Applied Nuclear ReactorTechnology - Energiteknik | KTH€¢ Pressure Distribution in Channels with Single-Phase Flow Henryk Anglart Nuclear Reactor Technology Division Department of

Henryk AnglartNuclear Reactor Technology Division

Department of Energy Technology, KTH

Applied Reactor Technology and Nuclear Power Safety - Lecture 7 Slide No 1

Lectures on Applied Reactor Technology and Nuclear Power Safety

Lecture No 7

Title: Thermal-Hydraulic Analysis of Single-Phase Flows in Heated Channels


Department of Energy TechnologyKTH

Spring 2005




Outline of the Lecture• Clad-Coolant Heat Transfer in Channels with Single

Phase Flows

• Coolant Enthalpy Distribution in Heated Channels

• Temperature Distribution in Fuel Elements

• Pressure Distribution in Channels with Single-Phase Flow




Clad-Coolant Heat Transfer in Channels with Single Phase Flows (1)

Wall temp Tw

Bulk liquid temp Tf

zsc

znb

supT∆

subT∆

G

Saturation temp

Single phase (non-boiling) flow region

Subcooled flow region





• In Light Water Reactors, coolant is subcooled at the inlet to the reactor core

• The subcooling is defined as the difference between the saturation temperature and the actual coolant temperature

• For example, if the inlet temperature and pressure of the water coolant are 549 K and 7 MPa, respectively, then the inlet subcooling is equal to 559 K – 549 K = 10 K, since the saturation temperature of water at 7 MPapressure is equal to 559 K





• In the single-phase region, when 0 < z < znb, the surface temperature Tw of the heated wall and the liquid bulk temperature Tf are related to each other as follows,

• where h is the heat transfer coefficient and is the temperature difference between the surface of the heated wall and the bulk liquid

hqTTT wffw ′′=∆≡−

wfT∆





• The heat transfer coefficient h is evaluated from correlations, which, in turn, are based on experimental data and are using the principles of the dimensionless analysis

• The following general relationship are employed

Nu = f (Re, Pr, …), where: Nusselt numberReynolds number

, Prandtl number

λhD

=Nu

µGD

=Re

TT

pw

p cc

=

==λµ

λµ

Pr,Prw





• Many different correlations have been proposed

• Example of a correlation for laminar flows (Re < 2000) is as follows

1.0

2

2325.043.033.0

PrPrPrRe17.0Nu

∆

⋅=

µβρ TgD

w





• For turbulent flows (Re > 2000) the following correlation, proposed by Dittus-Boelter, is used:

33.08.0 PrRe023.0Nu ⋅=




Coolant Enthalpy Distribution in Heated Channels (1)

• Assume a heated channel as shown in the figure to the right. The channel is uniformly heated along its length with heat flux q’’[W/m2], it has a flow cross-section area A and heated perimeter Ph.

• The energy balance for a portion of channel in green is as follows:

z

z+dz

W, Hci

Hc(z)+dHc

z

q’’Hc(z)

[ ]ccHc dHzHWdzzPzqzHW +⋅=⋅⋅′′+⋅ )()()()(

WzPzq

dzzdH Hc )()()( ⋅′′

=





• Thus, the enthalpy distribution of coolant is described by the following differential equation:

• Integration yieldsz

z+dz

W, Hci

Hc(z)+dHc

z

q’’Hc(z)

WzPzq

dzzdH Hc )()()( ⋅′′

=

dzzPzqW

HzHz

Hcic ⋅⋅′′+= ∫0 )()(1)(





• For small temperature and pressure changes the enthalpy can be expressed as a linear function of temperature:

dH = cp*dT

• Using W = G A and assuming constant channel area and heat flux distribution, the coolant temperature can be found as,

GAczPqTzT

p

Hfif

′′+=)(





• The temperature is thus linearly distributed between the inlet and the outlet of the assembly

• The outlet temperature becomes

z=L

z=0Tfi

Tfo

q’’

GAcLPqTT

p

Hfifo

′′+=

TWi, Tfi





q’’ z=L

z=0

• Usually the axial power distribution is not uniform. In a bare reactor the axial power distribution is given by the cosine function – or with z = 0 at the inlet – sinus function:

The differential equation for the enthalpy (temperature) distribution is now

⋅′′=′′

Lzqzq πsin)( 0

TWi, Tfi

⋅⋅′′

=

⋅′′

=Lz

cWzPq

dzzdT

orLz

WzPq

dzzdH

p

HfHc ππ sin)()(,sin)()( 00





• After integration, the coolant enthalpy (temperature) distribution is as follows

The outlet enthalpy (temperature) can be found as:

fip

Hf

ciH

c

TLz

cWLPqzT

orHLz

WLPqzH

+

−⋅

⋅⋅⋅⋅′′

=

+

−⋅

⋅⋅⋅′′

=

ππ

ππ

cos1)(

,cos1)(

0

0

fip

Hffoci

Hcco T

cWLPqLTTorH

WLPqLHH +

⋅⋅⋅⋅′′

==+⋅

⋅⋅′′==

ππ00 2)(,2)(

q’’ z=L

Tz=0

Tf(z)

Tfo

Wi, Tfi




Temperature Distribution in Fuel Elements (1)

• The stationary heat conduction equation for an infinite cylindrical fuel pin is as follows

integration yields

Coolant

Clad

Gap

Fuel

rF

qdrdTr

drd

r F ′′′=− λ1

qrdrdTrF ′′′−=

2

2

λ





• For constant fuel conductivity the equation can be readily integrated as follows

• However, the fuel conductivity is a function of a temperature, and the integration has to be performed as follows:

( ) CqrrTF

+′′′−=λ4

2

qrrdrqdTdrqrdT F FT

T

r FFF ′′′−=

′′′−=⇒′′′−= ∫ ∫

0 0

2

422λλ





• Introducing the average fuel conductivity given as:

• The total temperature drop in the the fuel can be found as:

• Or, using the so-called linear power density

∫−= 0

0

1 T

T FF

FF

dTTT

λλ

F

FFF

rqTTT

λ4

2

0

′′′=−≡∆

qrq F ′′′≡′ 2π

FF

qTλπ4′

=∆





• The temperature drop across the gap can be found assuming that the conductivity is the major heat transfer mechanisms:

• The integration constant C1 is found as:

• And the temperature drop in the gap is found as:

21

1 ln01 CrC

TCdrdTr

drdTr

drd

r GGG +=⇒=⇒=− λλ

ππλ

22 11 qC

rq

rC

drdT

FFrrG

F

′−=⇒

′=−=−

=

F

G

GGFG r

rqrTrTT ln2

)()(λπ′

=−=∆

λ





• Similar derivation for the clad region yields the following temperature drop:

here rc and λc are the outer clad radius and clad material conductivity, respectively

• Heat convection from the clad surface to the coolant is given as

G

C

CCGC r

rqrTrTT ln2

)()(λπ′

=−=∆

( )fC TThq −=′′





• Since

temperature drop in the coolant thermal boundary layer is as follows:

• The total temperature drop is thus

( )Crqq π2′=′′

hrqTTT

CfCf π2

′=−≡∆

+++

′=∆+∆+∆+∆=∆

hrrr

rrqTTTTT

CG

C

CF

G

GFfCGF

1ln1ln12

12 λλλπ





• For non-uniform (”cosine”) heat flux distribution

Substituting the above to

yields

⋅′′=′′

Lzqzq πsin)( 0 fi

p

Hf T

Lz

cWLPqzT +

−⋅

⋅⋅⋅⋅′′

=π

πcos1)( 0

( )hqTTTThq fCfC′′

+=⇒−=′′

⋅

′′++

−⋅

⋅⋅⋅⋅′′

=Lz

hqT

Lz

cWLPqzT fip

HC

πππ

sincos1)( 00





q’’ z=L

Tz=0

TC(z)

TCmax

• Figure to the right shows the clad temperature distribution assuming the ”cosine” axial power distribution

• It should be noted that the clad temperature gets its maximum value TCmax at a certain location zmax

Wi, Tfi





• The location of the maximum clad temperature can be found as:

• Substituting z = zmax in the equation for the clad temperature yields the maximum clad temperature

⋅⋅⋅⋅

−=

=

⋅⋅′′

+

⋅⋅′′

⇒=

−

H

p

p

HC

PLhcWLz

Lz

Lhq

Lz

cWPq

dzdT

ππ

πππ

1max

00

tan

0cossin:0

fip

HC T

Lz

cWLPqT +

−⋅

⋅⋅⋅⋅′′

= max0max sec1 π

π





• It can be shown that the maximum fuel temperature at the centerline is as follows,

fip

H

G

C

CF

G

GF

zCF

TL

zcWLPq

rr

rr

LzqrT

+

−⋅

⋅⋅⋅⋅′′

+

++

⋅′′⋅=

max0

max0max

sec1

ln1ln12

1sin

ππ

λλλπ




Pressure Distribution in Channels w/Single Phase Flow (1)

• Consider mass and momentum balance in the following channel

dFw

W U A p

W+d W U+dU A+dA p+dp

ϕ

dz

g

dFg





• The balance of forces projected on the channel axis is as follows:

– Pressure force

– Wall friction force dFw = PW dz τw

– Gravity force dFg = -ρ A dz g sinφ

– Momentum change -(W + dW)(U+ dU) + W U → -dW U - W dU

• The resultant equation is:

( )( ) AdppdAdAAdpppA −−⇒++−

ϕρτρ

sin1 2

gA

PdzdA

ApAG

dzd

Adzdp wW +++

=−





• For vertical channel with constant cross-section:

• Integration along channel of length L yields:

or

gA

Pdzdp wW ρτ

+=−

[ ] gLLA

PdzgA

PpLpdzdzdp wWL wWL

ρτρτ+=

+=−−=− ∫∫ 00

)0()(

[ ] gLLA

PpLpp wWtot ρτ

+=−−≡∆− )0()(





• In a straight channel with constant cross-section area the only irreversible pressure loss is caused by friction:

thus

LA

Pp wWfric

τ=∆−

ρτ

2

2GC fw ≡

ρ2

2GA

PCdzdp W

ffric

⋅=

−

ρ2

2GA

LPCp Wffric ⋅=∆−

where





• In a straight channel with constant cross-section area the only irreversible pressure loss is caused by friction:

thus

LA

Pp wWfric

τ=∆−

ρτ

2

2GC fw ≡

ρ2

2GA

PCdzdp W

ffric

⋅=

−

ρ2

2GA

LPCp Wffric ⋅=∆−

where





• For local obstacles, additional irreversible pressure losses appear

• Consider sudden expansion of a channel

– Mass balance

– Momentum balance”1”

”2”

”0”

2211 AUAU =

( ) 2211222111 ApApUAUUAUF −+−= ρ





• Solving mass and momentum equations for pressure difference yields

equation):

• And combining the two yields:

( )

−=−=− 1

1

22221212 A

AUUUUpp ρρ

IppUpU ∆++=+ 2221

21 2

121 ρρ

2

1

222 1

21

−=∆−

AAUpI ρ

• Applying mechanical energy balance (Bernoulli





• It is customary to express the local loss coefficient in the following form:

• That is, for the sudden channel enlargement, the local pressure loss is

ρξρξ

22

22 GUp losslossloss ==∆−

2

2

121

2

2

1 1;2

1

−=⋅

−=∆−

AAG

AAp enlI ξ

ρ





• In a similar manner, the local pressure loss for a sudden area contraction is found as:

”1”

”2””c”

vena contracta

2

2

22

2

2

1

;2

1

−=

⋅

−=∆−

ccont

I

AA

GAAp

ξ

ρ

12 AA

2AAc

( )22 1−cAA

0.00.070.210.360.450.5

1.00.7900.6860.6250.5980.586

1.00.80.60.40.20

c





• The total pressure drop in a channel with length L, constant cross-section area and local obstacles along the channel can be calculated as:

ϕρρ

ξ sin2

4 2

gLGD

LCpppp

ii

felevlocfrictot +

+=∆−∆−∆−=∆− ∑




Exercises (1)

• Exercise 18: A pipe with the inner diameter 10 mm and the outer diameter 12 mm is uniformly heated on the

-2

with water flowing inside the pipe. The inlet mass flux of water is 1200 kg m-2 s-1, the inlet enthalpy 1215.6 kJ kg-1

inlet the water will become saturated?

outer surface with heat flux q’’ = 1 MW m and cooled

and the inlet pressure 70 bar. At what distance from the




Exercises (2)

• Exercise 19: In a pipe as in Exercise 18 assume that the single-phase flow exists from the inlet to the pipe

the wall temperature on the outer surface of the pipe at the point where water becomes saturated if the wall

-1 -1

until the point where water becomes saturated. Calculate

conductivity is equal to 15 W m K .




Exercises (3)

• Exercise 20: For a fuel rod as in Example 7.3.1 (see Compendium, Chapter 7, page 7-8) assume that the fuel

T [K],[W m-1 K-1]

What will be the total temperature drop difference when the heat transfer coefficient drops from 5000 to 1300 W m-2 K-1?

4151034.0130

40 TF ⋅⋅+= −λ

heat conduction is the following function of temperature,

T+




Exercises (4)

• Exercise 21: Water flows in a horizontal pipe with a sudden expansion. Calculate the total pressure drop

diameter suddenly increases from A1 = 10 cm2 to A2 = 12 cm2. Assume flow of water in direction from smaller to

740 kg m-3.

(reversible and irreversible) in a place where the pipe

larger flow area with mass flow rate 1 kg/s and density




Exercises (5)

• Exercise 22: Solve the problem as in Exercise 21, but assume that the water flows in a direction from the larger

Explain why the pressure drops in Exercises 21 and 22 are different.

to the smaller cross-section area (sudden contraction).

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Applied Nuclear ReactorTechnology - Energiteknik | KTH€¢ Pressure Distribution in Channels with Single-Phase Flow Henryk Anglart Nuclear Reactor Technology Division Department of