Thermodynamics, Heat Transfer, And Fluid Flow. V.1. Thermodynamics, страница 12

The estimated heat loss is 5 Btu/lbm of steam. The flow entersthe turbine at 164 ft/sec at a point 6.5 ft above the discharge and leaves the turbine at 262ft/sec. Determine the work of the turbine.Solution:ṁin(hinpeinkein)1) Divide by ṁ since, ṁinQ̇ṁoutmȯut (houtpeoutkeout)Ẇṁ.(hin + pein + kein) + q = (hout + peout + keout) + wwhere:q= heat added to the system per pound (Btu/lbm)w= work done by the system per pound (ft-lbf/lbm)2) Use Joule’s constant J = 778 ft-lbf/Btu for conversions and substitute known values.1349 Btu/lbm + (6.5/778) Btu/lbm + [(164)2/2(32.17)(778)] Btu/lbm +(-5 Btu/lbm) = 1100 Btu/lbm + 0 peout + [(262)2/2(32.17)(778)] Btu/lbm + wNote: The minus sign indicates heat out of the turbine.3) Solve for work, w.1349 Btu/lbm + 8.3548 x 10-3 Btu/lbm + 0.5368 Btu/lbm - 5 Btu/lbm = 1100Btu/lbm + 1.37 Btu/lbm + w1344.54 Btu/lbm = 1101.37 Btu/lbm + ww = 1344.54 Btu/lbm - 1101.37 Btu/lbmw = 243.17 Btu/lbmHT-01Page 60Rev.

0ThermodynamicsFIRST LAW OF THERMODYNAMICSThis example demonstrates thatpotential and kinetic energy terms areinsignificant for a turbine, since the∆pe and ∆ke values are less than 1Btu/lbm.When the system (the fluid beingstudied) changes its properties(temperature, pressure, volume) fromone value to another as a consequenceof work or heat or internal energyexchange, then it is said that the fluidhas gone through a "process." Insome processes, the relationshipsbetween pressure, temperature, andvolume are specified as the fluid goesfrom one thermodynamic state toanother.

The most common processesare those in which the temperature,pressure, or volume is held constantduring the process. These would beclassified as isothermal, isobaric, orisovolumetric processes, respectively.Iso means "constant or one." If thefluid passes through various processesand then eventually returns to thesame state it began with, the system issaid to have undergone a cyclicprocess. One such cyclic process usedis the Rankine cycle, two examples ofwhich are shown in Figure 19.The processes that comprise the cycleare described below.Figure 19Rev.

0T-s Diagram with Rankine Cyclesab:Liquid is compressedwith no change inentropy (by ideal pump).bc:Constant pressure transfer of heat in the boiler. Heat is added to the compressedliquid, two-phase, and superheat states.cd:Constant entropy expansion with shaft work output (in ideal turbine).da:Constant pressure transfer of heat in the sink. Unavailable heat is rejected to theheat sink (condenser).Page 61HT-01FIRST LAW OF THERMODYNAMICSThermodynamicsNote the individual processes the fluid must go through before completing the complete cycle.Rankine cycles will be discussed in greater detail later in this module. Figure 20 shows a typicalsteam plant cycle. Heat is supplied to the steam generator (boiler) where liquid is converted tosteam or vapor.

The vapor is then expanded adiabatically in the turbine to produce a workoutput. Vapor leaving the turbine then enters the condenser where heat is removed and the vaporis condensed into the liquid state. The condensation process is the heat-rejection mechanism forthe cycle. Saturated liquid is delivered to the condensate pump and then the feed pump whereits pressure is raised to the saturation pressure corresponding to the steam generator temperature,and the high pressure liquid is delivered to the steam generator where the cycle repeats itself.Figure 20Typical Steam Plant CycleWith the example complete, it seems appropriate to discuss the various components of a typicalsteam plant system. Although such a system is extremely complex, only the major componentswill be discussed.

A typical steam plant system consists of: a heat source to produce the thermalenergy (e.g. nuclear or fossil fuel); a steam generator to change the thermal energy into steamenergy (a complete steam plant usually exists in connection with the steam generator inconverting the steam into eventual electrical energy); pumps to transfer the fluid back to the heatsource (reactor coolant pumps in a nuclear reactor); a pressurizer to ensure that the primarysystem maintains its desired pressure; and the necessary piping to ensure the fluid passes througheach stage of its cyclic process.

Of necessity, the steam plant is a large "closed" system.However, each component of the system is thermodynamically analyzed as an open system asthe fluid passes through it. Of primary importance is the process of dissipating the energycreated by the heat source. This process takes place in the steam generator, which acts as a gianttwo-phase heat generator.HT-01Page 62Rev. 0ThermodynamicsFIRST LAW OF THERMODYNAMICSThe hot fluid from the heat source passes through the primary side of the steam generator whereits energy is passed to the secondary side of the heat exchanger in such a manner as to createsteam.

The fluid, with its energy removed from the primary side, leaves the steam generator ata lower temperature, and is pumped back to the heat source to be "re-heated." Each majorcomponent of a steam plant can be treated as a separate open system problem. A thermodynamicanalysis, using the various forms of energies discussed, can be applied to the particularcomponent in studying its behavior. A simplified example of the thermodynamics involved inthe steam generator is shown below.Example 2: Open System - Steam Plant ComponentPrimary fluid enters the heat exchanger of a nuclear facility at 610°F and leaves at 540°F.The flow rate is approximately 1.38 x 108 lbm/hr.

If the specific heat of the fluid is takenas 1.5 Btu/lbm°F, what is the heat transferred out of the steam generator?Solution:ṁin(hpekein)Q̇ṁout(hpekeout)Ẇ1) Neglecting pe and ke and assuming no work is done on the system.Q̇ṁ(hin)Q̇ṁ(hout2) Substituting Q̇Q̇ṁ(hout)hin)ṁcp∆T where cp = specific heat capacity (Btu/lbm-°F).=ṁ (cp) (Tout - Tin)=1.38 x 108 lbm/hr (1.5 Btu/lbm-oF) (540 - 610oF)=-1.45 x 1010 Btu/hrThe minus sign indicating heat out of the heat exchanger, which is consistent with thephysical case. This example demonstrates that for a heat exchanger, the heat transfer ratecan be calculated using the equation Q̇ ṁ (hout-hin), or Q̇ ṁcp∆T.

It is importantto note that the later equation can only be used when no phase change occurs since ∆T= 0 during a phase change. The first equation can be used for a phase change heattransfer process as well as for latent heat calculations.Rev. 0Page 63HT-01FIRST LAW OF THERMODYNAMICSThermodynamicsThe pumps used for returning the fluid to the heat source can be analyzed as a thermodynamicsystem also. One such example is illustrated in Example 3.Example 3: Open System - CoolantA power pump is used to return the fluid from the heat exchanger back to the core. Theflow rate through the pump is about 3.0 x 107 lbm/hr with the fluid entering the pump assaturated liquid at 540°F.

The pressure rise across the pump is 90 psia. What is thework of the pump, neglecting heat losses and changes in potential and kinetic energy?Solution:ṁ (hin1)peinAssume Q̇ṁ(hin)2)ẆhinhoutQ̇ṁ (houtpeoutkeout)Ẇ0 and neglect changes in pe and keṁ(hout)Ẇṁ (hin - hout) where Ẇ is the rate of doing work by the pumpuinνPinuout νPout(hin - hout)3)kein)(uin - uout)(νPin - νPout) = ∆ u(νPin - νPout)Since no heat is transferred, ∆ u = 0 and the specific volume out of the pump isthe same as the specific volume entering since water is incompressible.(hin - hout) = ν(Pin - Pout)4)Substituting the expression for work, ẆẆ5)ṁ ν(Pinṁ (hin-hout) we have:Pout) .Using 0.01246 for specific volume.Ẇ = 3.0 x 107 lbm/hr (0.01246 ft3/lbm) (-90psia) (144 in2/ft2)/778 ft-lbf/BtuẆ = -6.23 x 106 Btu/hr or -2446 hpNote: The minus sign indicating work put into the fluid by the pump.

1 hp = 2545Btu/hr.HT-01Page 64Rev. 0ThermodynamicsFIRST LAW OF THERMODYNAMICSA thermodynamic balance across the reactor core gives an indication of the amount ofheat removed by the coolant that is given off by the fuel rods.Example 4: Thermodynamic Balance across Heat SourceIn a particular facility, the temperature leaving the reactor core is 612°F, while thatentering the core is 542°F. The coolant flow through the heat source is 1.32 x 108lbm/hr. The cp of the fluid averages 1.47 Btu/lbm°F.

How much heat is being removedfrom the heat source? The pe and ke energies are small compared to other terms andmay be neglected.Solution:ṁ(hQ̇peke) inṁ(houthin)1) Substituting Q̇Q̇ṁ(hpeke) outẆṁcp∆T where cp = specific heat capacity.Q̇=ṁ(cp) (ToutQ̇=1.32 x 108 lbm/hr (1.47 Btu/lbm -oF) (612 - 542oF)Q̇=1.36 x 1010 Btu/hrFor this example Q̇Tin)ṁcp∆T has been used to calculate the heat transfer rate since nophase change has occurred.

However, Q̇ ṁ (hout-hin) could also have been used hadthe problem data included inlet and outlet enthalpies.The individual principal components of a reactor system have been thermodynamicallyanalyzed. If all components were combined into an overall system, the system could beanalyzed as a "closed" system problem. Such an analysis is illustrated in the followingexample.Rev. 0Page 65HT-01FIRST LAW OF THERMODYNAMICSThermodynamicsExample 5: Overall Thermodynamic BalanceA nuclear facility (primary side) is to be studied as a complete system.

The heatproduced by the heat source is 1.36 x 1010 Btu/hr. The heat removed by the heatexchanger (steam generator) is 1.361 x 1010 Btu/hr. What is the required pump power tomaintain a stable temperature?Solution:Ẇp = pump work, Q̇c = heat produced by the heat source, Q̇into steam generatorṁ(h1)peke)ẆpQ̇cQ̇sgṁ(hpeheat transferredke)For a closed system, the mass entering and leaving the system is zero, therefore,ṁ is constant.