Unit code: H/ QCF level: 5 Credit value: 15 OUTCOME 1 - THERMODYNAMIC SYSTEMS TUTORIAL 2

Transcription

1 Unit 43: Plant and Process Princiles Unit code: H/60 44 QCF level: 5 Credit value: 5 OUCOME - HERMODYNAMIC SYSEMS UORIAL Understand thermodynamic systems as alied to lant engineering rocesses hermodynamic systems: closed systems; oen systems; alication of st Law to derive system energy equations; enthaly Proerties: system roerties e.g. intensive, extensive, two-roerty rule Polytroic rocesses: general equation v n = C; relationshis between index n and heat transfer during a rocess; constant ressure and reversible isothermal and adiabatic rocesses; exressions for work flow Relationshis: system constants for a erfect gas e.g. R = C - Cv and = γ= C/Cv When you have comleted tutorial you should be able to do the following. Exlain and use the First Law of hermodynamics. Solve roblems involving various kinds of thermodynamic systems. Exlain and use olytroic exansion and comression rocesses. D.J.Dunn

2 . ENERGY RANSFER here are two ways to transfer energy in and out of a system, by means of work and by means of heat. Both may be regarded as a quantity of energy transferred in Joules or energy transfer er second in Watts. When you comlete section one you should be able to exlain and calculate the following. Heat transfer. Heat transfer rate. Work transfer Work transfer rate (Power).. HEA RANSFER Heat transfer occurs because one lace is hotter than another. Under normal circumstances, heat will only flow from a hot body to a cold body by virtue of the temerature difference. here are 3 mechanisms for this, Conduction, convection and radiation. You do not need to study the laws governing conduction, convection and radiation in this module. Fig. A quantity of energy transferred as heat is given the symbol Q and it's basic unit is the Joule. he quantity transferred in one second is the heat transfer rate and this has the symbol and the unit is the Watt. An examle of this is when heat asses from the furnace of a steam boiler through the walls searating the combustion chamber from the water and steam. In this case, conduction, convection and radiation all occur together. Fig. D.J.Dunn

3 SELF ASSESSMEN EXERCISE No.. kg/s of steam flows in a ie 40 mm bore at 00 bar ressure and 400oC. i. Look u the secific volume of the steam and determine the mean velocity in the ie. (7.9 m/s) ii. Determine the kinetic energy being transorted er second. (3.3 W) iii. Determine the enthaly being transorted er second. (89 W) D.J.Dunn 3

4 .. WORK RANSFER Energy may be transorted from one lace to another mechanically. An examle of this is when the outut shaft of a car engine transfers energy to the wheels. A quantity of energy transferred as work is 'W' Joules but the work transferred in one second is the Power 'P' Watts. An examle of ower transfer is the shaft of a steam turbine used to transfer energy from the steam to the generator in an electric ower station. Fig.3 It is useful to remember that the ower transmitted by a shaft deends uon the torque and angular velocity. he formulae used are P = or P = N is the angular velocity in radian er second and N is the angular velocity in revolutions er second. D.J.Dunn 4

5 WORKED EXAMPLE No. A duct has a cross section of 0. m x 0.4 m. Steam flows through it at a rate of 3 kg/s with a ressure of bar. he steam has a dryness fraction of Calculate all the individual forms of energy being transorted. SOLUION Cross sectional area = 0. x 0.4 = 0.08 m. olume flow rate = mxvg at bar olume flow rate = 3 x 0.98 x =.6 m3/s. velocity = c = olume/area =.6/0.08 = 3.5 m/s. Kinetic Energy being transorted = mc/ = 3 x 3.5 / = 584 Watts. Enthaly being transorted = m(hf + x hfg) H= 3( x 0) = kW Flow energy being transorted = ressure x volume Flow Energy = x05 x.6 = Watts Internal energy being transorted = m(uf + x ufg) U = 3( x 05)= kw Check flow energy = H - U = = 50 kw D.J.Dunn 5

6 SELF ASSESSMEN EXERCISE No.. he shaft of a steam turbine roduces 600 Nm torque at 50 rev/s. Calculate the work transfer rate from the steam. (88.5 W). A car engine roduces 30 kw of ower at 3000 rev/min. Calculate the torque roduced. (95.5 Nm) D.J.Dunn 6

7 . HE FIRS LAW OF HERMODYNAMICS When you have comleted section two, you should be able to exlain and use the following terms. he First Law of hermodynamics. Closed systems. he Non-Flow Energy Equation. Oen systems. he Steady Flow Energy Equation.. HERMODYNAMIC SYSEMS In order to do energy calculations, we identify our system and draw a boundary around it to searate it from the surroundings. We can then kee account of all the energy crossing the boundary. he first law simly states that he net energy transfer = net energy change of the system. Energy transfer into the system = E(in) Energy transfer out of system = E (out) Fig. 4 Net change of energy inside system = E(in) - E (out) = E his is the fundamental form of the first law. hermodynamic systems might contain only static fluid in which case they are called NON-FLOW or CLOSED SYSEMS. Alternatively, there may be a steady flow of fluid through the system in which case it is known as a SEADY FLOW or OPEN SYSEM. he energy equation is fundamentally different for each because most energy forms only aly to a fluid in motion. We will look at non-flow systems first. D.J.Dunn 7

8 . NON-FLOW SYSEMS he rules governing non-flow systems are as follows. he volume of the boundary may change. No fluid crosses the boundary. Energy may be transferred across the boundary. When the volume enlarges, work (-W) is transferred from the system to the surroundings. When the volume shrinks, work (+W) is transferred from the surroundings into the system. Energy may also be transferred into the system as heat (+Q) or out of the system (-Q). his is best shown with the examle of a iston sliding inside a cylinder filled with a fluid such as gas. Fig.5 he only energy ossessed by the fluid is internal energy (U) so the net change is U. he energy equation becomes Q + W = U his is known as the NON-FLOW ENERGY EQUAION (N.F.E.E.) D.J.Dunn 8

9 .3 SEADY FLOW SYSEMS he laws governing this tye of system are as follows. Fluid enters and leaves through the boundary at a steady rate. Energy may be transferred into or out of the system. A good examle of this system is a steam turbine. Energy may be transferred out as a rate of heat transfer or as a rate of work transfer P. Fig.6. he fluid entering and leaving has otential energy (PE), kinetic energy (KE) and enthaly (H). he first law becomes + P = Nett change in energy of the fluid. + P = (PE)/s + (KE)/s + (H)/s his is called the SEADY FLOW ENERGY EQUAION (S.F.E.E.) Again, we will use the convention of ositive for energy transferred into the system. Note that the term means change of and if the inlet is denoted oint () and the outlet oint (). he change is the difference between the values at () and (). For examle H means (H-H). D.J.Dunn 9

10 WORKED EXAMPLE No.3 A steam turbine is sulied with 30 kg/s of suerheated steam at 80 bar and 400oC with negligible velocity. he turbine shaft roduces 00 knm of torque at 3000 rev/min. here is a heat loss of. MW from the casing. Determine the thermal ower remaining in the exhaust steam. SOLUION Shaft Power = N =(3000/60) x = 6.83 x 06 W = 6.83 MW hermal ower sulied = H at 80 bar and 400oC H = 30(339) = 9470 kw = 94.7 MW otal energy flow into turbine = 94.7 MW Energy flow out of turbine = 94.7 MW = SP + Loss + Exhaust. hermal Power in exhaust = = 30.4 MW SELF ASSESSMEN EXERCISE No.3. A non-flow system receives 80 kj of heat transfer and loses 0 kj as work transfer. What is the change in the internal energy of the fluid? (60 kj). A non-flow system receives 00 kj of heat transfer and also 40 kj of work is transferred to it. What is the change in the internal energy of the fluid? (40 kj) 3. A steady flow system receives 500 kw of heat and loses 00 kw of work. What is the net change in the energy of the fluid flowing through it? (300 kw) 4. A steady flow system loses kw of heat also loses 4 kw of work. What is the net change in the energy of the fluid flowing through it? (-6 kw) 5. A steady flow system loses 3 kw of heat also loses 0 kw of work. he fluid flows through the system at a steady rate of 70 kg/s. he velocity at inlet is 0 m/s and at outlet it is 0 m/s. he inlet is 0 m above the outlet. Calculate the following. i. he change in K.E./s (-0.5 kw) ii. he change in P.E/s (-3.7 kw) iii. he change in enthaly/s (.3 kw) D.J.Dunn 0

11 3. MORE EXAMPLES OF HERMODYNAMIC SYSEMS When we examine a thermodynamic system, we must first decide whether it is a nonflow or a steady flow system. First, we will look at examles of non-flow systems. 3. PISON IN A CYLINDER Fig. 7 here may be heat and work transfer. he N.F.E.E. is, Q + W = U Sometimes there is no heat transfer (e.g. when the cylinder is insulated). Q = 0 so W = U If the iston does not move, the volume is fixed and no work transfer occurs. In this case Q = U For a GAS ONLY the change in internal energy is U= m Cv. 3.. SEALED EAPORAOR OR CONDENSER. Fig. 8 Since no change in volume occurs, there is no work transfer so Q = U D.J.Dunn

12 WORKED EXAMPLE No.4 30 g of gas inside a cylinder fitted with a iston has a temerature of 5oC. he iston is moved with a mean force of 00 N so that that it moves 60 mm and comresses the gas. he temerature rises to oc as a result. Calculate the heat transfer given cv = 78 J/kg K. SOLUION his is a non flow system so the law alying is Q + W = U he change in internal energy is U = mcv = 0.03 x 78 x ( - 5) U = 9.4 J he work is transferred into the system because the volume shrinks. W = force x distance moved = 00 x 0.06 = J Q = U - W = 7.4 J Now we will look at examles of steady flow systems. D.J.Dunn

13 3.3. PUMPS AND FLUID MOORS he diagram shows grahical symbols for hydraulic ums and motors. he S.F.E.E. states, Fig.9 + P =KE/s + PE/s + H/s In this case, esecially if the fluid is a liquid, the velocity is the same at inlet and outlet and the kinetic energy is ignored. If the inlet and outlet are at the same height, the PE is also neglected. Heat transfer does not usually occur in ums and motors so is zero. he S.F.E.E. simlifies to P = H/s Remember that enthaly is the sum of internal energy and flow energy. he enthaly of gases, vaours and liquids may be found. In the case of liquids, the change of internal energy is small and so the change in enthaly is equal to the change in flow energy only. he equation simlifies further to P =FE/s Since FE = and is constant for a liquid, this becomes P = WORKED EXAMPLE No.5 A um delivers 0 kg/s of oil of density 780 kg/m3 from atmosheric ressure at inlet to 800 kpa gauge ressure at outlet. he inlet and outlet ies are the same size and at the same level. Calculate the theoretical ower inut. SOLUION Since the ies are the same size, the velocities are equal and the change in kinetic energy is zero. Since they are at the same level, the change in otential energy is also zero. Neglect heat transfer and internal energy. P= = m/ = 0/780 = m3/s = = 800 kpa P = x = W or 0.48 kw D.J.Dunn 3

14 WORKED EXAMPLE No.6 A feed um on a ower station ums 0 kg/s of water. At inlet the water is at bar and 0oC. At outlet it is at 00 bar and 40oC. Assuming that there is no heat transfer and that PE and KE are negligible, calculate the theoretical ower inut. In this case the internal energy has increased due to frictional heating. he SFEE reduces to P = H/s = m(h - h) he h values may be found from tables. h = 504 kj/kg his is near enough the value of hf at 0 o C bar in steam tables. h = 60 kj/kg P = 0 (60-504) = 969 kw or.969 MW If water tables are not to hand the roblem may be solved as follows. h = u + f.e. u = c where c = 4.8 kj/kg K for water u = 4.8 (40-0) =83.6 kj/kg f.e. = he volume of water is normally around 0.00 m3/kg. f.e. = 0.00 x (00 - ) x 05 = J/kg or 9.9 kj/kg hence h = u + fe = = 03.5 kj/kg P = mh = 0 x 03.5 = 070 kw or.07 MW he discreancies between the answers are slight and due to the fact the value of the secific heat and of the secific volume are not accurate at 00 bar. D.J.Dunn 4

15 3.4. GAS COMPRESSORS AND URBINES. Figure 0 shows the basic construction of an axial flow comressor and turbine. hese have rows of aerofoil blades on the rotor and in the casing. he turbine asses high ressure hot gas or steam from left to right making the rotor rotate. he comressor draws in gas and comresses it in stages. Fig. 0 Comressing a gas normally makes it hotter but exanding it makes it colder. his is because gas is comressible and unlike the cases for liquids already covered, the volumes change dramatically with ressure. his might cause a change in velocity and hence kinetic energy. Often both kinetic and otential energy are negligible. he internal energy change is not negligible. Figure shows grahical symbols for turbines and comressors. Note the narrow end is always the high ressure end. Fig. D.J.Dunn 5

16 WORKED EXAMPLE No.7 A gas turbine uses 5 kg/s of hot air. It takes it in at 6 bar and 900oC and exhausts it at 450oC. he turbine loses 0 kw of heat from the casing. Calculate the theoretical ower outut given that c = 005 J/kg K. First identify this as a steady flow system for which the equation is + P = K.E./s + P.E./s + H/s For lack of further information we assume K.E. and PE to be negligible. he heat transfer rate is -0 kw. he enthaly change for a gas is H = mc H = 5 x 005 x ( ) = W or -.6 MW P = H - = -6 - (-0) = -4 kw he minus sign indicates that the ower is leaving the turbine. Note that if this was a steam turbine, you would look u the h values in the steam tables. D.J.Dunn 6

17 3.5 SEADY FLOW EAPORAORS AND CONDENSERS A refrigerator is a good examle of a thermodynamic system. In articular, it has a heat exchanger inside that absorbs heat at a cold temerature and evaorates the liquid into a gas. he gas is comressed and becomes hot. he gas is then cooled and condensed on the outside in another heat exchanger. Fig.. For both the evaorator and condenser, there is no work transferred in or out. K.E. and P.E. are not normally a feature of such systems so the S.F.E.E. reduces to = H/s On steam ower lant, boilers are used to raise steam and these are examles of large evaorators working at high ressures and temeratures. Steam condensers are also found on ower stations. he energy equation is the same, whatever the alication. D.J.Dunn 7

18 WORKED EXAMPLE No.8 A steam condenser takes in wet steam at 8 kg/s and dryness fraction 0.8. his is condensed into saturated water at outlet. he working ressure is 0.05 bar. Calculate the heat transfer rate. SOLUION = H/s = m(h - h) h = hf + x hfg at 0.05 bar from the steam tables we find that h = (43) = 5 kj/kg h = hf at 0.05 bar = 38 kj/kg hence = 8(38-5) = kw he negative sign indicates heat transferred from the system to the surroundings. D.J.Dunn 8

19 SELF ASSESSMEN EXERCISE No.4. Gas is contained inside a cylinder fitted with a iston. he gas is at 0oC and has a mass of 0 g. he gas is comressed with a mean force of 80 N which moves the iston 50 mm. At the same time 5 Joules of heat transfer occurs out of the gas. Calculate the following. i. he work done.(4 J) ii. he change in internal energy. (- J) iii. he final temerature. (9.9 o C) ake cv as 78 J/kg K. A steady flow air comressor draws in air at 0oC and comresses it to 0oC at outlet. he mass flow rate is 0.7 kg/s. At the same time, 5 kw of heat is transferred into the system. Calculate the following. i. he change in enthaly er second. (70.35 kw) ii. he work transfer rate. (65.35 kw) ake c as 005 J/kg K. 3. A steady flow boiler is sulied with water at 5 kg/s, 00 bar ressure and 00oC. he water is heated and turned into steam. his leaves at 5 kg/s, 00 bar and 500oC. Using your steam tables, find the following. i. he secific enthaly of the water entering. (856 kj/kg) ii. he secific enthaly of the steam leaving. (3373 kj/kg) iii. he heat transfer rate. (37.75 kw) 4. A um delivers 50 dm3/min of water from an inlet ressure of 00 kpa to an outlet ressure of 3 MPa. here is no measurable rise in temerature. Ignoring K.E. and P.E, calculate the work transfer rate. (.4 kw) 5. A water um delivers 30 dm3/minute (0.3 m3/min) drawing it in at 00 kpa and delivering it at 500 kpa. Assuming that only flow energy changes occur, calculate the ower sulied to the um. (860 W) D.J.Dunn 9

20 6. A steam condenser is sulied with kg/s of steam at 0.07 bar and dryness fraction 0.9. he steam is condensed into saturated water at outlet. Determine the following. i. he secific enthalies at inlet and outlet. (33 kj/kg and 63 kj/kg) ii. he heat transfer rate. (4336 kw) kg/s of gas is heated at constant ressure in a steady flow system from 0oC to 80oC. Calculate the heat transfer rate. (37.4 kw) C =. kj/kg K kg of gas is cooled from 0oC to 50oC at constant volume in a closed system. Calculate the heat transfer. (-6.8 kj) Cv = 0.8 kj/kg. D.J.Dunn 0

21 4. POLYROPIC PROCESSES. When you comlete section four you should be able to do the following. Use the laws governing the exansion and comression of a fluid. State the names of standard rocesses. Derive and use the work laws for closed system exansions and comressions. Solve roblems involving gas and vaour rocesses in closed systems. We will start by examining exansion and comression rocesses. 4. COMPRESSION AND EXPANSION PROCESSES. A comressible fluid (gas or vaour) may be comressed by reducing its volume or exanded by increasing its volume. his may be done inside a cylinder by moving a iston or by allowing the ressure to change as it flows through a system such as a turbine. For ease of understanding, let us consider the change as occurring inside a cylinder. he rocess is best exlained with a ressure - volume grah. When the volume changes, the ressure and temerature may also change. he resulting ressure deends uon the final temerature. he final temerature deends on whether the fluid is cooled or heated during the rocess. It is normal to show these changes on a grah of ressure lotted against volume. (- grahs). A tyical grah for a comression and an exansion rocess is shown in fig.3. Fig. 3 It has been discovered that the resulting curves follows the mathematical law n = constant. D.J.Dunn

22 Deending on whether the fluid is heated or cooled, a family of such curves is obtained as shown (fig.4). Each grah has a different value of n and n is called the index of exansion or comression. Fig.4 he most common rocesses are as follows. CONSAN OLUME also known as ISOCHORIC A vertical grah is a constant volume rocess and so it is not a comression nor exansion. Since no movement of the iston occurs no work transfer has taken lace. Nevertheless, it still fits the law with n having a value of infinity. CONSAN PRESSURE also known as ISOBARIC A horizontal grah reresents a change in volume with no ressure change (constant ressure rocess). he value of n is zero in this case. CONSAN EMPERAURE also known as ISOHERMAL All the grahs in between constant volume and constant ressure, reresent rocesses with a value of n between infinity and zero. One of these reresents the case when the temerature is maintained constant by cooling or heating by just the right amount. When the fluid is a gas, the law coincides with Boyle's Law = constant so it follows that n is. When the fluid is a vaour, the gas law is not accurate and the value of n is close to but not equal to. D.J.Dunn

23 ADIABAIC PROCESS When the ressure and volume change in such a way that no heat is added nor lost from the fluid (e.g. by using an insulated cylinder), the rocess is called adiabatic. his is an imortant rocess and is the one that occurs when the change takes lace so raidly that there is no time for heat transfer to occur. his rocess reresents a demarcation between those in which heat flows into the fluid and those in which heat flows out of the fluid. In order to show it is secial, the symbol is used instead of n and the law is = C It will be found that each gas has a secial value for (e.g..4 for dry air). POLYROPIC PROCESS All the other curves reresent changes with some degree of heat transfer either into or out of the fluid. hese are generally known as olytroic rocesses. HYPERBOLIC PROCESS he rocess with n= is a hyerbola so it is called a hyerbolic rocess. his is also isothermal for gas but not for vaour. It is usually used in the context of a steam exansion. WORKED EXAMPLE No.9 A gas is comressed from bar and 00 cm3 to 0 cm3 by the law.3=constant. Calculate the final ressure. SOLUION. If.3 = C then.3 = C =.3 hence x 00.3 = x 0.3 x (00/0).3 = = 8. bar D.J.Dunn 3

24 WORKED EXAMPLE No.0 aour at 0 bar and 30 cm3 is exanded to bar by the law. = C. Find the final volume. SOLUION.. = C =. 0 x 30. = x. = (59.3) /. = 04.4 cm3 WORKED EXAMPLE No. A gas is comressed from 00 kpa and 0 cm3 to 30 cm3 and the resulting ressure is MPa. Calculate the index of comression n. SOLUION. 00 x 0 n = 000 x 30 n (0/30) n = 000/00 = 5 4 n = 5 nlog4 = log 5 n = log5/log4 =.6094/.3863 =.6 Note this may be solved with natural or base 0 logs or directly on suitable calculators. D.J.Dunn 4

25 SELF ASSESSMEN EXERCISE No. 5. A vaour is exanded from bar and 50 cm3 to 50 cm3 and the resulting ressure is 6 bar. Calculate the index of comression n. (0.63).a. A gas is comressed from 00 kpa and 300 cm3 to 800 kpa by the law.4=c. Calculate the new volume. (.4 cm 3 ).b. he gas was at 50oC before comression. Calculate the new temerature using the gas law / = C. (07 o C) 3.a. A gas is exanded from MPa and 50 cm3 to 50 cm3 by the law.5 = C. Calculate the new ressure. (506 kpa) 3.b. he temerature was 500oC before exansion. Calculate the final temerature. (34 o C) D.J.Dunn 5

26 D.J.Dunn COMBINING HE GAS LAW WIH HE POLYROPIC LAW. For gases only, the general law may be combined with the law of exansion as follows. and so Since for an exansion or comression n n n Substituting into the gas law we get n and further since n substituting into the gas law gives n o summarise we have found that n n In the case of an adiabatic rocess this is written as For an isothermal rocess n = and the temeratures are the same. WORKED EXAMPLE No. A gas is comressed adiabatically with a volume comression ratio of 0. he initial temerature is 5oC. Calculate the final temerature given =.4 SOLUION C K o or (0).4

27 D.J.Dunn 7 WORKED EXAMPLE No.3 A gas is comressed olytroically by the law. = C from bar and 0oC to bar. calculate the final temerature. SOLUION K n n WORKED EXAMPLE No.4 A gas is exanded from 900 kpa and 00oC to 00 kpa by the law.3 = C. Calculate the final temerature. SOLUION K n n

28 SELF ASSESSMEN EXERCISE No. 6. A gas is exanded from MPa and 000oC to 00 kpa. Calculate the final temerature when the rocess is i. Isothermal (n = ) (000 o C) ii Polytroic (n =.) (594 o C) iii. Adiabatic ( =.4) (386 o C) iv. Polytroic (n=.6) (64 o C). A gas is comressed from 0 kpa and 5oC to 800 kpa. Calculate the final temerature when the rocess is i. Isothermal (n = ) (5 o C) ii. Polytroic (n =.3) (73 o C) iii Adiabatic ( =.4) ( o C) iv. Polytroic (n =.5) (69 o C) 3. A gas is comressed from 00 kpa and 0oC to. MPa by the law.3=c. he mass is 0.0 kg. c=005 J/kg K. c v = 78 J/kg K. Calculate the following. i. he final temerature. (434 K) ii. he change in internal energy (.03 kj) iii. he change in enthaly (.84 kj) 4. A gas is exanded from 900 kpa and 00oC to 0 kpa by the law.4 = C. he mass is 0.05 kg. c=00 J/kg K cv = 750 J/kg K Calculate the following. i. he final temerature. (88 K) ii. he change in internal energy (-7.5 kj) iii. he change in enthaly (-0.7 kj) D.J.Dunn 8

29 4.3. EXAMPLES INOLING APOUR Problems involving vaour make use of the formulae n = C in the same way as those involving gas. You cannot aly gas laws, however, unless it is suerheated into the gas region. You must make use of vaour tables so a good understanding of this is essential. his is best exlained with worked examles. WORKED EXAMPLE No.5 A steam turbine exands steam from 0 bar and 300oC to bar by the law.= C. Determine for each kg flowing: a. the initial and final volume. b. the dryness fraction after exansion. c. the initial and final enthalies. d. the change in enthaly. SOLUION he system is a steady flow system in which exansion takes lace as the fluid flows. he law of exansion alies in just the same way as in a closed system. he initial volume is found from steam tables. At 0 bar and 300oC it is suerheated and from the tables we find v= 0.55 m3/kg Next aly the law. = C. =. 0 x = x. Hence =.53m3/kg Next, find the dryness fraction as follows. Final volume =.53 m3/kg = xvg at bar. From the tables we find vg is.694 m3/kg hence.53 =.694x x = We may now find the enthalies in the usual way. h at 0 bar and 300oC is 305 kj/kg h = hf + xhfg at bar (wet steam) h = 47 + (0.899)(58) = 447 kj/kg he change in enthaly is h - h = -578 kj/kg D.J.Dunn 9

30 SELF ASSESSMEN EXERCISE No.7. 3 kg/s of steam is exanded in a turbine from 0 bar and 00oC to.5 bar by the law.=c. Determine the following. i. he initial and final volumes. (0.68 m 3 and 3 m 3 ) ii. he dryness fraction after exansion. (0.863) iii. he initial and final enthalies. (89 kj/kg and 388 kj/kg) iv. he change in enthaly. -34 kw)..5 kg/s of steam is exanded from 70 bar and 450oC to 0.05 bar by the law.3 = C. Determine the following. i. he initial and final volumes. (0.066 m 3 /kg and 7.4 m 3 /kg) ii. he dryness fraction after exansion. (0.4) iii. he initial and final enthalies. (387 kj/kg and 35 kj/kg) iv. he change in enthaly. (-38 kw) 3. A horizontal cylindrical vessel is divided into two sections each m3 volume, by a non-conducting iston. One section contains steam of dryness fraction 0.3 at a ressure of bar, while the other contains air at the same ressure and temerature as the steam. Heat is transferred to the steam very slowly until its ressure reaches bar. Assume that the comression of the air is adiabatic (=.4) and neglect the effect of friction between the iston and cylinder. Calculate the following. i. he final volume of the steam. (.39 m 3 ) ii. iii. he mass of the steam. (.97 kg) he initial internal energy of the steam. (053 kj) iv he final dryness fraction of the steam. (0.798) v. he final internal energy of the steam. (47 kj) vi. he heat added to the steam. (9 kj) D.J.Dunn 30

31 4.4. CLOSED SYSEM WORK LAWS EXPANSION OF PRESSURE WIH OLUME We will start by studying the exansion of a fluid inside a cylinder against a iston which may do work against the surroundings. A fluid may exand in two ways. a) It may exand raidly and uncontrollably doing no useful work. In such a case the ressure could not be lotted against volume during the rocess. his is called an UNRESISED EXPANSION b) It may exand moving the iston. he movement is resisted by external forces so the gas ressure changes in order to overcome the external force and move the iston. In this case the movement is controlled and the variation of ressure with volume may be recorded and lotted on a - grah. Work is done against the surroundings. his rocess is called a RESISED EXPANSION. Consider the arrangement shown in fig. 5. Assume that there is no ressure outside. If the string holding the weight was cut, the gas ressure would slam the iston back and the energy would be dissiated first by acceleration of the moving arts and eventually as friction. he exansion would be unresisted. Fig. 5 If the weights were gradually reduced, the gas would ush the iston and raise the remaining weights. In this way, work would be done against the surroundings (it ends u as otential energy in the weights). he rocess may be reeated in many small stes, with the change in volume each time being d. he ressure although changing, is at any time. D.J.Dunn 3

32 his rocess is characterised by two imortant factors.. he rocess may be reversed at any time by adding weights and the otential energy is transferred back from the surroundings as work is done on the system. he fluid may be returned to its original ressure, volume, temerature and energy.. he fluid force on one side of the iston is always exactly balanced by the external force (in this case due to the weights) on the other side of the iston. he exansion or comression done in this manner is said to be REERSIBLE and CARRIED OU IN EQUILIBRIUM WORK AS AREA UNDER HE - DIAGRAM If the exansion is carried out in equilibrium, the force of the fluid must be equal to the external force F. It follows that F = A. When the iston moves a small distance dx, the work done is dw dw = - F dx =- Adx = - d. he minus sign is because the work is leaving the system. For an exansion from oints to it follows that the total work done is given by W We must remember at this stage that our sign convention was that work leaving the system is negative. It should be noted that some of the work is used to overcome any external ressure such as atmosheric and the useful work is reduced. Consider the system shown in fig.5 again but this time suose there is atmosheric ressure on the outside a. In this case It follows that F + a A = A. F = A. - a A d When the iston moves a small distance dx, the the useful work done is F dx - F dx = - (Adx - a Adx) = - ( - a )d. For an exansion from oints to it follows that the useful work done is given by W ( a ) d D.J.Dunn 3

33 WORK LAWS FOR CLOSED SYSEMS If we solve the exression W d we obtain the work laws for a closed system. he solution deends uon the relationshi between and. he formulae now derived aly equally well to a comression rocess and an exansion rocess. Let us now solve these cases. CONSAN PRESSURE W W d d W = - ( - ) CONSAN OLUME If is constant then d = 0 W = 0. HYPERBOLIC his is an exansion which follows the law = C and is ISOHERMAL when it is a gas. Substituting = C - the exression becomes W d C d C ln Since = C then W ln since W ln In the case of gas we can substitute = mr and so W mr ln mr ln D.J.Dunn 33

34 POLYROPIC In this case the exansion follows the law n = C. he solution is as follows. W W W C d n n n but C d n W C n Since C or For gas only we may substitute = mr and so ADIABAIC -n W mr n Since an adiabatic case is the secial case of a olytroic exansion with no heat transfer, the derivation is identical but the symbol is used instead of n. W For gas only we may substitute = mr and sow mr his is the secial case of the olytroic rocess in which Q=0. Q 0 W mr Substituting for Q and U in the NFEE we find Q W U Since R C Cv mr 0 mcv C Cv Cv( ) C Cv R Cv his shows that the ratio of the rincial secific heat caacities is the adiabatic index. It was shown earlier that the difference is the gas constant R. hese imortant relationshis should be remembered. C Cv = R = C/Cv D.J.Dunn 34

35 WORKED EXAMPLE No.5 Air at a ressure of 500 kpa and volume 50 cm3 is exanded reversibly in a closed system to 800 cm3 by the law.3 = C. Calculate the following. SOLUION a. he final ressure. b. he work done. = 500 kpa = 50 x 0-6 m3 = 800 x 0-6 m3.3 W 3.6x0 n.3 W 47Joules 3 or 500x kPa 3.6x0 (50x0 3 6 ).3 (800x0 6 x800x0 500x ).3 x50x0 6 WORKED EXAMPLE No.6 Steam at 6 bar ressure and volume 00 cm3 is exanded reversibly in a closed system to dm3 by the law. = C. Calculate the work done. SOLUION = 6 bar = 00 x 0-6 m 3 = x 0-3 m3.. 00x0 6x x x0 xx0 6x0 x00x0 W n W 35.Joules bar. D.J.Dunn 35

36 SELF ASSESSMEN EXERCISE No.8. 0 g of steam at 0 bar and 350oC exands reversibly in a closed system to bar by the law.3=c. Calculate the following. i. he initial volume. (0.008 m 3 ) ii. he final volume. ( m 3 ) iii. he work done. (-.9 kj). 0 g of gas at 0oC and bar ressure is comressed to 9 bar by the law.4 = C. aking the gas constant R = 87 J/kg K calculate the work done. (Note that for a comression rocess the work will turn out to be ositive if you correctly identify the initial and final conditions). (3.67 kj) 3. Gas at 600 kpa and 0.05 dm3 is exanded reversibly to 00 kpa by the law.35 = C. Calculate the work done. (-3.8 kj) 4. 5 g of gas is comressed isothermally from 00 kpa and 0oC to MPa ressure. he gas constant is 87 J/kg K. Calculate the work done. (.9 kj) 5. Steam at 0 bar with a volume of 80 cm3 is exanded reversibly to bar by the law =C. Calculate the work done. (-84. kj) 6. Gas fills a cylinder fitted with a frictionless iston. he initial ressure and volume are 40 MPa and 0.05 dm 3 resectively. he gas exands reversibly and olytroically to 0.5 MPa and dm 3 resectively. Calculate the index of exansion and the work done. (.463 and -3.4 kj) 7. An air comressor commences comression when the cylinder contains g at a ressure is.0 bar and the temerature is 0 o C. he comression is comleted when the ressure is 7 bar and the temerature 90 o C. (.4 and 944 J) he characteristic gas constant R is 87 J/kg K. Assuming the rocess is reversible and olytroic, calculate the index of comression and the work done. D.J.Dunn 36

37 WORKED EXAMPLE No.7 0. kg of gas at 00 oc is exanded isothermally and reversibly from MPa ressure to 00 kpa. ake Cv = 78 J/kg K and R = 87 J/kg K. Calculate i. he work transfer. ii. he change in internal energy. iii. he heat transfer. SOLUION W ln mr ln mr ln 6 x0 W 0.x87x373ln J or kj 5 0 x he work is leaving the system so it is a negative work transfer. Since is constant U = 0 Q = 0 Q = 49.3 kj Note that 49.3 kj of heat is transferred into the gas and 49.3 kj of work is transferred out of the gas leaving the internal energy unchanged. WORKED EXAMPLE No.8 Reeat worked examle 7 but for an adiabatic rocess with =.4 Calculate SOLUION 3 00 x 0 γ 373 x 93 K 6 x W mr 0. x 87 x 0.4 W 5830 J For an adiabatic rocess Q 0 Q W U Check U mc hence U J 0. x 78 x (93-373) J D.J.Dunn 37

38 WORKED EXAMPLE No.9 Reeat worked examle 7 but for a olytroic rocess with n=.5 Calculate SOLUION 00 x x 6 x 0 U mc Q U - W W mr W 3605 J Q (-3603) 83.3 J 3 n 35.3 K 0. x 87 x 0. x 78 x ( ) J D.J.Dunn 38

39 SELF ASSESSMEN EXERCISE No.9 ake Cv = 78 J/kg K and R = 87 J/kg K throughout.. dm3 of gas at 00 kpa and 0oC is comressed to. MPa reversibly by the law. = C. Calculate the following. i. he final volume. (0.6 dm 3 ) ii. he work transfer. (57 J) iii. he final temerature. (70oC) iv. he mass. (.89 g) v. he change in internal energy. (8 J) vi. he heat transfer. (-8 J) kg of gas at 0 bar and 00oC is exanded reversibly to bar by the law.3 =C in a closed system. Calculate the following. i. he initial volume. (9.85 dm 3 ) ii. he final volume. (58 dm 3 ) iii. he work transfer. (-7 kj) iv. he change in internal energy. (-0.3 kj) v. he heat transfer. (6.7 kj) kg of air at 700 kpa and 800oC is exanded adiabatically to 00 kpa in a closed system. aking =.4 calculate the following. i. he final temerature. (65.4 K) ii. he work transfer. (6.3 kj) iii. he change in internal energy. (-6.3 J) D.J.Dunn 39

40 4. A horizontal cylinder is fitted with a frictionless iston and its movement is restrained by a sring as shown (Figure 6.) Figure 6 a. he sring force is directly roortional to movement such that F/x = k Show that the change in ressure is directly roortional to the change in volume such that / = k/a b. he air is initially at a ressure and temerature of 00 kpa and 300 K resectively. Calculate the initial volume such that when the air is heated, the ressure volume grah is a straight line that extends to the origin. (0.5 dm3 ) c. he air is heated making the volume three times the original value. Calculate the following. i. he mass. (0.58 g) ii. he final ressure. (300 kpa) iii. he final temerature. (700 K) iv. he work done. (-00 kj) v. he change in internal energy. (97 J) vi. he heat transfer. (. kj) D.J.Dunn 40

41 4.5. CLOSED SYSEM PROBLEMS INOLING APOUR he solution of roblems involving steam and other vaours is done in the same way as for gases with the imortant roviso that gas laws must not be used. olumes and internal energy values should be obtained from tables and roerty charts. his is best illustrated with a worked examle. WORKED EXAMPLE No.0 kg of steam occuies a volume of 0. m3 at 9 bar in a closed system. he steam is heated at constant ressure until the volume is m3. Calculate the following. i. he initial dryness fraction. ii. he final condition. iii. he work transfer. iv. he change in internal energy. v. he heat transfer. SOLUION First find the initial dryness fraction. = 0. = mx v g at 9 bar x = 0./( x 0.49) x = 0.93 (initial dryness fraction). Now determine the secific volume after exansion. = 9 bar (constant ressure) = m 3. = mv v = 0.344/ = 0.344m 3 /kg First, look in the suerheat tables to see if this value exists for suerheat steam. We find that at 9 bar and 350oC, the secific volume is indeed m3/kg. he final condition is suerheated to 350oC. Note that if v was less than v g at 9 bar the steam would be wet and x would have to be found. Next find the work. W = -( - ) = -9x 05 ( ) = J W = kj (Energy leaving the system) D.J.Dunn 4

42 Next determine the internal energy from steam tables. U = m u and u = u f + x u fg at 9 bar u fg at 9 bar = u g - u f = = 839 kj/kg U = { (839) } = 454 kj U = m u and u = u at 9 bar and 350 o C = 877 kj/kg U = m u = (877) = 877 kj. he change in internal energy = U - U = 43 kj (increased) Finally deduce the heat transfer from the NFEE Q + W = U hence Q = U - W = 43 - (-0.95) Q = 56 kj (energy entering the system) D.J.Dunn 4

43 SELF ASSESSMEN EXERCISE No kg of dry saturated steam at 0 bar ressure is exanded reversibly in a closed system to bar by the law. =C. Calculate the following. i. he initial volume. (38.9 dm 3 ) ii. he final volume. (64 dm 3 ) iii. he work transfer. (-6 kj) iv. he dryness fraction. (0.779) v. he change in internal energy. (-08 kj) vi. he heat transfer. (-46 kj). Steam at 5 bar and 50 o C is exanded reversibly in a closed system to 5 bar. At this ressure the steam is just dry saturated. For a mass of kg calculate the following. i. he final volume. (0.375 m 3 ) ii. he change in internal energy. (-65 kj) iii. he work done. (-87 kj) iv. he heat transfer. (. kj) D.J.Dunn 43