Thermodynamics FIRST 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 10^{8} 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:m_{in}(hpeke_{in})Qm_{out}(hpeke_{out})W1) Neglecting pe and ke and assuming no work is done on the system.m(h_{in})Qm(h_{out})Qm(h_{out}h_{in})2) Substituting where c_{p} = specific heat capacity (Btu/lbm-°F).Qmc_{pD}T= (c_{p}) (T_{out} - T_{in})m= 1.38 x 10^{8} lbm/hr (1.5 Btu/lbm-^{o}F) (540 - 610^{o}F)= -1.45 x 10^{10} Btu/hrQThe 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 (h_{out}-h_{in}), or It is importantQmQmc_{pD}T.to note that the later equation can only be used when no phase change occurs since DT= 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. 0 Page 63 HT-01

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