TURBOMACHINESor

‘FLUID CARRYING MACHINES’

PUMPS & TURBINES

TURBOMACHINES‘FLUID CARRYING MACHINES’

There are two broad categories of turbo-machinery:

PUMPS

TURBINES

PUMPS

Power is REQUIRED to obtain this HEADIn ordinary life;Pumps are referring to those fluid machines that give pressure tomove the liquids.Fans, blowers and compressors are referring to those fluidmachines that add pressure to move the gases.

PUMP:

It is a general term used for any fluid machine that ADDS energy to a fluid. Some authors even

call pumps ENERGY ABSORBING DEVICES since energy is supplied to them so as to transfer that energy

to the fluid mainly by increasing its pressure.

Pump is a machine that is used to rise the available head in the system by giving energy head to the

system along the flow direction. A pump in practice, is utilizing constant energy (power) to carry discharge

(Q) to some elevation (head) (Hp). Depending on the amount of the discharge (Q) the elevation (head) (Hp)

given by the pump varies.

P OR

HUMAN HEART IS A GOOD EXAMPLE OF A PUMP

PUMPS

Some fundamental parameters used to analyse the performance of a pump.

1) Volume Flow Rate ‘Q’It is the simplified and the more commonly used parameter of the Mass Flow Ratedue to incompressible flow. This term is also called discharge. [m3/s]

2) Net Head ‘HP’It is the performance of the pump. It is defined as the change in Bernoulli (manometric)head between the inlet and outlet of the pump. [m]

3) Power ‘P’It is the useful power actually delivered to the fluid. Traditionally this power is called water horsepower even if the fluid pumped is not water. [Watt]

4) Pump Efficiency ‘η’All pumps suffer from irreversible losses due friction, internal leakages, flow separation on blade surfaces, turbulent dissipation etc. These losses are presented by efficiency.

PUMPS [PUMPS ARE HYDRAULIC MACHINES THAT CONVERT

MECHANICAL ENERGY TO FLUIDS AS AN ENERGY ]

PUMPS

PLEASE REMEMBER THAT:*PUMP DOES NOT INCREASE THE SPEED OFTHE FLUID PASSING THROUGH THE PUMP.

*INFACT INCREASES THE PRESSURE OF THEFLUID PASSING WITHIN THE PUMP.

Suct

ion

he

ad

inlet

outlet

LAKE

suct

ion

de

liver

y

Total head = Suction head + Delivery head

Total head = Suction head + Delivery head

WHY DO WE HAVE PUMPS WITHIN THE PIPELINES?

I. To INCREASE THE DISCHARGE WITHIN THE SYSTEMalong the flow direction for the cases where:

Total Energy ‘HA’ > Total Energy ‘HB’.

II. To ALLOW THE DISCHARGE WITHIN THE SYSTEMby overcoming the total energy needs for the caseswhere:

Total Energy ‘HA’ < Total Energy ‘HB’.

PUMP CHARACTERISTIC (PERFORMANCE) CURVE

It is interesting to note that, each pump type has its own head(Hp) - discharge (Q) curve characteristic that is given by the pumpmanufacturer, since the same pump delivers different dischargesunder different heads. This relationship is usually referred as ‘PumpCharacteristic Curve’

• The maximum discharge ‘Qmax’ occurs when the net Head ‘HP ’ iszero. This is called free delivery. This is only possible when thepump is not generating ANY work.[ Qmax @ HP = 0 η = 0 ].

• The net Head is maximum ‘HPmax ’ when the discharge ‘Q’ is zero.This is called shutoff head. This is only possible when the pump isnot generating USEFULL work.[ HPmax @ Q = 0 & η = 0 ].

PUMPS

PUMP CHARACTERISTIC (PERFORMANCE) CURVE

• Any pumps net Head changes non-linearly with respectto discharge between these two extremes due to thatpumps specification. This relationship is referred as‘Pump Characteristic Curve’.

Pumps efficiency is discharge dependent parameteronly. This relationship is referred as ‘Pump EfficiencyCurve’ supplied by the manufacturer.

PUMPS

Q

η

Pump Efficiency Curve

PUMPS

free delivery

shutoff head

Since for any given discharge (Q), a certain head (Hp),will be generated to maintain that flow, hence;once a pump is fixed into the given piping system, thispump will work as a part of that system, by rising theexisting head of the system under that systemsdischarge.

This relationship will be obtain from the given pipelinesystems energy equation.

But this relationship in general is non-linear and referredas System Characteristics Curve.

PUMPS

PUMPS‘System Characteristic Curve’ isgiven by the pump manufacturersas a relationship between discharge‘Q’ and Head ‘H’

Tabular form or

EquationH=A-BQn.

So the solution will be reached, once the pumpscharacteristic curve and the pipeline characteristicscurves coinsides. This point is called operating point.

For this reason, the characteristic curve of the givensystem i.e. ‘System Characteristic Curve (based on thevalues got from the Bernoulli equation)’ and ‘PumpCharacteristic Curve (based on the information given bythe manufacturer)’ should be drawn on a COMMONSCALE millimetric graph where their intersection willgive that required point.

Even the efficiency of the pump can be drawn on thesame graph so as to have all the required informationwithin that graph.

PUMPS

PUMPS

The operating (required duty) point of the piping system is obtained once thedischarge ‘Q’ gives the same ‘Hp’ from the systems characteristic curve (theequation) and from the pumps characteristic (performance) curve.

Hp from pipe (system) characteristic curve ≈ Hp from pump characteristic curve

Using this operating discharge (Q), the efficiency (η) of that pump can be determined.

PUMPS

Example 4.1a) Determine the elevation difference between reservoirs A and B,

if water at 20 °C flows through this pipe system (from A to B)with an average discharge Q=0.145 m3/s? Consider the minorlosses as well.

PUMPS

f1= 0.0121f2= 0.0118Δz= 71.08 m

• pvc • pvc

sharp edged sharp edged

Pump in catagory 1 system (HP= 143.25m 239.2 kW)

PUMPS b) Calculate the required Hp and the pump power (P) to maintainthe same discharge (Q=0.145 m3/s) in the opposite direction (from Bto A). Take the efficiency of the pump =0.85. Note that, the headloss due pump fittings is estimated to be 1.20 m! (the given value is

in unit ‘m’ implying (𝑘𝑣2

2𝑔) not k).

* Use water at 20 °C. Consider the minor losses as well.

• pvc • pvc

sharp edged sharp edged

Two reservoirs of water surface elevation difference Δz = 3.0 m are connectedby a single pipe of L=840 m; φ= 0.15 m of ks = 6.0 x 10-5 m. There is a pumpwithin the system that pumps water from the upper to the lower reservoir.Neglecting the minor losses, determine the power consumed (Pp) by thepump. Take μ = 1.14 x 10-3 Ns m-2, γ = 9810 N/m3.

The characteristics of this pump handling water are:

Q (lt/s) 5 22 30 37 41 44 60

H (m) 11.2 10.5 8.75 7.45 6.3 5.4 0

η (%) 32 65 81 83 68 22 4

(Pump in Category 2 system) Answer Pp=3.89 kW

Example 4.2 PUMPS

the

9.25 m

3.29 kW 3288.7 Watt

Q=2

7 lt

/s74.5

9.26

MULTIPLE PUMP PIPE-SYSTEMSMULTIPLE PUMP SYSTEMS

It is sometimes necessary to use more than one pump ‘n conjunction with a given system. So the

combination of these pumps may be:

a) Series Operation: In this configuration, where the inlet of the second pump is connected to the outlet of

the first pump, the pumps of usually having the same characteristics that are fixed one after the other so as to

increase the energy head. i.e. Carries the same discharge Q where the total head is added to the system.

HT =H1 + H2 + H3 +…

b) Parallel Operation: In this configuration, where the inlets of the pumps as well as their outs are coupled

together and any number of pumps can operate simultaneously where the objective is to increase the

discharge under the common head. i.e. Under the same head H the total discharge is increased within the

system. QT =Q1 + Q2 + Q3 +…

İ

Generating the MULTI-PUMPS Characteristic Curve• For Pumps in Series:

In practice this type of connection is used whenever asingle pump can not carry the discharge ‘Q’ to therequired Head ‘H’.At a constant discharge (Q) ADD the individuallygenerated heads of each pump (H1+H2+H3...) For Tabular form, keep the given discharges ‘Q’

constant and increase the head ‘H’ values dependingon the required number of series connection. 2pumps in series 2H, 3 pumps in series 3H...

For Equation form, multiply the whole equation bythe required number of pumps in series.

# pumps in series: #*(H=A-BQn),3 pumps in series: (3*H=3*A-3*BQn).

MULTIPLE PUMP PIPE-SYSTEMS

• For Pumps in Parallel:In practice this type of connection is used whenever a

single pump under the given pumps head can not give thedesired discharge ‘Q’.At a constant Head (H) ADD the individually generateddischarges of each pump (Q1+Q2+Q3...)

MULTIPLE PUMP PIPE-SYSTEMS

For Tablular form, keep the given head ‘H’ valuesconstant and increase the discharge ‘Q’ valuesdepending on the required number of parallel pumps.2 pumps parallel 2Q, 3 pumps parallel 3Q...

For Equation form, reorginize the given formuladepending on the required number of parallel pumps:

# pumps parallel: (𝐇 = 𝐀 −𝐁

#𝐧𝐐𝐧),

3 pumps parallel: (𝐇 = 𝐀 −𝐁

𝟑𝐧𝐐𝐧).

Two reservoirs are connected by a single galvanized iron pipe ofL=3470 m; φ= 25 cm to discharge water @20 oC between reservoirA (elevation 70 m) to reservoir B (elevation 84.50 m). Theestimated total minor loss coefficient due fittings, elbows, valvesetc... is k=19.80. Definitely pumping system should be installed toallow this flow to occur within the system so as to help for carryingwater from the lower reservoir ‘A’ to the upper reservoir ‘B’.i- check the possibility of a single pump?ii- determine the power (Pp) consumed by the suggested pumpingsystem (2 identical pumps in series).iii- if an other identical pump will be added in series (3 identicalpumps) to the system determined in part ii, what is the % increasein Q and in P in this case.iv- if two other identical pumps will be added in parallel to thesystem determined in part ii, what is the % increase in Q and in P inthis case (i.e. 4 pumps: 2 pumps series connected parallel with another 2 pumps in series.) (T)

Example 4.3 MULTIPLE PUMP PIPE-SYSTEMS

Q (lt/s) 5 15 30 40 50 60 80

H (m) 12.00 12.50 11.50 10.25 8.50 7.15 4.50

η (%) 15 45 75 85 70 55 20

Example 4.3 MULTIPLE PUMP PIPE-SYSTEMS

The characteristics of the selected pump are:

AP

B

70 m

84.5 m

Part i

Part ii

34

21.74

79

Part iii

A

P

B

P P

27.7 m

% increase Q=((46.5-34)/34)x100=36.75%% increase P=( (16.165-9.159)/9.159)x100=76.49%

Q lt/s

Q (lt/s) 10 30 60 80 100 120 160

H (m) 24.00 25.00 23.00 20.50 17.00 14.30 9.00

Part iv:

So the re-calculated curve details are: (2Q and 2H of the single pump)

Note: For power calculation use the efficiency ‘η’ of the single pumps Q;i.e. (Q found/2 (due 2 parallel pump set)

Q= 39.9 lt/s H= 24.34 mη= % 54.9P= 17.316 kW

% increase Q=((39.9-34)/34)x100=17.35%% increase P=((17.316-9.159)/9.159)x100=89.06%

Hp(single) ≈ 12.5 – 362 Q2.36 for H (m) & Q (m3/s)

The pump characteristic values can be approximately given by:

TURBINES

Power is GENERATED from the available HEAD.

TURBINE:

Turbine is the general term used for any fluid machine that ABSORBS energy from the fluid.

Sometimes referred as ENERGY PRODUCING DEVICE since it extracts energy from the fluid and

transfers most of that energy to mechanical energy.

OR

WATER METER IS A GOOD EXAMPLE OF A TURBINE

TURBINES

TURBINESTURBINE: Turbine is a machine that is used to generate energy from the available head in the system

(m) turbineby the absorbed HeadEnergy :H

/s)(m )(discharge rate flow the:Q

for water) N/m (9810 fluid theoft unit weigh :

turbine theof efficiency the:

(Watts) Turbine by the gereratedPower :P ere wh ηγQHP

T

3

3

TTT

γ

η

PLEASE REMEMBER THAT:TURBINE DOES NOT DECREASE THE SPEED OF THE FLUID PASSING THROUGH THE TURBINE. INFACT DECREASES THE PRESSURE OF THE FLUID WHILE PASSING WITHIN THE TURBINE.

TURBINESExample 4.4

The Turbine shown below figure, facilitates (generates) apower of 72.8 hp (horse power). Determine the cast iron pipediameter if the length of the pipe is L = 1225 m, the averagedischarge is 10.5 m3/s and the efficiency of the turbine is η = 87%.It is estimated that the entrance loss coefficient is k=0.25. Thewater elevation at A is zA=38.0 m and elevation of pipe at its end Bis zB=5.2 m. Ignore the other minor losses. Note that 1 hp = 745Watt.

A

Pipe is cast ironLength 1225 m

average discharge Q = 10.5 m3/sDiameter ϕ = ?

P = 72.8 hpηT = 0.87 water @ 20 °C

TB

Φ = 1.40 m

Mathematical Expessions for Different Identical Pump Connections

Single Pump: Hp(single)=13.2 – 420 Q2.3

3 Pumps in Series: Hp(3S)=39.6 – 1260 Q2.3

2 Pumps in Parallel: Hp(2P)=13.2 –𝟒𝟐𝟎

𝟐𝟐.𝟑Q2.3

6 Pumps (2Series-3Parallel): Hp(6: 2S-3P)=26.4 –𝟖𝟒𝟎

𝟑𝟐.𝟑Q2.3

Reminder

Not possible!

3

i) Fixing a pump system on pipe 3

3

Possible pipe and pump connections within any system

ii) Fixing a pump system on pipe 2 & 1

The symbol of pump may represent a connection of:- a single pump- pumps in series - pumps in parallel.

Based on the ENERGY EQUATION, for the parallel piping system,having pump and considering the major and minor losses, thegeneral parallelity equation is:

CV

CV

221 2

P 1 1 minor 15 4

1 1

221 1

1 1 minor 1 P5 4

1 1

222 2

2 2 minor 25 4

2 2

22 21 2

1 1 minor 1 P 2 25 4 5

1 1 2

L QH + f Q k

12.1 D 12.1 D

L Qf Q k - H

12.1 D 12.1 D

L Qf Q k

12.1 D 12.1 D

L LQ f Q k - H f Q

12.1 D 12.1 D 12.1 D

M N

M N

M N

H H

H H

H H

2

2minor 2 4

2

Qk

12.1 D1

Hp that satisfies Q1

Previous Exam QuestionQuestion 3: (65 p)

Two reservoirs connected to each other by means of the pipeline and a pump system shown below.

Water @ 20 °C is expected to be pumped from reservoir W to reservoir K. Considering the minor losses

and also using the below details for the pump characteristics that was obtained from its manufacturer,

determine:

a) the discharge passing from pipe 1, (40 p)

b) the power required for this pumping system ‘Pp’, (15 p)

c) check for the cavitation occurrence at point S (100 m before reservoir K) (10 p)

Note that the atmospheric pressure is 101.300 kPa.

Q (lt/s) 5 10 15 20 25 30 32.5

H (m) 14.7 13.5 11.6 9.1 6.2 2.8 0

η (%) 55.3 67.2 74.7 86.1 79.6 56.9 24.8

Three pipes A, B and C are interconnected with each other as shown where the water is flowing within this system

at 22.5 °C. Over the figure topographic elevations ‘z’ of some points are also given. Along the pipe A, there exists a

single (one) water-pump (its charateristics are detailed in the below appropriate table). The pipe characteristics and

the total minor loss coefficients for each pipe is given in the relevant table below. The working regions atmospheric

pressure is 101.300 kPa (absolute). Find the flow rate ‘Q’ passing in each pipe?

Pipe Diameter ‘Φ’ (mm) Length ‘L’ (m) Σ kminor (-) Material

A 150 4820 14.6 Rough riveted steel

B 120 680 23.5 PVC

C 250 1735 18.9 Galvanized iron

Q (lt/s) 0 5 10 15 20 25 30 35 40 45

Hp (m) 9.00 8.90 8.55 8.00 7.20 6.20 4.95 3.50 1.80 0

η (%) 0 34.3 57.8 73.5 82.6 75.2 65.3 49.8 33.9 0

[The equation of the pump ‘Hp – Q’ characteristic is: Hp = 9.0 - 0.0045Q2 (where Q is in lt/s)]

Previous Exam Question

QA=14.756 lt/s; QB=41.341 lt/s; Qc=56.097 lt/s

Previous Exam QuestionThe elevation differences between reservoirs A and B is H = 21.0 m. It is expected to find the discharge of water @

25°C passing in each pipe that is flowing from reservoir A to B.

a) Write the necessary and the appropriate set of equations with their names using proper letters, (25 p)

b) Solve this problem. (75 p)

Pipe Length

(m)

Cross-section Material Total Minor loss coefficient

Σk

1 510 Circular: ϕ=200 mm Rough Concrete 23.6

2 320 Circular: ϕ=250 mm Galvanized iron 12.8

3 270 Square L=500 mm Cast iron 6.9

4 480 Circular: ϕ=300 mm PVC 19.4

5 630 Circular: ϕ=320 mm Asphalted cast iron 4.8

Equation of a Single pump on pipe 1: Hp = 6.75 - 16900Q2 (where Q in [m3/s] and Hp in [m])