ME421Lec6

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ME421

Heat Exchanger andSteam Generator Design

Lecture Notes 6

Double-Pipe Heat Exchangers

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Introduction

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Introduction

• DP HEX: one pipe placed concentrically inside another

• One fluid flows through inner pipe, the other through the

annulus• Outer pipe is sometimes called the shell

• Inner pipe connected by U-shaped return bends

enclosed in a return-bend housing to make up a hairpin,so DP HEX = hairpin HEX

• Hairpins are based on modular principles: they can be

arranged in series, parallel, or series-parallelcombinations to meet pressure drop and MTDrequirements; add-remove as necessary

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Usage Areas / Advantages

• Sensible heating / cooling, small HT areas (up to 50 m2)

• High pressure fluids, due to small tube diameters

• Suitable for gas / viscous liquid (small volume fluids)

• Suitable for severe fouling conditions (easy to clean andmaintain)

• Finned tubes can be used to increase HT surface per unit

length, thus reduce length and Nhp

• Outside-finned inner tubes most efficient when low h fluid (oilor gas) flows through annulus

• Multiple tubes can be used inside the shell

• Used as counterflow HEX, so they can be used as analternative to shell-and-tube HEX

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Thermal / Hydraulic Design

Inner Tube• Use correlations to find HT coefficient and friction factor

• Total pressure drop

2

uN

d

2L4f p

2m

hpi

ρ=∆

Annulus• Same procedure as above, but use

– Hydraulic diameter, Dh = 4Ac/Pw for Re calculation

– Equivalent diameter, De = 4Ac/Ph for Nu calculation

• For a hairpin HEX with Bare Inner Tube,

– Dh = Di - do

– De = (Di2 - do

2)/do

Study Example 6.1 (detailed analysis)

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Thermal / Hydraulic Design (continued)

• For a hairpin HEX with Multitube Longitudinal FinnedInner Tubes

– Get Dh and De using

– Unfinned, finned, and total outside HT surface areas

( )

( ) f tf t2o2ic

f tf toh

f tf toiw

NNHNdD4A

)formulaincorrectionnote(NNH2NdP

NNH2NdDP

δ−−π

=

+π=++π=

( )( )

( )f f of ut

f f tf

f otu

HN2dLtN2AAA

H2LNN2A

LNLdN2A

+π=+=

δ+=δ−π=

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– Overall HT coefficient based on outer area of inner tubes

whereis the overall surface efficiency

Area ratios At/Ai and Af /At are neededRw is for bare tube wall

ηf is the efficiency of a rectangular continuouslongitudinal fin (for other types of fins, use references)

* h affects fin efficiency; have the fluid with the poorestHT properties on the finned side

ooo

fowtfi

i

t

ii

tf

h

1RRAR

A

A

hA

A1

U

η+

η+++

=

( ) ⎥⎦

⎤⎢⎣

⎡η−−=η

t

f f o

A

A11

( )*

f f

f

f k

h2

m,mH

mHtanh

δ==η


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• The heat transfer equation is (heat duty equation)

• The design problem, in general, includes determining the totalouter surface area of the inner tubes from the above equation.

• If the length of hairpins is fixed, then Nhp can be calculated.

• U can also be based on the inner area of the inner tubes, Ai

• For counterflow and parallel flow arrangements, no correction

is necessary for ∆ Tm. However, if hairpins are arranged in

series/parallel, a correction must be made (later).• Study Example 6.2 (detailed analysis)

mthp TAUNQ ∆=

( )

mihpi

tii

TANUQ

LNd2A

∆=

π=

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Parallel / Series Arrangement of Hairpins

• If the design indicates large Nhp, it may not be practical toconnect them all in series for pure counterflow. A large

quantity of fluid through pipes may result in ∆p >∆pallowable

• Solution: Separate mass flow into parallel streams, thenconnect smaller mass flow rate side in series. This is aparallel-series arrangement.

• If such a combination is used, the temperature difference of the inner pipe fluid will be different for each hairpin.

• Thus, in each hairpin section, different amounts of heat will

be transferred and true mean temperature difference, ∆ Tm

will be different from the LMTD.

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• The true mean temperature difference in

becomes

dimensionless quantity S is

• For n hairpins, S depends on the number of hot-cold streamsand their series-parallel arrangement.

• Simplest case is to either divide the cold fluid equallybetween n hairpins in parallel or to divide the hot fluidequally between n hairpins in parallel.

mthp TAUNQ ∆=

( )1c1hm T TS T −=∆

( )

( )1c1ht

2h1hp

T TUA

T TcmS

−

−=

&

F i h t fl id d ll l ld t

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• For one-series hot fluid and n1-parallel cold streams,

• For one-series cold fluid and n2-parallel hot streams,

• Then, the total heat transfer rate is

( )1c2c1

2h1h1

1c1h

1c2h1

1

n/1

11

1

1

11

1

T Tn T TR,

T T T TP

R1

P1

R1Rln

1RRn

P1S

1

−−=

−−=

⎥⎥

⎦

⎤⎢⎢

⎣

⎡ +⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ −⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

−

−=

( )

( )

( )1c2c

2h1h22

1c1h

2c1h2

2

n/1

2

2

2

2

2

T T

T TnR,

T T

T TP

R

P

1R1ln

R1

n

P1S2

−

−=

−

−=

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −⎟⎟

⎠

⎞⎜⎜

⎝

⎛

−

−=

( )1c1h T TUASQ −=

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• In the previous equations, it is assumed that U and cp of thefluids are constant, and the heat transfer rates of the two

units are equal.

• Graphs are available in literature for LMTD correction factorF as well.

• If number of tube-side parallel paths is equal to the numberof shell-side parallel paths, regular LMTD should be used.

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Total Pressure Drop

• Total pressure drop includes frictional pressure drop,entrance and exit pressure drops, static-head, and themomentum-change pressure drop.

• Frictional pressure drop is

• For frictional pressure drop, use correlations from Chapter 4or Moody diagram. Add equivalent length of the U-bend to

the L in tube-side (Dh = di) pressure drop.• You may need to account for the effect of property variations

on friction factor.

2

uN

D

2L4f p

2m

hp

h

ρ=∆

T t l P D ( ti d)

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• Entrance and exit pressure drops through inlet and outlet

nozzles is evaluated from

where K c = 1.0 at the inlet and 0.5 at the outlet nozzle.

• Static head is ∆pf =ρ∆H, where ∆H is the elevationdifference between inlet and outlet nozzles.

• For fully developed conditions, momentum-change pressuredrop is

• In all pressure drop calculations for design, allowable ∆pmust be considered.

• Cut-and-twist technique increases h in longitudinal finned-

tube HEX. See book for ∆p details.

Total Pressure Drop (continued)

2

uK p

2m

cn

ρ=∆

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

ρ−

ρ=∆

io

2m

11Gp

D i d O ti l F t

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• In hairpin HEX, two double pipes are joined at one end by a

U-tube bend welded to the inner pipes, and a return bendhousing on the shell-side. The housing has a removablecover to allow removal of inner tubes.

• Double-pipe HEX have four key design components– shell nozzles

– tube nozzles

– return-bend housing and cover plate on U-bend side

– shell-to-tube closure on other side of hairpin(s)

• The longitudinal fins made from steel are welded onto the

inner pipe. Other materials can be joined by soldering.

• Multiple units can be joined by bolts and gaskets.

• For low heat duty applications, simple constructions, easyassembly, lightweight elements and minimum number of parts contribute to minimizing costs.

Design and Operational Features

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IPS: inch per second (unit system)NFA: net flow area

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ME421Lec6