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Problem 1 (20 points) : List the release points of gaseous and liquid radioactive materials from a typical PWR-type nuclear power plant in effluents.

Solution

1) Gaseous effluents (10 points)

□ Containment purge or ventilation exhausts

□ PAB vent exhausts

□ Gaseous waste processing system discharge (controllable)

□ Steam leakage from the secondary systems

□ Condenser air-ejector off-gas

□ Steam generator blowdown system vent

□ Radwaste and fuel building vent

2) Liquid effluents (10 points) (all effluents are controllable)

□ BRS discharge (for H-3 control and water balance)

□ Dirty and miscellaneous wastes (containment sump, PAB floor drains, etc.)

□ S/G blowdown treatment system discharge

□ Chemical wastes (laboratory, sampling) small volume

□ Condensate demineralizer regeneration system discharges

□ Turbine building floor drain sump

□ Detergent wastes (decontamination); based on experiences

□ Clean waste (equipment and RCDT drains)

※ Acronyms

□ PAB : Primary Auxiliary Building

□ BRS : Boron Recycle System

□ RCDT : Reactor Coolant Drain Tank

Problem 2 (20 points) : What are the possible methods of treating or processing gaseous and liquid radioactive wastes in a nuclear power plant.

Solution

1) Gaseous waste treatment systems (10 points)

□ Decay tanks for holdup and decay : Decay tanks could be used for the purpose of holdup and

decay of radioactive gases before their discharges to the atmosphere.

□ Charcoal delay systems for delayed discharge : Charcoal delay systems are used for the purpose of

holdup and decay of radioactive gases. Charcoal adsorbers are used for delay of their releases.

□ Cover gas recycle to VCT(volume control tank) for recycle back to the primary coolant

□ Cryogenic distillation for volume reduction using low temperature separation : Inlet gases are

liquefied under liquid-N2 temperature. And the liquid is boiled off using distillation process.

According to the different boiling point of each gas, it will be separated via distillation process.

2) Liquid waste treatment methods (10 points)

□ Demineralization (ion exchange)

□ Evaporation

□ Reverse osmosis

□ Filtration

Problem 3 (20 points) : Lead, iron, concrete, water are common materials for gamma radiation shielding. Explain the characteristics of each material and its engineering constraints when it is used as the gamma-shield material.

Solution

1) The characteristics (10 points)

Material	Lead	Iron	Concrete	Heavy Concrete	Water
Density [g/cm3]	11.35	7.87	2.2∼2.4	3.7∼4.8	1
Yield Strength [psi]	860	25,000	-	-	-
Tensile Strength [psi]	1,900	40,000	< 500		-
Max. Temp. [℃]	327 (melting)	1,535 (melting)	93 (dehydration)		100 (boiling)
Radiation Damage	×	small	negligible		small

2) Engineering constraints (10 points)

□ Lead : shrinks about 4% in volume on solidification

□ Iron : strong inelastic & prompt capture □'s on neutron

□ Concrete : careful curing required (7∼28 days)

□ Water : Container required

Problem 4 (20 points) : The average primary coolant leakage into the auxiliary building is 80 kg/day.

1)  If there is no ventilation in the building,

find the equilibrium concentration of Xe-131m (Ci/m3) in the building air.

2)  Find the required ventilation rate

to keep the concentration of Xe-131m under DAC (2×107 Bq/m3).

3)  Find the annual release rate of Xe-131m to the environment via this ventilation exhaust.

Data :

□ Volume of the auxiliary building = 500 m3

□ Xe-131m concentration in the primary coolant during power operation = 23 □Ci/g

□ Half-life of Xe-131m = 11.8 days

Solution

1) (8 points)

The auxiliary building encloses the several systems which handle the primary coolant. Typical systems are CVCS (chemical and volume control system) and BRS (boron recycle system).

The rate of change in airborne activity of nuclide i in the building is given by:

[eqn: dP_i over dt `=`l_pp `C_i `-`lambda_i `P_i `-`R_v `P_i]

where, [eqn: P_i] = activity of nuclide i in the PAB, [□Ci]

[eqn: l_pp] = primary leak rate into the auxiliary building [eqn: =` {(80kg)(1000g/kg)} over 24hr/day `=`3333`g/hr]

[eqn: C_i] = activity of nuclide i in the reactor coolant

[eqn: =` (23 mu Ci/g)(3.7times10^4 Bq/mu Ci) `=`8.51times10^5 Bq/g]

[eqn: lambda_i `=`ln2 over {(11.8days)(24hrs/day)} `=`2.45times10^-3 /hr]

[eqn: R_v ~left(`= F_v over V_PAB `right)] = vent exhaust rate of the auxiliary building, [vol/hr]

[eqn: P_v] = vent exhaust rate of the auxiliary building, [m3/hr]

[eqn: V_PAB] = volume of auxiliary building = 500 m3

In an equilibrium state, the airborne concentration of radioactivity in the auxiliary building atmosphere is : [eqn: P_i `=`{l_pp `C_i} over {(lambda_i + R_v )V_PAB}]

If there is no ventilation, the concentration becomes :

[eqn: P_i `=`{l_pp `C_i} over {lambda_i `V_PAB} `=` {(3333g/hr)(8.51times10^5 Bq/g)} over {(2.45times10^-3 /hr)(500m^3 )} `=`bold{2.32times10^9 Bq/m^3}]

2) (7 points)

To keep the concentration of Xe-131m under DAC, the required ventilation rate becomes :

[eqn: P_i `=`{l_pp `C_i} over {(lambda_i + R_v )V_PAB} `□`DAC]

[eqn: R_v `V_PAB ~□& {l_pp `C_i} over {DAC} `-`lambda_i `V_PAB ## &=`` {(3333g/hr)(8.51times10^5 Bq/g)} over {2times10^7 Bq/m^3} `-`(2.45times10^-3 /hr)(500m^3 ) ## &=``1.41times10^2 `m^3 /hr]

[eqn: bold{therefore~R_v `V_PAB ~□~1.41times10^2 `m^3 /hr `=`3.37times10^3 `m^3 /day}]

3) (5 points)

All the leaked Xe-131m will be eventually vented out to the environment. Hence, the annual release of Xe-131m becomes :

[eqn: R_i ~=&l_pp `C_i `T ## =&(3333g/hr)(8.51times10^5 Bq/g)(365days/yr)(24hrs/day) ## =&bold{2.49times10^13 ``Bq/year `=`6.72times10^2 ``Ci/year}]

Problem 5 (20 points) : The mean concentration of the effluent at a point (x, y, z) (Ci/m3) based upon the "continuous Gaussian plume model" for atmospheric dispersion is given by :

[eqn: bold{barchi ``(x,y,z) `=`Q {1 over {2pi sigma_y `sigma_z `bar u}}exp left[{{- 1 over2}left({y^2 over sigma_y^2}+{z^2 over sigma_z^2}right)}right]}]

where, Q = continuous and constant point release of radioactivity

with no change in downwind direction, [Ci/sec]

[eqn: bold{bar u}]= downwind velocity, [m/sec]

[eqn: bold{sigma_i ``(i`=`y,z)}] = plume dispersion parameters in y & z - directions.

Show that the atmospheric dispersion factor that should be used for calculating the maximum concentration at the ground from a ground level release becomes as follows :

[eqn: bold{{bar chi over Q}(x) `=`{1 over {pi `sigma_y (x) `sigma_z (x) `bar u}}}]

Solution

[eqn: barchi ``(x,y,z) `=`Q {1 over {2pi sigma_y `sigma_z `bar u}}exp left[{{- 1 over2}left({y^2 over sigma_y^2}+{z^2 over sigma_z^2}right)}right]]

It is a common practice to present the concentration as "atmospheric dispersion factor" and often called "chi-by-cue" or "chi-over-cue" which is defined as sec/m3 as follows :

[eqn: {barchi} over Q (x,y,z)`=`{1 over {2pi sigma_y `sigma_z `bar u}}exp left[{{- 1 over2}left({y^2 over sigma_y^2}+{z^2 over sigma_z^2}right)}right]]

For an elevated continuous release of effluent at height h ; If the release point is h meter high from the ground, we can assume an imaginary source located at z = -h symmetrically reflected by the ground to account for reflected plume. In this case, the chi-by-cue becomes

□ With real source at z = h ; [eqn: {barchi} over Q (x,y,z)`=`{1 over {2pi sigma_y `sigma_z `bar u}}exp left[{{- 1 over2}left({y^2 over sigma_y^2}+{(z-h)^2 over sigma_z^2}right)}right]]

□ With imaginary source at z = -h ; [eqn: {barchi} over Q (x,y,z)`=`{1 over {2pi sigma_y `sigma_z `bar u}}exp left[{{- 1 over2}left({y^2 over sigma_y^2}+{(z+h)^2 over sigma_z^2}right)}right]]

Summing both equations,

[eqn: {barchi} over Q (x,y,z)`=`{1 over {2pi sigma_y `sigma_z `bar u}} exp left[-{1 over 2}left({y^2 over sigma_y^2}right)right] left{ exp left[{{- 1 over2}left({(z-h)^2 over sigma_z^2}right)}right] + exp left[{{- 1 over2}left({(z+h)^2 over sigma_z^2}right)}right] right}]

Therefore, for the centerline concentration (at y = z = 0) ; [eqn: bold{{barchi} over Q (x)`=`{1 over {pi sigma_y `(x) `sigma_z `(x) `bar u}}}]