I

NBSIR 74-375

REFRIGERATION FOR AN 8 K TO 14 K SUPERCONDUCTING

TRANSMISSION LINE

David E. Daney

Cryogenics Division

Institute for Basic Standards

National Bureau of Standards

Boulder, Colorado 80302

OCTOBER, 1974

Prepared for.

Stanford University

820 Quarry Road

Palo Alto, California 94304

NBSIR 74-375

REFRIGERATION FOR AN 8 K TO 14 K SUPERCONDUCTING

TRANSMISSION LINE

David E. Daney

Cryogenics Division

Institute for Basic Standards

National Bureau of Standards

Boulder, Colorado 80302

OCTOBER, 1974

Prepared for:

Stanford University

820 Quarry Road

Palo Alto, California 94304

U.S. DEPARTMENT OF COMMERCE, Frederick B. Dent, Secretary

NATIONAL BUREAU OF STANDARDS Richard W Roberts Director

CONTENTS

Page

1. INTRODUCTION 1

2. REFRIGERATION ANALYSIS - LOW TEMPERATURE

END 3

2. 1 Heat Exchanger Analysis 7

2. 2 Brayton Cycle with Expander Before the Load,

Scheme 1 9

2. 3 Brayton Cycle with Expander After the Load,

Scheme 2 10

2. 4 Brayton Cycle with a Separate Supercritical

Helium Loop and Circulating Pump 11

2. 5 Comments on the Computations 12

3. REFRIGERATION ANALYSIS - HIGH TEMPERATURE

END 13

4. COLD END RESULTS 17

5. DISCUSSION 41

6. COST ESTIMATES 45

6. 1 Capital Costs 45

6. 2 Optimum Refrigerator Spacing 48

6. 3 Operating Costs 50

7. ACKNOWLEDGMENTS 53

8. NOMENCLATURE 54

9. REFERENCES 56

iii
LIST OF FIGURES

Page

Figure 2. 1 Three schemes of refrigeration using

supercritical helium . 4

Figure 2. 2 One candidate refrigerator using scheme 1 at

the cold end 5

Figure 2. 3 Specific heat of helium as a function of

temperature for several pressures 6

Figure 3. 1 Three possible warm end arrangements ... 15

Figure 4. 1 through 4. 25 - Refrigerator Performance Curves

8 K Load Exit Temperature . 19

10 K Load Exit Temperature 22

12 K Load Exit Temperature 33

14 K Load Exit Temperature 37

Figure 6. 1 Minimum installed refrigerator cost/km for a

10 km refrigerator spacing (excluding site and

site development costs) 46

Figure 6. 2 Annual refrigeration cost based on table 6, 2 . . 51

LIST OF TABLES

Table 6. 1 Optimum refrigerator performance for

AU/m = 1100 and = 0. 7 47

e Table 6. 2 Assumed operating cost per refrigerator ... 52

iv
REFRIGERATION FOR

AN 8 K TO 14 K SUPERCONDUCTING TRANSMISSION LINE

David E. Daney

1. INTRODUCTION

This study was sponsored by the Stanford University Hansen

Laboratories in support of their superconducting transmission line

project. The Stanford investigation focuses on low-loss Nb^Sn as

the superconductor, and consequently the temperature range of interest

is about 8 to 14 K. Because supercritical helium is the only practical

refrigerant available in this temperature range, this report investigates its use as a refrigerant for temperatures from 8 to 14 K and

for pressures from 3 to 20 atm.

A number of variables and problems must be considered in

optimizing the transmission line coolant passage configuration and

size. However, in this report we restrict ourselves to one question-

what is the cost of refrigeration for a superconducting transmission

line for a specific set of conditions in the line and for a particular

refrigerator configuration ? Although refrigerator cost correlations

such as Strobridge's [1,2] are quite useful for general estimating

purposes, these studies do not differentiate between various refrigeration schemes or sets of conditions in the transmission line. They

simply give average values for existing refrigerators. If we wish to

compare the costs of various arrangements for cooling a transmission

line, and of various line pressures, temperatures, and pressure drops,

then a more detailed analysis is required.

In making a comparison of the efficiencies of various methods of

refrigeration, one must be careful to make realistic and fair comparisons by holding the efficiencies of components such as heat exchangers
and expanders constant in the comparison. Unfortunately, this has

not always been the case with some of the work in the literature, and

much confusion as to which schemes are preferred has resulted.

Considering all of the various combinations of expanders, heat

exchangers, temperature levels, and pressure levels, there is a myriad of schemes for producing refrigeration with supercritical

helium. To reduce the problem of refrigeration analysis to manageable

proportions, we have chosen to divide the refrigerator into two parts:

a low temperature end (below 30 K) where most of the irreversibilities

occur, and a high temperature end (30 K to 300 K) where helium behaves

nearly as an ideal gas, and where the irreversibilities are only a fraction of those at the cold end. The advantages of this approach are

clarity of comparison, reduced computation time, and greater generality.

For the low temperature end, we make detailed calculations for

three refrigerator arrangements and for a number of combinations of

temperature, pressure, and pressure drop in the transmission line.

The major effort of this study addresses this low temperature end.

For the high temperature end, we made an estimate of the

irreversibilities which depend only on the flow rate. The total work

required is then the sum of work required to run the cold end and the

warm end.

Finally, based on the first two parts, estimates of refrigerator

cost are made as a function of the transmission line conditions, heat

leak, and conductor losses.

Refrigerator reliability is of paramount importance in choosing

a refrigeration scheme since transmission line outage due to refrigeration failure simply cannot be tolerated. However, this facet would

require an extensive experimental study, which is outside the scope

of this report.

2
2. REFRIGERATION ANALYSIS - LOW TEMPERATURE END

We have chosen to analyze the three low temperature refrigeration schemes shown in figure 2. 1. These cycles, which were selected

as being the most practical, are the Brayton cycle with the expander

before the load, scheme 1; the Brayton cycle with the expander after

the load, scheme 2; and the Brayton cycle with a separate supercritical

helium loop and circulating pump, scheme 3. Figure 2. 2 is an example of what a complete practical refrigerator might look like with scheme 1 attached to the bottom. The expanders

at intermediate temperatures are required to reduce the upper heat

exchangers to reasonable size and to compensate for any adverse

imbalance in the specific heat of the helium refrigerant. The fraction

of the compressor flow which must be diverted through the intermediate

expanders is identical to the fraction of the total irreversibilities which

occur in the upper portion of the refrigerator.

As mentioned in the introduction, certain conditions of constancy

must be observed in making comparisons between refrigerators. For

the transmission line, we hold the pressure and temperature at the

exit and the pressure drop along the line constant. The temperature

at the exit, rather than some mean temperature in the line, is held

constant since it is the highest temperature that will determine whether

or not the line will go normal.

For the expander, the proper condition of constancy is constantexpander efficiency. In practice, expander efficiency may not remain

constant. For instance for turbo-expanders the efficiency generally

increases with size, so that for the same flow rate, the efficiency

tends to decrease with temperature. Rather than tie this study to a particular type or brand of expander, however, we adopt the simplest

and most generauL criterion of constant efficiency. For the heat

3
ono

LlJ Q

<

O-X

LlJ

4
Figure 2. 2 One candidate refrigerator using scheme 1 at the cold end.

5
Figure 2. 3 Specific heat of helium as a function of temperature for

several pressures.
exchanger, we have chosen to hold the quantity AU/m constant, where

A is the heat transfer area on one side of the heat exchanger, U is the

o overall heat transfer coefficient, and m is the mass flow rate.

2. 1 Heat Exchanger Analysis

Typically, in refrigeration cycle analyses, the warra end AT

of the heat exchanger is held constant, or in some cases the heat

exchanger effectiveness is held constant. If the relative values of the

specific heats of the helium in the two sides of the heat exchanger

remain constant, then the size of the heat exchanger (for a given flow

rate) will be constant for the constant effectiveness condition. For the

constant AT condition, the size for a given flow rate will be proportional

to the heat transferred, or approximately proportional to the temperature span of the heat exchanger.

Unfortunately, for helium the specific heat is very ill-behaved

at low temperature (see figure 2. 3) so that we can no longer hold the

heat exchanger size constant by use of these simple and convenient

relationships. Instead, we must numerically solve the differential

equations of the heat exchanger.

Consider a differential element of a counterflow heat exchanger.

1 3 Hot Side

<

dq

f

2

Then —d A-^ 4 Cold Side

dq = U(T,he- T ) dA, (i:

and

dq = dH^ = m^ dH^ . (2]

7
Equating (1) and (2), we obtain:

ATI dH = — (T, - T ) dt , (3) h m n c h

and

whe re

dH = -0— dH , (4) c m h

c

dt = ^ . (5]

Because T and T are difficult functions of H and H , eqs (3)

n c he

and (4) must be integrated numerically. The problem is similar to

an initial value problem or propagation problem in that initial values

at one end (e. g. , Tl and T2) are used to march through the heat

exchanger and obtain the values at the other end (T3 and T4). In our

cycle analysis, points at opposite ends of the heat exchanger are known

(e. g. , Tl and T3 or T2 and T3) so that iteration is required. If Tl

and T3 are specified together with the heat exchanger parameter,

o AU/rri^, then a value of T2 is estimated and eqs (3) and (4) are

numerically integrated to obtain a value for T3 . The process is

repeated until the calculated T3 agrees with the specified T 3 to within

the allowed error (0. 01 K). Convergence is usually obtained in 3 to 7

steps.

o To obtain a feeling for the significance of the quantity AU/m,

one should note that it is related to the commonly used dimensionless

heat exchanger size [3], NTU, by the expression AU/m = (NTU)(C ) ATT ^

since NTU = 4^ . m C Q

P

^
For a balanced counter flow heat exchange with constant specific heat,

NTU is related to the heat exchanger effectiveness, e, by the

exp re s sion

NTU

e =

1 + NTU

A more detailed analysis would consider the overall heat transfer

coefficient U as a function of temperature and flow rate. For our

purpose here, which is a comparative study of refrigeration cycles,

such an added complication is not justified.

2. 2 Brayton Cycle with Expander Before the Load, Scheme 1 Because we specify the conditions at the transmission line, the

pressure before the expander, P2, is not known. It must be found by

iteration. Referring to figure 2. 1 , the procedure is as follows.

The temperature at point 1, the state at 4, and the AP and

AT along the line are fixed.

From T4, P4, and AP line, and AT line, state 3 is obtained.

An initial estimate of temperature, T2, is made for use in

step 4 by the approximation

T2 r: T4 + -^H(P4,T4)

(AU/m)

The pressure at 2 is estimated by the perfect gas relationship

P2 = P3

(

T3 - T2\ iC /R

+ 1

The pressure at 1 is obtained for this trial from

PI = P2 (1 + AP/P),

where AP/P is the relative pressure drop in the heat

exchanger.

9
6. Since states 1 and 4 are now known (PI is only estimated).

routine.

7. Using the definition of the isentropic engine efficiency (see

nomenclature), we can now check if we obtain the correct

value for T3.

H3 = H2 (1 - Tl ) + Tl H (P3, S2)

e e

T3 = T (H3, P3).

8. If the value calculated for T3 in step 7 does not agree with

the value obtained in step 2, then iterate using steps 5 to 7

with a new value for P2. The solution is usually obtained

in 3 to 5 iterations.

2. 3 Brayton Cycle with Expander After the Load, Scheme 2

For this scheme, the unknown pressure which must be found by

iteration is P4. The computational scheme is as follows.

1. The temperature at point 1, the state at 3, and the AP and

. AT along the line are fixed.

2. From T3, P3, AP line, and AT line, state 2 is obtained.

3. The pressure at 1 is obtained from

we can obtain states 2 and 5 using the heat exchanger

PI = P2 (1 + AP/P).

4. Estimate a value of T4 for step 5 using the approximation

T4 = T2 - (HI - H2)

(AU/m)

5. Calculate the first trial value of P4 using the ideal gas

expression

P4 = P3 T4 - T3

Tl T 3 + 1

-I C /R

P

e

10
6. The enthalpy at point 4 may now be obtained from the

definition of the isentropic engine efficiency,

H4 = H3 (1 - 11 ) - T] H (P4, S3),

e e

7. Using states 2 and 4, we now solve the heat exchanger

equations and obtain a value for Tl.

8. We iterate with steps 6 and 7 until we find a value for P4

which gives a Tl that agrees with the given T 1 to within

the allowable error (.01 K). The solution is usually found in

5 to 9 iterations.

2, 4 Brayton Cycle with a Separate Supercritical Helium Loop and

Circulating Pump

This arrangement is simply scheme 1 with a supercritical loop

added to the bottom. As soon as states 3 and 4 are determined, the

solution proceeds as described in section 2. 2. We begin at the load.

1. We specify Tl, P4, P6, T6, AP line and AT line.

2. State 8 is determined from state 6, AT line, and AP line.

P7 is calculated from

P7 = P8 (1 + AP/P)

where AP/P is the relative pressure drop through

the pump loop heat exchanger ,

3. The enthalpy at 7 is determined using the definition of

isentropic pump efficiency (see nomenclature)

H7=H6^ H(P7 S6) -H6

and

T7 = T(H7,P7) . We use the isentropic pump efficiency because it is simpler

and less ambiguous than the efficiency based on fluid

11
horsepower. If the fluid horsepower efficiency is

adequately defined for a compressible fluid, then it can

be related to the isentropic pump efficiency.

4. Since we now know state 7, state 8, and P3, we have the

information required to solve the pump loop heat exchanger

problem. We assume a value of T4 and march through the

heat exchanger and obtain a T8. As before, we iterate until

the T8 obtained in this fashion agrees with our previously

calculated value. , .

5. We calculate P3 from

P3 = P4 (1 + AP/P),

and states 3 and 4 are determined.

The remainder of the problem is identical with that of

scheme 1 and we proceed as before.

Z, 5 Comments on the Computations

The ideal work of compression, Wc, was calculated from the

difference in the available work for a steady flow process, H - T S, o between states 1 and 5, This is not identical to the isothermal work o compression in going from P5 to PI at temperature T^, although the

difference is generally less than a few percent. The availability

function is used because it gives the minimum work for a reversible

process.

Two sources were used for the helium properties. For all

properties except H(P,S) and T (P,H), the programs of McCarty,

based on his equation of state [4, 5], were used. For H (P, S) and

T (P,H), we used equations developed by Arp [6].

McCarty' s work will undoubtedly stand for some time as the

fundamentally accurate helium state equation, giving both the PVT

12
surface and properties such as specific heats, compressibilities, etc. , which are related to the derivatives of the PVT surface. For practical

analysis of refrigeration and flow problems , however, the independent

state parameters should be the enthalpy (related to heat input), pressure

(related to flow friction factors), and sometimes the entropy (related to

system reversibility), rather than the density and the temperature

which are the independent parameters of McCarty's equations. This is

in no sense a criticism of McCarty's work since all fundamental theories

and equations of fluid properties are based on density and temperature

expansions. But as a result, the equations are very cumbersome to use

in engineering analysis. For example, a double iteration must be done

to find density and temperature and then other properties of interest

from a given pressure and enthalpy. Such a procedure is time consuming and very costly, and has been a significant limitation to the refrigeration and fluid flow analysis to date.

Consequently, Arp developed four new state equations for helium

based on McCarty's work. These are S (P,H), which calculates entropy

from pressure and enthalpy; H (P,S), which calculates enthalpy from

pressure and entropy; T (P, H), which calculates temperature from

pressure and enthalpy; and V (P, H) , which calculates the specific volume.

Use of the function T (P, H) resulted in a reduction in computer time of

two orders of magnitude in the heat exchanger calculations. Without this

function, interpolation in a table of properties would have been required

since McCarty's equations required several minutes of computer time

for each heat-exchanger calculation.

3. REFRIGERATION ANALYSIS - HIGH TEMPERATURE END

In turning our attention now to the warmer portion of the refrigerator, it is instructive to ask why more than a simple counter flow heat

exchanger may be required. There are two reasons. First, the size of

13
the heat exchanger necessary to achieve the required value of Tl - T5

would be enormous. Second, in some cases the slight imbalance in

the specific heats of the two helium streams is such that the heat

exchanger becomes pinched and negative temperature differences would

result.

Grassmann and Kopp [7] have pointed out that the heat exchanger

losses are minimized if the AT is proportional to the temperature. Thus,

several heat exchangers spanning the range from 300 K to 30 K are

desirable if the losses are to be kept low.

Three possible arrangements for the warm end of the refrigerator

which use two balanced flow-counterflow heat exchangers are shown in

figure 3. 1, Scheme A is the most common arrangement. To have low

losses, this arrangement must use a sufficiently high pressure ratio

across the expander so that the temperature span along the unbalanced

flow heat exchanger is large compared to the temperature difference

across the exchanger.

Scheme B is less common, although it has seen some use in

liquefiers. Its advantage is that, in principle, it can be used in a cycle

of any pressure ratio. Its practical disadvantage is that very low pressure ratios are usually required across the expanders, limiting its use

to specially designed turbo-expanders.

Scheme C is shown primarily as an aid in the analysis, although

it has seen some use in hybrid cycles employing Stirling -type machines.

Let us consider the problem of minimizing the losses or irreversibilities of the upper part of the refrigerator. This is equivalent to minimizing the total work required to run the refrigerators at temperatures

Tl and T2 in scheme C. '

Referring to figure 3. IC and taking an approach similar to

Kapitza's in his analysis of liquefiers [8], we may write that the heat

14
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LU

CJ3

=1:

AAA/WW

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LlJ Q

a.

•vwww

A/www

3 CD o z:

—I <:o Q X LU LlJ OZ I—

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-VWWW VWWW

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15
removed at Tl is

" ' D, = C (ATI - AT 2) . 1 1 p

We assume ATI and AT2 to be constant along each heat exchanger.

The work required to run this refrigerator is

Q, T - Tl

or

1 Tl T 1 c

m C ' T - T 1 W, =— 2-(^Tl-AT2)( °

1 Tl

' ^ ; V -J. J

. c An analogous expression may be written for . An assumption is required regarding the heat exchangers, and we

assume equal effectiveness of the two exchangers. This

implies that

, T - Tl

AT 1 o AT 2 Tl - T2 •

Eliminating AT 2 from the expression for the sum of the work, and

B W

setting

^ ^ ^ equal to zero, the resulting expression for the value of

Tl which gives the minimum work is

^ ' = l+T T2 • o JJt + T2) T2

' o For T - 300 K, T2 = 30 K, and the heat exchanger effectiveness

o

= 98 percent, the results are

Tl = 81 K

ATI = 4. 4 K

AT2 = 1. 0 K . 16
Changing Tl by ± 20 K increases the work by only about 5 percent, but

significantly larger variations in Tl cause a rapid increase in the work.

To determine what value might be reasonable for the efficiency,

Tj , of the external refrigerator, we compared scheme C to scheme B

c for a typical set of conditions using the values of Tl, ATI, and AT 2

listed above. We find that a 70 percent expansion engine efficiency in

scheme B is equivalent to a 60 percent of Carnot refrigerator efficiency

(36 percent for a 60 percent compressor efficiency) in scheme C. Since

these values of efficiency seem reasonable, the work calculated in this

example (659 J/mol) is taken as typical for this study.

The total work required to run the refrigerator is the sum of the

work required for the two parts. It is given by

/Wc\ /Wc \ /^^\ (7) \Q )

total \ Q /cold end I Q ) ( j warm end '

Note that the warm end work per unit of refrigeration is proportional to the flow rate. Thus, when other factors are nearly equal, it

o is advantageous to select a cycle with a low value of m/Q . Low mass

flow rates are also desirable from the standpoint of load pressure drop,

heat exchanger size, and compressor size.

The results of this section are approximate in that it may not

always be possible to construct a practical refrigerator with warm end

losses as low as we estimate. However, we feel that the advantage

gained due to the simplicity and generality of the approach more than

compensates for this difficulty.

4. COLD END RESULTS

The results of our calculations for the cold end of the refrigerator are presented as plots -of Wc/Q vs. the pressure at the exit from

the load for constant values of AT across the load. The values of Wc/Q

given assume reversible compression, so they must be divided by an

17
isothermal compressor efficiency to obtain the work of an actual

refrigerator. The work per unit of refrigeration for a complete refrigerator is obtained from eq (7) as discussed in the previous section.

Unless indicated otherwise on the plots, the following parameter

values were used in the computations:

Refrigerator heat exchanger size

AU/m = 1100 J/mol-K

Pump loop heat exchanger size

AU/m = 220 J/mol-K

Relative pressure drop in heat exchanger

AP/P = 0. 01

Temperature at top of refrigerator heat exchanger

Tl = 30 K

Expander efficiency

Tl = 0. 70 . :

. e Pump efficiency

T] = 0. 50 . P

Pressure drop across the load

AP load = 1.0 atm

Results are given for load exit temperatures of 8 K, 10 K,

12 K, and 14 K and the figures are arranged in that order. Plots of

om/Q are given for each temperature as an aid to calculating the total

work for a complete refrigerator. A few results are given for

parameter values different from those listed above in order to indicate

the effect of these parameters on Wc/Q.

The value of AU/m chosen for the heat exchanger corresponds

to an NTU of about 50, In a balanced flow exchanger, this is equivalent

to an effectiveness of 98 percent. The pump loop heat exchanger size

18
8 K

LOAD EXIT TEMPERATURE

19
110

100

90

80

70

60

1 30

SCHEME 1 = 8 K

0 5 10 15 20 25

P4 (ATM)

Figure 4. 1 Scheme 1 performance for an 8 K load exit temperature.

1 20

110

100

90

80

^^^1,3,= Vo K

SCHEME 2

T3 = 8K

10 15

P3 (ATM)

20 25

Figure 4, 2 Scheme 2 performance for an 8 K load exit temperature.

20
CO

I—t—

o-o

1 30

1 20

no

100

80

1 1

oLHLrlL

1

5

1

- \

T6 = 8 K

- \

1 1 1

-

1 10 1 5 20 25

Pg (ATM)

Figure 4. 3 Scheme 3 performance for an 8 K load exit temperature.

1 r

1 1

0.06 -

0.05 -

- 0.04 h o

^ 0.03

0.02 -

0.01

AT, .= 1 . 0 K load

"^LOAD " ^

^

10 1 5 20 25

Figure 4. 4 Flow rate vs. load pressure for an 8 K load exit temperature.

21
10 K

LOAD EXIT TEMPERATURE

22
100

— 90

I—

I—

o

80

70

60 -

50

SCHENE 1 = 10 K

10 15 20 25

(ATM)

Figure 4. 5 Scheme 1 performance for a 10 K load exit temperature.

100

C/1

o

90

80

70

60 -

50

T

SCHEFIE 2

T3 = 10 K

AT, .= 1 . 0 K load

3 K

1 0 1 5 20 25

(ATM)

Figure 4. 6 Scheme 2 performance for a 10 K load exit temperature.

23
0 5 10 15 20 25

' Pg (ATM)

Figure 4. 7 Scheme 3 performance for a 10 K load exit temperature.

24
CO

100

90

80

70

60

50

AT, .= 1 . 0 K

. > load

2.0 K

SCHEME 1 = 10 K

Au/m = 550 J/mol-K

10 15 20 25

P4 (ATM)

Figure 4. 8 Effect of reduced heat exchanger size on the perforraance

of s cheme 1

.

90

00

80

70

60

50

40

SCHEME 1

AP, .= 2 ATM

1 oad

1 ATM

0 ATM

10 15 20 25

P4 (ATM)

Figure 4. 9 Effect of load pressure drop on scheme 1 performcince.

25
26
130

50

10

P,

1 5 20 25

(ATM)

Figure 4. 1 1 Effect of load pressure drop on scheme 3 performance.

27
Figure 4, 12 Scheme 1 performance for a 10 K load exit temperature

with 80% engine efficiency.

90

80

— 70

HIcy 60

50 -

40

90

80

^ 70

J—

I—

cC

cx 60 u

50

40

T T

SCHEME 1 = 10 K

He = 0.80

10 15

P4 (ATM)

10 1 5 P3 (ATM)

20

20

25

25

Figure 4. 13 Scheme 2 performance for a 10 K load exit temperature

with 80% engine efficiency. 28
0 5 10 15 20 25

Pg (ATM)

Figure 4. 14 Scheme 3 performance for a 10 K load exit temperature

with 80% engine efficiency.

29
CO

I—

I—

100

90

80

^

70

60

50

10

SCHEME 3

T5 = 10 K

P5 = 10 ATM

riY/mp=0.67

20 30 40

Figure 4. 15 Effect of the refrigerator to pump mass flow ratio on the

performance of scheme 3.

30
Figure 4. 16 Scheme 1 performance as a function of heat exchanger size.

31
0 5 10 15 20 25

4. 17 Flow rate vs. load pressure for a 10 K load exit temperature.

32
12 K

LOAD EXIT TEMPERATURE

33
m

I—<

cru

90

80

70

60

50 -

40

SCHEME 1

T/i = 12 K

2 K

L

10 15 20 25

P4 (ATM)

Figure 4. 18 Scheme 1 performance for a 12 K load exit temperature.

100

CO

I-

«C

—

.

90

80

70

60

50

1 1 1 1

SCHEME 2

T3 = 12 K

A /AT, ,= 1 K _ \/ load

2 TO 0

X 1 1 I 1 10 15 20 25

P^CATM)

Figure 4. 19 Scheme 2 performance for a 12 K load exit temperature.

34
0 5 1 0 1 5 20 25

Pg (ATM)

Figure 4. 20 Scheme 3 performance for a 12 K load exit temperature.

35
5 10 1 5 20 25

ure 4. Zl Flow rate vs. load pressure for a 12 K load exit temperature.

36
14 K

LOAD EXIT TEMPERATURE

37
801 1 1 1 r

0 5 10 15 20 25

P4(ATM)

Figure 4, Z2 Scheme 1 performance for a 14 K load exit temperature.

PgCATM)

Figure 4. 23 Scheme 2 performance for a 14 K load exit temperature.

38

5 10 15 20 25

Pload'""'

Figure 4. 25 Flow rate vs. load pressure for a 14 K load exit temperature.

40
is equivalent to an NTU of about 8. The effect of AU/m on the refrigerator performance is shown in figures 4. 8 and 4, 16. Figure 4. 16 shows

there is little value in increasing the heat exchanger size much beyond

1100 J/mol-K, whereas substantial reductions in size below 1100 J/mol-K

result in a significant increase in the compressor work. Although by

no means a true optimum, 1100 J/mol-K appears to be a reasonable

heat exchanger size for an efficient refrigerator.

The choice of 30 K for the temperature at the top end of the

refrigerator heat exchanger was somewhat arbitrary. This temperature

is sufficiently high that helium behaves nearly as an ideal gas, while

at the same time, it is low enough that an excessively large heat exchanger

is not required. In future studies, some attention should be given to

optimizing this temperature. A simple approach such as that taken in

section 3 will probably not be adequate, however, because of the nonideal behavior of helium at low temperatures.

An expander efficiency of 0. 7 0 was chosen as being a value typical

of a good, but not necessarily excellent expander. For large turboexpanders efficiencies in excess of 0. 8 are possible [9].

The pump efficiency of 0. 5 is taken as typical of a high quality

centrifugal pump. One curve in figure 4. 7 is presented for a pump

efficiency of 0. 9, a value representative of a high quality reciprocating

pump.

Most of the curves are for a transmission line pressure drop of

1 atm, but figures 4. 9, 4. 10, and 4. 11 give results for 0, 1, and 2 atm,

5. DISCUSSION

The complex shapes of the performance curves in section 4

result from the interaction of several effects. The more important

ones are the work required to circulate the helium through the line,

41
the temperature difference at the lower end of the heat exchanger,

T2 - T4, and the Joule- Thomson effect.

The pump work required to circulate the fluid through the transmission line causes the large Wc/Q at low pressures. Because this

work is proportional to the specific volume of the helium, the high

specific volumes associated with lower pressures result in large

pumping losses. Figures 4.9, 4, 10, and 4. 11 illustrate this effect

for three load pressure drops.

For an adiabatic line, the minimum work required to circulate

the fluid is

W.^ = T AS . (8) ideal o

The minimum work differs from the work input to a reversible low

temperature pump, T^ AS. And eq (8) shows that all reversible

processes for circulating the helium require the same work input

regardless of the temperature of the circulating pump. For example,

a reversible low temperature pump requires less pump shaft work

than an ambient temperature pump because of the increased density of

the fluid, but this advajitage is cancelled by the refrigerator work

required to remove the pump work from the fluid. For a real pump

operating at low temperature, the irreversibilities must also be

removed by the refrigerator, and herein lies the disadvantage of the

low temperature pump. Thus, the pump losses must be small compared

to the load for this arrangement to be competitive.

This disadvantage is partially offset by the reduced compressibility,

(and hence reduced pump work) of helium at low temperatures. As

helium becomes less compressible at low temperature, the entropy

change (at constant enthalpy) between two given pressures is reduced,

and thus the ideal work required to circulate the helium is reduced.

An ambient temperature pump cannot take advantage of this reduction

in the ideal work because the concomitant imbalance in the specific heats

makes perfect heat exchange impossible.
The temperature difference at the lower end of the heat exchanger,

T2 - T4, represents lost refrigeration. Because of the imbalance in

the specific heats in the two sides of the heat exchanger, this temperature

difference may become as large as 2 K, resulting in substantial losses.

The bump in the zero atmosphere AP curve of figure 4. 10 is primarilydue to the variation of T2 - T4 with pressure.

At high pressures, the Joule- Thomson coefficient becomes

negative and the cooling produced by the expander decreases for a given

pressure ratio. For scheme 1 the truncation of the higher

AT curves at high pressure results from this effect. The expansion

engine simply cannot produce the cooling required in the load.

A change in a single transmission line variable might appear

simultajieously in several of the effects discussed above. For example,

an increase in the pressure might result in decreased pump work,

increased heat exchanger AT, and an adverse Joule-Thomson effect.

For this reason, it is necessary to make a complete cycle analysis,

as has been done here, rather than rely on simplified models that may

consider only one effect.

The results show that scheme 1 gives the lowest values of Wc/Q,

but that its useful pressure range becomes more limited as the

load temperature is reduced. An increasingly negative Joule-Thomson

coefficient with high pressure ajid low temperature is the cause of this

difficulty.

Scheme 2 will always produce refrigeration, but not necessarily

with pressures greater than atmospheric in the return side of the heat

exchanger. Only solutions with return pressures of one atmosphere

or greater are presented in section 4, and truncation of some of the

curves on the low pressure side results.

43
The work required for scheme 2 is significantly greater than for

scheme 1. The primary reason is that the temperature level of the

expansion engine is lowered by a value equal to the AT at the bottom

of the heat exchanger. Because of the imbalance in the heat exchanger

this AT can amount to as much as 2 K, which causes a significant

reduction in the cooling power of the expander.

Although scheme 3 might be expected to exhibit low efficiencies

because of the low temperature pump, our results show that its Wc/Q

is nearly identical to that of scheme 2 at 1 0 K and 12 K, At 8 K,

scheme 3 has a significantly lower Wc/Q than scheme 2 over much of

the pressure range investigated.

When we analyze the efficiency of a complete refrigerator,

using eq (7), we find that scheme 1 can be expected to yield efficiencies

(percent of Carnot) of about 25 percent and schemes 2 and 3 yield

efficiencies of about 20 percent. These values are typical for about

10 atm and the refrigerator parameters listed in section 4. Because

no attempt was made to optimize Tl or AT load, conclusions regarding

the efficiency as a function of temperature should be made with care.

44
6. COST ESTIMATES

A precise estimate of the cost of transmission line refrigeration

would require a complete specification of a number of parameters such

as the transmission line geometry, temperature, heat leak, and conductor losses. Because these parameters are not well defined at this

time, we estimate the capital and operating costs of the refrigeration

for a range of temperatures and line heat loads.

6. 1 Capital Costs

The initial or capital cost of the refrigerators in dollars is

estimated from the relationship [1]

C. ^ 6000 (Wc/e)°* ^, (8)

where Wc (kilowatts) is the isothermal compressor power calculated

from the cycle analysis, and e ( 0. 6) is the efficiency of a real

compressor compared to an ideal isothermal compressor.

Dividing eq (8) by the transmission line length, L, and factoring

the right-hand side, we obtain the capital cost per xinit length of transmission line in the form

Because the cycle analysis gives different values of Wc/Q for

different schemes and operating conditions, we choose the minimum

Wc/Q obtained from the cycle analysis at each temperature. These

minimum values of Wc/Q are given in table 6. 1 along with the corresponding Carnot efficiency. In all cases, the optimum was obtained

using scheme 1.

45
TEMPERATURE. K Figure 6. 1 Minimum installed refrigerator cost/km for a 10 km refrigerator

spacing (excluding site and site development costs).

46
Table 6. 1 Optimum refrigerator performance for

AU/m = 1100 and T] = 0, 7.

T(K) Wc/Q (Wc/Q)

Carnot

Percent Carnot Percent Carnot

8 88 35. 6 40. 5 24. 3

10 65 28. 3 43. 5 26. 1

12 58 23. 4 40. 4 24. 2

14 49 19. 9 38. 3 23. 0

It is important to note that a variety of cycle parameters must be specified

to obtain this minimum Wc/Q. The optimization process is a highly nonlinear function of these various parameters, and it may be possible to

improve (reduce) Wc/Q with subsequent work. But we feel that they

are realistic for this study. The falloff in efficiency with increasing

temperature is probably due to holding Tl constant at 30 K, This decrease

in efficiency should not be regarded as characteristic when the complete

cycle is optimized.

The thermal load per unit length, Q/L, may be divided into (1) a temperature and current independent component, Q^, which is the sum

of dewar losses and dielectric losses, and (2) losses in the superconductor,

Q . From previous studies reported by the Linde and the BNL groups,

c the fixed losses, Q^ , probably will be somewhere in the range 0. 05 to

2. 0 kilowatts per kilometer of line, depending on a variety of design

parameters. The superconductor losses may lie somewhere in the range

0. 2 to 2. 0 kilowatts per kilometer. Because of the emphasis on low-loss

materials in the Stanford project, a reasonable range of the total losses

to be considered here is 0. 2 to 3 kw/km of line.

For an assumed refrigerator spacing of 10 km, figure 6, 1 gives the

minimum installed refrigerator cost per km as a function of the temperature

for various values of the total thermal load (exclusive of coolant friction

losses). For spacings different from 1 0 km, or for refrigerator

47
efficiencies different from those given in table 6. 1, the cost should be

adjusted according to the expression

(C./L)' -

(a/L) [

(refrigerator efficiency) "| T L "] (refrigerator efficiency) 'J L'

A change in L by a factor of 2 will change by only about 23%, which

is within the accuracy of these cost calculations. It should be noted at

this point, that operation at the point of maximum refrigerator efficiency

is not advocated. The thermal and electrical requirements of the line

must naturally be considered, and the final selection of the operating

conditions should take account of both the line and the refrigerator

characteristics.

6. 2 Optimum Refrigerator Spacing

Optimizing the distance between refrigerators is a problem of

great complexity, so we will only touch on it here. The true optimum

spacing will result in lowest total cost (capital plus operating) for the

line and the refrigerator over their expected life. Thus, the coolant

passage flow area, which influences the line cost, line pressure drop,

line heat leak, refrigerator size, etc. , cannot necessarily be held

constant in optimization calculations.

If we consider only the refrigerator capital costs, then an optimum

spacing results because of the effect of spacing on the coolant pumping

losses, and hence on the Wc/Q required.

The compressor work per unit of refrigeration, Wc/Q, may be

approximated for cost estimating purposes by

W

Wc/Q = (Wc/Q) (1+—^) , (10) o Q

where the subscript o refers to zero pressure drop in the line and W^

is the pumping power required. It caji be shown that

W ~ m AP , P

48
and that

o L ~ m

since

(Q/L) L - m C AT

p line

Thus,

W -LAP,

P

and eq (10) may be rewritten as

or as

Wc/Q = (Wc/Q) (1 + a L AP) , o

Wc/Q = (Wc/Q)^ (1 + Pl"") . (11)

The value of the exponent n implies the following about the coolant

passage pressure drop and cross sectional area.

1. If n = 1 , AP/L ~ 1/L and large increases in the area

are required as the length increases.

2. If n = 2, AP/L = constant and moderate increases in the

o area are required as the length increases, because m ~ L .

3. If n = 4, the area is nearly constant for turbulent flow in

the line.

A reasonable value of n would seem to lie between 2 and 3. Combining eqs (9) and (11), we obtain

C./L = 6

49
When reasonable values for p and n are chosen, we find that eq (10)

gives an optimum refrigerator spacing on the order of 100 km. This

compares favorably with 80 km predicted by Cairns, Swift, Edney

and Steel flO].

If we consider the total cost (including operating costs) of the line

during its life, then some of these costs, such as power costs, will be

proportional to the first power of installed compressor power, and we

may write

0. 7

C /L^ (^ + ^^").^ +A(1 + Pl")

(L)

where is the total cost of the refrigeration. When A is of the order

of 1 (i. e. , the power costs equal the capital costs over the life of the

line) the optimum spacing is reduced to about 10 km, which is in

agreement with the length used in section 6. 1. In a precise analysis

the constant A should be weighted so as to take account of the diminished

present worth of future dollars. This consideration would tend to

increase the line length.

Labor and supervisory costs per refrigerator are relatively

independent of the refrigerator size, so that their cost per unit length

of line is proportional to 1/L. Consideration of these costs increase

the optimum length over the 10 km of the previous example.

Clearly, refrigerator spacing is a problem which will require

considerable study. We have touched on the problem here to demonstrate the complexity of the problem and to provide a rough estimate

for the cost analysis.

6. 3 Operating Costs

The estimated annual operating cost of refrigerating the line is

obtained by dividing the cost per refrigerator by the length of line

50
TEflPERATURE. K

Figure 6. 2 Annual refrigerator cost based on table 6. 2.

51
serviced by the refrigerator. Table 6. 2 gives the assumed operating

costs per refrigerator which were used to construct the curves of

operating cost per kilometer given in figure 6, 2.

Table 6. 2 Assumed operating cost per refrigerator.

Item

Electric Power

Labor - direct

Labor - overhead

Maintenance

Taxes, insurance

Supervision, misc.

Unit

kw-hr

man-hr

50% direct

Unit Cost

($/unit)

. 012

6. 0

3, 0

Yearly Cost

52, 600

23, 300

6% of installed cost per year

3% of installed cost per year

$20,000 per year, all cases.

The estimates in table 6. 2 are analogous to those made by

Winters and Snow TH] in an economic analysis of a 1. 5 kW helium

refrigerator. These estimates should be considered as approximations, not as the results of a detailed analysis.

The low sensitivity of the curves in figure 6. 2 to temperature

and thermal load results from the high fixed costs (Labor and

Supervision) per kW. Larger refrigerator spacings would reduce the

relative size of these fixed costs, and consequently increase the sensitivity of the total annual cost to temperature and thermal load. In a similar fashion, highly reliable refrigerators requiring an operator

only 8 hours a day, instead of the 24 hours assumed, would yield the

same advantages.

52
7. ACKNOWLEDGMENTS

The author wishes to thank V. D. Arp for a number of

suggestions and helpful discussions concerning all aspects of this

work. W. R. Parrish was always able to find time for my questions

on programming and numerical analysis. Both of these contributions

are gratefully acknowledged.

53
' 8. NOMENCLATURE

heat transfer surface area on one side of heat exchanger

installed cost of refrigerator, $

specific heat at constant pressure

isothermal compressor efficiency

specific enthalpyrefrigerator spacing, km

mass flow rate

o number of heat exchanger transfer units, AU/mC

P

pressure

heat transfer rate, watts

thermal load in the transmission line, watts

gas constant

entropy

absolute temperature

overall heat transfer coefficient

ideal work of compression

heat exchanger effectiveness

Carnot work/actual work for a refrigerator with given he;

expansion engine efficiency, actual work/isentropic work

pump efficiency, isentropic work/actual work

dimensionless heat exchanger length

54
Subscripts

c cold side of heat exchanger

h warm side of heat exchanger

o ambient conditions

p pump

55
9. REFERENCES

1. - Strobridge, T. R. , IEEE Transactions on Nuclear Science,

NS-16, No. 3 (1969).

2. Strobridge, T. R, , Cryogenic Refrigerators - an Updated Survey,

Nat. Bur. Stand. (U. S. ) Tech. Note 655, 1 1 pages ( 1974).

3. Rohsenow, W. M. , and Choi, H. Y. , Heat, Mass, and

Momentum T ransfer, p. 314 (Prentice -Hall, Englewood

Cliffs, New Jersey, 1961).

4. McCarty, R. D. , Thermophysical properties of helium-4 from

2 to 1500 K with pressures to 1000 atmospheres, Nat. Bur.

Stand. (U.S.), Tech. Note 631 (1972).

5. McCarty, R. D. , Thermophysical properties of helium from

o 2 to 1500 K at pressures to 1 x 10 pascals, to be published

(1973).

6. Arp, V. D. , Refrigeration of superconducting rotating machinery,

Nat. Bur. Stand. (U.S.), NBSIR 73-331 (1973),

7. Grassmann, P. , and Kopp, J. , Kaltetechnik 9^, 306 (1957).

8. Kapitza, P. L. , and Danilov, I. B. , Cryogenic engineering - present status and future development, Proceedings of the

First International Cryogenic Engineering Conference,

Tokyo, Japan, 1967, p 228 (Heywood- Temple Industrial

Publications, 1968).

9. Linhardt, H. D. , LNG/Cryogenics, 7 (February/March 1973).

10. Cairns, D. N. H. , Swift, D. A. , Edney, K. , and Steel, A. J. , Low temperatures and electric power, Proc. International

Institute of Refrigeration, London, p. 155 (1969).

11. Winters, A. R. , and Snow, W. A. , Advances in Cryogenic

Engineering p. Ill (Plenum Press, New York, N. Y. ,

1966).

56
NBS-114A (REV. 7-73)

U.S. DEPT. OF COMM. BIBLIOGRAPHIC DATA

SHEET

1. PUBLICATION OR REPORT NO. NBSIR 74-3752. Gov't Accession

No.

3. Recipient's Accession No.

4. TITLE AND SUBTITLE

REFRIGERATION FOR AN 8 K TO 14 K SUPERCONDUCTING

TRANSMISSION LINE

5. Publication Date October 1974

6. Performing Organization Code

7. AUTHOR(S)

David E, Daney

8. Performing Organ. Report No.

9. PERFORMING ORGANIZATION NAME AND ADDRESS

NATIONAL BUREAU OF STANDARDS, Boulder Labs

DEPARTMENT OF COMMERCE

Washington, D. C. 20234

10. Project/Task/Work Unit No. 2750575

11. Contract/Grant No. PR 1745

12. Sponsoring Organization Name and Complete Address (Street, City, State, ZIP)

Stanford University

820 Quarry Road

Palo Alto, California 94304

13. Type of Report & Period Covered

14. Sponsoring Agency Code

15. SUPPLEMENTARY NOTES

16. ABSTRACT (A 200-word or less factual summary of most sigpificant information. If document includes a significant

bibliography or literature survey, mention it here.)

The performance of three supercritical refrigeration cycles for cooling

superconducting transmission lines has been investigated for the temperature

range of 8 K to 14 K. This temperature range is applicable to the Stanford

superconducting (Nb3Sn) line. These cycles, which were selected as being the

most practical, are the Brayton cycle with the expander before the load, the

Brayton cycle with the expander after the load, and the Brayton cycle with a separate supercritical helium loop and circulating pump.

The results are presented as curves of Wc/Q (ratio of the work of

compression to the refrigeration load) vs the transmission line exit pressure

for various transmission line pressure and temperature drops. Contrary to

popular belief, all three cycles are competitive. Maximum efficiencies (percent

of Carnot) are about 25 percent.

Estimated capital and operating costs are presented as a function of

transmission line temperature for several values of thermal load per length

of line.

17. KEY WORDS (six to twelve entries; alphabetical order; capitalize only the first letter of the first key word unless a proper name; separated by semicolons)

Brayton cycle; refrigeration; superconducting transmission lines; superconductors.

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