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					     PLATE
HEAT EXCHANGER




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        GROUP MEMBERS


   Nadeem Akhtar (2006-chem-22)
   Matloob Ahmed (2006-chem-26)
   Zohaib Atiq Khan (2006-chem-40)




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           Introduction to PHE

   Second abundantly used HEX after STHE.
   Fall in the category of compact heat exchangers.
   Mostly used in food industry like milk, beverages
    and juices industry.
   Is usually comprised of a stack of corrugated or
    embossed metal plates in mutual contact.


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Facts and figures on Plate Heat
            Exchanger




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      1-Size range for unit and plates
(a)    For unit

          Size                         1540-2500m2
          Number of plates             Up to 700
          Port size                    Up to 39 cm




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  1-Size range for unit and plates
(b) For plates

         Thickness.                      0.5 – 1.2mm

         Size.                           0.03 – 2.2m

         Spacing.                        1.5 – 5mm

         Corrugation depth. 3 – 5mm


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  2-Standrad performance limits
Max. operating pressure.             30 bar. Or 360psi.




Max. operating temperature.          200oC.        Or 390 0F
Max. flow rate.                      3600 m3/hr. Or
                                     950,000USG/min.

Heat transfer coefficient.           3500 – 7500 W/m2 .oC
                                     Or 600 – 1300 BTU/ft2 hr of


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   2-Standrad performance limits
Heat transfer area.                  0.1 – 2200 m2 or
                                     2 – 24,000 ft 2

NTU.                                 0.3 – 0.4

Pressure drop.                       30kpa per NTU

Temperature approach.                As low as 2 oC

Heat recovery.                       As high as 93%

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          Mechanical parts of PHE

1.   Plates (provide heat transfer area)
2.   Gasket (prevents leakage of fluids).
3.   Frame (for enclosure, on front).
4.   Pressure plate (to press the plates on rare side).
5.   Support column (to support the exchanger).
6.   Splitter (plate dividing the PHE in parts in case of
     multi-streaming)
7.   Tightening bolts

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Multi streaming using splitter plate




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        Material of construction
(1) Plates
     As plates are very thin (0.5 – 1.2mm) So we can
     not compromise on material of construction
     plates are usually made of very strong materials,
     depending on operatin conditions
(a) stainless steel AISI 304
(b) stainless steel AISI 316
(c) Hastelloy B
(d) Hastelloy c-276
(e) alluminium brass 76/22/2
(f) incoloy 825
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              Material of construction
  (2) Gaskets
Nitrile rubber.                                Used up to 110 oC for mineral
                                               oils, dilute mineral acids, and
                                               aliphatic hydrocarbons.
EPDM.                                          Used up to 160 oC for mineral
(ethylene-propylene-diene monomer)             acids, or bases, aqeuous
                                               solutions or steam
Viton.                                         Used up to 100 oC for
( copolymer of vinylidine flouride and         hydrocarbons and chlorinated
hexafluoro-propylene)                          hydrocarbons


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        Classification of PHES
Plate heat exchangers can be classified based on
(1) Joints

(2) Plate corrugations.

(3) Flow arrangements.




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          Classification of PHES
     Based on joints PHES are classified in to three
      types
(1)   Gasketted.
(2)   Brazed.
(3)   Welded.




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        Classification of PHES
(1) gasketted.




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        Classification of PHES
(2) brazed.




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        Classification of PHES
(3) Welded plate.




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       Classification of PHES
Based on corrugation two types of PHES exist
(a) Wash board.
(b) Chevron.




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       Classification of PHES
Based on flow arrangement
(a) Series flow

(b) U-arrangement




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              Advantages of PHE
(1)   A PHE offers very high heat transfer coefficient. Increase
      in H.T coefficient is three to five times.
(2)   Is suitable even for a close approach temperature as low
      as 2 oC, and for a large temperature cross.
(3)   Offers ease of inspection, cleaning and maintenace.
(4)   Heat transfer area can be increased or decreased by
      adding or removing some plates.
(5)   Conveniently performs multiple heat exchange duties in a
      single exchanger.
(6)   Requires much less floor soace.
(7)   Costs less than shell and tube heat exchanger especially
      when expensive material of construction is used.

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                 Disadvantages
(1)   Effect of fouling because of scaling, deposition
      of solids by crystallization, corrosion, and even
      by biological materials is quite significant in
      PHES
(2)   Large over design is required. For example in an
      STHEX for a fouling resistance of 1.76*10-4 will
      increase the required surface area by 35% in case
      of STHEX but will increase the required surface
      area of a PHE by about 100%.
(3)   The allowable fouling resistance in PHE is one
      tenth of that in STHEX.
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                 Applications
(1)   Dairy industry
(2)   Pharmaceuticals
(3)   Food processing
(4)   Petroleum and chemical industries
(5)   Pulp and paper industry
(6)   Power generation
(7)   Reboiling or condensing services

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        Applications for which PHES are not
                   recommended

(1)   Gas-to-gas applications
(2)   Fluids with very high viscosities may pose to
      distribution problems, flow velocities less than
      0.1m/s are not used because of low H.T
      coefficient.
(3)   Less suitable for vapours condensing under
      vacuum


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Design of Plate
Heat Exchanger



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       Thermal Design Steps
                    Step # 1
   Calculate properties of fluids i.e
    density, viscosity, thermal conductivity,
    specific heat

   Also determine fluids unknown inlet and
    outlet temperatures and flow rates
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                     Step # 2

   Calculate heat duty, the rate of heat
    transfer required

            Qc   =        (mcp)c (t2-t1)
            Qh   =        (mcp)h (T2-T1)



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                  Step # 3
   Calculate the log mean temperature
    difference, LMTD

           LMTD =                  (T1-t2) – (T2-t1)
                                  ln(T1-t2)/(T2-t1)




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                     Step # 4
   Determine the log mean temperature
    correction factor, Ft
            NTU       =   ( To- Ti )
                           LMTD
   Where
      Ti    = stream inlet temperature °C
      To    = stream outlet temperature °C
    LMTD    = log mean temperature difference °C

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                     Step # 5

   Calculate the corrected mean temperature
    difference


                 ∆Tm= Ft x LMTD




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                      Step # 6
   Select specific construction of the plates suitable
    for the required service like

   Plate material
   Port diameter
   Gasket material




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              Step # 6 cont’d
   Corrugation type




                                          Washboard pattern   Chevron pattern



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              Step # 6 cont’d
   Effective length
   Width
   Plate pitch




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                    Step # 7
   Estimate the overall heat transfer coefficient




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                 Step # 8
   Calculate the surface area required

      Q    =    UA (Ft x LMTD )

      A    =          Q
                U (Ft x LMTD )



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                   Step # 9

   Determine the number of plates required

   Number of plates =              Total surface area
                                    Area of one plate




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      Area of one plate


A =   (L – D) x W




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                Step # 10
   Decide the flow arrangement and number
    of passes




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                   Step # 11
   Calculate the film heat transfer coefficients
    for each stream
              Nu = C Ren Prm (µ/µw)x
    Typical reported values are

       C     = 0.15-0.40
       n     = 0.65-0.85
       m     = 0.30-0.45 (usually 0.333)
       x     = 0.05-0.20
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           Step # 11 continued
   Most popular correlation for preliminary
    estimate of area is


    (hde/kf) =      0.26 Re0.65 Pr0.4 (µ/µw)0.14

   Also we can use general relation


     (hde/kf) =        Ch Ren Pr1/3 (µ/µw)0.17
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Ch & n values




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                     Step # 12
   Calculate the overall coefficient, allowing for
    fouling factors

       1     =      1 + 1 + t + Rfh + Rfc
       Ud           hh hc kw




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                      Step # 13
   Compare the calculated with the assumed overall
    coefficient.

   If satisfactory, say - 0% to + 10% error, proceed. If
    unsatisfactory return to step 7 and estimate another
    value of overall heat transfer coefficient U



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 Hydraulic Design
Pressure Drop Calculations




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                     Step # 14
   Check the pressure drop for each stream

   Channel pressure drop
                 ∆Pc =                   8 f Lp ρup2
                                             de 2

                 f          =           0.6 Re-0.3

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           Step # 14 continued
   We can also use


            ∆Pc =             4 f LpNp Gc2 (µ/µw)- 0.17
                                    de 2ρ

            f    =           Kp
                             Rem

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Kp & m values




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            Step # 14 continued
   Port pressure drop
                   ∆Pp =         1.3Npρupt2
                                        2
   upt = the velocity through the ports w/ρAp, m/s,
   w = mass flow through the ports, kg/s,
   Ap = area of the port = (3.14xd2pt)/4, m2,
   dpt = port diameter, m,
   Np = number of passes

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             Step # 14 continued

   Total pressure drop ∆Pt

       ∆Pt       =           ∆Pc +              ∆Pp




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Design Problem



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                   Statement

   Design a gasketed plate heat exchanger to cool
    methanol from 95 °C to 40°C. Flow-rate of
    methanol is 100,000 kg/h. Brackish water is used
    as coolant, with a temperature rise from 25° to
    40°C.




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                     Step # 1
Physical properties of            Methanol          Water
        fluids
Density (kg/m3)                         750          995

Viscosity (mNm-2s)                       3.4         0.8

Specific Heat Cp                        2.84         4.2
(kJ/kg oC)
Thermal conductivity                    0.19        0.59
(W/moC)
                         engineering-resource.com
                   Step # 2
   Heat duty of hot fluid (methanol)

       Qh   =    (mcp)h (T2-T1)

            =    100000 x 2.84 x (95 - 40)
                  3600
            =    4340 kW

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                 Step # 2 contd.
   Mass flow rate of cold fluid (Water)
   As             Qh =         Qc
   mc       =           Qh
                   Cpc x (t2 – t1)
             =            4340
                     4.2 x (40 - 25)

             =     68.9 kg/s

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                   Step # 3
                 LMTD Calculation
   (Methanol)         95 0C                   40 0C
   (Water)            40 0C                   25 0C
   LMTD =        (T1-t2) – (T2-t1)
                  ln(T1-t2)/(T2-t1)
            =    (95 – 40 ) – (40 - 25)
                 ln(95 – 40 )/(40 - 25)


            =    31 oC
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                       Step # 4
             Correction factor Ft
NTU          = ( To- Ti )                  =       95 - 40
                LMTD                                 31
          =    1.77
From figure 12.62

     try 1 : 1 pass

Ft           =        0.96
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0.96




       1.77


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               Step # 5
   ∆Tm   =   Ft x LMTD

          =   0.96 x 31

          =   29.76 0C




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                     Step # 6
   Estimate the overall heat transfer coefficient




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                 Step # 7
   Calculate the surface area required
   A      =          Q
                U (Ft x LMTD )
           =          4340000
                2000 x (0.96 x 31)

           =    72.92 m2


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                   Step # 8
   Select area of one plate

 Assuming
     effective Length of plate                =   1.5 m
     effective width of plate                 =   0.5 m
     plate spacing                            =   3 mm
then
     effective area of plate                  =   0.75 m2
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                 Step # 8
   Determine the number of plates required

   Number of plates =               total surface area
                                     area of one plate
                         =           72.92
                                      0.75
                         =           97

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                       Step # 8
   No need to adjust this, 97 will give an even number of
    channels per pass, allowing for an end plate

   Number of channels per pass (N)                = (97 - 1 )/2
                                                   = 48
   Channel cross-sectional area (Ac)              = 3 x 10-3 x 0.5
                                                   = 0.0015 m2
   hydraulic mean diameter (de)                   = 2 x 3 x 10-3
                                                   = 6 x 10-3 m
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                 Step # 9
   Selection of flow arrangement and number
    of passes

   No of passes     =               1-1
   Flow arrangement =               U - Arrangement




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                  Step # 10
   (Methanol) film heat transfer coefficient hh

   Channel velocity =                     m
                                       ρ x Ac x N
                           =                 27.8
                                       750 x 0.0015 x 48
                           =           0.51 m/s

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    (Methanol) film heat transfer coefficient hh



   Re      =      ρ x up x de
                       µ
            =     750 x 0.51 x 6x10-3
                      0.34x10-3
            =     6750



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    (Methanol) film heat transfer coefficient
                       hh


   hh     =   0.26 Re0.65 Pr0.4 (µ/µw)0.14kf/de

   Pr     =   µCp/k              =           5.1

   hh     =   0.26 (6750)0.65 (5.1)0.4 0.19/6x10-3
           =   4870 W/m2oC

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    (Water) film heat transfer coefficient hc

   Channel velocity =                    m
                                      ρ x Ac x N
                          =                 68.9
                                      995 x 0.0015 x 48

                          =           0.96 m/s



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    (Water) film heat transfer coefficient hc

   Re      =     ρ x up x d e
                      µ
            =    995 x 0.96 x 6x10-3
                     0.8x10-3

            =    6876



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         (Water) film heat transfer coefficient hc

   hc        =    0.26 Re0.65 Pr0.4 (µ/µw)0.14kf/de

   Pr        =    µCp/k             =             5.7

   hc        =    0.26 (6876)0.65 (5.7)0.4 0.59/6x10-3

              =    16,009 W/m2 0C

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                    Step # 11
 Overall heat transfer coefficient Ud
 1 =      1 + 1 + t + Rfh + Rfc
  Ud       hh hc        kw

       =   1 + 1 + 0.75x10-3 + 0.0001 + 0.00017
         4870 16,009     21
    Ud =   1754 W/m2 0C
                (too low than 2000 W/m2 0C )

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Fouling Factors




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                   Step # 12

   The value of design overall coefficient i.e
    1754 W/m2 0C is too low than assumed or
    estimated value i.e 2000 W/m2 0C )




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            Iterative procedure
   Again assume U(estimated)
   U(estimated)        =     1600 W/m2 0C
   Area (A)            =     91.94 m2
   Number of plates =        121
   hh                  =     4215 W/m2 0C
   hc                  =     13,846 W/m2 0C
   Ud                  =     1634 W/m2 0C
                 Ud ≈     U(estimated)

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    Hence

   Number of plate per pass =                 (121 – 1) / 2
                             =                 60




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                     Step # 13a
                Pressure drop calculations
   Channel pressure drop (Methanol)
             f     =     0.6 Re-0.3
             f     =     0.60(5400)-0.3 = 0.046
       ∆Pc         =      8 f Lp ρup2
                              de 2
   Path length Lp = plate length x number of passes
                   = 1.5 x 1 = 1.5 m.
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    Channel pressure drop (Methanol)


   up        =          0.41 m/s
   ρ         =          750 kg / m3
   de        =          6x10-3 m
   ∆Pc       =          5799 N / m2




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       Port pressure drop (methanol)
   ∆Pp              =            1.3Npρupt2
                                         2
   Port diameter dpt =           100 mm
   Port area         =           3.14xd2pt/4 = 0.00785 m2
   Port velocity     =           27.8/(750 x0.00785 )
                      =           4.72 ms-1
   ∆Pp               =           10,860 Nm-2

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       Pressure drop (Methanol)
   Total pressure drop ∆Pt

       ∆Pt        =           ∆Pc +              ∆Pp

                  =          5799 + 10,860
                  =          16,659 N/m2 0.16 bar



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                   Step # 13b
   Channel pressure drop (water)
             f     =     0.6 Re-0.3
             f     =     0.60(5501)-0.3 = 0.045
       ∆Pc         =      8 f Lp ρup2
                              de 2
   Path length Lp = plate length x number of
    passes
                   = 1.5 x 1 = 1.5 m.

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    Channel pressure drop (water)

   up       =          0.77 m/s
   ρ        =          995 kg / m3
   de       =          6x10-3 m
   ∆Pc      =          26,547 N / m2




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          Port pressure drop (water)
   ∆Pp               =            1.3Npρupt2
                                          2
   Port diameter dpt =            100 mm
   Port area         =            3.14xd2pt/4 = 0.0078 m2
   Port velocity     =            68.9/(995 x 0.0078 )
                      =            8.88 ms-1
   ∆Pp               =            50,999 Nm-2

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             Pressure drop (water)
   Total pressure drop ∆Pt

       ∆Pt        =           ∆Pc +              ∆Pp

                  =          26,547 + 50,999
                  =          77546 N/m2 0.78 bar



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Thank You

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