heat exchangers: rating & performance...
TRANSCRIPT
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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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
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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.
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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 )
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 4 / 35
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
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ǫ-NTU Method
T710
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ǫ-NTU Method
T711
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ǫ-NTU Method
T712
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ǫ-NTU Method
T713
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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.
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 10 / 35
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.
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 11 / 35
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.
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 12 / 35
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
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 13 / 35
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.
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 14 / 35
HTX: Performance Parameters
⊲ 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 .
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 15 / 35
HTX: Performance Parameters
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.
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HTX: Performance Parameters
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
Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 17 / 35
HTX: Performance Parameters
Effectiveness as a function of NTU
T782
ǫ as a function of NTU for various configurations (Cr = 1).
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HTX: Performance Parameters
T783
ǫ as a function of NTU for various configurations (Cr = 0.5).
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HTX: Performance Parameters
T784
ǫ as a function of NTU for various configurations (Cr = 0.25).
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HTX Rating Problems Double Pipe Heat Exchanger (DPHX)
DPHX: Double Pipe Heat Exchanger
T790
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HTX Rating Problems Double Pipe Heat Exchanger (DPHX)
T787T788
Hairpin heat exchanger.
T789
Multi-tube hairpin heat exchanger.
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HTX Rating Problems Double Pipe Heat Exchanger (DPHX)
T734
T759
T761
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HTX Rating Problems Double Pipe Heat Exchanger (DPHX)
� 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 )
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 24 / 35
HTX Rating Problems Double Pipe Heat Exchanger (DPHX)
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
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 25 / 35
HTX Rating Problems Double Pipe Heat Exchanger (DPHX)
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
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HTX Rating Problems Shell & Tube Heat Exchanger (STHX)
Shell & Tube Heat Exchanger (STHX)
T795
Shell-and-tube evaporator, flooded.
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HTX Rating Problems Shell & Tube Heat Exchanger (STHX)
T735
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HTX Rating Problems Shell & Tube Heat Exchanger (STHX)
T758
T760
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HTX Rating Problems Shell & Tube Heat Exchanger (STHX)
T794
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HTX Rating Problems Shell & Tube Heat Exchanger (STHX)
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
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 31 / 35
HTX Rating Problems Shell & Tube Heat Exchanger (STHX)
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
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 32 / 35
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
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 33 / 35
HTX Rating Problems Plate & Frame Heat Exchanger (PFHX)
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)
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Dr. Md. Zahurul Haq (BUET) Heat Exchangers: Rating & Performance ParametersME 307 (2017) 34 / 35
HTX Rating Problems Plate & Frame Heat Exchanger (PFHX)
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
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