DET: Mechanical Engineering Thermofluids (Higher)

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1 DET: Mechanical Engineering Thermofluids (Higher) 6485

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3 Spring 000 HIGHER STILL DET: Mechanical Engineering Thermofluids Higher Support Materials *+,-./

4 CONTENTS Section : Thermofluids (Higher) Student Notes Section : Self-Assessment Answers Section 3: Tutorials Section 4: Tutorials Marking Scheme Section 5: Appendix, Quantities used in Thermofluids (Higher) DET: Mechanical Engineering: Thermofluids Higher

5 DET: Mechanical Engineering: Thermofluids Higher

6 Section : Thermofluids (Higher) Student Notes DET: Mechanical Engineering: Thermofluids Higher Support Notes

7 DET: Mechanical Engineering: Thermofluids Higher Support Notes

8 OUTCOME : 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 p g, the actual or absolute pressure p in the vessel is given by: p = p g + p atm 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 - or Pa (PASCAL). The bar = 0 5 N m - = 00 kn m - is also commonly used. Standard atmospheric pressure = atm =.03 bar =.03 x 0 5 N m -. 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,000,000 = 0 6 Mega M,000 = 0 3 Kilo k e.g. 7,00,000 N m - = 7,00 kn m - or preferably 7. MN m -. DET: Mechanical Engineering: Thermofluids Higher Support Notes

9 Absolute temperature In problems involving the gas laws, the temperature of any gas is measured from absolute zero, which has been determined to be 73 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 73 to the Celsius temperature scale reading. i.e. Absolute temperature = Celsius temperature scale reading Hence 7 C = 300 K. Absolute temperature takes the SI base unit, the Kelvin (K), and has the symbol T. Note: A change in temperature of C = a change in temperature of 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: m 3. The recommended unit is the cubic metre. Subdivisions such as the cm 3 or litre (l) are also used, but as a general rule it is safer to convert data to give volumes in m 3 to avoid errors in calculations. Specific volume Symbol: v (small letter). Unit: m 3 kg -. This is the volume per unit mass and is the reciprocal of density. i.e. v = VOLUME MASS = V m 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes

10 Thus p. V e.g. or pv = constant 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 p V = p V = p n V n..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 V = constant T 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: V T = V T = V T n n..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: PRESSURE ABSOLUTE TEMPERATURE = constant p = C T (Provided neither mass or volume of gas changes) This is known as the Pressure Law, which may be stated algebraically in the form: p T = p T = p T n n DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

11 The Combined Gas Law If, during a process, the pressure, volume and absolute temperature of a gas are changed from p, V, and T to p, V and T 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: p V T = pv T = constant This is known as the Combined Gas Law. The Characteristic Gas Equation We have seen the combined gas law stated in the form: pv = constant C T 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 Or pv = mr T 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 - V = volume of the gas m 3 T = absolute temperature of the gas [(t + 73)] K m = mass of the gas kg R = gas constant J kg - K -. The table below lists the gas laws, together with appropriate equations for problem solving. DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

12 GAS LAW PROCESS CONDITION APPROPRIATE EQUATION Boyle s Law Constant Temperature p V = p V etc Charles s Law Constant Pressure V T = V T etc Pressure Law Constant Volume p p = T T etc Combined Gas Law Pressure Volume and Temperature all Change pv = T pv T 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. 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 = kj kg mol - K - 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 8. What is the value of R for nitrogen? R = Ro M = = 0.97 kj kg - 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. K - DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

13 SELF-ASSESSMENT Assignment. Test your knowledge of quantities, symbols and units covered so far by completing the table below. SPECIFIC VOLUME QUANTITY SYMBOL UNIT m m 3 ABSOLUTE PRESSURE K. Convert the undernoted temperature values from one scale to the other. o C K Boyle s Law deals with the behaviour of gases under isothermal conditions. What does the condition isothermal mean? Ans: DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

14 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: Identification of terms and units: DET: Mechanical Engineering: Thermofluids Higher Support Notes 7

15 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 A fixed mass of gas is compressed isothermally from a pressure of 0 bar and volume of 3. m 3 until it occupies 5% of its original volume. Calculate the final pressure of the gas. Known Data p = 0 x 0 5 N m - V = 3. m 3 p =? 5 V = 3. x 00 = 0.8 m 3 For an isothermal process, Boyle s Law applies and equation p V = p V can be utilised. From p V = p V p = V = p x V p = 8.0 x 0 6 N m - 0 x 0 5 x FINAL PRESSURE OF GAS, p = 8 MN m - or 80 bar. Exemplar Question A quantity of gas at a pressure of 80 kn m - and temperature 8 C, occupies a volume of.43 m 3. The gas is compressed until its pressure and temperature are 670 kn m - and 7 C respectively. If there is no loss of gas, what volume will it now occupy? Known Data From the Combined Gas Law p = 80 kn m - p = 670 kn m - T = (8 + 73) K T = (7 + 73) K V = 0.43 m 3 V =? p V T = p V T 3 p V T 80 x 0 x 0.43 x 400 V = = 3 p T 670 x 0 x 9 FINAL VOLUME OF GAS IS m 3 = m 3 DET: Mechanical Engineering: Thermofluids Higher Support Notes 8

16 Exemplar Question 3 An air receiver contains a fixed mass of air at a pressure of.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, p =.5 bar p = 7.8 bar the Pressure Law applies T = ( )K T =? Constant Volume Process V = V p p = T T T p = x T p x 0 = 5 x 357 =.768K.5 x 0 FINAL TEMP. OF AIR =.8 73 = -50. 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 m 3 and contains 4 kg of the gas at a pressure of 5 kn m - and a temperature of 0 C. Tank B has a volume of.0 m 3 and contains gas at a pressure of 340 kn m - and a temperature of 0 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 0 C. V = 3.0 V =.0 m 3 m = 4 TANK m =? p = 5 k B p = 340 kn m - T = 0 + TANK T = 93 K = 93 K A Char. Gas Constant from p V = m R T R = p V m T = 3 5 x 0 x x 93 GAS CONSTANT R = 57.4 J kg - K - DET: Mechanical Engineering: Thermofluids Higher Support Notes 9

17 MASS IN TANK B from p V = m R T m = p V R T x 0 x = 57.4 x 93 MASS OF GAS IN TANK B = kg When connecting valve is opened, the total volume V 3 = 5 m 3 and temperature remains at 93 K. FINAL PRESSURE from p3 V3 = m3 R T3 p 3 m3 R T = V 3 3 ( ) x 57.4 x 93 = 5 = N m - FINAL PRESSURE IN SYSTEM = 35 kn m - Exemplar Question 5 A fixed mass of gas contained in a closed system is initially at a pressure of 00 kn m -, a temperature of 5 C, and occupies a volume of 0.5 m 3. The gas is then compressed to a volume of 0.06 m 3 and a pressure of 300 kn m -. The gas is then expanded at constant pressure unit it reoccupies its original volume. If the characteristic constant for the gas is 89 J kg - K -, determine the mass of gas in the system and the temperature at the end of the compression and expansion processes. Known Data p = 00 x 0 3 N m - p = 300 x 0 3 N m - p 3 = p T = 5 C + 73 = 88 K T =? T 3 =? V = 0.5 m 3 V = 0.06 m 3 V 3 = V m =? R = 89 J kg - K - Mass of gas from p V = m R T m = p V R T 3 00 x 0 x 0.5 = = kg 89 x 88 MASS OF GAS IN SYSTEM = kg DET: Mechanical Engineering: Thermofluids Higher Support Notes 0

18 TEMPERATURE AFTER COMPRESSION from p V = m R T 3 p V 300 x 0 x 0.06 T = = = K m R x 89 TEMPERATURE AFTER COMPRESSION = = 7.56 C TEMPERATURE AFTER EXPANSION from p 3 V 3 = m R T 3 3 p3 V3 300 x 0 x 0.5 T 3 = = = K m R.756 x 89 TEMPERATURE AFTER EXPANSION T 3 = = C DET: Mechanical Engineering: Thermofluids Higher Support Notes

19 OUTCOME : 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 ( kg) of a substance and is identified by the symbol v (small letter). i.e. v = VOLUME V = UNIT : m 3 kg - MASS m Thermodynamic tables give the specific volume of dry saturated steam at a particular pressure under the heading v g. e.g. SPECIFIC VOLUME OF DRY SATURATED STEAM AT.4 bar =.36 m 3 kg - 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 50 C = m 3 kg - 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. When heat energy, which is a transient form of energy, is transferred to a fluid, the temperature and molecular activity of the fluid increases. These increases result in a corresponding increase in the store of internal energy within the fluid. DET: Mechanical Engineering: Thermofluids Higher Support Notes

20 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 - ). Thermodynamic tables give three values for the specific internal energy of steam as underlisted. u f = specific internal energy of saturated water u g = 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 V. For flow to occur, this volume must be displaced by an equal volume from outside the system. If the pressure in the fluid is p, then the work done on the fluid inside the system by the incoming fluid = force x distance the fluid is displaced. CROSS SECTIONAL AREA A X Y p p p VOLUME DISPLACED V p S S Flow energy DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

21 Now, Force = Pressure x Cross Sectional Area Therefore Work Done = p x A x s But, A s = Volume displaced V Therefore Work done on the system = p V In specific terms, i.e. per kg of mass, work done on system = p v 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 = p v (J kg - ) 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 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 - ). Thermodynamic property tables give four values for the specific enthalpy of steam as listed below: h f = specific enthalpy of saturated water h fg = specific enthalpy of evaporation h g = specific enthalpy of dry saturated steam h = specific enthalpy of superheated steam. DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

22 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 00 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

23 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 (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. Mass of Dry Steam Dryness fraction, x = Total mass of Steam and Moisture 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. P T vg u s f ug hf hfg h g 0 3 [ bar] [ C] m /kg kj/kg kj/kg The various symbols in the eight columns are identified with their quantities and units as below. SYMBOL QUANTITY UNIT p T s v g u f u g h f h fg h g 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 m 3 /kg kj/kg kj/kg kj/kg kj/kg kj/kg DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

24 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 bar = x 0 5 N m - or 00 kn m -. The second column headed T s is the temperature in C at which the water boils (saturation temperature). Note how T s changes relative to pressure. The third column v g is the specific volume in m 3 of kg of completely dry saturated steam. That is at pressure of bar, saturation temperature is 0. C and kg of dry steam occupies a volume of m 3. The fourth column u f is termed the specific internal energy of saturated liquid. That is the heat energy required to raise the temperature of kg of water from 0 C to saturation temperature (kj kg - ). This is a constant volume operation. The fifth column u g is the specific internal energy of kg of completely dry saturated steam. That is the heat energy required to raise the temperature of 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 h f is the specific enthalpy of saturated liquid i.e. the enthalpy of kg of water from 0 C to saturation temperature at constant pressure. Note that u f and h f are identical down to 4.5 bar and then they gradually drift apart since h f increases slightly faster than u f. The seventh column h fg is the specific enthalpy of evaporation i.e. the heat energy required to completely evaporate kg of water at saturation temperature to kg of dry saturated steam at constant pressure and at same temperature. The eighth column h g is the specific enthalpy of saturated vapour and is the enthalpy of kg of dry saturated steam at constant pressure measured from water at 0 C. On page 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 7

25 GRAPH OF HEAT ADDED AGAINST TEMPERATURE AT bar ABSOLUTE TEMP O C SENSIBLE HEAT PROCESS EVAPORATION PROCESS SUPERHEAT PROCESS D WATER WATER & STEAM SUPERHEATED STEAM SATURATED WATER DRY SATURATED STEAM Ts 9.6 O C B SATURATION TEMPERATURE C 0 C O A - hf hfg = 58 kj kg = 47 kj kg - hg = 675 kj kg - HEAT ADDED kj kg - From A to B, a heat transfer of 47 kj kg - raises the temperature from 0 C to saturation temperature (boiling point) of 99.6 C at the pressure of 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 58 kj kg -. 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 to 5 of the thermodynamic tables. These pages set out the properties and heat energy requirements for steam to be raised from water at 0 C to dry saturated steam at different pressures. DET: Mechanical Engineering: Thermofluids Higher Support Notes 8

26 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 Column as before, states the pressure (p) in bar but the figure in brackets under each pressure is the saturation temperature corresponding to that pressure. lists the properties still of dry saturated steam i.e. v g - Specific volume of dry saturated steam u g - Specific internal energy of dry saturated steam h g - 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 (T s ) for the corresponding pressure then superheated tables must be used. The difference between the superheated temperature (T) and the saturation temperature (T s ) is called the degree of superheat. Units in superheated steam tables: v in m 3 kg - u and h in kj kg - 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 9

27 Specific volume of wet steam The specific volume of steam with a dryness fraction x is given by: v x = v f + x v fg This is illustrated on the graph of temperature against specific volume shown below: TEMP Vx = Vf + X Vfg X Vfg Vf Vfg = Vg - Vf SPECIFIC VOLUME Vg Referring to above graph: v fg = v g - v f Therefore v x = v f + x(v g v f ) = v f + xv g x v f = ( x) v f + x v g Since v f is extremely small compared with v g, the term ( x) v f may be ignored. Hence, v x = x v g ** Example Determine the specific volume of wet steam having a pressure of.5 MN m - and dryness fraction MN m - =.5 bar, v x = x v g bar 3 bar v x = 0.9 = m 3 kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes 0

28 Specific internal energy of wet steam The specific internal energy of steam with a dryness fraction of x is given by: u x = u f + x u fg This is illustrated in the graph of temperature against specific internal energy shown below: TEMP Ux = Uf + X Ufg X Ufg Uf Ufg = Ug - Uf SPECIFIC INTERNAL ENERGY Ug Temperature against specific internal energy Referring to above graph: u fg = u g - u f Therefore u x = u f + x(u g u f ) = u f + xu g x u f Therefore u x = ( x) u f + x u g Example Determine the specific internal energy in wet steam at a pressure of 4 bar when it has a dryness fraction of Specific internal energy u x = ( x)u f + xu g u x = ( 0.87) x 554 u x = u x = kj kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes

29 Specific enthalpy of wet steam The specific enthalpy of steam with a dryness fraction x is given by: h x = h f + x h fg ** This is illustrated on the graph of temperature against specific enthalpy shown below: TEMP hx = hf + Xhfg Xhfg X hfg STEAM ( - x) kg WATER hf hg hfg = hg - hf SPECIFIC ENTHALPY Temperature against specific enthalpy Example Determine the specific enthalpy of wet steam at a pressure of 70 kn m - and having a dryness fraction of Pressure of 70 kn m - = 0.7 bar Specific enthalpy h x = h f + x h fg = x 83 = SPECIFIC ENTHALPY, h x = kj kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes

30 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 - ) when it is.8 dry. What is the saturate temperature? From steam tables at a pressure of 3 bar:- Saturation temperature T s = 33.5 C Specific volume v g =.6057 m 3 kg - Specific internal energy of sat. liquid u f = 56 kj kg - Specific internal energy of sat. vapour u g = 544 kj kg - Specific enthalpy of sat. liquid h f = 56 kj kg - Specific enthalpy of evaporation h fg = 64 kj kg - Specific internal energy of steam at 3 bar and.8 dry. u x = (- x) u f + xu g = ( -.8) x x 544 = u x = kj kg - Specific enthalpy of steam at 3 bar and.8 dry. h x = h f + xh fg = x 64 h x = kj kg - Specific volume at 3 bar and.8 dry. v x = xv g =.8 x.6057 v x =.4967 m 3 kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

31 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. Determine the specific volume of wet steam at 68 bar. at 65 bar, v g = m 3 kg - at 70 bar, v g = m 3 kg - Difference = at 68 bar, v g = (3/5 x.0035) = m 3 kg - or v g = (/5 x.0035) = m 3 kg -. Determine the specific enthalpy of dry saturated steam at 5 bar. at 50 bar, h g = 794 kj kg - at 55 bar, h g = 790 kj kg - Difference = 4 at 5 bar, h g = 794 (/5 x 4) = 79.4 kj kg - or h g = (3/5 x 4) = 79.4 kj kg - 3. Determine the specific internal energy of superheated steam at 4 bar and 35 C. at 0 bar and 35 C, u = at 5 bar and 35 C, u = = kj kg - = 86.0 kj kg - Difference = 8.5 kj kg - at 4 bar and 35 C, u = (4/5 x 8.5) = 87 kj kg - or u = (/5 x 8.5) = 87 kj kg - 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

32 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 and boils (evaporates) to a nearly dry state at point 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

33 Work Input Heat Rejection Q 3 Compressor Condenser Evaporator Heat Absorption Q Throttle Valve 4 High Pressure Side Low Pressure Side 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. The enthalpy at point 4 being equal to the enthalpy at point. This cycle of operations is repeated on a continuous basis in order to maintain a predetermined 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 R and R). The Freon refrigerants R and R are general purpose fluids specially manufactured for refrigeration and these are non-toxic and non-flammable. DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

34 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 00 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 Undercooled Liquid Refrigerant at 35 C and 9.6 bar Sat. Refrigerant Vapour at 40 C and 9.6 bar + Sensible Heat Superheated Refrigerant Vapour at 45 C and 9.6 bar DET: Mechanical Engineering: Thermofluids Higher Support Notes 7

35 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 R77 and R as listed on pages and 3 of the Rogers and Mayhew tables. Refrigerant quantities together with appropriate symbols and units are identified in the table below: SYMBOL QUANTITY UNIT T p s v g h f h g 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 m 3 kg - kj kg - kj kg - kj kg - As previously stated, a refrigerant is regarded as being in its saturated state when its saturation temperature, T, and its corresponding pressure, p s bear a `natural stable relationship to one another. The specific volume, v g, 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, v x = xv g. e.g. Spec. Volume of Refrigerant R at.004 bar and.96 dry v x =.96 x.594 =.530 m 3 kg - The heat energy required to change kg of saturated liquid refrigerant to a completely dry saturated vapour (gas) is called the enthalpy of evaporation. i.e. e.g. Enthalpy of evaporation h fg = h g - h f Enthalpy of evaporation of refrigerant R77 at.680 bar h fg = = kj kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes 8

36 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 00 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 R77 at a pressure of.908 bar and temperature of 0 C. From tables, the saturation temperature at.908 bar is 0 C. Hence (T T s ) = 30 C or 30 K. Specific Enthalpy, h = 55.7 kj kg - (From 50 K column). If the degree of superheat had been, say, 84K at the same pressure, then Specific Enthalpy, h = kj kg - (from 00 K column). DET: Mechanical Engineering: Thermofluids Higher Support Notes 9

37 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. WORKING FLUID ENTERING SURROUNDINGS (USUALLY THE ATMOSPHERE) WORK, IN OR OUT SYSTEM CONTAINING THE WORKING FLUID WORKING FLUID LEAVING HEAT, IN OR OUT BOUNDARY, e.g. THE WALL OF 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 30

38 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 - 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 - 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 WORK OUT Q IN ELECTRIC GENERATOR CONDENSER COOLING WATER PUMP Q OUT STEAM POWER PLANT -- SYSTEM DIAGRAM DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

39 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 HEAT LOSSQ -VE WET STEAM SPACE STEAM OUTPUT FEED WATER INPUT WATER HEAT INPUTQ +VE MASS FLOW AT = MASS FLOW AT 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 steam is required, the wet steam is removed from the steam space in the boiler and piped into an integrated superheater where it is further converted into dry or superheated steam by the addition of more heat energy. DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

40 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 Q FLUID INPUT TURBINE SYSTEM WORK OUTPUT FLUID OUTPUT MASS FLOW AT = MASS FLOW AT 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 33

41 Heat exchanger/condenser COOLING WATER STEAM INPUT CONDENSER OUTPUT 3 4 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 devices is shown below. COOLING WATER TURBINE STEAM CONDENSER 3 4 DET: Mechanical Engineering: Thermofluids Higher Support Notes 34

42 Rotary air compressor HEAT LOSS Q TO SURROUNDINGS LOW PRESSURE INTAKE WORK INPUT COMPRESSOR SYSTEM HIGH PRESSURE OUTPUT 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 - x m) = Nm = (J) and for unit mass of the fluid Specific potential energy = gz (J kg - ) DET: Mechanical Engineering: Thermofluids Higher Support Notes 35

43 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 mc (kg x m s - x m s - ) = Nm = (J) and for unit mass of the fluid Specific Kinetic Energy = C (J kg - ) 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. ENERGY IN FLUID ENTERING SYSTEM E FLUID IN SYSTEM W OUT Z HEAT Q IN ENERGY IN FLUID LEAVING SYSTEM E FLUID OUT DET: Mechanical Engineering: Thermofluids Higher Support Notes 36

44 The forms of energy associated with the moving fluid mass entering the system are: Potential energy = mgz (J) Kinetic energy = m C (J) Internal energy = U (J) Flow energy = p V (J) Hence, total energy of the moving fluid mass entering the system = mgz + m C + U + p V Also, total energy of the moving fluid mass leaving the system = mgz + m C + U + p V 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 mgz + m C + U + p V + Q = mgz + m C + U + p V + W This is known as the steady flow energy equation. For unit mass ( kg) of working fluid the equation becomes: gz + C + u + p v + Q = gz + C + u + p v + 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: gz + C + h + Q = gz + C + h + W for unit mass of working fluid. DET: Mechanical Engineering: Thermofluids Higher Support Notes 37

45 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:... C - C Q = W + m[g(z Z ) + + (u u ) + (p v p v )] OR Q = W + m[g(z Z ) + C - C + (h h )] 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:... C - C Q = W + m [ + (h h )] 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 38

46 SELF-ASSESSMENT Assignment 3. What practical purpose does a condenser serve in a steam power plant? Ans:. 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 WORK TRANSFER SPEC. INT. ENERGY. Q H DET: Mechanical Engineering: Thermofluids Higher Support Notes 39

47 PRACTICAL EXEMPLAR PROBLEMS Exemplar (SFEE) Pressurised feed water with a Specific Liquid Enthalpy of 97 kj kg - is supplied to a boiler facility at a Mass Flow Rate of 3.4 kg s -. 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 - (KW). Insert relevant given data into the systems diagram shown below and determine the required rate of heat input.. Q = 36 kj s - h = 97 kj kg Boiler p = 60 bar t = 450 C. m = 3.4 kg s - At pressure of 60 bar and temperature of 450 C, specific enthalpy, h from superheat tables is 330 kj kg -. Now, Energy input = energy output Inlet enthalpy + heat input = Exit enthalpy + heat loss.... m x h + heat input Q = m x h + Q... Required heat input rate, Q = m (h h ) + Q. Q = 3.4 (330 k 97 k) + 36 k = (3.4 x 39 k) + 36 k = k + 36 k. Q = kj s - i.e. REQUIRED RATE OF HEAT INPUT = kj s - or 7.95 MW DET: Mechanical Engineering: Thermofluids Higher Support Notes 40

48 Exemplar (SFEE) Steam enters a turbine with a pressure of 500 kn m - and leaves with a temperature of 0. C and dryness fraction 0.9. The power output from the turbine is 630 kw. If the mass flow rate of the steam is.5 kg s -, determine the temperature of the steam at entry to the turbine. W = 630 kw STEAM INPUT p = 500 kn m - h =? TURBINE m =.5 kg s - STEAM OUTPUT t = 0. C x = 0.9 Here, the PE and KE terms are assumed to be negligible. At exit, steam with a sat. temp. of 0. has a pressure of bar. Thus, spec. enthalpy at exit, h = h f + xh fg at bar. h = ( x 0) = kj kg - ENERGY INPUT = ENERGY OUTPUT... m x h = (m x h ) + W.5 x h = (.5 x 486.8) in kj s - = SPEC. ENTHALPY, h = = kj kg -.5 At a pressure of 500 kn m - (5 bar) and spec. enthalpy of steam is in superheated condition with temperature between 00 C and 50 C. From Superheat tables: AT 5 bar and 50 C, spec. enthalphy = 96 inlet enthalpy = AT 5 bar and 00 C, spec. enthalphy = 857 at 5 bar & 00 C = Difference = 05 and Thus, temperature of steam at inlet = 00 C + (50 C x 05 ) TEMPERATURE OF STEAM AT INLET = = 3.7 C DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

49 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 and in the tapered pipe shown in figure below which is full of steadily flowing fluid. At Section, the cross-sectional area is A, and the velocity of the fluid is C ; at section, the area and velocity are A and C respectively. CROSS-SECTIONAL AREA A LIQUID IN VELOCITY C VELOCITY C CROSS-SECTIONAL AREA A Continuity equation Steady flow conditions prevail when the rate at which the mass of fluid entering the pipe at datum position is the same as the rate at which it leaves at datum ; i.e. the mass flow rate is constant, Now, Volume of fluid passing = Volume of fluid passing Sect. per unit time Sect. per unit time A C = A C (Since, volume per unit time = C.S. Area x Velocity). Volumetric flow rate V = A C = A C (m x m s - = m 3 s - )** DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

50 Volumetric flow rates for fluids can be converted to mass flow rates by introducing density into the equation. In outcome, density was defined as follows: Density, ρ (RHO) = MASS UNIT VOLUME m = (units kg m -3 ) V.. Hence, mass flow rate, m = ρ x V = ρ x AC (kg m -3 x m x m s - = kg s - ). This equation m = ρ AC is known as the mass continuity equation ** Where. m = mass flow rate in kg s - ρ = density of fluid in kg m -3 A = cross-sectional area of pipe in m C = velocity of fluid in m s - Branched piping systems In many pipework systems, a single pipeline will split into two or more branches as shown in figure below: 3 Like the tapered pipe example already dealt with, steady flow conditions will apply when the volumetric and mass flow rates passing Section equal the combined total of volume and mass flow rates passing sections and 3, i.e.. V =.. V + V 3 Hence. V = A C = A C + A 3 C 3 (m 3 s - ) ** and. m = ρ A C = ρ (A C + A 3 C 3 ) (kg s - ) ** In our studies of incompressible fluid flow along pipes, the density of the fluid is assumed to remain constant throughout a process. DET: Mechanical Engineering: Thermofluids Higher Support Notes 43

51 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 mc (J) Internal Energy U (J) Flow Energy pv (J) Thus, in specific terms for a mass of kg, the total energy is: gz + C + u + pv (J kg - ) 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 m 3 kg - is the reciprocal of density, i.e. v = / ρ. Hence, by substitution, the above expression can be modified to give specific energy of moving fluid as: gz + C p + (J kg - ) ρ Further dividing throughout by g, gives: C p Z + + g J ( J kg x s m = Nm kg x s x m = kg x m x s x m x kg x s x m ) = m All three terms now have the dimension of length (metres) p is the pressure head of a fluid at pressure p, ρg and the sum of all three terms is called the total head. C p i.e. Total head = Z + + (m) g J DET: Mechanical Engineering: Thermofluids Higher Support Notes 44

52 Bernoulli s equation Consider unit mass of fluid flowing at a steady rate through the system/pipeline shown in the figure below. FLUID IN PRESSURE VELOCITY C p SYSTEM Z FLUID OUT PRESSURE VELOCITY C p Z DATUM 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 gz + At entry At exit C p C p + = gz + + = constant (J kg ) Dividing both sides of the equation by gravitational acceleration, g. Z + C p C p + = Z + + = g J g J 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). DET: Mechanical Engineering: Thermofluids Higher Support Notes 45

53 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 Z F = Frictional Resistance `Head Then, Bernoulli s Equation can be stated in the form: Z + C + g p J = Z C + + g p J + Z F As before, each term represents Head of Liquid in units of m. Z = potential head C = kinetic head g p J = pressure head DET: Mechanical Engineering: Thermofluids Higher Support Notes 46

54 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 mgz m 3 s - FLUID VELOCITY KINETIC HEAD kg s - 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: DET: Mechanical Engineering: Thermofluids Higher Support Notes 47

55 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 - and pressure 400 kn m -, and is discharged through an orifice 50 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. d = 500 mm 30 m d = 50 mm p =? C =? p = 400 kn m - C = 3 ms - A Velocity of oil at exit from A C = A C C = x C A C = Œ Œ 4 x.5 4 x.5 x 3 = OIL ρ = 800 kg m -3 Z F = 3.5 m.5.05 x 3 = m s VELOCITY, C AT POINT OF DISCHARGE = m s -. MASS FLOW RATE, m = ρ A C = 800 x π x.5 4 x = 47. kg s - Use Bernoulli s equation for pressure at exit - Z + C g + p!j = Z C + g + p!j + Z F x p + = x x 9.8 x x = p p 8.47 = p = ( ) x 7848 = 678 N m - DET: Mechanical Engineering: Thermofluids Higher Support Notes 48

56 PRESSURE p AT POINT OF DISCHARGE 67. kn m - 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. P C A P C A FLUID DENSITY ENTRY EXIT y h MANOMETER FLUID DENSITY X X After Section, 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 and allows the pressure difference to be measured. The higher pressure at Section causes the mercury in the manometer to be forced down the LH limb and up in the RH limb. DET: Mechanical Engineering: Thermofluids Higher Support Notes 49

57 Reference the figure and applying Bernoulli s equation between sections and of the Venturi gives: C p C p + = + g J g J (since Z = Z here) p - p C - C = (Equation ) J g. A The volumetric flow equation gives V = A C = A C C = x C A Substituting for C in equation gives: p - p J = A A C g - C p - p J = C A A g - (Equation ) 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 p + ρ g (Y + h) = p + ρ gy + ρ Hg gh p + ρ gy + ρ gh = p + ρ gy + ρ Hg gh p + ρ gh = p + ρ Hg gh p p = gh ( ρ Hg - ρ ) Divide by ρ p - p gh( = = gh Hg Hg Hg - - = gh - p - p Hg = h - J (Equation 3) DET: Mechanical Engineering: Thermofluids Higher Support Notes 50

58 Equating equations and 3: C A A g - = h Hg - Therefore velocity C = gh A A Hg - (Equation 4).. Volumetric flow rate, V, can now be deducted from equation, V = A C. Also, mass flow rate from equation m = ρ A C 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. 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

59 P A C P A C h P P L 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. DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

60 PRACTICAL EXEMPLAR PROBLEM Venturi meter A Venturi meter is used to establish the flow rate of water in a horizontal pipeline of 50 mm diameter. The throat diameter of the Venturi is 0 mm. The pressure drop between the entrance and throat section is 60 mm of mercury. Assuming no losses due to friction and taking the densities of water at 000 kg m -3 and mercury at 3600 kg m -3, determine: a) the velocity of the water in the 50 mm dia pipeline b) the volumetric flow rate of water c) the mass flow rate of water. Ø0mm h = 60mm Ø50mm a VELOCITY, C IN 50 PIPE FROM EQUATION, C = gh A A! Hg! - - = x 9.8x.06 Œ Œ 4 x.05 4 x (3.6 -) C = = = VELOCITY OF WATER IN 50 PIPELINE = 0.64 m s -. b VOLUMETRIC FLOW RATE, V = A C Œ = 4 x.05 x 0.64 =.5 x 0-3 m 3 s -. c MASS FLOW RATE m = ρ A C = 000 x.5 x 0-3 =.5 kg s - DET: Mechanical Engineering: Thermofluids Higher Support Notes 53

61 OUTCOME 5: SOLVE PROBLEMS ASSOCIATED WITH THE BEHAVIOUR OF LIQUIDS AT REST Outcomes 3 and 4 of this unit encompassed the behaviour, effects, and the solution of problems covering fluids in motion, i.e. thermodynamics In outcome 5 we are concerned with the solution of problems associated with the effects of liquids at rest. Hydrostatics is the study of the forces and pressures exerted by a fluid when the fluid system is in equilibrium. Gravitational force The derived SI unit of force is the Newton (N) defined as that force which, when applied to a body of mass one kilogram, gives it an acceleration of one metre per second squared. From Newton s Second Law of Motion:- FORCE = MASS x ACCELERATION F = m a (kg x m s - = N) When gravitational acceleration is considered, the above equation can be re-stated in the form:- F = m g (where g = 9.8 m s - ) Hence, the downward gravitational force exerted by a liquid column of mass 5 kg will be:- Pressure (symbol p) F = 5 x 9.8 = 45.5 N Pressure is defined as force (F) per unit cross-sectional area (A). i.e. Pressure p = FORCE CROSS - SECT. AREA The unit for pressure is the Newton per square metre (N m - ). The units Pascal (Pa) and bar are also commonly used and it should be noted that:- = F A Pa = N m - or kpa = kn m - and bar = 0 5 N m - or 00 kn m - DET: Mechanical Engineering: Thermofluids Higher Support Notes 54

62 Atmospheric pressure This is the pressure exerted on the surface of the earth by the gravitational pull of the mass of air in the earth s atmosphere. Atmospheric pressure is stated as an absolute pressure, i.e. relative to zero. Standard atmospheric pressure = atm =.03 bar =.03 x 0 5 N m - Gauge pressure and absolute pressure Various types of pressure gauges are commonly used to measure fluid pressures in vessels and pipelines and read pressures normally above or below atmospheric pressure. 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 p g, the true or absolute pressure p in the vessel is given by:- Absolute pressure p = gauge pressure + atmospheric pressure i.e. p = p g + p atm *** Static pressure characteristics The concept of pressure in a fluid and the manner in which it varies throughout the fluid mass is of fundamental importance in the science of hydrostatics. The experimentally established characteristics noted below need to be acknowledged.. The intensity of pressure in a fluid at rest is the same in all directions.. The pressure exerted by a static fluid always acts normally (perpendicular) to any boundary surface containing the fluid. 3. The pressure exerted by a static fluid is directly proportional to and increases with depth below its free surface. DET: Mechanical Engineering: Thermofluids Higher Support Notes 55

63 Pressure variation with depth in a liquid column C.S. AREA A Consider the forces acting on a vertical cylinder of liquid of height h and cross-sectional area A within a volume of static liquid. P FREE SURFACE DEPTH h FIGURE 5A P UPTHRUST = PA x Let p = pressure on free surface of liquid N m - p = pressure at depth h N m - A = C.S. Area of cylindrical liquid column m h = depth of liquid column m m = mass of liquid in the column kg ρ = density of the liquid kg m -3 The downward force exerted by the liquid column Now, mass of liquid m = density x volume (kg m -3 x m 3 = kg) and, volume of liquid V = CSA x height (m x m = m 3 ) Hence, mass of liquid in the column, m = m x g = ρ V = A h = ρ Ah Downward force exerted by the liquid column, F = ρ Ah x g Now, for the column of liquid to be in equilibrium, i.e. at rest UPWARD FORCES = DOWNWARD FORCES p A = p A + ρ Ahg Dividing through by A, p = p + ρ gh p p = ρ gh i.e. the pressure at any depth in a liquid can be derived from equation *** Pressure p = ρ gh (kg m -3 x m s - x m = N m - ) DET: Mechanical Engineering: Thermofluids Higher Support Notes 56

64 Hydrostatic pressure and thrust B C D FREE SURFACE h AREA A AREA A AREA A FIGURE 5B The above figure 5B illustrates three tanks having equal base areas A. Each vessel contains the same liquid to a common vertical depth h. Since the tanks hold different volumes of liquid, it might be thought that container C with the largest amount and greater mass of liquid would have the greatest force exerted 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 = ρ gh at all points on the bases since these are horizontal. Also, since the tanks have a common base area A, it follows that the forces acting on all three bases have equal magnitude. i.e. F = pa = ρ gha (all four quantities are constant) The arrows inside tanks B, C and D illustrate how pressure acts normal to retaining surfaces 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 the centroid of each base area. Our studies now extend into the determination of thrust exerted on submerged and partially submerged vertical plane surfaces and to fixing the point of application of this force at the centre of pressure. DET: Mechanical Engineering: Thermofluids Higher Support Notes 57

65 Thrust on a submerged vertical plane surface Consider a plane surface of area A immersed vertically in a liquid of density ρ. The pressure on one side acts at right angles to the surface and gives rise to a force or thrust on that side. FREE SURFACE y y df G CENTROID da FIGURE 5C By definition, Pressure = APPLIED FORCE C.S.A p = A F (N m - ) 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 = ρ gy Hence, force df on elemental strip = ρ gyda Total force on whole area A F = df Since both ρ and g are constants, Total force on whole area A F = ρ g yda But y da is the total moment of area about an axis through the free surface of the liquid 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 the depth of the centroid of the wetted area below the free surface and can be deduced from formula: Total thrust F = ρ ga y (N) *** DET: Mechanical Engineering: Thermofluids Higher Support Notes 58

66 Centre of pressure (or centre of force) We have seen that the intensity of pressure acting over any vertically submerged plane surface area increases proportionally with depth below the free surface of a liquid. Hence, for both rectangular and circular vertical surfaces, the pressure and forces applied above their horizontal geometric centre lines will be less than those acting below their centre lines. The combined action of all the elemental forces acting over the whole surface areas can 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 will obviously lie on a vertical centre line of area at some distance below the geometric centroid of the plane surface areas. FREE SURFACE b y y C Ø d CP G d G CP GC CIRCULAR PLANE SECTION RECTANGULAR PLANE FIGURE 5D For both configurations of immersed plane surfaces shown in FIG.5 above, the individual centroids (G) have a depth - y below the liquid free surface. Each centre of pressure (CP) is located vertically below the centroids by a distance designated GC. Setting aside its mathematical proof from first principles we can state the appropriate formula for calculating this distance, viz:- DISTANCE GC = RADIUS OF GYRATION ABOUT CENTROID SQUARED DEPTH OF CENTROID BELOW FREE SURFACE GC = k *** y Values for k can be calculated using formulae listed alongside figure 5E on following page. DET: Mechanical Engineering: Thermofluids Higher Support Notes 59

67 Area and k values for rectangular and circular plane surfaces BREADTHb DIA d AREA =bd AREA = 4 d DEPTH d G CP GC GC CP G k = d k = d 6 RECTANGLE CIRCULAR AREA FIGURE 5E For the rectangle, the geometric centroid, G, lies on the intersection of the vertical and 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 the centroid by a distance, say GC, and can be calculated using the stated formula for each listed below. FOR RECTANGULAR SURFACE FOR CIRCULAR SURFACE DISTANCE GC = k y d / = y GC = k y d = /6 y where y = depth of centroid below the liquid free surface. Partially submerged rectangular surfaces and wetted areas For rectangular plane surfaces which are partially submerged vertically in a liquid, the formulae developed for total thrust and centre of pressure remain applicable when the area of the wetted surface only is considered. DET: Mechanical Engineering: Thermofluids Higher Support Notes 60

68 The position of the geometric centroid for a partially submerged surface area can also be established for the wetted area only. TANK DAM WALL FS FS WETTED AREA WETTED SURFACE AREAS FIGURE 5F DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

69 PRACTICAL EXEMPLAR PROBLEMS. 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 the tank are 3 m in height. Determine the total hydrostatic thrust acting on the ends and sides of the tank when it is half full of oil of density 840 kg m -3. Determine also the position of the centre of pressure of the total thrust above the base. For the 4 m side: 3m.5m THRUST G C RESULTANT FORCE/TOTAL from eq n F = ρ ga y DISTANCE GC 4m. 5 F = 840 x 9.8 x 4 x.5 x F = kn For the 6 m side, total thrust F = ρ ga y. 5 F = 840 x 9.8 x 6 x.5 x F = kn The centre of pressure lies below the centroid of the wetted surface by the distance GC. Distance GC from eq n GC = k for rectangle k d = y d.5 GC = = = 0.5m x y x.75 Since centroid is located 0.75 m above the base level, the C of P will be = 0.5 m above base. This dimension 0.5 m above base applies for both sides and both ends since C of P is dependant on depth of wetted surface and independent of breadth b of the wetted surface. DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

70 . 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 on its horizontal centre line. The gate has a diameter of 4. m and its horizontal centre line 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.0 m above the centroid of the gate to just keep the gate closed against the hydrostatic thrust of the water. Total force acting on circular opening, say F from equation: F = ρ ga y PULLING FORCE = 000 x 9.8 x π x 4. 4 x 5 F = N Centre of pressure for force F lies at a distance GC below trunnion centre line. GC = GC = k and for circle k d = y 6 4. =.05 m 6 x 5 Equilibrium conditions for opening are: C.W. MOMENTS = A.C.W. MOMENTS Pulling force F x.0 = x.05 F = x.05 = 749 N MAGNITUDE OF PULLING FORCE = 74.9 kn DET: Mechanical Engineering: Thermofluids Higher Support Notes 63

71 Head of a liquid Whilst deriving Bernoulli s equation in Outcome 4, the total specific energy of a liquid stream was stated as: Potential energy + kinetic energy + flow energy gz + C p + (specific energy IN J kg - ) Dividing these terms by gravitational acceleration, g, each term can then be re-stated as:- Z + C p + g (head of liquid in metres, m) J 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 forcing a liquid along a pipeline in the presence of hydrostatic pressure. Although in common use, the term pressure energy tends to imply that a liquid may have its energy increased by pressurization. This is not the case since most liquids are practically incompressible. The term flow work may be considered more appropriate. Likewise, the three separate types of energy listed above for a liquid stream are often expressed more conveniently as heads of liquid. DET: Mechanical Engineering: Thermofluids Higher Support Notes 64

72 Static pressure head and hydrostatic pressure The hydrostatic pressure, p, at a point in a liquid stream, may be imagined to be due to its being at a depth, h, below the free surface of the same liquid. Thus, if a small hole be drilled into a pipeline carrying a flowing liquid and then fitted with a piezometer tube, the liquid would rise up the tube due to the pressure in the system and settle at a height when equilibrium was reached. PIEZOMETER TUBE PRESSURE p SECTION OF PIPELINE DENSITY FIGURE 5G Reference above figure, the height, h, to which the liquid rises in the tube provides a means whereby the hydrostatic pressure at that point in the pipeline can be determined. This height, h, is called the static pressure head of the liquid in the pipeline. Hence, Static pressure head, h = p J = FLOW WORK g (m) and hydrostatic pressure, p = ρ gh (N m - ) The static pressure head is easily measured and may be regarded as a head of liquid equivalent to the flow work. DET: Mechanical Engineering: Thermofluids Higher Support Notes 65

73 Pressure measurement Manometry may be defined as the science of utilising vertical columns (heads) of liquid to measure fluid pressures. Many different types of pressure measuring devices exist but our studies will be limited to the barometer, piezometer tube and U-tube manometers. Before considering these devices, however, it is appropriate to recap on definitions of pressure quantities already covered in this unit. Vacuum Atmospheric pressure, p atm Gauge pressure, p g Absolute pressure, p A perfect vacuum is a completely empty space and has zero pressure. The planet earth is surrounded by an atmosphere. The pressure due to this atmosphere depends upon the head of air above the earth s surface. At sea level atmospheric pressure is normally taken as.03 bar (0.3 kn m - ), equivalent to a head of 0.35 m of water or 760 mm of mercury, and decreases with altitude. is the intensity of pressure measured above or below atmospheric pressure. When a gauge shows a zero reading it means the pressure is atmospheric. is the intensity of pressure measured above absolute zero, which is a perfect vacuum. i.e. ABSOLUTE PRESSURE p = GAUGE PRESSURE p g + ATMOSPH. PRESS. p atm DET: Mechanical Engineering: Thermofluids Higher Support Notes 66

74 Barometer VACUUM MERCURY DENSITY P atm Hg AREA A h The figure alongside shows a very basic mercury barometer, a device suitable for measuring the pressure of the earth s atmosphere. The device consists of a small diameter glass tube about m long. The tube is filled with mercury and inverted with its open end in a dish of mercury. A vacuum is created at the top of the tube and the atmospheric pressure acting on the surface of the mercury in the dish supports a column of mercury in the tube of height h. B MERCURY P B FIGURE 5H If B is a point in the tube at the same level as the free surface of the mercury in the dish, then the pressure p B acting upwards at B will be equal to the atmospheric pressure p atm acting downwards, since, in a fluid at rest, the pressure is the same at all points at the same level. The column of mercury in the tube is in equilibrium due to the action of the force at B acting upwards against the force (m x g) of the column of mercury of height h acting vertically downwards. For equilibrium and summing forces in the vertical direction:- F = O Hence, and since = p B x A - ρ gha p B = ρ gh p B = p atm ATMOSPHERIC PRESSURE p atm = ρ Hg gh (kg m -3 x m s - x m = N m - ) When the height, h, of the liquid column is 760 mm and density of mercury taken at 3600 kg m -3. Atmospheric pressure p atm = 3600 x 9.8 x.76 = kn m - =.03 bar DET: Mechanical Engineering: Thermofluids Higher Support Notes 67

75 Piezometer tube Pipelines and vessels carrying liquid under pressure can have their pressure measured by manometers or pressure gauges employing liquid columns. The simplest type of manometer is the piezometer tube, which is a single vertical transparent open top tube fitted into the pipeline or vessel carrying pressurized liquid whose pressure is to be measured. Two types are shown below. LIQUID DENSITY A A FIGURE 5J Due to hydrostatic pressure in the system, the free surface of the liquid in the open tube will rise and stabilise at a height, h, above the centre line of the pipeline. The height, h, is the pressure head and allows gauge pressure to be calculated, viz. At position/level A. Gauge pressure, p g = ρ gh (kg m -3 x m s - x m = N m - ) For example, a pressure head of 40 mm of water converts to a pressure: 40 Gauge pressure, p g = ρ gh = 000 x 9.8 x 3 0 p g = 39.4 N m - PIEZOMETER TUBES The length of vertical tube which can be conveniently used limits the piezometer to measuring pressures in the lower ranges. For higher liquid pressures, U-tube manometers are often more appropriate. DET: Mechanical Engineering: Thermofluids Higher Support Notes 68

76 The U-tube manometer In order to measure higher pressure levels in pipelines, U-tube manometers employing mercury liquid columns are in widespread use. MERCURY DENSITY Hg ATER DENSITY C A B y X X U-TUBE MANOMETER (POSITIVE GAUGE PRESSURE) FIGURE 5K In above figure, the level of mercury in the LH limb of the U-tube is at section x-x and distance y below the axis of the pipeline. In the open limb, the mercury column is in equilibrium at a point C, height h above section x-x due to the pressure in the pipeline. B is a point in the LH limb on the same horizontal level as the pipe axis A. Working on the basis of gauge pressure then, from figure:- Pressure at C, p c = 0 (i.e. atmospheric) and pressure at x-x in RH limb = ρ Hg g h likewise pressure at x-x in LH limb = ρ Hg g h (same horizontal level) Thus, p B, pressure at B in LH limb = ρ Hg g h - ρ g y pressure in pipeline, p A = p B = g ( ρ Hg h - ρ y) *** DET: Mechanical Engineering: Thermofluids Higher Support Notes 69

77 Pressure difference between two pipes The pressure difference between two levels in a fluid can be measured directly by using a differential manometer. W A W X B h Y C D FIGURE 5L The U-tube differential manometer shown in figure above is to measure the pressure difference between levels A and B in pipelines carrying a liquid of specific weight w. The U-tube contains mercury of specific weight w. Since C and D are at the same level in a liquid at rest:- pressure at C = pressure at D p c = p D For the LH limb p c = p A + w X For the RH limb p D = p B + w (Y h) + w h p A + w X = p B + w Y w h + w h Hence, pressure difference p A - p B = w Y w h + w h w X = w (Y X) + h (w w ) Pressure difference between the two pipes can be deduced from the formula: *** p A p B = w (Y X) + h (w w ) DET: Mechanical Engineering: Thermofluids Higher Support Notes 70

78 Density to specific weight conversion When working through problems such as in manometry pressure calculations it is often convenient to convert one fluid quantity into another. Density This quantity has earlier been defined as mass per unit volume. MASS Density (RHO) ρ = = UNIT VOLUME Specific weight This is defined as weight per unit volume. m V (UNIT kg m -3 ) Specific weight, w = WEIGHT UNIT VOLUME Since weight = mass x gravitational acceleration = mg (N) Specific weight, w = mg (kg x m s - x m -3 = N m -3 ) V w = ρ g (kg m -3 x m s - = N m -3 ) Thus multiplying density by gravitational acceleration yields specific weight (symbol w, unit N m -3 ). Example Oil of density 800 kg m -3 has a specific weight of 800 x 9.8 = 7848 N m -3 DET: Mechanical Engineering: Thermofluids Higher Support Notes 7

79 DET: Mechanical Engineering: Thermofluids Higher Support Notes

80 SELF-ASSESSMENT ANSWERS DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers

81 DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers

82 Section : Thermofluids (Higher) Self-Assessment Answers DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers

83 DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers

84 SELF-ASSESSMENT ANSWERS Assignment. QUANTITY SYMBOL UNIT SPECIFIC VOLUME v m 3 kg - MASS m kg VOLUME V m 3 ABSOLUTE PRESSURE P N m - or Pa ABSOLUTE TEMP. T K. O C K An isothermal process is one which is carried out under constant temperature conditions. 4. Charles Law states that 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). EQUATION FOR CHARLES S LAW: V T = V T = Vn For The Mass of Gas T n DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers

85 5. When a mass of gas is cooled at constant volume its pressure and temperature will both decrease. 6. The characteristic gas equation for an ideal gas is: pv = mrt Where Unit p = absolute pressure of the gas N m - V = volume of the gas m 3 T = absolute temp. of the gas [(t + 73)] K m = mass of the gas kg R = characteristic or specific J kg - K - gas constant DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers

86 Assignment 3 A condenser is a thermodynamic device which is used to transfer heat energy from a hot fluid (e.g. steam) to a colder fluid. The four forms of energy associated with a moving fluid mass are: (a) POTENTIAL ENERGY (c) INTERNAL ENERGY (b) KINETIC ENERGY (d) FLOW/DISPLACEMENT ENERGY The symbol Z represents height in order to avoid confusion since the symbols H and h are attached to total enthalpy and specific enthalpy in the SFEE. The symbol C represents velocity in order to distinguish velocity from total volume V or specific volume v. The table should read as below: MASS FLOW RATE QUANTITY SYMBOL UNIT. m kg s - RATE OF HEAT TRANSFER. Q J s - or W WORK TRANSFER W J ENTHALPY H J SPEC. INTERNAL ENERGY u J kg - or kj kg - DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers 3

87 Assignment 4. QUANTITY SYMBOL/FORMULA UNIT CROSS-SECT. AREA A m PRESSURE HEAD p J m. VOLUMETRIC FLOW RATE V m 3 s - POTENTIAL ENERGY MgZ J FLUID VELOCITY C m s - KINETIC HEAD MASS FLOW RATE C m g. m kg s - BERNOULLI S EQUATION: C Z + + g J C + g p p = Z + J (UNITS m ' HEAD') SYMBOL QUANTITY UNIT Z HEIGHT ABOVE DATUM m C FLUID VELOCITY m s - g GRAVITATIONAL ACCEL N m s - p FLUID PRESSURE N m - ρ FLUID DENSITY kg m -3 MASS CONTINUITY EQUATION:. m = ρ AC DET: Mechanical Engineering: Thermofluids Higher Self-Assessment Answers 4

88 Section 3: Thermofluids (Higher) Tutorials DET: Mechanical Engineering: Thermofluids Higher Tutorials

89 DET: Mechanical Engineering: Thermofluids Higher Tutorials

90 OUTCOME : GAS LAWS AND CHARACTERISTIC GAS EQUATION. A mass of gas at a pressure of 3 MN m - occupies a volume of 0.66 m 3. If the pressure is increased isothermally to 36 bar, what volume will the gas occupy? (0.55 m 3 ) If a fixed mass of gas is expanded to five times its original volume at constant pressure, what will be the final temperature if the gas is initially at 50 C? (34 C) DET: Mechanical Engineering: Thermofluids Higher Tutorials

91 . A fixed mass of hydrogen is expanded at constant temperature from a pressure of 6 bar and a volume of 0.68 m 3 until it occupies 3 times its original volume. Calculate the final pressure of the hydrogen. ( bar) A gas occupies a volume of. m 3 at a temperature of 7 C and a pressure of 0.0 MN m -. What volume will the gas occupy at 0 C and.03 bar? (.56 m 3 ) DET: Mechanical Engineering: Thermofluids Higher Tutorials

92 A cylinder in an internal combustion engine is fitted with a piston. When the pressure on the piston is 3 bar and the temperature of the gas is 40 C, the volume of the gas is 0.7 m 3. When the piston pressure increases to bar the volume reduces to 0.57 m 3. Calculate the final temperature of the gas. (750 C) The nozzle of a burner in a hot air balloon delivers 48 m 3 of helium gas at a pressure of 90 kn m - and a temperature of 3 C. Given that the characteristic gas constant, R for helium is.079 kj kg - K -, what mass of gas has been supplied? (7.0 kg) DET: Mechanical Engineering: Thermofluids Higher Tutorials 3

93 3. A fixed mass of gas undergoes the following successive processes: a) expansion at constant temperature from a pressure and volume of bar and m 3 respectively to a volume of 0.0 m 3 b) heating at constant pressure until the volume is m 3 c) constant-volume heating until the gas attains its original pressure,i.e. bar. Determine the unknown pressure and temperature at the end of each process, given that the initial temperature is 50 C. (p = 4.58 bar; t 3 = 3 C; t 4 = 460 C) DET: Mechanical Engineering: Thermofluids Higher Tutorials 4

94 A rigid cylinder of internal volume 0.04 m 3 contains oxygen, initially at a pressure and temperature of 30 bar and 6 C respectively. Some of the oxygen is used in an oxy-acetylene welding process, reducing the pressure to 8 bar, the temperature remaining unchanged. If R for oxygen is 0.87 kj kg - K -, determine the mass of oxygen used. (.096 kg) DET: Mechanical Engineering: Thermofluids Higher Tutorials 5

95 A cylinder fitted with a movable piston contains 0. kg of oxygen at a pressure of 0 kn m - and temperature 5 C. The oxygen is heated at constant pressure until its volume becomes 0.7 m 3. Determine the final temperature and the original volume of the oxygen. Take R for oxygen as 0.6 kj kg - K -. (5.4 C; 0.48 m 3 ) DET: Mechanical Engineering: Thermofluids Higher Tutorials 6

96 4. Under conditions of s.t.p. 4 kg of hydrogen occupy 44.8 m 3. a) If the pressure is increased to.8 bar, what will the temperature of the gas reach if the volume occupied by the gas remains unchanged? (.09 C) b) What value for the specific gas constant for hydrogen can be deducted from this? (4.56 kj kg - K - ) Note: m 3 = 000 litres; bar = 0 5 N m - Standard temperature and pressure (s.t.p.) = 0 C and.03 bar DET: Mechanical Engineering: Thermofluids Higher Tutorials 7

97 OUTCOME : THERMODYNAMIC PROPERTY TABLES. At what pressure will water boil at a temperature of: a) 36.4 C b) 80.3 C c) 7. C. Use steam property tables to find values for underlisted quantities. a) Spec. internal energy of dry saturated steam at 85 bar. b) Enthalpy of evaporation of dry saturated steam at 0.05 bar. c) Spec. enthalpy of dry saturated steam at.9 bar. d) Spec. internal energy of boiling water at 35 bar. e) Volume per kg of dry saturated steam at 0.6 bar. f) Enthalpy per kg of boiling water at 48 bar. g) Spec. enthalpy of steam at 40 bar and 450 C. NOTE: State units for all the above values. DET: Mechanical Engineering: Thermofluids Higher Tutorials 8

98 3. State the saturation temperature of wet steam at 0.90 bar. 4. State the pressure of wet steam at a temperature of 09.8 C. 5. Use steam property tables to find the enthalpy of evaporation of wet steam at pressures listed below: Pressure Enthalpy of evaporation 3 bar 0.0 kn m - 95 bar 8500 N m bar.0 MN m - A tank contains superheated steam at a pressure of 7 bar and temperature of 350 C. a) State the degree of superheat for the steam. b) For above steam what is the spec. internal energy at the listed temperature? c) Determine the total enthalpy of 6 kg of steam in above listed condition. DET: Mechanical Engineering: Thermofluids Higher Tutorials 9

99 6. Use steam tables and appropriate formulae to determine the specific internal energy of wet steam at a pressure of 44 bar with a dryness fraction of Calculate the specific volume of wet steam at.095 bar and.95 dry. 8. Calculate the specific enthalpy of steam at a pressure of bar with a dryness fraction of Use steam tables and interpolation to determine the specific volume of superheated steam at a pressure of.5 bar and temperature of 450 C. DET: Mechanical Engineering: Thermofluids Higher Tutorials 0

100 . Use steam tables and interpolation to determine the specific internal energy of superheated steam at a pressure of 5 bar and 350 C.. Use steam tables and interpolation to determine the specific enthalpy of superheated steam at a pressure of 45 bar and a temperature of 475 C. 3. One kilogram of superheated steam 5 bar and 400 C has its pressure reduced to.5 bar at the same temperature. Determine the change in spec. enthalpy. 4. What is the difference in specific volume of superheated steam at 0.5 bar and 300 C when its pressure is increased to 5.0 bar at the same temperature? DET: Mechanical Engineering: Thermofluids Higher Tutorials

101 Refrigerants. At what temperature does refrigerant R77 boil at a pressure of.740 bar?. State the Specific Liquid Enthalpy of refrigerant R at 6.56 bar. 3. Refrigerant R has a Specific Enthalpy of 90.5 kj kg - at a pressure of.86 bar. State whether it is in the wet, dry, saturated or superheated condition. 4. Determine the Enthalpy of Evaporation of Refrigerant R77 at 4 C and.465 bar. 5. State the Specific Volume of dry saturated vapour for refrigerant R at 5 C and.37 bar. 6. Determine the Specific Volume of refrigerant R77 at a temperature of 0 C, pressure of 4.95 bar dryness fraction State the Specific Enthalpy of refrigerant R at.004 bar and temperature 30K. DET: Mechanical Engineering: Thermofluids Higher Tutorials

102 OUTCOME 3: STEADY FLOW ENERGY EQUATION/GASES AND VAPOURS. Air flows steadily at a rate of 0.4 kg s - through an air compressor, entering at 60 m s - with a pressure of bar and a specific volume of 0.85 m 3 kg - and leaving at 45 m s - with a pressure of 6.9 bar and a specific volume of 0.6 m 3 kg -. The internal energy of the air leaving is 88 kj kg - greater than that of the air entering. Cooling water in the jacket surrounding the cylinder absorbs heat from the air at the rate of 59 kj s -. Sketch a systems diagram for the compressor and attach the given data then calculate the power required to drive the compressor. (04 kw) DET: Mechanical Engineering: Thermofluids Higher Tutorials 3

103 The temperature of feed water entering a boiler is 3 C and steam is produced at 40 bar and 500 C. The rate of heat transfer to the surroundings is 35 kw. Determine the mass flow rate of the working fluid for a heat input to the boiler of 6 MW. (.98 kg s - ) DET: Mechanical Engineering: Thermofluids Higher Tutorials 4

104 . In a steady flow system, the working fluid flows at the rate of 4 kg s -. It enters at a pressure of 60 kn m -, a velocity of 300 m s -, internal energy 00 kj kg - and specific volume 0.37 m 3 kg -. It leaves the system at a pressure of 30 kn m -, a velocity of 50 m s -, internal energy 500 kj kg - and specific volume. m 3 kg -. During its passage through the system, the fluid has a heat transfer loss of 30 kj kg - to the surroundings. Attach the given data to the systems diagram shown below and determine the power of the system, stating whether it is from or to the system. Neglect any change in PE. (.7 MW) SYSTEM DET: Mechanical Engineering: Thermofluids Higher Tutorials 5

105 3. In a steady flow system, steam enters at a pressure of 3.6 MN m - and with negligible kinetic energy. It leaves at a pressure of 00 kn m -, 0.8 dry, and with a kinetic energy of 400 kj s -. If the steam flows at the rate of 5 kg s -, determine the condition of the steam at entry to the system, assuming there is no transfer of heat or of work energy. (.968 Dry) SYSTEM DET: Mechanical Engineering: Thermofluids Higher Tutorials 6

106 5. A steady flow of steam enters a turbine with a velocity of 6 m s - and specific enthalpy of 990 kj kg -. The steam leaves the turbine with a velocity of 37 m s - and specific enthalpy of 530 kj kg -. The heat lost to the surroundings as the steam passes through the turbine is 5 kj kg -. The steam flow rate is 9 kg s -. Insert the given data into the systems diagram shown and determine the power output from the turbine. (3.9 MW) STEAM INPUT TURBINE STEAM OUTPUT DET: Mechanical Engineering: Thermofluids Higher Tutorials 7

107 6. Exhaust steam from a turbine enters a condenser with a specific enthalpy of 64 kj kg - and a velocity of 75 m s -. During the cooling process, the heat loss to the cooling water per kg of working fluid is 680 kj kg -. The condensate exits the system with a velocity of m s -. Sketch a systems diagram for the condenser and determine the Specific Enthalpy of the working fluid at exit from the condenser. ( kj kg - ) DET: Mechanical Engineering: Thermofluids Higher Tutorials 8

108 7. An air/fuel mixture passes through a gas turbine system at a rate of 3 kg s -. It enters with a velocity of 30 m s - and a specific enthalpy of 600 kj kg -. At exit, the velocity is 05 m s - and the specific enthalpy is 950 kj kg -. In its passage through the turbine the working fluid has a heat transfer loss of 30 kj kg -. Sketch a systems diagram for the turbine and attach given data then determine the power developed by the turbine. (.87 MW) DET: Mechanical Engineering: Thermofluids Higher Tutorials 9

109 8. An air/fuel mixture enters a steady flow system with a velocity of 30 m s - and exits with a velocity of 40 ms -. The mass flow rate is 9 kgs -. The properties of the fluid at entry are: a) pressure 3.8 bar b) specific volume 0. m 3 kg - c) specific internal energy 4 kj kg -. At exit, the properties of the fluid are: d) pressure.035 bar e) specific volume m 3 kg - and f) specific internal energy 08 kj kg -. The heat transfer rate from the system is 4. kj s -. Produce a systems diagram and attach given data, then determine the rate of work transfer from the system. (.6 MW) DET: Mechanical Engineering: Thermofluids Higher Tutorials 0

110 9. boiler Turbine pump Cond. Steam Power Plant Systems Diagram Rankine cycle Liquid Condensate The conditions listed below apply to the turbine element in the systems diagram shown. At Entry At Exit Steam Pressure 80 bar.8 bar Steam Temperature 45 C - Fluid Velocity 36 m s - 85 m s - Dryness Fraction Mass Flow Rate of Working Fluid = 5.8 kg s - Rate of Heat Loss to Surroundings = 3 KW Using appropriate thermodynamic symbols, insert the given data into the diagram and use directional arrows for heat/work transfers. Also, calculate the work rate (power) of the turbine. (4.4 MW) DET: Mechanical Engineering: Thermofluids Higher Tutorials

111 0. A steady flow of wet steam enters a condenser with a pressure of 4.5 bar, a dryness fraction of 0.86, and a velocity of 00 m s -. The liquid condensate exits the condenser at a pressure of. bar and with a velocity of m s -. Determine: a) The change of Specific Enthalpy across the system b) The heat loss to the cooling water per kg of the working fluid COOLING WATER (.008 MJ kg - ;.03 MJ kg - ) STEAM INPUT CONDENSER OUTPUT 3 4 DET: Mechanical Engineering: Thermofluids Higher Tutorials

112 OUTCOME 4: APPLICATION OF THE MASS CONTINUITY AND BERNOULLI S EQUATIONS TO INCOMPRESSIBLE FLOW THROUGH PIPES. Oil of density 935 kg m -3 flows steadily through a pipe similar to that in a Venturi meter as shown below. From data given, determine: a) the volumetric flow rate (.006 m 3 s - ) b) the fluid velocity at 45 mm diameter section (.643 m s - ) c) the diameter d in millimeters (55.3 mm) d) the mass flow rate of the fluid (.444 kg s - ) Ø 80 Ø 45 Ø d 0.5ms - C ms -.09ms - DET: Mechanical Engineering: Thermofluids Higher Tutorials 3

113 . A 00 mm diameter pipe splits into branch pipes, one 60 mm diameter and the other 80 mm diameter. The system carries oil of density 930 kg m -3 in steady flow. If the velocity of the oil in the largest pipe is 3 m s - and in the smallest 4 m s -, determine the following: a) the volumetric flow rate in the 00 mm pipe (0.094 m 3 s - ) b) the mass flow rate in the 60 mm pipe (0.58 kg s - ) c) the velocity of the oil in the 80 mm pipe (6.49 m s - ) 3 3 ms - 80 mm 00 mm 60 mm 4 ms - DET: Mechanical Engineering: Thermofluids Higher Tutorials 4

114 3. A pipeline. m diameter at its upper end tapers to 0.6 m diameter at its lower end over a length of 300 m with datum levels as shown in figure. The pressure at the upper end is 69 kn m -. Water flows steadily downwards at a mass flow rate of 400 kg s -. Frictional losses in the pipeline are equivalent to a head of 0.85 m per 00 m run. Determine the velocity at each end of the pipeline and the water pressure at the lower end. 70m 300m 73m (velocity at 0.6 m end = 4.95 m s - ) (velocity at. m end =.4 m s - ) (pressure at 0.6 m end =.9 kn m - ) DET: Mechanical Engineering: Thermofluids Higher Tutorials 5

115 4. A horizontal pipeline carrying water in steady flow, tapers from 00 mm diameter at point A to 00 mm diameter at B. The volumetric flow rate of the water is m 3 s -. Determine: a) The fluid velocities at points A and B (.75 and 7.0 m s - ) b) The mass flow rate of the water (54.98 kg s - ) c) The pressure difference between points A and B. Neglect frictional losses. (.97 kn m - ) A B C =? C =? d = 00mm d = 00mm V = O.O55ms 3 - DET: Mechanical Engineering: Thermofluids Higher Tutorials 6

116 5. Oil of density 900 kg m -3 flows steadily through a pipe from point A to point B. At A the pipe diameter is 5 mm and the pressure is 60 kn m - and at point B which is m below A the diameter is 50 mm and the pressure is 0 kn m -. Determine: a) the velocity of the oil at point A (8.756 m s - ) b) the mass flow rate of the oil (96.7 kg s - ) A B m DET: Mechanical Engineering: Thermofluids Higher Tutorials 7

117 6. Oil of density 930 kg m -3 enters a horizontal pipeline with a pressure of.08 MN m - and a velocity of 7.5 m s -. The oil leaves the pipeline with a pressure of 0.64 MN m - and a velocity of 30 m s -. Apply Bernoulli s equation and determine the loss of head due to frictional resistance. (5.3 m) DET: Mechanical Engineering: Thermofluids Higher Tutorials 8

118 7. A horizontal pipe, 00 mm in diameter, full of water in steady flow, has a reduced section 50 mm in diameter forming a Venturi tube. The difference in pressure between the 00 mm and 50 mm sections is measured by a mercury filled U-tube in which the level difference is 445 mm. Taking the densities of water as 000 kg m -3 and mercury as 3600 kg m -3, calculate: a) the velocity of the water in the 00 mm section (.708 m s - ) b) the mass flow rate of the water (.7 kg s - ) Ø50 mm = 445mm Ø00mm DET: Mechanical Engineering: Thermofluids Higher Tutorials 9

119 8. A pipeline, full of water in steady flow, tapers from 0.7 m diameter at point A to 0.35 m diameter at B which is 5 m vertically below A. If at point A the pressure and velocity are 600 kn m - and 5 m s - respectively, calculate the velocity at B. If the frictional head loss is 3.6 m, what will be the pressure at point B? (0 m s - ; 6.4 kn m - ) A 5m B DET: Mechanical Engineering: Thermofluids Higher Tutorials 30

120 OUTCOME 5: SOLVE PROBLEMS ASSOCIATED WITH THE BEHAVIOUR OF LIQUIDS AT REST. An oil storage tank has vertical sides and is of rectangular section 5.8 m by 3.6 m. If the tank contains oil of density 90 kg m -3 to a depth of.5 m, calculate the total thrust due to the oil: a) on the bottom b) on the 5.8 m side c) on the 3.6 m wide end. (47. kn, kn, 0.53 kn). A horizontal pipe of.80 m bore is full of oil of density 890 kg m -3. The pipe is closed at both ends. Calculate the force acting on an end plate. (0 kn) DET: Mechanical Engineering: Thermofluids Higher Tutorials 3

121 3. A cylindrical vessel 0.6 m diameter contains oil of density 935 kg m -3 to a depth of 0.5 m. Calculate the maximum force exerted on the horizontal base of the vessel due to the liquid column. What depth of mercury, density 3600 kg m -3, would generate the same force on the base of the vessel? (9. N, 34.4 mm) DET: Mechanical Engineering: Thermofluids Higher Tutorials 3

122 4. A rectangular storage tank measures m long x 8 m wide and contains a liquid of density 900 kg m -3. A circular inspection hatch in the horizontal base of the tank is subjected to a force of 30.9 kn when the liquid depth is 7.0 m. Determine: a) the diameter of the hatch (0.798 m) b) the total force acting on the base (5.933 MN at mid point) c) the total thrust acting on the 8 m wide end (.73 MN) DET: Mechanical Engineering: Thermofluids Higher Tutorials 33

123 5. A total force of 65.6 kn acts on each vertical side of a square based storage tank containing oil of density 95 kg m -3. The wetted surface area on each vertical side is 8.9 m. Determine: a) the depth of oil in the tank (4.8 m) b) the height above the base where total force acts (.607 m) c) the total thrust on the base (.558 MN m - ) DET: Mechanical Engineering: Thermofluids Higher Tutorials 34

124 6. The figure shows a sectional view of a vertical gate in a dam wall. The rectangular gate measures.8 m deep by. m wide and is designed to pivot about its upper end which lies 8. m below the free surface. Take the density of water as 000 kg m -3 and determine: a) the magnitude of the total force acting on the gate (580 kn) b) the position of the centre of pressure below the pivot (.468 m) c) the force F required at bottom of gate to just keep the gate closed (304 kn) FS 8.m.8m EPTH PIVOT F DET: Mechanical Engineering: Thermofluids Higher Tutorials 35

125 7. A circular opening in the vertical wall of a storage tank is closed by a gate mounted on trunnions on its horizontal centre line. The gate has a diameter 0.5 m and its horizontal centre line lies 6.5 m below the fluid free surface. The tank contains oil of density 95 kg m -3. Determine the magnitude of the pulling force required to be applied at a point.60 m vertically above the centroid of the gate to just keep it shut against the hydrostatic force exerted by the oil. (4.78 kn) FS 6.5m Ø 3.5m.6m PULLING FORCE DET: Mechanical Engineering: Thermofluids Higher Tutorials 36

126 8. A mercury U-tube manometer as shown in figure is to measure the pressure of oil in the pipeline at A. Alongside the pipeline system is a mercury barometer with a column height of 756 mm of mercury. Determine: a) atmospheric pressure. b) the absolute pressure in the pipeline at A when dimension y is 0.6 m and dimension h is 0.98 m. Take the densities of oil and mercury as 900 kg m -3 and 3600 kg m -3 respectively. (00.86 kn m -, 6.3 kn m - ) OIL C M ERCURY D ENSITY H g A B y X X DET: Mechanical Engineering: Thermofluids Higher Tutorials 37

127 9. The figure shows a mercury manometer linked to a pipeline carrying water of density 000 kg m -3. If the gauge pressure at A is 0 kn m - and dimension y =.48 m, what will be the height h in the RH limb. Take the density of mercury as 3600 kg m -3. (.683 m) M ERCURY Hg WATER C A B y X X DET: Mechanical Engineering: Thermofluids Higher Tutorials 38

128 0. The figure shows a differential manometer connecting two pipelines carrying water of density 000 kg m -3. The U-tube contains mercury of density 3600 kg m -3. If the pressure difference between A and B is 48 kn m -, dimension X = 0.86 m, dimension Y = 0.5 m, what is the difference in level h? (4.5 cm) DET: Mechanical Engineering: Thermofluids Higher Tutorials 39

129 DET: Mechanical Engineering: Thermofluids Higher Tutorials

130 Section 4: Thermofluids (Higher) Tutorials Marking Scheme DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

131 DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

132 OUTCOME : GAS LAWS AND CHARACTERISTIC GAS EQUATION. A mass of gas at a pressure of 3 MN m - occupies a volume of 0.66 m 3. If the pressure is increased isothermally to 36 bar, what volume will the gas occupy? (0.55 m 3 ) For an isothermal process Boyle s Law applies Known data: p = 3 x 0 6 N m - p V = p V p = 36 x 0 5 N m - V = pv p V = 0.66 m 3 V =? = 6 3 x 0 x x 0 = 0.55 m 3 FINAL VOLUME OF GAS = 0.55 m 3. If a fixed mass of gas is expanded to five times its original volume at constant pressure, what will be the final temperature if the gas is initially at 50 C? (34 C) For a constant pressure process Known data: Charles Law applies V = V = 5 T = = 33 K T =? V V = T T T = = VT V 5 x 33 = 65 K FINAL TEMPERATURE OF GAS IS = 34 C DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

133 3. A fixed mass of hydrogen is expanded at constant temperature from a pressure of 6 bar and a volume of 0.68 m 3 until it occupies 3 times its original volume. Calculate the final pressure of the hydrogen. ( bar) For a constant temperature process Known data: Boyle s Law applies V =.68 m 3 V = (.68 x 3)m 3 p = 6 bar p =? p V = p V T = T p = p x V V = 5 6 x 0 x.68 3 x.68 = x 0 5 Nm - FINAL PRESSURE OF THE HYDROGEN = bar 4. A gas occupies a volume of. m 3 at a temperature of 7 C and a pressure of 0.0 MN m -. What volume will the gas occupy at 0 C and.03 bar? (.56 m 3 ) From the Combined Gas Law pv T pv = T Known data: V =. m 3 V =? T = (7 + 73)K T = 73 K p =.0 x 0 6 N m - p =.03 x 0 5 N m - V = pv T P T = 6.0 x 0 x. x x 0 x 300 =.56 m 3 GAS OCCUPIES A VOLUME OF.56 m 3 DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

134 5. A cylinder in an internal combustion engine is fitted with a piston. When the pressure on the piston is 3 bar and the temperature of the gas is 40 C, the volume of the gas is 0.7 m 3. When the piston pressure increases to bar the volume reduces to 0.57 m 3. Calculate the final temperature of the gas. (750 C) From the Combined Gas Equation pv T = T = = pv T pvt p V 5 x 0 x.57 x x 0 x.7 Known Data: p = 3 bar T = ( ) K V = 0.7 m 3 p = bar V = 0.57 m 3 T =? = 03 K FINAL TEMPERATURE OF GAS = = 750 C 6. The nozzle of a burner in a hot air balloon delivers 48 m 3 of helium gas at a pressure of 90 kn m - and a temperature of 3 C. Given that the characteristic gas constant, R for helium is.079 kj kg - K -, what mass of gas has been supplied? (7.0 kg) From equation p V = mrt m = = pv RT 3 90 x 0 x x 0 x 96 Known data: p = 90 x 0 3 N m - V = 48 m 3 T = = 96 K R =.079 x 0 3 J kg - K - = 7.0 kg MASS OF GAS SUPPLIED = 7.0 kg DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 3

135 7. A fixed mass of gas undergoes the following successive processes: a) expansion at constant temperature from a pressure and volume of bar and m 3 respectively to a volume of 0.0 m 3 b) heating at constant pressure until the volume is m 3 c) constant-volume heating until the gas attains its original pressure, i.e. bar. Determine the unknown pressure and temperature at the end of each process, given that the initial temperature is 50 C. a) For constant temperature process, Boyle s Law applies: (p = 4.58 bar; t 3 = 3 C; t 4 = 460 C) p V = p V p = p x V.005 =.0 x 0 5 x = 4.58 bar V.0 PRESSURE AFTER EXPANSION AT CONSTANT TEMPERATURE = 4.58 bar b) For constant pressure process, Charles Law applies: V T = V T = V V 3 x T = x ( ) T 3 = K = = 3 C TEMPERATURE AFTER HEATING AT CONSTANT PRESSURE = 3 C c) During constant volume heating, Known Data is p 3 = 4.58 bar p 4 = bar T 3 = 3 C + 73 = 305 K T 4 =? V 3 =.007 m 3 = V 4 V 4 = V 3 p3 V3 p4 V4 p4 From equation = 4 = x T3 T T p 3 5 x 0 0 T 4 = 5 x 305 = 73.5 K - 73 = 460 C 4.58 x 0 4 i.e. TEMPERATURE AFTER CONSTANT VOLUME HEATING = 460 C 3 DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 4

136 8. A rigid cylinder of internal volume 0.04 m 3 contains oxygen, initially at a pressure and temperature of 30 bar and 6 C respectively. Some of the oxygen is used in an oxy-acetylene welding process, reducing the pressure to 8 bar, the temperature remaining unchanged. If R for oxygen is 0.87 kj kg - K -, determine the mass of oxygen used. (.096 kg) Known data: V =.04 m 3 p = 30 bar T = 6 C + 73 = 89 K R = 0.87 kj kg - K - V = V =.04 m 3 (Rigid cyl) p = 8 bar T = T = 89 K Initial mass from p V = m RT m = p V R T = 5 30 x 0 x x 0 x 89 m =.4467 kg Final mass from p V = m RT m = p V R T = 5 8 x 0 x x 0 x 89 m =.3503 kg MASS OF GAS USED = m m = = kg DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 5

137 9. A cylinder fitted with a movable piston contains 0. kg of oxygen at a pressure of 0 kn m - and temperature 5 C. The oxygen is heated at constant pressure until its volume becomes 0.7 m 3. Determine the final temperature and the original volume of the oxygen. Take R for oxygen as 0.6 kj kg - K -. (5.4 C; 0.48 m 3 ) Known data: p = 0 x 0 3 N m - p = p T = 5 C + 73 = 88 K T =? V =? V = 0.7 m 3 m = 0. kg, R = 0.6 x 0 3 J kg - K - Final temp. from p V = m R T 3 p V 0 x 0 x 0.7 T = = 3 = 54.4 K m R 0. x 0.6 x 0 FINAL TEMP. T = = 5.4 C Original volume from p V = m R T 3 m R T 0. x 0.6 x 0 V = = 3 p 0 x 0 x 88 ORIGINAL VOLUME OF OXYGEN = 0.48 m 3 DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 6

138 0. Under conditions of s.t.p. 4 kg of hydrogen occupy 44.8 m 3. a) If the pressure is increased to.8 bar, what will the temperature of the gas reach if the volume occupied by the gas remains unchanged? (.09 C) b) What value for the specific gas constant for hydrogen can be deducted from this? (4.56 kj kg - K - ) a) Known data: p =.03 x 0 5 N m - p =.8 x 0 5 N m - V = 44.8 m 3 V = V = 44.8 m 3 T = 0 C + 73 = 73 K T =? m = 4 kg T from equation p V T = p V T 5 p.8 x 0 T = x T = 5 x 73 = K p.03 x 0 AFTER PRESSURE INCREASE, T = =.09 C b) Spec. gas constant, R from p V = m R T R = p V m T = 5.8 x 0 x x = J kg - K - SPECIFIC or CHAR. GAS CONSTANT, R = 4.56 kj kg - K - Note: m 3 = 000 litres; bar = 0 5 N m - Standard temperature and pressure (s.t.p.) = 0 C and.03 bar DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 7

139 OUTCOME : THERMODYNAMIC PROPERTY TABLES. At what pressure will water boil at a temperature of: a) 36.4 C 90 bar b) 80.3 C 0.48 bar c) 7. C bar. Use steam property tables to find values for the quantities listed below. a) Spec. internal energy of dry saturated steam at 85 bar. 565 kj kg - b) Enthalpy of evaporation of dry saturated steam at 0.05 bar. 45 kj kg - c) Spec. enthalpy of dry saturated steam at.9 bar. 74 kj kg - d) Spec. internal energy of boiling water at 35 bar. 530 kj kg - e) Volume per kg of dry saturated steam at 0.6 bar m 3 kg - f) Enthalpy per kg of boiling water at 48 bar. 4 kj kg - g) Spec. enthalpy of steam at 40 bar and 450 C kj kg - NOTE: State units for all the above values. DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 8

140 3. State the saturation temperature of wet steam at 0.90 bar C 4. State the pressure of wet steam at a temperature of 09.8 C. 9 bar 5. Use steam property tables to find the enthalpy of evaporation of wet steam at pressures listed below: Pressure Enthalpy of evaporation 3 bar 778 kj kg kn m - 39 kj kg - 95 bar 639 kj kg N m kj kg bar 369 kj kg -.0 MN m - 94 kj kg - 6. A tank contains superheated steam at a pressure of 7 bar and temperature of 350 C. a) State the degree of superheat for the steam. (350 65) = 85 C b) For above steam what is the spec. internal energy at the listed temperature? 880 kj kg -. c) Determine the total enthalpy of 6 kg of steam in above listed condition. 364 x 6 = 8984 kj.. DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 9

141 7. Use steam tables and appropriate formulae to determine specific internal energy of wet steam at a pressure of 44 bar with a dryness fraction of u x = ( x)u f + xu g = ( -.89) x 600 = = kj kg - 8. Calculate the specific volume of wet steam at.095 bar and.95 dry. v x = xv g =.95 x 5.4 = 4.63 m 3 kg - 9. Calculate the specific enthalpy of steam at a pressure of bar with a dryness fraction of 0.9. h x = h f + xh fg = x 000 = 58 kj kg - 0. Use steam tables and interpolation to determine the specific volume of superheated steam at a pressure of.5 bar and temperature of 450 C. Spec. volume, v at.5 bar and 400 C =.067 m 3 kg - Spec. volume, v at.5 bar and 500 C =.376 m 3 kg - Spec. volume, v at.5 bar and 450 C = =.5 m 3 kg - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 0

142 . Use steam tables and interpolation to determine the specific internal energy of superheated steam at a pressure of 5 bar and 350 C. AT 0 bar and 350 C, u = 86 kj kg - AT 30 bar and 350 C, u = 845 kj kg - Spec. internal energy at 5 bar and 350 C = = 853 kj kg -. Use steam tables and interpolation to determine the specific enthalpy of superheated steam at a pressure of 45 bar and a temperature of 475 C. At 40 bar and 475 C, h = At 50 bar and 475 C, h = At 45 bar and 475 C, h = = kj kg - = kj kg - = 338 kj kg - 3. One kilogram of superheated steam 5 bar and 400 C has its pressure reduced to.5 bar at the same temperature. Determine the change in spec. enthalpy. At.5 bar and 400 C, h = 377 kj kg - At 5 bar and 400 C, h = 356 kj kg - Change/increase in spec. enthalpy = kj kg - 4. What is the difference in specific volume of superheated steam at 0.5 bar and 300 C when its pressure is increased to 5.0 bar at the same temperature? At 0.5 bar and 300 C, v = 5.84 m 3 kg - At 5.0 bar and 300 C, v =.56 m 3 kg - Difference in specific volume = m 3 kg - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

143 Refrigerants. At what temperature does refrigerant R77 boil at a pressure of.680 bar? - C. State the Specific Liquid Enthalpy of refrigerant R at 0.84 bar kj kg - 3. Refrigerant R has a Specific Enthalpy of 04.0 kj kg - at a pressure of 4.95 bar. State whether it is in the wet, dry, saturated or superheated condition. SUPERHEATED h g h f 4. Determine the Enthalpy of Evaporation of ( ) Refrigerant R77 at 0 C and.90 bar. = 330. kj kg - 5. State the Specific Volume of dry saturated vapour for refrigerant R at 45 C and bar m 3 kg - v x = xv g =.88 x Determine the Specific Volume of refrigerant R77 at a temperature of C, pressure of bar dryness fraction m 3 kg - 7. State the Specific Enthalpy of refrigerant R at bar and temperature 30 K kj kg - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

144 OUTCOME 3: THE STEADY FLOW ENERGY EQUATION/GASES AND VAPOURS. Air flows steadily at a rate of 0.4 kg s - through an air compressor, entering at 60 m s - with a pressure of bar and a specific volume of 0.85 m 3 kg - and leaving at 45 m s - with a pressure of 6.9 bar and a specific volume of 0.6 m 3 kg -. The internal energy of the air leaving is 88 kj kg - greater than that of the air entering. Cooling water in the jacket surrounding the cylinder absorbs heat from the air at the rate of 59 kj s -. C = 60 m s - p = 00 kn m - v = 0.85 m 3 kg - Sketch a systems diagram for the compressor and attach the given data then calculate the power required to drive the compressor. COMPRESSOR Q = 59 kj s- C = 45 m s - p = 690 kn m - v = 0.6 m 3 kg - (04 kw) m = 0.4 kg s - (u u ) = 88 kj kg - Energy input = Energy output m (KE + IE + FE ) + W = m (KE + IE + FE ) + Q W = m [(KE KE ) + (IE IE ) + (FE FE ) + Q C - C = m 3 + (u - u ) + (pv - pv ) + Q x 0 = m x 0 + (88 k) + (690k x. 6-00k x.85) + 59k = m [-0.79 k + 88 k k] + 59 k = (0.4 x.6 k) + 59 k W = k + 59 k W = kj s - WORK INPUT RATE/POWER TO COMPRESSOR = 04 kw DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 3

145 . The temperature of feed water entering a boiler is 3 C and steam is produced at 40 bar and 500 C. The rate of heat transfer to the surroundings is 35 kw. Determine the mass flow rate of the working fluid for a heat input to the boiler of 6 MW. (.98 kg s - ) WATER AT t = 3 C m =? BOILER p = 40 bar t = 500 C Q = 6 MW Q = 35 kw Energy input = Energy output m x h + Q = m x h + Q Q = m (h - h) + Q Spec. enthalpy of water at 3 C = 408 kj kg - Spec. enthalpy of steam at 40 bar and 500 C = 3445 kj kg - Above values: (408 from h f Wet Tables) and (3445 from h Superheat Tables) 6 x 0 6 = m ( ) k + 35 k 6000 = m (037) + 35 kj s - m = =.98 kg s - MASS FLOW RATE, m =.98 kg s - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 4

146 3. In a steady flow system, the working fluid flows at the rate of 4 kg s -. It enters at a pressure of 60 kn m -, a velocity of 300 m s -, internal energy 00 kj kg - and specific volume 0.37 m 3 kg -. It leaves the system at a pressure of 30 kn m -, a velocity of 50 m s -, internal energy 500 kj kg - and specific volume. m 3 kg -. During its passage through the system, the fluid has a heat transfer loss of 30 kj kg - to the surroundings. Attach the given data to the systems diagram shown below and determine the power of the system, stating whether it is from or to the system. Neglect any change in PE. Q = 30 kj kg - (.7 MW) SYSTEM p = 60 kn m - C = 300 m s - u = 00 kj kg - v = 0.37 m 3 kg - - m = 4 kg s p = 30 kn m - C = 50 m s - u = 500 kj kg - v =. m 3 kg - Energy input = Energy output Neglecting the PE term, the SFEE per kg of fluid takes the form:- C + u + p v C = + u + pv + Q + W C - C W = 3 + (u - u) + (pv - pv) x 0 kj kg = 3 + (00-500) + (60 x x.) - 30 x 0 = Work done per kg = kj kg - Hence, power = x 4 kj kg - x kg s - = kj s - = kj s - POWER OUTPUT FROM THE SYSTEM =.7 MW DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 5

147 4. In a steady flow system, steam enters at a pressure of 3.6 MN m - and with negligible kinetic energy. It leaves at a pressure of 00 kn m -, 0.8 dry, and with a kinetic energy of 400 kj s -. If the steam flows at the rate of 5 kg s -, determine the condition of the steam at entry to the system, assuming there is no transfer of heat or of work energy. (.968 Dry) I = 3.6 MN m - i.e. = 36 bar KE = 0 h =? SYSTEM p = 00 kn m - i.e. = bar x = 0.8 K.E. = 400 kj s - - m = 5 kg s Q = 0 and W = 0 h = h f + x h fg Now, energy input to system = Energy output from system PE + KE + h = PE + KE + h h = ( x 0) 5 kj kg - h = SPEC. ENTHALPY AT ENTRY, h = kj kg - With reference to spec. enthalpy values in superheat tables for steam at 30 and 40 bar we can deduce that entry steam at kj kg - is in wet condition. Hence, h at 36 bar = h f + x h fg = x * 744 x = = STEAM IS IN WET CONDITION at 36 bar and DRY DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 6

148 5. A steady flow of steam enters a turbine with a velocity of 6 m s - and specific enthalpy of 990 kj kg -. The steam leaves the turbine with a velocity of 37 m s - and specific enthalpy of 530 kj kg -. The heat lost to the surroundings as the steam passes through the turbine is 5 kj kg -. The steam flow rate is 9 kg s -. Insert the given data into the systems diagram shown and determine the power output from the turbine. C = 6 m s - h = 990 kj kg - (3.9 MW) Q = 5 kj kg - STEAM INPUT m - = 9 kg s TURBINE W STEAM OUTPUT C = 37 m s - h = 530 kj kg - Energy input = Energy output KE + h = KE + h + Q + W kj kg - W = (KE KE ) + (h h ) Q C W - C = (h - h) x 0 kj s - = x 0 + ( ) - 5 = 9 [ ] = 9 x W = kj s - i.e. POWER OUTPUT FROM TURBINE = 3.9 MW DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 7

149 6. Exhaust steam from a turbine enters a condenser with a specific enthalpy of 64 kj kg - and a velocity of 75 m s -. During the cooling process, the heat loss to the cooling water per kg of working fluid is 680 kj kg -. The condensate exits the system with a velocity of m s -. Sketch a systems diagram for the condenser and determine the Specific Enthalpy of the working fluid at exit from the condenser. ( kj kg - ) C 3 = 75 m s - h 3 = 64 kj kg - COOLING WATER CONDENSER Q = 680 kj kg - C 4 = m s - h 4 =? 3 4 From the Principle of Conservation of Energy Energy input = Energy output and in a condenser no work is done W = O KE 3 + ENTHALPY 3 = KE 4 + ENTHALPY 4 + HEAT LOSS C 3 + h 3 = C 4 + h 4 + Q Specific enthalpy at exit, h 4 = C3 - C x h 3 Q = 75-3 x k 680k =.74 k + 64 k 680 k = kj kg - SPECIFIC ENTHALPY OF FLUID AT EXIT = kj kg - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 8

150 7. An air/fuel mixture passes through a gas turbine system at a rate of 3 kg s -. It enters with a velocity of 30 m s - and a specific enthalpy of 600 kj kg -. At exit, the velocity is 05 m s - and the specific enthalpy is 950 kj kg -. In its passage through the turbine the working fluid has a heat transfer loss of 30 kj kg -. Sketch a systems diagram for the turbine and attach given data then determine the power developed by the turbine. C = 30 m s - h = 600 kj kg - Q = 30 kj kg - (.87 MW) TURBINE m - = 3 kgs C = 05 m s - h = 950 kj kg - Energy input = Energy output The SFEE per kg of fluid can be stated as: C + h = C + h + Q + W Re-arranging, W = C - C + (h h ) Q = x 0 + ( ) 30 = = kj kg - Hence power W = x 3 (kj kg - x kg s - ) = kj s - POWER DEVELOPED BY TURBINE =.87 MW DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 9

151 8. An air/fuel mixture enters a steady flow system with a velocity of 30 m s - and exits with a velocity of 40 m s -. The mass flow rate is 9 kg s -. The properties of the fluid at entry are: a) pressure 3.8 bar b) specific volume 0. m 3 kg - c) specific internal energy 4 kj kg -. At exit, the properties of the fluid are: d) pressure.035 bar e) specific volume m 3 kg - and f) specific internal energy 08 kj kg -. The heat transfer rate from the system is 4. kj s -. Produce a systems diagram and attach given data, then determine the rate of work transfer from the system. (.6 MW) Q W C = 40 m s - C = 30 m s - p = 3.8 bar v = 0. m 3 kg - u = 4 kj kg - SYSTEM p =.035 bar (03.5 kn m - ) v = m 3 kg - u = 08 kj kg - Energy input = Energy output m (KE + IE + FE ) = m (KE + IE + FE ) + Q + W W = m [(KE KE ) + (IE IE ) + (FE FE )] - Q = m C - C + (u - u)+ (pv - pv) - Q IN kjs - W = m x 0 + (4-08) + (380 x x.805) - Q = m [ ] 4. = 9 [89.69] 4. W = kj s - RATE OF WORK TRANSFER FROM SYSTEM = W =.6 MW DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 0

152 9. Boiler Turbine W =? Pump h = 307 kj kg - SUPERHEAT TABLES h = h f + h fg = x 7 = kj kg - P = 80 bar t = 45 C C = 36 m s - Cond. Q = 3 kw p =.8 bar x =.87 C = 85 m s - Steam Power Plant Systems Diagram Rankine cycle Liquid Condensate The underlisted conditions apply to the turbine element in the systems diagram shown. At Entry At Exit Steam Pressure 80 bar.8 bar Steam Temperature 45 C - Fluid Velocity 36 m s - 85 m s - Dryness Fraction Mass Flow Rate of Working Fluid = 5.8 kg s - Rate of Heat Loss to Surroundings = 3 KW Using appropriate thermodynamic symbols, insert the given data into the diagram and use directional arrows for heat/work transfers. Also, calculate the work rate (power) of the turbine. (4.4 MW) Energy input = Energy output m (KE + h ) = m (KE + h ) + Q + W W = m [(KE KE ) + (h h )] - Q C - C = m 3 + (h - h) - x 0 Q = x 0 + ( ) - 3 = 5.8 [ ] 3 W = kj s - POWER DEVELOPED BY TURBINE = 4.4 MW kj s - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

153 0. A steady flow of wet steam enters a condenser with a pressure of 4.5 bar, a dryness fraction of 0.86, and a velocity of 00 m s -. The liquid condensate exits the condenser at a pressure of. bar and with a velocity of m s -. Determine: a) The change of Specific Enthalpy across the system b) The heat loss to the cooling water per kg of the working fluid p 3 = 4.5 bar C 3 = 00 m s - x = 0.86 COOLING WATER Q OUT (.008 MJ kg - ;.03 MJ kg - ) STEAM INPUT CONDENSER OUTPUT p 4 =. bar C 4 = m s a) Spec. enthalpy of wet steam at 4.5 bar and.86 dry, h x = h f + xh fg Spec. enthalpy at entry, h 3 = x h 3 = kj kg - Spec. liquid enthalpy at exit, h 4 = 439 kj kg - (h f on tables) (h 3 h 4 ) = CHANGE OF SPEC. ENTHALPY ACROSS SYSTEM =.008 MJ kg - b) Energy input = Energy output KE 3 + ENTHALPY h 3 = KE 4 + ENTHALPY 4 + Q Heat transfer per kg, Q = (KE 3 KE 4 ) + (h 3 h 4 ) C3 - C4 = x 0 kj kg = = x 0 HEAT TRANSFERRED TO COOLING WATER =.03 MJ kg - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

154 OUTCOME 4: APPLICATION OF MASS CONTINUITY AND BERNOULLI S EQUATIONS TO INCOMPRESSIBLE FLOW THROUGH PIPES. Oil of density 935 kg m -3 flows steadily through a pipe similar to that in a Venturi meter as shown below. From data given, determine: a) the volumetric flow rate (.006 m 3 s - ) b) the fluid velocity at 45 mm diameter section (.643 m s - ) c) the diameter d in millimeters (55.3 mm) d) the mass flow rate of the fluid (.444 kg s - ) Ø 80 Ø 45 Ø d 0.5ms - C ms -.09ms - a) at 80 mm diameter section, V = A C = 4 Œ x.08 x 0.5 volumetric flow rate, V =.006 m 3 s - b) velocity at 45 mm diameter section from A C = A C A velocity C = x C A Œ = 4 x.08 x 0. 5 Œ 4 x.045 fluid velocity at 45 mm diameter, C =.643 m s - c) diameter d from A C = A 3 C 3 A 3 = A x A 3 = Œ 4 x d = Œ 4 x.08 x C C 3 d = m diameter d = =.0555 m = 55.3 mm d) Mass flow rate m =! A C = 935 x Œ 4 x.08 x 0.5 =.4439 kg s - mass flow rate of oil =.444 kg s - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 3

155 . A 00 mm diameter pipe splits into branch pipes, one 60 mm diameter and the other 80 mm diameter. The system carries oil of density 930 kg m -3 in steady flow. If the velocity of the oil in the largest pipe is 3 m s - and in the smallest 4 m s -, determine the following: a) the volumetric flow rate in the 00 mm pipe (0.094 m 3 s - ) b) the mass flow rate in the 60 mm pipe (0.58 kg s - ) c) the velocity of the oil in the 80 mm pipe (6.49 m s - ) 3 3 ms - 80 mm 00 mm 60 mm 4 ms - a) Volumetric flow rate in 00 mm pipe, V = A C V = Π4 x 0. x 3 V = m 3 s - b) Mass flow rate in 60 mm pipe, m =! A C m = 930 x Π4 x.06 x 4 m = 0.58 kg s - c) Volumetric flow rate in 60 mm pipe, V = A C V = Π4 x.06 x 4 V = 0.03 m 3 s - Now V = V + V 3 V 3 = V - V V = = m 3 s - 3 Velocity in 80 mm pipe, V 3 = A 3 C 3 C 3 = V 3 A x 4 = Π[ 08 = 6.49 m s VELOCITY OF OIL in 80 mm diameter pipe = 6.49 m s - - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 4

156 3. A pipeline. m diameter at its upper end tapers to 0.6 m diameter at its lower end over a length of 300 m with datum levels as shown in figure. The pressure at the upper end is 69 kn m -. Water flows steadily downwards at a mass flow rate of 400 kg s -. Frictional losses in the pipeline are equivalent to a head of 0.85 m per 00 m run. Determine the velocity at each end of the pipeline and the water pressure at the lower end. 300m 70m 73m (velocity at 0.6 m end = 4.95 m s - ) (velocity at. m end =.4 m s - ) (pressure at 0.6 m end =.9 kn m - ) Fluid velocities at each end from equation m = AC C = $ m at lower end C = m $ and at upper end C = m $ 400 x x Œ[ Œ0.6 = 400 x x Œ[ Œ. = C = 4.95 m s - C =.4 m s - Find pressure from Bernoulli s equation: Z + C g + p!j = Z + C + g p!j + Z F p.4 69 x = x x x 9.8 x x 9.8 p = p = p =.409 x 980 = 9.3 N m - WATER PRESSURE AT LOWER END =.9 kn m - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 5

157 4. A horizontal pipeline, carrying water in steady flow, tapers from 00 mm diameter at point A to 00 mm diameter at B. The volumetric flow rate of the water is m 3 s -. Determine: a) The fluid velocities at points A and B (.75 and 7.0 m s - ) b) The mass flow rate of the water (54.98 kg s - ) c) The pressure difference between points A and B. Neglect frictional losses. (.97 kn m - ) A B C =? C =? d = 00mm d = 00mm V = O.O55m 3 s - a) Find velocities at points A and B from V = = A C = A C Velocity C = V and velocity C = A V A x 4 = Π[ x 4 = Π[ Velocity at A =.75 m s - and Velocity at B = 7.0 m s - b) Mass flow rate of water m =! A C Π= 000 x 4 x. x.75 Mass flow rate of the water = kg s - c) pressure difference from Bernoulli s equation: Z + C p C p + = Z + + g!j g!j p - p C = (Z = Z HORIZ.)!J g!j & - C ) Pressure difference p p = g 000 ( ) = = N m - PRESSURE DIFFERENCE BETWEEN A and B =.97 kn m - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 6

158 5. Oil of density 900 kg m -3 flows steadily through a pipe from point A to point B. At A the pipe diameter is 5 mm and the pressure is 60 kn m - and at point B, which is m below A the diameter is 50 mm and the pressure is 0 kn m -. Determine: a) the velocity of the oil at point A (8.756 m s - ) b) the mass flow rate of the oil (96.7 kg s - ) A B a) From equation A C = A C velocity at point A, m ΠA C = x C = 4 x.5 x C A Π4 x. 5 VELOCITY C = 4C Assuming no frictional losses, Bernoulli s equation gives: Z + C p + g J = Z + C p + g J C - C g = (4C) - C g = (Z Z ) + p - p J 6C - C x 9.8 = (0 ) + (0-60) x C 9.6 = Velocity C = 9.6 x =.89 m s - Velocity of oil at point A = 4C = 4 x.89 = m s - b) Mass flow rate of oil, m = A C = 900 x Mass flow rate of oil = 96.7 kg s - Π4 x.5 x DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 7

159 6. Oil of density 930 kg m -3 enters a horizontal pipeline with a pressure of.08 MN m - and a velocity of 7.5 m s -. The oil leaves the pipeline with a pressure of 0.64 MN m - and a velocity of 30 m s -. Apply Bernoulli s equation and determine the loss of head due to frictional resistance. (5.3 m) Bernoulli s equation gives: Energy input = Energy output Z + C p + g J C p = Z ZF g J For a horizontal pipeline, the Z terms are dropped Re-arranging: Frictional resistance, Z F = C - C p - p + g J Z F = x ( ) x x = x = Z F = 5.3 m LOSS OF HEAD DUE TO FRICTIONAL RESISTANCE = 5.3 m DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 8

160 7. A horizontal pipe, 00 mm in diameter, full of water in steady flow, has a reduced section 50 mm in diameter forming a Venturi tube. The difference in pressure between the 00 mm and 50 mm sections is measured by a mercury filled U-tube in which the level difference is 445 mm. Taking the densities of water as 000 kg m -3 and mercury as 3600 kg m -3, calculate: a) the velocity of the water in the 00 mm section (.708 m s - ) b) the mass flow rate of the water (.7 kg s - ) Ø50mm = 445mm Ø00mm a) Velocity in 00 mm diameter section from eauation. Velocity C = gh A A Hg - - C = x 9.8 x.445 Œ Œ 4 x. 4 x = x.6 5 VELOCITY OF WATER IN 00 mm SECTION =.708 m s - b) MASS FLOW RATE m =! A C = 000 x Œ 4 x. x.708 =.7 kg s - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 9

161 8. A pipeline, full of water in steady flow, tapers from 0.7 m diameter at point A to 0.35 m diameter at B which is 5 m vertically below A. If at point A the pressure and velocity are 600 kn m - and 5 m s - respectively, calculate the velocity at B. If the frictional head loss is 3.6 m, what will be the pressure at point B? (0 m s - ; 6.4 kn m - ) A Velocity at point B from A C = A C A C = x C A 5m Velocity at B = ΠΠ4 x x 0.35 x 5 = 0m s B Now Energy at level A = Energy at level B and using Bernoulli s equation we have: Z + C p + g J C p = Z ZF g J x x x p = x x = p p = p = ( ) x 980 = N m - PRESSURE AT SECTION B = 6.4 kn m - DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 30

162 OUTCOME 5: SOLVE PROBLEMS ASSOCIATED WITH THE BEHAVIOUR OF LIQUIDS AT REST. An oil storage tank has vertical sides and is of rectangular section 5.8 m by 3.6 m. If the tank contains oil of density 90 kg m -3 to a depth of.5 m, calculate the total thrust due to the oil (a) on the bottom (b) on the 5.8 m side (c) on the 3.6 m wide end. (47. kn, kn, 0.53 kn) a) Total thrust/force on bottom, F = ρ g A h = 90 x 9.8 x 5.8 x 3.6 x.5 Total thrust on bottom of tank = 47. kn b) Total thrust on 5.8 m side, F = ρ g A y Total thrust on 5.8 m side = kn = 90 x 9.8 x 5.8 x.5 x.5 c) Total thrust on 3.6 m side, F = ρ g A y Total thrust on 3.6 m side = 0.53 kn = 90 x 9.8 x 3.6 x.5 x.5. A horizontal pipe of.80 m bore is full of oil of density 890 kg m -3. The pipe is closed at both ends. Calculate the force acting on an end plate. (0 kn) Force acting on end plate, F = ρ g A y = 890 x 9.8 x = N Force acting on end = 0 kn Œ 4 x.8 x 0.9 DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 3

163 3. A cylindrical vessel 0.6 m diameter contains oil of density 935 kg m -3 to a depth of 0.5 m. Calculate the maximum force exerted on the horizontal base of the vessel due to the liquid column. What depth of mercury, density 3600 kg m -3, would generate the same force on the base of the vessel? (9. N, 34.4 mm) Force acting on base, F = ρ g A h = 935 x 9.8 x = 9. N Œ 4 x.6 x.5 For mercury, force F = ρ Hg g A h h = = Hg F A g 9. x x Œ[ Œ 6 x 9.8 = m depth of mercury = 34.4 mm DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 3

164 4. A rectangular storage tank measures m long x 8 m wide and contains a liquid of density 900 kg m -3. A circular inspection hatch in the horizontal base of the tank is subjected to a force of 30.9 kn when the liquid depth is 7.0 m. Determine: a) the diameter of the hatch (0.798 m) b) the total force acting on the base (5.933 MN at mid point) c) the total thrust acting on the 8 m wide end (.73 MN) a) pressure acting on hatch, p = ρ gh = 900 x 9.8 x 7 = 6803 N m - now, pressure on hatch, p = FORCE ON HATCH C.S.A OF HATCH = F A C.S.A. = F P = 30.9 x = m C.S.A. = Œ 4 d = x 4 diameter of hatch, d = = m Œ b) total force acting on base = pressure x C.S.A. of base total force on base = MN = 6803 x x 8 = N c) total thrust on 8 m side, F = ρ g A y = 900 x 9.8 x 8 x 7 x 3.5 = N total thrust on 8 m side =.73 MN DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 33

165 5. A total force of 65.6 kn acts on each vertical side of a square based storage tank containing oil of density 95 kg m -3. The wetted surface area on each vertical side is 8.9 m. Determine: a) the depth of oil in the tank (4.8 m) b) the height above the base where total force acts (.607 m) c) the total thrust on the base (.558 MN) a) force acting on vertical side F = J y 65.6 x 0 3 = 95 x 9.8 x 8.9 x y x 0 depth of centroid, y = =.4 m 95 x 9.8 x 8.9 depth of oil in tank =.4 x = 4.8 m b) depth of centre of pressure below centroid = = d y 4.8 x.4 =.803 m height of C of P above base = 4.8 ( ) =.607 m c) Total thrust on base, F = ρ g A h now, area of wetted vertical surface = 8.9 m = length of side x 4.8 length of side = l = 8.9 = 6 m 4.8 F = 95 x 9.8 x 6 x 4.8 = N TOTAL THRUST ON BASE =.558 MN DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 34

166 6. The figure shows a sectional view of a vertical gate in a dam wall. The rectangular gate measures.8 m deep by. m wide and is designed to pivot about its upper end which lies 8. m below the free surface. Take the density of water as 000 kg m -3 and determine: a) the magnitude of the total force acting on the gate (580 kn) b) the position of the centre of pressure below the pivot (.468 m) c) the force F required at bottom of gate to just keep the gate closed. (304 kn) 8.m FS a) depth of centroid below FS, y = 8. + y = 9.6 m total force, say F acting on gate from.8.8m EPTH PIVOT F F = J y = 000 x 9.8 x.8 x. x 9.6 TOTAL FORCE = 580 kn b) Distance, GC, centre of pressure lies below centroid = For rectangular vertical plane surface k d = k y (d = depth of gate) GC =.8 =.0680 m x 9.6 position of C of P below pivot = =.468 m.8m GATE DEPTH F F c) to find force F, take moments about pivot. Σ CM = Σ ACM F x.8 = F x.468 F = x 0 x FORCE, F = 304 kn DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 35

167 7. A circular opening in the vertical wall of a storage tank is closed by a gate mounted on trunnions on its horizontal centre line. The gate has a diameter 3.5 m and its horizontal centre line lies 6.5 m below the fluid free surface. The tank contains oil of density 95 kg m -3. Determine the magnitude of the pulling force required to be applied at a point.60 m vertically above the centroid of the gate to just keep it shut against the hydrostatic force exerted by the oil. FS (4.78 kn).5m.5m.6m PULLING FORCE Ø Total force acting on gate, say, F = J y F = 95 x 9.8 x Œ 4 x 3.5 x 6.5 F = kn This force F acts below centroid at a distance GC = k y also for a circle k d = where d = diameter of gate. 6 Hence GC = 3.5 = 0.5 m 6 x 6.5 Find pulling force, say F, to just keep door closed by taking moments about trunnion centre line Σ CM = Σ ACM F x.6 = x 0.5 F = x 0. 5 = 4.78 kn.6 PULLING FORCE = 4.78 kn DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 36

168 8. A mercury U-tube manometer as shown in figure is to measure the pressure of oil in the pipeline at A. Alongside the pipeline system is a mercury barometer with a column height of 756 mm of mercury. Determine: a) atmospheric pressure b) the absolute pressure in the pipeline at A when dimension y is 0.6 m and dimension h is 0.98 m. Take the densities of oil and mercury as 900 kg m -3 and 3600 kg m -3 respectively. (00.86 kn m -, 6.3 kn m - ) OIL C M ERCURY D ENSITY H g A B y X X a) Atmospheric pressure from barometer reading, p ATM = Hg p ATM = 3600 x 9.8 x.756 = kn m - b) Pressure at C in RH limb = atmospheric or zero gauge Absolute pressure at x-x in RH limb = x Hg g h = x x 9.8 x 0.98 = 3.6 kn m - Absolute pressure at x-x in LH limb = 3.6 kn m - (same level) Absolute pressure at B in LH limb = 3.6 x ρ OIL g y OIL = 3.6 x x 9.8 x 0.6 = 6.3 kn m - Alternative solution Gauge pressure in pipeline at A = g ( ρ Hg H - ρ OIL Y) = 9.8 (3600 x x 0.6) = 5.45 kn m - Absolute pressure at A = gauge pressure at A + atmospheric pressure = ( ) k ABSOLUTE PRESSURE AT A = 6.3 KN M - g h DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 37

169 The figure shows a mercury manometer linked to a pipeline carrying water of density 000 kg m -3. If the gauge pressure at A is 0 kn m - and dimension y =.48 m, what will be the height h in the RH limb. Take the density of mercury as 3600 kg m -3. (.683 m) MERCURY Hg WATER C A B y X X Pressure in pipeline at A = g ( Hg h - y) 0 x 0 3 = 9.8 (3600 h 000 x.48) 0 x 0 3 = 3346 h h = height h in RH limb =.683 m DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 38

170 9. The figure shows a differential manometer connecting two pipelines carrying water of density 000 kg m -3. The U-tube contains mercury of density 3600 kg m -3. If the pressure difference between A and B is 48 kn m -, dimension X = 0.86 m, dimension Y = 0.5 m, what is the difference in level h? (4.5 cm) W A W X B h Y C D Pressure difference formula states: p A p B = w (Y X) + h (w w ) p A - p B = ρ H0 g (Y X) + h ( Hg g - H0 g) = 000 x 9.8 ( ) + h (3600 x x 9.8) = 980 (- 0.34) + h ( ) 48 x 0 3 = h (3606) h = = m DIFFERENCE IN LEVEL h = 4.5 cm DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme 39

171 DET: Mechanical Engineering: Thermofluids Higher Tutorials Marking Scheme

172 Section 5: Appendix, Quantities used in Thermofluids (Higher) DET: Mechanical Engineering: Thermofluids Appendix (Higher)

173 DET: Mechanical Engineering: Thermofluids Appendix (Higher)

174 Quantities Used in Unit THERMOFLUIDS (HIGHER) Quantity Symbol Unit Cross-sectional area A m Velocity C m s - Force F N Specific enthalpy h J kg - Enthalpy H J Mass m kg Mass flow rate m. kg s - Gauge pressure p g N m - Atmospheric pressure p atm N m - Absolute pressure p N m - Heat transfer Q. J Rate of heat transfer Q J s -, W Characteristic gas constant R J kg - K - Universal gas constant R o J kg - mol K - Temperature t C Absolute temperature T K Specific internal energy u J kg - Internal energy U J Specific volume v m 3 kg - Volume V m. 3 Volumetric flow rate V m 3 s - Density (Rho) ρ kg m -3 Work Transfer W J Work rate/power W J s -, Watt Specific Weight w N m -3 Dryness Fraction x DET: Mechanical Engineering: Thermofluids Appendix (Higher)

175 DET: Mechanical Engineering: Thermofluids Appendix (Higher)