heat exchangers: rating & performance...

Heat Exchangers: Rating & Performance Parameters

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 & Performance ParametersME 307 (2017) 1 / 35

HTX 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.

T791

c


ǫ-NTU Method

ǫ-NTU Method

� If the inlet & outlet temperatures of the hot & 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 ≡

qqmax

=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 , ǫmax for Cr = 0, and ǫmin 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 )

T792

� 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


HTX: Performance Parameters

HTX: Behaviour as Cr → 0

� The effect of the heat exchanger configuration on its performance

is related to the interaction between the temperature changes of

the two fluid streams.

� The limit of Cr → 0 is approached in many practical heat

exchangers; e.g., the interaction of a flowing fluid with a constant

temperature solid (as in a cold-plate) or a well-mixed tank of fluid

or the situation that occurs when one of the two streams is

undergoing constant pressure evaporation or condensation.

c



⊲ If Cr → 0, no such interaction. For all configurations,

limCr→0

ǫ = 1 − exp(−NTU )

T779

Temperature distribution within a (a) counter-flow and (b) parallel-flow

heat exchanger as the capacity ratio approaches zero because Cc >> Ch .

c



HTX: Behaviour as NTU → 0

� If NTU is small, heat exchanger is under-sized & rate of heat

transfer is not sufficient to change the temperature of either fluid

substantially. Hence, ∆T = (Th ,in − Tc,out )

� NTU → 0: q = UA(TH ,in − TC ,in ) , ǫ ≡q

Cmin (TH ,in−TC ,in )

limNTU→0

ǫ = NTU

T780

Temperature distribution within a (a) counter-flow and (b) parallel-flow

heat exchanger as NTU approaches zero.

c



HTX: Behavior as NTU → ∞

T781

Temperature distribution within a balanced (i.e., Cr = 1) (a)

counter-flow and (b) parallel-flow heat exchanger for a finite NTU. The

temperature distribution within a balanced (c) counter-flow and (d)

parallel-flow heat exchanger as NTU → ∞.c



Effectiveness as a function of NTU

T782

ǫ as a function of NTU for various configurations (Cr = 1).

c



T783

ǫ as a function of NTU for various configurations (Cr = 0.5).

c



T784

ǫ as a function of NTU for various configurations (Cr = 0.25).

c


HTX Rating Problems Double Pipe Heat Exchanger (DPHX)

DPHX: Double Pipe Heat Exchanger

T790

c



T787T788

Hairpin heat exchanger.

T789

Multi-tube hairpin heat exchanger.

c



T734

T759

T761

c



� Laminar: Nu = 1.86(Gz )1/3(

µbµw

)0.14; Gz = RePr

L/D

� Turbulent: Nu = 0.023Re0.8Prn ; n =

{0.4 : heating

0.3 : cooling

� Friction factor

f =

{ξcorr [64/Re ] for laminar flow

(1.82 log10 Re − 1.64)−2 for turbulent flow

Tube area:

� Equivalent diameter, Det = Dht = IDt

� Pressure drop, ∆Pt = ftL

IDt(12ρtV

2t ), ξcorr = 1

Annular area:

� Dha = IDa −ODt , Dea =ID2

a−OD2t

ODt

�

1ξcorr

= 1+κ2

(1−κ)2+ 1+κ

(1−κ) ln(κ) , κ = ODt/IDa

� ∆Pa = (faL

Dha+ 1)(1

2ρaV2a )

c



DPHX Rating ⊲ Oil is to be heated from 30oC using hot water at 100oC.

Oil flow rate is 0.06 kg/s in the annulus, while wate flow rate is 0.5 kg/s.The

heat exchanger is made of 2 x 1 1/4 std type M cupper tubing that is 5.0 m

long. Use appropriate fouling factors, rate the DPHX.

T763

c



DPHX Sizing ⊲ Oil is to be heated to 45 oC from 30oC using hot water at

100oC. Oil flow rate is 0.05 kg/s in the annulus, while wate flow rate is 0.5

kg/s.The heat exchanger is made of 2 x 1 1/4 std type M cupper tubing. Use

appropriate fouling factors, size the DPHX.

T764

c


HTX Rating Problems Shell & Tube Heat Exchanger (STHX)

Shell & Tube Heat Exchanger (STHX)

T795

Shell-and-tube evaporator, flooded.

c



T735

c



T758

T760

c



T794

c



Tube area:

� At = Ntπ(ID2t )/4Np, Nt = no. of tubes, Np = no. of pass

� For smooth pipes,

ft =

{0.316Re−0.25 : Re ≤ 2 × 104

0.184Re−0.20 : 2 × 104 ≤ Re ≤ 3 × 105

� Pressure drop, ∆Pt = Np

(

ftL

IDt+ 4

)

(12ρtV

2t )

Shell area:

� Dhs = 4AP =

4

[

P2T−

πOD2t

4

]

πODt=

4P2T

πODt− ODt : square pitch

4

[

√

34

P2T−

πOD2t

8

]

πODt=

2√

3P2T

πODt− ODt : triangular pitch

� As = DsCBPT

, Ds = shell inside diam., B = baffle spacing

� Nu = 0.36Re0.55s Pr1/3 , Res = VsDhs/νs

� fs = exp [0.576 − 0.19 ln(Res)]

� ∆Pa = fs(Nb + 1)DsDe

(12ρsV

2s ), Nb = number of baffles

c



STHX Rating ⊲ 20.0 kg/s water at 50oC is to be cooled using 20.0 kg/s

water available at 25 oC. Nt = 200, Np = 2, L = 5.0, Nb = 15, B = 304.8

mm, Ds = 438.2 mm, IDt = 16.55 mm, ODt = 19.07 mm. Assume 25.4 mm

triangular pitch.

T762

c


HTX Rating Problems Plate & Frame Heat Exchanger (PFHX)

Plate & Frame Heat Exchanger (PFHX)

T765

T766

� NTU = UoAoNs( _mCp)min

, Ao = bl

� Effective NTU = FcorrNTU , Fcorr = 1 − 0.0166NTU

c



b

sT767

� Flow velocity, V =

{2 _m

ρsb(Ns+1) : odd number of plates2 _m

ρsbNs, 2 _m

ρsb(Ns+2) : even number of plates

� Heat transfer & pressure loss correlations are based on the

hydraulic diameter, Dh of the passage formed by the two plates.

� Dh =4(cross−sectional area)

wetted perimeter = 4sb2s+2b ≃

4sb2b = 2s , Re = VDh

ν

� Laminar flow: Nu = 1.86(Gz )1/3(

µbµw

)0.14; Gz = RePr

L/D Re < 100

� Turbulent flow: Nu = 0.374Re0.668Pr1/3; Re > 100

� Friction factor, f =

280Re : 1 < Re < 10

100Re0.589 : 10 < Re < 100

12Re0.183 : Re > 100

� Pressure drop, ∆P = fLDh

(12ρV

2) + 1.3(12ρV

2p ) ≃

fLDh

(12ρV

2)

c



PFHX Rating ⊲ 0.975 kg/s water at 25oC is to be cooled using 1.0 kg/s

water available at 7oC. If b = 457 mm, L = 914 mm, s = 5.08 mm, t = 1.016

mm, k = 14.3 W/mK, Ns = 11, fouling coefficients for both fluids = 3.52 x

10−6, rate the heat-exchanger.

T768

c


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