How To Design Plate Type Heat Exchanger

After design and calculate a heat transfer area following the parameter (flow rate, temperature, Heat load), you can use the following step to check if this heat transfer area is correct.

Product Selection Calculation:

1.Technical parameter of selection equipment



    Heat Exchanger Plate:SUS304

    Plate Heat Exchanger  Gasket:EPDM

    Frame:Carbon Steel

   Heat transfer area per plate:Ao=0.05m2  

   Channel cross-sectional area: As=0.000225m2

    Equivalent diameter:de=0.0044m

  Criterion equation:

     Nuh=0.381 Reh0.6273Pr0.3   (1400≤Reh≤7000)

     Nuc=0.381 Rec0.6273Pr0.4   (1200≤Reh≤600)

     Euh=652Reh-0.1465            (1400≤Reh≤7000)

     Euc=505Rec-0.1274            (1200≤Reh≤6000)

     △Ph= Euh.ω2.ρ.m

     △Pc= Euc.ω2.ρ.m

  Thickness of plate:δ=0.00045m

    Thermal conductivity of plate:λp=16.3W/(m.k)

  Fouling resistance: 

     Hot side:R1=0.00003m2.K/W

     Cold side:R2=0.00002m2.K/W

2.The parameter of liquid

 (1) Hot side:

     Media:Desalted water

     Inlet temperature:T1=60oC

     Outlet temperature:T2=35oC

     Flow rate:q1=28L/min


     Specific heat:Cp1=4.174kJ/(kg.K)

     Thermal conductivity:λ1=0.6445W/(m.K)

     Dynamic viscosity:μ1=0.575mPa.s

  (2) Cold Side:

     Media:Circulating cooling water

     Inlet temperature:t1=31oC

     Outlet temperature:t2=45oC (Tentative)


     Specific heat:Cp2=4.174kJ/(kg.K)

     Thermal conductivity:λ2=0.6308W/(m.K)

     Dynamic viscosity:μ2=0.6826mPa.s

3.Verifying calculation:

 (1)Heat Load  Q= Cp1.m. △T=4.174×0.028/60×989.125×(60-35)=48.2KW

 (2)Circulating cooling water flow rate




(4)No of plate:Ne=A/Ao=2/0.05=40pcs,Assembly plate N=41Pcs


 (6)elocity between two plates:



 (7)Reynolds number:




 (8)Bronte number:



  (9)Nusselt number:

         Nu1=0.381 Re10.6273Pr0.3 =0.381x1589.50.62733.720.3=57.58

        Nu2=0.381 Re10.6273Pr0.4 =0.381x23680.62734.520.4=91.15

   (10)The Euler number:

      Eu1=652Reh-0.1465 =652×1589.5-0.1465=221.5

      Euc=505Rec-0.1274 =505×2368-0.1274=187.6

   (11)Heat transfer coefficient:



Total heat transfer coefficient:




  (12)Heat transfer area


The equipment assemble area 2.0m2,Margin 25%,Then the heat transfer area fit for the heat load requirement

  (13)Pressure drop:

       △P1= Euh.ω2.ρ.m=221.5×0.212×989.125×2=19.32KPa

       △P2= Euc.ω2.ρ.m=187.6×0.372×992.9×2=51KPa

 Then ,the pressure drop fit for the application.