Mechanical Engineering Thermofluids

DET:Mechanical EngineeringThermofluids (Higher)

6485

Spring 2000

DET:Mechanical Engineering

ThermofluidsHigher

Support Materials

HIGHER STILL

DET: Mechanical Engineering: Thermofluids Higher

CONTENTS

Section 1: Thermofluids (Higher) Student Notes

Section 2: Self-Assessment Answers

Section 3: Tutorials

Section 4: Tutorials Marking Scheme

Section 5: Appendix, Quantities used in Thermofluids (Higher)

DET: Mechanical Engineering: Thermofluids Higher

DET: Mechanical Engineering: Thermofluids Higher Support Notes

Section 1:

Thermofluids (Higher) Student Notes

DET: Mechanical Engineering: Thermofluids Higher Support Notes

DET: Mechanical Engineering: Thermofluids Higher Support Notes 1

OUTCOME 1: APPLY GAS LAWS

The wide use made of various gases in the field of engineering makes it necessary topredetermine their reactions when they are heated, cooled, expanded or compressed.

When a process takes place, the changes which will occur in the properties ofvolume, absolute pressure and absolute temperature of the gases are related by thegas laws.

When solving problems utilising the gas laws, both pressures and temperatures mustbe expressed in absolute terms and these are defined thus:

Absolute pressure (Symbol p)Pressure gauges are commonly used to measure pressures in vessels and pipelines andread pressures normally above atmospheric pressure. If a gauge shows a zero readingit means the pressure is atmospheric.

If the pressure in a vessel is increased above atmospheric to a gauge pressure pg, theactual or absolute pressure p in the vessel is given by:

p = pg + patm

i.e. Absolute pressure p = gauge pressure + atmospheric pressure.

In most practical problems, listed pressures will be in absolute terms unless otherwisestated.

The unit for pressure is the N m-2 or Pa (PASCAL).

The bar = 105 N m-2 = 100 kN m-2 is also commonly used.

Standard atmospheric pressure = 1 atm = 1.013 bar = 1.013 x 105 N m-2.

When working through problems, the stated or calculated values for pressure areoften high numbers. In order to express multiples of SI units concisely, theundernoted prefixes are used:

MULTIPLICATION FACTOR PREFIX SYMBOL1,000,000 = 106 Mega M1,000 = 103 Kilo k

e.g. 7,200,000 N m-2 = 7,200 kN m-2 or preferably 7.2 MN m-2.


Absolute temperatureIn problems involving the gas laws, the temperature of any gas is measured fromabsolute zero, which has been determined to be 273C below the zero point on theCelsius scale, i.e. absolute zero. At this point the internal energy of the substance isalso zero.

Absolute temperature is the temperature above absolute zero and is determined byadding 273 to the Celsius temperature scale reading.

i.e. Absolute temperature = Celsius temperature scale reading + 273.

Hence 27C = 300 K.

Absolute temperature takes the SI base unit, the Kelvin (K), and has the symbol T.Note: A change in temperature of 1C = a change in temperature of 1 K. Otherquantities encountered in our thermofluid studies are defined as follows:

MassThis is usually defined as the quantity of matter in a body.Symbol: m (small letter).Unit: kg (small letters).

VolumeSymbol: V (capital letter).Unit: m3.The recommended unit is the cubic metre. Subdivisions such as the cm3 or litre (l)are also used, but as a general rule it is safer to convert data to give volumes in m3 toavoid errors in calculations.

Specific volumeSymbol: v (small letter).Unit: m3 kg-1.This is the volume per unit mass and is the reciprocal of density.

i.e. v =m

V =

MASSVOLUME

Boyles LawThe Irish scientist Sir Robert Boyle investigated the behaviour of gases whenexpanded or compressed under constant temperature (isothermal) conditions. Inessence, Boyles Law states:

For any given mass of a gas, the absolute pressure will vary inverselywith the volume providing that the temperature remains constant.


Thus p V1

or pV = constant

e.g. Doubling the absolute pressure gives half the volume. Three times theabsolute pressure gives one-third of the volume.

Boyles Law can also be expressed algebraically in the form

p1V1 = p2V2 = pnVn..for the mass of gas

Charles LawThe French scientist Jacques Charles conducted experiments on gases where thepressure of a fixed mass of gas was kept constant while variations in the volume andtemperature were examined. In essence, Charless Law states:

During the change of state of any gas in which the mass and pressureremain constant, the volume varies in proportion with the absolutetemperature (Kelvin).

Thus V T orTV

= constant

e.g. At double the absolute temperature, the volume is doubled. At three times theabsolute temperature, the volume is trebled.

Charless Law can be expressed algebraically in the form:

n

n

2

2

1

1

TV

=TV

=TV

..for the mass of gas

Constant volume processWhen a given mass of gas is heated at constant volume, its temperature and pressurewill both increase. Conversely, if the gas is cooled, the temperature and pressure willboth decrease. At any stage of either process, the ratio of the pressure p to theabsolute temperature T of the gas will be a constant. Hence:

EMPERATUREABSOLUTE TPRESSURE

= constant

Tp

= C (Provided neither mass or volume of gas changes)

This is known as the Pressure Law, which may be stated algebraically in the form:

n

n

2

2

1

1

Tp

=Tp

=Tp


The Combined Gas LawIf, during a process, the pressure, volume and absolute temperature of a gas arechanged from p1, V1, and T1 to p2, V2 and T2 respectively, then, provided there is nochange in the mass of gas, Boyles Law, Charless Law and the Pressure Law may becombined to give the algebraic expression:

2

22

1

11

TVp

=TVp

= constant

This is known as the Combined Gas Law.

The Characteristic Gas EquationWe have seen the combined gas law stated in the form:

TpV

= constant C

For a perfect or ideal gas, this constant C = mR where m is the mass of the gas and Ris the Characteristic Gas Constant or Specific Gas Constant.

Hence T

pV = mR

Or pV = mRT

This is known as the Characteristic Gas Equation of an ideal gas. When using thisequation for solving problems, it is essential to express all the terms in appropriateunits which are:

p = absolute pressure of the gas N m-2V = volume of the gas m3T = absolute temperature of the gas [(t + 273)] Km = mass of the gas kgR = gas constant J kg-1 K-1.

The table below lists the gas laws, together with appropriate equations for problemsolving.


GAS LAW PROCESS CONDITION APPROPRIATE EQUATION

Boyles Law Constant Temperature p1V1 = p2V2 etc

Charless Law Constant Pressure

2

2

1

1

TV

= TV

etc

Pressure Law Constant Volume

2

2

1

1

Tp

=Tp

etc

Combined Gas Law Pressure Volume andTemperature all Change

2

22

1

11

TVp

=

TVp

etc

Characteristic Gas Equation Includes Mass and CharacteristicGas Constant R

pV = mRT

The gas constant R, which appears in the Characteristic Gas Equation, is identified as

the Characteristic Gas Constant or the Specific Gas Constant and its value variesfor different gases as will be apparent in the questions covered in the Tutorial forOutcome 1.

Universal Gas ConstantThe Universal Gas Constant takes into account the concept of molecular mass ofsubstances. This constant, symbol Ro, and also known as the Molar Gas Constant,is the product of the relative molecular mass, M, and the Characteristic Gas Constant,R, and has the same value for all gases:

Thus, Ro = MR = 8.3143 kJ kg mol-1 K-1

It follows that the value of the Characteristic Gas Constant R can be found from therelationship

R = MRo

e.g. The molecular mass of nitrogen is 28. What is the value of R for nitrogen?

R = 1-1- K kJ kg297.0 = 283143.8

=

MRo

The universal gas constant is frequently utilised in problems dealing with thecombustion of various gases and it appears in a version of the Characteristic GasEquation called the Ideal Gas Equation. Our studies, however, will be restricted tothe use of the Characteristic Gas Equation, which employs the Characteristic orSpecific Gas Constant.


SELF-ASSESSMENT

Assignment 1

1. Test your knowledge of quantities, symbols and units covered so far bycompleting the table below.

QUANTITY SYMBOL UNIT

SPECIFIC VOLUME

m

m3

ABSOLUTE PRESSURE

K

2. Convert the undernoted temperature values from one scale to the other.

oC K

-150

388

128

162

0

3. Boyles Law deals with the behaviour of gases under isothermal conditions.What does the condition isothermal mean?

Ans:


4. State Charles Law and express the law in the form of an algebraic equation.

Ans:

Equation:

5. When a mass of gas is cooled at constant volume, what effect has this process onits pressure and temperature?

Ans:

6. State the characteristic gas equation for an ideal gas. Identify all terms in theequation and state the correct units for each.

Equation:

Identification of terms and units:


PRACTICAL EXEMPLAR PROBLEMS

Having developed the various gas law equations, we can apply these to the solution ofproblems dealing with the behaviour of gases when subjected to different processes.

Problems may be simplified by employing a method of converting given informationinto symbols and units which can then be fitted into an appropriate equation.Specimen worked exemplar problems adopting this strategy now follow.

Exemplar Question 1A fixed mass of gas is compressed isothermally from a pressure of 20 bar and volumeof 3.2 m3 until it occupies 25% of its original volume. Calculate the final pressure ofthe gas.

Known Datap1 = 20 x 105 N m-2 V1 = 3.2 m3p2 = ? V2 = 3.2 x 100

25

= 0.8 m3

For an isothermal process, Boyles Law applies and equation

p1 V1 = p2 V2 can be utilised.

From p1 V1 = p2 V2 p2 =

p1 x 2

1

VV =

20 x 105 x 8.02.3

p2 = 8.0 x 106 N m-2

FINAL PRESSURE OF GAS, p2 = 8 MN m-2 or 80 bar.

Exemplar Question 2A quantity of gas at a pressure of 180 kN m-2 and temperature 18C, occupies avolume of .43 m3. The gas is compressed until its pressure and temperature are670 kN m-2 and 127C respectively. If there is no loss of gas, what volume will itnow occupy?

Known DataFrom the Combined Gas Law p1 = 180 kN m-2 p2 = 670 kN m-2

T1 = (18 + 273) K T2 = (127 + 273) KV1 = 0.43 m3 V2 = ?

1

11

TVp

=2

22

T Vp

V2 =12

211

TpTVp

=

291 x 10 x 670400 x 0.43 x 10 x 180

3

3

= 0.1588 m3

FINAL VOLUME OF GAS IS 0.1588 m3


Exemplar Question 3An air receiver contains a fixed mass of air at a pressure of 12.5 bar and temperature84C. After a period of time, the pressure is observed to be 7.8 bar. What will be thetemperature of the air?

Known DataFor constant volume process, p1 = 12.5 bar p2 = 7.8 barthe Pressure Law applies T1 = (84 + 273)K T2 = ?Constant Volume Process V1 = V2

2

2

1

1

Tp

=

Tp

T2 = 11

2 T x pp

= 222.768K=357 x 10 x 12.510 x 7.8

5

5

FINAL TEMP. OF AIR = 222.8 273 = -50.2C

Exemplar Question 4Gas is stored in two tanks, A and B, which are connected by a pipe fitted with a stopvalve which is initially closed. Tank A has a volume of 3.0 m3 and contains 14 kg ofthe gas at a pressure of 215 kN m-2 and a temperature of 20C. Tank B has a volumeof 12.0 m3 and contains gas at a pressure of 340 kN m-2 and a temperature of 20C.Determine the characteristic gas constant for the gas and the mass of gas in tank B.

If the stop valve connecting the tanks is then opened, determine the final pressure ofthe gas, assuming the temperature remains at 20C.

V1 = 3.0 V2 = 12.0 m3m1 = 14 m2 = ?p1 = 215 k p2 = 340 kN m-2T1 = 20 + 2 T2 = 293 K

= 293 K

Char. Gas Constant from p1 V1 = m1 R T1

R = 293 x 14

3.0 x 10 x 215TmVp 3

11

11=

GAS CONSTANT R = 157.24 J kg-1 K-1

TANKA

TANKB


MASS IN TANK B from 2222 TRmVp =

m2 = 293 x 157.2412 x 10 x 340

=

TRVp 3

2

22

MASS OF GAS IN TANK B = 88.56 kg

When connecting valve is opened, the total volume V3 = 15 m3 and temperatureremains at 293 K.

FINAL PRESSURE from 3333 TRm=Vp

( )

15293 x 157.24 x 14+88.56

=

VTRm

=p3

333

= 315004.97 N m-2

FINAL PRESSURE IN SYSTEM = 315 kN m-2

Exemplar Question 5A fixed mass of gas contained in a closed system is initially at a pressure of100 kN m-2, a temperature of 15C, and occupies a volume of 0.15 m3. The gas isthen compressed to a volume of 0.06 m3 and a pressure of 300 kN m-2. The gas isthen expanded at constant pressure unit it reoccupies its original volume. If thecharacteristic constant for the gas is 189 J kg-1 K-1, determine the mass of gas in thesystem and the temperature at the end of the compression and expansion processes.

Known Datap1 = 100 x 103 N m-2 p2 = 300 x 103 N m-2 p3 = p2T1 = 15C + 273 = 288 K T2 = ? T3 = ?V1 = 0.15 m3 V2 = 0.06 m3 V3 = V1

m = ? R = 189 J kg-1 K-1

Mass of gas from p1 V1 = m R T1

m = 1

11

TRVp

= kg0.2756=288 x 189

0.15 x 10 x 100 3

MASS OF GAS IN SYSTEM = 0.2756 kg


TEMPERATURE AFTER COMPRESSION from p2 V2 = m R T2

T2 = K345.56=189 x 0.27560.06 x 10 x 300

=

RmVp 322

TEMPERATURE AFTER COMPRESSION = 345.56 273 = 72.56C

TEMPERATURE AFTER EXPANSION from p3 V3 = m R T3

T3 = K863.9=189 x .27560.15 x 10 x 300

=

RmVp 333

TEMPERATURE AFTER EXPANSION T3 = 863.9 273 = 590.9C


OUTCOME 2: SOLVE PROBLEMS USING DATA EXTRACTEDFROM THERMODYNAMIC PROPERTY TABLES

In this outcome we are concerned with the interpretation and extraction of data onthermodynamic properties of working fluids listed in tables as arranged by MessrsRogers and Mayhew.

These tables, commonly known as steam tables, give tabulated values for theproperties of steam and refrigerants over an extensive range of pressures andtemperatures.

The ability to understand and extract data from the tables extends into the solution ofproblems in this outcome and also in Outcome 3 when the steady flow energyequation is dealt with.

Before examining the tables, however, definitions need to be attached to specificthermodynamic quantities listed in the range for this outcome.

Specific volumeThis is the volume occupied per unit mass (1 kg) of a substance and is identified bythe symbol v (small letter).

i.e.v =

m

V=

MASSVOLUME UNIT : m3 kg-1

Thermodynamic tables give the specific volume of dry saturated steam at aparticular pressure under the heading vg.

e.g. SPECIFIC VOLUME OF DRY SATURATED STEAMAT 1.4 bar = 1.236 m3 kg-1

For superheated steam the specific volume is read against the symbol v for differentpressures and temperatures.

e.g. SPECIFIC VOLUME OF SUPERHEATED STEAMAT 6 bar and 250C = 0.3940 m3 kg-1

Internal energyA fluid may be defined as a substance or a mixture of substances in the liquid orgaseous state. All fluids consist of large numbers of molecules that move in randomdirections at high speeds. Each molecule possesses a minute amount of kineticenergy and the total kinetic energy possessed by all the molecules is known as theinternal energy of the fluid.

When heat energy, which is a transient form of energy, is transferred to a fluid, thetemperature and molecular activity of the fluid increases. These increases result in acorresponding increase in the store of internal energy within the fluid.


As a result of experimental work on this subject, Joule concluded that the internalenergy of a fluid is a function of temperature only and is independent of pressure andvolume (Joules Law).

For there to be a change in the internal energy of a fluid there must be a change intemperature.

The symbol used for the total internal energy in a fluid is U and its unit is theJoule (J).

Generally, the internal energy of a fluid is quoted as per unit mass (per kg). Thisquantity is known as specific internal energy and takes the symbol u.

The unit for specific internal energy is the Joule per Kilogram (J kg-1).

Thermodynamic tables give three values for the specific internal energy of steam asunderlisted.

uf = specific internal energy of saturated water

ug = specific internal energy of dry saturated steam

u = specific internal energy of superheated steam.

Flow or displacement energyAny volume of fluid entering or leaving a system must displace an equal volumeahead of itself in order to enter or leave the system as the case may be.

Let the mass of fluid between X and Y in the figure below have a total Volume V1.For flow to occur, this volume must be displaced by an equal volume from outside thesystem. If the pressure in the fluid is p1, then the work done on the fluid inside thesystem by the incoming fluid = force x distance the fluid is displaced.

Flow energy

X Y

p1

p1

p1

p1

S S

VOLUMEDISPLACED

V1

CROSS SECTIONALAREA A1


Now, Force = Pressure x Cross Sectional Area

Therefore Work Done = p1 x A1 x s

But, A1 s = Volume displaced V1

Therefore Work done on the system = p1V1

In specific terms, i.e. per kg of mass,work done on system

= p1 v1 where v equals the specificvolume of the fluid.

This is variously called flow energy, displacement energy or pressure energy.

At entry energy is received by the system.At exit energy is lost by the system.

Hence, specific flow energy = p1 v1(J kg-1)

EnthalpyIn steady-flow thermodynamic systems, internal energy and flow energy are present inthe moving fluid. Accordingly, it is convenient to combine these energies into a singleenergy quantity known as enthalpy, thus

Total Enthalpy = Internal energy + flow energy

The symbol used for enthalpy is H.

Hence H = U + pV

The unit for total enthalpy is the Joule (J).

When considering 1 kg of working fluid, then:

SpecificEnthalpy

= Specific InternalEnergy

+ Specific FlowEnergy

Hence h = u + pv (v = specific volume)

Specific enthalpy h takes the unit The Joule Per kg (J kg-1).

Thermodynamic property tables give four values for the specific enthalpy of steam aslisted below: hf = specific enthalpy of saturated water hfg = specific enthalpy of evaporation hg = specific enthalpy of dry saturated steam h = specific enthalpy of superheated steam.


The formation of steamConsider a quantity of water initially at 0C being heated in a vessel fitted with amovable piston such that a constant atmospheric pressure can be maintained in thevessel. If the water is heated until it has all been converted to steam then thetemperature/time graph would be as illustrated.

During the stage A to B sensible heat energy flows to the water accompanied by a riseof temperature. At B the water boils at a temperature referred to as saturationtemperature. This temperature depends on the pressure in the vessel and is 100C atatmospheric pressure. The energy required to produce this temperature rise is calledthe liquid enthalpy.

During stage B to C steam is being formed whilst the temperature remains constantand the contents of the vessel will be a mixture of water and steam known as wetsteam. At point C the steam will have received all the heat energy required toconvert the water completely to dry steam. The energy required to produce the totalchange from all water to all dry steam is called the enthalpy of evaporation.

When completely dry saturated steam has been formed at saturation temperature,further transfer of heat energy will produce superheated steam which will beaccompanied by a rise in temperature. The amount of heat energy in the superheatphase is called the superheat enthalpy.

Steam, therefore, can exist in three states: wet, dry, or superheated. Values forspecific enthalpy, specific internal energy, and specific volume may be obtaineddirectly from thermodynamic property tables for dry and superheated steam. For wetsteam, it is necessary to know the degree of dryness, or the dryness fraction, of thesteam before the various properties can be calculated.


Dryness fractionThe degree of dryness, or dryness fraction, of steam is that proportion of a givenmass of water which has been evaporated to form steam.

The dryness fraction may have any value from 0 (corresponding to boiling water) to 1(corresponding to dry saturated steam). For example, steam with a dryness fraction0.6 means that for each kg of water, 0.6 will be steam and 0.4 kg of saturated liquid.

The symbol x is used to represent dryness fraction.

Dryness fraction, x = MoistureandSteamofmassTotal

SteamDryofMass

Layout and use of thermodynamic tables for water and steamThe figure below duplicates the information given at the top of page 4 in the Rogers& Mayhew tables. The s figures in the right hand columns of the actual tables relateto entropy values and these are not required for this unit.

[ ] [ ] /kgmv

CT

barP

3g

0s

kJ/kguu gf

kJ/kghhh gfgf

1.0 99.6 1.694 417 2506 417 2258 2675

The various symbols in the eight columns are identified with their quantities and unitsas below.

SYMBOL QUANTITY UNITp

Tsvg

uf

ug

hfhfghg

Absolute pressure

Saturation temperature relating to value of p

Specific volume of dry saturated steam

Specific internal energy of saturated water

Specific internal energy of dry saturated steam

Specific enthalpy of saturated waterSpecific enthalpy of evaporation

Specific enthalpy of dry saturated steam

barC

m3/kg

kJ/kg

kJ/kg

kJ/kg

kJ/kg

kJ/kg


Thermodynamic property tablesWith reference to pages 3, 4 and 5 of the Rogers & Mayhew tables:

The first column headed p is the absolute pressure measured in bar, where1 bar = 1 x 105 N m-2 or 100 kN m-2.

The second column headed Ts is the temperature in C at which the water boils(saturation temperature). Note how Ts changes relative to pressure.

The third column vg is the specific volume in m3 of 1 kg of completely dry saturatedsteam. That is at pressure of 2 bar, saturation temperature is 120.2C and 1 kg of drysteam occupies a volume of 0.8856 m3.

The fourth column uf is termed the specific internal energy of saturated liquid. Thatis the heat energy required to raise the temperature of 1 kg of water from 0C tosaturation temperature (kJ kg-1). This is a constant volume operation.

The fifth column ug is the specific internal energy of 1 kg of completely dry saturatedsteam. That is the heat energy required to raise the temperature of 1 kg of water at0C to saturation temperature plus the heat energy required to completely evaporate it(specific enthalpy of evaporation) as if the operation were carried out at constantvolume.

The sixth column hf is the specific enthalpy of saturated liquid i.e. the enthalpy of1 kg of water from 0C to saturation temperature at constant pressure. Note that ufand hf are identical down to 4.5 bar and then they gradually drift apart since hfincreases slightly faster than uf.

The seventh column hfg is the specific enthalpy of evaporation i.e. the heat energyrequired to completely evaporate 1 kg of water at saturation temperature to 1 kg ofdry saturated steam at constant pressure and at same temperature.

The eighth column hg is the specific enthalpy of saturated vapour and is the enthalpyof 1 kg of dry saturated steam at constant pressure measured from water at 0C.

On page 2 of the Rogers & Mayhew tables the same properties are given but are setout against the reference of saturated water temperature (TC) in the first column.

The graph below illustrates three phases of steam formation and incorporatesenthalpy values from tables.


From A to B, a heat transfer of 417 kJ kg-1 raises the temperature from 0C tosaturation temperature (boiling point) of 99.6C at the pressure of 1 bar.

From point B, if more heat is added, the boiling water will evaporate to form steam atthe same temperature and pressure. At point C, the enthalpy of evaporation processis completed by a further heat energy transfer of 2258 kJ kg-1. At point C the steam isin a completely dry saturated state.

In the region C to D, further heat addition produces superheated steam.

Superheated steamSo far we have considered pages 2 to 5 of the thermodynamic tables. These pages setout the properties and heat energy requirements for steam to be raised from water at0C to dry saturated steam at different pressures.

TEMP

hf= 417 kJ kg

-1

hfg = 2258 kJ kg-1

hg = 2675 kJ kg-1

SENSIBLEHEATPROCESS

EVAPORATION PROCESS SUPERHEATPROCESS

WATER & STEAM

SATURATIONTEMPERATURE

WATER

SATURATEDWATER

DRYSATURATED

STEAM

SUPERHEATEDSTEAM

HEAT ADDED

kJ kg-1

Ts9.6 CO

C O

0 C O

A

B C

D

GRAPH OF HEAT ADDED AGAINST TEMPERATURE AT 1 bar ABSOLUTE


When steam has a temperature higher than its saturation temperature for a statedpressure, then the steam is in a superheat state in which case we use pages 6 to 9 ofthe steam tables.

The following explanation of the columns on these pages will enable their use:

Column 1 as before, states the pressure (p) in bar but the figure in bracketsunder each pressure is the saturation temperature corresponding tothat pressure.

Column 2 lists the properties still of dry saturated steam i.e.vg - Specific volume of dry saturated steamug - Specific internal energy of dry saturated steamhg - Specific enthalpy of dry saturated steam.

The remaining columns list these same properties corresponding to various degrees ofsuperheat.

RuleIn order to define the condition of superheated steam it is necessary to state both thepressure and temperature of the steam. Thus, if a temperature is quoted for steam in aproblem, check it against the tables and if it is higher than (Ts) for the correspondingpressure then superheated tables must be used.

The difference between the superheated temperature (T) and the saturationtemperature (Ts) is called the degree of superheat.

Units in superheated steam tables: v in m3 kg-1u and h in kJ kg-1

Where exact values of the condition of steam are not listed in the tables, linearinterpolation for both pressure and temperature may therefore be required. Classexemplars and tutorials on the use of steam tables cover this aspect.


Specific volume of wet steam

The specific volume of steam with a dryness fraction x is given by:

vx = vf + x vfg

This is illustrated on the graph of temperature against specific volume shown below:

Referring to above graph: vfg = vg - vf

Therefore vx = vf + x(vg vf)= vf + x vg x vf= (1 x) vf + x vg

Since vf is extremely small compared with vg, the term (1 x) vf may be ignored.

Hence, vx = x vg **

ExampleDetermine the specific volume of wet steam having a pressure of 1.25 MN m-2 anddryness fraction 0.9.

1.25 MN m-2 = 12.5 bar, vx = x vg

12 bar 13 bar

vx = 0.9 20.1512+0.1632

= 0.04148 m3 kg-1

TEMP

SPECIFICVOLUME

Vx = Vf + X Vfg

X Vfg

Vf Vfg = Vg - Vf

Vg


Specific internal energy of wet steamThe specific internal energy of steam with a dryness fraction of x is given by:

ux = uf + x ufg

This is illustrated in the graph of temperature against specific internal energy shownbelow:

TEMP

SPECIFICINTERNAL ENERGY

Ux = Uf + X Ufg

X Ufg

Uf Ufg = Ug - Uf

Ug

Temperature against specific internal energy

Referring to above graph: ufg = ug - uf

Therefore ux = uf + x(ug uf)= uf + x ug x uf

Therefore ux = (1 x) uf + x ug

ExampleDetermine the specific internal energy in wet steam at a pressure of 4 bar when it hasa dryness fraction of 0.87.

Specific internal energy ux = (1 x)uf + xug

ux = (1 0.87) 605 + 0.87 x 2554

ux = 78.65 + 2221.98

ux = 2300.63 kJ kg-1


Specific enthalpy of wet steamThe specific enthalpy of steam with a dryness fraction x is given by:

hx = hf + x hfg **

This is illustrated on the graph of temperature against specific enthalpy shown below:

Temperature against specific enthalpy

ExampleDetermine the specific enthalpy of wet steam at a pressure of 70 kN m-2 and having adryness fraction of 0.85.

Pressure of 70 kN m-2 = 0.7 bar

Specific enthalpy hx = hf + x hfg

= 377 + 0.85 x 2283

= 377 + 1940.55

SPECIFIC ENTHALPY, hx = 2317.55 kJ kg-1

TEMP

SPECIFICENTHALPY

hx = hf + Xhfg

X hfg STEAM(1 - x) kgWATER

hf hfg = hg - hf

hg

Xhfg


Using steam tables and wet steam formulae

ExampleDetermine the specific internal energy, specific enthalpy and specific volume ofsteam at a pressure of 3 bar (300 kN m-2) when it is .82 dry. What is the saturatetemperature?

From steam tables at a pressure of 3 bar:-

Saturation temperature Ts = 133.5CSpecific volume vg = .6057 m3 kg-1Specific internal energy of sat. liquid uf = 561 kJ kg-1Specific internal energy of sat. vapour ug = 2544 kJ kg-1Specific enthalpy of sat. liquid hf = 561 kJ kg-1Specific enthalpy of evaporation hfg = 2164 kJ kg-1

Specific internal energy of steam at 3 bar and .82 dry.

ux = (1- x) uf + xug

= (1 - .82) x 561 + .82 x 2544

= 100.98 + 2086.08

ux = 2187.06 kJ kg-1

Specific enthalpy of steam at 3 bar and .82 dry.

hx = hf + xhfg

= 561 + .82 x 2164

hx = 2335.48 kJ kg-1

Specific volume at 3 bar and .82 dry.

vx = xvg

= .82 x .6057

vx = .4967 m3 kg-1


Interpolation of steam tablesWhen quantities cannot be extracted directly from tables, intermediate values need tobe interpolated between the nearest listed values above and below the required value.

Examples

1. Determine the specific volume of wet steam at 68 bar.

at 65 bar, vg = 0.02972 m3 kg-1at 70 bar, vg = 0.02737 m3 kg-1Difference = 0.00235

at 68 bar, vg = 0.02972 (3/5 x .00235) = 0.02831 m3 kg-1or vg = 0.02737 + (2/5 x .00235) = 0.02831 m3 kg-1

2. Determine the specific enthalpy of dry saturated steam at 52 bar.

at 50 bar, hg = 2794 kJ kg-1at 55 bar, hg = 2790 kJ kg-1Difference = 4

at 52 bar, hg = 2794 (2/5 x 4) = 2792.4 kJ kg-1or hg = 2790 + (3/5 x 4) = 2792.4 kJ kg-1

3. Determine the specific internal energy of superheated steam at 14 bar and 325C.

at 10 bar and 325C, u = 22875+2794

= 2834.5 kJ kg-1

at 15 bar and 325C, u = 22868+2784

= 2826.0 kJ kg-1

Difference = 8.5 kJ kg-1

at 14 bar and 325C, u = 2834.5 (4/5 x 8.5) = 2827 kJ kg-1or u = 2826.0 + (1/5 x 8.5) = 2827 kJ kg-1

RefrigerationIn general, refrigeration may be defined as any process of heat removal. Morespecifically, refrigeration is defined as that branch of science that deals with theprocess of reducing and maintaining the temperature of a space or body below thetemperature of its surroundings.


If a space or body is to be maintained at a temperature lower than its surroundingambient temperature, heat must be removed from the space or body being refrigeratedand transferred to another body or substance whose temperature is below that of therefrigerated body.

Mechanical refrigeration is primarily an application of thermodynamics wherein thecooling substance goes through a cycle in which it is recovered for re-use. Athermodynamic cycle can be operated in the forward direction to produce mechanicalpower from heat energy, or it can be operated in the reverse direction to produce heatenergy from mechanical power. The reversed cycle is essentially utilised for thecooling effect that it produces during a portion of the cycle and is thus called arefrigeration cycle.

Vapour-compression refrigeration cyclesThe most widely used domestic refrigerators function on a vapour-compression cyclewhich operates between two pressure levels using a two-phase working substance orrefrigerant which alternates cyclically between the liquid and vapour phases incontinuos circulation.

In order to convert a liquid into a vapour, an energy transfer is required. This energyis acquired by the vapour molecules in the evaporation process and is termed theenthalpy of evaporation. If this energy is subsequently transferred from the vapour,the energy of the molecules is diminished and liquid is formed during the process ofcondensation.

The evaporation and condensation processes take place when the refrigerant isabsorbing and rejecting heat, and these are essentially constant temperature andconstant pressure processes.

Commonly, the vapour compression cycle system within a refrigerator comprises fourmain devices, viz: Evaporator Compressor Condenser Expansion or throttle valve.

These individual elements are illustrated and have their functions examined withreference to the following systems diagram of a refrigeration unit.

Referring to the diagram below, a wet low-pressure, low temperature refrigerantenters the evaporator at point 1 and boils (evaporates) to a nearly dry state at point 2by absorbing heat from a controlled refrigerated space thereby producing therefrigerating effect.

The vaporised refrigerant then enters the compressor in which it is compressed, by awork input, ideally to a dry saturated state at a higher pressure and temperature topoint 3. The refrigerant next passes through a condenser at constant pressure andtemperature until it is completely liquid at point 4 by transferring heat to thesurroundings.


Systems Diagram for Vapour Compression Cycle

The cycle is completed when the refrigerant is expanded through a throttle valveback to its original low pressure, low temperature, wet state at point 1. The enthalpyat point 4 being equal to the enthalpy at point 1.

This cycle of operations is repeated on a continuous basis in order to maintain a pre-determined sub-zero temperature within the controlled space.

RefrigerantThe working fluid that circulates in a refrigeration system is called a refrigerant andmay be defined as a substance that, by undergoing a change in phase (liquid to gas,gas to liquid), absorbs or releases a large quantity of heat in relation to its volume,thereby producing a considerable cooling effect.

A refrigerant is a fluid that absorbs heat during evaporation at a low temperature andpressure, and rejects heat by condensing at a higher temperature and pressure.Examples of refrigerants are ammonia, sulphur dioxide, and methyl chloride,although these are no longer widely used, having been largely replaced byfluorocarbons such as Freon (refrigerants R12 and R22).

The Freon refrigerants R12 and R22 are general purpose fluids speciallymanufactured for refrigeration and these are non-toxic and non-flammable.

Compressor

3 2

Condenser Evaporator

Throttle Valve

4 1

High PressureSide

Low PressureSide

Q1Heat Rejection Q2

HeatAbsorption

Work Input


Apart from the ability to boil (evaporate) at a low temperature, refrigerants shouldpossess other desirable characteristics such as: low cost and commercially available in quantity chemical stability non-explosive suitable working pressures and temperatures low specific volume in order to keep pipe sizes small the liquid enthalpy should be low and evaporation enthalpy high in order to

achieve a high refrigeration effect per kilogram of refrigerant.

There is no refrigerant with all these properties, so the choice of a suitable fluid forany particular application must represent some form of compromise. The R family ofrefrigerants is the safest group and most widely used. All new refrigerants in the Rfamily should have zero ozone depletion potential and be user friendly. Propertytables and charts are produced for the various refrigerants similar to those producedfor water and steam.

The behaviour of refrigerants is akin to that of water when subjected to heat. Waterboils at 100C when heat energy is supplied at atmospheric pressure. Evaporationthen takes place at constant temperature until the vapour is completely dry and in thegaseous state. Further heating raises the temperature and the fluid is in thesuperheated condition.

When the temperature and pressure of a refrigerant bear a `natural stable relationshipto each other, the refrigerant is regarded as being in its saturated state.

A refrigerant liquid in its saturated state can be further cooled at the same pressure. Itwill then become subcooled or undercooled.

A refrigerant vapour in its saturated state can be heated further at the same pressure.It will then become superheated.

Saturated LiquidRefrigerant

at 40Cand 9.6 bar

- Sensible Heat

+ Sensible Heat

Sat. RefrigerantVapourat 40C

and 9.6 bar

Undercooled LiquidRefrigerant

at 35Cand 9.6 bar

SuperheatedRefrigerant Vapour

at 45Cand 9.6 bar


Use of thermodynamic tables for refrigerantsIn this outcome we have already dealt with the use of property tables for water andsteam. With the exception of entropy (s values), our studies now extend into theinterpretation and extraction of information covering the ammonia and fluorocarbonrefrigerants R717 and R12 as listed on pages 12 and 13 of the Rogers and Mayhewtables.

Refrigerant quantities together with appropriate symbols and units are identified inthe table below:

SYMBOL QUANTITY UNITT

psvg

hfhgh

Saturation Temperature

Corresponding Saturation Pressure

Specific Volume of Saturated Vapour

Specific Enthalpy of Saturated LiquidSpecific Enthalpy of Saturated Vapour

Specific Enthalpy of Superheated Vapour

Cbar

m3 kg-1

kJ kg-1

kJ kg-1

kJ kg-1

As previously stated, a refrigerant is regarded as being in its saturated state when itssaturation temperature, T, and its corresponding pressure, ps bear a `natural stablerelationship to one another.

The specific volume, vg, of a saturated refrigerant vapour, (i.e. completely dry) can beread directly from the tables against any given pressure.

For a wet vapour, the total volume of the mixture is given by the volume of liquidpresent plus the volume of dry vapour present. The volume of liquid is usuallynegligibly small compared to the volume of dry saturated vapour, hence for mostpractical problems, vx = xvg.

e.g. Spec. Volume of Refrigerant R12 at 1.004 bar and .96 dryvx = .96 x .1594 = .1530 m3 kg-1

The heat energy required to change 1 kg of saturated liquid refrigerant to acompletely dry saturated vapour (gas) is called the enthalpy of evaporation.

i.e. Enthalpy of evaporation hfg = hg - hf

e.g. Enthalpy of evaporation of refrigerant R717 at 2.680 bar

hfg = 1430.5 126.2= 1304.3 kJ kg-1


The specific enthalpy, h, of a refrigerant in the superheat condition can be extracteddirectly from the tables from either of two column headings (50 K and 100 K)dependent on the degree of superheat which is obtained by the difference between thesuperheat temperature and the saturation temperature at the specified pressure.

ExampleDetermine the specific enthalpy of refrigerant R717 at a pressure of 2.908 bar andtemperature of 20C.

From tables, the saturation temperature at 2.908 bar is 10C.Hence (T Ts) = 30C or 30 K.

Specific Enthalpy, h = 1551.7 kJ kg-1 (From 50 K column).

If the degree of superheat had been, say, 84K at the same pressure, thenSpecific Enthalpy, h = 1665.3 kJ kg-1 (from 100 K column).


OUTCOME 3: SOLVE PROBLEMS ASSOCIATED WITH STEADY FLOWENERGY EQUATION APPLICATIONS FOR GASES AND VAPOURS

In this outcome we are concerned with the solution of problems utilising the steadyflow energy equation as applied to typical thermodynamic devices such as boilers,steam turbines, compressors, etc.

Thermodynamics deals with the relationships between energy transfers within suchdevices/systems in the form of heat and work, and the related changes in theproperties of the working fluid.

Steady flow thermodynamic systemsThe steady flow energy equation (SFEE) is applicable to open two-flow systemswhere the working fluid may be a gas or vapour. Steady flow conditions prevailwhen an equal mass of fluid per unit time is both entering and leaving the system.

In order to analyse specific situations where thermodynamic principles are involved,we adopt a systems approach and make use of diagrams to illustrate the system, itsboundary, its surroundings, together with input, output and process data.

The figure below identifies these elements of a two-flow open system.

Only three things can cross the system boundary:a) energy in the form of heatb) energy in the form of workc) a mass of fluid which will possess certain forms of energy.

Heat and work transfers across the system boundary are shown by two-way arrowssince both quantities can either enter or leave the system.

SURROUNDINGS(USUALLY THE ATMOSPHERE)

WORK, IN OR OUT

HEAT, IN OR OUT

WORKING FLUIDLEAVING

WORKING FLUIDENTERING

SYSTEM CONTAININGTHE WORKING

FLUID

BOUNDARY,e.g. THE WALL OF


Heat received or rejectedIn any system a fluid can have a direct reception or rejection of heat energytransferred through the system boundary. This is designated by Q (Unit J),or if the rate of heat energy is given, by Q (Unit J s-1 or Watt).

Thus if heat is received, then Q is positive heat is rejected, then Q is negative.

If heat is neither received or rejected, then Q = 0.External work doneIn any system a fluid can do external work or have external work done on ittransferred through the system boundary. This is designated by W (Unit J)or if the rate of work done is given by W (Unit J s-1 or W).

Thus if external work is done by the fluid, then W is positive external work is done on the fluid, then W is negative.

If no external work is done on or by the fluid, then W = 0.

In order to satisfy performance criteria (a) in this outcome, students are required toconvert common thermodynamic devices into representative input/output sub-systemdiagrams. Examples of these follow in pages 3 to 7 of the notes and also in thetutorial questions.

Steam power plantAn important industrial application for vapours is the steam power plant asrepresented in the system diagram shown below:

BOILERTURBINE

PUMP

WORKOUT

ELECTRICGENERATOR

COOLINGWATER

CONDENSER

Q OUT

Q IN

STEAM POWER PLANT -- SYSTEM DIAGRAM


Feed water from the pump enters the boiler which is supplied with fuel to provideheat input Q +ve.

Wet or superheated steam from the boiler rotates the turbine and the work outputW +ve drives an electric generator via the turbines output shaft.

Exhaust steam from the turbine flows to the condenser where heat energy Q ve isremoved by the cooling water. The steam becomes water again (condensate) thenreturns to the feed pump where the cycle is continued.

Each of the devices identified in the above diagram can be categorised as sub-systemsof the integrated whole steam power plant. Each of the items is an example of asteady-flow system to which the steady flow energy equation can be applied.

Boiler

MASS FLOW AT 1 = MASS FLOW AT 2

Input/output system

In a steam power plant facility, the boiler is the device/sub-system in which steam isgenerated. In essence, a conventional boiler consists of a water container togetherwith some heating device.

The boiler is supplied with a steady flow of water which is converted into wet steamusing the heat released by burning a fuel such as coal, oil or gas. If superheatedsteam is required, the wet steam is removed from the steam space in the boiler andpiped into an integrated superheater where it is further converted into dry orsuperheated steam by the addition of more heat energy.

1 2

WATER

HEAT LOSSQ-VE

HEAT INPUTQ+VE

FEED WATERINPUT

STEAMOUTPUTWET STEAM SPACE


In a boiler no work is done, hence W = 0.

Heat input Q is required to generate steam in a boiler which can also have a heat lossthrough the boiler casing to the surroundings.

In a steam power plant the boiler provides wet or superheated steam to the turbine inthe system.

The steam turbine

MASS FLOW AT 1 = MASS FLOW AT 2

Input/output system

In the steam turbine, inlet steam is supplied to the system with a high energy leveland impinges across curved blades causing the turbine to rotate. An output shaftcoupled to the blade mechanism delivers external work. The exhaust steam exitsfrom the system with a low energy level.

Heat may be lost from the system to the surroundings or additional heat may betransferred into the system. In this case work is done by the system.

In an integrated steam power plant the turbine element may be used to drive anelectric generator.

In a steam power plant, the boiler supplies high energy steam to the turbine element.A simple integrated systems diagram for these two devices is shown below.

1

2

Q

TURBINESYSTEM

WORKOUTPUT

FLUIDINPUT

FLUIDOUTPUT


Heat exchanger/condenser

MASS FLOW AT 3 = MASS FLOW AT 4MASS FLOW OF COOLING WATER IN = MASS FLOW OF COOLING WATER OUT

Input/output system

A heat exchanger is a device that transfers heat energy from a hot fluid to a colderfluid, e.g.

oil coolers in engines and turbines where hot oil is cooled by a flow of cold water condensers in steam power plants. Exhaust steam from the turbine is cooled and

condensed by cold water.Normally the two fluids interacting are separated by tube walls.In a condenser the work transfer is zero, i.e. W = 0.

In a steam power plant, exhaust steam from a turbine is fed into a condenser forcooling into condensate. A simple integrated systems diagram for these 2 devices isshown below.

CONDENSER

COOLING WATER

STEAM

TURBINE

2 3 4

1

CONDENSER

COOLING WATER

STEAMINPUT

3 4

OUTPUT


Rotary air compressor

Input/output system

In the rotary type compressor, atmospheric air is induced to a cylinder where it iscompressed by an offset rotor and blade mechanism or rotary screw typearrangement. The high-pressure air is subsequently delivered to a storage tank fromwhere it can be tapped off and used to operate pneumatic tools such as rock drills,demolition tools and riveting hammers.

Portable compressors usually have a diesel engine as the power source and an inputshaft drives the rotor. In this case work is done on the system.

Having developed system diagrams for various thermodynamic devices, we extendour studies in this outcome into the solution of practical problems involving thesteady flow energy equation for both gases and vapours.

In outcome one, we defined and generated formulae for internal energy, flow energyand enthalpy. Two other energy forms, potential energy and kinetic energy are alsopresent in a moving fluid and these are dealt with below.

Potential energyThis is the energy possessed by a mass of fluid, m, by virtue of its height Z above agiven datum position, thus:

Total potential energy = mgZ (kg x m s-2 x m) = Nm = (J)

and for unit mass of the fluid

Specific potential energy = gZ (J kg-1)

1

2

HEAT LOSS Q TOSURROUNDINGS

COMPRESSORSYSTEM

WORKINPUT

LOW PRESSUREINTAKE

HIGH PRESSUREOUTPUT


Kinetic energyIf a fluid is in motion then it possesses kinetic energy. Thus, for a mass of fluid m,flowing with velocity C.

Total Kinetic Energy = mC2 (kg x m s-1 x m s-1) = Nm = (J)

and for unit mass of the fluid

Specific Kinetic Energy = 2

C2 (J kg-1)

Various energy forms exist in thermodynamic systems. In certain systems they mayall be present. In other systems only some may be present. Not infrequently, energyforms of insignificant value may be ignored in the solution of problems.

The steady flow energy equationThe figure below represents an open system in which a steady-flow process is takingplace. At entry to the system, the working fluid possesses potential, kinetic andinternal energy and entry flow work is done. During its passage through the systemthe working fluid is considered to take in a quantity of heat Q and do external workW.

At exit from the system the working fluid will again possess potential, kinetic andinternal energy and will do flow work to leave the system.

1

SYSTEM

ENERGY IN FLUIDENTERING SYSTEM E1

FLUID

IN

HEAT QIN

W OUT

FLUID OUT

ENERGY IN FLUIDLEAVING SYSTEM E

Z1

2


The forms of energy associated with the moving fluid mass entering the system are:

Potential energy = mgZ1 (J)

Kinetic energy = m 2

C21 (J)

Internal energy = U1 (J)

Flow energy = p1V1 (J)

Hence, total energy of the moving fluid mass entering the system

= mgZ1 + m 2C21

+ U1 + p1V1

Also, total energy of the moving fluid mass leaving the system

= mgZ2 + m 2C22

+ U2 + p2V2

In a steady-flow system it is considered that the mass flow rate and the total energy ofthe working fluid remains constant throughout the process.

Applying the principle of conservation of energy to the steady-flow open systemthen:

Initial energy + Energy entering = Final energy + Energy leavingof the system the system of the system the system

mgZ1 + m 2C21 + U1 + p1V1 + Q = mgZ2 + m 2

C22 + U2 + p2V2 + W

This is known as the steady flow energy equation.

For unit mass (1 kg) of working fluid the equation becomes:

gZ1 + 2C21 + u1 + p1v1 + Q = gZ2 + 2

C22 + u2 + p2v2 + W

Also, the combination of properties of internal and flow energies is called enthalpyand these may be combined and designated by the specific enthalpy symbol h.

Hence, the steady flow energy equation can be expressed in the form:

gZ1 + 2C21

+ h1 + Q = gZ2 + 2C22 + h2 + W

for unit mass of working fluid.


When the mass flow rate of working fluid (m) and rates of heat input (Q) and

work output (W) are given then the steady-flow energy equation can be rearranged asfollows:

Q = W + m[g(Z2 Z1) +

2C-C 21

22

+ (u2 u1) + (p2v2 p1v1)]

OR

Q = W + m[g(Z2 Z1) +

2C-C 21

22

+ (h2 h1)]

Frequently in thermodynamic problems, changes in potential energy are smallcompared with other energy changes or even non-existent when there is no differencebetween entry and exit datum levels. The gZ terms can therefore be neglected ordropped and the equation shortens to:

Q = W + m [

2C-C 21

22

+ (h2 h1)]

It is important to note that, in thermofluids, the symbol H represents total enthalpyand h represents specific enthalpy. For this reason we identify height in the PEformula by the symbol Z.

Similarly, for the KE formula, the symbol C is used for fluid velocity in order todistinguish velocity from total volume, V or specific volume, v.

. . .

. . .

. . .

. .

.


SELF-ASSESSMENT

Assignment 3

1. What practical purpose does a condenser serve in a steam power plant?

Ans:

2. List the four forms of energy associated with a moving fluid mass.

a) .. b) .......c) . d) .

State an appropriate formula which gives the total energy for the energy formslisted.

Ans: Total energy =

3. Why do we use the symbol Z to represent height in the potential energy formula?

Ans:

4. In the kinetic energy formula velocity is given the symbol C. Why not use V or v?

Ans:

5. Test your knowledge of quantities, symbols and units by completing the table below:

QUANTITY SYMBOL UNIT

MASS FLOW RATE

.

Q

WORK TRANSFER

H

SPEC. INT. ENERGY



Exemplar 1 (SFEE)Pressurised feed water with a Specific Liquid Enthalpy of 972 kJ kg-1 is supplied to aboiler facility at a Mass Flow Rate of 3.4 kg s-1. Superheated steam is produced at apressure of 60 bar and Temperature of 450C. During the process, Heat is lost to thesurroundings at a rate of 36 kJ s-1 (KW).

Insert relevant given data into the systems diagram shown below and determine therequired rate of heat input.

At pressure of 60 bar and temperature of 450C, specific enthalpy, h2 from superheattables is 3301 kJ kg-1.

Now, Energy input = energy output

Inlet enthalpy + heat input = Exit enthalpy + heat loss . . . .

m x h1 + heat input Q1 = m x h2 + Q2 . . .

Required heat input rate, Q1 = m (h2 h1) + Q2.

Q1 = 3.4 (3301 k 972 k) + 36 k

= (3.4 x 2329 k) + 36 k

= 7918.6 k + 36 k.

Q1 = 7954.6 kJ s-1

i.e. REQUIRED RATE OF HEAT INPUT = 7954.6 kJ s-1 or 7.95 MW

2

Boiler

1

h1 = 972 kJ kg

.

Q2 = 36 kJ s-1

p2 = 60 bart2 = 450C

.

m = 3.4 kg s-1


Exemplar 2 (SFEE)Steam enters a turbine with a pressure of 500 kN m-2 and leaves with a temperatureof 120.2C and dryness fraction 0.9. The power output from the turbine is 630 kW.If the mass flow rate of the steam is 1.5 kg s-1, determine the temperature of the steamat entry to the turbine.

Here, the PE and KE terms are assumed to be negligible. At exit, steam with a sat.temp. of 120.2 has a pressure of 2 bar. Thus, spec. enthalpy at exit, h2 = hf + xhfg at2 bar.

h2 = (505 + 0.9 x 2202) = 2486.8 kJ kg-1

ENERGY INPUT = ENERGY OUTPUT. . .

m x h1 = (m x h2) + W 1.5 x h1 = (1.5 x 2486.8) + 630 in kJ s-1 = 4360.2

SPEC. ENTHALPY, h1 = 1.54360.2

= 2906.8 kJ kg-1

At a pressure of 500 kN m-2 (5 bar) and spec. enthalpy of 2906.8 steam is insuperheated condition with temperature between 200C and 250C.

From Superheat tables:

AT 5 bar and 250C, spec. enthalphy = 2962 inlet enthalpy = 2906.8AT 5 bar and 200C, spec. enthalphy = 2857 at 5 bar & 200C = 2857.0

Difference = 105 and 49.8

Thus, temperature of steam at inlet = 200C + (50C x 105

8.49 )

TEMPERATURE OF STEAM AT INLET = 200 + 23.7 = 223.7C

W

STEAM

OUTPUT

STEAM

INPUTTURBINE

2

1

= 630 kW

t2 = 120.2Cx = 0.9

p1 = 500 kN m-2h1 = ?

m = 1.5 kg s-1


OUTCOME 4: APPLY THE MASS CONTINUITY AND BERNOULLISEQUATIONS TO FLOW THROUGH PIPES

This outcome covers the development and utilisation of the mass continuityequation and Bernoullis equation as applied to the steady flow of incompressiblefluids (liquids) through pipes.

Mass continuity equationConsider sections 1 and 2 in the tapered pipe shown in figure below which is full ofsteadily flowing fluid. At Section 1, the cross-sectional area is A1, and the velocity ofthe fluid is C1; at section 2, the area and velocity are A2 and C2 respectively.

Continuity equation

Steady flow conditions prevail when the rate at which the mass of fluid entering thepipe at datum position 1 is the same as the rate at which it leaves at datum 2; i.e. themass flow rate is constant,

Now,

Volume of fluid passing = Volume of fluid passingSect. 1 per unit time Sect. 2 per unit time

A1C1 = A2C2

(Since, volume per unit time = C.S. Area x Velocity)

Volumetric flow rate V = A1C1 = A2C2 (m2 x m s-1 = m3 s-1) ** .

C R O S S -S E C TIO N ALA R E A A 1

C R O S S -S E C TIO N ALA R E A A 2

L IQ U ID INV E LO C IT Y C 1 V E LO C IT Y C 2


Volumetric flow rates for fluids can be converted to mass flow rates by introducingdensity into the equation. In outcome 1, density was defined as follows:

Density, (RHO) = Vm

VOLUMEUNITMASS

= (units kg m-3)

Hence, mass flow rate, m = x V = x AC (kg m-3 x m2 x m s-1 = kg s-1)

This equation m = AC is known as the mass continuity equation **

Where m = mass flow rate in kg s-1 = density of fluid in kg m-3A = cross-sectional area of pipe in m2C = velocity of fluid in m s-1

Branched piping systemsIn many pipework systems, a single pipeline will split into two or more branches asshown in figure below:

Like the tapered pipe example already dealt with, steady flow conditions will applywhen the volumetric and mass flow rates passing Section 1 equal the combined totalof volume and mass flow rates passing sections 2 and 3, i.e.

V1 = V2 + V3

Hence V = A1C1 = A2C2 + A3C3 (m3 s-1) **

and m = A1C1 = (A2C2 + A3C3) (kg s-1) **

In our studies of incompressible fluid flow along pipes, the density of the fluid isassumed to remain constant throughout a process.

. .

.

.

.

.

. . .

1

2

3


Energy of a flowing fluidIn outcome 3 we determined that the total energy of a mass, m, of flowing fluid hadfour components.

Potential Energy mgZ (J)

Kinetic Energy2

2mC (J)

Internal Energy U (J)

Flow Energy pV (J)

Thus, in specific terms for a mass of 1 kg, the total energy is:

gZ + 2

C2 + u + pv (J kg-1)

where u is the specific internal energy of the fluid and v is the specific volume of thefluid.

The specific internal energy term, u, depends wholly on fluid temperature. In mosthydrodynamic situations the change in fluid temperature is very small, so the internalenergy term has little significance and can be neglected.

The specific volume v of a fluid in m3 kg-1 is the reciprocal of density,i.e. v = 1/ .

Hence, by substitution, the above expression can be modified to givespecific energy of moving fluid as:

gZ + p

2C2

+ (J kg-1)

Further dividing throughout by g, gives:

( ) m=m xs x kg x m x s x m x kg=m x s x kgNm=ms x kgJ

p+

2g

2C+Z 1-21-2-1-21-1-21-

All three terms now have the dimension of length (metres)

gp

is the pressure head of a fluid at pressure p,

and the sum of all three terms is called the total head.

i.e. Total head = Z +

p+

2gC2

(m)


Bernoullis equationConsider unit mass of fluid flowing at a steady rate through the system/pipelineshown in the figure below.

Applying the Principle of Conservation of Energy we have:

Specific energy in the = Specific energy in thefluid entering system fluid leaving system

Specific Specific Specific Specific Specific SpecificPotential + Kinetic + Flow = Potential + Kinetic + FlowEnergy Energy Energy Energy Energy Energy

At entry At exit

gZ1 +

p+

2C

+gZ=

p+

2C 2

22

21

21

= constant (J kg 1)

Dividing both sides of the equation by gravitational acceleration, g.

Z1 + =

p+

2gC

+ Z=

p+

2gC 222

21

21

constant (m)

This is known as Bernoullis Equation **

Each quantity in the Bernoulli Equation is measured in terms of head of liquid, i.e.height of liquid above a given datum. The Unit for each quantity is the metre (m).

2

1

DATUM

Z1

Z2

FLUID IN

FLUID OUT

PRESSUREVELOCITY C

p11

PRESSUREVELOCITY C

p22

SYSTEM


Frictional resistance to flow (loss of energy)Bernoullis equation assumes there to be no frictional resistance to the flow of anincompressible fluid/liquid through a system/pipeline. In practical applicationshowever, frictional resistance to flow is always present and reduces the availableenergy in the fluid at exit from the system. Thus:

Specific Energy in Specific Energy in Specific EnergyFluid Entering the = Fluid Leaving the + To OvercomeSystem System Frictional Resistance

Let ZF = Frictional Resistance `Head

Then, Bernoullis Equation can be stated in the form:

Z1 + F2

22

21

21

Z+

p+

2gC

+ Z=

p+

2gC

As before, each term represents Head of Liquid in units of m.

Z = potential head

2gC2

= kinetic head

p= pressure head


SELF-ASSESSMENT

Assignment 4

In relation to the quantities and energies established in this outcome for a movingfluid, complete the table below.

QUANTITY SYMBOL/FORMULA UNIT

A

PRESSURE HEAD

m3 s-1

mgZ

FLUID VELOCITY

KINETIC HEAD

kg s-1

State Bernoullis equation. Identify each symbol in the equation and list units foreach on one side only.

EQUATION:

SYMBOL QUANTITY UNIT

State the Mass Continuity Equation.

EQUATION:


PRACTICAL EXEMPLAR PROBLEM

Bernoulli & Mass Continuity EquationsOil of density 800 kg m-3 enters a 500 mm diameter pipeline with a velocity of 3 m s-1and pressure 400 kN m-2, and is discharged through an orifice 150 mm diameter and30 m below the entry point. If the frictional losses are equivalent to a head of 3.5 mof the oil, determine the velocity, the mass flow rate, and the pressure of the oil at thepoint of discharge.

Velocity of oil at exit from A1C1 = A2C2 C2 = 12

1 C x AA

C2 = 1-22

sm33.33=3 x .0225.25

=3 x .15x4

.5 x 4

VELOCITY, C2 AT POINT OF DISCHARGE = 33.33 m s-1

MASS FLOW RATE, m = A2C2 = 800 x 4

x .152 x 33.33 = 471.2 kg s-1

Use Bernoullis equation for pressure at exit

Z1 + F2

22

21

21

Z+

p+

2gC

+ Z=

p+

2gC

30 + 3.5+9.81 x 800p

+9.81 x 233.33

+0=9.81 x 80010 x 400

+9.81 x 23 2

232

30 + .4587 + 50.9684 = 56.62 + 3.5+7848

p2

81.4271 = 60.12 + 7848

p2

p2 = (81.4271 60.12) x 7848

= 167218 N m-2

.

d2 = 150 mmp2 = ?C2 = ?

OIL = 800 kg m-3ZF = 3.5 m

d1 = 500 mm

p1 = 400 kN m-2C1 = 3 ms-1

30 m


PRESSURE p2 AT POINT OF DISCHARGE 167.2 kN m-2

Venturi meterIn many engineering applications it is necessary to measure and control the flow ratesof fluids. The Venturi meter is a device which allows pressure changes to bemeasured at different sections in a pipeline carrying a flowing fluid. Once pressuredifferentials are obtained, velocity and flow rates can be calculated using BernoullisEquation.

The figure below illustrates a typical horizontally mounted Venturi meter where fluidenters at Section 1.

After Section 1, the bore of the device converges to a small diameter known as the`throat then gradually diverges back to its original cross-sectional area. As the fluidpasses through the restricted throat section, its velocity, and consequently its kineticenergy will increase. Since the total energy of the steadily flowing fluid remainsconstant, it follows that the pressure and flow energy will decrease at the throatsection.

A U-tube manometer, with say mercury as the denser fluid, fitted between sections 1and 2 allows the pressure difference to be measured. The higher pressure at Section 1causes the mercury in the manometer to be forced down the LH limb and up in theRH limb.

h

y

FLUID DE NSITY

EN TRY EX IT

M AN O M ETE R FLU ID DE NSITY

X X

1 2

P 1C 1A 1

P 2C 2A 2


Reference the figure and applying Bernoullis equation between sections 1 and 2 ofthe Venturi gives:

p+

2gC

=

p+

2gC 222121 (since Z1 = Z2 here)

2gC-C

=

p-p 212221

(Equation 1)

The volumetric flow equation gives V = A1C1 = A2C2 C2 = 12

1 CxAA

Substituting for C2 in equation 1 gives:

2g

C-CAA

=

p-p21

2

12

1

21

2g

1-AA

C=

p-p

2

2

121

21 (Equation 2)

The pressure at level x x is the same in both limbs of the manometer. Also, thepressure at a depth in a liquid is obtained from the formula; pressure, p = gh.

Now, Pressure in LH Limb at x x = Pressure in RH Limb at x x

p1 + g (Y + h) = p2 + gY + Hggh p1 + gY + gh = p2 + gY + Hggh p1 + gh = p2 + Hggh p1 p2 = gh ( Hg - )

Divide by

-gh(=

p-p Hg21

=

-

gh Hg

= gh 1-

Hg

1-

h=

p-p Hg21 (Equation 3)

.


Equating equations 2 and 3:

1-

h=2g

1-AA

CHg

2

2

121

Therefore velocity C1 =1

AA

1-

2gh2

2

1

Hg

(Equation 4)

. .

Volumetric flow rate, V, can now be deducted from equation, V = A1C1 .

Also, mass flow rate from equation m = A1C1

The above equation 4 for calculating velocity involving a Venturi meter is aderivation of Bernoullis equation. Students would not be expected to memorise orreproduce this formula from first principles but would be required to apply formula inan open book situation.

The orifice plateA less sophisticated and much less costly device for measuring incompressible fluidflow rates in pipelines is the orifice plate. This consists of a circular plate with aconcentric orifice which is inserted into the pipeline.

In passing through the orifice the liquid stream contracts and in so doing convertsstatic pressure head into velocity head as in a Venturi meter. The reduction inpressure head can be measured using a manometer with tapping points sited eitherside of the orifice plate. Upstream, the pressure should be tapped at a position wherethe flow pattern is not influenced by the presence of the orifice plate. The throatpressure tapping point is situated downstream from the orifice plate where theeffective diameter of the converging liquid stream is at a minimum.

Downstream turbulence causes orifice plates to have a typical co-efficient ofdischarge of around 0.65 whilst a Venturi meter is closer to 1. Generally orificeplates give less accurate results for flow rates mainly due to high turbulence levels.They are, however, much less expensive than the Venturi meter and easier to install.


The diagram above shows an orifice plate installed in a pipeline.

Our studies do not extend into problem solving or calculations on orifice plates.Students should be capable of sketching an orifice plate and describing its function.

P 1

P L

PA 1C 1

P 2A 2C 2

h


PRACTICAL EXEMPLAR PROBLEM

Venturi meterA Venturi meter is used to establish the flow rate of water in a horizontal pipeline of50 mm diameter. The throat diameter of the Venturi is 20 mm. The pressure dropbetween the entrance and throat section is 60 mm of mercury. Assuming no lossesdue to friction and taking the densities of water at 1000 kg m-3 and mercury at 13600kg m-3, determine:

a) the velocity of the water in the 50 mm dia pipelineb) the volumetric flow rate of waterc) the mass flow rate of water.

a VELOCITY, C1 IN 50 PIPE FROM EQUATION,

C1 =

1-.02 x 4

.05 x 4

1-100013600

.06 x 9.81 x 2=

1-AA

1-

2gh2

2

22

2

1

Hg

C1 = .3896932=38.062514.83272

=

1.0004.0025

1)-(13.61.17722

VELOCITY OF WATER IN 50 PIPELINE = 0.624 m s-1.

b VOLUMETRIC FLOW RATE, V = A1C1 = 4 x .052 x 0.624

= 1.225 x 10-3 m3 s-1.

c MASS FLOW RATE m = A1C1 = 1000 x 1.225 x 10-3= 1.225 kg s-1

h =

60m

m

50

mm

20m m


OUTCOME 5: SOLVE PROBLEMS ASSOCIATED WITH THE BEHAVIOUROF LIQUIDS AT REST

Outcomes 3 and 4 of this unit encompassed the behaviour, effects, and the solution ofproblems covering fluids in motion, i.e. thermodynamics

In outcome 5 we are concerned with the solution of problems associated with theeffects of liquids at rest. Hydrostatics is the study of the forces and pressures exertedby a fluid when the fluid system is in equilibrium.

Gravitational forceThe derived SI unit of force is the Newton (N) defined as that force which, whenapplied to a body of mass one kilogram, gives it an acceleration of one metre persecond squared. From Newtons Second Law of Motion:-

FORCE = MASS x ACCELERATION F = m a (kg x m s-2 = N)

When gravitational acceleration is considered, the above equation can be re-stated inthe form:-

F = m g (where g = 9.81 m s-2)

Hence, the downward gravitational force exerted by a liquid column of mass 25 kgwill be:-

F = 25 x 9.81 = 245.25 N

Pressure (symbol p)

Pressure is defined as force (F) per unit cross-sectional area (A).

i.e. Pressure p = AF

=

AREASECT.-CROSSFORCE

The unit for pressure is the Newton per square metre (N m-2).

The units Pascal (Pa) and bar are also commonly used and it should be noted that:-

1 Pa = 1 N m-2 or 1 kPa = 1 kN m-2

and 1 bar = 105 N m-2 or 100 kN m-2


Atmospheric pressureThis is the pressure exerted on the surface of the earth by the gravitational pull of themass of air in the earths atmosphere.

Atmospheric pressure is stated as an absolute pressure, i.e. relative to zero.

Standard atmospheric pressure = 1 atm = 1.013 bar = 1.013 x 105 N m-2

Gauge pressure and absolute pressureVarious types of pressure gauges are commonly used to measure fluid pressures invessels and pipelines and read pressures normally above or below atmosphericpressure. If a gauge displays a zero reading it means the pressure is atmospheric.

If the pressure in a vessel is increased above atmospheric to a gauge pressure pg, thetrue or absolute pressure p in the vessel is given by:-

Absolute pressure p = gauge pressure + atmospheric pressure

i.e. p = pg + patm ***

Static pressure characteristicsThe concept of pressure in a fluid and the manner in which it varies throughout thefluid mass is of fundamental importance in the science of hydrostatics. Theexperimentally established characteristics noted below need to be acknowledged.

1. The intensity of pressure in a fluid at rest is the same in all directions.

2. The pressure exerted by a static fluid always acts normally (perpendicular) to anyboundary surface containing the fluid.

3. The pressure exerted by a static fluid is directly proportional to and increaseswith depth below its free surface.


Pressure variation with depth in a liquid column

Consider the forces acting ona vertical cylinder of liquidof height h and cross-sectionalarea A within a volume of staticliquid.

FIGURE 5A

Let p1 = pressure on free surface of liquid N m-2p2 = pressure at depth h N m-2A = C.S. Area of cylindrical liquid column m2h = depth of liquid column mm = mass of liquid in the column kg = density of the liquid kg m-3

The downward force exerted by the liquid column = m x g

Now, mass of liquid m = density x volume (kg m-3 x m3 = kg) = V

and, volume of liquid V = CSA x height (m2 x m = m3) = A h

Hence, mass of liquid in the column, m = Ah

Downward force exerted by the liquid column, F = Ah x g

Now, for the column of liquid to be in equilibrium, i.e. at rest

UPWARD FORCES = DOWNWARD FORCES

p2 A = p1 A + Ahg

Dividing through by A, p2 = p1 + gh

p2 p1 = gh

i.e. the pressure at any depth in a liquid can be derived from equation

*** Pressure p = gh (kg m-3 x m s-2 x m = N m-2)

C.S. AREA A

P1

P2

FREE SURFACE

DEPTH h

UPTHRUST = xPA2


Hydrostatic pressure and thrust

FIGURE 5B

The above figure 5B illustrates three tanks having equal base areas A. Each vesselcontains the same liquid to a common vertical depth h.

Since the tanks hold different volumes of liquid, it might be thought that container Cwith the largest amount and greater mass of liquid would have the greatest forceexerted on its base. Tank D might be considered as having the least force on its base.

For all three tanks, however, the base pressure p = gh at all points on the basessince these are horizontal. Also, since the tanks have a common base area A, itfollows that the forces acting on all three bases have equal magnitude.

i.e. F = pA = ghA (all four quantities are constant)

The arrows inside tanks B, C and D illustrate how pressure acts normal to retainingsurfaces and increases in intensity with depth below the free surface.

The total force/thrust acting on the horizontal bases of tanks B, C and D will act at thecentroid of each base area.

Our studies now extend into the determination of thrust exerted on submerged andpartially submerged vertical plane surfaces and to fixing the point of application ofthis force at the centre of pressure.

AREA A AREA A AREA A

B C DFREE SURFACE

h


Thrust on a submerged vertical plane surfaceConsider a plane surface of area A immersed vertically in a liquid of density . Thepressure on one side acts at right angles to the surface and gives rise to a force orthrust on that side.

FIGURE 5C

By definition, Pressure = C.S.AFORCEAPPLIED

p = AF (N m-2)

Re-arranging, Force = pressure x C.S.A F = pA(N)

Reference above figure:

Pressure p on one side of elemental strip of area dA = gy

Hence, force dF on elemental strip = gydA

Total force on whole area A F = dF

Since both and g are constants,

Total force on whole area A F = g yda

But y dA is the total moment of area about an axis through the free surface of theliquid and is also = A x y

Where, y is the vertical depth of the centroid, G below the free surface of the liquid.

Thus the total thrust on an immersed plane vertical surface is proportional to thedepth of the centroid of the wetted area below the free surface and can be deducedfrom formula:

Total thrust F = gA y (N) ***

FREE SURFACE

yy

dAdF

GCENTROID


Centre of pressure (or centre of force)We have seen that the intensity of pressure acting over any vertically submergedplane surface area increases proportionally with depth below the free surface of aliquid. Hence, for both rectangular and circular vertical surfaces, the pressure andforces applied above their horizontal geometric centre lines will be less than thoseacting below their centre lines.

The combined action of all the elemental forces acting over the whole surface areascan be replaced by a single resultant force, equal in magnitude to the total thrust,and acting at a point on the surface called the centre of pressure. This point willobviously lie on a vertical centre line of area at some distance below the geometriccentroid of the plane surface areas.

FIGURE 5D

For both configurations of immersed plane surfaces shown in FIG.5 above, theindividual centroids (G) have a depth y below the liquid free surface. Each centre ofpressure (CP) is located vertically below the centroids by a distance designated GC.

Setting aside its mathematical proof from first principles we can state the appropriateformula for calculating this distance, viz:-

DISTANCE GC = SURFACEFREEBELOWCENTROIDOFDEPTHSQUAREDCENTROIDABOUTGYRATIONOFRADIUS

GC =yk2

***

Values for k2 can be calculated using formulae listed alongside figure 5E onfollowing page.

-

FREE SURFACE

y y

G GC GC

CP CP

d

CIRCULAR PLANE SECTION RECTANGULAR PLANE

b

d


Area and k2 values for rectangular and circular plane surfaces

FIGURE 5E

For the rectangle, the geometric centroid, G, lies on the intersection of the verticaland horizontal centre lines.

For the circular area the centroid, G, lies at centre of circle.

For either immersed surface, the centre of pressure, C, will lie vertically below thecentroid by a distance, say GC, and can be calculated using the stated formula foreach listed below.

FOR RECTANGULAR SURFACE FOR CIRCULAR SURFACE

DISTANCE GC = y/12d

=

yk 22

GC = y16/d

=

yk 22

where y = depth of centroid below the liquid free surface.

Partially submerged rectangular surfaces and wetted areasFor rectangular plane surfaces which are partially submerged vertically in a liquid,the formulae developed for total thrust and centre of pressure remain applicable whenthe area of the wetted surface only is considered.

CIRCULAR AR EARECTANGLE

GGC

CP

BREADTHb

DEPTHd

GGC

CP

DIA d

AREA =bd

k2 = d2

12k2 = d

2

16

AREA =4

d2


The position of the geometric centroid for a partially submerged surface area can alsobe established for the wetted area only.

FIGURE 5F

WETTED AREA

FS

FS

TANK DAM WALL

WETTED SURFACE AREAS



1. Total thrust and centre of pressure (vertical rectangular surface)A rectangular storage tank has a horizontal base 6 m x 4 m. The vertical sides of thetank are 3 m in height. Determine the total hydrostatic thrust acting on the ends andsides of the tank when it is half full of oil of density 840 kg m-3. Determine also theposition of the centre of pressure of the total thrust above the base.

For the 4 m side:

RESULTANT FORCE/TOTALfrom eqn F1 = gA y

F1 = 840 x 9.81 x 4 x 1.5 x 25.1

F1 = 37.082 kN

For the 6 m side, total thrust F2 = gA y

F2 = 840 x 9.81 x 6 x 1.5 x 25.1

F2 = 55.623 kN

The centre of pressure lies below the centroid of the wetted surface by the distanceGC.

Distance GC from eqn GC =yk2

for rectangle k2 = 12d2

GC = m25.0 = 75 x .12

5.1 =

y x 12d 22

Since centroid is located 0.75 m above the base level, the C of P will be0.75 0.25 = 0.5 m above base.

This dimension 0.5 m above base applies for both sides and both ends since C of P isdependant on depth of wetted surface and independent of breadth b of the wettedsurface.

4m

G

C

DISTANCE GC

3m

1.5m

THRUST


2. Total thrust and centre of pressure (vertical circular surface)A circular opening in a vertical dam face is closed by a gate mounted on trunnions onits horizontal centre line. The gate has a diameter of 4.2 m and its horizontal centreline is 5.0 m below the water level in the dam.

Determine the magnitude of the pulling force required to be applied at a point 2.0 mabove the centroid of the gate to just keep the gate closed against the hydrostaticthrust of the water.

Total force acting on circular opening, say F1 from equation:

F = gA y

= 1000 x 9.81 x 4

x 4.22 x 5

F1 = 679559 N

Centre of pressure for force F1 lies at a distanceGC below trunnion centre line.

GC =yk2

and for circle k2 = 16d2

GC = 5 x 162.4 2

= .2205 m

Equilibrium conditions for opening are:

C.W. MOMENTS = A.C.W. MOMENTS

Pulling force F2 x 2.0 = 679559 x .2205

F2 = 22205 x .679559

= 74921 N

MAGNITUDE OF PULLING FORCE = 74.9 kN

PULLINGFORCE


Head of a liquidWhilst deriving Bernoullis equation in Outcome 4, the total specific energy of aliquid stream was stated as:

Potential energy + kinetic energy + flow energy

gZ +

p +

2C2 (specific energy IN J kg-1)

Dividing these terms by gravitational acceleration, g, each term can then be re-statedas:-

Z +

p +

g2C2 (head of liquid in metres, m)

i.e. POTENTIAL HEAD + KINETIC HEAD + PRESSURE HEAD

The energy quantity

p is also referred to as flow work or pressure energy and is the

energy which must be continuously expended by a pump (or its equivalent) in forcinga liquid along a pipeline in the presence of hydrostatic pressure.

Although in common use, the term pressure energy tends to imply that a liquid mayhave its energy increased by pressurization. This is not the case since most liquidsare practically incompressible. The term flow work may be considered moreappropriate.

Likewise, the three separate types of energy listed above for a liquid stream are oftenexpressed more conveniently as heads of liquid.


Static pressure head and hydrostatic pressureThe hydrostatic pressure, p, at a point in a liquid stream, may be imagined to be dueto its being at a depth, h, below the free surface of the same liquid. Thus, if a smallhole be drilled into a pipeline carrying a flowing liquid and then fitted with apiezometer tube, the liquid would rise up the tube due to the pressure in the systemand settle at a height when equilibrium was reached.

FIGURE 5G

Reference above figure, the height, h, to which the liquid rises in the tube provides ameans whereby the hydrostatic pressure at that point in the pipeline can bedetermined. This height, h, is called the static pressure head of the liquid in thepipeline.

Hence,

Static pressure head, h = gFLOW WORK

=

p (m)

and hydrostatic pressure, p = gh (N m-2)

The static pressure head is easily measured and may be regarded as a head of liquidequivalent to the flow work.

PIE ZOM ETERTUBE

PR ESSUR E

p

DE NSITY

SECTIO N O F PIPELINE


Pressure measurement

Manometry may be defined as the science of utilising vertical columns (heads) ofliquid to measure fluid pressures.

Many different types of pressure measuring devices exist but our studies will belimited to the barometer, piezometer tube and U-tube manometers.

Before considering these devices, however, it is appropriate to recap on definitions ofpressure quantities already covered in this unit.

Vacuum A perfect vacuum is a completely empty space and has zero pressure.

Atmospheric pressure, patm The planet earth is surrounded by an atmosphere. The pressure due to this atmosphere depends upon the head of air above the earths surface. At sea level

atmospheric pressure is normally taken as 1.013 bar (101.3 kN m-2), equivalent to a head of 10.35 m of water or 760 mm of mercury, and decreases with altitude.

Gauge pressure, pg is the intensity of pressure measured above or below atmospheric pressure. When a gauge shows a zero r

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