Fill Time Estimate
Prepared:
Checked:
Approved:
Jiawei Mei
H. Nann
M. Snow
In order to reduce the time to liquefy the hydrogen gas and fill the target, and
thereby contribute to the target safety by reducing the odds of operator error, a
mechanical refrigerator will be used to accelerated the filling by pre-cooling the
incoming gas. The pre-cooler will consist of a CTI-Cryogenics 1020 mechanical
refrigerator with a CTI-Cryogenics 8500 compressor. This refrigerator will be
mounted to the 6-way cross which connects the main vacuum line vertical and
horizontal sections, see figure 1. A copper sleeve with direct thermal contact to
the second stage of the refrigerator will surround the hydrogen line to cool down
the stainless steel tubing and the incoming gas flow, and an aluminum radiation
shield will be connected to the first stage of the refrigerator to surround a portion
of the cooled fill/vent tube. The cooling power is 35 W at 77 K and 12 W at 20 K.
The temperature of the refrigerator will be monitored using diode thermometry.
None of these elements change the integrity/safety of the hydrogen line. There
are no new welded or other type of joints.
Fig. 1 Connection of the pre-cooling refrigerator to the 6-way cross of the fill/vent
line
The refrigerator will cool the hydrogen gas under the maximum flow rate of 10
SLPM to a temperature of about 90K with a copper sleeve of about 20 inch
length surrounding the hydrogen line. The estimated target filling time for this
case is about 24 hours. The cooling rate calculations are shown below.
We present now estimates for target filling times with no pre-cooler, onerefrigerator pre-cooler, and two-refrigerator pre-cooler.
Assumptions:
(1) The temperature of the target vessel is 20K.
(2) Cooling power of both CVI and Cryomech refrigerators is 12 W at 20 K.
(3) The incoming H2 gas is liquefied and fully converted into para-H2 immediately,
and then reaches steady-state.
(4) q      A T14  T24  0.01 5.669 10 8  0.276(300 4  20 4 )  1.3J Radiation heat
of the target vessel at 20K


(5) The following formulas (1 - 6) are used to calculate the heat flux through 1.5
in OD pipe.

Q  kA(Ti  To )
1


kA  Am
………………………………
(1)
………………………………
(2)
  ro  ri
A  Ai
Am  o
ln Ao Ai
Ao  2ro l
Ai  2ri l
……………………………… (3)
……………………………… (4)
……………………………… (5)
……………………………… (6)

Q
ri
ro
Ti
To
Heat flux through the pipes in [W];
The inner radius of the pipe in [m];
The outer radius of the pipe in [m];
Temperature at ri in [K];
Temperature at ro in [K];

Heat conductivity in [W/(m K)], 12.3217 [W/(m K)] for stainless steel;
The length of the tube in [m];
l
The mean thermal conductivity of stainless steel can be represented by the
following equation:
  T1  T2 1   T dT ,
T1
.
T2
 T   4.7127  1.4280T  (5.1049  10 )T  (6.5181 10 )T
3
2
6
(6) Total 16L liquid H2 in the vessel  13520 STL H2 gas.
1. Without Pre-cooler:
3
Assuming the H2 gas flow rate is x SLPM. Then by solving the following equation
we can get the maximum flow rate.
Q1  Q2  Q3  P  q ;
x
Q1  T  C p  m  (300  20)  14.314   0.08988 , 300K H2 gas  20K H2 gas
60
released heat;
x
Q2  m  H 
 0.08988  904 , 20K H2 gas  20K H2 liquid released heat;
2  60
x
Q3  p  H  m  (1  0.25)  300   0.08988 , 25% Para-H2  100% Para-H2
60
released heat;
P : The total cooling power, 24W;
q : Radiation heat of the target vessel, 1.3W, and heat conduction through the
fill/vent line (300K  20K), 2.1W;
The maximum flow rate is 3.1 SLPM  filling time is 72.2 hours.
2. With one Refrigerator:
When our filling speed is 10SLPM and the temperature of the pre-cooler is
around 90 K, we get the most optimal combination. See the following calculation.
1) Heat released at the pre-cooler
When the temperature of the pre-cooler is ~90 K, the power of the pre-cooler
refrigerator (CTI-1020) is ~47 W (Fig.2).
Fig. 2. CTI-1020 Cryocooler’s refrigeration capacity at different temperatures.
The heat released per second for 10 SLPM H2 gas from 300K to 90K is:
H cooling  T  C p  m  (300  90)  14.314  10  0.08988 / 60  45 J .
The heat radiation from the pre-cooler wrapped with superinsulation is ~ 1 W.
And the heat conduction from 300K SST tube to 90K tube through 10 in long
small cross section of SST tubing is 2 W. The sum of heat is ~48 W.
2) Heat released in the cryostat
Similarly, the total heat released has three contributions, one from 90 K H 2 gas to
20 K gas, one from 20 K gas to 20 K liquid, another from 25% para-H2 to 100%
para-H2. The total heat released by 10 SLPM H2 gas is about 21.78 J, plus the
radiation heat of the target vessel, 1.3 W, and heat conduction through the
fill/vent line (90 K  20 K), 0.1 W; the total heat released is close to the sum of
the two refrigerators at 20 K, 24 W.
With a maximum flow rate of10 SLPM the filling time is 22.6 hours.
3. With two Refrigerators:
When our filling speed is 12.4 SLPM and the temperature of the pre-cooler is
around 70 K, we get the most optimal combination. See the following calculation.
1) Heat released at the pre-cooler
When the temperature of the pre-cooler is ~70 K, the total power of the two
refrigerators is ~64 W.
The heat released per second for 10 SLPM H2 gas from 300 K to 70 K is:
H cooling  T  C p  m  (300  70)  14.314  12.4  0.08988 / 60  61.15 J .
The heat radiation from the pre-cooler wrapped with superinsulation is ~ 1 W.
And the heat conduction from 300 K SST tube to 70 K tube through 10 in long
small cross section of SST tubing is 2 W. The sum of heat is ~64 W.
2) Heat released in the cryostat
Similarly, the total heat released has three contributions, one from 70 K H 2 gas to
20 K gas, one from 20 K gas to 20 K liquid, another from 25% para-H2 to 100%
para-H2. The total heat is around 22.67J, plus the radiation heat of the target
vessel, 1.3 W, and heat conduction through the fill/vent line (70 K  20 K), 0.1
W; the total heat released is close to the sum of the two refrigerators at 20 K, 24
W. With a maximum flow rate of12.4 SLPM the filling time is 18.2 hours.
Because of the extra complexity of adding two refrigerators which leads to only a
marginal increase in the cool-down time, we propose to include only one
refrigerator in the line.
Estimate of the length of cooled hydrogen line needed to cool the gas
Assumptions:
(1) Flow speed = 10 SLPM;
(2) when filling, the pressure in the vessel = 1.5 bar;
(3) the tubing wall temperature = 90 K;
(From the previous filling time calculation, we know the most optimal combination
is 10 SLPM flow rate plus 90 K pre-cooler. And the heat released from 10 SLPM
H2 gas at 300 K cooled down to 90 K is 45 J/s.)
Table. 1
Hydrogen properties at different temperatures
Hydrogen Properties:
T[i]
ρ
Cp
3
(K)
(kg/m ) [kJ/(kg ℃)]
k
[W/(m ℃)]
30
50
100
150
0.0228
0.0362
0.0665
0.0981
0.84722
0.50955
0.24572
0.16371
10.840
10.501
11.229
12.602
α x 104
(m2/s)
μ x 105
[kg/(m
s)]
0.02493 0.1606
0.0676 0.2516
0.2408 0.4212
0.475
0.5595
ν x 106
(m2/s)
Pr
1.895
4.880
17.14
34.18
0.759
0.721
0.712
0.718
200
250
300
0.12270 13.540
0.09819 14.059
0.08185 14.314
0.1282
0.1561
0.182
0.772
1.130
1.554
0.6813
0.7919
0.8963
55.53
80.64
109.5
Table 1 shows hydrogen properties at different temperatures which are used in
the calculation. When the temperature T is between T[i] and T[i+1], formula
(T  T [i]) X [i  1]  (T [i  1]  T ) X [i]
is used to determine the corresponding
X 
X [i  1]  X [i]
properties.
G
m
,
Ac
Re 
DG

,
D
),
L
0.0668Gz
N u  3.657 
2 ,
1  0.04Gz3
Gz  Re Pr(
…………………………
(7)
Nu k
,
D
Q  hc Aw (TAve  Tw ),
m
mass rate in [kg/s],
Ac
tubing cross section in [m2],
D
inside diameter of tube in [m],
μ
viscosity in [Pa-s],
Re
Reynold number,
Gz
Graetz number,
L
tubing length in [m],
Pr
Prandtl number,
Nu
Nusselt number,
k
thermal conductivity in [W/m-K],
hc
heat transfer coefficient in [W/m2-K],
Aw
inside surface area in [m2],
TAve average temperature of gas flow in [K],
Tw
tubing wall temperature in [K],
Q
heat transfer rate in [W].
hc 
Q  mC p (Tin  Tout ),
Cp
Tin
Tout
…………………………
heat capacity in [kJ/kg-K],
incoming gas temperature in [K],
outgoing gas temperature in [K].
(8)
0.719
0.713
0.706
Fig.3 H2 gas flow rates for different cold tube lengths
Assuming 10 SLPM H2 gas at 1.5 bar and 300 K flows inside a 1.5 in OD tubing
having an inside diameter of 1.37 in (34.8 mm) with a tube wall temperature of 90
K. By using formulas (7) and (8), we can get the outgoing gas temperature after
passing a given length of cold tubing. In the calculation, T in is used to be instead
of Tave. In order to get a more accurate estimation, the tubing length for the H2
gas passing at each step is chosen to be 1 in (25.4 mm). The T out of the previous
step is the Tin of the next step and so on. At each step, the hydrogen properties
at Tin of this step were used. Please see Table 2 and Fig 3.
Table. 2
L(in)
T(K)
L(in)
T(K)
0
300
8
101.15
1
212
9
97.49
2
176.72
10
95.63
3
148
11
93.77
4
132.14
12
92.27
5
118.95
13
91.52
6
110.88
14
90.76
7
104.77
After passing a 14 in long 90 K tube, the temperature of H2 gas is close to 90 K.
The heat transfer through the SST tube and copper cylinder is also estimated to
see how cold the cold head needs to be to maintain the SST tubing at 90 K. I
assumed a copper cylinder with 0.25 in wall thickness surround the 1.5 in outer
diameter SST tube. By using formula (9), I got the temperature the refrigerator
(9)
Q  Sk m (TH  TC ),
2L
,
S = Conduction shape factor, for hollow cylinder S 
 Do 

Ln
 Di 
By solving the equation
Q
TH  TC
,
1
1

Sk1 Sk 2
Q  45W ,
TH  90 K ,
L  14in  0.356m,
We get  Tc = 89.95 K.
The heat radiation from the 14 in long copper cylinder wrapped with
superinsulation is ~ 1 W. The heat conduction from the 300 K SST tube portion to
the 90 K tube portion through a 10 in long small cross section of SST tubing is 2
W. The sum of heat ((45+1+2) W) is close to the power of the refrigerator at 90 K
(47 W).
Conclusion:
1) The power of one refrigerator at 90 K is enough to cool down 10 SLPM H 2
from 300 K to 90 K.
2) An at least 14 in long 90 K tube with inside diameter 1.37 in is required to cool
down the gas flow to 90 K. To be safe, we will choose a longer length.