Thermodynamics FIRST LAW OF THERMODYNAMICS The hot fluid from the heat source passes through the primary side of the steam generator where its energy is passed to the secondary side of the heat exchanger in such a manner as to create steam.   The fluid, with its energy removed from the primary side, leaves the steam generator at a  lower  temperature,  and  is  pumped  back  to  the  heat  source  to  be  "re-heated."  Each  major component of a steam plant can be treated as a separate open system problem.  A thermodynamic analysis,  using  the  various  forms  of  energies  discussed,  can  be  applied  to  the  particular component in studying its behavior.   A simplified example of the thermodynamics involved in the steam generator is shown below. Example 2:   Open System - Steam Plant Component Primary 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 taken as 1.5 Btu/lbm°F, what is the heat transferred out of the steam generator? Solution: m_in(h pe ke_in) Q m_out(h pe ke_out) W 1)   Neglecting pe and ke and assuming no work is done on the system. m(h_in) Q m(h_out) Q m(h_out h_in) 2)   Substituting where c_p = specific heat capacity (Btu/lbm-°F). Q mc_pDT = (c_p) (T_out - T_in) m = 1.38 x 10⁸ lbm/hr (1.5 Btu/lbm-^oF) (540 - 610^oF) = -1.45 x 10¹⁰ Btu/hr Q The  minus sign indicating heat  out of  the heat exchanger, which  is consistent with the physical case.  This example demonstrates that for a heat exchanger, the heat transfer rate can be calculated using the equation (h_out-h_in), or It is important Q m Q mc_pDT. 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  heat transfer process as well as for latent heat calculations. Rev. 0 Page 63 HT-01