1 LOCA in a PWR

Example: A major loss-of-coolant accident occurs in a P^^. The reactor is tripped and goes subcritical after I s; dryout (see Section 3.3) occurs after 4 s when the heat trans­fer coefficient from the fuel pin drops from 50,000 W/m2K to a very low value; the blowdown is complete in 30 s. The fuel pins consist of 10-mm U0 fuel pellets in an 11-mm outside diameter Zircaloy can. The maximum rating R is 40 kW/m. The coolant is initially at 300°C; the temperature drops through the can and across the fuel-pellet — to-can gap are initially 50 K and 300 K, respectively. What is the can temperature at the end of the blowdown phase?

After 30 s the ECCS system operates and provides water at 30°C. The heat transfer coefficient during the refill stage is 50 W/m2 K. What is the maximum temperature recorded during the refill stage, and when is it recorded?

Solution. There are four sources of energy that could make the can temperature rise.

(1) Delay in shutting down the reactor. We will assume that from the start of the transient to the time when the reactor goes subcritical at I s, normal cooling is provided and this component makes no contribution to can temperature rise.

(2) Internal energy stored in thefuel.

(a)

Fuel pellet temperatures equalize at a time t, given by s d/к, where s is a shape factor (equal to 0.2 in this case), K is the thermal diffusivity (= k/pc), and a the fuel pellet radius. For U02, k = 2.5 W/m K, (! = 10,000 kg/m3, and cp = 350 J/kg K. Therefore, к =0.7 x 10~—Dm2/s, and for a 5-mm fuel pellet temperatures equalize in

(b) The thermal capacity of the fuel per unit length (C) is given by Cf = 1ta2Qcp

= x25x 10-6 X 10- x 350 = 275 J/mK

The thermal capacity of the can (C) is given by

Cc = (b2 — a2 )e, cc

where b is the outer can radius, Qc is the can density (6500 kg/m3), and cc is the can material specific heat (350 J/kg K). Thus

Cc = X (0.001 Є -0.00102) X 6500 X 350 = 350J/kg K

(c) The energy stored in the fuel pin above the can surface temperature is made up of two parts: that in the fuel pellet and that in the can.

(3) Stored energy is thefuel pellet. is the product of the thermal capacity of the fuel

per unit meter and the mean temperature of the fuel pellet above the can surface temperature (T) Since the time (30 s) is long compared with the time for equaliza­tion across the fuel pellet, the fully equalized pellet temperature can be used and is

(= 40,000 J = 637 K

81tk I 8x7tx2.5 J

To this must be added the temperature drop across the can and across the pellet/can gap, 50 K + 300 K = 350 K.

Therefore, stored energy in the pellet above Ts = C, (637 + 350) = 2.71 X 105 J.

(4) Stored energy in the can. This is the product of the thermal capacity of the can per unit meter and the mean temperature of the can above the can surface temperature 77

We take the mean temperature as half the temperature rise across the can (= 50/2) = 25 K. Therefore, the stored energy in the can above 7 = Cc x 25 = 937 J (which can be neglected in relation to the stored energy of the fuel pellets).

(5) Residual fission heating—as the neutron chain reaction dies away. This can be taken as

I.6x R = 1.6 x 40,000 = 64,000 J

(6) Fission product decay heating. Integrating the fission product decay heating for 29 s, we have

1.5 X R = 1. 5 X 40, 000 = 60, 000 J

(7) But some heat is removed by good cooling between I and 4 s. This is estimated at 120,000 J. If we now add the various energy contributions, we have

(1) + (3) + (5) + (6)-(7)

0 + 271 kJ + 64 kJ + 60 kJ -120 kJ = 275 kJ/m

The thermal capacity of the fuel (pellet and can) per meter length is

275 + 37.5 = 312.5 J/m K

Therefore, the temperature rise is 880 K.
The initial can temperature is

So at the end of the blowdown phase the final can temperature is 880 K + 323.1°C = 1203.1°С. However, the cbolant temperature does not remain at 300°C, and during the blowdown the saturation temperature falls to that for I atm (~100°C). So the actual can temperature will be significantly less than 1203.1°C (by as much as 200°C, i. e., around 1000°C) (see also Figure 4.1 for an approximate solution to this problem).

At the end of the blowdown phase, the decay heating is 0.04 R or 1600 W/m. The heat removed by the ECCS is

X 0Ш1 X 50 X (1000-30) = 1676 W/m

So the heat removal is about the same as or slightly higher than the heat released, and the can temperature will be at a maximum of 1000°C at about 30 s and will start to fall as the decay heat falls.

Problem: Repeat the calculations described in the example, but assume that dryout of the fuel occurs after 1 s (simultaneous with the reactor trip). [2]

delay in shutting down the reactor is

35,000 J/ms x 4=140,000 J/m

(2) Internal energy stored in the fuel.

(a) The metal fuel temperatures equalize at a time t given by sa2/R where Ris the thermal diffusivity (=kQc), s a shape factor (= 0.2), and a the free radius. For uranium metal k = 32 W/m K, (! = 19,000 kg/m3, and cp = 170 J/kg K. There­fore, k = 9.8 x 10-6m2/s, and for a= 15 mm, metal fuel temperatures equalize in

0.2 X 225 X 10-6

t =——————— = 4.5 s

9.8 X 10-6

(b) The thermal capacity of the fuel per meter length Cj. is given by ndQcp. = x 225 x 10-6 x 1.9 x 104 x 170 = 2283 J/mK.

For the clad we are not told the mass or the dimensions of the can. From reference sources we established that the mass of clad per unit length is 1 kg/m and that cc is 1200 J/kg K. Therefore, the thermal capacity of the can=

1 X 1200=1200 J/m K

(c) The energy stored in the fuel pin above the clad surface temperature is made of two parts: that in the metal fuel and that in the cladding (which we neglect).

Tbe stored energy in the metal fuel is the product of the thermal capacity of the fuel per unit meter and the mean temperature of the metal fuel above the clad sur­face temperature. Since the time (50 — 5) s is long compared with the time for equalization across the fuel, the fully equalized metal temperature can be used and is

R = ( 35, 000 81tK |^8x 1tx32

We can neglect any other temperature drop in the fuel element; therefore, the stored energy in the fuel above the external clad temperature is

43.5 x 2283 = 99.310 J/m

(3) Residualfission heating—as the neutron chain reaction dies away. This can be taken as

1.6 X R = 1.6 X 35, 000 = 56, 000 J/m

(4) Fission product decay heating. Integrating the fission product decay heating for (50 — 5) s, we have

2.1 X R = 2.1 X 35,000 = 73,500 J/m

If we now add the various energy contributions, we have

(1) + (2) + (3) + (4) =

140,000 + 93,310 + 56, 000 + 73,500 = 362,810 J/m

The thermal capacity of the fuel (metal and cladding) per unit meter is

2283 + 1200 = 3843 J/m K

Therefore, the temperature rise is 104.1 K.

This is the temperature rise assuming no cooling of the fuel. Let us now calculate the effect of radiation of heat to the moderator. The moderator is assumed to be at a tem­perature of 350°C; the fuel element is assumed to be a cylinder of 50 mm (fin outside diameter) in a fuel channel of 100 mm.

The average clad temperature during the transient is

104

450 + — = 502° C 2

The heat lost by radiation may be calculated from the formula

where £ and £2 are the emissivities of the fuel and graphite, respectively, and are assumed to be 0.6 in each case. cr = 5.67 x 10—8 W/m2 K4 (Stefan’s constant), and the surface areas of the fuel element (Aj) and the graphite channel (A) per meter are

A = 0.157 m2 A2 = 0.314m2

(This assumes a 5-cm-diameter can in a 10-cm-diameter channel.) Heat lost by radia­tion is thus

5.67 x 10~8 X 0.157 (7754 — 6234)
1.98

The decay heat rate at (50 — 5) s is 0.038 of full power = 0.038 x 35,000 = 1330 W/m. So the heat lost by radiation is less than the decay heat rate, and in the absence of any other cooling the fuel element will heat up at

1330-945

3483

until the heat removed by radiation matches the decay heat rate.

The total heat removed during 50 s by radiation is

945 x 50 = 47, 250 J/m

corresponding to a reduction of 13.4 K compared with the no-cooling temperature rise of 104.1 K.

Therefore, the maximum cladding temperature is

450 + 104.1 — 13.5 = 540.6°C

Problem: A new Magnox fuel design is being considered in which the fuel element di­ameter is to be increased from 5 cm to 6.5 cm and the uranium metal fuel pin diame­ter is increased to 4 em. Repeat the calculations given in the example, and evaluate the effect of this design change on the temperatures reached in the specified accident.

3 Pumps on orpumps off?

Problem: There has been considerable controversy about whether the opera­tors should leave the main circulating pumps operating or stop them during a small break loss-of-coolant accident for a PWR. With the aid of the accompanying diagram, discuss the events that occur when the pumps are stopped (path I) or left operating (path 2).

Bffi

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Barsell, A. W. (1981). A Study of the Risk Due to Accidents in Nuclear Power Plants. Ger­man Risk Study—Main Report, various pages.

Bergles, A. E., J. G. Collier., J. M. Delhaye, G. F. Hewitt, and F. Mayinger, (1981). Two — Phase Flow and Heat Transfer in the Power and Process Industries. Hemisphere, Washington, D. C., 719 pp.

Bowerman, J. M., and G. C. Dale, eds. (1982). The Safety of the AGR. CEGB and SSEB,

111 pp.

Fast Reactor Safety Technology. Proceedings of a Meeting, Seattle, August 19-23, 1979. American Nuclear Society, LaGrange Park, III., 5 vols.

Flowers, B. 0976). Nuclear Power and the Environment. Royal Conunission on Envi­ronmental Pollution, Sixth Report. HMSO, London, 237 pp.

Jones, O. C. (1981). Nuclear Reactor Safety Heat Transfer. Papers presented at the sum­mer school on nuclear reactor safety heat transfer, Dubrovnik, Yugoslavia, August 24-29, 1980. Hemisphere, Washington, D. C., 959 pp.

Judd, A. M. (1981). Fast Breeder Reactors: An Engineering Introduction. Pergamon, Elmsford, N. Y., 161 pp.

Menlo. M. (1983). Thermal-Hydraulics of Nuclear Reactors. Papers presented at the Second International Topical Meeting on Nuclear Reactor Thermal-Hydraulics, Santa Barbara, Calif., January 11-14, 1983, 1529 pp.

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