Thermodynamics FIRST 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 10^{8}lbm/hr. The c_{p} 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:m(hpeke)_{in}Qm(hpeke)_{out}WQm(h_{out}h_{in})1) Substituting where c_{p} = specific heat capacity.Qmc_{pD}T=Qm(c_{p}) (T_{out}T_{in})= 1.32 x 10^{8} lbm/hr (1.47 Btu/lbm -^{o}F) (612 - 542^{o}F)Q= 1.36 x 10^{10} Btu/hrQFor this example has been used to calculate the heat transfer rate since noQmc_{pD}Tphase change has occurred. However, (h_{out}-h_{in}) could also have been used hadQmthe 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. 0 Page 65 HT-01

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