Problem Set Topic 2 Solutions PDF

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Summary

Questions prefixed with (*) are adapted from examples worked through in the lectures. The solutions below are for guidance only and there may be other ways of achieving the correct or fully developed answers.Turbine Questions (i) Write Bernoulli’s equation for the water at (1) the top of the reservo...

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Questions prefixed with (*) are adapted from examples worked through in the lectures. The solutions below are for guidance only and there may be other ways of achieving the correct or fully developed answers.

Turbine Questions 1. (i) Write Bernoulli’s equation for the water at (1) the top of the reservoir and (2) as it exits the penstock into the bottom reservoir to show that the velocity of a jet may be expressed as Vj = √2gH. The above definition of jet velocity may also be found by equating the gravitational and kinetic energies of water. Show this alternative solution. If a simple hydroelectric plant consists of an upper and lower reservoir, both open to the atmosphere and connected by a frictionless penstock with a 10 m vertical height, what is the velocity of the water as it exits the penstock? [Answer: vj = 14 m/s]

Solution

𝐸𝑔𝑟𝑎𝑣 = 𝐸𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑚𝑔ℎ =

1

2

𝑚𝑣 2

∴ 𝑣 = √2𝑔ℎ

Write Bernoulli's equation for points 1 and 2, noting that the pressure terms cancel (as both are equal). The water at point 1 has zero velocity v1= 0, and the streamline at 1 is z1 = 10m with respect to point 2, which is zero z2 = 0.

𝑉2 = 𝑉𝑗 = √2𝑔𝑧1 = √2 𝑥 9.81 𝑥 10 = 14.0 𝑚/𝑠

ES4E0 Topic 2 Problem Set

2. Starting with an expression for gravitational potential energy, show how the relationship P = ηρgQH is derived. Define each of the terms.

Solution 𝐸𝑔𝑟𝑎𝑣 = 𝑚𝑔ℎ

𝑃 = 𝐸󰇗𝑔𝑟𝑎𝑣 = 𝑚󰇗𝑔ℎ

Where 𝑚 = mass, 𝑔 = 9.81 m/s, h = vertical height

For fluid flow 𝑚󰇗 = ρQ, h = H, η = turbine efficiency ∴ P = ηρgQH

3. A hydraulic turbine is to be sited on a dam on a river. The flow rate is 1500 m3/h, the available head is 24 m and the turbine is to rotate at 480 rpm. The efficiency is estimated at 89%. What type of turbine would be suitable for this location? Solution Convert flow rate to SI units: Q=

1500 𝑚3 /ℎ 3600

= 0.42 m3/s

Calculate the hydraulic power produced in the turbine: Phydraulic = 𝜌𝑔𝑄𝐻gross = 0.89 x 1000 x 9.81 x 0.42 x 24 = 88.0 x 103 W = 88KW Convert angular speed to SI units: 𝜔=

2𝑁𝜋 60

=

2 ∙ 480 ∙ 𝜋 60

= 50.3 rad/s

Calculate shape number: =

1

50.3 𝑥 (88 ∙ 1000 ⁄1000 )2 (9.81 ∙ 24)5/4

= 0.51

𝑁𝑠 > 2 – Kaplan turbine

0.3 < 𝑁𝑠 < 2 - Francis turbine 𝑁𝑠 < 0.3 – Pelton turbine

∴ A Francis turbine would be most suitable

ES4E0 Topic 2 Problem Set

4. At one potential site for a hydroelectric plant the gross head is 150 m, and the estimated volumetric flow rate of water through each turbine would be 1.2 m3/s. Estimate the ideal power production, per turbine, in MW. Assume ρwater = 1000 kg/m3. Solution Pideal = 𝜌𝑔𝑄𝐻gross = 1000 x 9.81 x 1.2 x 150 = 1.77 x 106 W = 1.77 MW 5. Briefly describe the differences between Impulse and Reaction turbines. In what situations do they work most efficiently, and why? Give an example of both an Impulse and Reaction turbine and include a sketch. Solution Impulse turbines generally uses the velocity of the water to move the runner and discharges to atmospheric pressure. The water stream hits each bucket on the runner. There is no suction on the down side of the turbine, and the water flows out the bottom of the turbine housing after hitting the runner. An impulse turbine is generally suitable for high head, low flow applications. Reaction turbines develop power from the combined action of pressure and moving water. The runner is placed directly in the water stream flowing over the blades rather than striking each individually. Reaction turbines are generally used for sites with lower head and higher flows than compared with impulse turbines.

Examples of impulse turbines: Pelton, Turgo and Crossflow. - A Pelton wheel has one or more free jets discharging water at atmospheric pressure. The buckets are double hemispherical in shape. The water strikes the bucket in the centre and flows out at both sides, essentially making a U-turn. The surface inside the buckets is polished and smooth to reduce hydraulic losses. Bronze or stainless steel is generally used for the buckets, which are usually detachable.

Examples of reaction turbines: Kaplan (propeller) and Francis. - Kaplan, where both the blades and the wicket gates are adjustable, allowing for a wider range of operation. ES4E0 Topic 2 Problem Set

- Francis, which has a runner with fixed buckets (vanes), usually nine or more. Water is introduced just above the runner and all around it and then falls through, causing it to spin. Besides the runner, the other major components are the scroll case, wicket gates, and draft tube.

The diagrams above taken from the lecture slides for guidance only. Sketches should be labelled, show the inlet and outlet flow direction, and also include a cross-sectional view. 6. What must happen to the shape number of a turbine in order to increase power but maintain the same angular velocity at a lower head? How can this be achieved with a Pelton Wheel? How else can even more power be gained from water? Solution

The shape number must increase. Shape number = 𝑁𝑠 =

1

2 𝜔(𝑃 ⁄𝜌 )

(𝑔𝐻)5/4

ES4E0 Topic 2 Problem Set

Methods of increasing the power from a given size of machine, working at the same water pressure:

1. Increase the number of jets 𝑁𝑠 =

𝑟 √η n 𝑅 0 .68

The above relation is for a Pelton Wheel, and increasing n (number of jets) increased shape number. 2. Make the entire periphery of the wheel one large ‘slot’ jet for water to enter and then turn in a vortex to push against the rotor vanes (i.e. use a Francis Turbine). Francis turbines consist of fixed guide vanes called stay vanes, adjustable guide vanes called wicket gates, and rotating blades called runner blades. Flow enters tangentially at high pressure, is turned toward the runner by the stay vanes as it moves along the spiral casing or volute, and then passes through the wicket gates with a large tangential velocity component. Momentum is exchanged between the fluid and the runner as the runner rotates, and there is a large pressure drop. Unlike the impulse turbine, the water completely fills the casing of a reaction turbine.

3. Make the incoming water ‘jet’ almost as large in cross-section as the wheel itself (i.e. use a propeller/Kaplan Turbine). This concept leads to a turbine in the form of a propeller, with the flow mainly along the axis of rotation (e.g. the Kaplan turbine) - essentially a propeller that is run backwards. With the propeller of a ship, mechanical energy is supplied to the propeller causing it to turn and drive water backward and thus move the ship forward. Here in the reverse situation, the motion of water through the propeller causes it to acquire mechanical energy of rotation that can be harnessed to produce electricity. Although many propeller turbines have nonadjustable blades, the Kaplan turbine blades do have an adjustable pitch angle, allowing it to cope with varying flow rates. In addition to being highly efficient, Kaplan turbines, by decreasing the blade angle, can cope with extremely low heads—as little as a meter and still have a sufficiently large ω.

ES4E0 Topic 2 Problem Set

7. Draw a Q-H diagram and show the approximate locations of Pelton, Francis and Propeller turbines. Solution

ES4E0 Topic 2 Problem Set

8. At what jet deflection angle is the maximum theoretical efficiency of a Pelton Wheel? Why is this not achievable in practice? If this maximum theoretical angle of deflection is reduced by 20 degrees, by what percentage does the efficiency reduce to? Solution Maximum deflection occurs at 180°, however this could result in the water exiting one bucket hitting the back of the next bucket, opposing wheel motion.

The efficiency of a Pelton Wheel is given by 𝜂𝑤ℎ𝑒𝑒𝑙 =

2(V𝑗 – u)(1 − kcosβ)u 𝑉𝑗 2

Dividing ( 𝜂𝑤ℎ𝑒𝑒𝑙 , β) by (𝜂𝑤ℎ𝑒𝑒𝑙 , 180°) simplifies to 𝜂β = When β = 160°, 𝜂β = 0.97 (a reduction of 3%).

1 − cosβ 1 − cos(180°)

9. A Pelton turbine develops 67.5 kW under a head of 60 m of water. It rotates at 400 rev/min. The diameter of the penstock is 200 mm, the ratio of bucket speed to jet velocity is 0.46 and the overall efficiency of the installation is 83%. Calculate: • the volumetric flow rate • the diameter of the jet • the turbine wheel diameter • the shape number of the turbine Solution Overall efficiency 𝜂 = ∴Q=

67.5 𝑥 1000

0.83 × 9800 × 60

𝑃

𝜌𝑔𝑄𝐻

; 𝑄 =

𝑃

𝜂𝜌𝑔 𝐻

= 0.138 m3/s

Velocity of the jet, 𝑉𝑗 = √2𝑔𝐻 = √2 𝑥 9.81 𝑥 60 = 34.2 𝑚/𝑠 Q = area of nozzle x velocity of jet 𝜋

∴ d2 =

Q = 4 x d 2 x Vj 0.138 𝑥 4 𝜋 𝑥 34.2

= 5.14 x 10-3

d = 71.7 mm 𝑢

Speed ratio = = 0.46; u = 0.46 x 34.2 = 15.7 m/s 𝑉𝑗

Wheel speed = 𝑢 =

𝜋𝐷𝑁 60

∴D=

60 × 𝑢 𝜋×𝑁

=

60 × 15.7 𝜋 × 400

ES4E0 Topic 2 Problem Set

= 0.75 m

Shape number = 𝑁𝑠 =

𝜔=

2𝑁𝜋 60

=

2 𝑥 400 𝑥 𝜋 60

12

) 𝜔(𝑃⁄𝜌5/4 (𝑔𝐻)

= 41.8 rad/s

P = 67.5 x 103 W 𝜌 = 1000kg/m3 H = 60m ∴ 𝑁𝑠 =

1

⁄ )2 41.8 𝑥 (67.5 𝑥 1000 1000 = (9.81 𝑥 60)5/4

0.11

𝑁𝑠 < 0.3 – Pelton turbine

∴ confirming that a Pelton turbine is the most suitable turbine for this location 10. In a proposed hydroelectric scheme the output power required is 60 MW. The gross head at the reservoir is 300 m and head lost in the pipeline will not exceed 20 m. It is intended to use several Pelton wheels, each with the following characteristics: operating speed 400 rpm; turbine shape number 0.3; overall efficiency 80%; 6 jets; Cv of nozzles 0.97; speed ratio (i.e. bucket velocity / jet velocity) 0.46. Assuming operation at maximum efficiency, calculate: (a) the maximum output power per wheel; (b) the number of wheels required; (c) the velocities of jets and buckets; (d) the diameter of each wheel; (e) the output power per jet; (f) the quantity of flow per jet; (g) the diameter of each jet; (h) the hydraulic efficiency if the buckets deflect water through 165° and reduce the relative velocity by 15% so that k = 0.85.

Solution Calculate total available head: H = Hgross – Hlosses = 300 – 20 = 280 m Convert angular speed to SI units: 𝜔=

2𝑁𝜋 60

=

2 × 400 × 𝜋 60

= 41.9 rad/s

Use the equation for shape number to give the power per wheel ES4E0 Topic 2 Problem Set

∴ P = ((

𝑁𝑠 =

5 2

𝑤𝑇

12

𝜔(𝑃 ⁄𝜌5/4 ) (𝑔𝐻)

5

2

) x (9.81 𝑥 280)4 ) x 1000 = 20.3 MW ) x (𝑔𝐻)4 ) x 𝜌 = (( 0.3 41.9 𝜔

Calculate number of wheels required: Total power / Power per wheel = 60 MW/20.3 𝑀𝑊 = 2.96 ≈ 3 wheels Calculate jet velocity:

Vj = 0.97√2𝑔𝐻 = 0.97 x √2 𝑥 9.81 𝑥 280 = 71.9 𝑚/𝑠

Calculate bucket velocity: u = 0.46 x Vj = 0.46 x 71.9 = 33.1 m/s Calculate Pelton wheel diameter: u = 𝜔r ∴ r =

𝑢

=

33.07

= 0.79 m D = 2r = 2 x 0.79 = 1.58 m 𝜔

41.89

Calculate power per jet: Power per jet =

𝑝𝑜𝑤𝑒𝑟 𝑝𝑒𝑟 𝑤ℎ𝑒𝑒𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑗𝑒𝑡𝑠

Calculate flow rate per jet:

=

20300000 6

Qjet = ∴Q=

= 3.38 x 106 W = 3.38 MW

𝑃𝑗𝑒𝑡

η𝜌𝑔𝐻

3.38 𝑥 1000000

0.8 × 1000 × 9.81 𝑥 280

= 1.56 m3/s

Calculate jet diameter: Q = jet area x jet velocity = 4𝑄

∴ d = √𝜋 𝑉 = √ 𝑗

4 𝑥 1.56

𝜋 𝑥 71.9

𝜋

4

x d 2 x Vj

= 0.166 m

Calculate hydraulic efficiency: Hydraulic efficiency = 𝑢(Vj – u)(1 − kcosβ) 𝑔𝐻

=

𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑤ℎ𝑒𝑒𝑙 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑝𝑜𝑤𝑒𝑟

=

𝑢𝜌𝑄(Vj – u)(1 – kcosβ)

33.07 𝑥 (71.9 – 33.07)(1 − (0.85 x cos165)) 9.81 𝑥 280

∴ Hydraulic efficiency = 85%

ES4E0 Topic 2 Problem Set

𝑄𝜌𝑔𝐻

= 0.851

11. (*) Show using of diagrams and velocity triangles, that the power induced in a Pelton Wheel is equal to 𝑃 = (V𝑗 – u)(1 − kcosβ)𝑢𝜌𝑄.

Hence, derive an expression for wheel efficiency and show that maximum efficiency occurs when the speed ratio, 𝜙, is equal to 0.5. Solution *Based on derivation found in Mechanics of Fluids, Massey, 7th edition, Chapter 13.3: Rotodynamic Machines

Side-on Pelton Wheel diagram showing one bucket (left) and top-down view of jet striking one hemispherical bucket (right)

Part 1: Draw velocity triangles at inlet and outlet INLET

Write an equation linking wheel velocity, absolute jet velocity and velocity of oncoming jetrelative to bucket, noting that all three vectors are collinear (shown as not collinear above for clarity):

R 1 = V1 – u We require the velocity tangential to the Pelton Wheel (as only the tangential component does work) also known as the whirl velocity which in this case in equal to the incoming jet velocity:

∴ V1,whirl = V1 = R1 + u

OUTLET ES4E0 Topic 2 Problem Set

We define |𝑅1 | = 𝑘|𝑅2 |, where k is a value less than 1 that takes into account losses as the jet is deflected and the fact that a small amount of water hits the space between both hemispherical cups. We require the whirl velocity at the outlet, which is the component of V2 in the direction of bucket movement: V2,whirl = u – R2cos(𝜋 − β) The change in whirl velocity is therefore: ∆Vwhirl = V1,whirl – V2,whirl This can be expressed as: = V1 - u + R2cos(𝜋 − β)

= (R1 + u) - u + kR1cos(𝜋 − β) = R1 + kR1cos(𝜋 − β)

= R1 - kR1cosβ [note the relation used here is -cos β = cos(π − β)] = R1(1 - kcosβ)

= (V1 – u)( 1 - kcosβ)

∴ ∆Vwhirl = (V1 – u)( 1 - kcosβ)

The rate of mass flow in the jet is 𝜌𝑄 and so the rate of change of momentum is 𝜌𝑄(∆Vwhirl), which corresponds to the force driving the wheel round. This gives the torque as 𝜌𝑄(∆Vwhirl)r and the power output as 𝜌𝑄(∆V whirl)r𝜔 or 𝜌𝑄(∆Vwhirl)u. Substituting in our expression for ∆Vwhirl above we get power out = 𝜌𝑄(V1 – u)(1 - kcosβ)u = 𝜌𝑄(Vj – u)(1 - kcosβ)u (Note the replacement of V1 with 𝑉𝑗 )

Part 2: Deriving an expression for wheel efficiency The energy arriving at the wheel is in the form of kinetic energy of the jet and is given by wheel efficiency = 𝜂𝑤ℎ𝑒𝑒𝑙 =

𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒

=

𝜌𝑄(V1 – u)(1 − kcosβ)u 1 𝜌𝑄𝑉1 2 2

1

2

𝜌𝑄𝑉1 2. Therefore

This simplifies to: 𝜂𝑤ℎ𝑒𝑒𝑙 =

2(V1 – u)(1 − kcosβ)u 𝑉1 2

We now introduce the speed ratio, 𝜙, which equals u/v1 (wheel velocity divided by incoming jet velocity), giving: ES4E0 Topic 2 Problem Set

u

𝜂𝑤ℎ𝑒𝑒𝑙 = 2( )(1 - u )(1 − kcosβ) 𝑉1 𝑉1 = 2𝜙(1 - 𝜙)(1 − kcosβ)

In reality k and β are assumed constant (which would be the case for a particular Pelton Wheel). The wheel efficiency represents the effectiveness of the wheel in converting the kinetic energy of the jet into mechanical energy of rotation. The maximum value of 𝜂𝑤ℎ𝑒𝑒𝑙 can be found by differentiating the above expression with respect to 𝜙, and setting this to zero: 𝑑(𝜂𝑤ℎ𝑒𝑒𝑙) 𝑑(𝜙)

=

𝑑(2𝜙(1 − 𝜙)(1 − kcosβ)) 𝑑(𝜙)

= 2(1 − kcosβ)

= 2(1 − kcosβ)

𝑑(𝜙 − 𝜙 2 ) 𝑑(𝜙)

𝑑(𝜙(1 − 𝜙)) 𝑑(𝜙)

= 2(1 − kcosβ)(1 − 2𝜙)

As 2(1 − kcosβ) is a constant, we only need to set 1 − 2𝜙 to zero to find the maximum point: 1

1 − 2𝜙 = 0 ∴ 𝜙 = 2 or u

𝑉1

1

= = 0.5 2 or

𝑉1 = 2𝑢 at 𝜂𝑤ℎ𝑒𝑒𝑙,𝑚𝑎𝑥 . This means that the wheel is operating at maximum efficiency when the incoming jet is at twice the velocity of the wheel. In practice, overall efficiency occurs at a value closer to 𝜙 = 0.46, due to frictional losses, however we will assume this theoretical value of 0.5, unless otherwise stated.

Note that maximum wheel efficiency occurs when the speed ratio is 0.5.

ES4E0 Topic 2 Problem Set

12. (*) Water from a river flows at 100 m3/s down a smooth pipe, falling 50m into a turbine. How much power is available? If 10% of the power is lost due to frictional losses, how many houses having a daily electricity use of 12 kWh would this power supply? Solution As the pipe is smooth it is assumed there are no frictional losses, and therefore the head is taken to be 50m. P = 𝜌𝑔𝑄𝐻 = 1000 x 9.81 x 100 x 50 = 49.1 x 106 W = 49.1 MW 49.1MW * 0.9 = 44.1MW Convert daily energy use into average power: 12kWh = 0.5kW 𝑝𝑜𝑤𝑒𝑟 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 (𝑘𝑊)

∴ number of houses supplied = 𝑝𝑜𝑤𝑒𝑟 𝑝𝑒𝑟 ℎ𝑜𝑢𝑠𝑒 (𝑘𝑊) =

44.1 x 103 0.5

= 88,200 houses

13. (*) A Pelton wheel is used to produce hydroelectric power. The average radius of the wheel is 1.83 m, and the jet velocity is 102 m/s from a nozzle of exit diameter equal to 10.0 cm. The turning angle of the buckets is 165°. Assume 𝜌water = 1000 gm/m3. Calculate: • • •

the volumetric flow rate through the turbine in m3/s the optimum rotation rate (in rpm) of the wheel for maximum power the output shaft power in MW if the efficiency of the turbine is 82 percent

Solution Volumetric flow rate = jet area x jet velocity Q = Vj x πd2/4 = 102 x π x (0.12)/4 = 0.801 m3/s At maximum theoretical power, Vj = 2u (where u = 𝜔𝑅 ) ∴𝜔=

𝑣𝑗

2𝑅

102

= 2 × 1.83 = 27.87 rad/s

27.87 x

1

2π

x 60 = 266 rpm

P = 𝑢𝜌𝑄(Vj – u)(1 - cosβ);

u = 𝜔𝑟

∴ Ptheoretical = 27.87 x 1.83 x 1000 x 0.801 x (102 – (27.87 x 1.83)) x (1 – cos165°) = 4.09 x 106 W = 4.09 MW Poutput = Ptheoretical x ηturbine = 4.09 x 0.82 = 3.35 MW

ES4E0 Topic 2 Problem Set

14. (*) A hydroelectric plant has turbines each providing 447 kW at 400 rpm. The available head is 9 m. What type of turbines are these? Solution Convert angular speed to SI units:

Calculate shape number: 𝑁𝑠 =

𝜔=

2𝑁𝜋 60 1

𝜔(𝑃 ⁄𝜌 ) 2 (𝑔𝐻)5/4

=

=

2 ∙ 400 ∙ 𝜋 60

= 41.9 rad/s

1

41.9 ∙ (447 ∙ 1000 ⁄1000 )2 (9.81 ∙ 9)5/4

= 3.27

∴ These are Kaplan turbines

15. (*) A Pelton Wheel develops 4.5 MW under a head of 120 m at a speed of 200 rpm. The wheel diameter is 8 times the jet diameter. Use the experimental data (dashed line) on the figure at maximum efficiency to determine the flow rate, diameter of each jet, the number of jets required and the specific speed.

ES4E0 Topic 2 Problem Set

Solution

From the figure, 𝜙 = 0.42 and 𝜂𝑚𝑎𝑥 occurs at 0.8. We can express flow rate as Q =

𝑃

η𝜌𝑔𝐻

4.5 × 1000000

∴ Q = 0.8 × 1000 × 9.81 × 120 = 4.78 m3/s Ideal jet velocity Vj = √2𝑔𝐻 = √2 × 9.81 × 120 = 48.5 m/s Speed ratio 𝜙 =

u

𝑉𝑗

= 0.42, therefore u = (0.42)(48.5) = 20.4 m/s 𝜋𝐷𝑁

Wheel speed = 𝑢 =

60

∴D=

60 × 𝑢 𝜋×𝑁

=

60 × 20.4 𝜋 × 200

= 1.95 m

Jet diameter = D/8 = 1.98/8 = 0.243 m 𝜋

∴ 𝑛=

Therefore 𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑛𝑎𝑗 𝑣𝑗 = 𝑛( 4 d2)𝑣𝑗 4𝑄

𝜋d2 𝑣𝑗

=

𝜔=

𝑁𝑠 =

4 × 4.78

𝜋 × ...