PLATE AND FRAME HEAT EXCHANGER
PRESENTED BY • HAFEERA SHABBIR
06-CHEM-19
• MUBASHRA LATIF
06-CHEM-23
• PAKEEZA TARIQ MEER 06-CHEM-65 • MAHPARA MUGHAL
06-CHEM-69
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 Applications • Advantages • Limitations of Operation • Comparison of with STH • Design steps with Solved example
Introduction • It is a type of compact heat exchanger 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 –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
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
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
PHE
STHE
•Multiple duty
Possible
Impossible
•Hold up volume Low On each plate •Gaskets •modifications
High On flanged joints
Easy by adding impossible or removing plates
FEATURES
PHE
•Repair
Easy to replace Requires tube plates and plugging gaskets
•Detection of leakage • Access for inspection
Easy to detect
Difficult to detect
On each side of Limited plate
•Time reqd. for 15 min opening •Fouling
STHE
15 to 20 % of STHE
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 stainlesssteel plate [k w=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 Re0.64Pr 0.4 • 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
Heat capacity CP
1.622*103 J/kg.K
Thermal conductivity k
0.138 J/.m.K
Viscosity µ Density ρ
Water at 77◦C 4.198*103J/kg.K 0.668J/s.m.K
2.70*10-3Pa.s
3.72*10-4Pa.s
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 N p 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)m 2
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 DUTY AND FLOW RATES TOTAL RATE OF HEAT TRANSFER : FOR DOWTHERM q= (mCpΛT)c =4(1.622*103)(70-10) =3.89*10 5 W 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*10 5 /n This indicates that Reynold number is greater than 100 and correlation for Nu can be used. Pr NUMBER: Pr h = Cpµ/k = (4.198*103)(3.72*10-4)/0.668 =2.34 • hh = (0.4)(kh/De)Re0.64Pr 0.4 =[0.668/0.01][1.139*105/n]0.64(2.34)0.4 =6.467*10 4 /n 0.64 W/m 2 .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*10 4 /n Pr c=(1.622*103)(2.70*10-3)/(0.138) =31.73 This also indicates that Re>100 hc=(0.4)(kc/De)Re0.64Pr 0.4 =(0.4)(0.138/0.01)(237*104/n)0.64 (31.73)0.4 =1.388*10 4 /n 0.64 W/m 2 .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 x w 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 -5 n 0.647 +7.11*10 -5 m. K/W
USING THE NTU METHOD • A NTUmin for cold stream with a minimum mc p is defined NTUmin=UA/(Mcp)min =Tc,out –Tc,in/ f ΛT◦,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 1.664
= (
)( 8.751*105n0.64+7.11*10-5
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*10 2)(0.0853)2/2
•
=984N/m2
•
=984Pa
