Transcript Document 7176800

5/20/2016
MAJED AL-SHAHRANI
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5/20/2016
MAJED AL-SHAHRANI
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King Fahd University of Petroleum & Minerals
Mechanical Engineering Department
COOP PROGRAM
ME351
MAJED AL-SHAHRANI
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ADVISOR
Dr. MUMMER KALYON
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CONTENTS
•INTRODUCTION
•WORK ENVIRONMENT
•HEAT EXCHANGERS
•2 CASE STUDIES
•CONCLUSION
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Cooperative Training Program
in
Saudi Aramco
Consulting Services Department
(CSD)
Heat Exchangers
Heat Transfer Group
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Work Environment
Consulting Services Department (CSD)
ESD
M&CED
ME&CCD
RED
The mission of consulting services department (CSD) is
to provide technical consultations on field problems
PEU
P&VUstandards.
CEU
HUG
and maintaining
quality
The responsibility of Process
Unit (PEU)
PressureEquipment
Vessels & Storage
Tanksis to offer technical
consultations on the parts related to pressure vessels, storage tanks
and
boilers
heat transfer equipment.
Heat Transfer Equipments
Fired heaters
Heat exchangers
Insulation & refractory materials
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HEAT EXCHANGERS
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HEAT EXCHANGERS
HEAT EXCHANGERS transfer heat from a hot fluid to a colder
fluid through the combined mechanisms of conduction and
forced convection.
ALL HEAT EXCHANGERS are similar in their principle of
operation; however, heat exchangers may differ in the specific
fluids that are used in the heat transfer process, the layout of
the metal tubes, and the configuration of the enclosure.
THE PURPOSES of the heat exchanger are:
1. change the temperature of fluid.
2. change the phase of a fluid.
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HEAT EXCHANGERS
2.Air
cooled
heat
exchanger.
1.
Shell
tube
heat
exchangers.
4.Plate
and
frame
heat
exchanger.
3.Double
pipe
heat
exchanger.
1.
2.
3.
4.
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Shell and tube heat exchangers.
Air cooled heat exchanger.
Double pipe heat exchanger.
Plate and frame heat exchanger.
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shell-and-tube heat exchanger:
The shell-and-tube heat exchanger is the type that is most
commonly used in process plants.
SHELL
CHANAL
THE BUNDEL
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Air-Cooled Heat Exchangers:
FANS
Types of Air-Cooled Heat Exchangers
1.
2.
3.
Forced draft.
Induced draft.
Humidified forced draft.
BUNDEL
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CASE STUDY
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CASE STUDY#1
Tube insertion:
Tube insertion is a new technology that used to
repairing
the tube
failures
by insert
a tube
with Aramco air cooled heat
This case study
is about
a problem
in one
of Saudi
thin
wall inside the damaged tube. This method can
exchangers.
be done without replacing the existing tube.
Problem:The tube
of a will
fin fan
This new
method
leadcooler
to: at Shaybah Producing Facilities has a problem
which is tube failure.
1. Protecting damaged tube inlets.
2. Restoring plugged leaking tubes to active service.
3. Restoring original compressive strength to weakened tube
to tube sheet joints.
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CASE STUDY#1
INDUCED DRAFT
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staggered type
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THREE BAYS
ONE BUNDLE
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ONE BAYS
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Process
Side
Name
Flow Rate
Temp. In
Temp Out
Tube Side
Fluid
Process Side
Name
Water
62603(lb/h
r)
7.88798
(kg/sec)
28396
(kg/hr)
130 F
54.44 C
327.6 K
36.667C
309.817
K
98 F
This table
represents the
Specific
0.999(Btu/l
HeatPropertiesbof
F) air at 90 ºF
Viscosity
0.769(cP)
Density
62.116(lb/ft
3)
Thermal
Conductivit
y
0.3585(Btu
/(hr ft F))
Fouling
Resistance
0.001(ft²
°F hr/Btu)
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Air Side
1.86028(
Ib/ft.hr)
137151(ft³/
min)
9924.4(Ib/
min)
595463.9
9(lb/hr)
Temp. In
90 F
32.22 C
305.4 K
Temp Out
103.94 F
39.97C
313.12K
Flow Rate
This table0.2407(Btu/
represents the
Specific Heat
lb F)
Properties of water at 130 ºF
Viscosity
0.185(cP)
Density
0.07248(lb/
ft3)
Thermal
Conductivity
0.015206(B
tu/(hr ft F))
Fouling
Resistance
0.002(ft²
°F hr/Btu)
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0.0447(Ib/
ft.hr)
0.000746(
Ib/ft.min)
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THE TUBE AND BUNDLE PARAMETER
in
ft
2.25
0.1875
Tube (OD)
1
0.0833
Tube (ID)
0.87
0.0725
Wall Thickness ∆x
0.065
0.005417
360
30
Cooler Width
273.96
22.83
Tube Layout
staggered
Fin diameter (Df)
Tube Length L
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Pitch of Tubes in Plane Perpendicular
to Air Flow (SL)
3
0.25
Pitch of Tubes in Direction of Air Flow
(ST)
3
0.25
Fin Height (H)
0.625
0.0521
Fin Width (w)
0.016
0.00133
(w+s)
0.116
0.00967
Fin Spacing (s)
0.1
0.00833
# of tubes(N)
264
# of rows(NL)
6
# of Tubes in one row(NT)
44
# of passes
6
Thermal cond.(Fin material)
110
Thermal cond.(Tube material)
8.2
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Area calculations
Surface area of fins: (AF)
AF= ((N*L*Pi)/(s+w))*(0.5*(Df²-Dr²) +Df*w)
36944.02 ft²
Surface area b/w fins: (Aw)
Aw= ((N*L*Pi)/(s+w))*(Dr*s)
Surface area of tubes with fins: (A)
A= Aw+AF
38729.4 ft²
Surface area of tubes without fins: (AT) AT= N*L*Pi*Dr
Frontal area: (Ao)
1785.4 ft²
2072.623 ft²
Ao= L * cooler width
684.9 ft²
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Minimum flow area: (Amin)
A1= NT*L*(St-Dr-((2*w*h)/ (w+s)))
= 201.13 ft²
Or
A2=2*Nt*L*(SD-Dr-((2*w*h)/ (w+s)))
SD= (SL²+ (ST/2)²)½
= (0.25²+ (0.25/2)²)½
= 0.2792 ft
= 479.254 ft²
Then
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(Amin) =
201.13 ft²
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To find the air side Coefficient of heat transfer: ha
The maximum velocity
The Mass flow rate
The Reynolds number
V max = V approach/б
=595463.99 Ib/hr
Ma = V max* Amin * ρa
Rea = (V max* Dr * ρa)/ µa
The Prandtl number
Pr = (µ Cp/ kt)
=681.12 ft/ min
= 5513.3743
= 0.708
The Nusselt number Nu = 0.242*(Re ^0.658)*((s/h) ^0.297)*((St/Sl) ^-0.091)*(Pr^ (1/3))*F1*F2
= 34.428
F1= factor of fluid property variation. Assume it F1=1.
When the number
of tube
rows=6
F2= factor
of number
of tubeand
rows.staggered type.
The Coefficient of heat transfer: ha
F2 =0.95.
ha = Nu * kt / Dr
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= 6.285 (Btu/ft².hr. ºF)
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To Find Fin efficiency: ηf
ηf = (tanh (((2*ha/ (w*λf)) ½)* ψ)/ (((2*h/ (w*λf)) ½)* ψ)
ψ= (Dr/2)*((Df/Dr)-1)*(1+0.35*ln (Df/Dr))= 0.0669
ηf = 0.8891
Effective Air Side Heat Transfer Coefficient Based on
Total Surface Area
h'a = ((ηf*AF+Aw)/A)*(ha)
= 5.628 (Btu/hr.ft². ºF)
Effective Air Side Heat Transfer Coefficient Based on
External Surface Area without Fins
ha'r = h'a *(A/AT)
= 105.14 (Btu/hr.ft². ºF)
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To find the tube side Heat Transfer Coefficient: ht
The Crosse sectional area: At
= 92.475 ft/min
Ub, t = Gt/ρt
Tube Side Velocity: Ub,t
= 13431.9
Ret = (Gt* ID)/µt
Tube Side Reynolds Number: Ret
= 5.184
Pr = µt*CP/K
The Prandtl number
Friction factor: ft
= 0.006873
ft = 0.046*((Ret) ^ (-0.2))
Nut = 0.023*((Ret) ^ (0.8))*((Pr) ^ (0.4))
Tube Side Heat Transfer Coefficient: ht
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= 95.736 Ib/s. ft²
Gt= mt/At
Mass flux: Gt
Tube Side Nusselt Number: Nut
= 26.1566 in²
At= (pi*ID²/4)*(N/Ntp)
ht = Nu*(k/ID)
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= 89.15134
= 440.8534(Btu/hr.ft².F)
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To find the Overall heat transfer coefficient: Ur
1/Ur = I/har + (OD/ (2*λf))*(ln (OD/ID)) + (1/ht)*(OD/ID)
1/Ur = 0.012826 ft².F.hr/Btu
Uclean
Ur = 77.965 Btu/hr. ft².F
rf, d =Ff, t+ (ID/OD)*Ff, s= 0.002741 ft².F.hr/Btu
1/Ud = (I/Ur) + rf, d= 0.015567 ft².F.hr/Btu
Uservice
Ud = 64.238 Btu/hr. ft².F
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OD=1 in
ID= 0.87in
Thick= 0.065in
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OD=1in
NEW ID=0.8149in
Insert tube thick=0.02756in
NEW thick=0.09255in
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AFTER
BEFORE
Crosse sectional area: At
Crosse sectional area: At
At=26.16in²
At=22.95in²
Mass flux: Gt
Mass flux: Gt
Gt= 5744.16 Ib/min. ft²
Gt= 6547.18 Ib/min. ft²
Tube Side Velocity: Ub,t
Tube Side Velocity: Ub,t
Ub, t =92.475 ft/min
Ub, t =105.403 ft/min
Tube Side Reynolds Number: Ret
Tube Side Reynolds Number: Ret
Uservice
Ret =Uservice
13431.9
Ret = 14340.034
Ud = 63.87 Btu/hr. ft².F
Ud
= 64.238 Btu/hr. ft².F
Pr =5.184
Pr =5.184
Friction factor: ft
Friction factor: ft
ft = 0.006873
ft = 0.0067834
The tube insertion would cost only one third of the
Tube Side Nusselt Number: Nut retubingTube Side Nusselt Number: Nut
Nut = 89.15
Nut = 93.94
Tube Side Heat Transfer
Tube Side Heat Transfer Coefficient:
Coefficient: ht
ht
ht = 440.85(Btu/hr.ft².F)
ht = 495.9 (Btu/hr.ft².F)
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CASE STUDY#2
This case study is about a new channel cylinder of shell and tube heat exchanger.
The out side diameter of the channel cylinder is 44.75” then the outside radius is 22.375”.
Thechannel
inside diameter
the44channel
cylinder
is 38.500”
the inside radius is 19.25”.
This
cylinder of
has
3/4"O.D,
32" length
and 3 then
1/8" wall.
The thickness of the channel cylinder is 3.125”.
machining
process the
inspection
TheAfter
internal
design pressure
is 2296
Psi. found different thicknesses
hetemperature
measures itisby875ºF.
using ultra sonic
Thewhen
design
tmin = (P*R)/(S*E-(0.6*P)) 3.084” and 3.078”
The stress of this material: SA-336F22CL3 forging at875ºF
P=
internal
design
pressure,
psiMechanical
can beS=16100psi
found
by
using @
The
American
Society of
S=
stress
T=875ºF
EngineersR=
(ASME)
section radius,
II part D.in
the inside
S= maximum allowable stress value, psi
E= joint efficiency.
3.084” and 3.078”
tmin = (2296*19.25)/(16100*1-(0.6*2296)) = 3.002 in
MWAP = S*E*t/(R+0.6t) = 16100*1*3.125/ (19.25+0.6*3.125) = 2381.66 psi
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Conclusion:
I Got hands-on experience on heat exchangers.
In my first case staduy, the method of tube inserts is acceptable from thermal
point of view.
I did some Field Trips:
1. AlQatif Project
2. Uthmaniyah Gas Plant
3. Juaymah Gas Plant
4. The Saudi Aramco Shell Refinery Company
5. Juaymah Heat Exchanger Shop
The co-operative training program (co-op) is very important issue through
university studying. It is helpful for the applied engineering students because it
gives them an idea and background about the field work and that will be helping
students after the graduation.
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DONE BY: MAJED SAAD AL-SHAHRANI
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