Toolbox 8 PUMPS and FANS - Wiley€¦ · PUMPS and FANS PUMPS 1. There are two types of modern pump...

APPLIED INDUSTRIAL ENERGY AND ENVIRONMENTAL MANAGEMENT

Z. K. Morvay, D. D. Gvozdenac

Part III:

FUNDAMENTALS FOR ANALYSIS AND CALCULATION OF ENERGY AND

ENVIRONMENTAL PERFORMANCE

1

Applied Industrial Energy and Environmental Management Zoran K. Morvay and Dusan D. Gvozdenac © John Wiley & Sons, Ltd

Toolbox 8

PUMPS and FANS

PUMPS 1. There are two types of modern pump applied in industry: (a) Velocity head and (b) Positive

displacement. The following categorization of pumps could be applied:

(a) Velocity head

Centrifugal:

– Axial flow (single or multistage)

– Radial flow (single or double suction)

– Mixed flow (single or double suction)

- Peripheral (single or multistage)

Special effect:

– Gas lift

– Jet

– Hydraulic ram

– Electromagnetic

(b) Positive displacement:

Reciprocating:

– Piston plunger

– Diaphragm (mechanically or fluid driven, simplex or

multiplex)

Rotary:

– Single rotor (vane, piston, screw, flexible member,

peristaltic)

– Multiple rotor (gear, lobe, screw, circumferential

piston)

Centrifugal pumps are used in more industrial applications than any other kind of pump. The main

reason is than these pumps offer low investment and maintenance costs. These pumps have been

limited to low-pressure-head applications, but modern pumps are designed for quite high pressures.

These pumps have a smooth flow and the ability to tolerate non-flow conditions.

The main parts of the centrifugal pump are the impeller and volute (Fig. 8.1). An impeller can

take many forms. This device imparts a radial velocity to the fluid that has entered the pump

perpendicular to the impeller. The volute performs the function of slowing the fluid and increasing the

pressure. There may be one or more more volutes. The number of stages in the pump greatly affects

the pump’s output characteristics. Several stages can be incorporated into the same casing, with an

associated increase in pump output. Multistage pumps are often used for applications with a total

developed head of over 50 bar.

Whether or not a pump is self-priming can be important. If a centrifugal pump is filled with air

when it is turned on, the initiation of pumping action may not be sufficient to bring the fluid into the

pump. Pumps can be specified with features that can minimize priming problems.

Part III – Toolbox 8:

PUMPS and FANS 2

Impeller

Volute

Rotation

Figure 8.1: Schema of a Centrifugal Pump

PistonSuction

valve

Discharge

valve

H

Figure 8.2: Scheme of Positive-Displacement Pump

(Piston Type)

Positive-displacement pumps demonstrate high

discharge pressure and low flow rates and, usually, this

is accomplished with some flow pulsation. A piston

pump (Fig. 8.2) is a classic example of this type of

pump. Reciprocating pumps offer very high

efficiencies. For the larger sizes of these pumps it

reaches 90 %. These types of pump are more

appropriate for pumping abrasive liquids than are

centrifugal pumps.

Rotary pumps (Fig. 8.3) are one type of positive-

displacement pump that does not impart pulsations to

the exit flow.

Figure 8.3: Scheme of Rotary pump

Positive-displacement pumps require special seals to contain the fluid. The costs are higher for

both investment and for maintenance compared with most pumps that operate on a velocity head

basis. Positive-displacement pumps demonstrate an efficiency that is nearly independent of the flow

rate, in contrast to the velocity head types.

For positive-displacement it is important to stress that the output flow is proportional to pump

speed. This allows this type of pump to be used for metering applications. Also, a positive aspect of

these pumps is that they are self-priming, except at initial start-up.

If the downstream flow is blocked, high head pressure can be developed in a positive-

displacement pump and the pump could be damaged. For this reason, a pressure relief valve bypass

must be used when a positive-displacement pump is used.

2. Pumps are applied mainly for the following purposes:

circulation of fluids;

supplying the processes;

transferring the fluids.

Typical percentages for industrial pump application are presented in Fig. 8.4.


PUMPS and FANS 3

Circulation

58%

Supply

36%

Transfer

6%

Figure 8.4: Pump Application in Industrial Practice

3. The number of pumps versus a pump's power in industrial practice is presented in Fig. 8.5. Peak

pump powers range from 3.5 to 15.0 kW.

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

0.2

5

0.4

0

0.7

5

1.5

0

2.2

0

3.7

0

5.5

0

7.5

0

11

.00

15

.00

18

.50

22

.00

30

.00

37

.00

45

.00

55

.00

75

.00

90

.00

Output power of the motor [kW]

Perc

enta

ge [%

]

Analysis was performed on the sample of 3500 pumps

Figure 8.5: Pump Power Occurrences in Industry

4. The cost of electricity for running the pump is shown in Table 8.1. The calculation is performed for

4600 and 8500 operating hours of the pump per year and for the most frequently used power of the

pump. The life cycle of a pump can be estimated as 10 or 15 years. In this case the investment and

maintenance cost of the pump can be estimated as 5 % of the total cost for running the pump in that

period with operating hours of between 4600 and 8500 per year. That means that if some reasonable

investment is made for improving the energy efficiency of the pump, the pay back period will be very

short.


PUMPS and FANS 4

Table 8.1: Cost of Electrical Energy of Pump [US$] (1kWh = 0.045 US$) Power of

pump

kW

Annual operating hours of pump

4600 h/year

Annual operating hours of pump

8500 h/year

1 year 5 years 10 years 15 years 1 year 5 years 10 years 15 years

2.2 455 2277 4554 6831 842 4208 8415 12623

3.7 766 3830 7659 11 489 1415 7076 14 153 21 229

5.5 1139 5693 11 385 17 078 2104 10 519 21 038 31 556

7.5 1553 7763 15 525 23 288 2869 14 344 28 688 43 031

11.0 2277 11 385 22 770 34 155 4208 21 038 42 075 63 113

15.0 3105 15 525 31 050 46 575 5738 28 688 57 375 86 063

18.5 3830 19 148 38 295 57 443 7076 35 381 70 763 106 144

22.0 4554 22 770 45 540 68 310 8415 42 075 84 150 126 225

5. The relation that determines the total head developed by the pump is as follows:

hHg

ppH g

12 (8.1)

where:

H = Total head developed by pump in meters of the column of the liquid being pumped,

[m]

p2, p1 = Pressure in delivery and suction space, respectively, [Pa]

= Density of the liquid being pumped, [kg/m3]

Hg = Geometrical height to which liquid is lifted, [m] h = Head of the pump required for creating a velocity and for overcoming friction

resistance in the pipes and local obstacles in both suction and delivery lines, [m]

g = Acceleration due to gravity, = 9.81 [m/s2] commonly

Eq. (1) is used when designing pump installation.

The notations used in Eq. (1) are presented in Fig. 8.6.

H

H g

p d

p s

p 1

p 2

Figure 8.6: Pump Installation


PUMPS and FANS 5

6. Total head (H) can be also calculated by using the network equation:

2

wwH

g

ppH

2s

2d

0sd (8.2)

where:

pd = Delivery pressure at the outlet of the pump, [Pa]

ps = Suction pressure of the pump, [Pa]

H0 = Vertical distance between the measuring points of pressure pd and ps, [m]

wd = Velocity of fluid in the delivery (supply) pipeline, [m/s]

ws = Velocity of fluid in the suction pipeline, [m/s]

This equation is used for network characteristic determination.

7. The possible piping arrangements. Some possible piping arrangements are presented in Fig. 8.7.

Total static head is defined as the vertical distance from the surface of the source of the liquid supply

to the free surface of the liquid in the discharge receiver or to the point of free discharge from the

discharge pipe. When both the suction and discharge surfaces are open to the atmosphere, the total

static head equals the vertical difference in elevation. For calculations one must use the free-surface

elevations that cause the maximum suction lift and discharge head. That means that the lowest

possible level in the supply tank and the highest possible level in the discharge tank or pipe have to be

used. If the supply source is below the pump shaft line, the vertical distance is called static suction

head. With supply above the pump shaft line, the vertical distance is called static suction head. If the

liquid level varies during the pump operation, the lowest liquid level has to be used for calculating the

total static head.


PUMPS and FANS 6

(a) Suction lift and

submerged discharge

Pump shaft

Static discharge

head

Static suction

head

To

tal sta

tic h

ead

(b) Suction lift and free

discharge

Pump shaft

Static suction

head

To

tal sta

tic h

ead

Static discharge

head

(c) Suction lift and varying

discharge head

Pump shaft

Static discharge

head

Static suction

head

To

tal sta

tic h

ead

(d) Static suction head and

submerged discharge

Pump shaftStatic suction

head

Total static head

Static discharge

head

Free discharge

Maximum level

(e) Static suction head and

discharge head due to elevation

and pressure in the tank

Pump shaft

Static discharge head

(Tank pressure has to

be added)

Static suction

head

To

tal sta

tic h

ead

(f) Static suction head and

varying discharge

Pump shaftStatic suction

head

Total static head

Static discharge

head

Pressurized tank

Maximum level

Figure 8.7: Typical Pump Suction and Discharge Piping Arrangements

8. There are numbers of combinations of how the elements of network and pumps can be connected.

A number of characteristic connections are as follows:

friction losses without static head (Fig. 8.8);

friction losses with static head (Fig. 8.9);

system with two discharge heads (Fig. 8.10);

negative lift (gravity head) (Fig. 8.11).

He

ad

Flow rate

Friction

losses

System

friction curve

Figure 8.8: Friction Losses without Static Head


PUMPS and FANS 7

He

ad

Flow rate

Friction

losses

System

friction curveHs

Hs

Figure 8.9: Friction Losses with Static Head

He

ad

Flow rate

Friction

losses

System

friction curveHs

Hs

Figure 8.10: System with Two Different Discharged Heads

He

ad

Flow rate

Hs

Hs

Figure 8.11: Negative Lift (Gravity Head)

9. Plotting Eqs (8.1) and (8.2), the intersection of the two curves gives the Working Point of the

system formed by the network and pump which serve the given network.

Generally, the characteristics of the network can be expressed by the following equation:

2

g VaH p (8.3)


PUMPS and FANS 8

where:

p = Pressure drop in meters of liquid is being pumped, [m]

a = Characteristics of the network, [m s2/m

6]

Hg = Geometrical height to which liquid is lifted, [m]

An example of pump and network characteristics for Hg = 0 is presented in Fig. 8.12.

0

5

10

15

20

25

30

35

40

0 1 2 3 4 5 6 7 8 9

Volume flow rate [m3/s]

He

ad

an

d p

ressu

re lo

sse

s o

f n

etw

ork

[m

]

Head of pump

Network characteristic

Figure 8.12: Head and Network’s Pressure Losses versus Flow Rate

10. Mass and volume flow rates relation:

V m (8.4)

where:

m = Mass flow rate of fluid, [kg/s]

= Density, [kg/m3]

V = Volume flow rate of fluid, [m3/s]

11. The power of the electrical motor of the pump is:

1000

HgVN (8.5)

where:

V = Volume output (delivery) of the pump, [m3/s]

H = Total head developed by pump in meters of the column of the liquid being pumped, [m]

= Overall efficiency of the pump installation equal to the product of the efficiencies of the

pump p, the transmission tr, and the electrical motor m (= p· tr· m)

The motor installed for a pump has a somewhat greater power Ninst than that needed by the pump in

order to provide a reserve for possible overloading:

N Ninst (8.6)


PUMPS and FANS 9

The power reserve factor is taken depending on the value of N and is given in Table 8.2.

Table 8.2: Power reserve factor N [kW]

< 1 2.0–1.5

1–5 1.5–1.2

5 –50 1.2–1.15

> 50 1.1

12. Power of three-phase electrical motors. Asynchronous three-phase electrical motors are the

most common motors used in industry. The power taken from grid for these motors (and synchronous

also) is as follows:

]W[cosIU3N ll (8.7)

where:

Ul = line voltage, [V]

Il = line current, [A]

cos = power factor, [-]

or,

]W[cosIU3N pp (8.8)

where:

Up = phase voltage, [V]

Ip = phase current, [A]

An example of a three-phase symmetric circuit is presented in Fig. 8.13.

220 V

220 V

220 V

Phase 1

Phase 2

Phase 3

Up = 220 V

Ul = 380 VZero

Figure 8.13: Three-Phase Symmetric Circuit

13. By changing the speed of a centrifugal pump within a limited range, the changes in its flow rate,

head and power consumed are defined by the following relations:

2

1

2

1

n

n

V

V (8.9)

2

2

1

2

1

n

n

H

H (8.10)


PUMPS and FANS 10

3

2

1

2

1

n

n

N

N (8.11)

Where:

V = Volume output (delivery) of the pump, [m3/s]

H = Total head developed by the pump in meters of the column of the liquid being pumped,

[m]

N = Power of electrical motor of pump, [kW]

n = Speed of pump, [rpm]

14. The suction height of a centrifugal pump is calculated by using the following relation:

cavs,1satas hhhPH (8.12)

where:

Pa = Atmospheric pressure

hsat = Saturated vapor pressure of the liquid being sucked in at the pumping temperature

h1,s = Hydraulic resistance of the suction line including the energy needed for imparting a

velocity to the flow of liquid

hcav = Cavitation correction (a reduction in the suction height to avoid cavitation) depending on

the volume flow rate of the pump [m3/s] and the speed [rpm].

The cavitation correction can be determined by using the following equation:

67.02

cav nV00125.0h (8.13)

All values in Eq. (8.12) are expressed in meters of the column of the liquid being pumped.

15. Cavitation is a local condition that allows liquid to boil and form a vapor. The pump application

shown in Fig. 14 is a once-through system. However, the leg of piping though pressure drop 1 shown

there can have some important implications related to net positive suction head. Net positive suction

head (NPSH) is the difference between the local absolute pressure of a liquid and the thermodynamic

saturation pressure of the liquid based upon the temperature of the liquid. If NPSH = 0, the liquid can

vaporize, and this can result in a variety of outcomes from noisy pump operation to outright failure of

components. Cavitation, if it occurs, will take place at the lowest pressure point and this point is

located at the inlet of the pump or inside the pump. Most manufacturers specify how much NSPH is

required for the satisfactory operation of their pumps. Actual net positive suction head (NPSHA) is

the NPSH at the given state of operation of a pump and must be larger than the required net positive

suction head (NPSHR) specified by the manufacturer for a given application.


PUMPS and FANS 11

PUMPPressure

drop 2

Pressure

drop 1

FLOW

TANK

Figure 8.14: Typical Pump Application

16. The output of piston pump V [m3/s] is:

Single acting and differential (trim) pumps:

60

nsAV V (8.14)

Double-acting pumps:

60

nsAA2V r

v (8.15)

where:

V = Delivery factor (from 0.8 to 0.9), [-]

A = Cross-sectional area of the piston, [m2]

Ar = Cross-sectional area of the road, [m2]

s = Piston stroke, [m]

n = Speed of rotation (number of double piston strokes per minute), [rpm]

17. The output of a gear pump V [m3/s] is:

60

nzbA2V 1

v (8.16)

where:

V = Delivery factor, [-]

A1 = Cross-sectional area of a tooth restricted by addendum circle of the matrix gear wheel,

[m2]

b = Width of a tooth, [m]

z = Number of teeth on a gear wheel, [-]

n = Speed of rotation (number of double piston strokes per minute), [rpm]

18. There are many possibilities for reducing the energy consumption of pumps. All of these energy

conservation measures can be specified in two groups (Fig. 8.15):


PUMPS and FANS 12

a) Effective operation

b) Increasing the efficiency of the motor and/or pump

However, the effect of the possible technical measures which can be applied is different and

always has to be analyzed simultaneously with: (a) operational conditions of pump, (b) economical

aspects and (c) possible influence of its application on process. Adjusting pump performance to meet

process requirements is the most effective way of reducing energy cost (Fig. 8.16) and improving the

efficiency of both motor and pump (Fig. 8.17) can result only in energy cost reduction by a small

percentage.

Reduce Impeller Diameter

Adjust Speed

Change Model

Select Running Pump(s)

Reduce Pipe Loss

Effective

Operation

Raise Up

Efficiency

Pump

Efficiency

Motor

Efficiency

Improve

System

Adjust Pump

Performance

Figure 8.15: Ways to Reduce the Energy Consumption of a Pump

He

ad

[m

WC

] a

nd

Sh

aft P

ow

er

[kW

]

Volume flow rate [m3/s

Losses

HEAD

POWER

Figure 8.16: Adjustment of Pump Performance

He

ad

[m

WC

],

Effic

ien

cy [%

] a

nd

Sh

aft P

ow

er

[kW

]

Volume flow rate [m3/s]

HEAD

POWER

EFFICIENCY

Figure 8.17: Increasing Efficiency of Pump

and/or Motor


PUMPS and FANS 13

19. Check list of centrifugal pump faults and their causes:

SYMPTOM POSSIBLE CAUSE OF TROUBLE

FAILS TO DELIVER LIQUID

1. Wrong direction of rotation

2. Pump not primed

3. Suction line not filled with liquid 4. Air or vapor pocket in suction line

5. Inlet to suction pipe not sufficiently submerged

6. Available net positive suction head (NPSH) not high enough 7. Height from suction liquid level to pump shaft too great

8. Distance from suction liquid level to pump shaft too small

9. Difference between suction pressure and vapor pressure too small 10. Pump not up to rated speed

11. Total head greater that head for which pump is designed

PUMP DOES NOT DELIVER

RATED CAPACITY

1. Wrong direction of rotation

2. Suction line not filled with liquid 3. Air or vapor pocket in suction line

4. Air leaks in suction line or through stuffing boxes

5. Suction pipe intake not submerged enough 6. Available NPSH not sufficient

7. Height from liquid level to pump shaft too great 8. Distance from suction liquid level to pump shaft too small

9. Difference between suction pressure and vapor pressure too small

10. Pump not up to rated speed 11. Total head greater than head for which pump was designed

12. Foot valve too small

13. Foot valve clogged with trash 14. Viscosity of liquid greater than that for which pump was designed

15. Mechanical defects (wearing rings worn, impeller damaged, internal leaks caused by

defective gaskets)

PUMP’S DISCHARGE

PRESSURE LOW

1. Gas or vapor in liquid 2. Pump not up to rated speed

3. Greater discharge pressure needed than that for which pump was designed

4. Liquid thicker than that for which pump was designed 5. Wrong rotation

6. Mechanical defects (wearing rings worn, impeller damaged, internal leaks caused by

defective gaskets)

STUFFING BOXES

OVERHEAT

1. Packing too tight

2. Packing not lubricating

3. Wrong grade of packing 4. Not enough cooling water to jacket

5. Stuffing box improperly packed

PUMP LOSSES PRIME

AFTER STARTING

1. Suction line not filled with liquid

2. Air leaks in suction line or through stuffing boxes 3. Gas or vapor in liquid

4. Air or vapor pocket in suction line 5. Inlet to suction line not submerged far enough

6. Available NPSH not sufficient

7. Height from liquid level to pump shaft too great 8. Distance from suction liquid level to pump shaft too small

9. Difference between suction and vapor pressure too small

10. Liquid seal piping to lantern ring plugged 11. Lantern ring not properly placed in stuffing box

PUMP OVERLOADS

DRIVER

1. Speed too high

2. Total head lower than rated head

3. Either the specific gravity or viscosity of liquid or both different from that for which pump is rated

4. Mechanical defects (misalignment, shaft bent, rotating elements dragging, packing too tight)

VIBRATION

1. Starved suction (gas or vapor in liquid, available not positive suction head not high enough, inlet to suction line not submerged enough, gas or vapor pockets in suction line)

2. Misalignment

3. Warn or loose bearings 4. Rotor out of balance (impeller being plugged or damaged)

5. Shaft bent

6. Control valve in discharge line improperly placed 7. Foundation not rigid

BEARINGS OVERHEAT

1. Oil level too low

2. Improper or poor grade of oil

3. Dirt in bearings 4. Dirt in oil

5. Moisture in oil 6. Oil cooler clogged or scaled


PUMPS and FANS 14

7. Any failure of oiling system

8. Not enough cooling water

9. Bearing too tight 10. Oil seats fitted too closely on shaft

11. Misalignment

BEARINGS WEAR RAPIDLY

1. Misalignment 2. Shaft bent

3. Vibration

4. Excessive thrust resulting from mechanical failure inside the pump 5. Lack of lubricant

6. Bearings improperly installed

7. Dirt in bearings 8. Moisture in oil

9. Excessive cooling of bearings

FANS

20. Generally, fans are classified according to how the gas flows through the impeller. These flows

may be axial, radial, mixed and cross.

The typical characteristics of fans are shown in Fig. 8.18. Fan efficiency variations indicate a

sharp maximum value at the design point. That means that choosing the fan must be tuned carefully to

the required conditions.

Volume flow rate

Total p

Efficiency

Power

Figure 8.18: Typical Characteristics of a Centrifugal Fan

Two typical fans, most frequently used in industry, are presented in Figs 8.19 and 8.20. There are

a number of vane types and these types can also be used for fan classification. Axial fans (Fig. 8.20)

usually have vanes of an airfoil shape or vanes of uniform thickness. Some types of vane that can be

found on centrifugal fans are presented in Fig. 8.19.

Most industrial applications use centrifugal type of fans. Compared with axial fans they have a

higher pressure rise, but a lower flow rate.


PUMPS and FANS 15

Rotation

Impeller

Van

Backward inclined

Some possible vane types:

Forward curved

Radial

Airfoil

Backward curved

Tubular

Volute

Figure 8.19: Some Vane Types that Might be Used on a Centrifugal

Fan

Impeller

Figure 8.20: Axial Fan

21. The pressure increase created by a fan (Fig. 8.21) or total head of fan (H) is:

zg)(2

w)pp(ppp air

2

ds12 (8.17)

or

2

wp

2

wpp

2s

s,st

2d

d,st (8.18)

where:

p1 = Pressure in the volume from which the fan takes the gas, [Pa]

p2 = Pressure in the volume into which the fan delivers the gas, [Pa]

ps, pd = Pressure drops in suction and delivery lines, [Pa]

w = Velocity of gas at the outlet from the installation, [m/s]

pst,d, pst,s = Static pressures directly after the fan and before it, [Pa]

wd, ws = Velocities of gas in delivery and suction pipelines, [m/s]

= Density of gas, [kg/m3]

air = Density of ambient air, [kg/m3]

z = Difference between heights of the delivery and suction places, [m]

22. The power of the fan is:

]kW[1000

PVN (8.19)

where:

V = Volume flow rate, [m3/s]

P = Total head of fan, [Pa]

= Overall efficiency of the fan installation equal to the product of pump efficiency p, the

transmission tr, and the electrical motor m (= p tr m)


PUMPS and FANS 16

P2

P1

sP

pP

w

z

Figure 8.21: Fan Installation

23. Fan flow control is a very important design and operation concern. There are five types of controls

used for forced- and induced- draft fans:

damper in the duct with constant-speed fan drive;

two-speed fan driver;

variable inlet vanes or inlet louvers with a constant-speed drive;

multiple-step variable fan drive;

variable speed drive.

Dampers are the least expensive to install, but also the most inefficient in terms of energy use. A

variable speed drive control system is the preferred control system, when a combination of initial and

operating costs is considered.

24. EXAMPLES

Example 1: A solution (3m/kg1120 ) has to be fed at a flow rate of 115 m

3/h from a tank

into an apparatus at a height of 10.8 m above the liquid level in the tank. The gauge pressure in the

apparatus is 0.4 barg and the pressure in the tank is atmospheric. The pipeline has diameter of

mm5.4140 . Its total equivalent length (the actual length plus the equivalent length of the local

obstacles) is 140 m. The coefficient of friction resistance in the pipeline is 03.0 .

The centrifugal pump envisaged for this job was tested at a speed of 1200 rpm and the following

results were obtained:

V [l/s] 0 10.8 21.2 29.8 40.4 51.1

H [m] 23.5 25.8 25.4 22.1 17.3 11.9

P [kW] 5.16 7.87 10.1 11.3 12.0 18.5

The liquid density during the testing of the pump was 3m/kg1120 .

It needs to be checked as to whether is it possible to use this pump for the proposed pumping

under the prescribed conditions.


PUMPS and FANS 17

Solution:

Pump

The efficiency of the pump is (see Eq. (8.3)):

P1000

HgV (8.20)

V [l/s] 0 10.8 21.2 29.8 40.4 51.1

[-] 0.000 0.389 0.586 0.640 0.640 0.361

The pump’s characteristics are presented in Fig. 8.22 and the poly trend line are as follows:

11223446 102.3480 + V103.6455 + V101.1604 -V101.5357 - V103.0219H (8.21

)

5.1756 +V109.1191 + V102.3995 + V101.0747 - V101.3111N 2223345 (8.22

) -42233547 108.1892 - V105.2564 + V101.8843 - V104.1539 + V10-4.3201 (8.23

)

0

5

10

15

20

25

30

35

0 10 20 30 40 50 60

Volume flow rate [l/s]

Head [m

] and P

ow

er

[kW

]

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

Effic

iency [-]

n=1200 rpm

Density = 1120 kg/m3

Efficiency [-]

Power [kW]

Head [m]

Figure 8.22: Pump Characteristics

Pump and system characteristics

The fluid velocity is:

];s/m[37.2

4

131.0

3600

115

w2

]m[2863.081.92

37.2h

2

w (8.24

)

The loss of head for overcoming the pipe resistance and local obstacles is:


PUMPS and FANS 18

]m[1791.92863.0131.0

14003.0h

D

)LL(h w

eq

lf (8.25

)

The required total head of the pump is (Eq. (8.1)):

]m[9060.232863.01791.98.10112081.9

000,1004.0H

(8.26

)

This total head is calculated for a volume flow rate of 31.94 [l/s].

The tested pump has to be used for a described duty, but, plotting the working point (V = 31.94

l/s; H = 23.906 m) in Fig. 8.22 one can see that this point is above the curve of the pump

characteristics and, consequently, the pump at n = 1200 rpm cannot ensure the required output. At H =

23.906 m the pump delivered only 25.67 l/s (poly trend curve H = f(V) can be used too). However, by

increasing the pump’s speed, it will be possible to reach the desired working point. Using Eqs (8.6)

and (8.7) the experimental data can be recalculated for a new speed of pump. If the speed is nnew =

1260 rpm, the recalculated pump characteristics are as follows:

V [l/s] 0.00 11.34 22.26 31.29 42.42 53.66

H [m] 25.91 28.44 28.00 24.37 19.07 13.12

P [kW] 5.97 9.11 11.69 13.08 13.89 21.42

[-] 0.000 0.389 0.586 0.640 0.640 0.361

The working point and new head-volume flow rate curve for a speed of pump of 1260 rpm are

presented in Fig. 8.23.

0.00

5.00

10.00

15.00

20.00

25.00

30.00

35.00

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

Volume flow rate [l/s]

Head [m

]

Working pointn = 1200 rpm

n = 1260 rpm

Figure 8.23: Pump Characteristics for 1200 and 1260 rpm and Working Point

By increasing the speed of pump from 1200 to 1260 rpm, the pump can ensure the required flow

rate of 31.94 l/s (115 m3/h) and head of 23.91 m. The power of the pump is, taking the power from the

Fig. 8.22 or calculating the power from the poly trend line for real volume flow rate (31.94 l/s):

]kW[96.121200

1260194.11

n

nPP

33

newnew

(8.27

)


PUMPS and FANS 19

Example 2: A centrifugal fan was tested at a speed of n = 1440 rpm. The following results

were obtained:

V m3/h 100 350 700 1000 1600 2000

p Pa 449 424 432 427 387 316

The resistance of the network calculations shows that when air flows through it at a rate of 1350 m3/h,

the pressure losses are Pa85pw and Pa288pp lf . The difference between the pressures in

the delivery and the suction spaces for the calculated network is Pa128pp .

If the testing fan is installed in a described network, the volume flow rate of air has to be

calculated.

Solution:

Network

A parabola expresses the characteristics of the network:

2Vbap (8.28

)

The first addend in the right-hand side does not depend on the flow rate and is the difference between

the pressures in the delivery and suction spaces. The second is equal to the sum of pressure losses

lfw ppp and changes proportionally to the square of the flow rate. That means that

characteristics of the network are:

22

t 1350

V373128

V

V)28885(128p

(8.29

)

where p is in Pa and (V/1350) is the ratio of the actual flow rate and the calculated value of flow

rate based on the resistance of the network.

0

100

200

300

400

500

600

700

800

900

1000

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Volume flow rate [m3/h]

Pre

ssure

dro

p [P

a]

Network

Working

pointFan

Figure 8.24: Fan and Network Head versus Flow Rate


PUMPS and FANS 20

The working point is the intersection of two characteristics (fan and network) and it shows that in

operation for the given network the fan will deliver 1187 m3/h. The head of fan will be 416 Pa.

Example 3: The motor driving a fan is rated at a current of 15 A. During the energy audit the

current of 11 A was measured and the speed of the fan was 900 rpm. The airflow rate delivered by the

fan is to be increased as much as possible. What is the permissible speed of the fan within the rating

of the motor, and what percentage increase in airflow can be expected?

Solution:

By Eq. (10) one can find that speed of the fan can be increased up to:

rpm99811

15900

I

Inn 33

old

newoldnew

(8.30

)

The power of a three-phase electrical motor is cosIV3N LL . In the equation above it is

assumed that line voltage and cos are constant. The power of the electrical motor will increase by

36.4 % as compared to the present conditions for driving the fan.

Since flow rate is proportional to the fan’s speed, the percentage increase in flow rate is:

%9.10100900

900998100

V

VVV

old

oldnew (8.31

)

The fan’s speed can be increased, for example, by simply replacing the fan’s or electrical motor’s

pulley.

Example 4: A centrifugal pump feeds 50 m3/h of

water temperature 60 oC into the tank. The inner

diameter of the suction pipeline is 100 mm and its

length is 6 m. The friction factor is 0.028 and the sum

of loss obstacles is 2.5. The speed of the pump is 1450

rpm.

The suction head (height) has to be calculated

(Fig. 8.25).

PUMP

Pump shaft

FLOW

Hs

din

Figure 8.25: Pump Application


PUMPS and FANS 21

Solution:

The velocity of water in the suction pipe is:

]s/m[768.1

4

1.0

3600

50

4

d

Vw

22s (8.32

)

The hydraulic resistance of the suction line including the energy needed for imparting a velocity to the

flow of liquid is as follows:

]m[8256.080665.92

768.1

1.0

6028.05.21

g2

w

d

L1h

22s

s,1 (8.33

)

The cavitation correction can be determined by using Eq. (13):

]mWC[2267.114503600

5000125.0nV00125.0h

67.0

267.02cav

(8.34

)

or, bar120.01080665.92267.1h 2cav .

The suction head of a centrifugal pump is calculated by using the following relation:

]m[11.62267.18256.00980665.0

19917.0

0980665.0

0.1hhhPH cavs,1satas

(8.35

)

where:

Pa = 1.0 bar (atmospheric pressure)

hsat = 0.19917 bar (saturated vapor pressure of the water at the pumping temperature – see

Toolbox III-4 (Eq. 4.6) or Software 5).

Example 5: The delivery factor of a gear pump has to be determined for a speed of 440 rpm.

The number of teeth on a gear wheel is 12, the width of a tooth is 42 mm, and the cross-sectional area

of a tooth restricted by the addendum circle of the mating gear wheel is 960 mm2. The actual flow rate

of the pump is 0.312 m3/min.

Solution:

The theoretical flow rate of the pump is:

]s/m[007096.060

44012042.000096.02V 3

(8.36

)

The actual (measured) flow rate is ]s/m[0052.060

312.0V 3

a . The delivery factor is now:

733.0007096.0

0052.0V

(8.37

)


PUMPS and FANS 22

Example 6: The characteristics of a centrifugal fan are:

V m3/s 0 2 4 6 8 10 12

p Pa 165 170 170 160 135 110 55

- 0 0.36 0.49 0.51 0.44 0.32 0.16

This fan operates in a network defined by the following equation:

2

n V20p (8.38

)

Where:

V = Flow rate of air, [m3/s]

pn = Pressure losses in network, [Pa]

The comparison of power has to be analyzed for the following cases:

a) defined working point;

b) when speed of fan is reduced to meet the flow rate of 6 m3/s.;

c) when throttling is used to reduces the flow rate.;

d) when two fans are connected in line. The resulting flow rate has to be found.

Solution:

By fitting the values from fan characteristics given in the table above, the following analytical

equations are obtained:

3

234-2

101.64556 + V106.77489

+ V101.88920 - V1.46780 + V10-8.28598p

(8.39

)

-223-14-3 103.75190 - V7.32905+ V1.63695 - V101.99293 + V10-7.48625N (8.40

)

-3-12-23-4 106.66667 + V102.19623 + V102.76190 - V108.68056 (8.41

)

The table's values and values calculated from the above mentioned equations are presented in Fig.

8.26.


PUMPS and FANS 23

0

300

600

900

1200

1500

1800

0 2 4 6 8 10 12 14

Flow rate [m3/s]

Head [P

a]

0

10

20

30

40

50

60

Pow

er

[kW

] and E

ffic

iency [%

]

Head

Power

Efficiency

Figure 8.26: Characteristics of Fan versus Flow Rate

a. The working point is defined as a cross-section of head curve and network curve (Fig. 8.27). The

following values are obtained at the working point:

Va = 8.25 m3/s

Ha = 1360 Pa

Na = 26.22 kW

a = 0.43

The power per unit flow rate is:

)]s/m/(kW[178.325.8

22.26

V

N 3

a

(8.42

)

The working point is presented in Fig. 27.

0

500

1000

1500

2000

2500

0 2 4 6 8 10 12

Flow rate [m3/s]

He

ad

an

d P

ressu

re lo

sse

s o

f n

etw

ork

[P

a]

Working point

V = 8.25 [m3/s]

H = 1360 [Pa]

Head

Network

Figure 8.27: Working Point for Case a


PUMPS and FANS 24

b. In this case the flow rate of air is 6 m3/s. The network pressure drop curve is not changed. By

reducing the speed of the fan from na to nb the power of the fan will also be reduced. From Eqs

(8.9)–(8.11) the following relation can be obtained:

bb

aa

b

a

HV

HV

N

N (8.43

)

where subscripts a and b are related to two different speeds of fan. Comparing this relation to the fan

power relation (Eq. (8.5)), it is obvious that this relation is valid only if the efficiency for these two

speeds are the same ( 43.0ba ).

For flow rate 6 m3/s the pressure losses of network are:

]Pa[720620p 2b,n (8.44

)

The power of the fan for a reduced speed and flow rate is as follows:

]kW[05.1043.01000

7206Nb

(8.45

)

The power per unit flow rate in this case is:

)]s/m/(kW[675.100.6

05.10

V

N 3

b

(8.46

)

c. By throttling the flow without changing the speed of the fan, the new flow rate will be 6 m3/s. The

head and efficiency of the fan can be calculated by using analytical expressions. The results are:

Vc = 6.00 m3/s

Hc = 1582 Pa

Nc = 18.35 kW

c = 0.52


)]s/m/(kW[058.300.6

35.18

V

N 3

c

(8.47

)

d. By connecting two fans in a line (serial connection), the head is doubled for the same flow rate

(Fig. 8.28). The working point in this case is as follows:

Vd = 11.66 m3/s

Hd = 2719 Pa

Nd = 2 40.42 = 80.82 kW

d = 0.19


PUMPS and FANS 25


)]s/m/(kW[933.666.11

42.402

V

N 3

d

(8.48

)

0

500

1000

1500

2000

2500

3000

3500

4000

0 2 4 6 8 10 12 14

Flow rate [m3/s]

He

ad

[P

a]

(a)(b)

(c)

(d)

Network

Head (one fan)

Head (two fans in line)

V = 8.25 [m3/s]

H = 1360 [Pa]

N = 26.22 kW

= 0.43

V = 11.66 [m3/s]

H = 2719 [Pa]

N = 80.82 kW

= 0.19

V = 6.00 [m3/s]

H = 1852 [Pa]

N = 18.35 kW

= 0.52

V = 6.00 [m3/s]

H = 720 [Pa]

N = 10.05 kW

= 0.43

Figure 8.28: Working Points for Case a, b, c and d

Comparing the results one can conclude that regulating flow by changing the speed of the fan is more

efficient than throttling the flow. The power per unit of flow rate is, in the first case, 1.675

[kW/(m3/s)] and, in the second, 3.058. By using two fans to increase the flow rate, the power per unit

flow rate is quite high (6.933). That means that some other solution has to be analyzed (for example,

increasing speed and replacing the electrical motor or replacing the fan).

References Chopey, N.P., Hicks, T.G. (Eds) (1984) Handbook of Chemical Engineering Calculations,

McGraw-Hill Book Co.

Elonka, S.M. (1980) Standard Plant Operators' Manual, McGraw-Hill.

Pavlov, K.F., Romankov, P.G., Noskov, A.A. (1979) Examples and Problems in the Course of Unit

Operations of Chemical Engineering, Moscow.

Stoecker, W.F., Jones, J.W. (1984) Refrigeration and Air Conditioning, McGraw-Hill.

Toolbox 8 PUMPS and FANS - Wiley€¦ · PUMPS and FANS PUMPS 1. There are two types of modern pump...

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Transcript of Toolbox 8 PUMPS and FANS - Wiley€¦ · PUMPS and FANS PUMPS 1. There are two types of modern pump...