The purpose of this evaluation is to calculate the cooldown time for
upstream and downstream strings and to determine the limiting factors
for the cooldown time.[]{#assumptions .anchor}

(Click here to go the [**top**](#assumptions), or
[**conversion**](#conversion) calculations, [**cooldown**](#cooldown)
calculations, or [**conclusions**](#conclusion))

[Cooldown Mode Specification]{.underline}

  ---------------------------------------------------- ------------------
  Refrigerator Capacity at 2K                          2400 W

  Refrigerator Capacity at 6K                          1500 W

  Refrigerator Capacity at 50K                         8000 W

  Liquefaction coefficient                             \~ 5%

  Temperature Primary                                  2.1 K

  Temperature Secondary                                5 K

  Temperature Shield                                   35 K

  Pressure Primary                                     0.041 bar

  Pressure Secondary                                   3.0 bar

  Pressure Shield                                      4.0 bar

  Total Heat Load at 2K                                1400 W

  Upstream String Low Beta cavities                    16 (2 x8)

  Upstream String Medium Beta cavities                 24 (3 x8)

  Upstream String High Beta cavities                   56 (7 x8)

  Upstream String TESLA cavities                       80 (10 x8).

  Refrigerator Feed Box                                1

  Downstream String TESLA cavities                     240 (30 x8)

  Upstream string length                               311 m

  Upstream string length                               480 m

  Average cavity mass                                  250 lb/m

  Static heat load per cryo cavity for Low/Medium/Beta 2.0 W/m @ 2K

  Static heat load per cryo cavity for TESLA           2.0 W/m @ 2K
  ---------------------------------------------------- ------------------

The assumption is that the cryo flow schematics will be similar of the
TESLA design (Figure 1).

![](media/image1.png){width="6.497916666666667in" height="3.40625in"}

> Figure 1

In this design, helium at variable temperatures is supplied from the
refrigeration plant to the 2.2K helium supply line. There are two
possible scenarios for cooldown configuration:

-   Expand helium through JT valves to 2-phase line and flow it to each
    cavity. Though this scenario has the least flow resistance, it may
    be inefficient since helium boil-off gas has to exit from cavities
    through the same connection and then exit to 300 mm GRP.

-   []{#scenario .anchor}Flow helium to warm-up/cooldown line, and then
    to the cavities via thin 3 mm capillary tubes. In this scenario, the
    flow resistance is much greater, but the flow is unidirectional and
    uniformly distributed for all cavities connected to the cold line.
    The 500 mm long and 3 mm diameter tube can support up to 7 g/s flow
    of 2.2K helium from 1.3 Bar to 1 Bar, or 2.5 g/s flow of 4.5K
    helium, or 0.45 g/s of 80K helium, or 0.2 g/s of 300K helium. There
    are 176 cavities in the upstream string and 248 cavities in the
    downstream string. As shown later, the maximum available flow from
    the refrigeration plant is less then 50 g/s at 4.5K. Therefore the
    flow per cavity is expected to be less then 0.3 g/s. Therefore, we
    will consider this scenario feasible and presenting no restriction
    for the cooldown flow. (Click here to go the [top](#assumptions), or
    [cooldown](#cooldown) calculations, or [conclusions](#conclusion))

![](media/image2.wmf)

> Figure 2

The quintessence of the model is that the cooldown wave of P~1~ = 1.3
Bar and 4.5K or 2.2K goes from the refrigerator plant through the 2.2K
supply line, cooldown line, and then through *n* cryo modules (CM) via
cavities to 2-phase line, 300 mm GRP, and finally 6" suction header. The
cooldown valve is located at the end of the section of n cryo modules
and vents the 300K helium directly to the 6" return suction header. The
\[4.5K -- 300K\] cooldown wave is normally \~5 m long. In this
configuration, the cooldown line must be cooled down first through a
valve located on its end, and then all cavities will be simultaneously
cooled down via capillary lines. If there is no valve installed at the
end of the cooldown line, then the cavities will be cooled down
semi-sequentially. The pressure drop related to cooldown flow within the
cavities and GRP is considered to be negligible. Thus the only limiting
factor may be a flow through the cooldown valve. [(click here to jump to
cooldown results)](#cooldown)

The cooldown time can be calculated if an available cooldown flow from
the refrigeration plant is known. That flow is limited with the plant
capacity at given temperature and with the pressure drop in the cooldown
path. At this point of the design study, we can make a few assumptions
for both, the plant capacity and the pressure drop.

1\) We assume that the cooldown flow will be limited with the pressure
drop in the "warm" section of the strings. The length of the "warm"
section will be decreasing with the cold wave propagating from the plant
supply to the end of the cooldown path. The first estimate is that the
cooldown valve capacity may be the limiting factor for the cooldown
flow. Thus we calculate the flow through a cooldown valve installed at
the end of *n* modules. The flow at P~1~ = 1.5 Bar and 300K to the 1 Bar
suction header can be estimated as:

Equation 1

![](media/image3.wmf)![](media/image4.wmf) 1,000 kg/hr, or
![](media/image5.wmf) g/s

where: P~1~ in Bar Abs

C~v~ = 34 (1-1/2" CryoLab 21000 Series), 50 (2")

G~f~ = 0.138, specific gravity He \@300K

C~f~ = 0.8, critical flow factor for split-body globe valve

[For 2" valve and P~1~ = 1.5 Bar -\> flow = 110 g/s; for 1.5" valve -\>
flow = 75 g/s]{.underline}. These results are close to the results
calculated with Crane^©^ DFS Software.

2\) In our initial estimate, we assume that the cryo plant is designed
to provide 2,400 W of refrigeration @ 2K. To convert 2,400 W of
refrigeration at 2K to *Watts* of refrigeration at 4.5K, we use equation
2:

Equation 2

![](media/image6.wmf) Watts

In order to provide a cooldown wave, the refrigeration plant must work
as a liquefier supplying 4.5K helium to the magnets and accepting 100%
of the mass flow back to the compressor suction at 300K. The following
flow and T-S diagrams for the ideal refrigeration (Figure 1) and
liquefaction cycles (Figure 2) will serve to clarify the method of
conversion between *g/s* of liquefaction and *Watts* of refrigeration.
[(click here for conclusions)](#conclusion)

For an ideal refrigeration cycle between T~h~ and T~x~ designed to
remove heat from a thermo-load by using its latent heat, a minimal
specific work (J/kg) is described with equation 3 (for an ideal Carnot
cycle) or equation 4 (see Figure 3):

Equation 3

![](media/image7.wmf), J/kg

Equation 4

![](media/image8.wmf), J/kg

![](media/image9.wmf)

> Figure 3

For an ideal liquefaction cycle between T~h~ and T~x~, a minimal
specific work (J/kg) is described with equation 5 (see Figure 4):

Equation 5

![](media/image10.wmf), J/kg

![](media/image11.wmf)

> Figure 4

Then equation 6 is used to describe a ratio of specific work
requirements for ideal liquefaction and refrigeration cycles with
identical ![](media/image12.wmf)described by equation 7:

Equation 6

![](media/image13.wmf)

Equation 7

![](media/image14.wmf), J/kg

Finally, we calculate an ideal power requirement for a liquefaction of 1
g/s between T~h~ = 300K and T~x~ = var.

[]{#conversion .anchor}(Click here to go the [top](#assumptions), or
[conversion](#conversion) calculations, [cooldown](#cooldown)
calculations, or [conclusions](#conclusion))

![](media/image16.wmf)Finally, assuming that conversion coefficient from
watts of refrigeration to liquefaction capacity is 120 W per 1 g/s of
flow, we have the following liquefaction capacity at 4.5K for the
refrigeration plant:

Equation 8

![](media/image17.wmf) g/s

That is probably the maximum available liquid helium flow that could be
supplied by the cryo plant. Therefore, a single 2" cooldown valve does
not restrict all mass flow from the refrigeration plant.

In order to calculate a cooldown time, we assume a simple model for the
cooldown of a string of cryo modules.

1.  2

![](media/image18.wmf) Material mass = m~w~ ![](media/image19.wmf)
![](media/image20.wmf)

![](media/image21.wmf) ![](media/image22.wmf)

Assuming that the cooldown goes from t~0~=0 to t~ss~, which is required
to be estimated. The inlet of the system is denoted as "1", the outlet
as "2". The initial conditions are denoted as "0", the final steady
state conditions are "ss". The 1^st^ law applied to the system within
the cryo modules:

Equation 9

![](media/image23.wmf)

where: Q = total heat transferred to the system (from metal to cryogen),
J

m = mass, kg

![](media/image24.wmf) = mass flow, kg/s

u = internal energy, J/kg

h = enthalpy, J/kg

c~w~ = specific heat of the material

At the same time the same total heat can be written as:

Equation 10

![](media/image25.wmf)

Now we assume that the rate of the heat leak is one-half of the final
rate of the heat leak (in our case leak = 2W for TESLA modules, and 4.5W
for the rest of the modules at 2K; we assume that the rate will be the
same at 4.5K). Therefore, equation 10 can be written as equation 11:

Equation 11

![](media/image26.wmf)

Conservation of mass produces:

Equation 12

![](media/image27.wmf) , where: ![](media/image28.wmf) is the mean mass
flow rate of vapor from the string.

Therefore, equation 9 can be written in the following way:

Equation 13

![](media/image29.wmf)

The mass of the liquid into the string can be eliminated using equation
12. Then equation 13 is transformed into equation 14:

Equation 14

![](media/image30.wmf)

The initial and final mass of the cryogen within the string can be
written in terms of the volume V of the line and density:

![](media/image31.wmf) and ![](media/image32.wmf), where:
![](media/image33.wmf)

Finally solving for the cooldown time t~ss~, we have:

Equation 15

![](media/image34.wmf)

Finally, using equation 15, it is possible to estimate the cooldown time
for upstream and downstream cryo modules. The cooldown assumes that 4.5K
helium at 1.5 Bar is flown via supply line to the distribution box and
further through the cavities back to GRP and refrigerator suction at 1
Bar. The assumption is that the cold wave will be propagating along the
string with most of the helium boiling and expanding via GRP as outlined
[earlier](#scenario) in this write-up. As the cavities get cold, the
fraction of the helium will flow through 3mm capillary tubes to the
warm-up line and back to GRP. The first approximation for the cooldown
time consists of three stages:

-   Cooldown from 300K to 4.5K. Refrigeration plant is in the liquefier
    mode. All boil-off gas returns back via GRP and 6" suction header to
    the plant suction at 1 Bar (24 hrs)

-   Fill at 4.5K and 1 Bar. Refrigeration plant is in the refrigeration
    mode (3 hrs)

-   Cooldown and fill at 2.1K and 0.043 Bar. Refrigeration plant is in
    the refrigeration mode (13 hrs)

> []{#cooldown .anchor}The calculations show that both, upstream and
> downstream cryo modules can be cooled down from 300K to 4.5K within 24
> hours with additional 16 hours to cool and fill the strings with 2.1K
> saturated helium at 0.043 Bar. It is interesting to note that the time
> needed to cooldown \~700 tons of TESLA cryo modules to 2.1K was
> calculated to be 240 hrs (10 days) with additional 60 hours to fill
> the helium vessels with liquid[^1]. A direct scaling from 700,000 kg
> of TESLA to 92,000 kg of 8 GeV Linac produces cooldown time of 40 hrs
> (2.1K). Seems to be a close match, though TESLA has a much lower ratio
> of cooldown to fill time compare to the ratio calculated for the 8 GeV
> linac. (Click here to go the [top](#assumptions), or
> [conversion](#conversion) calculations, or [conclusions](#conclusion))

![](media/image35.wmf)

The cooldown valve flow capacity does not seem to be a limiting factor.
It is the cooldown flow from the cryo plant that would limit the speed
of the cooldown. If the cooldown plant has 45 g/s at 4.5K, then each
string (upstream and downstream) requires a 1-1/2" cooldown valve at the
end of a string to provide for simultaneous cooldown of both strings. It
is possible to assume, though, that a cooldown wave of 35K could be set
to speed up an overall initial cooldown of the accelerator. In this
case, higher flow may be provided from the cryo plant. Therefore, it is
prudent to install two 2" cooldown valves for each string.

A cooldown scenario could possibly be considered utilizing LN2-to-GHe
heat exchanger to precool full compressor capacity of the refrigeration
plant (approximately 0.4 kg/s). The ratio of LN2 flow required to cool
He from 300K to 82K is 2.7:1. Therefore, \~1.1 kg/s of LN2 is needed for
precool (or equivalent of Fermilab NRL production). The cost of such
precooler can be close to \$1M capital cost[^2], and the LN2 cost from
vendor is \~\$6,200/day (at \$0.00236/scf).

The warmup and cooldown scenario for one string (upstream or downstream)
is evaluated using initial assumptions for the cryo system and assumed
cooldown capacity for the plant at 2.1K. The assumption is that the
refrigeration plant is switched to 4.5K refrigeration mode for the
duration of the warmup, repairs and initial cooldown. Therefore, the
warmup and cooldown consist of four stages:

-   Warmup of both strings from 2.1K to 4.5K.

-   Warmup of the string to be repaired from 4.5K to 300K.

-   Cooldown of the repaired string from 300K to 4.5K.

-   ![](media/image36.wmf)Cooldown and fill of both strings from 4.5K to
    2.1K.

![](media/image37.wmf)

Assuming the warmup time is equal to the cooldown time, and additional
15% contingency for plant reconfiguration, the total time for warmup and
repair is estimated as 75 hours, or 3 days. The repair time certainly
has to be added to calculate a total time needed to replace an element
of a cryo module.

If a warmup, repair and cooldown of a string are conducted while keeping
the other string and the plant at 2.1K with cold compressors running
(GRP should be at 0.043 Bar and warmup flow should be directed to the
suction manifold), then the cooldown is estimated as 36 hours. Then the
total time is again \~75 hours, or 3 days.

[]{#conclusion .anchor}**[Conclusions]{.underline}:** [(click here to go
to the top of the document)](#assumptions)

-   **Upstream and downstream cryo modules can be cooled down from 300K
    to 4.5K within 24 hours with additional 16 hours to cool and fill
    the strings with 2.1K saturated helium at 0.043 Bar.**

-   **It is recommended to install two 2" cooldown valves for each
    string.**

-   **A total time for warmup and repair of a cryo module of either
    string is estimated as 75 hours.**

[^1]: The TESLA Cryogenic Distribution. S. Wolff et al, H. Quack, TESLA
    Report 2001-37.

[^2]: Specification for RHIC 80K Cooler, SPEC.CAD-1121, 2001, BNL