This articles aims to explain how one can relatively easily calculate the pressure drop within a condenser or an evaporator, where two-phase flow occurs and the Navier-Stokes equation becomes very tedious.

1. 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
𝑑𝑢
𝑑𝑡
𝜋𝐷2
4
𝛿𝑧 = −
𝑑𝑃
𝑑𝑧
𝜋𝐷2
4
𝛿𝑧 − 𝜏𝜋𝐷𝛿𝑧 − 𝜌𝑔𝑠𝑖𝑛𝜃
𝜋𝐷2
4
𝛿𝑧
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
𝜋𝐷2
4
𝜌
𝑑𝑢
𝑑𝑡
=
𝜋𝐷2
4
𝜌
𝑑𝑢
𝑑𝑧
𝑑𝑧
𝑑𝑡
=
𝜋𝐷2
4
𝜌𝑢
𝑑𝑢
𝑑𝑧
=
𝜋𝐷2
4
(
𝑑[ 𝜌𝑢2]
𝑑𝑧
− 𝑢
𝑑[ 𝜌𝑢]
𝑑𝑧
)
=
𝜋𝐷2
4
𝑑[ 𝜌𝑢2]
𝑑𝑧
(#1) = −
𝑑𝑃
𝑑𝑧
𝜋𝐷2
4
− 𝜏𝜋𝐷 − 𝜌𝑔𝑠𝑖𝑛𝜃
𝜋𝐷2
4
−
𝑑𝑃
𝑑𝑧
=
4𝜏
𝐷
+
𝑑
𝑑𝑧
[ 𝜌𝑢2] =
4𝜏
𝐷
+ 𝐺2
𝑑
𝑑𝑧
[
1
𝜌
] (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
=
𝑓𝐺2
2𝜌

2. 𝜏 =
𝐷
4
(
𝑑𝑃
𝑑𝑧
)
𝑓
=
𝑓𝐺2
2𝜌
, 𝑓 =
𝜌𝐷
2𝐺2
(
𝑑𝑃
𝑑𝑧
)
𝑓
& (
𝑑𝑃
𝑑𝑧
)
𝑓
=
2𝑓𝐺2
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 (𝑅𝑒,
𝜀
𝐷
), 𝑅𝑒 ≡
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
1
𝜌ℎ
=
𝑥
𝜌 𝑔
+
1 − 𝑥
𝜌𝑙
Where 𝜌 𝑔 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:
1
𝜇ℎ
=
𝑥
𝜇 𝑔
+
1 − 𝑥
𝜇𝑙
Or, from Beattie and Whalley, 1982[1], the homogeneous viscosity can be obtained
from the following equation:
𝜇ℎ = 𝜇𝑙(1 − 𝛽)(1 + 2.5𝛽) + 𝜇 𝑔 𝛽

3. 𝛽 =
𝜌𝑙 𝑥
𝜌𝑙 𝑥 + 𝜌 𝑔(1 − 𝑥)
Thus the Navier-Stokes can be written as
−
𝑑𝑃
𝑑𝑧
= (
𝑑𝑃
𝑑𝑧
)
𝑓
+ 𝐺2
𝑑
𝑑𝑧
[
1
𝜌
] = 𝐺2
(
2𝑓 (𝑅𝑒,
𝜀
𝐷
)
𝜌ℎ 𝐷
+ (
1
𝜌 𝑔
−
1
𝜌𝑙
)
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( 𝑅𝑒) =
16
𝑅𝑒
=
16𝜇ℎ
𝐷𝐺
(𝑒𝑞. 4)
For turbulence flow (Re>3000), an implicit form of f is given by Beattie and Whalley:
1
√ 𝑓
= 3.48 − 4𝑙𝑜𝑔10 (2(
𝜀
𝐷
) +
9.35
𝑅𝑒√ 𝑓
) (𝑒𝑞.5)
There is a more convenient equation in hand, the Blasius equation
𝑓 = 0.079𝑅𝑒−0.25
𝑓𝑜𝑟 𝑅𝑒 > 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]

4. 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
−
𝑑𝑃
𝑑𝑧
=
4𝜏
𝐷
+ (𝛼𝜌 𝑔 + (1 − 𝛼) 𝜌𝑙)𝑔𝑠𝑖𝑛𝜃 + 𝐺2
𝑑
𝑑𝑧
[
𝑥2
𝛼𝜌 𝑔
+
(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:
(
𝑑𝑃
𝑑𝑧
)
𝑓
= 𝜙𝑙
2
(
𝑑𝑃
𝑑𝑧
)
𝑙
= 𝜙 𝑔
2
(
𝑑𝑃
𝑑𝑧
)
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:
(
𝑑𝑃
𝑑𝑧
)
𝑓
= 𝜙𝑙𝑜
2
(
𝑑𝑃
𝑑𝑧
)
𝑙𝑜
= 𝜙 𝑔𝑜
2
(
𝑑𝑃
𝑑𝑧
)
𝑔𝑜
(𝑒𝑞. 9)
(
𝑑𝑃
𝑑𝑧
)
𝑙
=
2𝑓𝑙 𝐺2
(1 − 𝑥)2
𝜌𝑙 𝐷
, (
𝑑𝑃
𝑑𝑧
)
𝑔
=
2𝑓𝑔 𝐺2
𝑥2
𝜌 𝑔 𝐷
, (
𝑑𝑃
𝑑𝑧
)
𝑙𝑜
=
2𝑓𝑙𝑜 𝐺2
𝜌𝑙 𝐷
, (
𝑑𝑃
𝑑𝑧
)
𝑔𝑜
=
2𝑓𝑔𝑜 𝐺2
𝜌 𝑔 𝐷
The indexes 𝜙𝑙, 𝜙 𝑔, 𝜙𝑙𝑜, 𝜙 𝑔𝑜 are obtained with different formulas for different
situations:
1. For 𝜇𝑙/𝜇 𝑔 < 1000, the Friedel correlation should be used.
𝜙𝑙𝑜
2
= 𝐸 + 3.23𝐹𝐻𝐹𝑟0.045
𝑊𝑒0.035
(𝑒𝑞. 10)
𝐸 = (1 − 𝑥)2
+ 𝑥2
𝜌𝑙 𝑓𝑔𝑜
𝜌 𝑔 𝑓𝑙𝑜
𝐹 = 𝑥0.78(1 − 𝑥)0.224
𝐻 = (
𝜌𝑙
𝜌 𝑔
)
0.91
(
𝜇 𝑔
𝜇𝑙
)
0.19
(1 −
𝜇 𝑔
𝜇𝑙
)
0.7
𝐹𝑟 =
𝐺2
𝜌ℎ
2 𝑔𝐷
𝑊𝑒 =
𝐺2
𝐷
𝜌ℎ 𝜎
𝜌ℎ =
𝜌 𝑔 𝜌𝑙
𝑥𝜌𝑙 + (1 − 𝑥)𝜌 𝑔

5. Where 𝜎 is the surface tension of the liquid-gas interface.
.
2. For 𝜇𝑙/𝜇 𝑔 > 1000 & 𝐺 > 100 𝑘𝑔/𝑚2
∙ 𝑠, the Chisholm correlation should be
used.
𝜙𝑙𝑜
2
= 1 + ( 𝑌2
− 1)(𝐵𝑥
2−𝑛
2 (1− 𝑥)
2−𝑛
2 + 𝑥2−𝑛
) (𝑒𝑞.11)
𝑌 = (
(
𝑑𝑃
𝑑𝑧
)
𝑔𝑜
(
𝑑𝑃
𝑑𝑧
)
𝑙𝑜
)
1
2
𝐹𝑜𝑟 0 < 𝑌 < 9.5, 𝐵 =
{
55
√ 𝐺
, 𝐺 ≥ 1900 𝑘𝑔/𝑚2
∙ 𝑠
2400
𝐺
, 500 ≤ 𝐺 ≤ 1900 𝑘𝑔/𝑚2
∙ 𝑠
4.8, 𝐺 < 500 𝑘𝑔/𝑚2
∙ 𝑠
𝐹𝑜𝑟 9.5 < 𝑌 < 28, 𝐵 =
{
520
𝑌√ 𝐺
, 𝐺 ≤ 600 𝑘𝑔/𝑚2
∙ 𝑠
21
𝑌
, 𝐺 > 600 𝑘𝑔/𝑚2
∙ 𝑠
𝐹𝑜𝑟 𝑌 > 28, 𝐵 =
15000
𝑌2
√ 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/𝜇 𝑔 > 1000 & 𝐺 < 100 𝑘𝑔/𝑚2
∙ 𝑠, the Martinelli correlation should be
used.
𝜙𝑙
2
= 1 +
𝐶
𝑋𝑡𝑡
+
1
𝑋𝑡𝑡
2
𝜙 𝑔
2
= 1 + 𝐶𝑋𝑡𝑡 + 𝑋𝑡𝑡
2
(𝑒𝑞. 11)
𝑋𝑡𝑡 = (
(
𝑑𝑃
𝑑𝑧
)
𝑙
(
𝑑𝑃
𝑑𝑧
)
𝑔
)
1
2
𝐶 = {
20, 𝑓𝑜𝑟 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 − 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤
12, 𝑓𝑜𝑟 𝑣𝑖𝑠𝑐𝑜𝑢𝑠(𝑔𝑎𝑠) − 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡(𝑙𝑖𝑞𝑢𝑖𝑑) 𝑓𝑙𝑜𝑤
10, 𝑓𝑜𝑟 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡(𝑔𝑎𝑠) − 𝑣𝑖𝑠𝑐𝑜𝑢𝑠(𝑙𝑖𝑞𝑢𝑖𝑑) 𝑓𝑙𝑜𝑤
5, 𝑓𝑜𝑟 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 − 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑙𝑜𝑤

6. 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 − 𝑥
𝑥
𝜌 𝑔
𝜌𝑙
)
(𝑒𝑞. 12)
𝑆 = (𝑥 (
𝜌𝑙
𝜌 𝑔
− 1) + 1)
1
2
(𝑒𝑞. 13)
In order to gain a better approximation, one can do a CISE correlation for S:
𝑆 = 1 + 𝐸1 (
𝑦
1 + 𝑦𝐸2
− 𝑦𝐸2)
0.5
(𝑒𝑞. 14)
𝑦 =
𝛽
1 − 𝛽
𝛽 =
𝜌𝑙 𝑥
𝜌𝑙 𝑥 + 𝜌 𝑔(1 − 𝑥)
𝐸1 = 1.578𝑅𝑒−0.19
(
𝜌𝑙
𝜌 𝑔
)
0.22
𝐸2 = 0.0273𝑊𝑒𝑅𝑒−0.51
(
𝜌𝑙
𝜌 𝑔
)
−0.08
𝑅𝑒 =
𝐺𝐷
𝜇𝑙
𝑊𝑒 =
𝐺2
𝐷
𝜎𝜌𝑙
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)

7.  An appropriate transformation of the Moody Chart from 𝑓 (𝑅𝑒,
𝜀
𝐷
)
to 𝑓 (𝑥,
𝜀
𝐷
); 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, denoting 𝑧𝑖,
assign the quality on its midpoint to it, denoting 𝑥 𝑖.
2. Calculate the corresponding (
𝑑𝑃
𝑑𝑧
)
𝑖
for each segment.
3. Sum all the (
𝑑𝑃
𝑑𝑧
)
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 (𝑅𝑒,
𝜀
𝐷
)
to 𝑓 (𝑥,
𝜀
𝐷
)
 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 :

8. 1. Divide the length of tube into N segments. For each segment, denoting 𝑧𝑖,
assign the quality on its midpoint to it, denoting 𝑥 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
+
𝜕[ 𝜌𝑢2]
𝜕𝑥
= 𝑢 (
𝜕[ 𝜌]
𝜕𝑡
+
𝜕[ 𝜌𝑢]
𝜕𝑥
) + 𝜌 (
𝜕[ 𝑢]
𝜕𝑡
+ 𝑢
𝜕[ 𝑢]
𝜕𝑥
) = 𝜌
𝐷
𝐷𝑡
[ 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( 𝑅𝑒) =
16
𝑅𝑒
2. 2300 ≤ Re ≤ 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 (𝑅𝑒,
𝜀
𝐷
) = 2 ((
8
𝑅𝑒
)
12
+
1
(𝐴 + 𝐵)1.5
)
1
12
𝐴 =
(
2.457𝑙𝑛(
1
(
7
𝑅𝑒
)
0.9
+ 0.27
𝜀
𝐷
)
)
16

9. 𝐵 = (
37530
𝑅𝑒
)
16
3. For Re>4000, Colebrook equation is applied
1
√ 𝑓
= 3.48 − 4𝑙𝑜𝑔10 (2 (
𝜀
𝐷
) +
9.35
𝑅𝑒√ 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.