Question

In: Chemistry

a) Estimate the decay heat load from a 3000 MWt reactor at one day, one week,...

a) Estimate the decay heat load from a 3000 MWt reactor at one day, one week, one month, and one year after shutdown from a 2-year run. b) Estimate the rate of heat up in a reactor core that has lost all of its coolant. Consider the decay heat loads found above. Assume the heat capacitance of the dry reactor core is 16,500 Btu/°F.

Expert Solution

a. E(t) = 7.6*10^-3* P[(t)^0.8 - ( t+T)^0.8 + (T)^0.8 ] where

E(t) - Energy after period of t days in MWt.day

P - reactor power in MWt,

T - Period for which the reactor has run, in days

After 1 day:

E(1) = 7.6X10^-3X3000 [ 1^0.8- (1+2X365 )^0.8 + (2X365)^0.8] = 22.8 [ 0.79] = 18.01 MWt.day = 18.01 X 10⁶ X 24 X 60 X 60 Joules = 1556064 x10⁶ Joules = 1474.86 X 10⁶ Btu.

After 1 week ( 7 days ) :

E(7) = 22.8 [ 7^0.8 - ( 7+ 2x365 )^0.8 + ( 2X365 ) ^0.8 ] = 74.1 MWt.day = 74.1 X 10⁶ X 24 X 60 X 60 Joules = 6402240 X 10 ⁶ Joules = 6068.13 X 10⁶ Btu.

After 1 month ( 30 days ) :

E(30) = 22.8 [ 30^0.8 - ( 30 + 2X 365 ) ^0.8 + ( 2 X 365 )^0.8] = 200.64 MWt day = 16430.62 X 10⁶ Btu

After 1 year ( 365 days ):

E (365 ) = 22.8 [ 365^0.8 - ( 365 + 2X365 )^0.8 + ( 2X365 ) ^0.8 ] = 851.35 MWt day = 69718.13 X 10⁶ Btu

b. The rate of heat of up

After 1 day: decay heat / heat capacitance = 1474.86 X 10⁶ / 16500 = 89.3 X 10³ ^oF

After 1 week: 6068.13 X 10⁶ / 16500 = 367.7 X10^3
oF

After 1 month: 16430.62 X 10⁶/ 16500 = 995.8 X 10^{3 o}F

After 1 year: 69718.13 X 10⁶/16500 = 4.22 X 10^{6 o}F

queen_honey_blossom answered 1 year ago

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