MIET2072_C1

7/21/2019 MIET2072_C1

http://slidepdf.com/reader/full/miet2072c1 1/38

School of Aerospace, Mechanical and Manufacturing Engineering

of 38

Topic 1: DESIGNING MECHANICAL

ENGINEERING SUPPLY SYSTEMS INVOLVING

CONSIDERATIONS OF LOAD PROFILE,STORAGE AND LOAD FACTOR

Learning Outcomes

Upon successful completion of this topic you will be able to:

• describe the effects of load profile and storage on component sizing inelectricity, water, air, and mechanical rotational energy supply systems

•

describe load factor and its effects on correctly estimating the load on asupply system

• determine a factory’s compressed air load profile, including consideration ofthe load factors of partially used tools and large peak demands

• determine the minimum required air delivery capacity of the compressor

• determine the required compressed air storage (“receiver”) size

• determine a flywheel’s required moment of inertia for energy storage

• describe the stresses in a simple disc flywheel.

Introduction to the Topic

Engineers of all sorts (mechanical, civil, electrical, chemical) are called on to designsystems to supply some reticulated commodity such as compressed air, steam,water or electricity. Mechanical engineers are also called on to design systemsdelivering rotational mechanical energy in which flywheels are frequentlyincorporated. In this topic the particular focus is on a compressed air system, butby considering other supply systems we can see some useful commonality betweenthem.

In several topics, including this topic, the work is at a level of complexity that youwill be able to map out your own problem solving strategy. You will be asked todo this, where appropriate, in the activities section.

For this topic considerable help in defining and refining your strategy will beobtained in reading analogous system design problems in this topic.

In some later topics where complex legally required standards are being dealt with,you will be given more explicit step by step problem solving strategies.

7/21/2019 MIET2072_C1



of 38

Background Skills and Knowledge

Students will require the following:

• Ability to carry out basic integration of areas under graphs

•

Familiarity with and ability to use the perfect gas equation

• Familiarity with and ability to use the principles of conservation of mass andconservation of energy

Acti vi ty 1A – Reading

1. Read all of Chapter 1 below i.e. section 1.1 “Load Profile andStorage”, section 1.2 “Load factor” and section 1.3 “Flywheels”

Chapter 1: Designing Mechanical Engineering

Supply Systems Involving Considerations of Load

Profile, Storage and Load Factor

Section 1.1: Load Prof ile and Storage

In many engineering, and indeed household supply systems, the demand on thesystem is variable. Peaks in demand may occur that would be difficult to meet fromthe ultimate supply source if a reserve quantity was not held in a store somewherein the system. Some examples of such systems are:

(a) Domestic hot water supply systems

Hot water is required in small or large quantities for dishwashing, clothes washing,showers and baths, separated by periods of no usage. For the purposes ofperforming tests on hot water systems, there are standard draw off patternsavailable.

If the water is heated as it is required, then the thermal energy flow involved inheating the water for filling a bath or washing machine is quite high as can be seenfrom the following basic calculation:

7/21/2019 MIET2072_C1


7/21/2019 MIET2072_C1



of 38

To help analyse storage requirements it is useful to draw graphs of rates ofrequirement (demand) and rates of supply versus time. For the situation justdescribed this is done in a simplified form in Figures 1.1a and 1.1b. Figure 1.1awhich shows the rate of “usage” for thermal energy (including heat losses from thestorage tank) could be described as a load profile.

Fig. 1.1a: Rate of Energy “Usage” or Demand from Tank © RMIT University, 2013

Fig. 1.1b: Rate of Energy Supply to Tank © RMIT University, 2013

7/21/2019 MIET2072_C1


7/21/2019 MIET2072_C1



of 38

(i) a reduction in capital cost of the alternator

(ii)

a reduction in the amount of engine power consumed by the alternator,thereby leaving more engine power available for propelling the vehicle.

The power out and power in characteristics for the car are very similar to those forthe off-peak domestic hot water system in two important respects namely that:

(i)

the main demand occurs at a time when the recharging device is inoperative,hence the storage device must have sufficient capacity to meet the maindemand

(ii) the recharging device can operate for more units of time than the time takenby the main discharging operations, hence the recharging device (electric

heating element or alternator) can be of smaller capacity and so save oncapital cost.

(c) Supply of water to a city

City water supply systems typically have a distant large primary reservoir in somerainy hilly area, from which very large pipes distribute the water to smallersecondary reservoirs in the city, in some circumstances requiring a pump if areas ofthe city are higher than the primary reservoir. From these secondary reservoirs

smaller pipes distribute the water to the consumers as shown schematically inFigure 1.3.

Fig. 1.3: Schematic of City Water Supply © RMIT University, 2013

Flow rate to the consumers from the secondary reservoirs is variable and mostly

during the hours when people are awake. A load profile graph could be drawnsuch as the simplified one in Figure 1.4a. At the secondary reservoir small peaks

7/21/2019 MIET2072_C1



of 38

caused by somebody showering in one house would tend to be smoothed out byother small troughs, however an overall increase in flow will occur as people get upin the morning. Part of the flow to the consumers in evening is any loss fromdripping taps they may have. The flow rate from the primary reservoir canhowever be at a steady smaller rate than the combined peaks of the outflow fromthe secondary reservoirs, as it has 24 hours in which to occur. This is shown inFigure 1.4b. There are definite advantages, as regards capital expenditure, achievedby having the flow rate from the primary reservoir smaller than the peak flow rateto the consumers. These advantages include: minimising pipe sizes on the routefrom the primary to secondary reservoirs, which is often a long route; andminimising the size of pumps and associated electrical connections.

Fig. 1.4a: City Water Demand = Flow from Secondary Reservoirs © RMIT University, 2013

Fig. 1.4b: Flow from Primary Reservoirs = Supply to Secondary Reservoirs © RMIT University, 2013

7/21/2019 MIET2072_C1



of 38

Integrating to obtain the area under each graph, which have axes of volume/time(or mass/time) and time, yields volume (or mass) demand and supply from graphs1.4a and 1.4b respectively. From conservation of mass it can be said that the areaunder one graph must equal the area under the other if the secondary reservoirshave returned to their original level after 24 hours.

If one graph is overlaid on top of the other as is shown in Figure 1.4c, it can be seenthat the time where the demand exceeds the supply is that during which thesecondary reservoirs are being depleted; and conversely when the demand is lessthan the supply the secondary reservoirs are being refilled. It can therefore bereasoned that the area D representing the depletion of volume (or mass) stored inthe secondary reservoirs must be equal to the areas F 1 + F 2 representing the volume(or mass) that refills them.

Fig. 1.4c: Water Flow Rate to (- - - -) and from ( ) Secondary Reservoirs © RMIT University, 2013

It can also be reasoned that the minimum size of the secondary reservoir systemthat could be tolerated is the volume represented by D, or F 1 + F 2, since D = F 1 + F 2.However, in reality the system must be able to go on supplying the city in the eventof an interruption to the flow in the pipes and/or pumps between the primary and

secondary reservoir caused by mishap or for maintenance, and therefore it isprudent to have secondary storage capacity greater than the minimum indicated.

(d) Electricity supply for a state

Electric generators may be driven by a variety of prime movers such as steamturbines, gas turbines, water (hydro) turbines and wind turbines. In the state ofVictoria the largest portion of power generation is by steam turbines supplied bybrown coal fired boilers.

7/21/2019 MIET2072_C1



of 38

The electricity load profile may be usefully plotted against hour of day, as in Figure1.5.

Fig. 1.5 Power output in summer and winter © RMIT University, (Dixon C., Marchiori G.) 2013

The peaks in demand, that occur in morning and evening and a re more obvious in

the winter graph, result from increased cooking, heating and use of electric trainsat these times. In summer the running of air conditioning equipment contributessignificantly to peaks. To meet these peaks, sufficient generating plant must havebeen purchased and installed, so it is desirable, if capital expenditure is to beminimised, that the load by shifted where possible from the peaks to the troughs.Moving load to the troughs also has the desirable effect of making fuller use of thepurchased generating plant.

To encourage consumers to move some of their electricity demand to the trough or

“off-peak” period, electricity supply companies may offer the electricity at a lowertariff during this period. Economically this is regarded as having a net advantagedue to the significant reduction in capital borrowing required. Techniques for usingoff-peak power usually involve storage of heat or “coolth” such as domestic hotwater tanks, cast iron block storage heaters for space heating, and chilled water orice storage for space cooling. These stores are “charged” during the off-peak periodand “discharged” when the hot water or space heating / cooling are required.Figure 1.6 contrasts, in simplified form, the generating capacity requirements withand without off-peak storage of heat and “coolth”.

7/21/2019 MIET2072_C1



of 38

Fig. 1.6 Generating capacity requirements with and without off-peak storage © RMIT University, 2013

The reduction in required generating capacity at the power station, achieved by the

incorporation of these storage devices, is analogous to the reduction in the requiredsize of pipes and pumps installed between the primary and secondary reservoirs,achieved by the use of adequately sized secondary reservoirs in the city watersupply system. In both cases a reduction in capital expenditure has been achievedby the supply authority/company, and the capital is being more profitablyemployed since the primary plant is being well utilised (e.g. in the case of thepower stations, rather than generators switching off during the off-peak period,some of them are providing energy to re-charge the off-peak hot water heaters).

The area F1 + F2 represents the energy put into storage as heat or “coolth” and the

area D1 represents the energy leaving the storage (both usefully and as losses).From conservation of energy F1 + F2 = D1.

Coal fired power stations require some 16 hours to reach capacity from a cold start,so it is best that they be running steadily to meet the base demand. Gas turbinepower stations are more flexible and hydro-electric power stations are much moreflexible, being able to go from zero to full capacity in about 3 minutes, so these twotypes can be used to help deal with the fluctuating demand. Another form ofstorage is therefore possible in this context: water can be pumped back up to the

hydro reservoirs during the off-peak period using spare output from the coal fired

7/21/2019 MIET2072_C1



of 38

stations to provide the electricity to run the pumps. Figure 1.7 shows such hydropump/turbine-motor/generator machinery.

Fig. 1.7 Water turbine © RMIT University, (Dixon C., Marchiori G.) 2013

Figure 1.8 graphs, in simplified form, power for the two scenarios of with andwithout pumped storage.

Fig. 1.8 Power generation requirements with and without pumped storage © RMIT University, 2013

Areas F 3 + F 4 represents the energy pumped into the hydro storage, area D2 represents this same energy removed from the hydro storage, and area D3

7/21/2019 MIET2072_C1



of 38

represents the consumption of hydro stored energy that was supplied by rainfallreplenishment of the dams. The required capacity of the coal fired power stationshas gone up from C 1 to C 2, and in both cases they are running steadily, however thegas turbine capacity previously used to help with the peak is no longer required.

(e) Compressed air supply systems

Compressed air is a popular medium for powering labour saving tools, and controlequipment, in factories and workshops. The compressed air is commonly deliveredat pressures around 700 kPa gauge (approximately 100 psi gauge) having beingraised to this level from atmospheric pressure by some sort of compressor.Obviously the volume of the air is reduced at it is compressed. In order to havesome degree of standardisation when describing air volume flows used by toolsand delivered by compressors it is customary to use units of “free” air flow rate

which is here given the symbol f V

. This is the volume flow rate that would bemeasured if the air in question was at atmospheric pressure and temperature. Unitsin every day use are litres/sec., m3/hour, and cubic feet/min. [cfm]. It is of coursealso possible to use mass flow rate. Consider now a pipe containing air at 700kPa.g,at atmospheric temperature T amb, with a compressed (“restrained” – not “free”)

flow rate r V . From the perfect gas law:

pV = m RaT

Where:

p is absolute pressure [N/m2]

V is volume flow rate at that pressure [m3/sec]

m is mass flow rate [kg/sec]

Ra is the gas constant for air [Joule/(kg.K)]

T is the absolute temperature [K]

With subscript f for free and r for restrained (not free).So, treating air as a perfect gas, from conservation of mass:

m =

pipein

restrained

amba

r r

T R

V p

=

pressure

catmospheri

atfreeif

amba

f f

T R

V p

eqn. 1.1

∴ f V = r

f

r V p

p eqn. 1.2

7/21/2019 MIET2072_C1



of 38

When designing a compressed air system for a factory an assessment must be madeof the likely rate of usage of air in the factory taking into consideration theknowledge that it is most unlikely that all the tools will be used at full capacity allof the time, but nevertheless certain peaks in demand will occur. (The issue ofallowing for the usual case that not all the tools will be used at full capacity all ofthe time will be discussed further in the next section of this chapter.) Having donesuch an assessment, a load profile graph can be drawn showing demand versustime in a similar manner to those for the various supply systems hitherto described.It is left for the student to do the load assessment and graph as part of the ProjectPart A for this course.

The existence of brief peaks in demand could be met by having extra compressorcapacity for those brief times, but it is more usual to cater for brief peaks byincorporating storage into the system in the form of an air “receiver” which

supplies the extra air the compressor is not big enough to supply at that moment. Itis left to the student to contemplate the capital cost and electric usage implicationsof these two strategies.

When the peak is passed, if the compressor is appropriately sized, it will havesufficient capacity to recharge the receiver in the time available before the nextpeak. The minimum possible size compressor would have just completed thisrecharging process the instant the next peak in demand commenced, i.e. thecompressor would be running continuously. As was the case in the previoussupply examples that we have discussed, graphs of demand, and supply from thebasic source, can be usefully drawn on common axes and consideration given toareas under graphs and between graphs. It is left for the student to do this as partof the Project.

During the storage discharge process, when the air demand exceeds that beingsupplied by the compressor, the pressure in the air receiver falls, and obviously itmust not fall too low or some of the air powered tools in the system will not workproperly. Conversely, when the peak is past and the compressor has spare capacity,it can recharge the receiver, which it does by building up the pressure therein.

Obviously the pressure must not exceed the design pressure of the air receiver forlegal and safety reasons as explosion of such vessels can be very destructive. Thedesign of such pressure vessels is the subject of Chapter 5 and builds on your SolidMechanics 3 studies.

The mathematical inter-relationships of pressure in the receiver, time elapsed,storage volume, and inflow and outflow from the receiver, are important fordetermining the necessary size of the receiver and analysing its behaviour. Anequation will now be developed to express these inter-relationships. Shown in

Figure 1.9 is the receiver at the beginning and end of a time interval Δtime. Thereceiver has an actual volume, occupied by the restrained air and represented by

7/21/2019 MIET2072_C1



of 38

Vrr. Air flows into the receiver at mass flow rate im and out at om . At the start of the

time interval the mass of air contained in the receiver, and its absolute pressure arem1 and p1 respectively and at the end of the time interval m2 and p2

Fig. 1.9 Receiver at the beginning and end of a time interval © RMIT University, 2013

From conservation of mass:

Change in stored mass = Net inflow of mass

12 mm − = ( ) time x ∆− oi mm

eqn. 1.3

Taking air to be a perfect gas:

rr V p1

= 11 T Rm a ∴

1m =

1

1

T R

V p

a

rr eqn. (a)

rr V p2

=22

T Rma

∴ 2m =

2

2

T R

V p

a

rr eqn. (b)

The mass flows may be related to the free air volume flows by the perfect gas law,as shown in equation 1.1.

i.e. im =amba

fi f

T R

V p

eqn. (c)

7/21/2019 MIET2072_C1



of 38

ando

m =amba

fo f

T R

V p

eqn. (d)

Substituting from equations (a), (b), (c) and (d) into equation 1.3:

⇒ 1

1

2

2 T R

V p

T R

V p

a

rr

a

rr − =

−

amba

fo f

amba

fi f

T R

V p

T R

V p

timex ∆

Assuming the temperature of the air in the receiver is constant, and equal toambient, i.e.

ambT T T == 21

⇒ rr rr V pV p

12 − = ( )

fo f fi f V pV p − time∆

∴ time∆ =( )

( )

12

fo fi f

rr

V V p

V p p

−

− eqn. 1.4

Equation 1.4, or re-arrangements of it, are extremely useful for calculating receiversize and analysing time dependent behaviour of receiver pressures.

In addition to providing reserve compressed air to deal with peaks in demand,receivers also perform the following useful functions:

• In the situation where compressor capacity exceeds demand, pressure buildsup and the compressor cycles off, the stored compressed air in the receiverwill go on supplying the system thereby allowing the compressor to stay offfor a reasonable period and not cycle back on virtually immediately. This isparticularly important if electric motors are being stopped and started eachtime, since too frequent starting of motors causes them to overheat.

•

To reduce pressure pulsations in the piping system.

• To help cool the air and to help separate water and oil mist from the air. [Thetopic of moisture condensing out of compressed air will be considered in thenext chapter.]

7/21/2019 MIET2072_C1



of 38

Supply systems (c) Water for a city, (d) Electricity for a state,

and (e) Compressed air for a factory

For these three supply systems there are five important common points to begained from the above discussions:

(i)

It is extremely useful to draw graphs showing load (demand) and basic supplycapability as a function of time. Such graphs highlight peak requirements andgive insight into storage requirements and possibilities.

(ii) Consideration of areas on the graphs, and application of the principles ofconservation of mass or energy as appropriate, assist with quantification of

aggregate requirements and storage requirements.

(iii) If the demand peak is not completely out of phase with the basic source ofsupply then storage devices such as secondary reservoirs, pumped hydrostorage, or air receivers, do not in principle have to be of sufficient size to meetthe whole peak demand on their own since the basic source of supply is inoperation during the peak period as well, i.e. the primary reservoir, the coalfired power station or the compressor.

(iv)

Use of storage can lower the peak capacity requirement of the basic source ofsupply, e.g. incorporation of adequately sized secondary reservoirs reduced therequired pipe and pump sizes feeding from primary to secondary reservoirs;incorporation of off-peak hot water stores and “coolth” stores reduced theamount of power generating plant that had to be installed; and incorporation ofthe air receiver reduced the size of air compressor required. Reducing therequired capacity of basic supply items such as: pipes and pumps; boilers,turbines and generators; air compressors and associated motors should reducethe overall capital cost of the systems since they are expensive items, andstorage, being inherently simpler, should be comparatively cheaper.

(v)

Use of storage, as well as reducing the capital spent, implies that the machinerythat has been purchased is utilised more of the time, because during periods oflow demand it is usefully engaged in recharging the stores. Hence if themachinery is being used for a large fraction of the time the capital expended onit is engaged profitably for a greater part of the time in producing a usefulproduct.

7/21/2019 MIET2072_C1



of 38

Section 1.2: Load Factor

It has just been noted that for a machine on which capital has been expended,minimising the idle time, and/or the time in which it is working at less than fullcapacity, will usually maximise the profit the machine contributes to.

The term load factor is used in this context to describe the ratio of the averagegenerator load, or compressor load during a given period of time (say a week or ayear) to the maximum rated load of the machine e.g. generator produced 50,000

joules per year but if running at max rated load continuously it would produce100,000 joules per year ∴load factor is 50%. To make good use of the capitalexpended on the machine this fraction should be as near as 1 or 100% as is possibleconsistent with sufficient downtime to maintain the machine at a level that ensures

safety, reliability and acceptable economic life.

At the demand end of the system the concept of load factor is important in thecontext of good utilisation of capital as has just been discussed but is also importantfor the designer of a supply system to know just how often and how heavily theconsumers are really going to use their water taps, electric toasters, etc., orcompressed air tools. It is usually the case that all the items are not used at fullpower and all at once, and if a system was designed on the assumption that theywere, it would be radically oversized and involve a gross waste of capital.

For example in the case of domestic house wiring, if an electrician connected thefour 10 amp outlets requested in one bedroom using a 40 ampere rated cable backto the switchboard then, for most, if not all of the time, the copper in this cablewould be underutilised because it is extremely unlikely that the combinationwould occur of four items all requiring the full 10 amps and all being operated atthe same time. A more sensible strategy would be to (a) estimate the likelycombinations that can reasonably be expected to occur, say at any one time twooutlets are in use (i.e. 50%) and those two are each delivering 8amps (i.e. 80%oftheir individual capacity) giving 16 amperes in total (i.e. 80% of 50% of 4

x10amperes), (b) fit a cable rated to carry this 16amps safely, and (c) incorporate afuse or circuit breaker to halt the current if it should try to go above the cable’srating.

Electrical wiring Standards give advice on these matters. Such strategies make goodeconomic sense, and are acceptable as long as some costly or dangerous outcomewill not result from the occasional interruption to the power. Clearly for situationssuch as power supply to artificial heart units in intensive care wards of hospitals, amuch more conservative approach is appropriate. In a similar vein to this heart unit

example, civil engineers when designing the floor of a building, or a bridge, mustanticipate the worst possible load that can possibly be put on the structure and

7/21/2019 MIET2072_C1



of 38

design it to withstand said load with a factor of safety greater than 1, because theconsequences of failure, however “occasional”, are unacceptable. On the other handin the design of building heating and cooling systems it is usually regarded as awaste of capital to fit equipment large enough to hold the building temperaturewithin exactly the desired limits on the worst day in winter or summer. Rather,equipment is selected so that it can hold the temperature within the desired limitsfor say 97.5% of the time and for the remainder lets it drift slightly outside thedesired limits. In most buildings no significant ill effects will result from such astrategy, whereas the savings in capital outlay may be substantial. Once again, ifthe ill effects are going to be significant, than a more conservative strategy must beadopted and larger plant installed.

In the case of compressed air systems the designer must, much like the electricianmentioned earlier, make a realistic assessment of the number of tools likely to be

used at any one time and the level of power at which they will operate when beingused. In a factory there may be many compressed air tools. If we assume that theyare all running at any one time, and all running at their full power rating we willtypically radically overestimate the air demand and consequently waste capitalbuying a compressor and motor far bigger than is really necessary. Suchoverestimating will also result in unnecessary capital expenditure on wiring,including the wiring and transformers the power supply company may have toorganise in the street. A more sensible approach combines information on (a) whatpercentage of tools are running at any one time (e.g. “time factor” of 50%) and (b)how heavily those that are running are being worked ( e.g. “work factor” of 80%).

Bringing these 2 factors together gives us a “load factor” of 80% of 50% equals 40%.Using this analysis we see that a compressor and motor 40% of the capacitypreviously thought necessary is what is actually required.

7/21/2019 MIET2072_C1



of 38

Acti vi ty 1B – Pneumatic hand tools

Step 1: Read the e-reserve reference

Please refer to pages 292 to 295 of the Pneumatic Handbook (PDF 197kB) 1

The extract shows typical air consumption rates (in free air units) of various airtools when the tool is supplied with air at 6 bar (600kPag) and used at its fullcapacity.

Step 2: Watch the pneumatic hand tools video

Select and watch either the High or Low resolution video options below.

Topic 1 Hand Tools (high) 2

Topic 1 Hand Tools (low) 3

Note: Right click (Google Chrome browser) and choose Save Video As… to download the video.

Section 1.3: Flywheels

Mechanical engineers are sometimes called on to design flywheels as parts ofsystems delivering rotational mechanical energy. The flywheels store and releaseenergy as the torque fed into or out of them varies with angular position. Thetechniques used for analysing and designing such flywheel systems have much incommon with those used for air receivers. Examples of flywheel energy storageoccur in both work consuming machinery and work producing machinery.

Fig 1.10 and 1.11 illustrate two cases of work consuming machines in which a flywheel helps an electric motor to get through the peak torque demands imposed onit. During troughs in torque requirement the electric motor has spare torque

capacity which can be used to accelerate the flywheel slightly thereby storing someextra kinetic energy. During peaks in the torque requirement the flywheeldecelerates thereby giving up some kinetic energy and hence helping the electricmotor do the necessary rotational work to get through the peak. It may hence be

1 Warring, RH., 1982, pp.292-295, Pneumatic handbook, 6th ed., Trade & Technical Press, Morden, Surrey, England, viewed

on 28th August 2013 <https://equella.rmit.edu.au/rmit/file/56debd2e-caad-4184-bfcc-d7f94597bce3/1/130828_3_077.pd>

2 https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ht-high.mp4 3 https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ht-low.mp4

https://equella.rmit.edu.au/rmit/file/56debd2e-caad-4184-bfcc-d7f94597bce3/1/130828_3_077.pdf



https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ht-high.mp4


https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ht-low.mp4





7/21/2019 MIET2072_C1



of 38

possible to reduce the size of the electric motor and associated electric supplywiring. (The motor of course must still be of a sufficient size such that it has enoughtorque to be able to start the machine, i.e. overcome the inertia of the moving parts.In the case of the compressor the required starting torque can be reduced by thetechnique of holding the suction valves open until the compressor has beenbrought up to its operating speed. In the case of the press no work piece should beput into the press until the machine has been brought up to speed.)

Fig. 1.10 Motor flywheel compressor combination © RMIT University, (Dixon C., Marchiori G.) 2013

Fig. 1.11 Motor flywheel metal punch or press combination © RMIT University, (Dixon C., Marchiori G.) 2013

7/21/2019 MIET2072_C1



of 38

Fig 1.12 illustrates a case of a work producing machine in which a fly wheel helps areciprocating engine to get through troughs in torque production and therebytogether they supply a relatively smooth torque to the electric generator at a moresteady rotational speed than would be the case if a flywheel was not present. Duringpeaks in engine torque production the engine has spare torque capacity which can beused to accelerate the flywheel slightly thereby storing some extra kinetic energy.During troughs in the torque production the flywheel decelerates thereby giving upsome kinetic energy and hence helping the engine do the necessary rotational work toget through the trough and together supply a relatively steady torque to the generatorat a relatively constant speed.

Fig. 1.12 Engine flywheel generator combination © RMIT University, (Dixon C., Marchiori G.) 2013

Acti vi ty 1C – Video on Recip rocating Engines, Compressors &Flywheels

Refer to the video chapter on Small reciprocating compressors (11.05).


Topic 1 Reciprocating Engines, Compressors & Flywheels (high) 4

Topic 1 Reciprocating Engines, Compressors & Flywheels (low) 5

4 https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ecf-high.mp4 5 https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ecf-low.mp4

https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ecf-high.mp4


https://www.dlsweb.rmit.edu.au/set/Videos/MIET2349/Topic1-ecf-low.mp4




7/21/2019 MIET2072_C1



of 38

Note: Right click (Google Chrome browser) and choose Save Video As… to download the video.

Acti vi ty 1D – Video on Recip rocating Engines, Compressors &Flywheels

Refer to the video chapter on Otto cycle engine (4.35) and Car flywheelpressure plate (8.52).


Topic 1 Reciprocating Engines, Compressors & Flywheels (high) 6

Topic 1 Reciprocating Engines, Compressors & Flywheels (low) 7

Note: Right click (Google Chrome browser) and choose Save Video As… to download video

Fig 1.13 illustrates another case of a work producing reciprocating engine in whicha fly wheel could be used to help the engine get through troughs in torque production and thereby together supply a relatively smooth torque to the drivenmachine whatever that might be (assuming it is of a type that requires a smoothtorque and a steady speed). A steam engine represents one of the earliestapplications of flywheels in mechanical engineering and the RMIT has a fineexample of such a machine. Following figure 1.13 are some links to videos about

the action of the fly wheel in this machine.








7/21/2019 MIET2072_C1



of 38

Fig. 1.13 Single cylinder double acting reciprocating steam engine © RMIT University, (Dixon C., Marchiori G.)2013

Acti vi ty 1E – Video on Reciprocating Engines, Compressors &Flywheels

Refer to the video chapter on Steam engine (0.5).


Topic 1 Reciprocating Engines, Compressors & Flywheels (high) 8 video

Topic 1 Reciprocating Engines, Compressors & Flywheels (low) 9 video

Note: Right click (Google Chrome browser) and choose Save Video As… to download video

The first step in analysing these systems (with the ultimate aim of determiningwhat amount of flywheel storage is needed) is to draw a torque versus crank anglediagram. The torque should be in Nm and the crank angle should be expressed inradians. Note that a radian measurement is a non dimensional unit being a ratio ofarc length to radius. Using this unit system means that integration of the torquecurve will give rotational work in Nm, i.e. Joules. Such a diagram for a singlecylinder double acting steam engine is illustrated in Fig 1.14 and will now beconsidered. This same general technique can be used for other engine/flywheelsystems.

Since the engine is single cylinder no torque will be generated by steam pressure onthe piston when the piston is at top or bottom dead centre. The net torque includesthe combined effect of steam pressure on the piston, friction and inertia of thepiston and connecting rod system. Regarding inertia-in the first part of a singlecylinder steam engine’s motion, as the piston moves away from top dead centre,the piston’s inertia is opposing acceleration and thus reduces the amount of steam

force getting through to the crank. Conversely, towards the end of the stroke thepiston resists deceleration and effectively adds to the steam force getting through tothe crank.








7/21/2019 MIET2072_C1



of 38

Fig. 1.14 Torque (components) Vs Crank angle © RMIT University, (Dixon C., Marchiori G.) 2013

7/21/2019 MIET2072_C1



of 38

Fig. 1.15 Net Torque Vs Crank angle © RMIT University, (Dixon C., Marchiori G.) 2013

The next step is to find the equivalent steady torque that would do the sameamount of work over one complete cycle that is done by the variable torque. To dothis we integrate the variable torque curve, thereby finding the amount of workdone in one cycle (in Joules). We then divide that work quantity by the angulardistance turned by the crank in one cycle (Π in this case) to obtain the value of thesteady torque that would do the same amount of work in one cycle. This value canthen be drawn as a horizontal line on torque versus crank angle diagram asillustrated in Fig 1.15.

7/21/2019 MIET2072_C1



of 38

Fig. 1.16 Flywheel energy exchanges © RMIT University, (Dixon C., Marchiori G.) 2013

The next step is to superimpose the graph of net steady torque onto the graph of

net variable torque, as illustrated in fig 1.16.Consider now that part of the cycle where the torque from the engine is greater

than the steady torque that it is desired to feed out to an electric generator. Duringthis phase the surplus torque accelerates the flywheel. The increase in kineticenergy of the flywheel ΔE is represented by the area contained between thevariable torque curve and steady torque line This ΔE is the work done on theflywheel by the surplus torque during the acceleration phase. By carrying out thisarea determination i.e. carrying out the integration:∫ [Variable Torque – Steady Torque]dθ we thus obtain ΔE in Joules.

Having obtained ΔE by this integration we then equate it to the expression for thechange in kinetic energy, being ½ Iωmax 2 - ½ Iωmin 2 where ωmin and ωmax are the

7/21/2019 MIET2072_C1



of 38

rotational speeds at the beginning and end of the acceleration phase, and I is themoment of inertia of the flywheel.

Similarly consider now that part of the cycle where the torque from the engine isless than the steady torque that it is desired to feed out to an electric generator.

During this phase the flywheel decelerates, feeding work out to the generator tocompensate for the inadequacy of the engine’s torque. The decrease in kineticenergy of the flywheel ΔE during this phase is represented by the area containedbetween the steady torque line and the variable torque curve below it.

During the next part of the cycle when there is again a surplus of engine torque theflywheel is recharged.

Returning now to the value of ΔE obtained from the integration:

ΔE = ∫ [Variable Torque – Steady Torque]dθ

And substituting it, together with the tolerable lower and upper rotational speeds

ωmin and ωmax, into the equation:½ Iωmax 2 - ½ Iωmin 2 = ΔE

we can obtain the required moment of inertia of the flywheel I.

It is worth observing that the tolerable variation in rotational speed is analogous tothe tolerable variation in air pressure in the air receiver previously studied, and themoment of inertia of the flywheel I is analogous to the receiver volume Vrr .

Having obtained the value of required moment of inertia of the flywheel I, one candecide on its proportions taking into consideration the radial and axial spaceavailable and the tolerable mass. For a disc of uniform thickness, with a central

hole, with external radius b, internal radius a and mass m , the moment of inertia Iabout the central axis is ½ m(b2 +a2)

7/21/2019 MIET2072_C1



of 38

In the case of a pressure vessel used to store gaseous elastic energy, the designerneeds to make sure the vessel is strong enough to withstand the stresses caused bythe contained pressure. Similarly in the case of a flywheel used to contain kineticenergy the designer needs to make sure the flywheel is strong enough to withstandthe stresses caused by centrifugal force. Figure 1.17 shows a flywheel treated in asimilar manner to the thick walled pressure vessels studied in Solid Mechanics 3

Figure 1.17 Rotating Disk © RMIT University, (Dixon C., Marchiori G.) 2013

By writing the condition of equilibrium of element ABDC isolated from the disc bytwo meridian sections and two concentric cylinder surfaces we must take intoaccount inertial force shown in Fig 1.17 as “di”. The disc is of unit depth into thepage.

( ) ( ) [ ]

( )

2

22 2 2

3

3

11

:

1

i

kgd w r dr d m m N

m S

where

kgdensity of materialm

w angular velocityS

ρ

ρ

= ⋅ ⋅ ⋅ ⋅ Θ ⋅ ⋅ =

=

=

In this case the equilibrium equation obtained in Solid Mechanics 3 for a thickwalled pressure vessel can be replaced by the relation

2 20t r

d r r w r

dr

σ σ σ ρ − − − ⋅ ⋅ =

7/21/2019 MIET2072_C1



of 38

Acti vi ty 1F(Optional)– Read the onl ine resource

For a detailed development of the solution read the PDF excerpt fromStrength of materials pp. 214-223 (PDF 420KB). 10

Making the consolidation:

; ;a r

x bwb b

α ν = = =

leads to:

( )2

2 2 2

2

22 2 2

2

3 1 31

8 3

31

8

t r

r

p x f xx

xx

µ µ α σ ν α σ

µ

µ α σ ρν α

+ += + − + = +

+= + − −

Max value of r σ is for:

( )22

max

3

18r

ax r ab

bα

µ

σ ρν α

= = ⇒ =

+

= −

Stress tσ is also positive for all values of x and becomes maximum at the surface of

internal hole (x = a)

2 2

max

3 11

4 3t

µ µ σ ρν α

µ

+ −= +

+

From comparison of equations for r σ and tσ we can see that tσ is always greater

than r σ .

Forming even the smallest hole in a disc significantly increases the stresses.In a solid disc (without the hole) r σ and tσ are equal and max at the centre where

r=0

10 Timoshenko, S, 1956, Strength of materials. Part II, Advanced theory and problems, pp. 214-223, 3rd ed.,Huntington, New York. <https://equella.rmit.edu.au/rmit/file/c1e58d32-26fb-4454-8928-

1a07dd7e1251/1/130823_7_1.pdf> viewed 27th August 2013.

https://equella.rmit.edu.au/rmit/file/c1e58d32-26fb-4454-8928-1a07dd7e1251/1/130823_7_1.pdf










7/21/2019 MIET2072_C1



of 38

Thus for a solid disc:

( )2 2

2 2

31

8

3 1 31

8 3

r

t

x

x

µ σ ρν

µ µ σ ρν

µ

+= −

+ += −

+

as beforer

xb

=

Both stresses are positive for all values of x and increase towards the centre where

2

max

3

8t r mac

µ σ σ ρν

+= =

Thus in a disc with a very small central hole, stress tσ at the edge of the hole is twotimes greater than at the centre of solid disc because of stress concentration, (SeeFig. 1.15)

Figure 1.18 © RMIT University, (Dixon C., Marchiori G.) 2013

7/21/2019 MIET2072_C1



of 38

Acti vi ty 1G – Graphs and Calculations

Having read Chapter 1, carry out the following three short practice exercises,checking your answer against the answer given. Consult your tutor if, aftermaking a serious attempt, you are still having difficulties.

Exercise 1: Read the scenario then complete calculations for (a), (b), and (c)

A building has a cooling load of 200 kW (thermal) from 7 am to 5 pm. Therefrigeration equipment used to do the chilling has a coefficient of performanceof 4. Electricity is available at 2 tariffs:

Off peak from midnight to 6 am $0.03 per kilowatt hourOn peak (day rate) 6 am to midnight on $0.12 per kilowatt hour

(a) For the case where no storage is used, calculate:

(i) the power (in kilowatts) of the electric motor driving the chiller

(ii) the daily running cost of the electric motor.

(b) For the case where the chiller only runs during the off-peak period calculate:

(i) the size of the thermal store (in kilojoules)

(ii) the power (in kilowatts) of the electric motor driving the chiller

(iii) the daily running cost of the electric motor.

(c) For the case to the smallest possible chiller is used, i.e. runs for 24-hours,calculate:

(i) the size of the thermal store (in kilojoules)

(ii) The power (in kilowatts) of the electric motor driving the chiller

(iii) The daily running cost of the electric motor

Discuss how you would determine which option was economically superior.

Exercise 2 : Read the scenario then complete the calculations

An inventor has developed a single cylinder double acting piston engine, thepiston of which is driven back and forth by an organic vapour boiled off within a

7/21/2019 MIET2072_C1



of 38

solar collector. The crankshaft of the engine is to be connected to an electricgenerator. As currently running the crankshaft rpm is too variable forsatisfactory electricity generation. A mean speed of 1000rpm is desired and themaximum and minimum are to be 0.3 % above and below this respectively. Thevariation in crankshaft speed is caused by the fact that the torque delivered bythe piston to the crankshaft is not constant. You work for a consultingengineering firm which has been approached by the inventor wanting some helpwith designing a flywheel to act as an energy store and hence help smooth thetorque output downstream of the flywheel.

The torque delivered to the crankshaft, by the action of the high pressure vapouron the piston, may be approximated as rising from 0Nm to 100Nm as it passesthrough top dead centre, then fixed at 100Nm while the shaft turns through anangular displacement of π / 3 from top dead centre, then steadily falling from100 to 20Nm as the shaft turns through the next π / 3, then fixed at 20Nm as the

shaft turns through the nextπ

/ 3 to arrive at bottom dead centre, where thetorque momentarily falls to zero. The approximate pattern then repeats with thetorque rising from 0Nm to 100Nm as it passes through bottom dead centre.

Exercise 2 calculations:

Recalling that the energy stored in a flywheel is ½ I ω2 determine the requiredsecond moment of mass I that the fly wheel must have. Include a graph ofTorque versus angular position, and discussion of it, as part of your solution.

Exercise 3: Read the scenario then complete the calculations

A punching operation, performed by a cranked punch, requires 300J of workwhich is consumed during 30° of crankshaft movement. The torque pattern isrising steadily (from 0Nm of torque) for 10°, constant for 10° and then fallingsteadily (back to 0Nm of torque) for a further 10° of crankshaft movement. Theremaining 330° of crankshaft movement involves no work. The shaft on whichthe flywheel is fitted rotates at a mean speed of 200rpm, and is permitted to vary± 5% in this application.

Exercise 3 calculations:

Determine the necessary moment of inertia the flywheel must have, given thatthe other parts of the machine contribute 1.5 kg.m2 referred to the flywheel shaft.

7/21/2019 MIET2072_C1



of 38

Acti vi ty 1G Answers

Exercise 1 Answers:

(a) (i) 50kW (ii) $60

(b) (i) 7,200,000kJ (ii) 83.3kW (iii) $15

(c) (i) 4,200,000kJ (ii) 20.83kW (iii) $48.75

Exercise 2 Answer:

Answer 0.793kg.m2

Exercise 3 Answer:

Answer 4.75 kg.m2

7/21/2019 MIET2072_C1



of 38

Activ ity 1H - Reading, Reflect ions and Graphs and Calculations

Step 1: Revision

Re-read the summarisation at the end of section 1.1 which lists the importantcommon points and analogous behaviour and design techniques described forwater supply systems, electric supply systems and compressed air supplysystems.

Step 2: Refer to Assessment Section

Refer to Project Part A in the Assessment section and read the Introduction,LOAD DEFINITION and Question 1. Note particularly the point made in the lastparagraph of the introduction. Design frequently involves reflection on issues,prior and subsequent to making calculations, which enables you to makeinformed decisions

Step 3: Strategy and action for completion of Question 1 of Project Part A

Discuss with your partner your strategy for tackling question 1, and then:complete question 1 in cooperation with your partner.

Feedback

Feedback will be provided on your submitted project documentation by theengineering lecturer/tutor responsible for marking it.

7/21/2019 MIET2072_C1



of 38

Activ ity 1I – Receiver Size - Reading, Reflections and Graphs

and Calculations

Step 1: Refer to Project Part A in Assessment

Refer to Project Part A in the Assessment Section. After you have completedQuestion 2 (which is done as part of topic 2) read: the preamble to question 3entitled RECEIVER SIZE and then read Question 3

Step 2: Revise Chapter 1 Section 1.1(e)

Re-read Chapter 1 Section 1.1(e) noting particularly equation 1.4 derived from

considerations of the change in the stored mass of air in a receiver.

Step 3: Complete Question 3

Complete Question 3 consulting with your partner on that part of the questionyou are doing jointly.

Feedback

Feedback will be provided on your submitted project documentation by the

engineering lecturer/tutor responsible for marking it.

7/21/2019 MIET2072_C1



of 38

Summary and Outcome Checklis t

Engineers of all sorts (mechanical, civil, electrical, chemical) are called on to designsystems to supply some reticulated commodity such as compressed air, steam,water or electricity. Mechanical engineers are also called on to design systemsdelivering rotational mechanical energy. In all of these systems it is likely that thedemand will be variable and storage of some sort can be usefully incorporated tohelp meet that demand.

For these systems there are five important common points:

(i)

It is extremely useful to draw graphs showing load (demand) and basic supply

capability as a function of time. Such graphs highlight peak requirements andgive insight into storage requirements and possibilities. When making anassessment of demand it is important to take into consideration realistic loadfactors so that demand is not seriously overestimated.

(ii) Consideration of areas on the graphs, and application of the principles ofconservation of mass or energy as appropriate, assist with quantification ofaggregate requirements and storage requirements.

(iii) If the demand peak is not completely out of phase with the basic source ofsupply then storage devices such as secondary reservoirs, pumped hydrostorage, air receivers or flywheels, do not in principle have to be of sufficient

size to meet the whole peak demand on their own since the basic source ofsupply is in operation during the peak period as well, i.e. the primary reservoir,the coal fired power station, the compressor or engine.

(iv) Use of storage can lower the peak capacity requirement of the basic source ofsupply, e.g. incorporation of adequately sized secondary reservoirs reduced therequired pipe and pump sizes feeding from primary to secondary reservoirs;incorporation of off-peak hot water stores and “coolth” stores reduced theamount of power generating plant that had to be installed; and incorporation ofthe air receiver reduced the size of air compressor required. Reducing the

required capacity of basic supply items such as: pipes and pumps; boilers,turbines and generators; air compressors and associated motors should reducethe overall capital cost of the systems since they are expensive items, andstorage, being inherently simpler, should be comparatively cheaper.

(v) Use of storage, as well as reducing the capital spent, implies that the machinerythat has been purchased is utilised more of the time, because during periods oflow demand it is usefully engaged in recharging the stores. Hence if themachinery is being used for a large fraction of the time the capital expended onit is engaged profitably for a greater part of the time in producing a usefulproduct.

7/21/2019 MIET2072_C1


Tick the box for each statement with which you agree:

I am able to

describe the effects of load profile and storage on component sizing inelectricity, water, air, and mechanical rotational energy supply systems

describe load factor and its effects on correctly estimating the load on asupply system

determine a factory’s compressed air load profile, including consideration ofthe load factors of partially used tools and large peak demands

determine the minimum required air delivery capacity of the compressor

determine the required compressed air storage (“receiver”) size

determine a flywheel’s required moment of inertia for energy storage

describe the stresses in a simple disc flywheel.

Assessment

This topic will be assessed as part of the Project Part A and the end of semesterexamination (see: Assessment section of the Course Introduction for more detail),

MIET2072_C1

Documents

Transcript of MIET2072_C1