Nuclear Thermal Hydraulic

System Experiment

NUCLEAR AND HYDROGEN SYSTEM LABORATORY, KAIST

2015.05.28

Fukushima accident• Earth quake Reactor shut down

• Electricity supply from grid is failed

• Emergency diesel generator is also failed• Submerged by Tsunami

• Reactor Coolant Pump is turned off

• Thermal energy couldn’t be removed• Most of safety systems were de-

pendent on electricity

• Fuel melt / Hydrogen explosion / heat up of spent fuel pool

Nuclear Safety SystemsSystems that required to remove the decay heat under the postulated accident

ex ) Safety Injection system (Inject water to the primary loop during accient)

After Fukushima, following requirement were added to the safety systems System should remove the decay heat even without electricity We call this system as “Passive safety system”

Usually gravity is used for the driving force of passive safety system

Nuclear Power Plant accidentFu

el ro

ds

ThermalEnergyFrom

Fuel rod

(from atomic decay)

“Decay heat”

A number of systems(Thermal hydraulic systems)

ElectricEnergy

Safety system

Heat should be removed

Passive Containment Cooling System

Passive system that removes decay heat by cooling the steam/air in the containment

Decay heat

Decay heat (Steam)

How can we measure and calculate the performance of this system Experiment for PCCS

PCCS Test Purpose

◦ To calculate heat removal capacity and heat transfer coefficient of the condensing tube

◦ Check the effect of the Air/Steam Fraction ◦ Check the effect of the Pressure

PCCS Test Facility Main test section

◦ Chambers : Heat exchangers ◦ Steam Generator◦ PCCT and Cooling water pipes

SG

Chamber : Air/Steam mixture

PCCT

Experiments Procedure 1. Set up test condition

◦ Steam pressure of the steam generator◦ Angle of the condensing tube (Already Fixed)◦ Temperature of cooling water (inlet)◦ Mass flow rate of cooling water

2. Open the valve of SG connected to the chamber

3. Measure or set the properties of system◦ System pressure and temperature◦ Outlet temperature of cooling water

4. Calculate heat capacity and heat transfer coefficient

5.Open the venting valve on the chamber for 1 min and close it◦ Air/Mass Fraction of the System changed ( for next Exp. Case.)

6.Repeat previous steps

Experiments Procedure Steam generator

◦ 2 ,3 and 4 bar saturated condition (one pressure condition for one group)

Cooling water◦ Pressure: 1 bar(atmosphere)

What you measure? Or check? ◦ Temperature of cooling water in / out ◦ Mass flow rate◦ Twall

Channel geometry information◦ Outer diameter of pipe = 49.1 mm◦ Inner diameter of pipe = 40 mm◦ Length of pipe = 0.7 m

Heat transfer rate and heat transfer coefficient

** Mass flow rate,

Heat transfer rate [W]Q

minT

moutT

Single-phase fluid

( ) ( )out in p out inQ m h h mC T T

Heat transfer rate : Amount of heat transferred per time

m V uA

3

3

2

: Density of fluid [kg/m ]

: Volumetric flow rate [m /s]

u : velocity of fluid [m/s]

A : Cross-sectional area of flow channel [m ]

V

Heat transfer rate and heat transfer coefficient

Heat transfer coefficient is affected by

: Geometry / Angle / Materials / Air-mass fraction / Pressure ….

minT

moutT

2 [W/m .K]( )sat wall

Qh

A T T

Heat transfer coefficient : Amount of heat transferred per time, per area, and per temperature difference

Tbulk

Twall

Heat transfer rate [W]Q

Heat transfer rate and heat transfer coefficient

Heat transfer rate [W]Q

minT

moutT

Single-phase fluid

Heat transfer rate : Amount of heat transferred per time

31/ 4( )

0.728[ ]( )

L L G LG L

L sat wall

gi kh

D T T

Notes Your Report Should includes

◦ Clear Conditions for each case◦ Air / Steam Fraction of each experiments (Through the calculation)◦ Heat Transfer Coefficient of PCCS for each case and the reason for the difference between each case.