Как выбрать гостиницу для кошек
14 декабря, 2021
1 Decay heat removal using PORVs
Example: Following the TMI accident, a utility was considering the possibility of increasing the number of PORVs in its 4000-MW(t) PWR system to allow release (in the form of steam) of the full decay energy at 100 s from shutdown. Assuming a flow area for each valve of 0.002 m2, how many valves would be required?
Solution: After 100 s, the decay heat rate may he estimated from Table 2.2 and is
3.2 X 4000 I 100 = 128 MW
The flow area required can be estimated assuming a release rate of 17,000 MW/rn2 (see Section 4.3.2).
128
Thus Flow area =——— = 0.0075 m2
17,000
and four PORVs would be required.
Problems: A 3000-MW(t) PWR has two PORVs, each with a flow area of 0.0015 m2.
Would these valves be sufficient to allow release of decay energy from the reactor ves
sel in the form of steam, and consequent maintenance of fuel cooling by “feed-and — bleed" operation, at 1 h from shutdown?
2 Evaporation ojcoolant
Example: Following a small-break loss-of-coolant accident, the fuel of a 3800-MW(t)
P’^TC has become uncovered and the top half of the fuel is dry. What is the rate at which the core is becoming uncovered at 1 h after shutdown, assuming a mean void fraction in the wetted region of 0.5? Also assume that the fuel occupies 40% of the core volume, that the core diameter is 3.6 m, and the core length 4 m, and that the heat flux is uniform in the core. The system pressure during the uncovery period was 85 bars. Solution: The volume of water per meter length of the core in the wetted region is given by
Cross-sectional area of corex ( 1 — void fraction)
x(l — fractional area occupied by fuel)
= | 2Lx 3.6x 3.6 X (1-0.5) x (1-0.4) = 3.054m3/m
The heat release rate to water from the submerged half of the fuel at 1 h from shutdown is given (using Table 2.2) by
1 4
x—= 2.66 X107 W 1QQ
Evaporation rate of water
heat release rate
latent heat of evaporation of water at 85 bars 266 x!07 W
1.40 x 106 J/kg
It is now necessary to iterate to ensure consistency with only half of the core being uncovered in 1 h.
Problem: For the reactor core described in the example, the heat flux would not in
practice be uniformly distributed. Rather the flux profile along the core length follows a law that would typically be of the following form:
1t(z + a)
L + 2a
where q is the local heat flux, if. av the average heat flux, z is the distance from the bottom of the core, L is the core length, and a is a constant. F is a form factor (ratio of peak to average heat flux). Assuming F = 1.4 and a = 0.3, calculate the total time required to totally uncover the core described in the example. Assume a constant heat input equivalent to that occurring 1 h after shutdown, that the core is initially just filled with a steam mixture water with 50% void fraction, and that the void fraction remains constant during the uncovery. Also, plot the movement with time of the mixture level.
3 Fuel blockage in a fast reactor
Example: Calculate the location and magnitude of the peak dad temperature in the peak rated channel of a fast reactor under normal flow conditions. Would a blockage leading to a 50% reduction in flow lead to the fuel elements exceeding the creep limit of 670°C, above which ballooning of the cans would occur? In the calculations, assume a 3300-MW(t) reactor having hexagonal fuel assemblies, which each have 325 fuel pins 5.84 ^m in diameter with the distance across the faces of the hexagon being 135 ^m. The normal mass rate of flow through each subassembly Щ) is 39kg/s, and the core length is I m. Liquid sodium enters the core region at 370°C. In the core region the peak fuel rating in the highest-rated fuel assembly is 44 kW/m’ and (for the purposes of this present calculation*"’), assume that the local rating r is given by
..
r = rmax sin — = 44 sin — max L L
where z is the distance from the beginning of the core and L is the core length. Assume a heat transfer coefficient x between the fuel and the sodium of 55,000 W/m2 K at the full flow conditions and 32,000 W/m2 K at 50% flow. Assume that the sodium has a specific heat capacity (c) of 1275 J/kg K.
Solution: The total heat generation rate (QJ in this assembly is:
• The difference between this value and the value of 27 k W/m given in Table 2.3 is that the figure in the table was an average rating including those parts of the fuel in the blanket and outer core regions.
••Note: The equation for flux profile implies that the flux goes to zero at the bottom and top of the core. This simplifies the calculation, but the actual profile would go to a finite rating at the extremities of the
ГПГР ЯПГІ indpprl rhprp i<.: finitP opnprsrinn nf hp-:.it in thp hlon’L-pr rpoinn -.1 ant] hplnu. r thp rnrp
The temperature rise Д across the element under full flow conditions is thus
and for this condition
325. 1
—— sin — =—— cos —
Mcp L Da L
= tan"1 (-0.4763)
= 0.8585 m
At z = 0.8585 m, the maximum pin surface temperature for normal flow conditions is given by
= 370 + 1-cos
Mep1t ^ L J
rxL
+ max sin —
7tDa L
= 370 + 325 x 44,°0° X 1 a + 0.9028)
39 x 1275x
______ 44, 000 X 1_____
X 5.84 X 10-3 x 5.5 x 10^
= 370 + 174.2 + 18.7 = 562.9
Thus, the peak clad temperature is normally well below the creep limit of 670°C. For a f1ow reduction of 50%, the peak clad temperature occurs at a distance z from the inlet given by
The clad temperature at this position is given by
325rZ
rL
+ —^ sin 1tDa L
325 x 44,000 X 1
39x 1275x
= 370 + 352.6 + 28.4 = 751.0°C
Thus, a flow blockage leading to a 50% reduction in flow would lead to the peak clad temperature in excess of the creep limit of 670°C and would be unacceptable. Problem: If, for the fast reactor described in the example above, the flow reduction due to blockage was even greater than 50%, boiling of the sodium would ultimately begin at the fin surfaces. The boiling point of sodium would be required to initiate boiling; calculate the flow reduction that would be required to cause boiling to start on the peak-rated fuel assembly. Also calculate the position on the fuel assembly at which such boiling would be initiated.