Analogies among Mass, Heat, and Momentum Transfer

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Analogies among Mass, Heat,
and Momentum Transfer
Analogies
Heat  Mass  (sometimes) Momentum
Analogies are useful tools
1. An aid to understand transfer phenomena
2. A sound means to predict behavior of systems for
which limited quantitative data are available
Molecular Transport Equations
RECALL:
driving force
rate of transport =
resistance
d(v x  )
 yx  
dy
MOMENTUM
Newton’s law
qy
A
 
d(  c p T)
dy
HEAT
Fourier’s law
dcA
J  DAB
dy
*
Ay
MASS
Fick’s law
Analogous quantities
in transport phenomena
Reynolds Analogy
The general transport equation can be written in the form
𝑑Γ
𝜓 =− 𝛿+𝐸
𝑑𝑥
where ψ = flux of a property at any value of x
δ = molecular diffusivity
E = eddy diffusivity
Г = volume concentration of transferent property
Turbulent diffusion equations
Transfer coefficient for momentum
In cylindrical geometry,
𝑑Γ
𝜓 =− 𝛿+𝐸
𝑑𝑟
Integrating the above equation and multiplying by A to get a rate
equation,
𝛿+𝐸
𝜓𝐴 = −4
(Γ1 − Γ)𝐴
𝛾𝐷
where A = cross-sectional area perpendicular to flow
𝐸 = mean eddy diffusivity
Γ −Γ
𝛾 = 1 = ratio of the difference in concentration of transferent property
Γ1 −Γ0
between the wall and the mean value and the mean value of the
fluid to the maximum difference between the wall and the center
D = diameter
Transfer coefficient for momentum
𝛿+𝐸
𝜓𝐴 = −4
(Γ1 − Γ)𝐴
𝛾𝐷
The transfer coefficient is then defined as
𝛿+𝐸
𝜑 = −4
𝛾𝐷
Substituting and rearranging,
Γ1 − Γ
𝜓𝐴 = −
1
𝜑𝐴
Transfer coefficient for momentum
𝛿+𝐸
𝜓𝐴 = −4
(Γ1 − Γ)𝐴
𝛾𝐷
The transfer coefficient is then defined as
𝛿+𝐸
𝜑 = −4
𝛾𝐷
Substituting and rearranging,
Γ1 − Γ
𝜓𝐴 = −
1
𝜑𝐴
Transfer coefficient for momentum
Γ1 − Γ
𝜓𝐴 = −
1
𝜑𝐴
For momentum transfer,
𝜓=𝜏
Γ = 𝜌𝑣
𝜏 = −𝜑 [𝜌𝑣1 − 𝜌𝑣]
Transfer coefficient for momentum
𝜏 = −𝜑 [𝜌𝑣1 − 𝜌𝑣]
At the wall, v1 = 0 so that,
𝜏 = 𝜑 𝜌𝑣
𝜏
𝜑=
𝜌𝑣
If we divide by 𝑣,
𝑓 𝜑
𝜏
= = 2
2 𝑣 𝜌𝑣
The Reynolds analogy
For turbulent transport,
For heat transfer,
For momentum transfer,
We assume that α and μ/ρ are negligible, and that 𝛼𝑡 = 𝜖𝑡
The Reynolds analogy
Dividing the momentum equation by the heat equation then gives
𝜏
𝑞 𝑐𝑝
𝐴
𝑇
𝑑𝑇 =
𝑇𝑖
𝑣𝑎𝑣
0
𝑑𝑣
The Reynolds analogy
𝑞
𝐴
2
Substituting = ℎ 𝑇 − 𝑇𝑖 and 𝜏𝑠 = 𝑓𝑣𝑎𝑣
𝜌/2
The Reynolds analogy
Stanton number
𝑁𝑁𝑢
f
𝑁𝑆𝑡 =
=
𝑁𝑅𝑒 𝑁𝑃𝑟 2
Dimensionless Groups
Dim. Group
Ratio
Equation
Prandtl, Pr
molecular diffusivity of momentum /
molecular diffusivity of heat
Schmidt, Sc
momentum diffusivity/ mass diffusivity
Lewis, Le
thermal diffusivity/ mass diffusivity
Stanton, St
heat transferred/ thermal capacity
𝑐𝑃 𝜇
𝑘
ν
𝐷𝐴𝐵
𝛼
𝐷𝐴𝐵
ℎ
𝑐𝑝 𝜌𝑣
The Reynolds analogy
f
h
=
2 cp 𝜌 𝑣
Experimental results show that the above equation
1. Correlate data approximately for gases in turbulent flow
2. DOES NOT correlate experimental data for liquids in turbulent flow
3. DOES NOT correlate experimental data for any fluids in laminar flow
* 0.6 < NPr for gases < 2.5
It was concluded that the Reynolds analogy is valid ONLY at NPr = 1
The Reynolds analogy
In a similar manner,
we can relate mass transfer with momentum transfer
For turbulent transport
And the complete Reynolds analogy is
The Reynolds analogy
f 𝑘𝑐′
=
2
𝑣
Experimental results show that the above equation
1. Correlate data approximately for gases in turbulent flow
2. DOES NOT correlate experimental data for liquids in turbulent flow
3. DOES NOT correlate experimental data for any fluids in laminar flow
* NSc for gases ~ 1.0
It was concluded that the Reynolds analogy is valid ONLY at NSc = 1
The Reynolds analogy
CONCLUSIONS
1. At NPr = NSc = 1, the mechanisms for mass, heat, and momentum
are identical
2. For other fluids, transfer processes differ in some manner
functionally related to the Pr and Sc numbers.
The Reynolds analogy
Note that the Reynolds analogy assumes that
1. the turbulent diffusivities are equal and
2. the molecular diffusivities are negligible.
When are these assumptions not valid?
1. For other fluids, where 𝑁𝑃𝑟 ≠ 𝑁𝑆𝑐 ≠ 1
 usually the case for liquids
2. We CANNOT neglect molecular diffusivities
 in the boundary layer where diffusion, conduction, and viscosity
are important
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