Heat Transfer

Double Pipe Heat Exchanger Design

This article shows how to do design for Double Pipe Heat Exchanger and estimate length of double pipe required.

Determine Heat Load

Obtain flowrate (W ), inlet, outlet temperatures and fouling factor for both hot and cold stream. Calculate physical properties like density (ρ), viscosity (μ), specific heat (Cp) and thermal conductivity (k) at mean temperature. Determine heat load by energy balances on two streams.

Q = mH.CpH(THot In - THot Out)
  = mC.CpC(tCold Out - tCold In)

where,

  • mH , mC: Mass flow rate of Hot and Cold Stream
  • CpH , CpC: Specific Heat of Hot and Cold Stream
  • THot In , THot Out: Inlet and outlet temperature of Hot Stream
  • tCold In , tCold Out: Inlet and outlet temperature of Cold Stream

Calculate Logarithmic Mean Temperature Difference (LMTD)

LMTD = (ΔT1 - ΔT2)/ln( ΔT1 / ΔT2)

For Counter-current flow

ΔT1 = THot In - tCold Out
ΔT2 = THot Out - tCold In

For Co-current flow

ΔT1 = THot In - tCold In
ΔT2 = THot Out - tCold Out

Calculate Film Coefficient

Allocate hot and cold streams either in inner tube or annular space. General criteria for fluid placement in inner tube is corrosive fluid, cooling water, fouling fluid, hotter fluid and higher pressure stream. Calculate equivalent diameter (De) and flow area (Af) for both streams.

Inner Tube

De = Di
Af = π Di²/4

Annular Space

De = D1 - Do
Af = π (D1² - Do²)/4

where,

  • Di : Inside Pipe Inner Diameter
  • Do : Inside Pipe Outer Diameter
  • D1 : Outside Pipe Inner Diameter

Calculate velocity (V), Reynolds No. (Re) and Prandtl No. (Pr) number for each stream.

V = W / ( ρ Af )
Re = De V ρ / μ
Pr = Cp μ / k

For first iteration a Length of double pipe exchanger is assumed and heat transfer coefficient is calculated. Viscosity correction factor (μ / μw)0.14 due to wall temperature is considered 1.

For Laminar Flow (Re <= 2300), Seider Tate equation is used.

Nu = 1.86 (Re.Pr.De/L )1/3(μ/ μw)0.14

For Transient & Turbulent Flow (Re > 2300), Petukhov and Kirillov equation modified by Gnielinski can be used.

Nu = (f/8)(Re - 1000)Pr(1 + De/L)2/3/[1 + 12.7(f/8)0.5(Pr2/3 - 1)]*(μ/μw)0.14
f  = (0.782* ln(Re) - 1.51)-2

where,

  • L : Length of Double Pipe Exchanger
  • μw : Viscosity of fluid at wall temperature
  • Nu : Nusselts Number (h.De / k)

Estimate Wall Temperature

Wall temperature is calculated as following.

TW = (hitAve + hoTAveDo/Di)/(hi + hoDo/Di)

where,

  • hi : Film coefficient Inner pipe
  • ho : Film coefficient for Annular pipe
  • tAve : Mean temperature for Inner pipe fluid stream
  • TAve : Mean temperature for Annular fluid stream

Viscosity is calculated for both streams at wall temperature and heat transfer coefficient is multiplied by viscosity correction factor.

Overall Heat Transfer Coefficient

Overall heat transfer coefficient (U) is calculated as following.

1/U = Do/hi.Di + Do.ln(Do/Di)/2kt + 1/ho+ Ri.Do/Di + Ro

where,

  • Ri : Fouling factor Inner pipe
  • Ro : Fouling factor for Annular pipe
  • kt : Thermal conductivity of tube material

Calculate Area and length of double pipe exchanger as following.

Area = Q / (U * LMTD )
L = Area / π * Do

Compare this length with the assumed length, if considerable difference is there use this length and repeat above steps, till there is no change in length calculated.

Number of hair pin required is estimated as following.

N Hairpin = L / ( 2 * Length Hairpin )

Calculate Pressure Drop

Pressure drop in straight section of pipe is calculated as following.

ΔPS = = f.L.G²/(7.5x1012.De.SG.(μ/ μw)0.14)

where,

  • ΔP : Pressure Drop in PSI
  • SG : Specific Gravity of fluid
  • G : Mass Flux ( W / Af ) in lb/h.ft²

For Laminar flow in inner pipe, friction factor can be computed as following.

f = 64/Re

For Laminar flow in annular pipe.

f = (64 / Re) * [ (1 - κ²) / ( 1 + κ² + (1 - κ²) / ln κ) ]
κ = Do / D1

For turbulent flow in both pipe and annular pipe

f = 0.3673 * Re -0.2314

Pressure Drop due to Direction Changes

For Laminar Flow

ΔPR = 2.0x10-13. (2NHairpin - 1 ).G²/SG

For Turbulent Flow

ΔPR = 1.6x10-13. (2NHairpin - 1 ).G²/SG

Total Pressure Drop

ΔPTotal = ΔPS + ΔPR

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