me 306 part 3 turbomachinery.pdf

3-1

ME 306 Fluid Mechanics II

Part 3

Turbomachinery

These presentations are prepared by

Dr. Cüneyt Sert

Department of Mechanical Engineering

Middle East Technical University

Ankara, Turkey

[email protected]

Please ask for permission before using them. You are NOT allowed to modify them.

3-2

Fluid Machinery

• Fluid machinery is used to convert hydraulic energy to mechanical energy or vice versa.

Power absorbing

Work is done on the fluid

Mechanical Energy → Hydraulic Energy

Power producing

Work is done by the fluid

Hydraulic Energy → Mechanical Energy

Pump Turbine

3-3

Classification of Fluid Machinery

• Fluid machinery can be classified based on the motion of moving parts.

1 ) Positive Displacement Machines

• Fluid is directed into a closed volume.

• Energy transfer is accomplished by movement of the boundary of the closed volume.

• Closed volume expands and contract, sucking the fluid in or pushing it out.

Human heart

http://speakeasies.biz

Water well pump

http://www.bicycleaccessories.us

Tire pump

http://en.wikipedia.org

Gear pump

3-4

Classification of Fluid Machinery (cont’d)

2 ) Turbomachines

• Turbo means ‘‘spin’’ or ‘‘whirl’’ in Latin.

• Turbomachines use rotating shafts with attached blades, vanes, buckets, etc.

• In ME 306 we’ll study turbomachines (mostly pumps) as black boxes, i.e. without analyzing the details of flow fields inside them.

Axial fan

http://en.wikipedia.org

Centrifugal pump

http://www.noehill.com/nevada_county_california

/cal1012.asp

Pelton wheel Kaplan type

hydraulic turbine

http://www.britannica.com

3-5

Classification of Turbomachines

• Pumps increase the pressure of a liquid without changing its velocity considerably.

• Shown centrifugal (radial) pump is the most common type.

• Visit www.standartpompa.com to get more information on sizes and capacities.

Turbomachines

Power absorbing Power producing

Incompressible Compressible Incompressible Compressible

Pump

http://www.standartpompa.com/

• Propellers are very similar to fans, but they are used to generate thrust.

• Marine propellers work with incompressible water water and aircraft propellers work with compressible air.

• Pressure difference between the front and back surfaces of the blades create the thrust.

3-6

Classification of Turbomachines (cont’d)

Turbomachines



Pump Propeller

3-7

• The main difference between fans, blowers and compressors is the pressure difference they create.

• Fans create small pressure difference. Their main purpose is to put high amount of fluid into motion.

• Shown is axial fan of a wind tunnel.

Turbomachines



Pump Propeller Fan


3-8

Turbomachines



Pump Propeller Fan Blower

• Compared to compressors, blowers work with higher amouts of fluid at lower pressure ratios.

• They are mostly centrifugal type.

• Shown is an industrial type blower.


3-9

• Compressors work with smaller flow rates, but create very high pressure ratios.

• Shown is a multi-stage axial compressor.

• Compressors are used in gas and steam turbines, natural gas pumping stations, turbochargers, etc.

Turbomachines



Pump Propeller Compressor Fan Blower


3-10

Turbomachines



Pump Propeller Compressor Pelton wheel

Fan Blower

• Pelton wheels are impulse type turbines commonly used to generate electricity.

• They convert kinetic energy of a high speed liquid into mechanical energy.

• Largest ones used at hydraulic power plants have capacities up to 200 MW.


3-11

Turbomachines




Hydraulic turbine

Fan Blower

• Hydraulic turbines are used at dams to generate electricity using high pressure water.

• Common types are Francis and Kaplan.

• Shown are the runner blades of the Francis turbines used at Three Gorges Dam / China.

• Atatürk Dam has a capacity of 8 x 300 MW.


3-12

Turbomachines


Incompressible Compressible

Steam turbine



Hydraulic turbine

Fan Blower

• Steam turbines are used at power plants to generate electricity using high temperature and high pressure steam.

• 80 % of world’s electricity is produced by steam turbines.

• Afşin-Elbistan thermal power plant has a capacity of 4 x 344 MW.


3-13

Turbomachines



Steam turbine



Hydraulic turbine

Gas turbine

Fan Blower

• Gas turbines are similar to steam turbines, but they use high temperature and high pressure combustion gases.

• A Boeing 777 is powered by 2 turbofan engines, each generting a thrust of ~500 kN.

• To learn how a turbofan engine operates visit http://www.youtube.com/watch?v=GpklBS3s7iU


http://www.youtube.com/watch?v=GpklBS3s7iU



3-14

Turbomachines



Steam turbine



Hydraulic turbine

Gas turbine

Wind turbine

Fan Blower

• As of 2011 Turkey’s wind energy production is 1415 MW. (http://www.ruzgarenerjisibirligi.org.tr)

• World’s total wind energy capacity is 175 GW, which is about 2 % of all electricity usage.

• There are wind turbines with more than 120 m rotor diameter, producing 6 MW of electricity (enough for 4500 homes) (http://www.youtube.com/watch?v=LQxp6QTjgJg)


http://www.ruzgarenerjisibirligi.org.tr/



http://www.youtube.com/watch?v=LQxp6QTjgJg

http://www.youtube.com/watch?v=LQxp6QTjgJg

3-15

Turbomachines

Uncased Cased Impulse Reaction

Axial Flow Propeller

• Uncased turbomachines do not have a solid casing around them.

Another Classification of Turbomachines


3-16

Turbomachines



Axial

• Fluid enters an axial flow turbomachine parallel to the axis of rotation.

• Fluid leaves the machine also in axial direction.

Another Classification of Turbomachines (cont’d)


In

Out

3-17

Turbomachines



Axial Radial

• In radial flow machines fluid intake is parallel to the axis of rotation.

• Rotating impeller blades push the fluid in radial direction.

• Fluid leaves the machine perpendicular to the rotation axis.



In Out

3-18

Turbomachines

Uncased Cased

Axial (Kaplan)

Impulse Reaction


Mixed Pelton wheel

Wind turbine

Axial Radial

• Kaplan turbines are axial flow machines.

• They are preferred for low head and high flow rate configurations.

• Their capacities are less than Francis type, less than 200 MW.

• They can provide efficiencies higher than 95 %.



3-19

Turbomachines

Uncased Cased

Axial (Kaplan)

Impulse Reaction


Mixed Pelton wheel

Wind turbine

Radial (Banki)

Mixed (Francis)

Axial Radial



• Francis turbines are the most widely used turbomachines for hydropower.

• They are of mixed flow type.

• They can provide more than 800 MW power.

• For more information http://www.voithhydro.com

http://www.voithhydro.com/



Parts of a Centrifugal Pump

• Centrifugal pump is the most commonly used type of turbomachine.

• They are responsible for ~5% of all electricity comsumption of USA.

http://pdf.directindustry.com

3-20

3-21

Parts of a Centrifugal Pump (cont’d)

• For centrifugal pump details watch

http://www.youtube.com/watch?v=9nL1XhKm9q8 (Principles and parts) http://www.youtube.com/watch?v=Vq3hEe5jzSM (Pump Parts) http://www.youtube.com/watch?v=UrChdDwHybY (Impeller animation) http://www.youtube.com/watch?v=RvOzKhUDmJM (Computer aided blade design)

• Most important part is the impeller. It may have different designs such as

• Backward-curved, radial or forward-curved

• Closed (shrouded) or open

Backward curved

Forward curved

Open

Closed (Shrouded) Open, radial

http://www.youtube.com/watch?v=9nL1XhKm9q8



http://www.youtube.com/watch?v=Vq3hEe5jzSM



http://www.youtube.com/watch?v=UrChdDwHybY



http://www.youtube.com/watch?v=RvOzKhUDmJM



3-22

Pump Head (𝑕)

• Consider the BE between the inlet (suction) and outlet (discharge) of a pump.

𝑝𝑖𝑛𝜌𝑔+𝑉𝑖𝑛2

2𝑔+ 𝑧𝑖𝑛 =

𝑝𝑜𝑢𝑡𝜌𝑔+𝑉𝑜𝑢𝑡2

2𝑔+ 𝑧𝑜𝑢𝑡 − 𝑕

• Pump head is the difference between the total heads at

pump inlet and outlet.

• Pump head is a positive quantity with units of length.

Pump head

𝑝𝑖𝑛

𝑉𝑖𝑛

𝐴𝑖𝑛

𝑝𝑜𝑢𝑡

𝑉𝑜𝑢𝑡

𝐴𝑜𝑢𝑡

𝑄

𝑄

𝑧

Datum

3-23

Pump Head (cont’d)

• Elevation difference between inlet and outlet is generally negligibly small.

• If suction and discharge pipe diameters are the same → 𝑉𝑖𝑛 = 𝑉𝑜𝑢𝑡

• For this simplified case

𝑕 = 𝑝𝑜𝑢𝑡 − 𝑝𝑖𝑛𝜌𝑔

= ∆𝑝

𝜌𝑔

i.e. pump head is the pressure rise across the pump expressed as a head.

• Pump head is directly related to the power delivered to the fluid, known as water horsepower

𝒫𝑓 = 𝜌𝑔𝑄 𝑕

• Pump head can be defined as the power delivered to the fluid per weight of the fluid flowing through the pump in unit time (weight flow rate).

Weight flow rate

𝑓 for fluid

3-24

Pump Efficiency (cont’d)

• Power necessary to run the pump, known as brake horsepower (bhp), is larger than power delivered to the fluid due to

• mechanical and fluidic frictional losses

• flow separation on impeller blade surfaces

• misalignment of inlet flow velocity with impeller blade geometry

• internal leakage, etc.

𝒫𝑝 = 𝑇𝑠𝑕𝑎𝑓𝑡 𝜔 > 𝒫𝑓

• Efficiency of the pump is defined as

𝜂𝑝 = 𝒫𝑓

𝒫𝑝 < 1

Rotational speed of the pump Torque supplied to

the pump shaft

𝑝 for pump

3-25

Performance of Turbomachines

• Important quantities of interest for a turbomachine are

• Volumetric flow rate (discharge, capacity) 𝑄

• Head 𝑕

• Size of the machine 𝑑

• Rotational speed 𝑁 or 𝜔

• Power consumption or generation 𝒫

• Efficiency 𝜂

• For a pump, fundamental characteristic is a plot of 𝑕 vs. 𝑄 at a given rotational speed 𝜔. It is customary to plot 𝒫 and 𝜂 on the same figure.

• For a turbine, fundamental characteristic is a plot of 𝒫 vs. 𝜔 at a given head of 𝑕. It is customary to plot 𝑄 and 𝜂 on the same figure.

3-26

Performance Curve of a Centrifugal Pump

• 𝑕 − 𝑄 curve of a pump is known as its performance (characteristic) curve.

• For steady conditions, a pump can operate only on its performance curve.

• At a given rotational speed (𝜔) a centrifugal pump’s typical performance curve is

Shutoff head

• Outlet of the pump is blocked and 𝑄 = 0.

• Pump is not doing any useful work.

Free delivery

• There is no load on the pump and 𝑕 = 0.

• Pump is not doing any useful work

𝑄

𝑕

3-27

Best Efficiency Point (BEP)

• The exact operation point of a pump depends on the system it is working in.

• Pumps are designed to work at their maximum efficiency, but this is not always possible.

Note that these curves are for a given rotational speed 𝜔.

Best Efficiency Point (BEP) (or design point)

𝑄

Head (𝑕) Efficiency (𝜂)

Power (𝒫𝑝)

𝑕, 𝒫, 𝜂

𝑄 for maximum efficiency

𝑕 for maximum efficiency

3-28

Similarity Laws for Pumps

• Similitude analysis for a pump is useful

• to predict the performance of geometrically similar pumps with different sizes, i.e. with different impeller diameters.

• to predict the performance of a pump working at different speeds.

• Perform a Buckingham-Pi analysis with the following parameters

𝑄 , 𝑔𝑕 , 𝐷 , 𝒫𝑝 , 𝜔 , 𝜌 , 𝜇

to get the following non-dimensional groups

Flow coefficient : Head coefficient :

Power coefficient : Reynolds number :

𝜋𝑄 =𝑄

𝜔𝐷3 𝜋𝑕 =

𝑔𝑕

𝜔2𝐷2

𝜋𝒫 =𝒫𝑝

𝜌𝜔3𝐷5 𝜋𝜇 =

𝜇

𝜌𝜔𝐷2=1

𝑅𝑒

3-29

Similarity Laws for Pumps (cont’d)

• In most pump applications similarity of viscous forces are not as important as the other 𝜋 groups.

• Two geometrically similar pumps are said to be operating under similar conditions if the remaining three 𝜋 groups are equal.

𝜋𝑄1 = 𝜋𝑄2 , 𝜋𝑕1 = 𝜋𝑕2 , 𝜋𝒫1 = 𝜋𝒫2

where 1 and 2 refer to two different operating conditions of similar pumps.

• These three equalities are known as affinity laws.

Exercise : Show that when affinity laws are satisfied efficiencies of two similar operating points (homologous points) are equal.

Exercise : How do nondimensional performance curve (𝜋𝑕 vs. 𝜋𝑄) of two

geometrically similar pumps compare with each other?

3-30

Similarity Case 1 – Different Rotational Speeds

• Consider a pump with a known performance. We want to determine its operation at a different speed.

• Pump sizes are the same and size parameter 𝐷 drops from nondimensional groups, resulting in the following simplified affinity laws

𝑄1𝜔1=𝑄2𝜔2 ,

𝑕1

𝜔12 =𝑕2

𝜔22 ,

𝒫𝑝1𝜔13 =𝒫𝑝2𝜔23

Rotating at 𝜔1

Delivering 𝑄1 and 𝑕1

Using 𝒫𝑝1

Rotating at 𝜔2


Using 𝒫𝑝2

3-31

Similarity Case 1 (cont’d)

• According to the affinity laws, for similarity case 1

• 𝑄 is proportional with 𝜔.

• 𝑕 is proportional with the square of 𝜔.

• 𝒫𝑝 is proportional with the cube of 𝜔.

Exercise : What determines the operation speed of a pump? Is it something adjustable or should a pump be always run at the same speed?

Exercise : Performance data of a centrifugal pump, running at 750 rpm is given in the following table. We want to predict the performance of the same pump when it is running at a speed of 900 rpm.

1A 1B 1C 1D 1E 1F 1G 1H

𝑄(𝑚3/𝑠) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

𝑕 (𝑚) 40 41 41 40 38 34 26 15

𝒫𝑝 (𝑘𝑊) 0 114.9 134.1 152.9 175.4 196.2 204.1 187.3

𝜂 0 0.35 0.60 0.77 0.85 0.85 0.75 0.55

Point 1

(a point similar to point 2)

𝑄1 = ?

𝑕1 = ?

𝑁1 = 1500 𝑟𝑝𝑚 3-32


Exercise : Head and efficiency of a centrifugal pump running at 1500 rpm are given as

𝑕 = 50 – 200 𝑄 – 24000 𝑄2

𝜂 = 60 𝑄 – 1200 𝑄2

It is desired to deliver 0.03 m3/s of water against a head of 36 m. Determine

a) speed of the pump

b) efficiency of the pump

c) power consumption

of the pump

Point 2

(desired operating point)

𝑄2 = 0.03 𝑚3/𝑠

𝑕2 = 36 𝑚

𝑁2 = ?

0 0.01 0.02 0.03 0.04

60

𝑕 [𝑚]

𝑄[𝑚3/𝑠]

50

40

30

20

10

0

3-33

Similarity Case 2 – Different Sizes (Impeller Diameters)

• Consider a pump with a known performance. We want to determine the operation of a gemoetrically similar pump with a different size rotating at the same speed.

• Rotating speeds are the same. 𝜔 drops from nondimensional groups, resulting in the following simplified affinity laws

𝑄1

𝐷13 =𝑄2

𝐷23 ,

𝑕1

𝐷12 =𝑕2

𝐷22 ,

𝒫𝑝1𝐷15 =𝒫𝑝2𝐷25

Size is 𝐷1


Using 𝒫𝑝1

Size is 𝐷2


Using 𝒫𝑝2

3-34


Exercise : Characteristic curve of a centrifugal pump is given as

𝑕 = 100 – 1000 𝑄2

It is desired to deliver 0.1 m3/s of water against a head of 70 m. For these requirements it is thought that using a similar pump with a smaller impeller would be more efficient.

a) Determine the required percent reduction in impeller diameter.

b) Determine the percent decrease in power consumption.

Point 2

(desired operating point)

𝑄2 = 0.1 𝑚

3/𝑠

𝑕2 = 70 𝑚

𝐷2 = ?

Point 1

(a point similar to point 2)

𝑄1 = ?

𝑕1 = ?

𝐷1 = ?

0 0.1 0.2 0.3

120

𝑕 [𝑚]

𝑄[𝑚3/𝑠]

100

80

60

40

20

0

3-35

Similarity – Combined Case 1 and 2

Exercise : A pump manufacturing company produces a water pump (called pump A) with an impeller diameter of 6 𝑐𝑚. Its performance data at 𝑁 = 1725 𝑟𝑝𝑚 is shown below.

Marketing department of the company is recommending the design of a new larger pump (called pump B) that will be used to pump refrigerant R-134a at room temperature. Pump B should be designed such that its BEP occurs as close as possible to a flow rate of 2400 𝑐𝑚3/𝑠 at a head of 450 𝑐𝑚.

Perform a preliminary similarity analysis to determine if a scaled-up version of pump A will be suitable or not.

a) Plot performance curves of pump A using both dimensional and nondimensional parameters.

b) Calculate the required diameter, rotational speed and power for pump B operating at its BEP.

𝑄(𝑐𝑚3/𝑠) 𝑕 (𝑐𝑚) 𝜂

100 180 0.32

200 185 0.54

300 175 0.70

400 170 0.79

500 150 0.81

600 95 0.66

700 54 0.38

Data for pump A

3-36

System Characteristic

• A pump works at an operating point on its characteristic curve.

• But this operating point depends on the system that the pump is installed in.

• Following pump works between a suction reservoir (𝑠𝑟) and a discharge reservoir (𝑑𝑟).

• BE between points 𝑠𝑟 and 𝑝𝑟

𝑝𝑠𝑟𝜌𝑔+𝑉𝑠𝑟2

2𝑔+ 𝑧𝑠𝑟 =

𝑝𝑑𝑟𝜌𝑔+𝑉𝑑𝑟2

2𝑔+ 𝑧𝑑𝑟 − 𝑕 + 𝑕𝑓

• If 𝑝𝑠𝑟 = 𝑝𝑑𝑟 = 𝑝𝑎𝑡𝑚 and 𝑉𝑠𝑟 = 𝑉𝑑𝑟 = 0

𝑕 = 𝑕𝑔𝑡 + 𝑕𝑓

Total major and minor losses

Total geometric head (= 𝑧𝑑𝑟 − 𝑧𝑠𝑟)

𝑝𝑠𝑟

𝑝𝑑𝑟

𝑕𝑔𝑡

Pump

3-37

System Characteristic (cont’d)

• 𝑕𝑓 consists of major and minor losses calculated as

𝑕𝑓 = 𝑓𝑖 𝐿𝑖𝐷𝑖 𝑉𝑖2

2𝑔

𝑖

+ 𝑘𝑖 𝑉𝑖2

2𝑔

𝑖

• 𝑉𝑖’s are average velocities in suction and discahrge pipes.

• Using the continuity eqution 𝑉𝑖 = 𝑄/𝐴𝑖

and the total loss becomes

𝑕𝑓 = 𝑓𝑖 𝐿𝑖𝐷𝑖 𝑄2

2𝑔𝐴𝑖2

𝑖

+ 𝑘𝑖𝑄2

2𝑔𝐴𝑖2

𝑖

or simply 𝑕𝑓 = 𝐾𝑄2

Friction factor

Actual or equivalent pipe length

Head loss coefficient for minor losses

Pipe diameter

• Using this 𝑕𝑓 in the BE we get the following system characteristic equation

• Minimum head the pump should provide is equal to the total geometric head.

• Additional pump head is necessary to overcome frictional losses. This part increases with the square of the flow rate.

• Total geometric head (𝑕𝑔𝑡) is usually negligibly small for gas flow applications. 3-38


𝑄

𝑕

𝑕𝑔𝑡

𝑕𝑔𝑡 + 𝐾𝑄2

𝑕 = 𝑕𝑔𝑡 + 𝐾𝑄2

• It is possible to change system characteristic in two ways.

3-39


Total geometric head (𝑕𝑔𝑡 ,difference

between reservoir levels) can be

changed.

Friction loss can be changed by

changing 𝐾. For example by opening

or closing a valve.

𝑄

𝑕

𝑕𝑔𝑡 increases

𝑕𝑔𝑡1+ 𝐾𝑄2



𝑄

𝑕

𝐾 increases

𝑕𝑔𝑡 + 𝐾1𝑄2



• A pump installed on a system will not work at an arbitrary point.

• It will operate at the point where pump and system characteristics intersect.

3-40

Operating Point

𝑄

𝑕

Pump characteristic (Supply curve)

System characteristic (Demand curve)

𝑕𝑜𝑝

Operating point

𝑄𝑜𝑝

• Normally we want the operating point to be close to the BEP (design point).

• However BEP is not always the most economical operating point as far as the power consumption is concerned, i.e. BEP is not necessarily the minimum 𝒫𝑝 point.

3-41

System Characteristic, Operating Point & Similarity

Exercise : Characteristics of a centrifugal pump at 600 𝑟𝑝𝑚 is given below. The pump is used to elevate water by 32 𝑚 from a lake to an open tank. Flow rate is measured as 22 𝑙𝑝𝑠 while the delivery valve is fully open and the pump running at 600 𝑟𝑝𝑚.

Determine the power consumptions for the following operations.

a) Valve closure is increased such that the frictional loss is doubled.

b) Pump speed is increased to 720 rpm while keeping the valve fully open.

0 5 10 15 20 25 30 35 40 45 50

70

𝑕 𝑚 𝜂 [%]

𝑄 [𝑙𝑝𝑠]

65

60

55

50

45

40

35

30

25

20

15

10

5

0

3-42

Cavitation

• In a liquid flow cavitation occurs when the local satic pressure falls below the vapor pressure of the liquid.

• For a cavitating flow

• liquid locally vaporizes forming bubbles.

• bubbles collapse as they travel to higher pressure regions and cause erosion/surface pitting.

• flow becomes unsteady and noisy causing turbomachine to vibrate.

• performance of turbomachine drops.

• For a pump, critical low pressure region is the entrance, and for a turbine it is the exit.

• High speed regions like propeller blade tips are also critical.

• Listen to the sound of a cavitating pump : www.youtube.com/watch?v=Qw97DkOYYrg

• Watch propeller tip cavitation : www.youtube.com/watch?v=GpklBS3s7iU

http://www.youtube.com/watch?v=Qw97DkOYYrg


3-43

Damage on Francis turbine blades

Damage on centrifugal pump impeller

www.pumpfundamentals.com

en.wikipedia.org

Damage on propeller blades

Cavitation Damage

en.wikipedia.org

3-44

Cavitation of a Pump and 𝑁𝑃𝑆𝐻

• Cavitation possibility of a pump is checked using a variable called Net Positive Suction Head (𝑁𝑃𝑆𝐻).

• 𝑁𝑃𝑆𝐻 is the difference between the total head at the suction side and the head corresponding to the vapor pressure.

𝑁𝑃𝑆𝐻 = 𝑝𝑠𝜌𝑔 + 𝑉𝑠2

2𝑔 − 𝑝𝑣𝜌𝑔

• Suction side of the pump is the side where fluid enters into the pump. It is used because it is the critical region due to low pressures.

Vapor pressure Suction pressure

Suction velocity

Total head at the suction side of the pump (Datum can be arranged so that

𝑧𝑠 = 0)

3-45

NPSH (cont’d)

• There are two values of 𝑁𝑃𝑆𝐻 that we work with

• 𝑁𝑃𝑆𝐻 required (𝑁𝑃𝑆𝐻𝑅)

• 𝑁𝑃𝑆𝐻 available (𝑁𝑃𝑆𝐻𝐴)

• 𝑁𝑃𝑆𝐻𝑅 is the value that must be exceeded to prevent cavitation.

• It is measured by the manufacturer of the pump and provided as an extra curve on the pump characteristic plot.

𝑄

𝑕

𝑁𝑃𝑆𝐻𝑅

𝑕


3-46

NPSH (cont’d)

• 𝑁𝑃𝑆𝐻𝐴 is the value that we need to calculate for the problem of interest.

• Consider the BE for the suction side of a pump

𝑝𝑠𝑟𝜌𝑔+𝑉𝑠𝑟2

2𝑔+ 𝑧𝑠𝑟 =

𝑝𝑠𝜌𝑔+𝑉𝑠2

2𝑔 + 𝑧𝑠 + 𝑕𝑓𝑠

𝑝𝑠𝑟 = 𝑝𝑎𝑡𝑚

𝑉𝑠𝑟 = 0

𝑧𝑠 − 𝑧𝑠𝑟 = 𝑕𝑠

𝑕𝑓𝑠 : Frictional losses at the suction side

𝑝𝑠𝜌𝑔+𝑉𝑠2

2𝑔 = 𝑝𝑎𝑡𝑚𝜌𝑔− 𝑕𝑓𝑠 − 𝑕𝑠

𝑠

𝑠𝑟

𝑕𝑠

Pump

Suction pipe

3-47

NPSH (cont’d)

• Using the last equation in the definition of 𝑁𝑃𝑆𝐻𝐴

𝑁𝑃𝑆𝐻𝐴 = 𝑝𝑠𝜌𝑔 + 𝑉𝑠2

2𝑔 − 𝑝𝑣𝜌𝑔 → 𝑁𝑃𝑆𝐻𝐴 =

𝑝𝑎𝑡𝑚 − 𝑝𝑣𝜌𝑔

− 𝑕𝑓𝑠 − 𝑕𝑠

• To prevent cavitation

𝑁𝑃𝑆𝐻𝐴 ≥ 𝑁𝑃𝑆𝐻𝑅

Exercise : What can be done to make 𝑁𝑃𝑆𝐻𝐴 larger for the pump shown in the previous slide?

Exercise : How does the vapor pressure of water change with temperature? What does this information tell us as far as cavitation prevention is concerned?

𝑕


𝑁𝑃𝑆𝐻𝐴

Cavitation No cavitation

3-48

NPSH (cont’d)

Exercise : 184 mm impeller diameter centrifugal pump of Standart Pompa 32-160 series running at 1450 rpm is used to pump water at 25 ℃ from a reservoir whose surface is 1.2 m above the centerline of the pump inlet. Reservoir is open to atmospheric pressure.

The piping system from the reservoir to the pump consists of 3.2 m of cast iron pipe with a diameter of 10 cm and an average roughness of 0.05 cm. Minor losses at the suction side of the pump are; a sharp edged inlet (𝑘 = 0.5), three flanged smooth 90o elbows (𝑘 = 0.3 each) and a fully open flanged globe valve (𝑘 = 6).

Estimate the maximum flow rate that can be pumped without cavitation.

www.standartpompa.com

3-49

Suction Specific Speed (𝑆)

• Suction specific speed is a nondimensional version of 𝑁𝑃𝑆𝐻

𝑆 =𝜔 𝑄

(𝑔 𝑁𝑃𝑆𝐻𝑅)3/4

• It is usually evaluated at the BEP of a pump.

• For a family of geometrically similar pumps 𝑆 (evaluated at BEP) should have a fixed value.

• To prevent cavitation 𝑆 should be less than a critical 𝑆𝑐, which is around 3 for centrifugal pumps.

• In United States it is a common practice to use the following dimensional form of 𝑆

𝑆𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 =𝜔 𝑟𝑝𝑚 𝑄[𝑔𝑝𝑚]

(𝑁𝑃𝑆𝐻𝑅 𝑓𝑡 )3/4

Exercise : For the pump given in the previous slide calculate the suction specific speed at BEP and show that it is much larger than the critical value of 3. What does this calculation tell us?

3-50

Specific Speed (𝑁𝑠)

• Suction specific speed (𝑆) that we use for cavitation check is actually a specific form of a more general nondimensional number called specific speed.

• 𝑁𝑠 is obtained by combining 𝜋𝑄 and 𝜋𝑕 as follows

𝑁𝑠 = 𝜋𝑄1/2

𝜋𝑕3/4 = 𝜔 𝑄

(𝑔𝑕)3/4

• 𝑆 is a special version of 𝑁𝑠 obtained by replacing 𝑕 with 𝑁𝑃𝑆𝐻𝑅.

• 𝑁𝑠 is useful to classify and compare different types of pumps at their BEP.

• For example radial and axial pumps work efficiently at low and high Ns values, respectively.

• 𝑁𝑠 is mainly used for preliminary pump selection.

• Similar to 𝑆, there are dimensional forms (more than one) of 𝑁𝑠 commonly used in the industry.

3-51

Specific Speed (cont’d)

0.1 0.2 0.4 0.6 1 2 4 6 10

Adapted from Aksel’s textbook

Radial Mixed Axial

0.1 0.2 0.4 0.6 1 2 4 6 10

100

𝑁𝑠

𝜂 [%]

95

90

85

80

75

70

𝑁𝑠

Radial Mixed Axial

3-52

Series Combination of Pumps

• Series combination is used if the head provided by a single pump is not enough.

• Same 𝑄 goes through both pumps.

• Total head provided is the sum of individual heads.

• Pumps can be identical or different.

• More than two pumps can be combined in series.

Pump A provides 𝑕𝐴

Pump B Provides 𝑕𝐵

𝑄 = 𝑄𝐴 = 𝑄𝐵

𝑕 = 𝑕𝐴 + 𝑕𝐵

3-53

Series Combination of Pumps (cont’d)

Exercise : Show that for two pumps combined in series overall efficiency is

𝜂 =𝑕𝐴 + 𝑕𝐵𝑕𝐴𝜂𝐴+𝑕𝐵𝜂𝐵

• To get combined pump characteristic, individual pump characteristics are added vertically.

• If the pumps are identical

𝑄

𝑕

System characteristic

Operating point

Pump A or B

Pump A+B

3-54


• If the pumps are NOT identical

• Above a certain 𝑄𝑐 pump B is forced to operate above its free delivery point. For such a case it just creates extra loss and should be shut off and bypassed.

𝑄

𝑕 System characteristic

Operating point

Pump B

Pump A+B

Pump A

𝑄𝑐

3-55


Exercise : Two identical pumps, with shown characteristics, are combined in series and

used to transport water between two reservoirs with an elevation difference of

𝑧𝑑𝑟 – 𝑧𝑠𝑟 = 50 𝑚. Total length of suction and discharge pipes is 120 𝑚. Pipe

diameters are 0.12 𝑚. Friction factor inside the pipes is 0.022. Neglecting the minor

losses, determine the power required to drive both pumps.

0 0.01 0.02 0.03 0.04 0.05

90

𝑄 [𝑚3/𝑠]

𝑕 [𝑚] 𝜂 [%]

70

50

30

10

• Parallel combination is used if the flow rate provided by a single pump is not enough.

• Each pump provides the same head 𝑕.

• Total flow rate is the sum of individual flow rates.

• Pumps can be identical or different.

• More than two pumps can be combined in parallel.

3-56

Parallel Combination of Pumps

Pump A

Pump B

𝑄𝐴

𝑕 = 𝑕𝐴 = 𝑕𝐵

𝑄𝐵

𝑄 = 𝑄𝐴 + 𝑄𝐵

3-57

Parallel Combination of Pumps (cont’d)

Exercise : Show that for two pumps combined in parallel overall efficiency is

𝜂 =𝑄𝐴 + 𝑄𝐵𝑄𝐴𝜂𝐴+𝑄𝐵𝜂𝐵

• To get combined pump characteristic, individual pump characteristics are added horizontally.

• If the pumps are identical

𝑄

𝑕 System

characteristic

Operating point

Pump A or B

Pump A+B

3-58

Parallel Combination of Pumps (cont’d)

• If the pumps are NOT identical

• Above a certain 𝑕𝑐 pump B is forced to operate above its shutoff head. For such a case it just creates extra loss and should be shut off and its branch should be blocked with a valve.

𝑄

𝑕

System characteristic

Operating point

Pump B

Pump A+B

𝑕𝑐

Pump A

3-59

Pump Selection

• Two main inputs for pump selection are

• required head

• required flow rate

• Additional consideration for pump selection are

• pump speed (depends on the speed of the electric motor)

• type of fluid (highly viscous, muddy, etc.)

• available space, vertical placement limitations that will affect 𝑁𝑃𝑆𝐻

• maximum allowable noise level

• etc.

• For preliminary pump selection specific speed (𝑁𝑠) and suction specific speed (𝑆) are two commonly used variables.

3-60

Pump Selection (cont’d)

• Using the required head and flow rate we can first select a pump family from a manufacturer’s catalogue. For example for the plot given below at 2900 𝑟𝑝𝑚, ‘‘65-160’’ family is suitable for 𝑄 = 100 𝑚3/𝑕 and 𝑕 = 30 𝑚.

5 10 15 20 30 40 50 100 200 300 600

𝑕 𝑚

𝑄 [𝑚3/𝑕]

120

100

80

60

40

30

20

15

10 www.standartpompa.com

2900 𝑟𝑝𝑚

3-61

Pump Selection (cont’d)

• These are the detailed performance curves of the the pumps in ‘‘65-160’’ family running at 2900 𝑟𝑝𝑚.

• There are three similar pumps with impeller diameters of 160 𝑚𝑚, 175 𝑚𝑚 and 184 𝑚𝑚.

• Black curves are iso-efficinecy lines.

• 𝑁𝑃𝑆𝐻𝑅 and 𝒫𝑝 curves are also

provided.

• One of these three pumps can be selected by considering cavitation possibility, efficiency and power consumption.

• The smallest pump can not provide the required head of 30 𝑚 at the desired flow rate of 100 𝑚3/𝑕.

0 20 40 60 80 100 120 140

𝑕 𝑚

𝑄 [𝑚3/𝑕]

20

N𝑃𝑆𝐻𝑅

𝑚

𝒫𝑝 𝑘𝑊

15

10

5

0

2

6

10

20

30

40

50

25

35

45

𝜙 184

𝜙 175

𝜙 160

𝜙 160 𝜙 184

50 % 60

65

70

73.5

70

65

60

50

𝜙 184

𝜙 175

𝜙 160

Adapted from www.standartpompa.com

3-62

Turbines

• Fundamental performance characteristic or a reaction type turbine is the power produced vs. rotational speed curve at a given head.

• Affinity laws used for pumps are valid for turbines too.

• Specific speed can be used for preliminary turbine selection, but it is defined in a slightly different way compared to pumps

𝑁𝑠𝑡 = 𝜋𝒫1/2

𝜋𝑕5/4 =

𝜔 𝒫𝑡𝜌 (𝑔𝑕)5/4

𝑁

𝒫𝑡 At a given head 𝑕

3-63

Turbine Specific Speed

0.1 0.2 0.4 0.6 1 2 4 6 10

Adapted from Aksel’s textbook

Impulse

Francis

Kaplan

0.1 0.2 0.4 0.6 1 2 4 6 10

100

𝑁𝑠𝑡

𝜂 [%]

95

90

85

80

75

70

𝑁𝑠𝑡

Impulse Mixed Axial Radial

3-64

Turbines (cont’d)

Exercise : Calculate the specific speed of the following turbines

a) Francis type radial flow turbine at the Round Butte hydroelectric power station in Madras rotating at 180 𝑟𝑝𝑚 and producing 119 𝑀𝑊 of power at a flow rate of 127 𝑚3/𝑠 from a head of 105 𝑚.

b) Francis type mixed flow turbine at the Smith Mountain hydroelectric power station in Roanoke, VA, rotating at 100 𝑟𝑝𝑚 and producing 194 𝑀𝑊 of power at a flow rate of 375 𝑚3/𝑠 from a head of 54.9 𝑚.

c) Kaplan type axial flow turbine at the Warwick hydroelectric power station in Cordele, GA, rotating at 100 𝑟𝑝𝑚 and producing 5.37 𝑀𝑊 of power at a flow rate of 63.7 𝑚3/𝑠 from a head of 9.75 𝑚.

Exercise : Learn the meaning of the following turbine related terms

Runner blade, wicket gate, stay vane, crown, penstock, draft tube, tail water

Exercise : How hydraulic power works ? http://www.youtube.com/watch?v=cEL7yc8R42k

Virtual turbines http://www.youtube.com/watch?v=HzQPNpP55xQ

http://www.youtube.com/watch?v=cEL7yc8R42k





me 306 part 3 turbomachinery.pdf

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