Heat Exchangers: Rating & Sizing - Iteacher.buet.ac.bd/zahurul/ME307/ME307_HeatExchangerRatingSizing … · Heat Exchangers: Rating & Sizing - I Dr. Md. Zahurul Haq Professor Department

Heat Exchangers: Rating & Sizing - I

Dr. Md. Zahurul Haq

ProfessorDepartment of Mechanical Engineering

Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh

[email protected]://teacher.buet.ac.bd/zahurul/

ME 307: Heat Transfer Equipment Design

http://teacher.buet.ac.bd/zahurul/ME307/

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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Sizing - I ME 307 (2017) 1 / 32

Heat Exchanger: Rating & Sizing

HE Rating is concerned with the determination of the heat

transfer rate, fluid outlet temperatures, and the pressure drop for

an existing heat exchanger or the one which is already sized.

HE Sizing is concerned with the determination of the matrix

dimensions to meet the specified heat transfer and pressure drop

requirements.

c


T718

Application of and contrast between (a) a thermodynamic and (b) a heat

transfer model for a typical shell-and-tube heat exchanger used in

chemical processing.

c


Heat Exchanger: Energy Balance Equation

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T666

_q = _mh (ih ,i − ih ,o) = _mhcp,h(Th ,i − Th ,o) = Ch(Th ,i − Th ,o)_q = _mc(ic,o − ic,i) = _mccp,c(Tc,o − Tc,i ) = Cc(Tc,o − Tc,i )

_q = UA∆Tm

◮ i is the fluid enthalpy, subscripts h and c refer to hot and cold

fluids, whereas the subscripts i and o designate fluid inlet and outlet

conditions.

c


Special Heat Exchanger Conditions

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T671

(a)T672

(b)

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T670

(c)

(a): Ch ≫ Cc or a condensing vapour.

(b): An evaporating liquid or Ch ≪ Cc.

(c): A counterflow heat exchanger with Ch = Cc.

c


Heat Exchanger: Fouling

During operation, heat exchangers become fouled with an

accumulation of deposits of one kind or another on the heat

transfer surface areas.

Major categories of fouling:1 scaling or precipitation fouling2 particulate fouling3 chemical reaction fouling4 corrosion fouling5 biological fouling6 solidification fouling

c


_q = UoAoTLM

1UoAo

= R = 1hiAi

+ Rd,iAi

+

ln(Do/Di )2πkL + Rd,o

Ao+ 1

hoAo

T719

c


Cengel Ex. 11-2 ⊲ Effect of fouling on overall heat transfer coefficient.

Double-pipe (shell-and-tube) heat exchanger made of steel, k = 15.1 W/m-K,

inner-diameter of outside shell is 3.2 cm.

T704

c


LMTD Method

LMTD: Parallel-Flow Heat Exchanger

T702

_q = UA∆TLM

d _q = − _mhcp,hdTh = −ChdTh

d _q = _mccp,cdTc = CcdTc

d _q = U (Th − Tc)dA

d(∆T ) = dTh − dTc

d∆T = −[

1Ch

+ 1Cc

]

d _q

d(∆T )∆T = −U

[

1Ch

+ 1Cc

]

dA

∆T1 = (Th ,i − Tc,i )

∆T2 = (Th ,o − Tc,o)

_q = UA[

∆T2−∆T1

ln(∆T2/∆T1)

]

= UA∆TLM

c


LMTD Method

LMTD: Counter-Flow Heat Exchanger

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T667

_q = UA∆TLM

∆T1 = (Th ,i − Tc,o)

∆T2 = (Th ,o − Tc,i )

_q = UA[

∆T2−∆T1

ln(∆T2/∆T1)

]

= UA∆TLM

c


LMTD Method

Cengel Ex. 11-4 ⊲ Heating Water in a Counter-Flow Heat Exchanger: If Uo

= 640 W/m2K, determine the length of the heat exchanger required to achieve

the desired heating. Re-estimate the length for Parallel-Flow configuration.

T707

c


LMTD Method

Correction for LMDT: Cross-Flow & Multipass HE

If a heat exchanger other than the double-pile type is used, _q is

calculated by using a correction factor, F applied to LMDT for

counter-flow arrangement with same hot and cold fluid

temperatures.

_q = FUA∆TLM

T ≡ shell-side temperatures

t ≡ tube-side temperatures

F = 1.0 for boiling and condensation.

F-values are presented in figures, using two dimensionless

parameters:1 P = t2−t1

T1−t1: 0 ≤ P ≤ 1.0: thermal efficiencies of tube-side flow.

2 R = T1−T2

t2−t1=

(mcp)tube−side

(mcp)shell−side: 0 ≤ R ≤ ∞:

R → 0: vapour condensation on the shell side

R → ∞: evaporation on the tube side.c


LMTD Method

T714

Correction-factor plot for exchanger with one shell pass and two, four,

or any multiple of tube passes.

c


LMTD Method

T715

Correction-factor plot for exchanger with two shell passes and four,

eight, or any multiple of tube passes.

c


LMTD Method

T716

Correction-factor plot for single-pass cross-flow exchanger, both fluids

unmixed.

c


LMTD Method

T717

Correction-factor plot for single-pass cross-flow exchanger, one fluid

mixed, the other unmixed.

c


LMTD Method

Holman Ex. 10-4 ⊲ In a counterflow double-pipe heat exchanger, water at

the rate of 68 kg/min is heated from 35 to 75oC by an oil having a specific

heat of 1.9 kJ/kgoC. Oil enters the exchanger at 110oC and leaves at 75oC.

Given that, Uo = 320 W/m2oC. Estimate heat transfer area, A. 15.82 m2.

Holman Ex. 10-5 ⊲ Instead of the double-pipe heat exchanger of Ex. 10-4,

it is desired to use a shell-and-tube exchanger with the water making one

shell pass and the oil making two tube passes. Calculate the area, A,

assuming that the overall heat-transfer coefficient remains at 320 W/m2oC.

c


LMTD Method

Holman Ex. 10-7 ⊲ A cross-flow heat exchanger, one fluid mixed and one

unmixed, is used to heat an oil in the tubes (c = 1.9 kJ/kg oC) from 15oC to

85oC. Steam (5.2 kg/s, c = 1.86 kJ/kgoC) blows across the outside of the

tube, enters at 130oC and leaves at 110oC. Uo = 275 W/m2 oC . Calculate A.

c


LMTD Method

Cengel Ex. 11-3 ⊲ Steam condenser with surface area of the tubes is 45 m2,

and the overall heat transfer coefficient is 2100 W/m2K.

T706

c


LMTD Method

Cengel Ex. 11-6 ⊲ Cooling of Water in an Automotive Radiator: The

radiator has 40 tubes of internal diameter 0.5 cm and length 65 cm in a

closely spaced plate-finned matrix. Hot water enters the tubes at a rate of 0.6

kg/s. Determine the overall heat transfer coefficient Ui of this radiator based

on the inner surface area of the tubes.

T709

c


LMTD Method

Ozisik Ex. 11-2 ⊲ A heat exchanger is to be designed to cool 8.7 kg/s an

ethyl alcohol solution [cp,h = 3840 J/kgoC] from 75oC to 45oC with cooling

water [cp,c = 4180 J/kgoC entering the tube side at 15oC at a rate of

9.6 kg/s. Given, Uo = 500 W/m2oC. Estimate heat transfer area, for:

1 parallel flow, shell and tube 85.7 m2

2 counter flow, shell and tube 57.3 m2

3 one shell pass and two tube pass 65.9 m2

4 cross-flow, both fluids unmixed 61.2 m2

5 cross-flow, one fluid unmixed 63.7 m2

c


ǫ-NTU Method

ǫ-NTU Method

If the inlet and outlet temperatures of the hot and cold fluid, and

the overall heat transfer coefficients are specified, the LMDT

method can be used to solve rating and sizing problem (e.g.

process, power and petrochemical industries).

If only the inlet temperatures and fluid (hor & cold) flow rates are

given LMDT estimation requires cumbersome iterations.

ǫ-NTU method is generally performed for the design of compact

heat exchangers for automotive, aircraft, air conditioning, and

various industrial applications where the inlet temperatures of the

hot and cold fluids are specified and the heat transfer rates are to

be determined.

c


ǫ-NTU Method

Maximum Heat Transfer Rate, qmax

Maximum HT could be achieved in counter-flow HE with x → ∞.

If Cc < Ch : |∆Tc | > |∆Th |, COLD fluid have ∆Tmax .

⇒ If x → ∞ ⇒ Tc,o = Th ,i ⇛ qmax = Cc(Th ,i − Tc,i )

If Cc > Ch : |∆Tc | < |∆Th |, HOT fluid have ∆Tmax .

⇒ If x → ∞ ⇒ Th ,o = Tc,i ⇛ qmax = Ch(Th ,i − Tc,i )

=⇒

qmax = Cmin(Th ,i − Tc,i ) = Cmin∆Tmax

=⇒

Effectiveness, ǫ ≡q

qmax=

Ch (Th,i−Th,o)

Cmin (Th,i−Tc,i )or Cc(Tc,o−Tc,i )

Cmin (Th,i−Tc,i )

c


ǫ-NTU Method

ǫ-NTU Method (Parallel-Flow Heat Exchanger)

Number of Transfer Unit, NTU ≡UA

Cmin

Capacity ratio, Cr ≡Cmin

Cmax

d(∆T )∆T = −U

[

1Ch

+ 1Cc

]

dA → ln(∆T2/∆T1) = −UA[

1Ch

+ 1Cc

]

=⇒∆T2∆T1

= Th ,o−Tc,oTh,i−Tc,i

= exp(−NTU (1 +Cr ))

If Cmin = Ch :, Cr = ChCc

=

_mhCp,h

_mccp,c= Tc,o−Tc,i

Th,i−Th,o

⇒ ǫ =Cc(Tc,o−Tc,i )

Cmin (Th,i−Tc,i )= 1

Cr

Tc,o−Tc,iTh,i−Tc,i

= Th,i−Th,oTh,i−Tc,i

⇒Th ,o−Tc,oTh,i−Tc,i

= −ǫ+ 1 −Crǫ → ǫ =1−exp(−NTU (1+Cr ))

1+Cr

Same result is obtained, if Cmin = Cc

ǫ =1−exp(−NTU (1+Cr ))

1+Cr: Parallel-Flow HTX

c


ǫ-NTU Method

T710

c


ǫ-NTU Method

T711

c


ǫ-NTU Method

T712

c


ǫ-NTU Method

T713

c


ǫ-NTU Method

The value of capacity ratio, Cr ranges between 0 and 1.0

For a given NTU , the effectiveness becomes maximum for Cr = 0,

and a minimum for Cr = 1.0

Cr = Cmin/Cmax → 0 corresponds to Cmax → ∞, which is realized

during a phase-change process in a condenser or boiler. Hence,

ǫ = ǫmax = 1 − exp(−NTU )

Cr → 0 is applicable where heat exchanger is in contact with the

ambient environment as Cmax is a very large number.

Effectiveness is lowest for Cr = Cmin/Cmax = 1, when heat rates

of two fluids are the same. Example, HVAC applications.

c


ǫ-NTU Method

Holman Ex. 10-7 ⊲ A cross-flow heat exchanger, one fluid mixed and one

unmixed, is used to heat an oil in the tubes (c = 1.9 kJ/kg oC) from 15oC to

85oC. Steam (5.2 kg/s, c = 1.86 kJ/kgoC) blows across the outside of the

tube, enters at 130oC and leaves at 110oC. Uo = 275 W/m2 oC . Calculate A.

Holman Ex. 10-9 ⊲ Investigate the heat-transfer performance of the

exchanger in Ex. 10-7 if the oil flow rate is reduced in half while the steam

flow remains the same. Assume U remains constant at 275 W/m2oC.

c


ǫ-NTU Method

Holman Ex. 10-4 ⊲ In a counterflow double-pipe heat exchanger, water at

the rate of 68 kg/min is heated from 35 to 75oC by an oil having a specific

heat of 1.9 kJ/kgoC. Oil enters the exchanger at 110oC and leaves at 75oC.

Given that, Uo = 320 W/m2oC. Estimate heat transfer area, A. 15.82 m2.

Holman Ex. 10-10 ⊲ The heat exchanger of Ex. 10-4 is used for heating

water. Using the same entering-fluid temperatures, calculate the exit water

temperature when only 40 kg/min of water is heated but the same quantity of

oil is used. Also calculate the total heat transfer under these new conditions.

c


ǫ-NTU Method

Cengel Ex. 11-8 ⊲ The overall heat transfer coefficient of the heat

exchanger is 640 W/m2K. Using the effectiveness-NTU method determine the

length of the heat exchanger required to achieve the desired heating.

T707

c


Documents

Heat Exchangers: Rating & Sizing - Iteacher.buet.ac.bd/zahurul/ME307/ME307_HeatExchangerRatingSizing … · Heat Exchangers: Rating & Sizing - I Dr. Md. Zahurul Haq Professor Department