Equation (14.111) for the minimum power of 0.0923 kW to produce 1 kg of separative work per year in uranium isotope separation was derived for cross flow on the low-pressure side of the barrier, with the composition of gas on that side y’ equal to the composition of the net flow u. The purpose of this section is to show that the minimum power requirement could be reduced further by having и greater than у by an appropriate amount and to derive an expression for the optimum difference between v and у and the corresponding power consump­tion per unit separative capacity. For this minimum-power case, pressures on the high-pressure and low-pressure sides of the barrier must be so low the only flow through the barrier is of the separating, molecular type, and the mixing efficiency on each side of the barrier is unity.

A stream containing x mole fraction light component flowing at molar rate N carries separative work at the rate

N(2x — 1) In = (Nt — N2) In ^ (14.118)

where N і and N2 are the molar flow rates of light and heavy components, respectively.

Consider the small element of barrier area dA shown in Fig. 14.9, at which the flow rates of light and heavy components are as follows:

	Molar velocity through barrier		Molar flow rates
	High-pressure side	Low-pressure side
Component	To dA	From dA	To dA	From dA
Light Heavy	<4	N2	І І 1 1	M M7	Mi + dMx Мг + dMi

Table 14.9 Design conditions and characteristics of gaseous diffusion stage designed for minimum unit cost of separative work
Barrier type	Anodized aluminum
Permeability у	15.6 X 10’5
Characteristic pressure pc	1.974 atm
Tube diameter d	0.014 m
Tube length L	4 m
Operating conditions
Barrier temperature T	358 К
High-side pressure p"	0.55 atm
Pressure ratio q = p’/p" Low-side pressure p	0.32
0.176 atm
Stage properties
Barrier efficiency Яв, Eq. (14.58)	0.57168
Mixing efficiency EM, Eq. (14.65)	0.88388
Separation factor a — 1, Eq. (14.95)	0.003004
For a separative capacity of 1 kg SWU/yr
Compressor capacity V/A, Eq. (14.113)	0.0098539 m3/s
Barrier area А /Д, Eq. (14.115)	0.81024 m2
Loss of availability Q/A, Eq. (14.117)	0.16776 kW
Contributions to capital cost, $/(kg SWU/yr)
Compressors and piping, 10,000 V/A	$ 98.54
Converters and barrier, 50 A/A	40.51
Electrical equipment and cooling system, (2)( 100) Q/A	33.55
Direct capital costs	$172.60
Indirect capital costs @50%	86.30
Total capital costs	$258.90
Unit costs, $/kg SWU
Capital charges, (0.2X258.90)	$ 51.78
Power 7[QjA) kW/(kg SWU/yr)] (8760 h/yr) (0.02 $/kWh)	58.78
Total	$110.56

High-pressure stream

Low-pressure stream

Heavy component —Gt—*~

(14.125)

In terms of the molar velocity G of both components through the barrier and the mole fraction v of light component in the net flow through the barrier,

For uranium isotope separation, є < 1 and и — x ” <в 1. Hence, to the second order in є and v — x”, Eq. (14.130) reduces to

dA e(v — x") є2 GdA~x"( 1-х") 2

For the present assumption of pure molecular flow, и — x" from Eq. (14.30) is

_ " s*"0 -*”) + д(х" — У) ~ 6<?У(1 ~У)

V X 1 + 5x" — q( 1 + §У)

where 5 = 0!o — 1 (14.133)

To the first order in 5 and У — x", Eq. (14.132) reduces to

v — x" = 5x"(l — x")———- — Cv’-x”) (14.134)

1 “<?

To the first order in є, Eq. (14.129) reduces to

У — X = ex (1 — X )

With Eqs. (14.134) and (14.135), Eq. (14.131) becomes

dA ca(l+<?)

GdA Є 2(1 — q)

The optimum value of є is the one that makes (14.136) a maximum, at which

de GdAJ 1 ~q

„ «О-<7)

Hence eop, = ————-

1 + q

„d (^)

GdA/ max 2(1 +q)

From (14.129), (14.134), and (14.138), it is found that

5x"(l — x")q
ГТ~q

The minimum power required to recompress GdA moles flowing through pressure ratio q is

(14.141)

From (14.139) and (14.141), the ratio of minimum power to maximum separative capacity is

dQ = 2RT0 (1 + g) ln(!Ajr) dA/„и,, (a0 — 1)J 1 ~q

where (do — 1) has been substituted for 6. Values of 2[(1 + q) ln(l/г?)]/(1 — q) are tabulated below.

Pressure ratio q 0.2 0.3 0.4 0.6 0.8 1.0

2[(1 +<?)ln(l/q)]/(l-<?)] 4.83 4.47 4.28 4.09 4.02 4.00

Hence the minimum value occurs at a ratio of 1.0, as the low-side pressure becomes equal to the high-side pressure. At this limiting condition, the minimum ratio of power to separative capacity is

,dQ =JRh_

q-*-0v^/min («0 1)

The coefficient 4 in Eq. (14.143) for optimum counterflow is to be compared with 5.11 in Eq. (14.110) for cross flow. The minimum possible power input to produce 1 kg of separative work per year is

ART0 _ (4)[0.002310 kWh/(kg-mol-K) (300 K)

m(a0 -1)2~ [238 kg U/(kg-mol)] (0.00429)J(8760 h/yr)

= 0.0722 kW/(kg SWU/yr)

This result is to be compared with 0.0923 kW for the minimum with cross flow.

Because this minimum value with counterflow is obtained in the limit of zero pressure difference across the barrier, it would require use of an infinite amount of barrier surface. This condition is analogous to the familiar thermodynamic condition that the loss in availability in a heat exchanger is a minimum when an infinite amount of surface is used.

The foregoing tabulation shows, however, that even at a practical pressure ratio of 0.3, the coefficient of RT0/(uo — l)2 would be 4.47, substantially less than 5.11 with cross flow. However, this result would be somewhat offset by mixing inefficiency on the low-pressure side when v differs from y’, and by the need to use a counterflow, p-up, one-down cascade to obtain the optimum difference at as many points as possible in the cascade.