DET: Mechanical Engineering Thermofluids (Higher)

DET:

Mechanical Engineering

Thermofluids (Higher)

Spring 2000

DET:

Mechanical Engineering

Thermofluids

Higher

HIGHER STILL

CONTENTS

Section 1: Thermofluids (Higher) Student Notes

Section 2: Self-Assessment Answers

Section 3: Tutorials

Section 4: Tutorials Marking Scheme

Section 1:

OUTCOME 1: APPLY GAS LAWS

The wide use made of various gases in the field of engineering makes it necessary to predetermine their reactions when they are heated, cooled, expanded or compressed. When a process takes place, the changes which will occur in the properties of

volume, absolute pressure and absolute temperature of the gases are related by the gas laws.

When solving problems utilising the gas laws, both pressures and temperatures must be 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 and read pressures normally above atmospheric pressure. If a gauge shows a zero reading it means the pressure is atmospheric.

If the pressure in a vessel is increased above atmospheric to a gauge pressure pg, the actual 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 otherwise stated.

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 are often high numbers. In order to express multiples of SI units concisely, the undernoted prefixes are used:

MULTIPLICATION FACTOR PREFIX SYMBOL

1,000,000 = 106 _{Mega M}

1,000 = 103 _{Kilo k}

Absolute temperature

In problems involving the gas laws, the temperature of any gas is measured from absolute zero, which has been determined to be 273°C below the zero point on the Celsius scale, i.e. absolute zero. At this point the internal energy of the substance is also zero.

Absolute temperature is the temperature above absolute zero and is determined by

adding 273 to the Celsius temperature scale reading.

i.e. Absolute temperature = Celsius temperature scale reading + 273. Hence 27°C = 300 K.

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

Mass

This is usually defined as the quantity of matter in a body. Symbol: m (small letter).

Unit: kg (small letters).

Volume

Symbol: 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 to avoid errors in calculations.

Specific volume

Symbol: 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 = MASS VOLUME Boyle’s Law

The Irish scientist Sir Robert Boyle investigated the behaviour of gases when expanded or compressed under constant temperature (isothermal) conditions. In essence, Boyle’s Law states:

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

Thus p. …

V 1

or pV = constant

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

Boyle’s Law can also be expressed algebraically in the form

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

Charles’ Law

The French scientist Jacques Charles conducted experiments on gases where the pressure of a fixed mass of gas was kept constant while variations in the volume and temperature were examined. In essence, Charles’s Law states:

During the change of state of any gas in which the mass and pressure remain constant, the volume varies in proportion with the absolute temperature (Kelvin).

Thus V. T or

T V

= constant

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

Charles’s Law can be expressed algebraically in the form:

n n 2 2 1 1 T V = T V = T V

………..for the mass of gas

Constant volume process

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

EMPERATURE ABSOLUTE T PRESSURE = constant T p

= 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 T p = T p = T p

The Combined Gas Law

If, during a process, the pressure, volume and absolute temperature of a gas are changed from p1, V1, and T1 to p2, V2 and T2 respectively, then, provided there is no change in the mass of gas, Boyle’s Law, Charles’s Law and the Pressure Law may be combined to give the algebraic expression:

2 2 2 1 1 1 T V p = T V p = constant

This is known as the Combined Gas Law.

The Characteristic Gas Equation

We have seen the combined gas law stated in the form:

T pV

= constant C

For a perfect or ideal gas, this constant C = mR where m is the mass of the gas and R is 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 this equation for solving problems, it is essential to express all the terms in appropriate units which are:

p = absolute pressure of the gas N m-2

V = volume of the gas m3

T = absolute temperature of the gas [(t + 273)] K

m = mass of the gas kg

R = gas constant J kg-1 K-1.

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

GAS LAW PROCESS CONDITION APPROPRIATE EQUATION

Boyle’s Law Constant Temperature p1V1= p2V2 etc

Charles’s Law Constant Pressure

2 2 1 1 T V = T V etc

Pressure Law Constant Volume

2 2 1 1 T p = T p etc

Combined Gas Law Pressure Volume and

Temperature all Change

2 2 2 1 1 1 T V p = T V p etc

Characteristic Gas Equation Includes Mass and Characteristic Gas 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 varies for different gases as will be apparent in the questions covered in the Tutorial for Outcome 1.

Universal Gas Constant

The Universal Gas Constant takes into account the concept of molecular mass of substances. 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 the relationship

R = M Ro

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

⇒ R = = 0.297 kJ kg-1 K-1 28 3143 . 8 = M Ro

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

SELF-ASSESSMENT

Assignment 1

1. Test your knowledge of quantities, symbols and units covered so far by completing 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. o C K -150 388 128 162 0

3. Boyle’s Law deals with the behaviour of gases under isothermal conditions. What does the condition ‘isothermal’ mean?

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 on its pressure and temperature?

Ans:

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

Equation:

PRACTICAL EXEMPLAR PROBLEMS

Having developed the various gas law equations, we can apply these to the solution of problems dealing with the behaviour of gases when subjected to different processes. Problems may be simplified by employing a method of converting given information into symbols and units which can then be fitted into an appropriate equation.

Specimen worked exemplar problems adopting this strategy now follow.

Exemplar Question 1

A fixed mass of gas is compressed isothermally from a pressure of 20 bar and volume of 3.2 m3 until it occupies 25% of its original volume. Calculate the final pressure of the gas. Known Data p1 = 20 x 105 N m-2 V1 = 3.2 m3 p2 = ? V2 = 3.2 x 100 25 = 0.8 m3

For an isothermal process, Boyle’s Law applies and equation

p1V1 = p2V2 can be utilised. From p1V1 = p2V2 ⇒ p2 = p1 x 2 1 V V = 20 x 105 x 8 . 0 2 . 3 ⇒ p2 = 8.0 x 106 N m-2

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

Exemplar Question 2

A quantity of gas at a pressure of 180 kN m-2 and temperature 18°C, occupies a volume of °.43 m3. The gas is compressed until its pressure and temperature are 670 kN m-2 and 127°C respectively. If there is no loss of gas, what volume will it now occupy?

Known Data

From the Combined Gas Law p1 = 180 kN m-2 p2 = 670 kN m-2 T1 = (18 + 273) K T2 = (127 + 273) K V1 = 0.43 m3 V2 = ? 1 1 1 T V p = 2 2 2 T V p

Exemplar Question 3

An air receiver contains a fixed mass of air at a pressure of 12.5 bar and temperature 84°C. After a period of time, the pressure is observed to be 7.8 bar. What will be the temperature of the air?

Known Data

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

⇒ ₁1 T₂2 p = T p ⇒ T2 = 1 1 2 T x p p = _{12.5 x 10} x 357=222.768K 10 x 7.8 5 5

∴ FINAL TEMP. OF AIR = 222.8 – 273 = -50.2°C Exemplar Question 4

Gas is stored in two tanks, A and B, which are connected by a pipe fitted with a stop valve which is initially closed. Tank A has a volume of 3.0 m3 and contains 14 kg of the gas at a pressure of 215 kN m-2 and a temperature of 20°C. Tank B has a volume of 12.0 m3 and contains gas at a pressure of 340 kN m-2 and a temperature of 20°C. 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 of the gas, assuming the temperature remains at 20°C.

V1 = 3.0 V2 = 12.0 m3

m1 = 14 m2 = ?

p1 = 215 k p2 = 340 kN m-2

T1 = 20 + 2 T2 = 293 K

= 293 K

Char. Gas Constant from p1V1 = m1 R T1

⇒ R = 293 x 14 3.0 x 10 x 215 T m V p 3 1 1 1 1 = ∴ GAS CONSTANT R = 157.24 J kg-1 K-1 TANK A TANK B

MASS IN TANK B from p₂ V₂ = m₂ RT₂ ⇒ m2 = 293 x 157.24 12 x 10 x 340 = T R V p 3 2 2 2

∴ MASS OF GAS IN TANK B = 88.56 kg

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

FINAL PRESSURE from p3V3 =m3RT3

⇒ ( ) 15 293 x 157.24 x 14 + 88.56 = V T R m = p 3 3 3 3 = 315004.97 N m-2

∴ FINAL PRESSURE IN SYSTEM = 315 kN m-2

Exemplar Question 5

A fixed mass of gas contained in a closed system is initially at a pressure of 100 kN m-2, a temperature of 15°C, and occupies a volume of 0.15 m3. The gas is then compressed to a volume of 0.06 m3 and a pressure of 300 kN m-2. The gas is then expanded at constant pressure unit it reoccupies its original volume. If the characteristic constant for the gas is 189 J kg-1 K-1, determine the mass of gas in the system and the temperature at the end of the compression and expansion processes.

Known Data

p1 = 100 x 103 N m-2 p2 = 300 x 103 N m-2 p3 = p2 T1 = 15°C + 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 p1V1 = m R T1

⇒ m = 1 1 1 T R V p = = 0.2756kg 288 x 189 0.15 x 10 x 100 3

TEMPERATURE AFTER COMPRESSION from p2V2 = m R T2 ⇒ T2 = = 345.56K 189 x 0.2756 0.06 x 10 x 300 = R m V p_{2 2} 3

∴ TEMPERATURE AFTER COMPRESSION = 345.56 – 273 = 72.56°C

TEMPERATURE AFTER EXPANSION from p3 V3 = m R T3

⇒ T3 = =863.9K 189 x .2756 0.15 x 10 x 300 = R m V p_{3 3} 3

OUTCOME 2: SOLVE PROBLEMS USING DATA EXTRACTED FROM THERMODYNAMIC PROPERTY TABLES

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

These tables, commonly known as ‘steam tables’, give tabulated values for the properties of steam and refrigerants over an extensive range of pressures and temperatures.

The ability to understand and extract data from the tables extends into the solution of problems in this outcome and also in Outcome 3 when the steady flow energy

equation is dealt with.

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

Specific volume

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

i.e. v = m V = MASS VOLUME UNIT : m3 kg-1

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

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

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

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

Internal energy

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

internal energy of the fluid.

As a result of experimental work on this subject, Joule concluded that the internal energy of a fluid is a function of temperature only and is independent of pressure and volume (Joule’s Law).

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

temperature.

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

Generally, the internal energy of a fluid is quoted as per unit mass (per kg). This quantity 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 as underlisted.

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 energy

Any volume of fluid entering or leaving a system must displace an equal volume ahead 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 the system. If the pressure in the fluid is p1, then the work done on the fluid inside the system by the incoming fluid = force x distance the fluid is displaced.

Flow energy X Y p1 p1 p1 p1 S S VOLUME DISPLACED V1 CROSS SECTIONAL AREA _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

= p1v1 where v equals the specific volume 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) Enthalpy

In steady-flow thermodynamic systems, internal energy and flow energy are present in the moving fluid. Accordingly, it is convenient to combine these energies into a single energy 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: Specific Enthalpy = Specific Internal Energy + Specific Flow Energy

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 as listed below:

• hf = specific enthalpy of saturated water

• hfg = specific enthalpy of evaporation

The formation of steam

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

During the stage A to B sensible heat energy flows to the water accompanied by a rise of temperature. At B the water boils at a temperature referred to as saturation

temperature. This temperature depends on the pressure in the vessel and is 100°C at atmospheric pressure. The energy required to produce this temperature rise is called the ‘liquid enthalpy’.

During stage B to C steam is being formed whilst the temperature remains constant and the contents of the vessel will be a mixture of water and steam known as ‘wet steam’. At point C the steam will have received all the heat energy required to convert the water completely to dry steam. The energy required to produce the total change 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 be accompanied by a rise in temperature. The amount of heat energy in the superheat phase is called the ‘superheat enthalpy’.

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

Dryness fraction

The degree of dryness, or dryness fraction, of steam is that proportion of a given mass 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 fraction 0.6 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 = Moisture and Steam of mass Total Steam Dry of Mass

Layout and use of thermodynamic tables for water and steam

The 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 relate to entropy values and these are not required for this unit.

[

]

[ ]

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

SYMBOL QUANTITY UNIT

p Ts vg uf ug hf hfg hg 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 water

Specific enthalpy of evaporation Specific enthalpy of dry saturated steam

bar °C m3/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg

Thermodynamic property tables

With reference to pages 3, 4 and 5 of the Rogers & Mayhew tables: The first column headed p is the absolute pressure measured in bar, where 1 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 saturated steam. That is at pressure of 2 bar, saturation temperature is 120.2°C and 1 kg of dry steam occupies a volume of 0.8856 m3.

The fourth column uf is termed the specific internal energy of saturated liquid. That is the heat energy required to raise the temperature of 1 kg of water from 0°C to saturation 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 saturated steam. That is the heat energy required to raise the temperature of 1 kg of water at 0°C to saturation temperature plus the heat energy required to completely evaporate it (specific enthalpy of evaporation) as if the operation were carried out at constant volume.

The sixth column hf is the specific enthalpy of saturated liquid – i.e. the enthalpy of 1 kg of water from 0°C to saturation temperature at constant pressure. Note that uf and hf are identical down to 4.5 bar and then they gradually drift apart since hf increases slightly faster than uf.

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

The eighth column hg is the specific enthalpy of saturated vapour and is the enthalpy of 1 kg of dry saturated steam at constant pressure measured from water at 0°C. On page 2 of the Rogers & Mayhew tables the same properties are given but are set out against the reference of saturated water temperature (T°C) in the first column. The graph below illustrates three phases of steam formation and incorporates enthalpy values from tables.

From A to B, a heat transfer of 417 kJ kg-1 raises the temperature from 0°C to

saturation temperature (boiling point) of 99.6°C at the pressure of 1 bar.

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

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

Superheated steam

So far we have considered pages 2 to 5 of the thermodynamic tables. These pages set

TEMP hf = 417 kJ kg-1 hfg= 2258 kJ kg-1 hg= 2675 kJ kg-1 SENSIBLE HEAT PROCESS

EVAPORATION PROCESS _SUPERHEAT PROCESS

WATER & STEAM

SATURATION TEMPERATURE WATER SATURATED WATER DRY SATURATED STEAM SUPERHEATED STEAM HEAT ADDED kJ kg-1 Ts 9.6 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 stated pressure, then the steam is in a superheat state in which case we use pages 6 to 9 of the 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 brackets under each pressure is the saturation temperature corresponding to that pressure.

Column 2 lists the properties still of dry saturated steam i.e.

vg - Specific volume of dry saturated steam

ug - Specific internal energy of dry saturated steam hg - Specific enthalpy of dry saturated steam.

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

Rule

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

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

Units in superheated steam tables: v in m3 kg-1

u and h in kJ kg-1

Where exact values of the condition of steam are not listed in the tables, linear interpolation for both pressure and temperature may therefore be required. Class exemplars 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 **

Example

Determine the specific volume of wet steam having a pressure of 1.25 MN m-2 and dryness fraction 0.9. 1.25 MN m-2 = 12.5 bar, vx = x vg 12 bar 13 bar ⇒ v = 0.9 0.1632+0.1512 = 0.04148 m3 kg-1 TEMP SPECIFIC VOLUME Vx = Vf + X Vfg X Vfg Vf Vfg = Vg - Vf Vg

Specific internal energy of wet steam

The 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 shown below: TEMP SPECIFIC INTERNAL 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

Example

Determine the specific internal energy in wet steam at a pressure of 4 bar when it has a 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

Specific enthalpy of wet steam

The 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

Example

Determine the specific enthalpy of wet steam at a pressure of 70 kN m-2 and having a dryness 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 SPECIFIC ENTHALPY hx = hf + Xhfg X hfg STEAM (1 - x) kg WATER hf hfg = hg - hf hg Xhfg

Using steam tables and wet steam formulae

Example

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

From steam tables at a pressure of 3

bar:-Saturation temperature Ts = 133.5°C

Specific volume vg = .6057 m3 kg-1

Specific internal energy of sat. liquid uf = 561 kJ kg-1 Specific internal energy of sat. vapour ug = 2544 kJ kg-1 Specific enthalpy of sat. liquid hf = 561 kJ kg-1 Specific 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

Interpolation of steam tables

When quantities cannot be extracted directly from tables, intermediate values need to be 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-1

at 70 bar, vg = 0.02737 m3 kg-1

Difference = 0.00235

∴ at 68 bar, vg = 0.02972 – (3/5 x .00235) = 0.02831 m3 kg-1 or 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-1 at 55 bar, hg = 2790 kJ kg-1

Difference = 4

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

3. Determine the specific internal energy of superheated steam at 14 bar and 325°C. at 10 bar and 325°C, u = 2 2875 + 2794 = 2834.5 kJ kg-1 at 15 bar and 325°C, u = 2 2868 + 2784 = 2826.0 kJ kg-1 Difference = 8.5 kJ kg-1 ∴ at 14 bar and 325°C, u = 2834.5 – (4/5 x 8.5) = 2827 kJ kg-1 or u = 2826.0 + (1/5 x 8.5) = 2827 kJ kg-1 Refrigeration

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

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

Mechanical refrigeration is primarily an application of thermodynamics wherein the cooling substance goes through a cycle in which it is recovered for re-use. A

thermodynamic cycle can be operated in the forward direction to produce mechanical power from heat energy, or it can be operated in the reverse direction to produce heat energy from mechanical power. The reversed cycle is essentially utilised for the cooling effect that it produces during a portion of the cycle and is thus called a

refrigeration cycle.

Vapour-compression refrigeration cycles

The most widely used domestic refrigerators function on a vapour-compression cycle which operates between two pressure levels using a two-phase working substance or refrigerant which alternates cyclically between the liquid and vapour phases in continuos circulation.

In order to convert a liquid into a vapour, an energy transfer is required. This energy is acquired by the vapour molecules in the evaporation process and is termed the

enthalpy 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 of

condensation.

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

Commonly, the vapour compression cycle system within a refrigerator comprises four main devices, viz:

• Evaporator • Compressor • Condenser

• Expansion or throttle valve.

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

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

refrigerating effect.

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

Systems Diagram for Vapour – Compression Cycle

The cycle is completed when the refrigerant is expanded through a throttle valve back to its original low pressure, low temperature, wet state at point 1. The enthalpy at 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.

Refrigerant

The working fluid that circulates in a refrigeration system is called a refrigerant and may 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 and pressure, 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 by fluorocarbons such as Freon (refrigerants R12 and R22).

Compressor 3 2 Condenser Evaporator Throttle Valve 4 1 High Pressure Side Low Pressure Side Q1 Heat Rejection Q2 Heat Absorption Work Input

Apart from the ability to boil (evaporate) at a low temperature, refrigerants should possess 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 for any particular application must represent some form of compromise. The R family of refrigerants is the safest group and most widely used. All new refrigerants in the R family should have zero ozone depletion potential and be user friendly. Property tables and charts are produced for the various refrigerants similar to those produced for water and steam.

The behaviour of refrigerants is akin to that of water when subjected to heat. Water boils at 100°C when heat energy is supplied at atmospheric pressure. Evaporation then takes place at constant temperature until the vapour is completely dry and in the gaseous state. Further heating raises the temperature and the fluid is in the

superheated condition.

When the temperature and pressure of a refrigerant bear a `natural’ stable relationship to 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. It will 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 Liquid Refrigerant at 40°C and 9.6 bar - Sensible Heat + Sensible Heat Sat. Refrigerant Vapour at 40°C and 9.6 bar Undercooled Liquid Refrigerant at 35°C and 9.6 bar Superheated Refrigerant Vapour at 45°C and 9.6 bar

Use of thermodynamic tables for refrigerants

In this outcome we have already dealt with the use of property tables for water and steam. With the exception of entropy (s values), our studies now extend into the interpretation and extraction of information covering the ammonia and fluorocarbon refrigerants R717 and R12 as listed on pages 12 and 13 of the Rogers and Mayhew tables.

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

SYMBOL QUANTITY UNIT

T ps vg hf hg h Saturation Temperature

Corresponding Saturation Pressure Specific Volume of Saturated Vapour Specific Enthalpy of Saturated Liquid Specific Enthalpy of Saturated Vapour Specific Enthalpy of Superheated Vapour

°C bar 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 its saturation temperature, T, and its corresponding pressure, ps bear a `natural’ stable relationship to one another.

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

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

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

The heat energy required to change 1 kg of saturated liquid refrigerant to a completely 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 extracted directly 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 the superheat temperature and the saturation temperature at the specified pressure.

Example

Determine the specific enthalpy of refrigerant R717 at a pressure of 2.908 bar and temperature of 20°C.

From tables, the saturation temperature at 2.908 bar is – 10°C. Hence (T – Ts) = 30°C 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, then Specific Enthalpy, h = 1665.3 kJ kg-1 (from 100 K column).

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

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

Thermodynamics deals with the relationships between energy transfers within such devices/systems in the form of heat and work, and the related changes in the

properties of the working fluid.

Steady flow thermodynamic systems

The steady flow energy equation (SFEE) is applicable to open two-flow systems where the working fluid may be a gas or vapour. Steady flow conditions prevail when 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, its boundary, 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 heat

b) energy in the form of work

c) a mass of fluid which will possess certain forms of energy.

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

SURROUNDINGS (USUALLY THE ATMOSPHERE)

WORK, IN OR OUT HEAT, IN OR OUT WORKING FLUID LEAVING WORKING FLUID ENTERING SYSTEM CONTAINING THE WORKING FLUID BOUNDARY, e.g. THE WALL OF

Heat received or rejected

In any system a fluid can have a direct reception or rejection of heat energy transferred 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 done

In any system a fluid can do external work or have external work done on it transferred 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 to convert common thermodynamic devices into representative input/output sub-system diagrams. Examples of these follow in pages 3 to 7 of the notes and also in the tutorial questions.

Steam power plant

An important industrial application for vapours is the steam power plant as represented in the system diagram shown below:

BOILER TURBINE PUMP WORK OUT ELECTRIC GENERATOR COOLING WATER CONDENSER Q OUT Q IN

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

Wet or superheated steam from the boiler rotates the turbine and the work output

W+ve drives an electric generator via the turbine’s output shaft.

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

Each of the devices identified in the above diagram can be categorised as sub-systems of the integrated whole steam power plant. Each of the items is an example of a steady-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 is generated. In essence, a conventional boiler consists of a water container together with some heating device.

The boiler is supplied with a steady flow of water which is converted into wet steam using the heat released by burning a fuel such as coal, oil or gas. If superheated

1 2

WATER HEAT LOSSQ-VE

HEAT INPUTQ+VE FEED WATER

INPUT

STEAM OUTPUT WET 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 loss through the boiler casing to the surroundings.

In a steam power plant the boiler provides wet or superheated steam to the turbine in the 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 level and impinges across curved blades causing the turbine to rotate. An output shaft coupled to the blade mechanism delivers external work. The exhaust steam exits from the system with a low energy level.

Heat may be lost from the system to the surroundings or additional heat may be transferred 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 an electric 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

TURBINE

SYSTEM

Heat exchanger/condenser

MASS FLOW AT 3 = MASS FLOW AT 4

MASS 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 colder fluid, 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 for cooling into condensate. A simple integrated systems diagram for these 2 devices is shown below. CONDENSER COOLING WATER STEAM TURBINE 1

CONDENSER

COOLING WATER

STEAM

INPUT

3

4

OUTPUT

Rotary air compressor

Input/output system

In the rotary type compressor, atmospheric air is induced to a cylinder where it is compressed by an offset rotor and blade mechanism or rotary screw type

arrangement. The high-pressure air is subsequently delivered to a storage tank from where 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 input shaft drives the rotor. In this case work is done on the system.

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

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

Potential energy

This is the energy possessed by a mass of fluid, m, by virtue of its height Z above a given 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 TO SURROUNDINGS

COMPRESSOR

SYSTEM

Kinetic energy

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

Total Kinetic Energy = ½ 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 may all be present. In other systems only some may be present. Not infrequently, energy forms of insignificant value may be ignored in the solution of problems.

The steady flow energy equation

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

W.

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

1 SYSTEM ENERGY IN FLUID ENTERING SYSTEM E1 FLUID IN HEAT Q IN W OUT FLUID OUT ENERGY IN FLUID LEAVING 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 C₁2 (J) • Internal energy = U1 (J) • Flow energy = p1V1 (J)

Hence, total energy of the moving fluid mass entering the system = mgZ1 + m

2 C₁2

+ U1 + p1V1

Also, total energy of the moving fluid mass leaving the system = mgZ2 + m

2 C22

+ U2 + p2V2

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

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

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

⇒ mgZ1 + m 2 C12 + 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 + 2 C12 + u1 + p1v1 + Q = gZ2 + 2 C22 + u2 + p2v2 + W

Also, the combination of properties of internal and flow energies is called enthalpy and 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 + 2 C12 + h1 + Q = gZ2 + 2 C22 + 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 as follows: Q = W + m[g(Z2– Z1) +     2 C -C2_{2 1}2 + (u2– u1) + (p2v2– p1v1)] OR Q = W + m[g(Z2– Z1) +     2 C -C2_{2 1}2 + (h2– h1)]

Frequently in thermodynamic problems, changes in potential energy are small

compared with other energy changes or even non-existent when there is no difference between entry and exit datum levels. The gZ terms can therefore be neglected or dropped and the equation shortens to:

Q = W + m [     2 C -C 12 2 2 + (h2– h1)]

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

Similarly, for the KE formula, the symbol C is used for fluid velocity in order to distinguish 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 forms listed.

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

PRACTICAL EXEMPLAR PROBLEMS

Exemplar 1 (SFEE)

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

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

At pressure of 60 bar and temperature of 450°C, specific enthalpy, h2 from superheat tables 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 bar t2 = 450°C . 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 temperature of 120.2°C 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 steam at 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 at 2 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 = 1.5 4360.2 = 2906.8 kJ kg-1

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

From Superheat tables:

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

Difference = 105 and 49.8 Thus, temperature of steam at inlet = 200°C + (50°C x

105 8 . 49

)

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

W STEAM OUTPUT STEAM INPUT TURBINE 2 1 = 630 kW t2 = 120.2°C x = 0.9 p1 = 500 kN m-2 h1 = ? m = 1.5 kg s-1

OUTCOME 4: APPLY THE MASS CONTINUITY AND BERNOULLI’S EQUATIONS TO FLOW THROUGH PIPES

This outcome covers the development and utilisation of the mass continuity

equation and Bernoulli’s equation as applied to the steady flow of incompressible

fluids (liquids) through pipes.

Mass continuity equation

Consider sections 1 and 2 in the tapered pipe shown in figure below which is full of steadily flowing fluid. At Section 1, the cross-sectional area is A1, and the velocity of the 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 the pipe at datum position 1 is the same as the rate at which it leaves at datum 2; i.e. the mass flow rate is constant,

Now,

Volume of fluid passing = Volume of fluid passing Sect. 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 = A C = A C (m2 x m s-1 = m3 s-1) **

. C R O S S -S E C TIO N AL A R E A A₁ C R O S S -S E C TIO N AL A R E A A₂ LIQ U ID IN V E LO C IT Y

C

_C

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

Density, ρ (RHO) = V m VOLUME UNIT MASS = (units kg m-3)

Hence, mass flow rate, m = ρ x V = ρ x AC (kg m-3 x m2 x m s-1 = kg s-1) This equation m = ρ AC is known as the mass continuity equation ** Where m = mass flow rate in kg s-1

ρ = density of fluid in kg m-3

A = cross-sectional area of pipe in m2

C = velocity of fluid in m s-1

Branched piping systems

In many pipework systems, a single pipeline will split into two or more branches as shown in figure below:

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

V1 = V2 + V3

Hence V = A1C1 = A2C2 + A3C3 (m3s-1) ** and m = ρ A1C1 = ρ (A2C2 + A3C3) (kg s-1) **

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

. . . . . . . . . 1 2 3

Energy of a flowing fluid

In outcome 3 we determined that the total energy of a mass, m, of flowing fluid had four components. Potential Energy mgZ (J) Kinetic Energy 2 2 mC (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 the fluid.

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

The specific volume v of a fluid in m3kg-1 is the reciprocal of density, i.e. v = 1/ρ .

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

gZ + ρ p 2 C2 + (J kg-1)

Further dividing throughout by g, gives:

(

)

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

Bernoulli’s equation

Consider unit mass of fluid flowing at a steady rate through the system/pipeline shown in the figure below.

Applying the Principle of Conservation of Energy we have: Specific energy in the = Specific energy in the fluid entering system fluid leaving system

⇒Specific Specific Specific Specific Specific Specific

Potential + Kinetic + Flow = Potential + Kinetic + Flow

Energy Energy Energy Energy Energy Energy

At entry At exit ∴gZ1 +  p + 2 C + gZ =  p + 2 C 2_{2 2} 2 1 2 1 = constant (J kg –1) Dividing both sides of the equation by gravitational acceleration, g.

Z1 + = J p + 2g C + Z = J p + 2g C 2 2 2 2 1 2 1 constant (m) This is known as Bernoulli’s 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 PRESSURE VELOCITY C p₁ 1 PRESSURE VELOCITY C p₂ 2 SYSTEM

Frictional resistance to flow (loss of energy)

Bernoulli’s equation assumes there to be no frictional resistance to the flow of an incompressible fluid/liquid through a system/pipeline. In practical applications however, frictional resistance to flow is always present and reduces the available energy in the fluid at exit from the system. Thus:

Specific Energy in Specific Energy in Specific Energy Fluid Entering the = Fluid Leaving the + To Overcome

System System Frictional Resistance

Let ZF = Frictional Resistance `Head’

Then, Bernoulli’s Equation can be stated in the form:

Z1 + F 2 2 2 2 1 2 1 Z + J p + 2g C + Z = J p + 2g C

As before, each term represents Head of Liquid in units of m. Z = potential head 2g C2 = kinetic head J p = pressure head

SELF-ASSESSMENT

Assignment 4

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

QUANTITY SYMBOL/FORMULA UNIT

A PRESSURE HEAD m3s-1 mgZ FLUID VELOCITY KINETIC HEAD kg s-1

State Bernoulli’s equation. Identify each symbol in the equation and list units for each on one side only.

EQUATION:

SYMBOL QUANTITY UNIT

State the Mass Continuity Equation. EQUATION:

PRACTICAL EXEMPLAR PROBLEM

Bernoulli & Mass Continuity Equations

Oil of density 800 kg m-3 enters a 500 mm diameter pipeline with a velocity of 3 m s-1 and pressure 400 kN m-2, and is discharged through an orifice 150 mm diameter and 30 m below the entry point. If the frictional losses are equivalent to a head of 3.5 m of the oil, determine the velocity, the mass flow rate, and the pressure of the oil at the point of discharge.

Velocity of oil at exit from A1C1 = A2C2⇒ C2 = 1 2 1 C x A A C2 = 1 -2 2 s m 33.33 = 3 x .0225 .25 = 3 x .15 x 4 Œ 4 x .5 Œ

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

MASS FLOW RATE, m = ρ A2C2 = 800 x π x .15₄ 2 x 33.33 = 471.2 kg s-1 Use Bernoulli’s equation for pressure at exit

Z1 + F 2 2 2 2 1 2 1 Z + !J p + 2g C + Z = !J p + 2g C 30 + +3.5 9.81 x 800 p + 9.81 x 2 33.33 + 0 = 9.81 x 800 10 x 400 + 9.81 x 2 32 3 2 ₂ 30 + .4587 + 50.9684 = 56.62 + +3.5 7848 p2 81.4271 = 60.12 + 7848 p₂ . d2 = 150 mm p2 = ? C2 = ? OIL ρ = 800 kg m-3 ZF = 3.5 m d1 = 500 mm p1 = 400 kN m-2 C1 = 3 ms-1 30 m

PRESSURE p2 AT POINT OF DISCHARGE 167.2 kN m-2

Venturi meter

In many engineering applications it is necessary to measure and control the flow rates of fluids. The Venturi meter is a device which allows pressure changes to be

measured at different sections in a pipeline carrying a flowing fluid. Once pressure differentials are obtained, velocity and flow rates can be calculated using Bernoulli’s Equation.

The figure below illustrates a typical horizontally mounted Venturi meter where fluid enters 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 fluid passes through the restricted throat section, its velocity, and consequently its kinetic energy will increase. Since the total energy of the steadily flowing fluid remains constant, it follows that the pressure and flow energy will decrease at the throat section.

A U-tube manometer, with say mercury as the denser fluid, fitted between sections 1 and 2 allows the pressure difference to be measured. The higher pressure at Section 1 causes the mercury in the manometer to be forced down the LH limb and up in the RH limb. h y FLU ID D E N SITY EN TR Y EX IT M AN O M ETE R FLU ID D E N SITY X X 1 2 P₁ C1 A₁ P₂ C₂ A2

Reference the figure and applying Bernoulli’s equation between sections 1 and 2 of the Venturi gives:

J p + 2g C = J p + 2g C 2 ₂ 2 1 2 1 (since Z1 = Z2 here) ⇒ 2g C -C = J p -p 12 2 2 2 1 (Equation 1)

The volumetric flow equation gives V = A1C1 = A2C2 ⇒ C2 = 1 2 1 C x A A

Substituting for C2 in equation 1 gives:

2g C -C A A = J p -p 2 1 2 1 2 1 2 1 ⇒ 2g 1 -A A C = J p -p 2 2 1 2 1 2 1 (Equation 2)

The pressure at level x – x is the same in both limbs of the manometer. Also, the pressure at a depth in a liquid is obtained from the formula; pressure, p = ρ gh. Now, Pressure in LH Limb at x – x = Pressure in RH Limb at x – x

⇒ p1 + ρ g (Y + h) = p2 + ρ gY + ρHggh ⇒ p1 + ρ gY + ρ gh = p2 + ρ gY + ρHggh ⇒ p1 + ρ gh = p2 + ρHggh ⇒ p1– p2 = gh (ρHg - ρ ) Divide by ρ ⇒   -gh( =  p -p_{1 2} Hg =   -  gh Hg = gh -1  Hg ∴ p1-p2 =h Hg -1 (Equation 3) .

Equating equations 2 and 3: 1 -  h = 2g 1 -A A C Hg 2 2 1 2 1 Therefore velocity C1 = 1 A A 1 -  2gh 2 2 1 Hg (Equation 4) . .

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

Also, mass flow rate from equation m = ρ A1C1

The above equation 4 for calculating velocity involving a Venturi meter is a

derivation of Bernoulli’s equation. Students would not be expected to memorise or reproduce this formula from first principles but would be required to apply formula in an open book situation.

The orifice plate

A less sophisticated and much less costly device for measuring incompressible fluid flow rates in pipelines is the orifice plate. This consists of a circular plate with a concentric orifice which is inserted into the pipeline.

In passing through the orifice the liquid stream contracts and in so doing converts static pressure head into velocity head as in a Venturi meter. The reduction in pressure head can be measured using a manometer with tapping points sited either side of the orifice plate. Upstream, the pressure should be tapped at a position where the flow pattern is not influenced by the presence of the orifice plate. The ‘throat’ pressure tapping point is situated downstream from the orifice plate where the effective diameter of the converging liquid stream is at a minimum.

Downstream turbulence causes orifice plates to have a typical co-efficient of discharge of around 0.65 whilst a Venturi meter is closer to 1. Generally orifice plates 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.

P1 PL P A1 C1 P2 A2 C2

h