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 (C_{p}) and thermal conductivity (k) at mean temperature. Determine heat load by energy balances on two streams.

`Q = m`

_{H}.Cp_{H}(T_{Hot In}- T_{Hot Out})`= m`

_{C}.Cp_{C}(t_{Cold Out}- t_{Cold In})

where,

m_{H} , m_{C}: Mass flow rate of Hot and Cold Stream

Cp_{H} , Cp_{C}: Specific Heat of Hot and Cold Stream

T_{Hot In} , T_{Hot Out}: Inlet and outlet temperature of Hot Stream

t_{Cold In} , t_{Cold 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 = T`

_{Hot In}- t_{Cold Out}`ΔT2 = T`

_{Hot Out}- t_{Cold In}

For Co-current flow

`ΔT1 = T`

_{Hot In}- t_{Cold In}`ΔT2 = T`

_{Hot Out}- t_{Cold Out}

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**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 (D_{e}) and flow area (A_{f}) for both streams.

*Inner Tube*

`D`

_{e}= D_{i}`A`

_{f}= π D_{i}²/4

*Annular Space*

`D`

_{e}= D_{1}- D_{o}`A`

_{f}= π (D_{1}² - D_{o}²)/4

where,

D_{i} : Inside Pipe Inner Diameter

D_{o} : Inside Pipe Outer Diameter

D_{1} : Outside Pipe Inner Diameter

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

`V = W / ( ρ A`

_{f})`Re = D`

_{e}V ρ / μ`Pr = C`

_{p}μ / k

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

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

`Nu = 1.86 (Re.Pr.D`_{e}/L )^{1/3}(μ/ μ_{w})^{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 + D`

_{e}/L)^{2/3}/[1 + 12.7(f/8)^{0.5}(Pr^{2/3}- 1)]*(μ/μ_{w})^{0.14}`f = (0.782* ln(Re) - 1.51)`

^{-2}

where,

L : Length of Double Pipe Exchanger

μ_{w} : Viscosity of fluid at wall temperature

Nu : Nusselts Number (h.D_{e} / k)

**Estimate Wall Temperature**

Wall temperature is calculated as following.

`T`_{W} = (h_{i}t_{Ave} + h_{o}T_{Ave}D_{o}/D_{i})/(h_{i} + h_{o}D_{o}/D_{i})

where,

h_{i} : Film coefficient Inner pipe

h_{o} : Film coefficient for Annular pipe

t_{Ave} : Mean temperature for Inner pipe fluid stream

T_{Ave} : 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 = D`_{o}/h_{i}.D_{i} + D_{o}.ln(D_{o}/D_{i})/2k_{t} + 1/h_{o}+ R_{i}.D_{o}/D_{i} + R_{o}

where,

R_{i} : Fouling factor Inner pipe

R_{o} : Fouling factor for Annular pipe

k_{t} : Thermal conductivity of tube material

Calculate Area and length of double pipe exchanger as following.

`Area = Q / (U * LMTD )`

`L = Area / π * D`

_{o}

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.

`ΔP`_{S} = = f.L.G²/(7.5x10^{12}.D_{e}.SG.(μ/ μ_{w})^{0.14})

where,

ΔP : Pressure Drop in PSI

SG : Specific Gravity of fluid

G : Mass Flux ( W / A_{f} ) 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 = D`

_{o}/ D_{1}

For turbulent flow in both pipe and annular pipe

`f = 0.3673 * Re `^{-0.2314}

**Pressure Drop due to Direction Changes**

For Laminar Flow

`ΔP`_{R} = 2.0x10^{-13}. (2N_{Hairpin} - 1 ).G²/SG

For Turbulent Flow

`ΔP`_{R} = 1.6x10^{-13}. (2N_{Hairpin} - 1 ).G²/SG

**Total Pressure Drop**

`ΔP`_{Total} = ΔP_{S} + ΔP_{R}