Massoud M. Engineering Thermofluids: Thermodynamics, Fluid Mechanics, and Heat Transfer

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682 Vc. Two-Phase Flow and Heat Transfer: Condensation

Solution: For this transient problem, we may estimate the plate temperature, from

an energy balance using the lumped capacitance approach:

()

sfilm

TThA

Tcd

−=

In this relation, T

film

≈ (T

sat

+ T

)/2 and h in Equation Vc.2.6 may be written as:

h =

sat

– T)

–1/4

where

()

25.0

)/()(943.0 Lhkg

ffgfgff

µρρρ

′

−

Assuming

= T

sat

– T

, the above energy balance simplifies to:

³³

−= dtcAd )V2/(/

4/3

ρζθθ

Integrating and setting

= 0, we find )V/2/(

4/1

cAt

ρζθ

= (T

sat

– T

) = 212 – 200 = 12 F, A/V = 12/0.125 ft,

c = 48.8 Btu/ft

·F

For the condensate layer, we find film properties at T

film

= 209 F

()

25.0

)1069.0/(97039.0)035.089.59(89.5936002.32943.0 ××−×× =

1772.5 Btu/ft

·h·F

3/4

t = (12)

1/4

/[2 × 1772.5 × 12/ (44.8 × 0.125)] = 14 s

3. Empirical Solution

Application of the empirical solutions depends on the value of the Reynolds num-

ber. If we substitute for mass flow rate, the Reynolds number can also be written

)/(4Re



= 4

V δ /

where the average film velocity is used and the

plate width (b) cancels out from the numerator and the denominator.

3.1. Condensation on Vertical Plates and Cylinders

By defining a condensation Nusselt number, Nu

also referred to as the condensa-

tion number, we may express the heat transfer coefficient in terms of the Reynolds

number. These are shown in the table below. In the wavy laminar region, being

the transition region between laminar and turbulent, the Kutateladze correlation

and in the turbulent region the Labuntsov correlation are recommended.

3. Empirical Solution 683

Example Vc.3.1. A vertical flat plate, 1.1 m long and 0.5 m wide maintained at

50 C is exposed to saturated steam at 0.5 bar. Find the total rate of heat transfer to

the plate and the condensate mass flow rate.

Solution: We first find T

sat

(0.5 bar) = 81.33 C then T

film

= (81.33 + 50)/2 =

65.67 C:

For steam at T

sat

= 81.33 C: h

= 2305.4 kJ/kg and

= 0.308 kg/m

For water at T

film

= 65.67 C: k

= 0.659 W/m·K,

= 429E-6 N·s/m

980 kgm/m

, c

= 4.2 kJ/kg·K

Since T

<< T

sat

, we calculate need to find

′

Ja = 4.2(81.33 – 50)/2305.4 = 0.057

′

= 2305.4(1 + 0.68 × 0.057) = 2395 kJ/kg

h =

25.0

1.1)5033.81(6E429

3E2395659.0)308.0980(9808.9

943.0

×−×−

××−×

= 4310 W/m

·K

)(

ssat

TTAhQ −=



= 4310 (1.1 × 0.5) × (81.33 – 50) = 74,268 W

hQm

′

= /



= 74,268/2.395E6 = 0.031 kg/s

Finding Re

= 4 m



b = 4 × 0.031/(429E-6 × 0. 5) = 578 shows the flow regime

is actually wavy laminar. We should then use the heat transfer coefficient based

on the Kutateladze correlation as shown below:

3/121.22

)/(2.51.08Re

)(4

)(Re

)(

TTA

ssat

fgf

ssat

−

′

−

′



Solving for Re

, we find Re

= 716.5. Using this in the Kutateladze correlation,

we find:

h = 716.5 × 0.659/{(1.08 × 716.5

1.22

–5.2)[(429E-6/980)

/9.8]

1/3

} = 5340

W/m

·K

Revised values for Q



and m



are:



= 5340 (1.1 × 0.5) × (81.33 – 50) = 92,016 W



= 92,016/2.395E6 = 0.0384 kg/s.

The same procedure used for vertical flat plates is applicable to vertical cylin-

ders if δ

<< 0.5D.

3.2. Condensation on Spheres, Horizontal Cylinders

and on Banks of Tubes

Correlations similar to Equation Vc.2.6 are obtained for condensation on radial

systems (Dhir & Lienhard):

684 Vc. Two-Phase Flow and Heat Transfer: Condensation

25.0

)(

−

′

−

DTT

hkg

ssatf

fgfgff

ρρρ

Vc.3.4

where C = 0.815 for condensation on spheres and C = 0.729 for condensation on

horizontal tubes. Equation Vc.3.4 is also applicable to condensers, which consist

of banks of horizontal tubes with cold fluid flows inside the tubes and vapor con-

denses on the tubes. For film condensation inside horizontal tubes, C = 0.555.

Example Vc.3.2. Find the condensate flow rate for the following data of a power

plant condenser: P = 2 in Hg, D = 1¼ in, L = 28.5 ft, N

tube

= 16500, T

= 75 F.

Solution: We first find T

sat

(2 in Hg) = 101 F then T

film

= (101 + 75)/2 = 88 F

For steam at T

sat

= 101 F: h

= 1036 Btu/lbm and

= 0.003 lbm/ft

For water at T

film

= 88 F: k

= 0.36 Btu/ft·h·F,

= 1.9 lbm/ft·h,

= 62 lbm/ft

= 1 Btu/lbm·F

Since T

<< T

sat

we need to find

′

= h

(1 + 0.68Ja)

Ja = 1 × (101 – 75)/1036 = 0.025 and

′

= 1054 Btu/lbm

e now use Equation Vc.3.4 to find the average heat transfer coefficient over a sin-

gle tube:

h = 0.729 × {(32.2 × 3600

) × 62

× 0.36

× 1054/[1.9 × (101 – 75) ×

(1/12)]}

0.25

= 1525 Btu/ft

·h·F

)(

ssatD

TTAhQ −=



= 1525 × (

× 28.5 × 1.25/12) × (101 - 75) = 369,800 Btu/h

hQNm

′

= /

tube



= 16,500 × 368,800/1054 = 5.8 Mlbm/h.

4. Condensation Degradation

The presence of even a small amount of noncondensable gas significantly de-

grades the rate of condensation heat transfer. As shown in Figure Vc.4.1, these

gases tend to migrate and accumulate near the colder surface, reducing the partial

pressure of the vapor and subsequently the corresponding saturation temperature

of the vapor. For containment response analysis, safety regulations require the use

of the Tagami correlation for the case of a LOCA and the Uchida correlation dur-

ing a MSLB analysis.

Tagami Correlation

Tagami, an empirical correlation, applies during the forced convection period fol-

lowing the blowdown phase of a LOCA. The key parameter in the Tagami corre-

Questions and Problems 685

lation is therefore time t

, which marks the end of the blowdown-induced forced

convection period. As such, the Tagami correlation is only applicable up to time

. The natural convection phase of a blowdown must be analyzed by using the

Uchida or the turbulent natural convection correlation depending on the value of

the Ra number. If h

maximum

is the heat transfer coefficient corresponding to time t

then the heat transfer coefficients at other times (t < t

) are obtained as

)/(

max p

tthh = . According to Tagami, h

maximum

itself is calculated from:

max

h =

()

0.62

CU t

ªº

¬¼

, where U is the total blowdown energy released during t

and V

is containment free volume. In SI units, U is in J, t

in s, V in m

, and C

is 0.607.

Time t

is not known beforehand rather it should be obtained by iteration. A typi-

cal value for t

given a large dry containment is about 13 seconds.

Uchida Correlation

The Uchida correlation is given as a table of heat transfer coefficient versus the ra-

tio of air to steam mass. The maximum value is 1590 W/m

·K and the minimum

value is 11.4 W/m

·K.

Total Pressure

Gas Partial

Pressure

Steam Partial

Pressure

Fluid

Temperature

Uchida Condensation M odel

200

400

600

800

1000

1200

1400

1600

1800

0.1 1 10 100

Air Mass To Steam Mass Ratio

Heat Transfer Coefficient (W/m2-K

)

(a) (b)

Figure Vc.4.1. (a) Effect of noncondensable gas on pressure and (b) on h (Uchida model)

QUESTIONS

– Find the condensation mode for the following examples, a power plant con-

denser, steam condensation on spray droplets, steam condensation in a suppres-

sion pool, rain.

– Which condensation mode is more efficient, dropwise or film?

– Does dropwise condensation occur if a liquid wets the surface

– Can the heat transfer coefficient in condensation reach 10,000 Btu/ft

·h·F?

– Consider two condensers, one using horizontal tubes the other vertical tubes.

Cold water flows inside the tubes. If all other parameters are identical which

condenser is more efficient?

686 Vc. Two-Phase Flow and Heat Transfer: Condensation

– Inside the condensate film, where can you find the minimum shear stress

– The flow of the condensate film beyond what Reynolds number becomes turbu-

lent?

PROBLEMS

1. A vertical plate, 3 ft long and 5 ft wide is insulated from one side and the other

side is exposed to steam at 15 psia. The plate temperature is maintained at 120 F.

At what rate is condensate produced? [Ans.: Ja = 0.033, Re = 287.5, Nu = 7004,

h = 878 Btu/ft

·h·F, Q



= 126 kW, and m



= 0.116 lbm/s].

2. steam at 4 psia is condensing on a vertical plate 1 m long and 2 m wide. The

plate is at 50 C. Find



. [Ans.: Ja = 0.03, Re = 297, Nu 7621, h = 4968

W/m

·K, Q



= 278 kW, and m



= 0.12 kgm/s].

3. A steam condenser consists of a square array of 529 horizontal tubes, 1 in di-

ameter and 12 ft long. Tubes are maintained at 95 F to condense steam at 1 psia.

Find the condensate production rate. [Ans.: 27 lbm/min.]

4. Find the condensation heat transfer coefficient for saturated Freon-12 at 50 C

on a horizontal tube, having a diameter of 3 cm and maintained at 40 C. [Ans.:

1244 W/m

·C].

5. Consider condensation of benzene vapor at 1 bar on a vertical flat plate of

height 0.3 m. If the plate is kept at 60 C, find the condensation heat transfer coef-

ficient. At 1 bar, T

sat

= 80 C,

= 823 kg/m

= 2.74 kg/m

, h

= 398 kJ/kg, c

p,l

= 1.88 kJ/kg·K,

= 321E-6 N·s/m

, k

= 0.131 W/m·K. [Ans. 1270 W/m

·k].

6. Saturated steam at 1 atm is condensing on a horizontal tube at a rate of

300 kg/h. The tube is 2 m long and is maintained at 60 C. Find the tube diameter.

[Ans.: 13 cm].

7. A steam power plant produces 2700 MWe at a thermal efficiency of 29%. Find

the surface area and the number of tubes

8. A steel plate having L = 3 m, b = 1.5 m, and thickness of 0.5 cm is placed in

steam at 1 atm. Initially, the plate is at 80 C. Plot the plate temperature versus

time. Identify the simplifying assumptions made.

9. Water flows at a rate of 296 kg/h in a horizontal thin-walled tube (d = 0.025 m) at

25 C. Benzene vapor condenses at 1 bar on the tube. Find the rate of condensation

per meter. [Ans.: By iteration 24 kg/m·h].

VI. Applications

In this chapter we will study such important topics as heat exchangers, flow-

meters, and turbomachines. We will also evaluate the thermal hydraulics response

of systems to transients and heat generation from nuclear energy.

VIa

. Heat Exchangers

Heat exchanger (HX) is a generic term applied to a wide range of mechanical sys-

tems, which are designed for the purpose of exchanging thermal energy between

two streams of fluids separated by a solid surface. Some heat exchangers are used

as heat sinks including automotive inter-coolers, containment air coolers, cooling

towers, automotive radiators, and power plant condensers. Some other heat ex-

changers are used as a heat source including boilers, radiators for space heating,

and steam generators. For example, a nuclear plant utilizes many heat exchangers.

In a typical plant, feedwater heaters are employed in the balance of plant to im-

prove plant thermal efficiency. Also, other heat exchangers, such as the compo-

nent cooling water and shutdown cooling (also known as the residual heat re-

moval) system, provide a heat sink for the reactor during the shutdown period.

Finally, the service water heat exchanger provides a heat sink for the balance of

plant equipment. Heat exchangers have a hot side and a cold side separated by

tubes or plates. Heat transfer between the fluids in each side takes place through

the surface dividing the hot side and the cold side. Heat transfer may take place

between liquid and liquid, liquid and gas, and gas and gas. Heat exchangers may

also carry two-phase flow resulting in boiling or in condensation. Heat exchang-

ers may be operating at steady-state or transient conditions. Due to such design

variations, heat exchangers require careful analysis in design optimization as well

as in performance evaluation. In this chapter we will deal primarily with the

thermal aspects of tubular heat exchangers.

1. Definition of Heat Exchanger Terms

Concentric parallel flow heat exchanger consists of a tube surrounded by a

shell. In parallel flow heat exchangers, both hot and cold streams flow in the same

direction as shown in Figure VIa.1.1.

688 VIa Applications: Heat Exchangers

Tube-side Flow

Shell-side Flow

Tube

Shell

Inlet

Plenum

Exit

Plenum

Baffle Plate

Tube Sheet

Cross Flow

Figure VIa.1.2. Schematic of a shell and tube heat exchanger

Parallel Flow HX

Counterflow HX

Cross flow HX

Figure VIa.1.1. Schematics of parallel, counterflow, and cross flow heat exchangers

Concentric counter flow heat exchanger consists of a tube surrounded by a

shell. In counter flow heat exchangers, the hot and cold streams flow in opposite

directions as shown in Figure VIa.1.1.

Cross flow heat exchangers are generally used in gas-liquid applications.

They include tubes with a stream flowing parallel to the xy-plane and cross flow

parallel to the the yz-plane and perpendicular to the tube axis. The cross flow may

be mixed, as shown in Figure VIa.1.1, or unmixed by passing cross flow through

parallel plates.

Shell and tube heat exchanger is similar to the concentric heat exchanger.

However, shell and tube heat exchangers consist of multi tubes held in place by

baffle plates. These plates prevent tube vibration and enhance the rate of heat

transfer by diverting the shell-side fluid in a cross flow manner, as shown in Fig-

ure VIa.1.2. Shell and tube heat exchangers may be arranged in series to obtain

1. Definition of Heat Exchanger Terms 689

several cascaded shells. Each heat exchanger has a minimum of four ports. As

shown in Figure VIa.1.2, there is an inlet port and an outlet port in the hot side.

Similarly, there is an inlet and an outlet port in the cold side. Tubes are either

straight or bent in the form of a U. Heat exchangers including straight tubes have

an inlet plenum and an outlet plenum. Tubes are installed inside tube sheets,

which also act as barriers between the plenum fluids and the shell-side fluid.

Fouling factor f, is a measure of cleanliness of heat exchangers. Fluid streams

generally carry impurities. Sedimentation of the impurities over time leads to a

layer of deposit. This poses a resistance to heat transfer across the solid surface.

The sedimentation may also consist of sludge that would pit and corrode solid sur-

faces. Maintaining a closely monitored chemistry of the fluid stream and periodic

cleaning of heat exchangers is essential in ensuring proper operation of the device.

In a shell and tube heat exchanger, for example, cleaning of the shell side is espe-

cially challenging due to the presence of the baffle plates. In such cases, the

cleaner stream flows in the shell side, as cleaning the inside of the tubes is by far

easier. In power plants using large bodies of water as the heat sink, steam always

condenses on the tubes with river, lake, bay, or ocean water flowing in the tubes.

Fouling may be categorized as particulate fouling, crystallization fouling, corro-

sion fouling, biofouling, and chemical reaction fouling (Kakac). One way to ac-

count for fouling in the design process of heat exchangers is to increase the sur-

face area for heat transfer. Fouling factor has units of thermal resistance

·h·F/Btu or m

·K/W. Some typical values are as follows:

Fluid f (m

·K/W) f (ft

·h·F/Btu)

Transformer oil 0.000176 0.0010

Engine lube oil 0.000176 0.0010

Crude oil 0.000352 0.0020

Heavy gas oil 0.000881 0.0050

Heavy fuel oil 0.001233 0.0070

Vegetable oil 0.000528 0.0030

Seawater 0.000176 0.0010

Brackish water 0.000528 0.0030

Muddy or silt water 0.000705 0.0040

River water (< 50 C) 0.001000 0.0060

Refrigerating liquid 0.000200 0.0011

From: Standards of Tubular Exchangers Manufacturers Association

Overall heat transfer coefficient is given in Equation IVa.6.8 for a clean con-

centric heat exchanger. To account for fouling resistance, Equation IVa.6.8

should be modified as follows:

)/ln(

−

++++=

ooo

oio

LhdLd

Lhd

πππππ

= U

VIa.1.1

690 VIa Applications: Heat Exchangers

where subscripts i and o stand for tube inside and tube outside, respectively. Tube

outside is also known as the tube bundle, shell side, or secondary-side. In Equa-

tion VIa.1.1, f

and f

are the tube-side and shell-side fouling factors, respectively.

Typical values of U for various streams are as follows.

Stream A / Stream B U (W/m

·K) U (Btu/ft

·h·F)

Condensing Steam / Water (Condenser & FWH) 1100 – 5500 200 – 1000

Freon 12 Condenser / Water 300 – 850 50 – 150

Water / Water 850 – 1700 150 – 300

Water / Oil 100 – 350 20 – 60

Gas / Gas 10 – 40 2 – 8

Cleanliness factor, C

is another measure of heat exchanger cleanliness and is

defined as the ratio of the overall heat transfer coefficient when a heat exchanger

is fouled, to the overall heat transfer coefficient of the clean heat exchanger, C

fouled

clean

. When defining C

, we need not to consider any fouling (f

and f

) in

Equation VIa.1.1, rather calculate U

clean

and find U

fouled

from U

fouled

= C

clean

Heat Capacity, C is the product of mass flow rate and specific heat, C =



having units of Btu/s·F or W/C.

2. Analytical Solution

In this section we discuss steady-state operation of heat exchangers. As shown in

Figure VIa.2.1, we can assign control volumes to the entire hot-side, the cold-side

and the solid surface separating the hot and cold sides. In steady state conditions,

flow rates at the inlet and outlet ports for each control volume are equal hence:

houthinh

mmm



VIa.2.1a

coutcinc

mmm



VIa.2.1b

inhh



incc



Hot-side CV

Cold-side CV

outhh



outcc



Tube CV

hpouth

cpoutc

hpinh

cpinc



Figure VIa.2.1. Control volumes for the primary and secondary sides

The hot stream enters at an enthalpy of h

and leaves at an enthalpy of h

h,out

h,in

. If the heat exchanger shell is fully insulated so that there is no heat loss to

the environment, then under steady-state operation:

2. Analytical Solution 691

QQQ



== VIa.2.2a

On the other hand, the cold side fluid enters at an enthalpy of h

c,i

and leaves at an

enthalpy of h

c,o

> h

c,i

. The gain in the cold stream energy is due to the transfer of

heat from the solid surface. In steady-state:

QQQ



== VIa.2.2b

Substituting for the rate of heat transfer while assuming negligible potential and

kinetic energies, we find:

)(

,, outhinhh

hhmQ −=



VIa.2.3a

)(

,, incoutcc

hhmQ −=



VIa.2.3b

For a special case where each stream exits at the same phase as it entered the heat

exchanger, we may replace ∆h § c

∆T. Assuming variations in specific heat are

small and using c

= f(T

+ T

out

)/2, we can write the axial energy equations as:

)(

,, outhinhh

TTCQ −=



VIa.2.4a

)(

,, incoutcc

TTCQ −=



VIa.2.4b

So far we used the conservation equations of mass and energy in the axial direc-

tion. In the next section we use the conservation equation of energy in the trans-

verse direction for elemental control volumes to tie the hot-side and cold-side

temperatures by applying the overall heat transfer coefficient. The reason for us-

ing elemental control volume is that axial temperature profiles are generally not

linear functions of the heat exchanger length. As a result, we cannot generally use

average temperatures for the hot and cold side as:

]

[

,,,, outcincouthinh

TTTT

−

=∆

VIa.2.5

If it were possible, the rate of heat transfer would have been calculated from

TUAQ ∆=



. The simple relation given in Equation VIa.2.5 provides only an es-

timation of ∆T. In the next section we will see that ∆T is a logarithmic function of

the inlet and outlet temperatures referred to as the logarithmic mean temperature

difference (LMTD).

2.1. LMTD Method of Analysis

The analysis performed below applies to parallel flow heat exchangers as shown

in Figures VIa.2.2(a). Similar analysis can be performed for counterflow heat ex-

changers, shown in Figure VIa.2.2(b).