreversible head losses in the system. In this ideal case, the pump would
function as a turbine when the water is allowed to flow from the upper reservoir
to the lower reservoir and extract 20 kW of power from the water.

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Water in a partially filled large tank is to be supplied to the
roof top, which is 8 m above the water level in the tank,
through a 2.5 cm internal diameter pipe by maintaining a
constant air pressure of 300 kPa (gauge) in the tank. If the
head loss in the piping is 2 m of water, determine the
discharge rate of the supply of water to the roof top.
EXAMPLE 8

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ANALYSIS:
We take point 1 at the free surface of the tank, and point 2 at the exit of the
discharge pipe. Noting that the fluid velocity at the free surface of the tank is
very low (V1≅ 0) and water discharges into the atmosphere (and thus P2 =
Patm), the energy equation written in the head form simplifies to
SOLUTION:
Water from a pressurised tank is supplied to a roof top. The discharge rate of water
from the tank is to be determined.
ASSUMPTIONS:
The flow is steady and incompressible.
The effect of the kinetic energy correction factor is negligible and thus α2 = 1 (we
examine the effect of this approximation in the discussion).
PROPERTIES:
We take the density of water to be 1000 kg/m3.

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Manipulating
Solving for U2 and substituting the discharge velocity is determined to

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DISCUSSION:
This is the discharge rate that will prevail at the beginning. The mean flow
velocity will decrease as the water level in the tank decreases. If we assume that
the flow in the hose at the discharge is fully developed and turbulent, α2 ≈ 1.05,
then the results change to V2 = 19.610 m/s ≈ 19.6 m/s, and the Volflowrate =
0.0096263 m3/s 9.63 L/s , a decrease (as expected since we are accounting for
more losses) of about 2.4%.
Then the initial rate of discharge of water becomes

KV
Water is pumped from an underground reservoir by a 78%
efficient 5 kW submerged pump to a pool whose free surface
is 30 m above the underground water level. The diameter of
the pipe is 7 cm on the intake side and 5 cm on the
discharge side. (Assuming an irreversible head of 4m)
Determined
• The maximum flow rate of the water
• The pressure difference across the pump
Assume the elevation difference better the pump inlet and
the outlet and the effect of the kinetic energy correction
factors to be negligible.
EXAMPLE 9 & 11

KV
ANALYSIS:
(a) The pump-motor draws 5 kW of power, and is 78% efficient. Then the
useful mechanical (shaft) power it delivers to the fluid is
SOLUTION:
Underground water is pumped to a pool at a given elevation. The maximum flow rate
and pressures at the inlet and outlet of the pump are to be determined.
ASSUMPTIONS:
•The flow is steady and incompressible.
•The elevation difference between the inlet and the outlet of the pump is
negligible.
•We assume the frictional effects in piping to be negligible since the maximum
flow rate is to be determined,
•The effect of the kinetic energy correction factors is negligible, α = 1.
PROPERTIES:

KV
ANALYSIS:
(a) The pump-motor draws 5 kW of power, and is 78% efficient. Then the
useful mechanical (shaft) power it delivers to the fluid is
We take point 1 at the free surface of underground water, which is also
taken as the reference level (z1 = 0), and point 2 at the free surface of
the pool. Also, both 1 and 2 are open to the atmosphere (P1 = P2 =
Patm), the velocities are negligible at both points (V1 ≅ V2 ≅ 0), and
frictional losses in piping are disregarded. Then the energy equation for
steady incompressible flow through a control volume between these two
points that includes the pump and the pipes reduces to

KV
In the absence of a turbine,
and
Thus,
Then the mass and volume flow rates of water become (NOTE - Solution
for 11 in Blue)

KV
ANALYSIS:
(b) We take point 3 and 4 at the inlet and the exit of the pump respectively,
were the flow velocities are
We take the pump as the control volume. Noting that z3 = z4 the energy
equation for this control volume reduces to

KV
Substituting,
DISCUSSION: In an actual system, the flow rate of water will be less
because of friction in the pipes. Also, the effect of flow velocities on the
pressure change across the pump is negligible in this case (under 2%) and
can be ignored.

KV
A utility company is selling electric power for €0.08/kWh at night and is willing to
pay €0.195/kWh for power produced during the day. To take advantage of this
opportunity an individual is considering building a large reservoir 40 m above
the level of a lake. Availing of the cheap electricity rates at night, this individual
intends to pump water from the lake to reservoir. During the day, the water will
be allowed to flow back through the pump-motor, which will act as a turbine-
generator (i.e. the pump-motor operating in reverse), thereby producing
electricity. Initial analysis shows that a water flow rate of 2 m3⋅s-1 can be used in
either direction, and the irreversible head loss of the piping system is 4 m. The
combined pump-motor and turbine-generator efficiencies are expected to be
75% each. Assuming the system operates for 10 hours each in pump and
turbine modes during a typical day, determine the potential revenue this pump-
turbine system can generate per year.
EXAMPLE 10

KV
ANALYSIS:
Choose points 1 and 2 at the free surfaces of the lake and the reservoir, and
take the surface of the lake as the datum. Both points are open to the
atmosphere (P1 = P2 = Patm); velocities at both are negligible (V1 = V2 = 0).
SOLUTION:
Water is pumped to a reservoir at a given elevation during the night to avail of night
rate electricity. During the day the water flows back through the pump operating in
reverse and generates electricity. Required to determine if this a cost making
venture.
ASSUMPTIONS:
• The flow in each direction is steady and incompressible
• The elevation difference between the lake and the reservoir can be taken to be
constant
• The elevation change of reservoir during charging and discharging is
disregarded
• Unit prices remain constant
• The system operates every day of the year for 10 hours in each mode
PROPERTIES:

KV
Then the energy equation in terms of heads for steady incompressible flow
through a control volume between these two points that includes the pump (or
the turbine) and the pipes reduces to;
The pump and turbine power corresponding to there heads are;

KV
Then the power cost of the pump, the revenue generated by the turbine, and
the net income (revenue minus cost) per year become;
It appears that this pump-turbine system has a potential annual income of about
€41,135.5 per annum. A decision on such a system will depend on the initial cost
of the system, its life, the operating and maintenance costs, the interest rate, and
the length of the contract period.

KV
VELOCITY
TRIANGLES
Radii of the impeller - ri
Absolute velocity vector - U
Tangential velocity vector - V
Relative velocity vector - Ur
Angles between the relative velocity
vector and tangential velocity α and
β
Angles between the relative velocity
vector and tangential velocity θ and
Φ

T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx

T2c - Centrifugal Pumps, turbines and Impeller calculations 2023.pptx

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