Methods to Determine Pressure Drop

In an Evaporator or a Condenser

In this article we will review two methods of determining pressure drop in a two phase flow. The two methods introduced will both be able to give a rough estimation of the real pressure drop in an evaporator or a condenser with relatively easy procedures, avoiding complicated numerical computation.

Method 1: Homogeneous Method

The homogeneous method treats the two phase flow as a homogeneous mixture of fluid, and thus the final equation of motion is a one dimension Navier-Stokes equation.

The Navier-Stokes equation of a single phase fluid inside a tube with steady state assumption is:

ρ dudt

π D2

4δz=−dP

dzπ D2

4δz−τπDδz−ρgsinθ π D2

4δz

Where u is the speed, D the diameter, δz the element length, -(dP/dz) the pressure gradient, τ the shear stress, g the acceleration due to gravity and θ the angle of inclination to the horizontal plane.

Neglecting the gravity term (which is usually legitimate if the initial and final height varies not much) and applying the chain rule to the left hand side we shall have

π D2

4ρ du

dt=π D2

4ρ du

dzdzdt

=π D2

4ρu du

dz=π D2

4(d [ ρ u2 ]

dz−u d [ ρu ]

dz)

¿ π D2

4d [ ρu2 ]

dz(¿1)=−dP

dzπ D2

4−τπD−ρgsinθ π D2

4−dP

dz=4 τ

D+ d

dz[ ρ u2 ]= 4 τ

D+G2 d

dz [ 1ρ ](eq .1)

Where G is the mass flow.

Now we turn to the stress term. It can be determined by the Fanning friction factor:

τ= fρu2

2= f G2

2 ρ

τ=D4 ( dP

dz )f= f G2

2 ρ, f = ρD

2G2 ( dPdz )

f∧( dP

dz )f=2 f G2

ρD(eq . 2)

The Fanning factor can be found under both laminar and turbulence conditions by the Moody Chart. It is a function of the Reynolds number and relative roughness.

f =f (ℜ , εD ) ,ℜ≡ DG

μ

Where μ is the viscosity.

The Moody Chart (from Wikipedia)#2

When dealing with a two phase flow, we assume the mixture of these two phases is homogeneous. The density of the mixture ρh is

1ρh

= xρg

+ 1−xρl

Where ρg is the density of gas, ρl the density of liquid, and x the quality.

The viscosity of the mixture μh can also be derived as similar method:

1μh

= xμg

+ 1−xμl

Or, from Beattie and Whalley, 1982[1], the homogeneous viscosity can be obtained from the following equation:

μh=μ l (1−β ) (1+2.5 β )+μg β

β=ρl x

ρl x+ρg(1−x )

Thus the Navier-Stokes can be written as

−dPdz

=( dPdz )

f+G2 d

dz [ 1ρ ]=G2( 2 f (ℜ , ε

D )ρh D

+( 1ρg

− 1ρl )dx

dz )(eq .3)

The last thing to do is to determine the quality distribution throughout the tube, i.e, to determine; we should be able to simply assume linearity.

x=x (z )

For laminar flow (Re<3000), we have the Stokes Law:

f =f (ℜ )=16ℜ =

16 μh

DG(eq . 4)

For turbulence flow (Re>3000), an implicit form of f is given by Beattie and Whalley:1√ f

=3.48−4 log10(2( εD )+ 9.35

ℜ√ f )(eq .5)

There is a more convenient equation in hand, the Blasius equationf =0.079 ℜ−0.25 for ℜ>2000 (eq .6)

Regardless of its simplicity, the method does yield a relatively reliable result. Below is how Beattie and Whalley method, aside with other methods, worked out compared with experimental data.

Comparison of the Method with Experimental Data

Adapted from Fan et al, 2016 [2]

Method 2: Two Phase Method

The second method is adapted from Sadik Kakac’s “Boilers, Evaporators, and Condensers” in 1991 [3]. It acknowledges the fact that there are two phases coexisting in the tube. Then the Navier-Stokes equation of the tube should be rewritten as

−dPdz =

4 τD +( α ρg+(1−α ) ρl ) gsinθ+G2 d

dz [ x2

α ρg+

(1−x )2

(1−α ) ρl ](eq . 7)

Again neglecting the gravity term, let us discuss the friction term and the inertial term respectively.

For the friction term, it is convenient to relate it to that of a single phase flow of the gas or liquid which has their actual respective mass flux:

( dPdz )

f=ϕl

2( dPdz )

l=ϕg

2( dPdz )

g(eq .8)

Or, in some cases, one should relate the friction term to that of a single phase flow of the gas or liquid which has the total mass flux:

( dPdz )

f=ϕlo

2( dPdz )

lo=ϕgo

2(dPdz )

go(eq . 9)

( dPdz )

l=

2 f lG2(1−x)2

ρl D,( dP

dz )g=

2 f gG 2 x2

ρg D,( dP

dz )lo=

2 f lo G2

ρl D,( dP

dz )go=

2 f go G2

ρg D

The indexes ϕl , ϕg , ϕ lo , ϕ go are obtained with different formulas for different situations:

1. For μl / μg<1000, the Friedel correlation should be used.

ϕlo2=E+3.23 FH Fr0.045We0.035(eq .10)

E=(1−x )2+ x2 ρl f go

ρg f lo

F=x0.78 (1−x )0.224

H=( ρl

ρg)

0.91

( μg

μ l)

0.19

(1−μg

μ l)

0.7

Fr= G2

ρh2 gD

We=G2 Dρh σ

ρh=ρg ρl

x ρl+(1−x) ρg

Where σ is the surface tension of the liquid-gas interface. .

2. For μl / μg>1000∧G>100 kg /m2 ∙ s, the Chisholm correlation should be used.

ϕlo2=1+(Y 2−1 ) (B x

2−n2 (1−x )

2−n2 +x2−n)(eq . 11)

Y=((dPdz )

go

( dPdz )

lo)

12

For 0<Y <9.5 ,B={55√G

,G ≥ 1900 kg/m2 ∙ s

2400G

,500 ≤ G≤ 1900 kg /m2 ∙ s

4.8 , G<500 kg /m2 ∙ s

For 9.5<Y <28 , B={ 520Y √G

,G ≤ 600 kg /m2 ∙ s

21Y

, G>600 kg /m2 ∙ s

For Y >28 , B=15000Y 2 √G

Where n is the exponent of the Reynolds number in friction factor relationship; for example, in equation 6, n=0.25.

3. For μl / μg>1000∧G<100 kg /m2 ∙ s, the Martinelli correlation should be used.

ϕl2=1+ C

X tt+ 1

X tt2

ϕ g2=1+C X tt+ X tt

2

(eq .11)

X tt=( ( dPdz )

l

( dPdz )

g)

12

C={ 20 , for turbulent−turbulent flow12 , for viscous (gas)−turbulent (liquid ) flow10 , for turbulent (gas)−viscous(liquid ) flow

5 , for viscous−viscous flow

After the frictional pressure loss is calculated, we now consider the inertial term due to mass change on the liquid-vapor interface. Two parameters will be used, the void fraction α (which we had introduced previously in equation 5) and the slip velocity ratio S:

α= 1

1+(S 1−xx

ρg

ρl)(eq .12)

S=(x ( ρl

ρg−1)+1)

12(eq . 13)

In order to gain a better approximation, one can do a CISE correlation for S:

S=1+E1( y1+ y E2

− y E2)0.5

(eq . 14)

y= β1−β

β=ρl x

ρl x+ρg (1−x )

E1=1.578 ℜ−0.19( ρl

ρg)

0.22

E2=0.0273We ℜ−0.51( ρl

ρg)−0.08

ℜ=GDμl

We=G2 Dσ ρl

Summarization: Suggested Procedure

Below we will summarize the procedure of obtaining the pressure drop by the two mentioned methods.

Method #1 Initial Parameters Independent of Device:

Density of gas and liquid phase Viscosity of gas and liquid phase An appropriate Moody chart (Available online; one can also create his/her

own chart)

An appropriate transformation of the Moody Chart from f (ℜ , εD ) to

f (x , εD ); for the concern of our application this can be done by point

plotting and linear interpolation. Initial Parameters Dependent of Device:

Length of tube Relative roughness of tube Diameter of the tube Initial and final quality of the condenser/evaporator Mass flux of the refrigerant

Steps :1. Divide the length of tube into N segments. For each segment, denotingz i,

assign the quality on its midpoint to it, denotingx i.

2. Calculate the corresponding ( dPdz )

i

for each segment.

3. Sum all the( dPdz )

i

, with either direct summation or other sophisticated

technique. Since each segment is calculated independent of each other, the result is guaranteed to converge.

Method #2 Initial Parameters Independent of Device:

Density of gas and liquid phase Viscosity of gas and liquid phase Surface tension on the gas liquid interface An appropriate Moody chart

An appropriate transformation of the Moody Chart from f (ℜ , εD ) to

f (x , εD )

Initial Parameters Dependent of Device:

Length of tube Relative roughness of tube Diameter of the tube Initial and final quality of the condenser/evaporator Mass flux of the refrigerant

Steps :1. Divide the length of tube into N segments. For each segment, denotingz i,

assign the quality on its midpoint to it, denotingx i.

2. Calculate the frictional term of pressure drop for each segment:i. If the Chisholm correlation is applied, determine the fanning factors of

all the segment first, empirically fit them with their corresponding Reynold number to find the parameter n.

ii. If the Martinelli correlation is applied, determine the state of flow, viscous or turbulent, of the gas and liquid for each segment to determine the parameter C.

3. Calculate the inertial term of pressure drop for each segment.4. Sum all the segments up.

Notes

#1:The continuity equation in one dimension tube states that

∂ [ ρ ]∂ t

+∂ [ ρu ]∂ x

=0

The momentum equation also states that, when applied with the continuity equation

∂ [ ρu ]∂ t

+∂ [ ρ u2 ]

∂ x=u( ∂ [ ρ ]

∂ t+

∂ [ ρu ]∂ x )+ ρ( ∂ [u ]

∂ t+u ∂ [ u ]

∂ x )=ρ DDt

[u ]=∑ F

Which is the origin of the Navier-Stokes equation.

#2:The chart can be drawn by dividing the Reynolds number into three regions:1. For Re<2300, Stokes Law is applied and

f (ℜ )=16ℜ

2. 2300 ≤ℜ≤ 4000 lies the critical region. Churchill equation can be used, and can give fine results for relative roughness smaller than 0.01, but keep in mind that

no general theory can yet describe this region.

f (ℜ , εD )=2¿¿

A=(2.457 ln( 1

( 7ℜ )

0.9

+0.27 εD ))

16

B=(37530ℜ )

16

3. For Re>4000, Colebrook equation is applied1√ f

=3.48−4 log10(2( εD )+ 9.35

ℜ√ f )This equation requires iteration.

Note that because of the factor four in front of the shear stress term in equation one, and because of the definition of Fanning friction factor we have taken here, the results is four times smaller than the Moody chart given by Wikipedia.

Reference

1. D. R. H. BEATTIEt, P.B.W., A simple two-phase frictional pressure drop calculation model. Int. I. Mtdtiphase Flow, 1982. 8(1): p. 5.

2. Xiaoguang Fan, X.M., Lei Yang, Zhong Lan, Tingting Hao, Rui Jiang, Tao Bai, Experimental study on two-phase flow pressure drop during steam condensation in trapezoidal microchannels. Experimental Thermal and Fluid Science 2016. 76: p. 12.

3. Kakac, S., ed. Boilers, Evaporators, and Condensers. 1991, John Wiley & Sons, Inc.: USA. 835.