Plate and Frame Heat Exchanger

PLATE AND FRAME HEAT EXCHANGER

engineering-resource.com

PRESENTED BY

• HAFEERA SHABBIR 06-CHEM-19

• MUBASHRA LATIF 06-CHEM-23

• PAKEEZA TARIQ MEER 06-CHEM-65

• MAHPARA MUGHAL 06-CHEM-69


OUTLINE

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


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


CONSTRUCTION

• Based on their construction plate and frame heat exchangers are classified into

• (a) Gasketed–plate

• (b) Welded-plate


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.




Construction of GPHE

• Plates

• Gaskets

• Plate frame


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


http://en.wikipedia.org/wiki/File:Plate_frame_2.png

PLATES PATTERNS1)Induce turbulence for high HT coefficient2)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 corrugations2)Chevron or herringbone corrugations


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


CHEVRON CORRUGATIONS



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



DIMPLE CORRUGATIONS


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


GASKET MATERIALS

• Typical gasket materials are

Natural rubber styrene

Resin cured butyl

Compressed asbestos fiber gaskets


FRAMES

• Materials

1)Carbon steel with a synthetic resin finish

2)stainless steel


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


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.



APPLICATIONS

3 major applications

• (1)liquid-liquid services

• (2)condensing and evaporative

• (3)Central cooling


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.


CONDENSATION AND VAPORIZATION

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


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


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


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


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


• Large difference b/w flow rates cant be handled

• High pressure drop

• Potential for leakage


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


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


FEATURES PHE STHE

Sensitivity to vibrations

Not sensitive sensitive


DESIGN STEPS WITH SOLVED EXAMPLE


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 [kw=45j/s.m.K]has 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.


DATA REQUIRED

• The performance characteristics for the chevron configuration selected for the plates are shown . For

• Re > 100,Nu and f can be represented by the following relationships:

• Nu = 0.4 Re0.64Pr0.4

• f = 2.78Re-0.18

• :


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.


Properties of each fluid at the mean temperature in the exchanger are

property Dowtherm at 40◦C Water at 77◦C

Heat capacity CP 1.622*103 J/kg.K4.198*103J/kg.K

Thermal conductivity k 0.138 J/.m.K 0.668J/s.m.K

Viscosity µ 2.70*10-3Pa.s 3.72*10-4Pa.s

Density ρ 1.044*102kg/m3 9.74*102kg/m3


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 NTUmin=UA/(mCp)min.the area of the plate and frame exchanger can be calculated once the overall heat transfer coefficient has been evaluated.

•


CALCULATION OF HT AREA

• For a single pass configuration with Np plates and NP+1 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


CALCULATION OF FLOW AREA

• The flow area for each stream with n/2flow passages is given by:

S=n/2(W)(b)

=n/2(0.25)(0.005)

=(6.25*10-4)n.


CALCULATION OF HEAT DUTYAND FLOW RATES

TOTAL RATE OF HEAT TRANSFER: FOR DOWTHERM q= (mCpΛT)c

=4(1.622*103)(70-10) =3.89*105W THE MASS FLOW RATE OF WATER : mh=q/(CPΛT)h

=3.89*105/(4.198*103)(95-60) =2.65 Kg/s VELOCITY OF WATER: Vh =mh /ρhS =2.65/(9.74*102)(6.25*10-4)n =(4.35/n)m/s


• EQUIVALENT DIAMETER:

De=2b

=0.01m


CALCULATION OF HOT SIDE HT COEFFICIENT

• REYNOLD NUMBER: Reh=DeVhρh /µh

=0.01(4.35/n)(9.74*102)/(3.72*10-4) =1.139*105/nThis indicates that Reynold number is greater than 100 and correlation

for Nu can be used. Pr NUMBER: Prh = Cpµ/k = (4.198*103)(3.72*10-4)/0.668 =2.34• hh = (0.4)(kh/De)Re0.64Pr0.4

=[0.668/0.01][1.139*105/n]0.64(2.34)0.4

=6.467*104/n0.64W/m2.K


CALCULATION OF COLD SIDE HT COEFFICIENT

The same calculations are repeated for cold stream. V=mc/ρc S

=4.0/(1.044*103)(6.25*10-4)n =6.13/n Re=DeVcρc/µc

=0.01(6.13/n)(1.044*103)/(2.70*10-3) =2.37*104/n Prc=(1.622*103)(2.70*10-3)/(0.138) =31.73 This also indicates that Re>100 hc=(0.4)(kc/De)Re0.64Pr0.4

=(0.4)(0.138/0.01)(237*104/n)0.64 (31.73)0.4

=1.388*104 /n0.64 W/m2.K


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/U=1/hh+xw/kw+1/hc

=n0.64/(6.467*104)+(0.0032)/(45)+n0.64/(1.388*104)

=8.751*10-5n0.647+7.11*10-5 m. K/W


USING THE NTU METHOD

• A NTUmin for cold stream with a minimum mcp is defined

NTUmin=UA/(Mcp)min

=Tc,out–Tc,in/ fΛT◦,log mean

LOG MEAN TEMPERATURE DIFF:

ΛT◦,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 ,F=1.


• NTU =70-10/36.067

=1.664

• To satisfy the other NTU definition of UA/(Mc) in terms of results in the relation

• 1

8.751*105n0.64+7.11*10-5 =1.664 ( )(

0.25(n-1)

4.0(1.622*103)

)


ITERATIVE METHOD

• This equation can be solved with itreration to indicate that n=51.Thus 50 plates are required to meet to the heat transfer needs to preheat 4kg/s of dowtherm from 10 to 70◦C.


HYDRAULIC DESIGN

• PRESSURE DROP IN WATER STREAM:

• Vh =4.35/51=0.0853m/s

• Reh=1.139*105/51=2233

• Since Re>100

• f =2.78Re-0.18

• =2.78(2233)-0.18

• =0.694


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:

• ΛP=4f(L/De)ρh Vh2 /2

• =4(0.694)(1/0.01) (9.74*102)(0.0853)2/2• =984N/m2

• =984Pa


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.


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