PLATE AND FRAME HEAT EXCHANGER

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PLATE AND FRAME HEAT EXCHANGER
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PRESENTED BY

• HAFEERA SHABBIR 06-CHEM-19
• MUBASHRA LATIF 06-CHEM-23
• PAKEEZA TARIQ MEER 06-CHEM-65
• MAHPARA MUGHAL 06-CHEM-69

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OUTLINE

• Introduction
• Construction
• Principle of Operation
• Applications
• Advantages
• Limitations of Operation
• Comparison of with STH
• Design steps with Solved example

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Introduction

• It is a type of compact heat exchanger
• A plate heat exchanger is a type of heat
  exchanger that uses metal plates to transfer heat
  between two fluids

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CONSTRUCTION

• Based on their construction plate and frame heat
  exchangers are classified into
• (a) Gasketedplate
• (b) Welded-plate

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GASKETED-PLATE HEAT EXCHANGER(GPHE)

• Parallel corrugated plates clamped in a frame
  with each plate sealed by gaskets and with four
  corners ports, one pair for each of the two
  fluids.
• The fluids are at all times separated by 2
  gaskets, each open to the atmosphere. Gasket
  failure cannot result in fluid intermixing but
  merely in leakage to atmosphere, hence a
  protective cover is there.

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Construction of GPHE

• Plates
• Gaskets
• Plate frame

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PLATES

• Plate thickness is 0.4 to 0.8 mm
• Channel lengths are 2-3 meters
• Plates are available in Stainless Steel,
  Titanium, Titanium-Palladium, Nickel

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PLATES

• PATTERNS
• 1)Induce turbulence for high HT coefficient
• 2)Reinforcement and plate support points that
  maintains inter-plate separation.
• TYPES OF PATTERNS
• Mainly 2 types of patterns (corrugations) are
  used
• 1)Intermating or washboard corrugations
• 2)Chevron or herringbone corrugations

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CHEVRON OR HERRINGBONE

• Most common type
• Corrugations are pressed to same depth as plate
  spacing
• Operate at High pressure
• Corrugation depth 3mm to 5mm
• Velocity 0.1 to 1 m/s

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CHEVRON CORRUGATIONS
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• INTERMATING TROUGH PATTERNS
• Pressed deeper than spacing
• Fewer connection points
• Operate at Lower pressure
• Max channel gap 3mm to 5mm
• Min channel gap 1.5 mm to 3 mm
• Velocity range in turbulent region is 0.2 to 3
  m/s

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DIMPLE CORRUGATIONS
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GASKETS

• They limit the maximum operating temperature for
  a plate heat exchanger. Material selection
  depends upon
• 1)Chemical resistance
• 2)Temperature resistance
• 3)Sealing properties
• 4)Shape over an acceptable period of time

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GASKET MATERIALS

• Typical gasket materials are
• Natural rubber styrene
• Resin cured butyl
• Compressed asbestos fiber gaskets

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FRAMES

• Materials
• 1)Carbon steel with a synthetic resin finish
• 2)stainless steel

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WELDED PLATE HEAT EXCHAGERS(WPHE)

• Developed to overcome the limitations of the
  gasket in GPHE
• Inabilty of heat transfer area inspection and
  mechanical cleaning of that surface

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OPERATION

• Channels are formed between the plates and corner
  ports are arranged so that the two media flow
  through alternate channels.
• The heat is transferred through the thin plate
  between the channels, and complete counter
  current flow is created for highest possible
  efficiency. No intermixing of the media or
  leakage to the surroundings will take place as
  gaskets around the edges of the plates seal the
  unit.

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APPLICATIONS

• 3 major applications
• (1)liquid-liquid services
• (2)condensing and evaporative
• (3)Central cooling

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LIQUID-LIQUID SERVICES

• It is well-suited to liquid/liquid duties in
  turbulent flow, i.e. a fluid sufficiently viscous
  to produce laminar flow in a smooth surface heat
  exchanger may well be in turbulent flow in PHE.
• It has major applications in the food industry.

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CONDENSATION AND VAPORIZATION

• Condensation of vapor (including steam) at
  moderate pressure, say 6 to 60 Psi, is also
  economically handled by PHEs, but duties
  involving large volumes of very low pressure gas
  or vapor are better suited to other forms of heat
  exchangers

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CENTRAL COOLING

• It is the cooling of a closed circuit of fresh
  non-corrosive and non-fouling water for use
  inside a plant, by means of brackish water.
  Central coolers are made of titanium, to
  withstand the brackish water

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ADVANTAGES

• Compactness
• Flexibility
• Very high heat transfer coefficients on both
  sides of the exchanger
• Close approach temperatures and fully
  counter-current flow
• Ease of maintenance. Heat transfer area can be
  added or subtracted with out complete dismantling
  the equipment

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CONTD..

• Ease of inspection on both sides
• Ease of cleaning
• Savings in required flow area
• Low hold-up volume
• Low cost
• No Local over heating and possibility of stagnant
  zones is also reduced
• Fouling tendency is less

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LIMITATIONS

• Low Pressure
• upto 300 psi
• Low temperature
• upto 300 F
• Limited capacity
• Limited plate size
• 0.02 sq.m to 1.5 sq.m

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• Large difference b/w flow rates cant be handled
• High pressure drop
• Potential for leakage

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COMPARISON BETWEEN PHE AND STHE
FEATURES Multiple duty Hold up volume Gaskets modifications PHE Possible Low On each plate Easy by adding or removing plates STHE Impossible High On flanged joints impossible
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FEATURES PHE STHE
Repair Detection of leakage Access for inspection Time reqd. for opening Fouling Easy to replace plates and gaskets Easy to detect On each side of plate 15 min 15 to 20 of STHE Requires tube plugging Difficult to detect Limited 60 to 90 min
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FEATURES PHE STHE
Sensitivity to vibrations Not sensitive sensitive
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• DESIGN STEPS WITH SOLVED EXAMPLE

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STATEMENT OF PROBLEM

• A plate heat exchanger was use to preheat 4 kg/s
  of dowtherm from 10 to 70?C with a hot water
  condensate that was cooled from 95 to
  60?C.Determine the number of plates required for
  a single-pass counter flow plate and frame
  exchanger. Assume that each mild stainless-steel
  plate kw45j/s.m.Khas a length of 1.0m and a
  width of 0.25m with a spacing between the plates
  of 0.005m.Also,estimate the pressure drop of the
  hot water stream as it flows through the
  exchanger.

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DATA REQUIRED

• The performance characteristics for the chevron
  configuration selected for the plates are shown .
  For
• Re gt 100,Nu and f can be represented by the
  following relationships
• Nu 0.4 Re0.64Pr0.4
• f 2.78Re-0.18

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ASSUMPTIONS

• The plate heat exchanger operates under steady
  state
• conditions.
• No phase change occurs both fluids are single
  phase and are unmixed.
• Heat losses are negligible the exchanger shell
  is
• adiabatic.
• The temperature in the fluid streams is uniform
  over
• the flow cross section.
• There is no thermal energy source or sink in the
  heat
• exchanger.
• The fluids have constant specific heats.
• The fouling resistance is negligible.

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Properties of each fluid at the mean temperature
in the exchanger are
property
Dowtherm at 40
?C
Water at 77?C
4.198103J/kg.K
Heat capacity CP
1.622103 J/kg.K
0.138
J/.m.K
Thermal conductivity k
0.668J/s.m.K
Viscosity µ
3.7210-4Pa.s
2.7010-3Pa.s
Density ?
1.044102kg/m3
9.74102kg/m3
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SOLUTION

• APPROACH TO THE PROBLEM
• To avoid an iterative calculation because of the
  interdependency between the heat transfer area
  and the total flow area, use the NTU approach to
  determine the NTUmin required, noting that
  NTUminUA/(mCp)min.the area of the plate and
  frame exchanger can be calculated once the
  overall heat transfer coefficient has been
  evaluated.

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CALCULATION OF HT AREA

• For a single pass configuration with Np plates
  and NP1 flow passages ,solution of the problem
  can be simplified mathematically by assuming n
  flow passages and n-1 plates ,since flow
  velocities involve flow passages and not plates.
  with this modification, the heat transfer surface
  area of the exchanger in terms of n is
• A(n-1)LW(n-1)(1)(0.25)0.25(n-1)m2

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CALCULATION OF FLOW AREA

• The flow area for each stream with n/2flow
  passages is given by
• Sn/2(W)(b)
• n/2(0.25)(0.005)
• (6.2510-4)n.

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CALCULATION OF HEAT DUTYAND FLOW RATES

• TOTAL RATE OF HEAT TRANSFER
• FOR DOWTHERM
• q (mCp?T)c
• 4(1.622103)(70-10)
• 3.89105W
• THE MASS FLOW RATE OF WATER
• mhq/(CP?T)h
• 3.89105/(4.198103)(95-60)
• 2.65 Kg/s
• VELOCITY OF WATER
• Vh mh /?hS
• 2.65/(9.74102)(6.2510-4)n
• (4.35/n)m/s

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• EQUIVALENT DIAMETER
• De2b
• 0.01m

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CALCULATION OF HOT SIDE HT COEFFICIENT

• REYNOLD NUMBER
• RehDeVh?h /µh
• 0.01(4.35/n)(9.74102)/(3.7210-4)
• 1.139105/n
• This indicates that Reynold number is greater
  than 100 and correlation for Nu can be used.
  
• Pr NUMBER
• Prh Cpµ/k
• (4.198103)(3.7210-4)/0.668
• 2.34
• hh (0.4)(kh/De)Re0.64Pr0.4
• 0.668/0.011.139105/n0.64(2.34)0
  .4
• 6.467104/n0.64W/m2.K

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CALCULATION OF COLD SIDE HT COEFFICIENT

• The same calculations are repeated for cold
  stream.
• Vmc/?c S
• 4.0/(1.044103)(6.2510-4)n
• 6.13/n
• ReDeVc?c/µc
• 0.01(6.13/n)(1.044103)/(2.7010-3)
• 2.37104/n
• Prc(1.622103)(2.7010-3)/(0.138)
• 31.73
• This also indicates that Regt100
• hc(0.4)(kc/De)Re0.64Pr0.4
• (0.4)(0.138/0.01)(237104/n)0.64
  (31.73)0.4
• 1.388104 /n0.64 W/m2.K

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CALCULATION OF OVERALL HT COEFFICIENT

• The overall heat transfer coefficient can now be
  determined in terms of n. Since the surface areas
  on either side of the plate are the same, no
  correction for area is required.
• Assume a thickness of the plate xw of 0.0032m
• 1/U1/hhxw/kw1/hc
• n0.64/(6.467104)(0.0032)/(45)n0.64/(1.38810
  4)
• 8.75110-5n0.6477.1110-5 m. K/W

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USING THE NTU METHOD

• A NTUmin for cold stream with a minimum mcp is
  defined
• NTUminUA/(Mcp)min
• Tc,outTc,in/ f?T?,log
  mean
• LOG MEAN TEMPERATURE DIFF
• ?T?,log mean
  
  (Th,in-Tc,out)-(Th,out
  Tc,in)/ln(Th,in-Tc,out)/Th,out-Tc,in)
• (95-70)-(60-10)/ln(95-70)/(60-10)
• 36.067 K.
• For a single pass counter flow plate and frame
  heat exchanger ,F1.

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• NTU 70-10/36.067
• 1.664
• To satisfy the other NTU definition of UA/(Mc) in
  terms of results in the relation

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0.25(n-1)


1.664
(
)(

)


8.751105n0.647.1110-5
4.0(1.622103)
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ITERATIVE METHOD

• This equation can be solved with itreration to
  indicate that n51.Thus 50 plates are required to
  meet to the heat transfer needs to preheat 4kg/s
  of dowtherm from 10 to 70?C.

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HYDRAULIC DESIGN

• PRESSURE DROP IN WATER STREAM
• Vh 4.35/510.0853m/s
• Reh1.139105/512233
• Since Regt100
• f 2.78Re-0.18
• 2.78(2233)-0.18
• 0.694

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CONTD..

• Neglecting friction due to entrance and exit
  losses as well as temperature effects on the
  viscosity between the wall and the bulk fluids.
• So pressure drop is calculated from the following
  equation
• ?P4f(L/De)?h Vh2 /2
• 4(0.694)(1/0.01) (9.74102)(0.0853)2/2
• 984N/m2
• 984Pa

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CONCLUSION

• Since the entrance and exit losses will be small,
  the pressure drop per plate
• is small, and a new configuration with modified
  dimensions should be considered.