Boston, MA 6/14 to 6/18/2010USPAS Short Course 1
3.1 Cryogenic
Fluid Mechanics
Fluid flow commonly occurs in most cryogenic systemsRefrigeration systems: gas cycles including flow through heat exchangersFluid distribution systems: transfer lines, actively cooled thermal shieldsNatural and forced circulation systems for large devices (magnets) Forced flow superconductors for large magnets
Cryogenic systems frequently experience a variety of fluid flow conditions:
Single phase, subcooled liquid (incompressible)Compressible fluid flow (gases)Two phase flow (liquid + vapor co-existing)
Most fluid dynamics issues are “classical” in nature although compressible and two phase flow are common
Boston, MA 6/14 to 6/18/2010USPAS Short Course 2
Typical 1 D flow problem
Input variables: m (mass flow rate), Q (heat load), inlet T & PPhysical dimensions: length (L), tube diameters, other components (flow meters, valves, tube bends, etc.)Output Variables: outlet pressure & temperature, phasesFluid dynamics problem: determine the pressure drop (Δp) and adiabatic temperature change under given conditionsHeat transfer problem: determine the temperature of the fluid and tube wall for given heat transfer rate, Q. The heat transfer coefficient (h) vs. flow conditionsIn most cases, these two problems are coupled (i.e. must be solved simultaneously)
Δp
Q
m.
m.
Valve Orifice flow meter
Elbow Contraction
.
Boston, MA 6/14 to 6/18/2010USPAS Short Course 3
One phase, incompressible flow
Pressure drop:
fd is the “Fanning Friction Factor”, which depends on Reynolds number:
for laminar flow ReD < 2000
for turbulent flow 2000 < ReD < 10,000 (Blausius)
for turbulent flow 10,000 < ReD
Rough tubes:
value depends on ratio (k/D = amplitude of roughness/diameter)
⎟⎠⎞
⎜⎝⎛=Δ 2
214 u
DLfp
hd ρ
PA
D flowh
4=where
Hydraulic diameter
Ddf Re
16=
41
Re
0791.0
D
df =
( ) 396.0Reln737.112
1
21 −= dD
d
ff Von Karman-Nikuradse
Smooth tube correlations:μ
ρ hD
uD≡Re
⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
CDC fDk
f Re25.1
7.3log41
10Colebrook correlation for Re > 10000 and rough tubes
Boston, MA 6/14 to 6/18/2010USPAS Short Course 4
Fanning Friction FactorFr
ictio
n Fa
ctor
(fd)
Reynolds number = ρuD/μ
From: Bird, Stewart & Lightfoot, Transport Phenomena
Note: Darcy friction factor (ME) = 4 x Fanning friction factor (ChE)
Boston, MA 6/14 to 6/18/2010USPAS Short Course 5
Loss coefficients for components
⎟⎠⎞
⎜⎝⎛=Δ 2
21 uKp ρ
Similar to the tube pressure drop expression, with “K”
incorporating all hydraulic losses
ComponentsValvesElbowsTeesFlow meters
Comparison to flow in tube: K ~ 4f(L/Dh)Equivalent length of 2” 90 elbow:Leq
~ KD/4f= 0.95x2/4x0.005~ 100”
(2.5 m)
Boston, MA 6/14 to 6/18/2010USPAS Short Course 6
Compressible fluid dynamics
Liquids are incompressible and in many cases one can approximate gas flows with incompressible expressions assuming average properties. When is this OK? (v/c << 1 and Δp/p << 1, ΔT/T << 1)For some problems, it is necessary to include compressibility effects, such that the the density, ρ = Function (p,T)
For example, as the heat is applied to the flow, the temperature will increase resulting in an acceleration of the fluid.Also, isenthalpic flow (Q = 0, W = 0) of real gases can increase (or decrease) the temperature of the fluid (Joule-Thomson Effect)
.m
Q D
Boston, MA 6/14 to 6/18/2010USPAS Short Course 7
Compressible fluid dynamics (cont.)
Pressure drop:
Density depends on T, p through the equation of state for the fluid. These are related as:
Since the pressure drop depends on u and dT/dx, we need a second equation to solve for 1-D flow. That is the “Stagnation Enthalpy”:
dxduu
Df
dxdp d ρρ 22
214 +⎟
⎠⎞
⎜⎝⎛=
dxdT
dxdp
dxd ρβρκρ
+=
( ) ( )dxdTuCuD
dxdpuCuDq pjp βρκμρ 22
44−++−=
Where,hp
j pT
C ⎟⎟⎠
⎞∂∂
−=1μ is the Joule-Thomson coefficient
(1)
(2)
Equs. (1) and (2) are simultaneous equations with unknowns T & pV. Arp, Adv. in Cryo Engn. Vol 17, 342 (1972)
friction acceleration
Boston, MA 6/14 to 6/18/2010USPAS Short Course 8
Simultaneous solution of fluid equations
Separation of the two simultaneous equations yields,
( )( )( )βφκρ
βρβρ+−
−+−= 2
22
142
GuCDqGDGf
dxdp pd
( )( )( )βφκρ
ρφβ+−
−−= 2
22
124
GDGfuCGDq
dxdT dp
βκμ
φ 2
2
uCuC
p
pj
−
+=
(1’)
(2’)
Where we have defined the parameter,
Note that the dominant term in Eq. (1’) is the friction (2fG2/ρD) and in Eq. (2’) is the enthalpy flux (4q/GDCp )
AmG&
≡and
Boston, MA 6/14 to 6/18/2010USPAS Short Course 9
Approximate solution for sub-sonic flow
For nearly ideal gases, κ ~ p-1 and β ~ -T-1 and for relatively slow velocity, u << c (sound speed) the above two equations simplify,
It is straightforward to prepare a program to calculate dp/dx and dT/dxin a 1-D channel containing a compressible fluid of known propertiesNote that the 2nd term in the dp/dx equation (acceleration) has the effect of increasing the pressure gradient relative to the incompressible form (note that β is negative)The 2nd term in the dT/dx equation (isenthalpic expansion) can either increase or decrease the gradient relative to the incompressible solution depending on the magnitude and sign of μj (κ is positive)
jd
p DGf
GDCq
dxdT μ
ρ
224−≈
p
d
DCqG
DfG
dxdp
ρβ
ρ42 2
+−≈ (1”)
(2”)
Boston, MA 6/14 to 6/18/2010USPAS Short Course 10
Example: Compressible flow in a tube (1)
Supercritical heliumTube length = 500 mD = 4.8 mmm_dot = 0.98 gm/sQ = 0.074 W/mSolid lines are computed using compressible fluid mechanics
f = 0.007
J.W. Dean, et al, Adv. Cryog. Engn. 23, 250 (1978)
Note: this value of f was used to best fit the data and is slightly different from value predicted from correlation.
To compare to incompressible solution, choose average values for
T & p.
Boston, MA 6/14 to 6/18/2010USPAS Short Course 11
Length = 500 mD = 4.8 mmm_dot = 3 gm/sQ = 0.062 W/mSolid lines are computed using compressible fluid mechanics
f = 0.005
Example: Compressible flow in a tube (2)
Example
Boston, MA 6/14 to 6/18/2010USPAS Short Course 12
Two phase flow
Two-phase flow is an important topic in cryogenics since the liquidsare volatile and are commonly distributed near saturation conditionsTwo-phase flow is a difficult problem as there are many variables that affect it:
Mass flowPressure & temperature relative to saturationGravity, system configurationHeat transfer, rate of phase change
This is mostly an empirical subject due to its complexityNumerous two-phase flow models exist, although much of this work has been carried out on conventional fluids (water/steam, water/air)
Boston, MA 6/14 to 6/18/2010USPAS Short Course 13
Flow Regimes (Baker)
All these flow regimes may occur in a horizontal pipe flow. The actual flow regime will depend on fluid velocities and properties.
Boston, MA 6/14 to 6/18/2010USPAS Short Course 14
Baker plot for air-water system
Boston, MA 6/14 to 6/18/2010USPAS Short Course 15
Two-phase flow definitions
1.
Void Fraction (α): Ratio of the local vapor volume to the total flow volume
2.
Flow Quality (χ): Ratio of the vapor mass flow rate to the total mass flow rate
3.
Slip Ratio (S): ratio of the vapor to liquid velocity
vl
v
AAA+
=α
vl
v
mmm&&
&
+=χ
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −
⎟⎟⎠
⎞⎜⎜⎝
⎛−
==v
l
l
v
uuS
ρρ
αα
χχ 1
1
Boston, MA 6/14 to 6/18/2010USPAS Short Course 16
Pressure drop in two phase flow
Due to change in elevation (h):
Acceleration of the fluid stream:
Friction pressure drop depends on model for flow regime:
fagrT pppp Δ+Δ+Δ=Δ
dxgph
gr ∫−=Δ0
ρ
( )( )
( )( )
2
222
1
222
11
11
⎥⎦
⎤⎢⎣
⎡−−
+−⎥⎦
⎤⎢⎣
⎡−−
+=Δlvlv
a GGpρα
χαρχ
ραχ
αρχ
( )( )l
l dxdpdxdp φφ 22 ≡Define the two phase friction multiplier:
Boston, MA 6/14 to 6/18/2010USPAS Short Course 17
Homogeneous two phase flow model
Assumptions (mixed flow model):S = 1; i.e. the liquid and vapor velocities are equalThermodynamic equilibrium exists between the two phasesThe friction contribution to the pressure drop reduces to the standard expression for each phaseModel works most effectively for helium at high Reynolds number (ρv/ρl is typ. ~ 0.1)
⎟⎠⎞
⎜⎝⎛=Δ 2
214 u
DLfp d ρ where fd
is given by an appropriate correlation
41
Re
0791.0
D
df =For example:The two phase flow friction multiplier:
lllll f
fuu
pp φφφ
ρρ
φ 22
2222 =
Δ
Δ=
41
2 1111−
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎥
⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+=
v
l
v
ll μ
μχρρχφ
Using Blausius
correlation and ul = uv
Boston, MA 6/14 to 6/18/2010USPAS Short Course 18
Lockhart Martinelli
Correlation
Assumptions (separated flow model)Static pressure drop of two phases are equalFluid volume is a linear combination of two phasesOnly applies to friction contribution to pressure dropWorks well for nitrogen & hydrogen at moderate flow rates
Correlating parameter( )( )
22 1
⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛==
χχ
ρρ
vl
lv
v
l
ff
dxdpdxdpX
( ) 41
41
0791.0
Re
0791.0
lllD
lDu
fμρ
==where, for example:
X2 is the ratio of the pressure drop for the pure liquid/vapor phases.
The two phase flow friction multiplier is then given as a correlation in terms of the factor X2
)) 2
222 1
XCXX
dxdpdxdp
ll
++== φφ Where C = 20 if both phases are turbulent
Boston, MA 6/14 to 6/18/2010USPAS Short Course 19
Homogeneous Model vs
LM Correlation
helium nitrogen
P. S. Shen & Y. Jao, Advances in Cryo. Engn. 15 , 378 (1970)
⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=Χ
χχ
μμ
ρρ 1
145.0571.0
v
L
L
vttNote:
Boston, MA 6/14 to 6/18/2010USPAS Short Course 20
Natural circulation loops
Coil cooling system for CMS detector magnet at CERN-LHC
Boston, MA 6/14 to 6/18/2010USPAS Short Course 21
Natural Circulation Cooling Loop
Natural circulation loops are very useful in many cryogenic applications since they do not require a pump or other device to force the fluid flowCommon applications include:
Cooling shields for large cryostats (NHMFL magnets)Indirect cooling of large superconducting magnets such as the detector magnets for particle accelerators
These cooling loops work based on the balance between the hydrostatic head difference as the driving pressure and the friction and acceleration pressure drop
For a simple loop (right), the driving pressure is given by the difference in hydrostatic head on the two legs of the loop
The density varies with z because Q generates vapor and Δpd increases with Q
afd ppp Δ+Δ=Δ
∫−=ΔH
ld dzzggHp0
)(ρρ
H
Q
m&
m&
Boston, MA 6/14 to 6/18/2010USPAS Short Course 22
Pressure drop through the loop
There are several contributions to the pressure drop in the natural circulation loop:
1.
Single phase on the liquid side:
2.
Adiabatic two phase flow:
3.
Diabatic
two phase flow:
4.
Acceleration pressure drop:
hl DALmfp 2
12
1 2ρ&
=Δ
( )exphl
Ap DA
Lmfp χρ 22
22
2 2Φ=Δ
&
( ) χχχρ
χ
dDA
Lmfpex
pexhl
Dp ∫Φ=Δ
022
32
21
2&
( ) exlga vvAmp χ−=Δ 2
2&
The above equations relate to the total mass flow through the loop. However, the vapor mass flow, mv is determined by the heat rate, Q. Where,
fgv h
Qm =& mmv
ex &&=χand
L2
Q
m&
m&
L3L1
vg and vl are the specific volumes of the gas and liquid
.
Boston, MA 6/14 to 6/18/2010USPAS Short Course 23
Natural circulation loop test (helium)
Flow loop is ~ 5 mm ID and heated over part of its length.Simultaneous analysis of fluid dynamic equationsObservation: mass flow almost independent of Q from 2 to 20 W
Increasing Q increases χ and Δpd
Δpf increases due to frictionVery stable operationModel predicts behavior
L2
Q
m&
m&
L3L1
Boston, MA 6/14 to 6/18/2010USPAS Short Course 24
Design of LN2
cooled shield for 900 MHz NMR magnet
900 MHz NMR magnet is a large superconducting system under development at the NHMFLThe cryostat is to have high thermal efficiency. A critical area is the inner warm bore, where the space between 300 K and 2 K (operating temperature) is smallTo reduce the heat load at 2 K, there is a thermal shield between 300 K and 2 K that is operating at LN2 temperature The heat load on the LN2 shield is sufficiently high that it must be actively cooled with liquid along its lengthLN2 is supplied to the shield from the LN2reservoirAnticipated heat load on the inner bore is about 20 W
Boston, MA 6/14 to 6/18/2010USPAS Short Course 25
LN2
shield design
Shield is cooled by two tubes running axially along the length, ~ 3 mThe supply line to the cooling tubes is large diameter, so one can neglect the pressure drop in the inlet sideCircumferential conduction heat transfer maintains uniform temperature around shield. Used copper screen.Physical dimensions:
Shield diameter 100 mmLength = 3 mThickness < 5 mm
A schematic of the simplified system is given:
Assumed hydrodynamic conditions:Turbulent flow throughoutTwo phase flow on return leg onlyHomogeneous flow conditions Q
3 mm_dot
LN2
Copperscreen
Coolingtubes
Boston, MA 6/14 to 6/18/2010USPAS Short Course 26
Model AnalysisFlow rate through the cooling loop is dependent on the balance between
hydrostatic head difference and friction pressure drop. The following equations apply:
1.
Mass flow of the vapor:
~ 20/200 J/g ~ 0.1 g/s and
2.
Driving pressure head:
where
and χ = mv
/m is the return vapor quality
substituting for the average density,
3.
Friction loss due to adiabatic two phase flow
where f is the friction factor, A = flow area and dh
And for the homogeneous model;
fgv h
Qm =&
dzggHpH
ld ∫−=Δ0
ρρ
vl ρχ
ρχ
ρ+
−=
11
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+−=Δ
−1
11 χρρρ
v
lld gHp
( )χρ p
hlf dA
mfHp 22
22Φ=Δ
&
41
22 1111
−
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎥
⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+=
Δ
Δ=Φ
v
l
v
l
l
pp p
pμμχ
ρρχ
fghQm
χ=&
Boston, MA 6/14 to 6/18/2010USPAS Short Course 27
Two phase flow summary
Clearly, two phase flow is a complicated process given the number of variables involved:
Mass flow rate, pressure drop, heat transfer rate, temperatureVoid fraction, flow quality, flow regime, orientation
“Words of Wisdom”Avoid creating “traps” that can collect vaporReturn two phase liquid above supply surface (phase separation)Avoid parallel tubes with different hydraulic characteristics
m1m4 m3 m2
Δp
is the same for all channels, but vapor mass flow will increase for the channels with higher heat load. This will decrease the mass flow and could lead to “dry out’
or vapor lock condition.
m
v
Boston, MA 6/14 to 6/18/2010USPAS Short Course 28
Cryogenic fluid distribution
Major componentsVacuum jacket lines & bayonet connectionsLow heat leak valvesHeat exchangers & vaporizersCirculation pumpsCompressors and expansion engines
Performance of each component affects overall thermodynamic efficiency of systemInefficiency (entropy generation) mostly due to:
Fluid friction Heat exchange over finite ΔT
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