0% found this document useful (0 votes)
71 views3 pages

Power Profile in Finite Cylindrical Reactor: F F F F

The document discusses the power profile and temperature distribution in a finite cylindrical reactor. (1) The power per unit volume (P'') in a cylindrical reactor is related to the neutron flux profile according to equations 1 and 2. Substituting the flux profile equation into the power equation results in equation 3, which relates power to distance from the center and bottom of the cylinder. (2) For a cylindrical fuel element with uniform heat generation, the one-dimensional steady-state heat balance equation in cylindrical coordinates is shown in equation 5. Solving this equation using the boundary conditions of maximum temperature at the center and a given surface temperature results in an equation for the temperature distribution across the fuel element cylinder shown in equation

Uploaded by

mohammed sallem
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
71 views3 pages

Power Profile in Finite Cylindrical Reactor: F F F F

The document discusses the power profile and temperature distribution in a finite cylindrical reactor. (1) The power per unit volume (P'') in a cylindrical reactor is related to the neutron flux profile according to equations 1 and 2. Substituting the flux profile equation into the power equation results in equation 3, which relates power to distance from the center and bottom of the cylinder. (2) For a cylindrical fuel element with uniform heat generation, the one-dimensional steady-state heat balance equation in cylindrical coordinates is shown in equation 5. Solving this equation using the boundary conditions of maximum temperature at the center and a given surface temperature results in an equation for the temperature distribution across the fuel element cylinder shown in equation

Uploaded by

mohammed sallem
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 3

Power profile in finite cylindrical reactor

Please recall the neutron flux profile for a cylindrical reactor

(1)

Recalling the relationship between reactor power per unit volume (P’’) and neutron
flux (), we have

P’’=Ef Nf (z,r) f f (2)

Substituting Eq. (1) in Eq. (2) we get,

(3)

Equation (3) may be simplified as:

(4)
(4
where K’’ = Ef Nf f f

Equation (4) relates the heat generated as a function of distance from the centre of the
cylinder and as function of distance from the bottom. Equation (1) is similar to Eq. (4)
with K’’ of Eq. (4) replaced by 0 in Eq. (1). Hence profile of reactor power as a
function of distance from the centre will be qualitatively similar to neutron flux
profiles as shown in Figure 1 of the previous lecture.
Temperature distribution in cylindrical fuel elements
Analogous to the prediction of temperature profile for an infinite slab reactor, the
temperature profile in cylindrical fuel elements can also be obtained by the solution of
one-dimensional, steady-state energy balance equation with heat generation in
cylindrical coordinates. This is shown as Equation (5)


(5)

To begin with, let us solve Eq. (5) for the case of uniform volumetric rate of heat
generation, Pavg”

Equation (5) becomes

(6)

The temperature will be maximum at the centre of the fuel element. Hence one of the
boundary conditions is

At r = 0; dT/dr = 0 (7)

Let the temperature on the outer surface of the fuel be TS. Accordingly, the second
boundary condition is

At r = R; T = TS (8)

To solve Eq. (6) using the boundary conditions given in Eq. (7) and Eq. (8), we may
rearrange Eq. (6) as follows:

(9)

Integrating Eq. (9),

(10)

Applying the first boundary condition (r =0; dT/dr = 0) gives C1 = 0 and Eq. (10)
(11)

Integrating Eq. (11), we get

(12)

Applying the second boundary condition (r = R; T = TS), we get

(13)

Substitution of Eq. (13) in Eq. (6) gives

(14)

Re-arranging Eq. (14),

You might also like