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

Model：BR0.05-1.0-2-E-Ⅶ

Material：

Heat Exchanger Plate：SUS304

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

Density：ρ1=989.125kg/m3

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)

Density：ρ2=992.9kg/m3

Specific heat：Cp2=4.174kJ/(kg.K)

Thermal conductivity：λ2=0.6308W/(m.K)

Dynamic viscosity：μ2=0.6826mPa.s

3.Verifying calculation：

(2)Circulating cooling water flow rate

(3)LMTD

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

(5)Pass：2x10/2x10

(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：

K=3666W/(m2.K)

(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.