Download - heat exchangers
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HEAT EXCHANGERS
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Heat Exchangers
• A heat exchanger is used to exchange heat between two fluids of different temperatures, which are separated by a solid wall.
• Heat exchangers are ubiquitous to energy conversion and utilization. They encompass a wide range of flow configurations.
• Applications in heating and air conditioning, power production, waste heat recovery, chemical processing, food processing, sterilization in bio-processes.
• Heat exchangers are classified according to flow arrangement and type of construction.
Heat Exchangers: Heat Transfer Process
• DIRECT CONTACT TYPE – in this type, two immiscible fluids at different temperatures come in direct contact. i.e. jet condensers, desuperheaters, open feed-water heaters, scrubbers and cooling towers.
• TRANSFER TYPE/RECUPERATORS - in this type, the cold and hot fluids flow simultaneously through the device, and the heat is transferred through the wall separating them. This type is commonly used in most fields of engineering.
• REGENERATORS/ STORAGE TYPE – in this type, hot and cold fluids flow alternately on the same surface. This type is used as preheaters for steam power plants, blast furnaces, and oxygen producers.
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Heat Exchangers: Construction Features• TUBULAR HEAT EXCHANGERS – also known as tube in
tube, concentric tube or double-pipe heat exchangers. • SHELL AND TUBE HEAT EXCHANGERS – also known as
surface condensers and are most commonly used for heating, cooling, condensation, or evaporation applications.
• FINNED TUBE TYPE HEAT EXCHANGERS – when a high operating pressure or an enhanced heat transfer rate is required, extended surfaces are used on one side of the heat exchangers. This type are used for liquid to gas heat exchange. i.e. gas turbines, automobiles, airplanes, heat pumps, refrigeration, electronics, cryogenics, air conditioning, humidification.
• COMPACT HEAT EXCHANGERS – used for high area density heat transfer
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Finned Exchangers
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Heat Exchanger Applications
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Concentric Tube Construction
• - :• :
Parallel Flow CounterflowParallel Flow Counterflow
Parallel Flow CounterflowParallel Flow Counterflow
Heat Exchangers: Flow Arrangement
• PARALLEL FLOW - also known as concurrent heat exchangers, hot and cold fluids enter through the same point and leaves at the other end.
• COUNTERCURRENT FLOW – with this type, hot and cold fluids enter at opposite ends of the heat exchanger.
• CROSS FLOW – here, two fluids flow at a right angle to each other. For this type, it can further be classified as unmixed flow or mixed flow.
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Parallel Flow
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Counter Current Flow
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Heat Exchanger Analysis• Expression for convection heat transfer for flow of a fluid inside a tube:
)( ,, imompconv TTcmq
• For case 3 involving constant surrounding fluid temperature:
lms TAUq
)/ln( io
iolm TT
TTT
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Heat Exchanger Analysis
In a two-fluid heat exchanger, consider the hot and cold fluids separately:
)(
)(
,,,
,,,
icoccpcc
ohihhphh
TTcmq
TTcmq
lmTUAq
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Tlm: 1. Parallel-Flow Heat Exchangers
where
Parallel Flow CounterflowParallel Flow Counterflow
lmTUAq
)/ln( 12
12
TT
TTTlm
ocoh
icih
TTT
TTT
,,2
,,1
T1 T2
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Tlm: 2. Counter-Flow Heat Exchangers
where
lmTUAq
icoh
ocih
TTT
TTT
,,2
,,1
Parallel Flow CounterflowParallel Flow Counterflow
T1 T2
)/ln( 12
12
TT
TTTlm
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Overall Heat Transfer Coefficient
• For tubular heat exchangers we must take into account the conduction resistance in the wall and convection resistances of the fluids at the inner and outer tube surfaces.
kL
DDR
AhR
AhUA
iocond
oocond
ii
2
111
)/ln(
Parallel Flow CounterflowParallel Flow Counterflow
where inner tube surface
outer tube surface LDA
LDA
oo
ii
ooii AUAUUA
111
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Shell-and-Tube Heat Exchangers
Baffles are used to establish a cross-flow and to induce turbulent mixing of the shell-side fluid, both of which enhance convection.
The number of tube and shell passes may be varied
One Shell Pass and One Tube Pass
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One Shell Pass,Two Tube Passes
Two Shell Passes,Four Tube Passes
Shell-and-Tube Heat Exchangers
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Multipass and Cross-Flow Heat Exchangers
To account for complex flow conditions in multipass, shell and tube and cross-flow heat exchangers, the log-mean temperature difference can be modified:
CFlmlm TFT ,
where F = correction factor
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Correction Factor
where t is the tube-side fluid temperature
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Fouling
• Heat exchanger surfaces are subject to fouling by fluid impurities, rust formation, or other reactions between the fluid and the wall material. The subsequent deposition of a film or scale on the surface can greatly increase the resistance to heat transfer between the fluids.
• An additional thermal resistance, can be introduced: The Fouling factor, Rf.
Depends on operating temperature, fluid velocity and length of service of heat exchanger. It is variable during heat exchanger operation.
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Overall Heat Transfer Coefficient
•The overall heat transfer coefficient can be written:
ooo
ofcond
i
if
iiooii AhA
RR
A
R
AhAUAUUA
11111
",
",
oofcondo
i
ifo
ii
o
o
hRRA
A
RA
AhA
U1
1
",
",
Need to determine hi and ho
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Determination of ho
• Approach 1: Using correlations Approach 2: Using chart by Kern
Typical values of baffle cuts 20-25% for liquids and 40-45% for vapor
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Determination of tube side film coefficient, hi
• Approach 1: Using correlations Approach 2: Sieder and Tate
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Determination of Conduction Resistance
• Recall that
)/ln(2
2
)/ln(
ioo
condo
iocond
DDk
DRA
kL
DDR
• or
)/ln(2 io
w
ocondow DD
k
DRAr
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Example
In a heat exchanger, hot fluid enters at 60OC and leaves at 48OC, where as the cold fluid enters at 35OC and leaves at 44OC. Calculate the mean temperature difference for
a) parallel flow,
b) counter flow,
c) single pass cross flow (both fluids unmixed)
d) single pass cross flow (hot side fluid mixed, cold side fluid unmixed)
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Thi = Tha = 60OC Tho = Thb = 48OCTci = Tca = 35OC Tco = Tcb = 44OC
)/ln( 12
12
TT
TTTlm
ocoh
icih
TTT
TTT
,,2
,,1
)4/25ln(
425 lmT
253560
44448
2
1
T
T
CTlmΟ5.11
Parallel Flow
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Thi = Tha = 60OC Tho = Thb = 48OCTci = Tca = 35OC Tco = Tcb = 44OC
icoh
ocih
TTT
TTT
,,2
,,1
)/ln( 12
12
TT
TTTlm
164460
133548
2
1
T
T)13/16ln(
1316 lmT
CT Olm 45.14
Countercurrent Flow
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Z = Thi – Tho = 60 – 48 = 1.33 Tcb – Tci 44 – 35
ŋH = Tco – Tci = 44 – 35 = 0.36 Thi – Tci 60 – 35
Thi = Tha = 60OC Tho = Thb = 48OCTci = Tca = 35OC Tco = Tcb = 44OC
Cross Flow
Single pass cross flow; both fluids unmixed
From Figure 15.7 (b), F = 0.94
LMTD = (0.94)(14.45OC) = 13.583OC
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Z = Thi – Tho = 60 – 48 = 1.33 Tcb – Tci 44 – 35
ŋH = Tco – Tci = 44 – 35 = 0.36 Thi – Tci 60 – 35
Thi = Tha = 60OC Tho = Thb = 48OCTci = Tca = 35OC Tco = Tcb = 44OC
Cross Flow
Single pass cross flow; hot fluid mixed, cold fluid unmixed
From Figure 15.7 (a), F = 0.98
LMTD = (0.98)(14.45OC) = 14.16OC
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Example
A 1-2 heat exchanger containing one shell pass and two tube passes heats 2.52 kg/s of water from 21.1 to 54.4OC by using hot water under pressure entering at 115.6 and leaving at 48.9OC. The outside surface area of the tubes in the exchanger is Ao = 9.30 m2.
a)Calculate the mean temperature difference LMTD in the exchanger and the overall heat transfer coefficient Uo.
b)For the same temperature but using a 2-4 exchanger, what would be the LMTD?
Thi = Tha = 115.6OC Tho = Thb = 48.9OCTci = Tca = 21.1OC Tco = Tcb = 54.4OC
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q = mCp,c (Tco –Tci) = 2.52 kg/s (4184 J/kg.K)(54.4 – 21.1OC)q = 351104.54 W
LMTD = (115.6 – 54.4) – (48.9 – 21.1) = 42.3261OC ln (115.6 – 54.4) (48.9 – 21.1)
CFlmlm TFT ,For multipass and crossflow heat exchangers
Example
LMTD for crossflow;LMTD = (Tha – Tcb ) – (Thb – Tca)
ln (Tha – Tcb ) (Thb – Tca)
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a) Single pass, shell fluid mixed, other fluids unmixed, use Figure 15.7 a ( 1 – 2 Heat Exchangers)
Z = Thi – Tho = 115.6 – 48.9 = 2.00 Tcb – Tci 54.4 – 21.1
ŋH = Tco – Tci = 54.4 – 21.1 = 0.352 Thi – Tci 115.6 – 21.1
From Figure 15.7 (a), F = 0.74 LMTD = (0.74)( 42.3261) = 31.3213OC
Uo = q = 351104.54 W = 1205.351 W/m2.K Ao (LMTD) (9.30 m2)(31.3213OC)
Example
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b) Single pass, both fluids unmixed, use Figure 15.7 b ( 2 – 4 Heat Exchangers)
Z = Thi – Tho = 115.6 – 48.9 = 2.00 Tcb – Tci 54.4 – 21.1
ŋH = Tco – Tci = 54.4 – 21.1 = 0.352 Thi – Tci 115.6 – 21.1
From Figure 15.7 (b), F = 0.94 LMTD = (0.94)( 42.3261) = 33.60693OC
Uo = q = 351104.54 W = 1123.375 W/m2.K Ao (LMTD) (9.30 m2)(33.60693OC)
Example
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Example
A stainless steel tube (k = 45 W/m.K) of inner and outer diameters of 22 mm and 27 mm respectively, is used in a cross flow heat exchanger. The fouling factors for the inner and outer surfaces are estimated to be 0.0004 and 0.0002 (m2.K)/W respectively. Determine the overall heat transfer coefficient based on the outside surface area of the tube.
Water at 75OC & 0.5 m/s
Di = 22 mm Do = 27 mm
Air at 15OC & 20 m/s
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Properties of water at 75OCk = 0.6715 W/m.Kν = 0.39 x 10-6 m2/sPr = 2.38
Nu = hiDi = 0.023 Re0.8 Pr0.4 k = 0.023 (28205.13)0.8 (2.38)0.4 = 118.2
hi = 118.2 (0.6715) = 3608 W/m2.K 0.022
Re = u D = 0.5 m/s (0.022 m) = 28205.13 (turbulent) ν 0.39 x 10-6 m2/s
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Properties of air at 15OCk = 0.0255 W/m.Kν = 14.16 x 10-6 m2/sPr = 0.704
Nu = hoDo = [0.04 Re0.5 + 0.06 Re0.67] Pr0.4 (μ/μw)0.25 k = [0.04(38135.59)0.5 + 0.06(38135.59)0.67 ](0.704)0.4(1)0.25
= 139.3021
ho = 139.3021 (0.0255) = 131.5631 W/m2.K 0.027
Re = u D = 20 m/s (0.027 m) = 38135.59 ν 14.16 x 10-6 m2/s
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ooo
ofcond
i
if
iiooii AhA
RR
A
R
AhAUAUUA
11111
",
",
oofcondo
i
ifo
ii
o
o
hRRA
A
RA
Ah
AU
1
1
",
",
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Ro = 1 = 1 = 0.0896 hoAo (131.5631 W/m2.K)(П)(0.027) (1)
Ri = 1 = 1 = 0.00401 hiAi (3608 W/m2.K)(П)(0.022) (1)
Rfi = Fi = 0.0004 = 5.787 x 10 -3
Ai (П)(0.022) (1)
Rfo = Fo = 0.0002 = 2.358 x 10 -3
Ao (П)(0.027) (1)
Rcond = ln Do/Di = ln (0.027/0.022) = 7.24 x 10 -3 2 П k L 2 П (45) (1)
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ooo
ofcond
i
if
iioo AhA
RR
A
R
AhAU
111"
,"
,
0.0896 10 x 2.358 10 x 7.24 10 x 5.78700401.01 3-3-3-
oo AU
0.1024891
oo AU
.K W/m115.0296 2oU
oo
UA
0.102489
1
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A shell and tube heat exchanger with one shell pass and two tube passes is used to heat water (flowing in the tubes) at a rate of 10 kg/s from 30OC to 45OC with steam condensing over the tubes at 160OC. If the overall heat transfer coefficient (based on the outside area) has a value of 2000 W/m2.K, determine the area required. If 20 tubes of 25 mm OD are used. Determine the length of tube required.
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ExampleA counterflow, concentric tube heat exchanger is used to cool the lubricating oil for a large industrial gas turbine engine. The flow rate of cooling water through the inner tube (Di=25 mm) is 0.2 kg/s, while the flow rate of oil through the outer annulus (Do=45 mm) is 0.1 kg/s. The oil and water enter at temperatures of 100 and 30°C respectively. How long must the tube be made if the outlet temperature of the oil is to be 60°C?
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Example A shell-and-tube heat exchanger must be designed to heat 2.5 kg/s of water from 15 to 85°C. The heating is to be accomplished by passing hot engine oil, which is available at 160°C, through the shell side of the exchanger. The oil is known to provide an average convection coefficient of ho=400 W/m2.K on the outside of the tubes. Ten tubes pass the water through the shell. Each tube is thin walled, of diameter D=25 mm, and makes eight passes through the shell. If the oil leaves the exchanger at 100°C, what is the flow rate? How long must the tubes be to accomplish the desired heating?
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Heat Exchanger Effectiveness
Defined as the ratio of actual rate of heat transfer in a given exchanger to the maximum possible amount of heat transfer if an infinite heat transfer area were available.
)(
)(
,,,
,,,
icoccpcc
ohihhphh
TTcmq
TTcmq
ch
ccpc
hhph
CCthen
Ccm
Ccm
;
,
,
The cold fluid undergoes a greater temperature change, hence CC will be Cmin as minimum heat capacity and if there is an infinite area available then Tco = Thi.
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Heat Exchanger Effectiveness
)(
)(
,,
,,
icihc
ohihh
TTC
TTCe
)min(
)max(
,,
,,
icih
ohih
TTC
TTCe
Actual heat transfer
)min( ,, icih TTeCq
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)min(
)(
,,
,,
icih
ohih
TTC
TTChe
Heat Exchanger Effectiveness
In the case of a single pass, counter current flow
)min(
)(
,,
,,
icih
icoc
TTC
TTCce
)]/()ln[(
)()()(
,,,,
,,,,,,
ocihicoh
ocihicohicoc
TTTT
TTTTUATTCcq
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e = 1 – exp – UA 1 – Cmin Cmin Cmax
1 – Cmin exp – UA 1 – Cmin Cmax Cmin Cmax
Where: Number of Transfer Units is NTU = UA Cmin
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e = 1 – exp – UA 1 + Cmin Cmin Cmax
1 + Cmin Cmax
For parallel flow
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Example
Water flowing at a rate of 0.667 kg/s enters a counter current heat exchanger at 308 K and is heated by an oil stream entering at 383 K at a rate of 2.85 kg/s (Cp = 1.89 kJ/kg.K). The overall U = 300 W/m2.K and the area A = 15.0 m2. Calculate the heat transfer rate and the exit water temperature.
Assuming Tco = 370 K; Tf = (308 + 370 K) = 339 K Cp,C = 4.192 kJ/kg.K
ṁ Cp,h = Ch = 2.85 kg/s ( 1.89 x 103 J/kg.K) = 5386.5 W/K
ṁ Cp,c = Cc = 0.667 kg/s ( 4.192 x 103 J/kg.K) = 2796.06 W/K
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For counter current flow exchanger, using Figure 15.9 (a) e = 0.71
q = e Cmin (Thi – Tci) = 0.71(2796.06)(383 – 308) = 148890.4 W
q = 148890.4 W = 2796.06 (Tco – 308) Tco = 361.25 K
Cmin = 2796.06 = 0.5191 Cmax 5386.5
NTU = UA = 300 (15.0) = 1.6094 Cmin 2796.06
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A double pipe, parallel flow heat exchanger uses oil (Cp=1.88 kJ/kgOC) at an initial temperature of 205OC to heat water, flowing at 225 kg/h, from 16OC to 44OC. The oil flow rate is 270 kg/h a)what heat exchanger area is required fro an overall heat transfer coefficient of 340 W/m2.K b)determine the number of transfer units c)calculate the effectiveness of HE
Example
(225 kg/h)(4.18 x 103 kJ/kg.OC)(44 – 16) OC = (270 kg/h)(1.88 x103 kJ/kg.OC) (205 – Tho) OC
Tho = 153.12OCLMTD for crossflow;
LMTD = (Tha – Tcb ) – (Thb – Tca) ln (Tha – Tcb ) (Thb – Tca)
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LMTD for crossflow;LMTD = (205 – 44 ) – (153.12 – 16) = 148. 7406OC
ln (205 – 44 ) (153.12 – 16)
A = 225 (4.18 x 103) (44 – 16) (1/3600) = 0.1446 m2
340 (148.7406OC)
Cmin = (mCp)H = 270 (1.88 x 103) = 507,600 J/h.OC
Cmax = (mCp)C = 225 (4.18 x 103) = 940,500 J/h.OC
NTU = UA = 340 (0.1446) = 0.3487 Cmin 507,600 (1/3600)
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e = 1 – exp – UA 1 + Cmin Cmin Cmax
1 + Cmin Cmax
e = 1 – exp – 0.3487 1 + 507,600 940,500
1 + 507,600
940,500
e = 0.2698 = 26.98 %
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In an open heart surgery, under hypothermic conditions, the patient’s blood is cooled before the surgery and rewarmed afterward. It is proposed that a concentric tube counterflow heat exchanger of 0.5 m length be used for this purpose with a thin-walled inner tube that has a diameter of 55 mm. If the water at 60OC and 0.10 kg/s is used to heat the blood entering the exchanger at 18OC and 0.05 kg/s, what is the temperature of the blood leaving the exchanger? The overall heat exchanger coefficient is 500 W/m2.K and the specific heat of the blood is 3500 J/kg.K.
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A steam condenser condensing at 70OC has to have a capacity of 100 kW. Water at 20OC is used and the outlet water temperature is limited to 45OC. If the overall heat transfer coefficient has a value of 3100 W/m2.K, determine the area required.
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Heat Transfer Coefficients
The heat transfer coefficient hi for the tube side fluid in a shell and tube exchanger can be calculated using equations in chapter 12. the coefficient for the shell side ho cannot be calculated because the direction of flow is partly parallel to the tubes and partly across them. An approximate but general useful equation for predicting shell side coefficients is the Donohue equation.
hoDo = 0.2 DoGe 0.6 Cpμ 0.33 μ 0.14
k μ k μw
Ge = √ Gb Gc Gb = ṁ / Sb Gc = ṁ / Sc
Mass velocity parallel with tubes
Mass velocity for crossflow
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Sb = ƒb π Ds2 – Nb π Do
2
4 4
ƒb = fraction of cross-sectional area of shell occupied by baffle windowDs = inside daimeter of shellNb = number of tubes in baffle windowDo = outside diameter of tubes
Total area of baffle window
Total area occupied by the tubes
Sc = P Ds ( 1 – Do/ƿ)
P = center to center distance between tubesǷ = baffle pitch
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Heat Transfer Coefficients
ho Cpμ 2/3 μw 0.14 = jH = 0.2 DoGe -0.4 CpGe k μ μ
j - factor form
Crossflow Exchangers
hoDo = 0.287 DoG 0.61 Cpμ 0.33 Fa k μ k
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Example A tubular exchanger with 35-in (889-mm) ID contains eight hundred and twenty eight ¾ -in (19-mm) OD tubes 12 ft (3.66-mm) long on a 1-in (25-mm) square pitch. Standard 25 percent baffles are spaced 12-in (305 mm) apart. Liquid benzene at an average bulk temperature of 60OF (15.6OC) is bing heated in the shell side of the exchanger at the rate of 100,000 lb/h (45,360 kg/h). If the outside surfaces of the tubes are at 140OF (60OC), estimate the individual heat transfer coefficient of the benzene.
Do = 0.75 / 12 = 0.0625 ft Ds = 35 / 12 = 2.9167 ft
Ƿ = 1/ 12 = 0.0833 ft P = 1 ft
Sc = P Ds ( 1 – Do/ƿ)
Sc = 1 x 2.9167 1 – 0.0625 0.0833
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Nb = 0.1955 x 828 = 161.8 ≈ 161 tubes
Sb = ƒb π Ds2 – Nb π Do
2
4 4
Sb = 0.1955 π (2.9167)2 – 161 π (0.0625)2
4 4 Sb = 0.8123 ft2
Ge = √ Gb Gc Gb = ṁ / Sb Gc = ṁ / Sc
Gc = 100,000 = 137,137 lb/ft2.h 0.7292
Gb = 100,000 = 123,107 lb/ft2.h 0.8123
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Ge = √ (137,137)(123,107) = 129,933 lb/ft2.h
Properties:μ @ 60OF = 0.70 cP μ @ 140OF =m0.38 cPCp = 0.41 Btu/lb.OF k = 0.092 Btu/h.ft.OF
hoDo = 0.2 DoGe 0.6 Cpμ 0.33 μ 0.14
k μ k μw
ho = 0.2 0.0625(129,933) 0.6 (0.41)(0.70)(2.42) 0.33 0.70 0.14 0.092
0.70 (2.42) 0.092 0.38 0.0625
ho = 101 Btu/h.ft2.OF (573 W/m2.OC)
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HEAT TRANSFER IN AGITATED VESSELS
Heat transfer surfaces, which may be in the form of heating or cooling jackets or coils of pipe immersed in the liquid, are often used in the agitated vessels.
hjDi = 0.76 Da2ŋρ 2/3 Cpμ 1/3 μ 0.24 k μ k μw
hcDc = 0.17 Da2nρ 0.67 Cpμ 0.37 Da
0.1 Dc 0.5 μ 0.24
k μ k Dt Dt μw
With helical coil and turbine impeller
To or from the jacket of a baffled tank
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hjDt = K Da2ŋρ a Cpμ 1/3 μ 0.18 k μ k μw
Anchored agitators for viscous liquid
K = 1.0 a = ½ for 10 < Re < 300
K = 0.36 a = 2/3 for 300 < Re < 40,000
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Example
A turbine agitated vessel 3 m in diameter contains 6600 kg of a dilute aqueous solution. the agitator is 0.75 m in diameter and turns at 150 rev/min. the vessel is jacketed with steam condensing at 38 psig; the heat transfer area is 16 m2. the steel walls of the vessel are 10-mm thick. If the heat transfer coefficient of the steam is 12 kW/m2.OC, what is the rate of heat transfer between the steam and liquid and the time to heat the vessel contents from 30OC to 90OC?
P = 52.7 psiaTs = 284.216OF = 140.12OC
Ŋ = 150 rev/min δ = 10 mmA = 16 m2 hc = 12 kW/m2.OCTa = 30OC Tb = 90OC
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hjDi = 0.76 Da2ŋρ 2/3 Cpμ 1/3 μ 0.24 k μ k μw
Properties at Tf = 30 + 90 = 60OC = 140 OF 2
k = 0.378 Btu/ft.h.OFμ = 0.470 cP = 0.47 x 10-3 kg/m.sρ = 61.38 lb/ft3 μw = 0.196 cP
hj (3) = 0.76 0.7522.5(983.65) 2/3 4184(4.7x10-3) 1/3 0.47 0.24 0.6542 4.7 x 10-3 0.6542 0.196
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hj = 6061.436 W/m2.OC
Ui = 1 1 + 1 3 + 10/1000 3 6061.436 12000 3.02 45 3.01
Ui = 2131.093 W/m2.OC
tT = 6600 (4184) ln (140.12 – 30)/(140.12 – 90) 2131.093 (16)
tT = 637.487 sec = 10.63 min
q = 2131.093 W/m2.OC (16 m2) ( 140.12 – 60)OC