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 cp 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(hout
hin)
1) Substituting
where cp = specific heat capacity.
Q
mcpDT
=
Q
m(cp) (Tout
Tin)
=
1.32 x 108 lbm/hr (1.47 Btu/lbm -oF) (612 - 542oF)
Q
=
1.36 x 1010 Btu/hr
Q
For this example
has been used to calculate the heat transfer rate since no
Q
mcpDT
phase change has occurred. However,
(hout-hin) 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
