The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 5
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Energy can enter and exit a control volumeby work and heat transfer. Energy also enters and exits with flowing streams of matter. Accordingly, fora control volume with one-dimensional flow at a single inlet and a single outlet,d (U + KE + PE )cvv2v2= Q˙ cv − W˙ + m˙ ui + i + gz i − m˙ ue + e + gz e dt22___________© 1999 by CRC Press LLC___________(2.21)2-15Engineering Thermodynamicswhere the underlined terms account for the specific energy of the incoming and outgoing streams.
Theterms Q̇cv and Ẇ account, respectively, for the net rates of energy transfer by heat and work over theboundary (control surface) of the control volume.Because work is always done on or by a control volume where matter flows across the boundary, thequantity Ẇ of Equation 2.21 can be expressed in terms of two contributions: one is the work associatedwith the force of the fluid pressure as mass is introduced at the inlet and removed at the exit. The other,denoted as Ẇcv , includes all other work effects, such as those associated with rotating shafts, displacement of the boundary, and electrical effects.
The work rate concept of mechanics allows the first of thesecontributions to be evaluated in terms of the product of the pressure force, pA, and velocity at the pointof application of the force. To summarize, the work term Ẇ of Equation 2.21 can be expressed (withEquation 2.20) asW˙ = W˙ cv + ( pe Ae )v e − ( pi Ai )v i(2.22)= W˙ cv + m˙ e ( pe ve ) − m˙ i ( pi vi )The terms ṁi (pvi) and ṁe (peve) account for the work associated with the pressure at the inlet andoutlet, respectively, and are commonly referred to as flow work.Substituting Equation 2.22 into Equation 2.21, and introducing the specific enthalpy h, the followingform of the control volume energy rate balance results:d (U + KE + PE )cvv2v2= Q˙ cv − W˙ cv + m˙ i hi + i + gz i − m˙ e he + e + gz e dt22(2.23)To allow for applications where there may be several locations on the boundary through which massenters or exits, the following expression is appropriate:d (U + KE + PE )cv= Q˙ cv − W˙ cv +dt∑iv2m˙ i hi + i + gz i −2∑ev2m˙ e he + e + gz e 2(2.24)Equation 2.24 is an accounting rate balance for the energy of the control volume.
It states that the timerate of accumulation of energy within the control volume equals the difference between the total ratesof energy transfer in and out across the boundary. The mechanisms of energy transfer are heat and work,as for closed systems, and the energy accompanying the entering and exiting mass.Control Volume Entropy BalanceLike mass and energy, entropy is an extensive property. And like mass and energy, entropy can betransferred into or out of a control volume by streams of matter.
As this is the principal differencebetween the closed system and control volume forms, the control volume entropy rate balance is obtainedby modifying Equation 2.17 to account for these entropy transfers. The result isdScv=dtQ˙ j∑ T + ∑ m˙ s − ∑ m˙ si ijjie e+ S˙gene_____ ______________________ _________rate ofentropychange© 1999 by CRC Press LLCrate ofentropytransferrate ofentropygeneration(2.25)2-16Section 2where dScv/dt represents the time rate of change of entropy within the control volume. The terms ṁi si andṁe se account, respectively, for rates of entropy transfer into and out of the control volume associatedwith mass flow.
One-dimensional flow is assumed at locations where mass enters and exits. Q̇ j representsthe time rate of heat transfer at the location on the boundary where the instantaneous temperature is Tj;and Q˙ j / Tj accounts for the associated rate of entropy transfer. Ṡgen denotes the time rate of entropygeneration due to irreversibilities within the control volume. When a control volume comprises a numberof components, Ṡgen is the sum of the rates of entropy generation of the components.Control Volumes at Steady StateEngineering systems are often idealized as being at steady state, meaning that all properties are unchanging in time. For a control volume at steady state, the identity of the matter within the control volumechange continuously, but the total amount of mass remains constant. At steady state, Equation 2.19reduces to∑ m˙ = ∑ m˙ii(2.26a)eeThe energy rate balance of Equation 2.24 becomes, at steady state,0 = Q˙ cv − W˙ cv +∑ m˙ h + 2 + gz − ∑ m˙ h + 2 + gz iv i2iieiv e2ee(2.26b)eAt steady state, the entropy rate balance of Equation 2.25 reads0=Q˙ j∑ T + ∑ m˙ s − ∑ m˙ si ijjie e+ S˙gen(2.26c)eMass and energy are conserved quantities, but entropy is not generally conserved.
Equation 2.26aindicates that the total rate of mass flow into the control volume equals the total rate of mass flow outof the control volume. Similarly, Equation 2.26b states that the total rate of energy transfer into thecontrol volume equals the total rate of energy transfer out of the control volume.
However, Equation2.26c shows that the rate at which entropy is transferred out exceeds the rate at which entropy enters,the difference being the rate of entropy generation within the control volume owing to irreversibilities.Applications frequently involve control volumes having a single inlet and a single outlet, as, forexample, the control volume of Figure 2.1 where heat transfer (if any) occurs at Tb: the temperature, ora suitable average temperature, on the boundary where heat transfer occurs. For this case the mass ratebalance, Equation 2.26a, reduces to m˙ i = m˙ e . Denoting the common mass flow rate by m˙ , Equations2.26b and 2.26c read, respectively, v 2 − v e2 + g(z i − z e )0 = Q˙ cv − W˙ cv + m˙ (hi − he ) + i 2 0=Q˙ cv+ m˙ (si − se ) + S˙genTb(2.27a)(2.28a)When Equations 2.27a and 2.28a are applied to particular cases of interest, additional simplificationsare usually made.
The heat transfer term Q̇cv is dropped when it is insignificant relative to other energy© 1999 by CRC Press LLC2-17Engineering ThermodynamicsFIGURE 2.1 One-inlet, one-outlet control volume at steady state.transfers across the boundary. This may be the result of one or more of the following: (1) the outersurface of the control volume is insulated; (2) the outer surface area is too small for there to be effectiveheat transfer; (3) the temperature difference between the control volume and its surroundings is smallenough that the heat transfer can be ignored; (4) the gas or liquid passes through the control volume soquickly that there is not enough time for significant heat transfer to occur. The work term Ẇcv drops outof the energy rate balance when there are no rotating shafts, displacements of the boundary, electricaleffects, or other work mechanisms associated with the control volume being considered.
The changesin kinetic and potential energy of Equation 2.27a are frequently negligible relative to other terms in theequation.The special forms of Equations 2.27a and 2.28a listed in Table 2.1 are obtained as follows: whenthere is no heat transfer, Equation 2.28a givesse − si =S˙genm˙≥0(2.28b)(no heat transfer)Accordingly, when irreversibilities are present within the control volume, the specific entropy increasesas mass flows from inlet to outlet.
In the ideal case in which no internal irreversibilities are present,mass passes through the control volume with no change in its entropy — that is, isentropically.For no heat transfer, Equation 2.27a gives v 2 − v e2 + g(z i − z e )W˙ cv = m˙ (hi − he ) + i 2 (2.27b)A special form that is applicable, at least approximately, to compressors, pumps, and turbines resultsfrom dropping the kinetic and potential energy terms of Equation 2.27b, leavingW˙ cv = m˙ (hi − he )(compressors, pumps, and turbines)© 1999 by CRC Press LLC(2.27c)2-18Section 2TABLE 2.1 Energy and Entropy Balances for One-Inlet, OneOutlet Control Volumes at Steady State and No Heat TransferEnergy balance v 2 − v e2 W˙ cv = m˙ (hi − he ) + i+ g(z i − z e ) 2 Compressors, pumps, and turbinesa(2.27b)W˙ cv = m˙ (hi − he )(2.27c)he ≅ hi(2.27d)v e = v i2 + 2(hi − he )(2.27f)ThrottlingNozzles, diffusersbEntropy balancese − si =aS˙gen≥0m˙(2.28b)For an ideal gas with constant cp, Equation 1′ of Table 2.7 allowsEquation 2.27c to be written as˙ p (Ti − Te )W˙ cv = mc(2.27c′)The power developed in an isentropic process is obtained with Equation5′ of Table 2.7 asb( k −1) k ˙ p Ti 1 − ( pe pi )(2.27c″)W˙ cv = mc(s = c)where cp = kR/(k – 1).For an ideal gas with constant cp, Equation 1′ of Table 2.7 allowsEquation 2.27f to be written asv e = v i2 + 2c p (Ti − Te )(2.27f′)The exit velocity for an isentropic process is obtained with Equation5′ of Table 2.7 asv e = v i2 + 2c p Ti 1 − ( pe pi )( k −1)k(s = c)(2.27f″)where cp = kR/(k – 1).In throttling devices a significant reduction in pressure is achieved simply by introducing a restrictioninto a line through which a gas or liquid flows.
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