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Helium cooling of fusion reactors
NUCLEAR ENGINEERING AND DESIGN 26 (1974) 215-230. © NORTH-HOLLANDPUBLISHINGCOMPANY
HELIUM COOLING OF FUSION REACTORS G.R. HOPKINS and G. MELESE-d'HOSPITAL Gulf General Atomic Company, San Diego, California 92138, USA Received 5 July 1973
Basic equations relating temperatures, pressure drop, power generation and geometry are derived for cooling the vacuum wall and blanket regions of a fusion reactor with pressurized helium. Numerical results are presented for two cases which are applicable to use of a direct cycle pressurized helium gas turbine for power conversion in a fusion power reactor system.
1. Introduction
Various designs have been proposed for the blanket of a fusion reactor where energy is deposited as 14 MeV neutrons (about 80%) and as particle energy. As an example, [1, 2], fig. 1 shows the cross section of such a blanket surrounding a cylindrical plasma separated from the blanket by a metallic vacuum wall.
I_
Energy is deposited in the wall by radiation from the plasma and internal heat generation in the metal struc ture and lithium. Tritium used for the d e u t e r i u m tritium fusion reaction is bred in lithium. Neutrons are moderated in a graphite region and captured in the last lithium region which is surrounded by thermal insulation and a radiation shield to protect the external low temperature superconducting coils
UNIT LENGTH, L
COLD HELIUM
VACUUM WALL
\ PLASMA
Li R E G I O N ~
GRAPHI T E ~ _
IOMI.D.
L,
SHIELD, COIL, AND EXTERNAL COMPONENTS~
12 M 0.D. Fig. 1. Conceptual fusion reactor blanket with helium cooling.
216
G.R. Hopkins, G. Melese.d'Hospital, Helium cooling or fusion reactors
i---~, ,
jNb
I
Li
~
><
~,
PLASMA
//
TUBE
L i/
UNIT CELLS
~ \\
He
[
VACUUM WALL
'\~
(~ ~__
Li
r
GRAPHITE
t
•,
D
LITHIUM
Fig. 2. Model for blanket calculations. Note that unit cell diameter increases as heat deposition decreases, d = i.d. of tube, and t = wall thickness. CALCULATION Nb
MODEL
STRUCTURE
Nb
Nb i
Nb TUBE Li
RADIATION
PLASMA (VACUUM)
(Q = I00 W/CM 2)
Li BLANKET REGION
;M 3 PLASMA (VACUUM)
20
W/CM3
I I
3 REGIONS
REGIONS
Fig. 3. Model for vacuum wall calculations. (which produce the magnetic field). All the energy must therefore be extracted in the blanket region. Various coolants have been proposed for the reactor blanket: molten lithium and molten LiF-BeF2, among others. Pumping an electrical conductor such as molten lithium perpendicular to the large magnetic fields creates large pressure drops besides the possible corrosion problems and problems in design of steam generators. The use of pressurized helium coolant leads to a large reduction in corrosion, chemical and magnetic interaction problems. Because o f its very small neutron cross section, heliu- does not perturb the nuclear reactions. Pressurized helium also opens
the option for use of a direct gas turbine cycle. A large amount o f experience on helium technology is available from fission reactors: the power densities involved in the fusion reactor blanket range from densities encountered in thermal reactors ( 5 - 2 0 W/cm 3 in HTGR) to densities in fast gas-cooled reactors ( 1 0 0 - 4 0 0 W/cm 3 in GCFR). Thus blanket cooling should be feasible with helium at pressures ranging from 20 to 100 arm. Direct cycle helium gas turbines are presently under development for HTGR fission reactors [3] at temperatures and pressures comparable to those obtainable from helium cooling of fusion reactor blankets.
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
IST
~
INTERCOOL~
STARTING MO~)
~T~I
IPCOMPRESSOR\/ 2ND INTERCOOL~
217
REGENr- l ~ERATOR
.
3
HP COMPRESSOR\/ HP TURBINE/
PO RO EL~ 'EIF C
LP
I TURBINE \
L
C ELECTRC I AL
LP COMPRESSORGENERATOR /
Fig. 4. Direct cycle gas turbine with a fusion reactor. The authors have described a proposed fusion reactor blanket cooled by helium flowing in tubes parallel to the axis of the plasma and distributed across the thickness of the blanket in such a way as to lead to the same helium condition for various local rates of heat generation (see fig. 2). Helium will also be used to cool the vacuum wall by flowing through tubes welded to the wall and surrounded by stagnant lithium (fig. 3). Limitations on maximum stress and temperature of the metal and on maximum lithium temperatures will be imposed. In view of the higher rate of heat deposition in the vacuum wall region the helium inlet temperature will be lower than for the blanket region. Thus a typical flow diagram for the direct helium cycle is shown in fig. 4 where a fraction of the helium out of the high pressure compressor flows directly into the wall region, while the main helium flow is heated in the regenerator before passing through the blanket region. These two flows mix at the olatlet of the fusion reactor before going through the high pressure turbine.
wards the inside. We shall consider a certain length of blanket L with helium flowing in and out at the extremities. Since the blanket thickness is expected to be small compared to the diameter of the vacuum wall, slab geometry approximations will sometimes be used. We shall assume limitations on the maximum temperature of the pressure tube wall Tw and the maximum temperature of the lithium TLi , both occurring at the channel outlet since the heat generation is constant in the axial direction. The calculations will be based on the hot channel which is closest to the vacuum wall and then extended to the remainder of this first blanket region by proper tube pitch (or diameter) dimensioning for the same gas outlet temperature and for the same maximum wall temperature. For a given volume fraction of metal, the relative thickness of the pressure tubes t/d is related to the coolant void fraction I1, while for given maximum tube stress, the coolant pressure is itself proportional to t/d. Thus coolant pressure and void are simply related. 2.1. Temperature drop in the lithium
2. Blanket cooling We shall first study the cooling of the inner part of the blanket, consisting of stagnant lithium, by helium flowing inside metal pressure tubes (about 75% of the heat is generated there). Since heat generation is nonuniform, the cooling must also be non-uniform, with a greater density of tubes of given diameter towards the inside or with larger tubes on a constant pitch to-
Let us consider a unit blanket cell of equivalent diameter D with a coolant void fraction V and a coolant tube of inner diameter d and thickness t. We have, by definition, V = d2/D 2 . t
(1)
Let qm be the amount of heat generated per unit m length and qm the power density in the cell closest to the vacuum wall. Assuming for simplicity the same
218
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
average heat generation in lithium outside the coolant tubes and in the metal tube walls, we have the relationship r
m
qm = (7r/4) qm D2(1 - V).
(2)
The temperature drop in the lithium with a thermal conductivity KLi may be written, using the equivalent cylindrical geometry, as
or, with eq. (2): ATiLi _- /~--qml) ,,' - 2 /16KLi)(1 -- V+ lnV).
(3)
The equivalent cell diameter may be found from eq. (3):
D = 4(KLiATLi/qm)~(--1 + V -
lnV)-~ .
(4)
Using eq. (1), the coolant tube diameter reads
\l-V/
\
,,, qm
t
m = 37 W/cm 3 and eq. (5) yields the following relationship between tube diameter and coolant void fraction for this particular case: ~ V_V_]'~ / - [ 1 + (d)c m ~ 4 . 5 \ l _ V ] / \
l n g ]/ l-~]J
1 (6)
2.2. Tube wall surface temperature
( lnV A TLi - 4~-KmLi 1 + 1---f~ / , ¢
d=4
q m
-
Since the rate of heat generation is uniform along the coolant tube (of length L and diameter d), all temperatures vary linearly axially and the maximum wall temperature T w occurs at the exit. N S being the flow Stanton number, one finds the following relationship between geometry and temperatures (see ref. [4] for derivation of a similar relationship): L d
1+ (s)
Therefore the cell and tube dimensions depend only upon the coolant void fraction V for given maximum temperature drop in the lithium ATLi and power density qm" Since all maximum temperatures occur at the coolant channel outlet and the maximum lithium temperature (TLi)m and metal temperature T w are fixed, we have simply A TLi = ( TLi)m -- T w •
At operating temperatures of about 1200°C for niobium and 1 2 0 0 - 1 3 0 0 ° C for lithium, their thermal conductivity is about the same: K ~ 0.625 W/cm °C. (Niobium is only used for calculational illustration. See section 4 for materials considerations.) We shall use a maximum lithium temperature of 1275 °C, close to its boiling point at atmospheric pressure. N i o b i u m - 1% zirconium alloy has a yield strength higher than 700 kg/cm 2 (10 000 psi) at 1200 °C and therefore the maximum wall temperature will be fixed at 1200 °C. Thus we will use the value ATLi = 75 °C. An energy flux through the first wall from the plasma of 10 MW/m 2 gives a maximum power density
1
Tex Tin
(7)
4N S T w - Tex
Knowing the coolant inlet and outlet temperatures Tin and Tex and the maximum wall temperature Tw, the tube length becomes known if we use eq.(5) for the diameter: 1
1
'
-5
x
_
1+
(s)
Taking N S ~- 3 × 10 -3 (it is only a slowly varying function of the Reynolds number), and using the previous numerical values, the expression for the tube length becomes in our special case:
Lx_
(L)c m ~ 3 7 5 \ l - V ]
+ l_V//
\r w-Tex/
•
t (9)
2.3. Coolant mass f l o w and pressure drop in the tubes
The helium mass flow M in the channel is obtained from the heat balance
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors MCp(Tex - Tin) = Lqm ,
(10)
where Cp is helium specific heat (5.2 W sec/g°C). With the relationship between linear heat rate and temperature drop in the lithium,
( lnV qm = --47rKLiATLi 1 + ~ ]
,
(11)
ductivity, respectively. The value N S = 3 × 10 -3 corresponds to a Reynolds number N R ~ 10 5. Substituting t MCp from eq. (1),L/d from eq. (7) and qm from eq. (I 1) yields t
4 L qm NR - rrNpK d Tex - Tin
-4
and L from eq. (8), the mass flow becomes
-NpNs M = _~c
~ _
+
G(Tw-
U R = up- 6
1
(14)
where f is the friction factor, R the gas constant and p the coolant pressure. T v = (Tex + Tin)~2 is in °K. The ratio of pumping power to thermal power reads
_
MAp~Pin
=
P t h - M C p ( T e x - Tin)
_ -~
_ Tex
Tin p
(t6)
The Stanton number for smooth tubes may be written
[a K ~ w _ - T ~ x ) 1 + The friction factor is derived from eq. (16). We could now substitute the previous expression for N S in eqs. (8) and (13) to find more exact values. As a first approximation, we could also compute an average value o f N s and simply use it in these previous equations, since it will not vary much for given design conditions Tw, ATLi. With our numerical values, N S = 3 × 10 -3 is a good approximation and therefore f = 4.5 × 10 -3. With M from eq. (10), f from eq. (16) and L/d from eq. (7), the relative pressure drop reads
A p _ 4 8 N R Tin+Tex / qm }2 p rr2 S p2 2 [ dCp(Tex - Tin) 1 T e x - Tin ]3
X (4ATS~ww- ~-eex] " With qm derived from eq. (11) and d from eq. (5), we find
( 1)1-v
N S = aNp] NR0'2 , with a = 0.023 and the Reynolds number reads
4M N R - rr tad
Tex)
(15)
Tin is also in °K; 3, = Cp/C v (for helium). For smooth tubes the friction factor is proportional to the Stanton number; Np -- laCp/K being the Prandtl number (about-~ for helium), we find approximately that f ~ 2N S (Np)~ -~ 1.5 N S .
1+
(
~42"5X 103 (1~)~{ - (1 + lln~Vv)}-~(13) ( T - Tex ) "
AP - 2 fM2 L ( ~ - ) - 2 - 32f R T v M2 L p OavP d 7r2 p2 d 5 '
_
"
--KLi6TL K ( Tw -
3
The relative helium friction pressure drop in the tube may be written
P
lnV ~-1
(12)
T ex). . [qm) .... {
With our numerical values, one finds, with the temperature in °C (or °K),
(M)g/s
KLiATLi [ ~ex)~l+l-V]
We can now find both Reynolds number and Stanton number:
(KLiATLi)~
X
219
4MCp 7rd NpK '
where/1 and K are helium viscosity and thermal con-
A p _ - -8N 3p 2 1--~ ttt
qm KLiATLi X P2G
V
#x- T2 (T_Tex)3
(1+ ~lnV]-1
(17)
220
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
The relative pumping power is then given by eq. (15). With the pressure p in atm and the previous numerical values, eq. (17) reads, since 3' = ~ for helium, 2 A_pp~ 55.5 1 -- V Tex p
V
T2 in
(
1
l-V]
(18) Tin, T w and Tex are in °K. The pumping power fraction becomes, with p in atm,
(1+ lnVll
PP ~ 11 1 - V Tin(Tex + Tin) P--~ V p2(T w - Tex)3
1 - V,t
(19) "
D2_d 2
4V
t
1-V
d'
X1-V
d
4
(21)
V
(22)
t/d = p/2o .
Therefore the coolant pressure is related to the coolant void by the expression oX 1 - V
V "
(23)
Using eqs. (5) and (21), the tube thickness could be written t =X
" ~ KLiATLi X ,,, /
-
(
1+
10 -2
-- 1 + - -
T2 P
.
o2C
"
(24)
Li
(1 + ~ln] -I .
(26)
P
With our numerical values, this pressure drop reads, with temperatures in °K, T2 - T'2 (1 lnV ~-1 2A ~ i-_Vv ex m + l_--f-~] . P (T w Tex)3
(27)
The ratio of pumping to thermal power now becomes (see eq. (15) with 3' = -~) PP ~ - 0.4 Pth
V 1 -- V
Tin(Tex + Tin)
1 + 1 lnV - ~ ] '~-1 .(28)
(T w _ Tex)3
Using eq. (23), eq. (26) yields the pressure drop (in ttt atm), with o in atm, qm in W/cm 3, KLi in W/cm °C and Cp in W sec/g°C, as follows: tlt
Ap
--0.75
X 10 -5 ~1 1 ~
{ T2
T2
qm
With our numerical values, this thickness is related to
2
7] 1 - V (Tw_Tex)3
X2N 2 ~
XN 2
~
(25)
and
'" X qm KLiA
This relative thickness is also related to the coolant pressure and the allowable stress in the tube wall o by the relationship
P = 2
"
p = 7.5 (1 - V ) / V ,
(20)
from which the relative tube thickness becomes simply related to the coolant void fraction t
1-~VJJ
Equation (23) for p may now be substituted into eqs. ( 1 7 ) - (19) to give the relative pressure drop
The ratio X of the amount of metal used for pressure tube to the amount of lithium in the cell may be written approximately for thin tubes -
-
With, for instance, o = 250 atm (or about 35% of N b - Z r yield strength at 1200 °C), and X -~ 6 × 10 -2 (i.e. 6% metal in the blanket), eqs. (23) and (25) become respectively, with p in atm,
tcm ~ 6 . 7 5 X
2. 4. Coolant tube thickness and coolant pressure
4dt X ~ - -
tcm ~ 1.125X
lnV ~_1
1+
(Tw _ Tex)3 (Patm)2
the coolant void by the expression
(T w - Tex)3
\ 3~
j\
T
qmKLi_____~ Li
3"]
.C p \ -1
-Vl i
(29)
G.R. Hopkins, G. Melese-d'Hospital, Helium coolingor fusion reactors 2.5. Hot channel performance The thermal power of the hot channel is simply t
Pth = L qm = MCp (Tex - Tin).
(30)
221
blanket to find the total number of tubes per unit blanket length L. Recall that L depends upon the coolant void fraction (V, eq. (8)), or pressure (p, eq. (23)). Conversely, V or p could be found if length L is given.
With M from eq. (12), one finds the power
2. 6. Distribution of coolant tubes across the blanket 47r (KLiATLi)~
-- 1 +
et"-N s (qm) rex-Tin x
rw
_
re x
,
(31)
which, with our numerical values, becomes
(Pth)M w
~
0.22
Lx- rin G-rex"
X - -
-
1+
(32)
We therefore find that the geometric and physical quantities of interest may be expressed as functions of the coolant void fraction V and of the coolant inlet and exit temperatures Tin, Tex. Recall that T w is the maximum permissible wall temperature. As a numerical example for the hot channel, let us take a coolant void fraction V of 20%, and coolant temperatures Tin and Tex of 400 °C and 1000 °C respectively with a maximum wall temperature T w of 1200 °C. The coolant tube dimensions d, L and t are obtained from eqs. (6), (8) and (25), respectively, while the cell dimension D is derived from eq. (4). The coolant mass flow M is given by eq. (13), while the pressure p is derived from eq. (23), and the total channel thermal power from eq. (32). Finally, the relative pressure drop Ap/p and pumping power Pp/Pth are obtained from eqs. (27) and (28). With 6% volume fraction X of pressure tube material with a useful stress o of 250 atm, one finds d ~ 2.25 cm, L ~ 560 cm, t 0.135 cm, p = 30 atm, D ~ 5 cm, M = 107.5 g/sec, Pth ~ 336 000 W ~ 0.336 MW, Ap/p ~ 3.6 X 10 -2, and Pp/P~ ~ 1.62 × 10 -2. Recall that Ap ~ 1.08 atm is only the friction pressure drop in the coolant tube and Pp ~ 5.5 kW the corresponding pumping power fraction for this channel. The linear rating derived t from eq. (11) is qm ~ 600 W/cm. All of these values would appear to be quite acceptable. We shall now look at the other coolant tubes in the
For simplicity, we shall design all coolant tubes for the same heat load, pressure drop, mass flow and temperatures. Therefore the tube diameter remains constant but the void fraction, or the tube pitch, varies along the radius of the blanket since the power density is a function of this radius P. The linear rating q', the tube diameter d, and the local void fraction V1 are related to the local power density q'" by the relationship (see eqs. (1) and (2)) FI
n d2
1 -- V 1 - 4
(33)
q"'(P)/q'"
With assumed constant values o f d and q', we find therefore that
1
V1(p)
1=
-
1) qm/q ......
(P),
(34)
where V is the maximum void fraction (in the hot channel). The distribution of coolant void fraction Vl(P) is therefore known from the knowledge of the distribution of power density q'"(p)/qm. As expected VI(p) decreases with increasing p since q'"(p) also decreases. For the required small values o f the maximum coolant void fraction V, eq. (34) may be approximated by the expression
_{ q'"(P) qm
V
VI(P)~ 1 - V
ttw
1 -VV
1
q'"(p)} qm
(35)
ttt
Since the blanket is relatively thin with respect to its radius, slab geometry will be used throughout. If T O is the required thickness of the inner (lithium) blanket without helium coolant, the actual thickness will be T0/(1 - Vav) where Vav may be determined from eq. (35). As a first approximation, we shall take T ~ TO 1
1-
--
'
(36)
222
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
tt¢
qav being the average power density. For instance, with T O = 60 cm, qa'v/qm = 0.5 and V= 0.20, we would find T = 68 cm, with an average coolant void fraction of 1 0 - 11%. With a 10 m diameter plasma, an energy production 11 MW/m 2 of vacuum wall corresponds to 345 MW/m of reactor length. About 15% of the energy is produced in the vacuum wall, 81% in the inner (lithium) region of the blanket (or 280 MW/m) and 4% in the outer region. The total power per unit inner blanket length could also be calculated directly since it reads Q' ~ 27rR'To qa'v '
lnV ~-1 q ' = --47rKLiATLi 1 + ~ ] , the total number of tubes per length L of inner blanket is simply qa'v ( lnV ~ -1+ KLiATLi 1 -- V 1"
(38)
With our numerical values, we find a number of tubes N = 2.8 X 106/0.6 X 103 ~ 4 7 0 0 . Since the approximate number of tubes along the circumference is 2nR' 4R' , m ~- - (zrV)i D0r/4) ~ d
X \KLiATLi /
-
=
R'(n(1 - V))~
(39)
1+
the number of radial rows of tubes becomes independent of the blanket radius N m
X
Q'd 4R'q'(zrV)~
-- 1 +
lnV ~/- ~(Pb)MW ~ 1050 ~ww~ Tex \1 - V]
_
TO 2Or(l-- V))~
i(qav/qm~ i \KLiATL~i]
(40)
-- 1 (41)
(37)
where R' is the average blanket radius. With R ' = tit 500 cm, T O = 60 cm and qav = 15 W/cm 3, one finds Q' ~ 2.8 X 106 W/cm. Since the linear rating of each coolant tube has been found to be (see eq. (11))
'-R'To N- Q q, 2
Again, with our numerical values, we find n ~ 7 and m ~ 670. The average tube pitch is therefore 68/7 = 9.7 cm, while the nfinimum pitch at the inner radius is about half: D ( n / 4 ) ' = 5(7r/4)-~ ~ 4.45 cm. Finally, the total thermal power generated by a length L of inner blanket is simply Pb = L Q', with L given by eq. (8). Again with our numerical values, eq. (9) yields a power
With L = 560 cm and Q' = 2.8 MW/cm, the thermal power in the inner blanket is 1570 MW. Since about 95% of the blanket power is produced in the inner blanket, the total blanket power is about 1650 MW. The power density in the outer regions of the blanket (graphite and lithium regions) is lower than in the inner (lithium) region. The tube pitch or void fraction is determined in the same manner as for the inner blanket, assuming the same tube diameter and same coolant conditions. As a first approximation, the tube pitch varies as the inverse of the square root of the power density. With a power per unit length of about 13.8 MW/m in this outer blanket region, the number of coolant tubes is about 230. Thus the total number of tubes is about 5000. About 3% of the heat is generated in the graphite region with a maximum power density of 1.5 W/cm 3 and a graphite conductivity approximately half that of lithium. About 1% of the heat is generated in the last lithium region with an average power density of about 3 W/cm 3. Thus, there are about 170 tubes in two rows in the graphite region and only one row of 60 tubes in the outer lithium region. The maximum temperatures are 1275 °C in lithium and about 1350°C in graphite.
3. Helium cooling of the vacuum wall Let us see whether we could also cool the metal vacuum wall with flowing helium rather than with lithium. The proposed design consists of metal pressure tubes (containing helium) welded to the metal vacuum wall, with stagnant lithium surrounding the tubes. The vacuum wall is assumed to receive a heat flux of about 100 W/cm 2 from the plasma and the
223
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors internal heat generation is approximately 120 W/cm 3. Since the lithium is not flowing, at least only at a very slow rate, the corrosion problems should be greatly eased. We shall keep the same volume composition as with lithium cooling except that we now add some helium whose volume fraction is determined by the cooling requirements. For a numerical example we shall take the same limitations as before on maximum metal temperature (1200 °C) and lithium temperature (1275 °C). To simplify the calculations, we shall use an equivalent slab geometry with the same volume fractions as with cylindrical cooling tubes whose thickness is determined by the required helium pressure for a given tube diameter. Furthermore, since heat generation rates and thermal conductivities are similar in niobium and in lithium, at operating conditions, for simplification we will assume that we only have niobium at the inner face in our slab approximation. The geometry we will consider is shown in fig. 3. The first region o f thickness x 1 has an incident flux 4) = 100 W/cm 2 on its left face and is cooled by helium on its right face with a heat transfer coefficient h; the internal heat generation is H = 120 W/cm 3 and the thermal conductivity is K 1 = 0.625 W/cm°C. The temperature in region 1 is T l ( x ). On the other side of the helium cooling region, o f thickness e with a bulk coolant temperature TO, we find the second solid region of thickness x 2 consisting of metal with the parameters H 2 = H, K 2 = K 1 = K and T2(x ) also cooled by helium with a heat transfer coefficient h. The last region of thickness x 3 consists o f lithium with heat generation H ' ( ~ 70 W/cm3), thermal conductivity K 3 = 0.625 W/cm °C = K and the temperature T3(x ). The outer boundary of this lithium region will be assumed to be adiabatic. The regular blanket cooling tubes considered in the previous section will be distributed on the other side of this boundary; their pitch could even be closer than determined before, if need be, if the internal heat generation is higher than the 37 W/cm 3 used previously for the hot channel. For instance, eq. (34) gives the local void fraction; using V = 20% as before for the blanket hot channel, we would find a required void fraction of about 28% for a heat generation rate of say 58 W/cm 3 at the outer boundary o f this third (lithium) region (since h e decreases nearly linearly).
3.1. Temperature distribution The temperature distribution in the first region is
given by the following relationship, x = 0 being at the plasma boundary,
rl(x)_ro=y
l_x2)+~_
1
x + (42)
The maximum value of T 1 is obviously at x = O:
() + Hx 1 (T1)m - TO = - ~ - + ~ 1 + - hwhich could also be written, introducing the ratio of heat fluxes @ = d?/Hx 1 and the Biot number B=hxl/K,
as (T1) m - T o = ~ - -
+ qj +
.
(43)
Similarly, the temperature distributions in regions 2 and 3 (niobium and lithium, respectively) may be written, with x = 0 being this time at the helium boundary, as
Hx H ' x x3 Hx2 + H'x3 T2(x)-To =~ (2x2-x) + g + h ' (44)
T3(x)- T o
,,X ll+X3. :--K- (2
( +x3
x 2 -H + n x 2
x2
H'(x - x2) +
~-
(2x 3 + x 2 - x ) .
(45)
The maximum temperatures in these two regions are obtained at x = x 2 and x = x 2 + x 3 respectively:
x3
(r2)max - - - = - K - ~2+~2
(l+X hx2
x2 (46)
(T3)ma x - (T2)raax = H'x~/2K.
(47)
Since the maximum metal wall temperature is given, and the coolant temperature increases linearly with distance along the cooling tube, the difference t between maximum inner wall temperature T w and
224
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling or fusion reactors
coolant exit temperature reads T' w
T' ex
=
K
and dimensions t', d ' and 14] 1 . Let us now recall that the length-to-diameter ratio of the tube is related to the coolant temperatures by eq. (7). Since we want to keep the same channel length, eq. (7) becomes
1 +~+
y t
from which the required Biot number (or dimensionless helium heat transfer coefficient) is derived as a function of wall and gas temperature:
{
B=~=(I+~O)
w
K
ex Hx~
1 2
/
-1
~
(48)
The total heat flux per unit surface area generated in regions 1,2 and 3 is simply (49)
Q" --- (a + H x 1 + H x 2 + H ' x 3 .
L
1
d'
_
y r
ex
in
,
(54)
4Ns T" - T' w
ex
where T w is now the outer wall temperature. This temperature corresponds to the inside of the pressure tube. Since we find from eq. (42) T"w
T'ex = Tl(Xl) - TO = (dp+HXl)/h'
(55)
the coolant tube diameter becomes d ' = 4 N s L ( ~ +HXl)h(T'ex
(56)
T~n) .
Substituting for h from eq. (48) and for L from eq. (7) (for the blanket region) 3.2. Coolant tube geometry
If the coolant tube inner diameter is d ' , its thickness t', if the thickness of the inner wall is W1 and if we assume that the coolant tubes are touching, we find an equivalent thickness of the first region
L-
d
Tex - Tin
4 N s T w - Tex '
we finally find a coolant tube diameter
XlO+.Xl( T m ~
X1
(1
7r~ d '
412w1+
~ t ' (1
1
d'=d K Tw-Tex
t')
.
(51)
while the thickness of the third (lithium) region will be determined by the limitation on the maximum lithium temperature• With a temperature drop ,
2
2
X (1 + ~)_ 1 T ex Tin T'ex - T'in
(57)
The ratio of tube diameters in the wall and blanket regions becomes ~W--
Tox - Yin
this thickness becomes
x - - .
(52)
The unit cell width being d' + 2t', the linear heat rate per coolant tube is Q~ = ( d ' + 2 t ' ) Q" = d'(1 + 2 t ' / d ' ) X (¢~ + H x 1 + H x 2 + H ' x 3 )
Hx~
)
1
ex
-~ = -(~+~)K(T _Tex)+Tw_T~
(AT)L i = n x 3 / 2 K ,
x 3 = { 2 K (A T ) I J H ' } ~ .
ex
(50)
Similarly the thickness of the second region is simply x 2 / W 1 ~ (zrt'/2W1) (1 - t ' / d ' ) ,
m
(53)
x 1, x 2 and x 3 from eqs. (50) - (52) could be substituted into eq. (53) to obtain the linear heat rate as a function of the power densities H, H', heat flux
(58)
T'ex - T'in x I may now be substituted from eq. (50) into eq. (58) to yield an equation for d ' . The tube thickness is related to the coolant pressure p ' by a relationship similar to eq. (22): t'/d' = p'/2 o .
(59)
Therefore the tube dimensions t ,~ d ~ and heat load Q] (eq. (53)) are known• The mass flow per tube M ' is determined by the relationship
G.R. Hopkins, G. Melese-d'Hospital,Helium coolingof fusion reactors
rCp(
-
=
or
La'i '
(60) rt
with the linear rating from eq. (53)
Q'i - (d' -
_
r t
ex in - - - 1 +
+ 2 t ' ) Q"
(61)
p'
RT~v M'2L
7r2
p,2
T-Tin in-
(62)
d,5
1
Y x-Y
The relative pressure drop is still given by eq. (14) which now reads A'p _ 48Ns
225
Finally the ratio of pressure drops in the two regions may be derived from eq. (62): in
3. 3. Determination of coolant temperature and mass flow Let us now find the required coolant temperatures in order to keep the same coolant tube dimensions and helium pressure in the wall region and in the blanket region. In this case d ' = d, t ' = t and the dimensions x 1 and x 2 are known (eqs. (50) and (51)) if w 1 is known. Similarly, the linear heat flux Q~ is known, eq. (53). The difference between maximum inner (vacuum) wall temperature and coolant exit temperature could be written
Xl [
(63)
(q~ + H x 1 ) ,
r 'ex- r '
in
rw- x
(64)
Combining these two equations yields
ox
ZXl: 1
+24']+*~-*i~'T-rex T'
/¢x1!
+H {Wl+(1-;)d+zrt ( l - d ) } ? ,
-'
).
r e x - tin t w
(65) If we assume the same maximum wall temperature, T w = T w and a given helium inlet temperature for the wall region (T.~), eq. (65) gives the corresponding gas outlet temperature (Tex) in the wall region: .=
in
rw-rin
2~b
2K(rw-Tin)
+
,
(69)
Recall that eq. (11) gives n
1+ - -
B
If Q~ is the total power generated per unit length of blanket and y is the fraction generated in the vacuum wall region, the total helium mass flow in the wall region is M'
' =y L Qt/Cp
T'e x - -
T~)
(70)
Similarly, the total mass flow in the remainder of the blanket is M = (1
- y ) LQ't/Cp(Tex - Tin).
(71)
Therefore, the mixed mean outlet temperature will be
y Tdx/(Fex- Tin) + ( l - y ) (rex)mix -
ex rex-tin
(68)
with
_tlm = -- 4rrKATLi
h Tex-rin q~+Hx I
r w' - r '
with (Tex - T~)/(Tex - Tin) given by eq. (66), and Q~ derived from eqs. (50) - (53):
,
with h given by eqs. (7) and (56): 1
(67)
xI=WI+ ( 1 - 4 ) d + z r t ( l - d ) 2
HXl~ 1
V w - rex = - ~ [¢ + - - ~ - ) +
(66)
fi 11J '
Y/(
T t
ex
m T t
Tex/(rex - Tin)
in) ÷ (1 - y ) / ( T e x - Tin )
which could also be written
G.R. Hopkins, G. Melese-d'Hospital, Hel&m cooling or fusion reactors
226
thickness of the first region becomes x 1 = 0.5 + 0.24 + 0.20 = 0.94 cm (see eq. (69)),
(1 -y)Tex + y T~x(Tex- Tin)/(Tex - Tin) (Tex)mix =
(1
- y ) + Y(Tex
T;n)
-
while we find from eq. (51)x 2 ~ 0.20 cm. Similarly with H ' = 70 W/cm 3, eq.(52) yields x 3 ~ 1.16 cm. The linear heat rate in the vacuum wall region is then
(72) with Tex given by eq. (66). Hence the mixed mean outlet temperature
G(Tex)mix = Tex - y
Q~=2.52(100+120×
r i n - a ' ( r e x - rin) ,
1+a'(1-y)
with H = 120 W/cm 3, ¢ = 100 W/cm 2 and H ' = 70 W/cm 3. t o With an inlet coolant temperature Tin = 100 C, the exit temperature is derived from eq. (66) with T w = 1200 °C, Tex = 1000 °C and Tin = 400 °C:
(73)
with the parameter
a-
Tw
Tin
( T'ex
+2¢
-r' 2K
-
M+M'
(ly,xT /l { 1+--
= l+lyY(l+a)
T"m)/(Tex - Tin) ~ 1.081, hence T'ex ~ 748 °C.
The ratio of pressure drops given by eq. (67) becomes
'
The ratio of mass flows is
1.14+70× 1.16)=800W/cm,
,}1 .
y
(75) 3. 4. Numerical example for vacuum wall cooling As a numerical example, let us compute the coolant temperatures and the pressure drop for the following conditions:
A ' p _ (800 1 ]2 1394 Ap 1,600 1.681! X 1 9 ~ 1 " 0 8 " For this particular case, the pressure drop in the tubes of the vacuum wall is only 8% higher than the pressure drop in the blanket tubes. The ratio of mass flows given by eq. (75) shows that 14% of the total flow goes through the vacuum wall. With 15% of the heat generated in the vacuum wall region, the mixed mean outlet temperature now becomes (Tex)mix ~ 0.14 × 748.5 + 0.86 × 1000 = 965 °C ~ 1240 °K ,
Tw=T'W =1200°C Tin = 400 °C, Tex = 1000 ° C , T'in = 100 °C,(TLi)m = 1275 °C. This means that helium entering the cooling tubes in the wall region at 100 °C bypasses the regenerator of the direct helium gas turbine cycle, while, in the remainder of the blanket, the inlet temperature is 400°C (see fig. 4). Only ~ 15% of the heat is in tile vacuum wall region and therefore the net cycle efficiency will not be much affected by the lower exit temperature, and possibly a different pressure drop in the vacuum wall region. We keep the same coolant tube diameter and helium pressure (30 atm) as for the blanket in our previous numerical example. With W 1 = 0.5 cm, d = 2.25 cm and t = 0.135 cm, the equivalent
which is only 35 °C lower than the helium temperature at the exit of the blanket.
3.5. Thermal stresses in the tubes We have derived in Appendix A a simple approximation for the maximum tensile thermal stress in the coolant tube which occurs at the inner surface
aEq'" td 1 - V °th ~ 8 K ( 1 - v ) V Using eq. (21) for t/d, eq. (5) for d, and since we have found KLi = KNb, we f'md the following approximation for the thermal stress: °th
-
E a A TLi X 2(l--v) ( 1 - 1 ) ( l + l nl_-~ff V ]~-1 .
(77)
G.R. Hopkins, G. Melese.d'Hospital, Helium cooling or fusion reactors Using the physical values for niobium E ~ 14 X 106 psi, a ~ 7 × 10 - 6 °C -1 , v ~ 0.4 and the numerical values ATLi = 75 °C and X = 0.06, the thermal stress reads
-
1--~T!
(78)
This stress varies from 7 5 0 - 1 7 5 0 psi when the void fraction decreases from 27.3 to 14%, or when the coolant pressure increases from 20 to 50 atm. Since we have used a mechanical stress of 3500 psi, the maximum tensile stress (thermal plus mechanical) varies therefore from about 4250 to 5250 psi. We have also shown in the Appendix how this total stress could be minimized by proper choice of tube thickness and coolant pressure for given coolant void fraction.
4. Numerical results The numerical examples given in the text consider a high temperature outlet helium system, 1000 °C, with a high power output from the plasma, i.e. 10 MW/cm 2 of neutron energy flux incident on the vacuum wall. There are considered upper limits at the present time both for the plasma power output and the gas turbine inlet temperature as discussed in ref. [3]. The resultant high temperatures and thermal stresses impose severe conditions on the structural materials and while niobium was used as a typical material, m o l y b d e n u m is suggested on the basis of its high temperature strength as a more likely candidate for such a high temperature metallic system. Many other factors must be considered in the materials selection but these are beyond the scope of this report. Numerical results for a helium-cooled fusion reactor blanket with a reduced helium outlet temperature of 810 °C (compatible with present HTGR technology) have also been computed. A helium inlet temperature of 510 °C was taken together with a pressure of 50 atrrL The neutron energy wall loading was also reduced to 5 MW/m 2 and the maximum metal temperature limited to 1000 °C. The pertinent results are presented in table 1. The parameters such as void ( ~ 9%) and metal fraction ( ~ 4%) in the blanket required for the addition of helium cooling appear reasonable. The required
227
pumping power for the helium flow in the blanket tubes is quite small and the net plant efficiency should be at least 36% (with regeneration). The vacuum wall cooling conditions provide a helium exit temperature of only 572 °C which would produce a mixed mean outlet temperature for the blanket and vacuum wall of 780 °C. A separate design of the vacuum wall to given an outlet temperature equal to that of the blanket appears possible and this consideration along with variation of the other parameters needs to be done for better system trade-off studies.
5. Conclusions Pressurized helium cooling appears to have a large potential for application to proposed fusion power reactor systems. The system parameters all appear within reasonable design requirements. The utilization of direct cycle gas turbines with fusion reactors seems to offer excellent prospects for m i n i m u m adverse environmental effects from waste heat rejection. Much of the helium technology presently under development for gas-cooled fission reactors should be available for fusion reactors once the scientific and technical problems related to fusion have been solved. Table 1. Helium-cooled fusion reactor design parameters for 5 MW/m2 neutron energy flux on vacuum wall. Plasma radius (R') Blanket section length (L) Helium pressure (p) Helium inlet temperature - blanket (Tin) Helium exit temperature - blanket (Tex) Helium inlet temperature - vacuum wall (T~n) Helium exit temperature - vacuum wall (Tex) Helium void fraction - blanket average (Vav) Tube metal fraction - blanket average Tube diameter (d) Tube wall thickness (t) Number of tubes in a blanket section (N) Thermal stress (maximum in vacuum wall) Mechanical stress in tube wall Thermal stress in blanket tubes Thermal power of a blanket section Helium pumping power (% of thermal power)
5m 4m 50 atm 510 °C 810 ° C 175 ~C 572 °C 8.8% 3.8% 3.0 cm 0.3 cm 2400 3200 psi 3625 psi 2300 psi 650 MW 0.4%
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling or fusion reactors
228
Appendix: Thermal stresses in the coolant tubes qex' Since there is some internal heat generation q'" in the metallic coolant tubes, the thermal stresses due to this heat have to be added to those due to the external i heat flux qex" For thin tubes cooled at the inside surface, the maximum stress is the tangential (or axial) tensile stress at the inner surface. If A T is the temperature drop in the tube due to the external heat flux, the corresponding maximum stress [5] is
EaAT
-1
°ex = 2 ( 1 - v ) ln(1 +2t/d) X {1
ln(l+2t/d)}
2
(i)
1 - ( 1 + 2t/d) -2 where E = Youngs' modulus, a = linear thermal expansion coefficient, v = Poisson's ratio, t = tube wall thickness, and d = tube diameter. The corresponding temperature drop in the tube reads
AT = (q'ex/27rK) In (l + 2t/d) ,
(ii)
where K is the metal thermal conductivity. Therefore this stress becomes aqex °ex - 4K(I - v)
ln(1 + 2t/d) - 1 - ( 1 + 2t/d) -2 '
4rrq ,,,d2 ~1 - V {1
4 -V dt (1 + t/d )} (vii) rr 1 V
The thermal stress given by eq. (iv) becomes °ex
Ea ,,, 1 V 8 K ( 1 - v ) q t d - ~ v (1 - t/3d)
The maximum tensile stress due to internal heat generation in the tube (of inside diameter d and thickness t) may be written
°in-
oLEq'" .(d
4K(1-v)\2
+
t) 2 { 1 2 s 2 + l n s 2 } - + 1 , (ix) 1- s 2
with the ratio of tube radii s = (1 + 2t/d) -1. For thin tubes (t/d ~ 1), eq. (ix) becomes
aE q'" t 2 °in ~ 3 K ( 1 - p ) (1 + t/d).
(x)
(The first expression on the right-hand side corresponds to a plate.) The total thermal stress is now derived from eqs. (viii) and (x):
(iii)
aEq'"td
° t h ~ 8K(1-u)
{1 V ( ~-
1-
3~)
4 t (2t) - - . d 1+ TJ
which for thin tubes could be written t
3 d
°ex'~2~'K(1-v) d
1-~
1+
.
(xi)
. With t/d ~ 1, we find approximately
The total linear heat flux q' is the sum of the external t heat flux qex and of the internal heat flux generated in the tube: i
q' = qex + zrq'" t(d + t), or !
i
fir
qex =q - zrq td(l + t/d).
(v)
If we assume for simplicity the same rate of heat generation in lithium and metal, we also have (see eqs. (i) and (ii))
q' ~ (7r/4) q " d2(1 - V)/V, and eq. (v) reads
(vi)
aEq"'td
°th ~ 8K(1 - v)
l-g
V
{l+d(1.4
g
1- V -
1 )} (xii)
The coefficient of t/cl in the correction term on the right-hand side of eq. (xii) varies from - 0 . 1 to +0.2 when the coolant void fraction goes from 0.14 to 0.273. Therefore, for thin coolant tubes, the maximum tensile thermal stress has the following simple approximate expression:
%
aEq'" td 1 - V
ffg(1-v)
v
(xiii)
With eq. (21) for t/d and eq.(5) for d, the thermal stress also reads
G.R. Hopkins, G. Melese-d'Hospital,Heliumcoolingof fusion reactors aEXATIA (K___~) 1/V-1 °th~
2(1--u)
(xiv) at= (2Pl_~)~+Xi~
--(1+ ll--~n~)
Since niobium and lithium have nearly the same thermal conductivity, in that case one has the value °th
_ E2 (~lx- v ) a A T L i ( 1 - 1 )
229
_i~/1V+---llv (xviii)
The optimum volume fraction of metal X 0 which minimizes the total stress is obtained when thermal and mechanical stresses are equal. Hence the value
(1+ -1lnV - S V !~-1 " (xv)
Besides Young's modulus E, Poisson's ratio v and thermal expansion coefficient a, the thermal stress only depends upon the volume fraction of metal X, the coolant volume fraction V and the temperature drop in lithium A TLi. If we use the following physical properties of niobium at high temperature ( ~ 1000°C): a ~ 7 × X 10-6/°C, E ~ 14 X 106 psi, v ~ 0.4, then eq. (xv) becomes
Oth(psi) =_81.5 XATL i ( 1 _ 1) (1+ inN ~-1 -1- S T I
' tEaATLi 1/V-1 I -~ XO= (2P l--~)~ 1 2 ( l - v ) _ ( 1 + lnW~[
(xix) '
with the corresponding minimum total stress
_(1:
-
If we now fix the total stress o 0 rather than the mechanical stress, we arrive at the following relationship between pressure and coolant void fraction derived from the previous equation:
"
With X = 6 × 10 -2 and ATLi = 75 °C, we find the expression of the thermal stress:
P=
4
EaATLi
with the corresponding tube volume fraction
With V - g, 1 the maximum thermal stress is about 1465 psi. The stress decreases from 1750 to 750 psi when the maximum void fraction increases from 14 to 27.3%. Recall the pressure decreases from 50 to 20 atm and that the mechanical stress on the tube is fixed at 3500 psi. The sum of mechanical and thermal tensile stresses thus varies from 5250 to 4250 psi when the void fraction approximately doubles from 14 to 27.3%. l e t us notice that, if the tube material (niobium) remains in the elastic range, the total stress could be written Ot
X0 = - ( 1
-u) EotATL i 1 -- V
+--
.
~xxii)
With the numerical values we have used previously, and with o0 = 5000 psi, we find the relationship between pressure and coolant void fraction as follows: p(psi) ~ - 500 (1 + lln-~Vv),
(xxiii)
with the optium volume fraction of tube material X0~-0-4
~
in]g" ~ I+I-V
(xxiv)
Recall that, since o m = 00[2, we find also from eq. (xvii) that
= Oth + O"m ,
with the mechanical (or pressure) stress
Om _ p d _ 2p
2t
4 V p. AO a 0 1 - V v
V
X1-V"
(xvii)
Using eq. (xv) for the thermal stress, the total stress reads
(xxv)
With X = 20%, we find p = 500 psi = 34 atm and X 0 =p/o 0 = 0.10. The thickness ratio of the tube is simply
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
230
NS ,
t_p_ d
o0
1-v 4
°0 EaATLi
(
lnV] 1 + 1--~/
p,p '
(xxvi)
Ap, A'p
In our case, we find t/d = 0.10. Recall that o m = 2500 psi. Those values are to be compared with t/d ~ 0.06 and p ~ 30 atm with 20% coolant void fraction when the mechanical stress is 3500 psi and the volume fraction o f n i o b i u m is 6%. Since eq. (xviii) may be written
P q' q'" Q~ Q7 R R' t, t'
°t _ 1
T
o0
X +X-
,
(xxvii)
2
Ti, T;
rw we see that the total stress is only increased by 25% if the a m o u n t o f tube material is half its o p t i m u m value ( X / X 0 - -~ 1 ). These previous results are also valid for m o l y b d e n u m tubes rather than n i o b i u m tubes since the ratio a E / K ( 1 - u) is very close for those two materials. But the ultimate strength o f m o l y b d e n u m is m u c h higher than that o f n i o b i u m and therefore higher coolant pressures are permissible, with correspondingly higher volume fraction o f tube material (see eqs. (xxi) and (xxii)).
AT i V
I4@ X i X y a 3' u p Pav o
= Stanton number = pressure o f helium coolant = coolant friction pressure drop = thermal power = linear tube heat flux = p o w e r density = linear heat flux = heat flux per unit area = gas constant = average blanket radius = thickness o f tube wall = thickness o f blanket = absolute t e m p e r a t u r e = = = = = = =
m a x i m u m wall temperature temperature difference coolant void fraction thicknesses, v a c u u m wall region ratio, metal-to-lithium volumes p o w e r fraction in vacuum wall linear expansion coefficient
= Cp/Cv = Poisson's ratio = radius --- average density = stress = radiation flux --- ratio o f heat fluxes on v a c u u m wall.
Nomenclature a, a
P
B
% Cv d, d' D E f h H,H'
Ki L m M,M' n
N
Np NR
= constants = Biot n u m b e r --- specific heat at constant pressure = specific heat at constant volume = diameter of coolant tube --- diameter of unit cell --- Young's modulus = friction factor = heat transfer coefficient = power density in vacuum wall regions = thermal conductivity = length o f blanket section = number o f coolant tubes in a circumference = mass flow = number of radial rows o f coolant tubes = number o f coolant tubes in a blanket section = Prandtl n u m b e r = Reynolds number
References [ 1 ] G.R. Hopkins and G. Melese-d'Hospital, Direct Helium Cooling Cycle for a Fusion Reactor, Proceedings of the Conference on Nuclear Fusion Reactors, Culham, UK, Sept. 17-19, 1969, British Nuclear Energy Society, London, 522 - 535. [2] G. Melse-d'Hospital and G.R. Hopkins, Gas Cooling for Fusion Reactor Blankets, ENERGY 70, Intersociety Energy Conversion Engineering Conference, Las Vegas, USA, Sept. 22-25, 1970, Vol. 1, American Nuclear Society, 1-65 to 1-71. [3] F.R. Bell and S.L. Koutz, American Society of Mechanical Engineers Winter Annual Meeting, New York, Nov. 26-30, 1972, ASME paper 72-WA/NE-8. i4] G. Melese-d'Hospital, Influence of the choice of coolant gas on reactor thermal performance. Nucl. Appl. 2 (1966) 205-212. [5] A.M. Freudenthal and C. Bonilla (Eds.), in Nuclear Engineering, Ch. 11, McGraw-Hill, New York (1957).
HELIUM COOLING OF FUSION REACTORS G.R. HOPKINS and G. MELESE-d'HOSPITAL Gulf General Atomic Company, San Diego, California 92138, USA Received 5 July 1973
Basic equations relating temperatures, pressure drop, power generation and geometry are derived for cooling the vacuum wall and blanket regions of a fusion reactor with pressurized helium. Numerical results are presented for two cases which are applicable to use of a direct cycle pressurized helium gas turbine for power conversion in a fusion power reactor system.
1. Introduction
Various designs have been proposed for the blanket of a fusion reactor where energy is deposited as 14 MeV neutrons (about 80%) and as particle energy. As an example, [1, 2], fig. 1 shows the cross section of such a blanket surrounding a cylindrical plasma separated from the blanket by a metallic vacuum wall.
I_
Energy is deposited in the wall by radiation from the plasma and internal heat generation in the metal struc ture and lithium. Tritium used for the d e u t e r i u m tritium fusion reaction is bred in lithium. Neutrons are moderated in a graphite region and captured in the last lithium region which is surrounded by thermal insulation and a radiation shield to protect the external low temperature superconducting coils
UNIT LENGTH, L
COLD HELIUM
VACUUM WALL
\ PLASMA
Li R E G I O N ~
GRAPHI T E ~ _
IOMI.D.
L,
SHIELD, COIL, AND EXTERNAL COMPONENTS~
12 M 0.D. Fig. 1. Conceptual fusion reactor blanket with helium cooling.
216
G.R. Hopkins, G. Melese.d'Hospital, Helium cooling or fusion reactors
i---~, ,
jNb
I
Li
~
><
~,
PLASMA
//
TUBE
L i/
UNIT CELLS
~ \\
He
[
VACUUM WALL
'\~
(~ ~__
Li
r
GRAPHITE
t
•,
D
LITHIUM
Fig. 2. Model for blanket calculations. Note that unit cell diameter increases as heat deposition decreases, d = i.d. of tube, and t = wall thickness. CALCULATION Nb
MODEL
STRUCTURE
Nb
Nb i
Nb TUBE Li
RADIATION
PLASMA (VACUUM)
(Q = I00 W/CM 2)
Li BLANKET REGION
;M 3 PLASMA (VACUUM)
20
W/CM3
I I
3 REGIONS
REGIONS
Fig. 3. Model for vacuum wall calculations. (which produce the magnetic field). All the energy must therefore be extracted in the blanket region. Various coolants have been proposed for the reactor blanket: molten lithium and molten LiF-BeF2, among others. Pumping an electrical conductor such as molten lithium perpendicular to the large magnetic fields creates large pressure drops besides the possible corrosion problems and problems in design of steam generators. The use of pressurized helium coolant leads to a large reduction in corrosion, chemical and magnetic interaction problems. Because o f its very small neutron cross section, heliu- does not perturb the nuclear reactions. Pressurized helium also opens
the option for use of a direct gas turbine cycle. A large amount o f experience on helium technology is available from fission reactors: the power densities involved in the fusion reactor blanket range from densities encountered in thermal reactors ( 5 - 2 0 W/cm 3 in HTGR) to densities in fast gas-cooled reactors ( 1 0 0 - 4 0 0 W/cm 3 in GCFR). Thus blanket cooling should be feasible with helium at pressures ranging from 20 to 100 arm. Direct cycle helium gas turbines are presently under development for HTGR fission reactors [3] at temperatures and pressures comparable to those obtainable from helium cooling of fusion reactor blankets.
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
IST
~
INTERCOOL~
STARTING MO~)
~T~I
IPCOMPRESSOR\/ 2ND INTERCOOL~
217
REGENr- l ~ERATOR
.
3
HP COMPRESSOR\/ HP TURBINE/
PO RO EL~ 'EIF C
LP
I TURBINE \
L
C ELECTRC I AL
LP COMPRESSORGENERATOR /
Fig. 4. Direct cycle gas turbine with a fusion reactor. The authors have described a proposed fusion reactor blanket cooled by helium flowing in tubes parallel to the axis of the plasma and distributed across the thickness of the blanket in such a way as to lead to the same helium condition for various local rates of heat generation (see fig. 2). Helium will also be used to cool the vacuum wall by flowing through tubes welded to the wall and surrounded by stagnant lithium (fig. 3). Limitations on maximum stress and temperature of the metal and on maximum lithium temperatures will be imposed. In view of the higher rate of heat deposition in the vacuum wall region the helium inlet temperature will be lower than for the blanket region. Thus a typical flow diagram for the direct helium cycle is shown in fig. 4 where a fraction of the helium out of the high pressure compressor flows directly into the wall region, while the main helium flow is heated in the regenerator before passing through the blanket region. These two flows mix at the olatlet of the fusion reactor before going through the high pressure turbine.
wards the inside. We shall consider a certain length of blanket L with helium flowing in and out at the extremities. Since the blanket thickness is expected to be small compared to the diameter of the vacuum wall, slab geometry approximations will sometimes be used. We shall assume limitations on the maximum temperature of the pressure tube wall Tw and the maximum temperature of the lithium TLi , both occurring at the channel outlet since the heat generation is constant in the axial direction. The calculations will be based on the hot channel which is closest to the vacuum wall and then extended to the remainder of this first blanket region by proper tube pitch (or diameter) dimensioning for the same gas outlet temperature and for the same maximum wall temperature. For a given volume fraction of metal, the relative thickness of the pressure tubes t/d is related to the coolant void fraction I1, while for given maximum tube stress, the coolant pressure is itself proportional to t/d. Thus coolant pressure and void are simply related. 2.1. Temperature drop in the lithium
2. Blanket cooling We shall first study the cooling of the inner part of the blanket, consisting of stagnant lithium, by helium flowing inside metal pressure tubes (about 75% of the heat is generated there). Since heat generation is nonuniform, the cooling must also be non-uniform, with a greater density of tubes of given diameter towards the inside or with larger tubes on a constant pitch to-
Let us consider a unit blanket cell of equivalent diameter D with a coolant void fraction V and a coolant tube of inner diameter d and thickness t. We have, by definition, V = d2/D 2 . t
(1)
Let qm be the amount of heat generated per unit m length and qm the power density in the cell closest to the vacuum wall. Assuming for simplicity the same
218
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
average heat generation in lithium outside the coolant tubes and in the metal tube walls, we have the relationship r
m
qm = (7r/4) qm D2(1 - V).
(2)
The temperature drop in the lithium with a thermal conductivity KLi may be written, using the equivalent cylindrical geometry, as
or, with eq. (2): ATiLi _- /~--qml) ,,' - 2 /16KLi)(1 -- V+ lnV).
(3)
The equivalent cell diameter may be found from eq. (3):
D = 4(KLiATLi/qm)~(--1 + V -
lnV)-~ .
(4)
Using eq. (1), the coolant tube diameter reads
\l-V/
\
,,, qm
t
m = 37 W/cm 3 and eq. (5) yields the following relationship between tube diameter and coolant void fraction for this particular case: ~ V_V_]'~ / - [ 1 + (d)c m ~ 4 . 5 \ l _ V ] / \
l n g ]/ l-~]J
1 (6)
2.2. Tube wall surface temperature
( lnV A TLi - 4~-KmLi 1 + 1---f~ / , ¢
d=4
q m
-
Since the rate of heat generation is uniform along the coolant tube (of length L and diameter d), all temperatures vary linearly axially and the maximum wall temperature T w occurs at the exit. N S being the flow Stanton number, one finds the following relationship between geometry and temperatures (see ref. [4] for derivation of a similar relationship): L d
1+ (s)
Therefore the cell and tube dimensions depend only upon the coolant void fraction V for given maximum temperature drop in the lithium ATLi and power density qm" Since all maximum temperatures occur at the coolant channel outlet and the maximum lithium temperature (TLi)m and metal temperature T w are fixed, we have simply A TLi = ( TLi)m -- T w •
At operating temperatures of about 1200°C for niobium and 1 2 0 0 - 1 3 0 0 ° C for lithium, their thermal conductivity is about the same: K ~ 0.625 W/cm °C. (Niobium is only used for calculational illustration. See section 4 for materials considerations.) We shall use a maximum lithium temperature of 1275 °C, close to its boiling point at atmospheric pressure. N i o b i u m - 1% zirconium alloy has a yield strength higher than 700 kg/cm 2 (10 000 psi) at 1200 °C and therefore the maximum wall temperature will be fixed at 1200 °C. Thus we will use the value ATLi = 75 °C. An energy flux through the first wall from the plasma of 10 MW/m 2 gives a maximum power density
1
Tex Tin
(7)
4N S T w - Tex
Knowing the coolant inlet and outlet temperatures Tin and Tex and the maximum wall temperature Tw, the tube length becomes known if we use eq.(5) for the diameter: 1
1
'
-5
x
_
1+
(s)
Taking N S ~- 3 × 10 -3 (it is only a slowly varying function of the Reynolds number), and using the previous numerical values, the expression for the tube length becomes in our special case:
Lx_
(L)c m ~ 3 7 5 \ l - V ]
+ l_V//
\r w-Tex/
•
t (9)
2.3. Coolant mass f l o w and pressure drop in the tubes
The helium mass flow M in the channel is obtained from the heat balance
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors MCp(Tex - Tin) = Lqm ,
(10)
where Cp is helium specific heat (5.2 W sec/g°C). With the relationship between linear heat rate and temperature drop in the lithium,
( lnV qm = --47rKLiATLi 1 + ~ ]
,
(11)
ductivity, respectively. The value N S = 3 × 10 -3 corresponds to a Reynolds number N R ~ 10 5. Substituting t MCp from eq. (1),L/d from eq. (7) and qm from eq. (I 1) yields t
4 L qm NR - rrNpK d Tex - Tin
-4
and L from eq. (8), the mass flow becomes
-NpNs M = _~c
~ _
+
G(Tw-
U R = up- 6
1
(14)
where f is the friction factor, R the gas constant and p the coolant pressure. T v = (Tex + Tin)~2 is in °K. The ratio of pumping power to thermal power reads
_
MAp~Pin
=
P t h - M C p ( T e x - Tin)
_ -~
_ Tex
Tin p
(t6)
The Stanton number for smooth tubes may be written
[a K ~ w _ - T ~ x ) 1 + The friction factor is derived from eq. (16). We could now substitute the previous expression for N S in eqs. (8) and (13) to find more exact values. As a first approximation, we could also compute an average value o f N s and simply use it in these previous equations, since it will not vary much for given design conditions Tw, ATLi. With our numerical values, N S = 3 × 10 -3 is a good approximation and therefore f = 4.5 × 10 -3. With M from eq. (10), f from eq. (16) and L/d from eq. (7), the relative pressure drop reads
A p _ 4 8 N R Tin+Tex / qm }2 p rr2 S p2 2 [ dCp(Tex - Tin) 1 T e x - Tin ]3
X (4ATS~ww- ~-eex] " With qm derived from eq. (11) and d from eq. (5), we find
( 1)1-v
N S = aNp] NR0'2 , with a = 0.023 and the Reynolds number reads
4M N R - rr tad
Tex)
(15)
Tin is also in °K; 3, = Cp/C v (for helium). For smooth tubes the friction factor is proportional to the Stanton number; Np -- laCp/K being the Prandtl number (about-~ for helium), we find approximately that f ~ 2N S (Np)~ -~ 1.5 N S .
1+
(
~42"5X 103 (1~)~{ - (1 + lln~Vv)}-~(13) ( T - Tex ) "
AP - 2 fM2 L ( ~ - ) - 2 - 32f R T v M2 L p OavP d 7r2 p2 d 5 '
_
"
--KLi6TL K ( Tw -
3
The relative helium friction pressure drop in the tube may be written
P
lnV ~-1
(12)
T ex). . [qm) .... {
With our numerical values, one finds, with the temperature in °C (or °K),
(M)g/s
KLiATLi [ ~ex)~l+l-V]
We can now find both Reynolds number and Stanton number:
(KLiATLi)~
X
219
4MCp 7rd NpK '
where/1 and K are helium viscosity and thermal con-
A p _ - -8N 3p 2 1--~ ttt
qm KLiATLi X P2G
V
#x- T2 (T_Tex)3
(1+ ~lnV]-1
(17)
220
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
The relative pumping power is then given by eq. (15). With the pressure p in atm and the previous numerical values, eq. (17) reads, since 3' = ~ for helium, 2 A_pp~ 55.5 1 -- V Tex p
V
T2 in
(
1
l-V]
(18) Tin, T w and Tex are in °K. The pumping power fraction becomes, with p in atm,
(1+ lnVll
PP ~ 11 1 - V Tin(Tex + Tin) P--~ V p2(T w - Tex)3
1 - V,t
(19) "
D2_d 2
4V
t
1-V
d'
X1-V
d
4
(21)
V
(22)
t/d = p/2o .
Therefore the coolant pressure is related to the coolant void by the expression oX 1 - V
V "
(23)
Using eqs. (5) and (21), the tube thickness could be written t =X
" ~ KLiATLi X ,,, /
-
(
1+
10 -2
-- 1 + - -
T2 P
.
o2C
"
(24)
Li
(1 + ~ln] -I .
(26)
P
With our numerical values, this pressure drop reads, with temperatures in °K, T2 - T'2 (1 lnV ~-1 2A ~ i-_Vv ex m + l_--f-~] . P (T w Tex)3
(27)
The ratio of pumping to thermal power now becomes (see eq. (15) with 3' = -~) PP ~ - 0.4 Pth
V 1 -- V
Tin(Tex + Tin)
1 + 1 lnV - ~ ] '~-1 .(28)
(T w _ Tex)3
Using eq. (23), eq. (26) yields the pressure drop (in ttt atm), with o in atm, qm in W/cm 3, KLi in W/cm °C and Cp in W sec/g°C, as follows: tlt
Ap
--0.75
X 10 -5 ~1 1 ~
{ T2
T2
qm
With our numerical values, this thickness is related to
2
7] 1 - V (Tw_Tex)3
X2N 2 ~
XN 2
~
(25)
and
'" X qm KLiA
This relative thickness is also related to the coolant pressure and the allowable stress in the tube wall o by the relationship
P = 2
"
p = 7.5 (1 - V ) / V ,
(20)
from which the relative tube thickness becomes simply related to the coolant void fraction t
1-~VJJ
Equation (23) for p may now be substituted into eqs. ( 1 7 ) - (19) to give the relative pressure drop
The ratio X of the amount of metal used for pressure tube to the amount of lithium in the cell may be written approximately for thin tubes -
-
With, for instance, o = 250 atm (or about 35% of N b - Z r yield strength at 1200 °C), and X -~ 6 × 10 -2 (i.e. 6% metal in the blanket), eqs. (23) and (25) become respectively, with p in atm,
tcm ~ 6 . 7 5 X
2. 4. Coolant tube thickness and coolant pressure
4dt X ~ - -
tcm ~ 1.125X
lnV ~_1
1+
(Tw _ Tex)3 (Patm)2
the coolant void by the expression
(T w - Tex)3
\ 3~
j\
T
qmKLi_____~ Li
3"]
.C p \ -1
-Vl i
(29)
G.R. Hopkins, G. Melese-d'Hospital, Helium coolingor fusion reactors 2.5. Hot channel performance The thermal power of the hot channel is simply t
Pth = L qm = MCp (Tex - Tin).
(30)
221
blanket to find the total number of tubes per unit blanket length L. Recall that L depends upon the coolant void fraction (V, eq. (8)), or pressure (p, eq. (23)). Conversely, V or p could be found if length L is given.
With M from eq. (12), one finds the power
2. 6. Distribution of coolant tubes across the blanket 47r (KLiATLi)~
-- 1 +
et"-N s (qm) rex-Tin x
rw
_
re x
,
(31)
which, with our numerical values, becomes
(Pth)M w
~
0.22
Lx- rin G-rex"
X - -
-
1+
(32)
We therefore find that the geometric and physical quantities of interest may be expressed as functions of the coolant void fraction V and of the coolant inlet and exit temperatures Tin, Tex. Recall that T w is the maximum permissible wall temperature. As a numerical example for the hot channel, let us take a coolant void fraction V of 20%, and coolant temperatures Tin and Tex of 400 °C and 1000 °C respectively with a maximum wall temperature T w of 1200 °C. The coolant tube dimensions d, L and t are obtained from eqs. (6), (8) and (25), respectively, while the cell dimension D is derived from eq. (4). The coolant mass flow M is given by eq. (13), while the pressure p is derived from eq. (23), and the total channel thermal power from eq. (32). Finally, the relative pressure drop Ap/p and pumping power Pp/Pth are obtained from eqs. (27) and (28). With 6% volume fraction X of pressure tube material with a useful stress o of 250 atm, one finds d ~ 2.25 cm, L ~ 560 cm, t 0.135 cm, p = 30 atm, D ~ 5 cm, M = 107.5 g/sec, Pth ~ 336 000 W ~ 0.336 MW, Ap/p ~ 3.6 X 10 -2, and Pp/P~ ~ 1.62 × 10 -2. Recall that Ap ~ 1.08 atm is only the friction pressure drop in the coolant tube and Pp ~ 5.5 kW the corresponding pumping power fraction for this channel. The linear rating derived t from eq. (11) is qm ~ 600 W/cm. All of these values would appear to be quite acceptable. We shall now look at the other coolant tubes in the
For simplicity, we shall design all coolant tubes for the same heat load, pressure drop, mass flow and temperatures. Therefore the tube diameter remains constant but the void fraction, or the tube pitch, varies along the radius of the blanket since the power density is a function of this radius P. The linear rating q', the tube diameter d, and the local void fraction V1 are related to the local power density q'" by the relationship (see eqs. (1) and (2)) FI
n d2
1 -- V 1 - 4
(33)
q"'(P)/q'"
With assumed constant values o f d and q', we find therefore that
1
V1(p)
1=
-
1) qm/q ......
(P),
(34)
where V is the maximum void fraction (in the hot channel). The distribution of coolant void fraction Vl(P) is therefore known from the knowledge of the distribution of power density q'"(p)/qm. As expected VI(p) decreases with increasing p since q'"(p) also decreases. For the required small values o f the maximum coolant void fraction V, eq. (34) may be approximated by the expression
_{ q'"(P) qm
V
VI(P)~ 1 - V
ttw
1 -VV
1
q'"(p)} qm
(35)
ttt
Since the blanket is relatively thin with respect to its radius, slab geometry will be used throughout. If T O is the required thickness of the inner (lithium) blanket without helium coolant, the actual thickness will be T0/(1 - Vav) where Vav may be determined from eq. (35). As a first approximation, we shall take T ~ TO 1
1-
--
'
(36)
222
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
tt¢
qav being the average power density. For instance, with T O = 60 cm, qa'v/qm = 0.5 and V= 0.20, we would find T = 68 cm, with an average coolant void fraction of 1 0 - 11%. With a 10 m diameter plasma, an energy production 11 MW/m 2 of vacuum wall corresponds to 345 MW/m of reactor length. About 15% of the energy is produced in the vacuum wall, 81% in the inner (lithium) region of the blanket (or 280 MW/m) and 4% in the outer region. The total power per unit inner blanket length could also be calculated directly since it reads Q' ~ 27rR'To qa'v '
lnV ~-1 q ' = --47rKLiATLi 1 + ~ ] , the total number of tubes per length L of inner blanket is simply qa'v ( lnV ~ -1+ KLiATLi 1 -- V 1"
(38)
With our numerical values, we find a number of tubes N = 2.8 X 106/0.6 X 103 ~ 4 7 0 0 . Since the approximate number of tubes along the circumference is 2nR' 4R' , m ~- - (zrV)i D0r/4) ~ d
X \KLiATLi /
-
=
R'(n(1 - V))~
(39)
1+
the number of radial rows of tubes becomes independent of the blanket radius N m
X
Q'd 4R'q'(zrV)~
-- 1 +
lnV ~/- ~(Pb)MW ~ 1050 ~ww~ Tex \1 - V]
_
TO 2Or(l-- V))~
i(qav/qm~ i \KLiATL~i]
(40)
-- 1 (41)
(37)
where R' is the average blanket radius. With R ' = tit 500 cm, T O = 60 cm and qav = 15 W/cm 3, one finds Q' ~ 2.8 X 106 W/cm. Since the linear rating of each coolant tube has been found to be (see eq. (11))
'-R'To N- Q q, 2
Again, with our numerical values, we find n ~ 7 and m ~ 670. The average tube pitch is therefore 68/7 = 9.7 cm, while the nfinimum pitch at the inner radius is about half: D ( n / 4 ) ' = 5(7r/4)-~ ~ 4.45 cm. Finally, the total thermal power generated by a length L of inner blanket is simply Pb = L Q', with L given by eq. (8). Again with our numerical values, eq. (9) yields a power
With L = 560 cm and Q' = 2.8 MW/cm, the thermal power in the inner blanket is 1570 MW. Since about 95% of the blanket power is produced in the inner blanket, the total blanket power is about 1650 MW. The power density in the outer regions of the blanket (graphite and lithium regions) is lower than in the inner (lithium) region. The tube pitch or void fraction is determined in the same manner as for the inner blanket, assuming the same tube diameter and same coolant conditions. As a first approximation, the tube pitch varies as the inverse of the square root of the power density. With a power per unit length of about 13.8 MW/m in this outer blanket region, the number of coolant tubes is about 230. Thus the total number of tubes is about 5000. About 3% of the heat is generated in the graphite region with a maximum power density of 1.5 W/cm 3 and a graphite conductivity approximately half that of lithium. About 1% of the heat is generated in the last lithium region with an average power density of about 3 W/cm 3. Thus, there are about 170 tubes in two rows in the graphite region and only one row of 60 tubes in the outer lithium region. The maximum temperatures are 1275 °C in lithium and about 1350°C in graphite.
3. Helium cooling of the vacuum wall Let us see whether we could also cool the metal vacuum wall with flowing helium rather than with lithium. The proposed design consists of metal pressure tubes (containing helium) welded to the metal vacuum wall, with stagnant lithium surrounding the tubes. The vacuum wall is assumed to receive a heat flux of about 100 W/cm 2 from the plasma and the
223
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors internal heat generation is approximately 120 W/cm 3. Since the lithium is not flowing, at least only at a very slow rate, the corrosion problems should be greatly eased. We shall keep the same volume composition as with lithium cooling except that we now add some helium whose volume fraction is determined by the cooling requirements. For a numerical example we shall take the same limitations as before on maximum metal temperature (1200 °C) and lithium temperature (1275 °C). To simplify the calculations, we shall use an equivalent slab geometry with the same volume fractions as with cylindrical cooling tubes whose thickness is determined by the required helium pressure for a given tube diameter. Furthermore, since heat generation rates and thermal conductivities are similar in niobium and in lithium, at operating conditions, for simplification we will assume that we only have niobium at the inner face in our slab approximation. The geometry we will consider is shown in fig. 3. The first region o f thickness x 1 has an incident flux 4) = 100 W/cm 2 on its left face and is cooled by helium on its right face with a heat transfer coefficient h; the internal heat generation is H = 120 W/cm 3 and the thermal conductivity is K 1 = 0.625 W/cm°C. The temperature in region 1 is T l ( x ). On the other side of the helium cooling region, o f thickness e with a bulk coolant temperature TO, we find the second solid region of thickness x 2 consisting of metal with the parameters H 2 = H, K 2 = K 1 = K and T2(x ) also cooled by helium with a heat transfer coefficient h. The last region of thickness x 3 consists o f lithium with heat generation H ' ( ~ 70 W/cm3), thermal conductivity K 3 = 0.625 W/cm °C = K and the temperature T3(x ). The outer boundary of this lithium region will be assumed to be adiabatic. The regular blanket cooling tubes considered in the previous section will be distributed on the other side of this boundary; their pitch could even be closer than determined before, if need be, if the internal heat generation is higher than the 37 W/cm 3 used previously for the hot channel. For instance, eq. (34) gives the local void fraction; using V = 20% as before for the blanket hot channel, we would find a required void fraction of about 28% for a heat generation rate of say 58 W/cm 3 at the outer boundary o f this third (lithium) region (since h e decreases nearly linearly).
3.1. Temperature distribution The temperature distribution in the first region is
given by the following relationship, x = 0 being at the plasma boundary,
rl(x)_ro=y
l_x2)+~_
1
x + (42)
The maximum value of T 1 is obviously at x = O:
() + Hx 1 (T1)m - TO = - ~ - + ~ 1 + - hwhich could also be written, introducing the ratio of heat fluxes @ = d?/Hx 1 and the Biot number B=hxl/K,
as (T1) m - T o = ~ - -
+ qj +
.
(43)
Similarly, the temperature distributions in regions 2 and 3 (niobium and lithium, respectively) may be written, with x = 0 being this time at the helium boundary, as
Hx H ' x x3 Hx2 + H'x3 T2(x)-To =~ (2x2-x) + g + h ' (44)
T3(x)- T o
,,X ll+X3. :--K- (2
( +x3
x 2 -H + n x 2
x2
H'(x - x2) +
~-
(2x 3 + x 2 - x ) .
(45)
The maximum temperatures in these two regions are obtained at x = x 2 and x = x 2 + x 3 respectively:
x3
(r2)max - - - = - K - ~2+~2
(l+X hx2
x2 (46)
(T3)ma x - (T2)raax = H'x~/2K.
(47)
Since the maximum metal wall temperature is given, and the coolant temperature increases linearly with distance along the cooling tube, the difference t between maximum inner wall temperature T w and
224
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling or fusion reactors
coolant exit temperature reads T' w
T' ex
=
K
and dimensions t', d ' and 14] 1 . Let us now recall that the length-to-diameter ratio of the tube is related to the coolant temperatures by eq. (7). Since we want to keep the same channel length, eq. (7) becomes
1 +~+
y t
from which the required Biot number (or dimensionless helium heat transfer coefficient) is derived as a function of wall and gas temperature:
{
B=~=(I+~O)
w
K
ex Hx~
1 2
/
-1
~
(48)
The total heat flux per unit surface area generated in regions 1,2 and 3 is simply (49)
Q" --- (a + H x 1 + H x 2 + H ' x 3 .
L
1
d'
_
y r
ex
in
,
(54)
4Ns T" - T' w
ex
where T w is now the outer wall temperature. This temperature corresponds to the inside of the pressure tube. Since we find from eq. (42) T"w
T'ex = Tl(Xl) - TO = (dp+HXl)/h'
(55)
the coolant tube diameter becomes d ' = 4 N s L ( ~ +HXl)h(T'ex
(56)
T~n) .
Substituting for h from eq. (48) and for L from eq. (7) (for the blanket region) 3.2. Coolant tube geometry
If the coolant tube inner diameter is d ' , its thickness t', if the thickness of the inner wall is W1 and if we assume that the coolant tubes are touching, we find an equivalent thickness of the first region
L-
d
Tex - Tin
4 N s T w - Tex '
we finally find a coolant tube diameter
XlO+.Xl( T m ~
X1
(1
7r~ d '
412w1+
~ t ' (1
1
d'=d K Tw-Tex
t')
.
(51)
while the thickness of the third (lithium) region will be determined by the limitation on the maximum lithium temperature• With a temperature drop ,
2
2
X (1 + ~)_ 1 T ex Tin T'ex - T'in
(57)
The ratio of tube diameters in the wall and blanket regions becomes ~W--
Tox - Yin
this thickness becomes
x - - .
(52)
The unit cell width being d' + 2t', the linear heat rate per coolant tube is Q~ = ( d ' + 2 t ' ) Q" = d'(1 + 2 t ' / d ' ) X (¢~ + H x 1 + H x 2 + H ' x 3 )
Hx~
)
1
ex
-~ = -(~+~)K(T _Tex)+Tw_T~
(AT)L i = n x 3 / 2 K ,
x 3 = { 2 K (A T ) I J H ' } ~ .
ex
(50)
Similarly the thickness of the second region is simply x 2 / W 1 ~ (zrt'/2W1) (1 - t ' / d ' ) ,
m
(53)
x 1, x 2 and x 3 from eqs. (50) - (52) could be substituted into eq. (53) to obtain the linear heat rate as a function of the power densities H, H', heat flux
(58)
T'ex - T'in x I may now be substituted from eq. (50) into eq. (58) to yield an equation for d ' . The tube thickness is related to the coolant pressure p ' by a relationship similar to eq. (22): t'/d' = p'/2 o .
(59)
Therefore the tube dimensions t ,~ d ~ and heat load Q] (eq. (53)) are known• The mass flow per tube M ' is determined by the relationship
G.R. Hopkins, G. Melese-d'Hospital,Helium coolingof fusion reactors
rCp(
-
=
or
La'i '
(60) rt
with the linear rating from eq. (53)
Q'i - (d' -
_
r t
ex in - - - 1 +
+ 2 t ' ) Q"
(61)
p'
RT~v M'2L
7r2
p,2
T-Tin in-
(62)
d,5
1
Y x-Y
The relative pressure drop is still given by eq. (14) which now reads A'p _ 48Ns
225
Finally the ratio of pressure drops in the two regions may be derived from eq. (62): in
3. 3. Determination of coolant temperature and mass flow Let us now find the required coolant temperatures in order to keep the same coolant tube dimensions and helium pressure in the wall region and in the blanket region. In this case d ' = d, t ' = t and the dimensions x 1 and x 2 are known (eqs. (50) and (51)) if w 1 is known. Similarly, the linear heat flux Q~ is known, eq. (53). The difference between maximum inner (vacuum) wall temperature and coolant exit temperature could be written
Xl [
(63)
(q~ + H x 1 ) ,
r 'ex- r '
in
rw- x
(64)
Combining these two equations yields
ox
ZXl: 1
+24']+*~-*i~'T-rex T'
/¢x1!
+H {Wl+(1-;)d+zrt ( l - d ) } ? ,
-'
).
r e x - tin t w
(65) If we assume the same maximum wall temperature, T w = T w and a given helium inlet temperature for the wall region (T.~), eq. (65) gives the corresponding gas outlet temperature (Tex) in the wall region: .=
in
rw-rin
2~b
2K(rw-Tin)
+
,
(69)
Recall that eq. (11) gives n
1+ - -
B
If Q~ is the total power generated per unit length of blanket and y is the fraction generated in the vacuum wall region, the total helium mass flow in the wall region is M'
' =y L Qt/Cp
T'e x - -
T~)
(70)
Similarly, the total mass flow in the remainder of the blanket is M = (1
- y ) LQ't/Cp(Tex - Tin).
(71)
Therefore, the mixed mean outlet temperature will be
y Tdx/(Fex- Tin) + ( l - y ) (rex)mix -
ex rex-tin
(68)
with
_tlm = -- 4rrKATLi
h Tex-rin q~+Hx I
r w' - r '
with (Tex - T~)/(Tex - Tin) given by eq. (66), and Q~ derived from eqs. (50) - (53):
,
with h given by eqs. (7) and (56): 1
(67)
xI=WI+ ( 1 - 4 ) d + z r t ( l - d ) 2
HXl~ 1
V w - rex = - ~ [¢ + - - ~ - ) +
(66)
fi 11J '
Y/(
T t
ex
m T t
Tex/(rex - Tin)
in) ÷ (1 - y ) / ( T e x - Tin )
which could also be written
G.R. Hopkins, G. Melese-d'Hospital, Hel&m cooling or fusion reactors
226
thickness of the first region becomes x 1 = 0.5 + 0.24 + 0.20 = 0.94 cm (see eq. (69)),
(1 -y)Tex + y T~x(Tex- Tin)/(Tex - Tin) (Tex)mix =
(1
- y ) + Y(Tex
T;n)
-
while we find from eq. (51)x 2 ~ 0.20 cm. Similarly with H ' = 70 W/cm 3, eq.(52) yields x 3 ~ 1.16 cm. The linear heat rate in the vacuum wall region is then
(72) with Tex given by eq. (66). Hence the mixed mean outlet temperature
G(Tex)mix = Tex - y
Q~=2.52(100+120×
r i n - a ' ( r e x - rin) ,
1+a'(1-y)
with H = 120 W/cm 3, ¢ = 100 W/cm 2 and H ' = 70 W/cm 3. t o With an inlet coolant temperature Tin = 100 C, the exit temperature is derived from eq. (66) with T w = 1200 °C, Tex = 1000 °C and Tin = 400 °C:
(73)
with the parameter
a-
Tw
Tin
( T'ex
+2¢
-r' 2K
-
M+M'
(ly,xT /l { 1+--
= l+lyY(l+a)
T"m)/(Tex - Tin) ~ 1.081, hence T'ex ~ 748 °C.
The ratio of pressure drops given by eq. (67) becomes
'
The ratio of mass flows is
1.14+70× 1.16)=800W/cm,
,}1 .
y
(75) 3. 4. Numerical example for vacuum wall cooling As a numerical example, let us compute the coolant temperatures and the pressure drop for the following conditions:
A ' p _ (800 1 ]2 1394 Ap 1,600 1.681! X 1 9 ~ 1 " 0 8 " For this particular case, the pressure drop in the tubes of the vacuum wall is only 8% higher than the pressure drop in the blanket tubes. The ratio of mass flows given by eq. (75) shows that 14% of the total flow goes through the vacuum wall. With 15% of the heat generated in the vacuum wall region, the mixed mean outlet temperature now becomes (Tex)mix ~ 0.14 × 748.5 + 0.86 × 1000 = 965 °C ~ 1240 °K ,
Tw=T'W =1200°C Tin = 400 °C, Tex = 1000 ° C , T'in = 100 °C,(TLi)m = 1275 °C. This means that helium entering the cooling tubes in the wall region at 100 °C bypasses the regenerator of the direct helium gas turbine cycle, while, in the remainder of the blanket, the inlet temperature is 400°C (see fig. 4). Only ~ 15% of the heat is in tile vacuum wall region and therefore the net cycle efficiency will not be much affected by the lower exit temperature, and possibly a different pressure drop in the vacuum wall region. We keep the same coolant tube diameter and helium pressure (30 atm) as for the blanket in our previous numerical example. With W 1 = 0.5 cm, d = 2.25 cm and t = 0.135 cm, the equivalent
which is only 35 °C lower than the helium temperature at the exit of the blanket.
3.5. Thermal stresses in the tubes We have derived in Appendix A a simple approximation for the maximum tensile thermal stress in the coolant tube which occurs at the inner surface
aEq'" td 1 - V °th ~ 8 K ( 1 - v ) V Using eq. (21) for t/d, eq. (5) for d, and since we have found KLi = KNb, we f'md the following approximation for the thermal stress: °th
-
E a A TLi X 2(l--v) ( 1 - 1 ) ( l + l nl_-~ff V ]~-1 .
(77)
G.R. Hopkins, G. Melese.d'Hospital, Helium cooling or fusion reactors Using the physical values for niobium E ~ 14 X 106 psi, a ~ 7 × 10 - 6 °C -1 , v ~ 0.4 and the numerical values ATLi = 75 °C and X = 0.06, the thermal stress reads
-
1--~T!
(78)
This stress varies from 7 5 0 - 1 7 5 0 psi when the void fraction decreases from 27.3 to 14%, or when the coolant pressure increases from 20 to 50 atm. Since we have used a mechanical stress of 3500 psi, the maximum tensile stress (thermal plus mechanical) varies therefore from about 4250 to 5250 psi. We have also shown in the Appendix how this total stress could be minimized by proper choice of tube thickness and coolant pressure for given coolant void fraction.
4. Numerical results The numerical examples given in the text consider a high temperature outlet helium system, 1000 °C, with a high power output from the plasma, i.e. 10 MW/cm 2 of neutron energy flux incident on the vacuum wall. There are considered upper limits at the present time both for the plasma power output and the gas turbine inlet temperature as discussed in ref. [3]. The resultant high temperatures and thermal stresses impose severe conditions on the structural materials and while niobium was used as a typical material, m o l y b d e n u m is suggested on the basis of its high temperature strength as a more likely candidate for such a high temperature metallic system. Many other factors must be considered in the materials selection but these are beyond the scope of this report. Numerical results for a helium-cooled fusion reactor blanket with a reduced helium outlet temperature of 810 °C (compatible with present HTGR technology) have also been computed. A helium inlet temperature of 510 °C was taken together with a pressure of 50 atrrL The neutron energy wall loading was also reduced to 5 MW/m 2 and the maximum metal temperature limited to 1000 °C. The pertinent results are presented in table 1. The parameters such as void ( ~ 9%) and metal fraction ( ~ 4%) in the blanket required for the addition of helium cooling appear reasonable. The required
227
pumping power for the helium flow in the blanket tubes is quite small and the net plant efficiency should be at least 36% (with regeneration). The vacuum wall cooling conditions provide a helium exit temperature of only 572 °C which would produce a mixed mean outlet temperature for the blanket and vacuum wall of 780 °C. A separate design of the vacuum wall to given an outlet temperature equal to that of the blanket appears possible and this consideration along with variation of the other parameters needs to be done for better system trade-off studies.
5. Conclusions Pressurized helium cooling appears to have a large potential for application to proposed fusion power reactor systems. The system parameters all appear within reasonable design requirements. The utilization of direct cycle gas turbines with fusion reactors seems to offer excellent prospects for m i n i m u m adverse environmental effects from waste heat rejection. Much of the helium technology presently under development for gas-cooled fission reactors should be available for fusion reactors once the scientific and technical problems related to fusion have been solved. Table 1. Helium-cooled fusion reactor design parameters for 5 MW/m2 neutron energy flux on vacuum wall. Plasma radius (R') Blanket section length (L) Helium pressure (p) Helium inlet temperature - blanket (Tin) Helium exit temperature - blanket (Tex) Helium inlet temperature - vacuum wall (T~n) Helium exit temperature - vacuum wall (Tex) Helium void fraction - blanket average (Vav) Tube metal fraction - blanket average Tube diameter (d) Tube wall thickness (t) Number of tubes in a blanket section (N) Thermal stress (maximum in vacuum wall) Mechanical stress in tube wall Thermal stress in blanket tubes Thermal power of a blanket section Helium pumping power (% of thermal power)
5m 4m 50 atm 510 °C 810 ° C 175 ~C 572 °C 8.8% 3.8% 3.0 cm 0.3 cm 2400 3200 psi 3625 psi 2300 psi 650 MW 0.4%
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling or fusion reactors
228
Appendix: Thermal stresses in the coolant tubes qex' Since there is some internal heat generation q'" in the metallic coolant tubes, the thermal stresses due to this heat have to be added to those due to the external i heat flux qex" For thin tubes cooled at the inside surface, the maximum stress is the tangential (or axial) tensile stress at the inner surface. If A T is the temperature drop in the tube due to the external heat flux, the corresponding maximum stress [5] is
EaAT
-1
°ex = 2 ( 1 - v ) ln(1 +2t/d) X {1
ln(l+2t/d)}
2
(i)
1 - ( 1 + 2t/d) -2 where E = Youngs' modulus, a = linear thermal expansion coefficient, v = Poisson's ratio, t = tube wall thickness, and d = tube diameter. The corresponding temperature drop in the tube reads
AT = (q'ex/27rK) In (l + 2t/d) ,
(ii)
where K is the metal thermal conductivity. Therefore this stress becomes aqex °ex - 4K(I - v)
ln(1 + 2t/d) - 1 - ( 1 + 2t/d) -2 '
4rrq ,,,d2 ~1 - V {1
4 -V dt (1 + t/d )} (vii) rr 1 V
The thermal stress given by eq. (iv) becomes °ex
Ea ,,, 1 V 8 K ( 1 - v ) q t d - ~ v (1 - t/3d)
The maximum tensile stress due to internal heat generation in the tube (of inside diameter d and thickness t) may be written
°in-
oLEq'" .(d
4K(1-v)\2
+
t) 2 { 1 2 s 2 + l n s 2 } - + 1 , (ix) 1- s 2
with the ratio of tube radii s = (1 + 2t/d) -1. For thin tubes (t/d ~ 1), eq. (ix) becomes
aE q'" t 2 °in ~ 3 K ( 1 - p ) (1 + t/d).
(x)
(The first expression on the right-hand side corresponds to a plate.) The total thermal stress is now derived from eqs. (viii) and (x):
(iii)
aEq'"td
° t h ~ 8K(1-u)
{1 V ( ~-
1-
3~)
4 t (2t) - - . d 1+ TJ
which for thin tubes could be written t
3 d
°ex'~2~'K(1-v) d
1-~
1+
.
(xi)
. With t/d ~ 1, we find approximately
The total linear heat flux q' is the sum of the external t heat flux qex and of the internal heat flux generated in the tube: i
q' = qex + zrq'" t(d + t), or !
i
fir
qex =q - zrq td(l + t/d).
(v)
If we assume for simplicity the same rate of heat generation in lithium and metal, we also have (see eqs. (i) and (ii))
q' ~ (7r/4) q " d2(1 - V)/V, and eq. (v) reads
(vi)
aEq"'td
°th ~ 8K(1 - v)
l-g
V
{l+d(1.4
g
1- V -
1 )} (xii)
The coefficient of t/cl in the correction term on the right-hand side of eq. (xii) varies from - 0 . 1 to +0.2 when the coolant void fraction goes from 0.14 to 0.273. Therefore, for thin coolant tubes, the maximum tensile thermal stress has the following simple approximate expression:
%
aEq'" td 1 - V
ffg(1-v)
v
(xiii)
With eq. (21) for t/d and eq.(5) for d, the thermal stress also reads
G.R. Hopkins, G. Melese-d'Hospital,Heliumcoolingof fusion reactors aEXATIA (K___~) 1/V-1 °th~
2(1--u)
(xiv) at= (2Pl_~)~+Xi~
--(1+ ll--~n~)
Since niobium and lithium have nearly the same thermal conductivity, in that case one has the value °th
_ E2 (~lx- v ) a A T L i ( 1 - 1 )
229
_i~/1V+---llv (xviii)
The optimum volume fraction of metal X 0 which minimizes the total stress is obtained when thermal and mechanical stresses are equal. Hence the value
(1+ -1lnV - S V !~-1 " (xv)
Besides Young's modulus E, Poisson's ratio v and thermal expansion coefficient a, the thermal stress only depends upon the volume fraction of metal X, the coolant volume fraction V and the temperature drop in lithium A TLi. If we use the following physical properties of niobium at high temperature ( ~ 1000°C): a ~ 7 × X 10-6/°C, E ~ 14 X 106 psi, v ~ 0.4, then eq. (xv) becomes
Oth(psi) =_81.5 XATL i ( 1 _ 1) (1+ inN ~-1 -1- S T I
' tEaATLi 1/V-1 I -~ XO= (2P l--~)~ 1 2 ( l - v ) _ ( 1 + lnW~[
(xix) '
with the corresponding minimum total stress
_(1:
-
If we now fix the total stress o 0 rather than the mechanical stress, we arrive at the following relationship between pressure and coolant void fraction derived from the previous equation:
"
With X = 6 × 10 -2 and ATLi = 75 °C, we find the expression of the thermal stress:
P=
4
EaATLi
with the corresponding tube volume fraction
With V - g, 1 the maximum thermal stress is about 1465 psi. The stress decreases from 1750 to 750 psi when the maximum void fraction increases from 14 to 27.3%. Recall the pressure decreases from 50 to 20 atm and that the mechanical stress on the tube is fixed at 3500 psi. The sum of mechanical and thermal tensile stresses thus varies from 5250 to 4250 psi when the void fraction approximately doubles from 14 to 27.3%. l e t us notice that, if the tube material (niobium) remains in the elastic range, the total stress could be written Ot
X0 = - ( 1
-u) EotATL i 1 -- V
+--
.
~xxii)
With the numerical values we have used previously, and with o0 = 5000 psi, we find the relationship between pressure and coolant void fraction as follows: p(psi) ~ - 500 (1 + lln-~Vv),
(xxiii)
with the optium volume fraction of tube material X0~-0-4
~
in]g" ~ I+I-V
(xxiv)
Recall that, since o m = 00[2, we find also from eq. (xvii) that
= Oth + O"m ,
with the mechanical (or pressure) stress
Om _ p d _ 2p
2t
4 V p. AO a 0 1 - V v
V
X1-V"
(xvii)
Using eq. (xv) for the thermal stress, the total stress reads
(xxv)
With X = 20%, we find p = 500 psi = 34 atm and X 0 =p/o 0 = 0.10. The thickness ratio of the tube is simply
G.R. Hopkins, G. Melese-d'Hospital, Helium cooling of fusion reactors
230
NS ,
t_p_ d
o0
1-v 4
°0 EaATLi
(
lnV] 1 + 1--~/
p,p '
(xxvi)
Ap, A'p
In our case, we find t/d = 0.10. Recall that o m = 2500 psi. Those values are to be compared with t/d ~ 0.06 and p ~ 30 atm with 20% coolant void fraction when the mechanical stress is 3500 psi and the volume fraction o f n i o b i u m is 6%. Since eq. (xviii) may be written
P q' q'" Q~ Q7 R R' t, t'
°t _ 1
T
o0
X +X-
,
(xxvii)
2
Ti, T;
rw we see that the total stress is only increased by 25% if the a m o u n t o f tube material is half its o p t i m u m value ( X / X 0 - -~ 1 ). These previous results are also valid for m o l y b d e n u m tubes rather than n i o b i u m tubes since the ratio a E / K ( 1 - u) is very close for those two materials. But the ultimate strength o f m o l y b d e n u m is m u c h higher than that o f n i o b i u m and therefore higher coolant pressures are permissible, with correspondingly higher volume fraction o f tube material (see eqs. (xxi) and (xxii)).
AT i V
I4@ X i X y a 3' u p Pav o
= Stanton number = pressure o f helium coolant = coolant friction pressure drop = thermal power = linear tube heat flux = p o w e r density = linear heat flux = heat flux per unit area = gas constant = average blanket radius = thickness o f tube wall = thickness o f blanket = absolute t e m p e r a t u r e = = = = = = =
m a x i m u m wall temperature temperature difference coolant void fraction thicknesses, v a c u u m wall region ratio, metal-to-lithium volumes p o w e r fraction in vacuum wall linear expansion coefficient
= Cp/Cv = Poisson's ratio = radius --- average density = stress = radiation flux --- ratio o f heat fluxes on v a c u u m wall.
Nomenclature a, a
P
B
% Cv d, d' D E f h H,H'
Ki L m M,M' n
N
Np NR
= constants = Biot n u m b e r --- specific heat at constant pressure = specific heat at constant volume = diameter of coolant tube --- diameter of unit cell --- Young's modulus = friction factor = heat transfer coefficient = power density in vacuum wall regions = thermal conductivity = length o f blanket section = number o f coolant tubes in a circumference = mass flow = number of radial rows o f coolant tubes = number o f coolant tubes in a blanket section = Prandtl n u m b e r = Reynolds number
References [ 1 ] G.R. Hopkins and G. Melese-d'Hospital, Direct Helium Cooling Cycle for a Fusion Reactor, Proceedings of the Conference on Nuclear Fusion Reactors, Culham, UK, Sept. 17-19, 1969, British Nuclear Energy Society, London, 522 - 535. [2] G. Melse-d'Hospital and G.R. Hopkins, Gas Cooling for Fusion Reactor Blankets, ENERGY 70, Intersociety Energy Conversion Engineering Conference, Las Vegas, USA, Sept. 22-25, 1970, Vol. 1, American Nuclear Society, 1-65 to 1-71. [3] F.R. Bell and S.L. Koutz, American Society of Mechanical Engineers Winter Annual Meeting, New York, Nov. 26-30, 1972, ASME paper 72-WA/NE-8. i4] G. Melese-d'Hospital, Influence of the choice of coolant gas on reactor thermal performance. Nucl. Appl. 2 (1966) 205-212. [5] A.M. Freudenthal and C. Bonilla (Eds.), in Nuclear Engineering, Ch. 11, McGraw-Hill, New York (1957).