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Calculation of Decay Heat | Up
Thermodynamics Heat Transfer and Fluid Flow Volume 2 of 3 | Next
Decay Heat Removal |
Heat Transfer
DECAY HEAT
where:
= decay heat (Btu/hr)
Q
m = mass of coolant (lbm)
cp
= specific heat capacity of coolant (Btu/lbm-oF)
DT = temperature change of coolant (oF)
Dt = time over which heatup takes place (hr)
Example:
Three days after a planned reactor shutdown, it is desired to perform maintenance on one
of two primary heat exchangers. Each heat exchanger is rated at 12,000 Btu/hr. To
check the current heat load on the primary system due to decay heat, cooling is secured
to both heat exchangers. The primary system heats up at a rate of 0.8°F/hr. The primary
system contains 24,000 lbm of coolant with a specific heat capacity of 0.8 Btu/lbm-°F.
Will one heat exchanger be sufficient to remove the decay heat?
Solution:
Q
m cp
DT
Dt
(24,000 lbm)æ
ç
è
ö
÷
ø
0.8
Btu
lbm °F
æ
ç
è
ö
÷
ø
0.8°F
1 hr
15,360Btu
hr
One heat exchanger removes 12,000 Btu/hr.
One heat exchanger will not be sufficient.
Decay Heat Limits
Reactor decay heat can be a major concern. In the worst case scenarios, it can cause melting of
and/or damage to the reactor core, as in the case of Three Mile Island. The degree of concern
with decay heat will vary according to reactor type and design. There is little concern about core
temperature due to decay heat for low power, pool-type reactors.
Rev. 0
Page 55
HT-02
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