Transcript Intro to Iron Dominated Magnets

INTRODUCTION to the DESIGN and
FABRICATION of IRONDOMINATED ACCELERATOR
MAGNETS
Cherrill Spencer, Magnet Engineer
HOMEWORK PROBLEM for
Lecture # 2 of 2: READ SLIDES 2
through 12 and then do problem
described on slides 13 and 14
Mexican Particle Accelerator School, October 2011
Deciding how many turns to wind in a
magnet coil- part 1
• ME has calculated the number of amp-turns , NI,
needed to produce the required field. Now ME has
to decide how to divide up the NI value between a
number of turns N and a current I amps
• Several parameters to consider as make this decision :
– If N is larger then I will be smaller, and a smaller
conductor can be used; but magnet resistance will be
higher, so voltage will be higher. Do the I and the V match an
available , off-the-shelf, power supply?
– If N is smaller then I will be larger, so larger conductor
needed, resistance smaller, voltage lower. Power cable
might be too large to be easily installed in the tunnel.
2
MePAS, Cherrill Spencer, Magnet Lecture #1
Guanajuato. 1st October 2011
Deciding how many turns to wind in a magnet
coil (continued)
• Suppose computer modeling shows need 1200 ampturns in
a magnet coil, NI=1200
– Could say 2 x 600 = 1200
or
6 x 200 = 1200
• The 600 amps is a large current by power cable
considerations so better to start with 6 turns and 200 amps.
• Choose a conductor size, is easier if must choose from a
restricted list of sizes; keep the current density below a limit
[see next slide] Also pay attention to the diameter of the cooling hole.
• Choose how to arrange the 6 turns, could be one layer of 6
turns, or 2 layers of 3 turns each for e.g.
– Calculate the total length of conductor in the chosen array, hence the
R and power and see if it can be cooled while following another rule
that we have in magnet design regarding increase in temperature of
the cooling water– see next page for details
3
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011
Magnet design standards & rules we
enforce in SLAC magnets, page 1/2
• Current density = current in amps/ conductor area
• Typical current densities we use in SLAC coils: 5000 Amps per
square inch ( 7.75A/sq mm) to 7500 Amps/ sq inch (11.63 A/sq mm)
– Applying some current density limit determines number of turns N in a
coil, because current density limit limits current – after choosing a
conductor size.
• Increase in temperature of cooling water to be less than 25°C.
Reason for this rule: turns in a coil are held together with special glue called
“epoxy”, it begins to soften around 90°C & will liquefy above ~ 120°C.
Use ΔT of exiting LCW as indication of maximum coil temp and say (Tin
+ ΔT) < 60°C
At SLAC the Tin of our LCW is about 35°C, so our max ΔT is 25° C.
Application of this rule determines how many separate water circuits need to
have in a coil, because if circuit is too long, then LCW passing through will
get too hot
4
MePAS, Cherrill Spencer, Magnet Lecture #1
Guanajuato. 1st October 2011
Magnet design standards & rules we
enforce in SLAC magnets, page 2/2
• LCW flow must be turbulent
• LCW water velocity in coil to be less than 15 feet per
second: important rule to avoid erosion of copper from
inside of conductor. Water travelling too fast can erode
away enough copper over time to create a hole in the
conductor wall -> water leak!
• NO internal brazes in a coil –cracked braze is most frequent
cause of magnet failures and if braze is inside coil is
impossible to mend. So insist on continuous conductor
between external brazes [this defines the length of a water
circuit]
5
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011
Some heat generated by the current in
the coil can be removed by flowing water
• Heat is transferred from the warm conductor to the LCW,
can develop some equations based on these assumptions:
– The water flow requirements are based on the heat capacity of the
water and assumes no temperature difference between the bulk
water and conductor cooling passage surface.
– With turbulent flow , the heat transfer through the thin film separating
the conductor surface and the bulk fluid flow is high
– The temperature of the cooling passage surface and the bulk
conductor temperature are the same. This is a good assumption
since we specify good thermal conduction for the electrical cond.
– With these assumptions need only compute the maximum exit temp
of the LCW to estimate the maximum coil temperature
– See next slide for equation that relates amount of water flowing
through a hot conductor and how much its temperature increases
6
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011
Equation relating power in coil, LCW flow and
increase in temperature
• The amount of heat (=electrical power) absorbed by
water flowing across a heated surface is given by
• P = m cp ΔT, where m = mass flow rate, cp = heat
capacity and ΔT= Temperature change
– Using values of cp and density for water, transform above
equation into one using a volume flow rate q:
• q (liters/sec) = 0.2388 P(kW)/ ΔT (°C)
• We want to know the ΔT for a given flow rate & P:
• ΔT °C = 3.8 x Power in kW /water flow in gallons/min
– Use this equation to validate parameters of coil design
7
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011
Example calculation of cooling a magnet with
LCW at a known pressure drop page 1/3
• NI = 1200 ampturns in a dipole coil
• Initial choice of N =6, and I =200 amps, calculate if
coil can be cooled while above rules are enforced?
• Arrange the 6 turns in 2 layers with 3 turns each.
• Choose 0.255” square conductor with 0.125”
diameter cooling hole. [=6.48mm sq with 3.175mm dia hole]
• Suppose length and width of pole-tip make total
length of conductor per coil = 12.2 feet. Not very
long so guess that need just one water circuit in the
whole magnet = 12.2 x 2 =24.4 ft
• .
MePAS, Cherrill Spencer, Magnet Lecture #2
8
Guanajuato. 3rd October 2011
Example calculation of cooling a magnet with
LCW at a known pressure drop page 2/3
• Suppose Facilities Department has told ME will be 80 psi
pressure drop across this dipole when its in the beam-line
• Total length of conductor in units of 100 ft = 24.4/100=0.244
• So pressure drop per 100 ft = 80/0.244 = 328≈ 330
• Now use the water flow curves to estimate how much water
will flow through this 24.4 ft of 0.125” diameter “tube”
• Look at 2nd water flow plot to find 330 on x axis, run finger
up that value until intersect the 0.125” diagonal line, then
run finger horizontally to left to see what water flow has
been reached: 0.5 gallons per minute
• Now calculate the resistance of the 2 coils from their
conductor length [continued on next slide].
9
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011
Example calculation of cooling a magnet with
LCW at a known pressure drop page 3/3
• Calculate resistance of magnet from resistance per ft, at 40
°C, of the 0.255”sq conductor, which is 170.9x 10-6 Ω per ft
• Magnet resistance = 12.2 x 2 x 170.9x 10-6 Ω = 0.0042 Ω
• Calculate power in the 2 magnet coils at 200 amps:
– Power = I2 R = 2002 x 0.0042 = 168 watts = 0.168 kW
• Calculate the increase in LCW temperature as it flows
through the 2 coils in series [use equation on slide 7]
– ΔT °C = 3.8 x Power in kW /water flow in gallons per minute
– ΔT °C = 3.8 x 0.168 / 0.5 = 1.28 °C
• Check if LCW temperature rise, ΔT, is less than 25 °C ?
– YES, in fact this dipole will run cool & hardly radiate any heat which is good.
– Check if water velocity is less than 15 ft per second :YES, is 13 ft/ second.
• SO this dipole’s coil design has been validated.
10
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011
1 gallon = 3.7854 liters.
1 atmosphere = 101352
Pascal = 14.696 pounds per
square inch [psi]
Water flow curves showing
relations between pressure
drop across pipes of
various diameters, the
volume of water flow per
minute and the velocity of
the water in the pipe.
Based on the Williams and
Hazen formula for smooth
drawn copper pipes.
Sorry, these are in imperial
units, tried to make some
metric curves: ran out of time
11
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011
1 gallon = 3.7854 liters.
1 atmosphere = 101352
Pascal = 14.696 pounds
per square inch [psi]
Water flow curves showing
relations between pressure
drop across pipes of
various diameters, the
volume of water flow per
minute and the velocity of
the water in the pipe.
Based on the Williams and
Hazen formula for smooth
drawn copper pipes.
Sorry, these are in imperial
units, tried to make some
metric curves: ran out of time
12
MePAS, Cherrill Spencer, Magnet Lecture #1
Guanajuato. 1st October 2011
Homework Problem for Magnet Lecture
#2 (read slides 2 – 12 first) pg 1/2
• Suppose have a quadrupole that needs 2250 ampturns per coil to produce the required gradient in
the aperture.
• As first attempt at designing the coils [all 4 will be
same style] decide to try using 24 turns/coil of
0.255”sq hollow copper conductor with a 0.125 inch
dia hole
• Suppose have calculated the total length of
conductor per coil to be 96 feet
• Suppose the Facilities Department will be providing
a pressure drop of 80 psi across the whole magnet
13
MePAS, Cherrill Spencer, Magnet Lecture #1
Guanajuato. 1st October 2011
Homework Problem for Magnet Lecture
#2 (read slides 2 – 12 first) pg 2/2
• Use the “WATER FLOW CURVES” sheet I handed out to
determine the volume of water per minute that will flow if
each coil received its own LCW [= 4 water circuits, one/coil].
• Calculate the ΔT of the water with the flow you determined
• Evaluate if the flow parameters and ΔT obey the rules on
slides 4 and 5, in which case this is a good coil and water
cooling design.
• Determine if this quadrupole magnet could be run with one
water circuit, i.e. the LCW flows through the 4 coils in series
before it exits? Evaluate if the new set of flow parameters
and new ΔT obey the rules on slides 4 and 5, or not.
14
MePAS, Cherrill Spencer, Magnet Lecture #2
Guanajuato. 3rd October 2011