# Important system characteristics

Introduction

 A = It is important to remember that the fan and its system have to be in balance. Thus if the system resistance has been over or under estimated, the absorbed fan power will vary according to its type. This in turn will affect the losses in the transmission mo­tor and controls.

Overall fan efficiency

The absorbed fan power is only a start to understanding how to calculate the energy consumption of a fan. An important step is to calculate the input power kW at a specific duty. Multiplying this input power by the hours run at the duty will determine the kWh.

 QxpF Thus input kW = —

Equ 19.13

TIf xt1r xtIm xrlc

 Where: Q = flowrate (m3/s) Pf = fan pressure (kPa) TlF = fan efficiency expressed as a decimal M = transmission efficiency at duty Expressed as a decimal TlM = motor efficiency at absorbed power Expressed as a decimal TIC = control efficiency at absorbed power Expressed as a decimal
 PF and t|f may be either both total or both static.

Demand variations

The primary cause of a low overall fan efficiency factor is due to the variation, with respect to time, of the desired flow through the ducts, due to insufficient consideration having been given to this aspect at the design stage. Normally the mean flow per an­num — total supplied volume of air divided by a calendar year

— is about 10% to 15% of the installed flow capacity Qmax. The efficiency factor calculated as a mean efficiency over the year

— theoretical air energy consumption/actual electrical energy consumption — usually gives a value in the region 5% to 40%, which should then be compared with the normal maximum mo­mentary value of 55% to 75%.

The reasons for production or demand variations and the low mean flow are many and typical examples are:

• Seasonal variations in industry production

• Variations in ambient temperature

• Variations in ambient humidity

• Variations in building occupancy

• Component and calculation tolerances

• Margins of safety over and above the normal tolerances

A survey of flow variations with respect to time is a necessary basic requirement for all economic optimisations of air trans­portation. It is desirable that both progressive curves and con­stancy diagrams are produced.

Modifications for the improvement of the efficiency factor and hence the costs can be:

• To improve the part load efficiency by means of suitable flow control and regulation equipment.

• To reduce extreme flow peaks.

Availability

Availability is generally defined by the relationship:

MTBF

Equ 19.14

MTBF+MTTR

Where

A = availability (decimal)

MTBF = Mean Time Between Failures (hours)

MTTR = Mean Time To Restore (hours)

The MTBF should be the summation of operating hours per an­num. The MTTR should be the summation of all repair times per annum. MTTR should include all those hours of routine maintenance or adjustment when the fan must be stopped.

Another concept frequently overlooked, MTBA, Mean Time Be­tween Adjustments, includes maintenance work performed while the fan is running. MTBA does not affect availability but it does affect running costs.

Centrifugal fans used within the process industry, for well-tried, normal operating conditions, achieve values of availability in the order of A = 0.99 to 0.9999, i. e. shut-down due to malfunc­tion is between 1 and 80 hours per annum for continuous oper­ation. These values also include the squirrel cage induction motor but not other electrical equipment. At the other extreme some high pressure conveying fans, have an availability of 0.88 on a daily basis. Some fans must have the seals replaced fre­quently.

Availability calculations are applied mainly to multi-fan sys­tems. The percentage of operating time which a given number
of fans in a fan station can be expected to be available for oper­ation, is calculated from the formula:

N!

 P = Equ 19.15

 Where P N K A I K!(N — K)!

= system availability (decimal)

= number of fans installed (integer)

= number of fans unavailable (integer)

= availability of individual fans (decimal)

 Where F L M = factorial (4N24, 0!=1)

All the fans are thus assumed to have the same availability val­ues. By using equation 19.15 the resultant availability of a fan station can be calculated and the effect of a stand-by fan can be demonstrated.

 Without reserve fan With reserve fan Function Requirement Resultant Availability Function Requirement Resultant Availability 1 of 1 0.9900 1 of 2 0.9999 2 of 2 0.9801 2 of 3 0.9997 3 of 3 0.9703 3 of 4 0.9994 4 of 4 0.9606 4 of 5 0.9990 5 of 5 0.9510 5 of 6 0.9985
 Table 19.4 Availability of fan station with and without stand-by fan

From Table 19.4 it can be seen that a fan station’s availability is considerably improved by the installation of a stand-by fan

 •4 Function requirement 3 of 4 means that 3 out of 4 will be avail­able for operation. The individual fan availability has been taken as 0.99.

 4fL 1 q/ Pl d Wd4 . 32fLpqv The user must define the availability required at the inquiry stage, as this parameter has a crucial effect on fan selection. If the user wishes to operate for 50 weeks every year the avail­ability would have to be 0.959. If the user wishes to operate for 154 weeks and allow two weeks for repair the availability would have to be 0.987.

 Pl For this type of availability calculation to apply in practice it is necessary to maintain the stand-by fan to ensure that it will op­erate when required. In cases where it is possible to calculate the cost of a non-functioning fan, for example, in the form of lost production, availability calculations offer a direct economic opti­misation possibility. Stand-by fans usually run a limited number of hours regularly, one week in four say, to ensure the fan is functioning correctly.

Air power

General

 Pl The useful air power from the fan is used to satisfy the ducting system (and terminal loads) resistance at the actual flow.

Pf =qvpf Equ 19.16

Where

 2a 3fLpqv 32a5 Pf = power absorbed by fan impeller W or kW

Qv = volumetric flowrate m3/s

Pf = fan pressure Pa or kPa

If fan pressure is given in Pa (pascals) then power will be given in W (Watts).

 Ttd = 6a or a = — 6 If fan pressure is given in kPa (kilopascals) then power will be in kW (kilowatts).

To achieve an optimal economic result, it is important the air power is kept as low as possible. Limiting the required air power is well worthwhile, especially with overall fan efficiency factors of as low as 5% to 40%. A reduction of Pf by 1 kW can reduce the electrical power consumption by as much as 2.5 kW to 20 kW because of the low overall fan system efficiency in the con­version of electricity into air power.

Duct pressure losses

The approximate loss of pressure in straight ducting, assuming fully turbulent flow and smooth materials, can be defined as:

FL 1

Pl

Equ 19.17

Pv

Friction factor length of duct (m) mean hydraulic depth (m) area of duct cross section (m2)

M 2

Periphery of duct (m) p = air density (kg. nr3)

V = mean duct air velocity (m2)

Recognising that

Q„ volumetric flowrate = mean velocity v x cross-section area, then for a circular cross section duct

TOC o "1-5" h z Ttd2 1 d

M =—————— = —

4 id 4

and

Qv _ qv

V = —

Area Ttd

Or

•16

I. e.

Equ 19.18

If American values of f are used (i. e. based on duct diameter) then the constant becomes 8 instead of 32, but the value is still proportional to d’5.

Thus for an increase in duct diameter of only 10% then the re­sistance, and therefore the air power absorbed, has been re-

1

Duced to —^ x previous value i. e. 62%.

1-15

Considering now a rectangular duct, whose sides have a 2:1 ra­tion, handling the same airflow and having the same metal con­tent and periphery (when the price may be approximately the same) then:

FL 1 2

M Цpv

M 2

FL fio 1 V

2 2 4a

2

But

 FLp D
 5 x qv x23822
 Aq Qvm J
 Pfm-t
 Pfan = constant (qv
 ^Aqv) Aqv
 = constant qvm 1 +
 Aqv
 Equ 19.21
 P = p ■ fan ‘ fan-rnean
 The energy consumption due to the fan and duct system alone becomes dependent upon the shape of the flow demand/sup­ply curve with time and is illustrated in Figure 19.6 for three typi­cal cases. The energy consumption due to the fan and ducting system, but excluding the effects of motor, transmission and control losses becomes:
 AQ/Qm Pf/Pfm Ef/Pfm Square Saw tooth Sinusoidal 0 1 1 1 0.1 1.33 1.03 1.01 1.02 0.2 1.73 1.12 1.04 1.08 0.5 3.37 1.75 1.25 1.38
 Table 19.5 Installed air power and energy consumption for different types of demand/supply variation

Table 19.5 shows that the demand/supply variations have a considerable influence upon both the installed air power and the energy consumption. From the economic point of view, re­ducing the flow peaks and troughs can give good returns.

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