Thermodynamic Balance across Heat Source

Open System - Coolant

Overall Thermodynamic Balance

Thermodynamics Heat Transfer and Fluid Flow Volume 1 of 3
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Thermodynamics FIRST LAW OF THERMODYNAMICS A thermodynamic balance across the reactor core gives an indication of the amount of heat removed by the coolant that is given off by the fuel rods. Example 4: Thermodynamic Balance across Heat Source In a particular facility, the temperature leaving the reactor core is 612°F, while that entering the core is 542°F. The coolant flow through the heat source is 1.32 x 10⁸ lbm/hr. The c_p of the fluid averages 1.47 Btu/lbm°F. How much heat is being removed from the heat source? The pe and ke energies are small compared to other terms and may be neglected. Solution: m(h pe ke)_in Q m(h pe ke)_out W Q m(h_out h_in) 1) Substituting where c_p = specific heat capacity. Q mc_pDT = Q m(c_p) (T_out T_in) = 1.32 x 10⁸ lbm/hr (1.47 Btu/lbm -^oF) (612 - 542^oF) Q = 1.36 x 10¹⁰ Btu/hr Q For this example has been used to calculate the heat transfer rate since no Q mc_pDT phase change has occurred. However, (h_out-h_in) could also have been used had Q m the problem data included inlet and outlet enthalpies. The individual principal components of a reactor system have been thermodynamically analyzed. If all components were combined into an overall system, the system could be analyzed as a "closed" system problem. Such an analysis is illustrated in the following example. Rev. 0 Page 65 HT-01

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