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LMTD Correction Factor Charts

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LMTD Correction factor charts

Heat transfer rate in the exchanger is represented by

q = U * A * F * LMTD

here F (< 1) is interpreted as a geometric correction factor, that when applied to the LMTD (Log Mean Temperature Difference) of a counter flow heat exchanger, provides the effective temperature difference of the heat exchanger under consideration.

It is a measure of the heat exchanger’s departure from the ideal behavior of a counter flow heat exchanger having the same terminal temperatures. The F-LMTD method is widely used in heat exchanger analysis, particularly for heat exchanger selection, (sizing problems) when as a result of the process requirements the temperatures are known and the size of the heat exchanger is required.

Log Mean Temperature Difference is defined as

LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)

where,

ΔT1 = T1 - t2
ΔT1 = T2 - t1

T1, T2 are inlet and outlet temperature of Fluid 1;  t1, t2 are inlet and outlet temperature of Fluid 2.




Log Mean Temperature Difference Correction Factor F is dependent on temperature effectiveness P and heat capacity rate ratio R for a given flow arrangement. Temperature effectiveness P is different for each fluid of a two fluid exchanger.

For fluid 1, it is defined as the ratio of the temperature range of fluid 1 to the inlet temperature difference.

P1 = ( T2 - T1 ) / ( t1 - T1 )

Heat Capacity Ratio R is defined as

R1 = ( t1 - t2 ) / ( T2 - T1 )

N (Shell) – 2M (Tube) Pass Tema E

Following general equation is used for shell and tube heat exchanger having N shell passes and 2M tube passes per shell.

S = (R1² + 1)0.5 / (R1 - 1)
W = [(1 - P1.R1)/(1 - P1)]1/N
F = S.ln(W)/ ln[( 1 + W - S + S.W) /( 1 + W + S - S.W)]

For limiting case of R1 = 1,

W' = (N - N.P1)/( N - N.P1 + P1 )
F = 20.5 [(1 - W')/ W' ]/ln[( W'/(1-W') + 1/20.5)/( W'/(1-W') - 1/20.5)]

For plotting the correction factor charts P1 values are listed from 0.01 to 1 with increment of 0.01 and then F values are calculated for each P1 and R1 based on above equations.

For following type of exchangers F values depend on NTU along with P1 and R1, where NTU is defined as number of transfer units. Range of NTU values are listed from 0.01 to 32 and F value is calculated for each value. LMTD Correction factor, F is determined as following –

F = [ 1/(NTU1.(1 - R1))].ln[(1 - R1.P1)/(1 - P1)]  ---- (1)

For limiting case of R1 = 1,

F = P1 / [NTU1 . (1 - P1)] ---- (2)

Cross Flow Fluid 1 Unmixed Tema X

Following relation is used to calculate P1 using NTU.

K = 1 - exp(-NTU1)
P1 = [1 - exp(-K.R1)]/ R1

F factor is calculated as per equations (1) & (2).



Cross Flow Fluid 2 Unmixed Tema X

Following relation is used to calculate P1 using NTU.

K = 1 - exp(-R1.NTU1)
P1 = 1 - exp(-K/R1)

F factor is calculated as per equations (1) & (2).

Cross Flow Both Fluid Mixed Tema X

Following relation is used to calculate P1 using NTU.

K1 = 1 - exp(-NTU1)
K2 = 1 - exp(-R1.NTU1)
P1 = 1/[1/K1 + R1/K2 - 1/NTU1]

F factor is calculated as per equations (1) & (2).

In a similar way, LMTD correction charts can be prepared for different type of geometries based on relation between NTU, P & R.

Spreadsheet

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References

  • Journal of Heat Transfer, Vol. 125, June 2003 – A General Expression for the Determination of the Log Mean Temperature Correction Factor for Shell and Tube Heat Exchangers – Ahmad Fakheri
  • Handbook of Heat Transfer 3rd Edition – Chapter 17 – Heat Exchangers by R.K. Shah and D.P. Sekulic
Jacketed Vessel Heat Transfer (Half Pipe Coil)

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Agitator equipped vessels with half pipe coil jackets are widely used in variety of process applications. This article shows how to calculate heat transfer in an agitated vessel provided with an external half pipe coil jacket.

Overall heat transfer coefficient, U is defined as

1/U = 1/hi + ffi + x/k + ffo + 1/ho

where,

  • hi : film coefficient process side
  • ho : film coefficient coil side
  • ffi : fouling factor process side
  • ffo : fouling factor coil side
  • x : vessel wall thickness
  • k : vessel wall thermal conductivity

Process Side, hi

Process side film coefficient, hi depends upon type of impeller, Reynold’s number (Re) and Prandtl number (Pr).

Re = D².N.ρ / μ
Pr = Cp.μ/k

where, D is impellor diameter, N is impellor rpm, ρ is fluid density, μ is fluid viscosity, Cp is fluid specific heat and k is fluid thermal conductivity at bulk fluid temperature. hi is defined as following :

Nui = C.Rea. Prb. (μ/ μw)c. Gc 
hi.DT/k = Nui

where constants C, a, b & c are available in literature for different type of impellors. DT is vessel diameter. Gc is geometric correction factor for non-standard geometries. μ/ μw is viscosity correction factor due to difference in viscosities at bulk fluid and wall temperatures. These constants are available in references mentioned below.



Coil Side, ho

Pipe coils are made with a 180° central angle or a 120° central angle. Equivalent diameter (De) and Flow area (Ax) is defined as following.

180° Coil

De = (π/2).dci
Ax = (π/8).dci²

120° Coil

De = 0.708 dci
Ax = 0.154 dci²

where, dci is inner diameter of pipe.

Reynold’s and Prandtl number are calculated based on jacket fluid properties and velocities.

Re = De.v.ρ / μ
Pr = Cp.μ / k

where, ρ is coil fluid density, μ is coil fluid viscosity and k is coil fluid thermal conductivity. v is fluid velocity in coil.

Turbulent Flow

For Re > 10000

Nuc = 0.027 Re0.8 Pr0.33 (μ/ μw)0.14 (1 +3.5 De/Dc)

where Dc is the mean or centerline diameter of the coil. Coil outer diameter (Do) is determined as following.

180° Coil : Do = DT + 2(dci/2) + 2.x
120° Coil : Do = DT + 2(dci/4) + 2.x
Dc = ( Do + DT) / 2

Laminar Flow

For Re < 2100

Nuc = 1.86 [ Re.Pr.De/L ]0.33 (μ/ μw)0.14

where, L is coil length along the vessel.



Transient Flow

For 2100 < Re < 10000
Calculate NuLaminar based on Re = 2100 and NuTurbulent based on Re = 10,000.

NuTransient = NuLaminar + (NuTurbulent - NuLaminar).(Re - 2100)/(10000 - 2100)

ho is determined as following

ho.De/k = Nuc

ho and hi thus calculated are used to get value of overall heat transfer coefficient U.

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References

  • Heat Transfer Design Methods – John J. McKetta (1992)
  • Heat Transfer in Agitated Jacketed Vessels – Robert F. Dream, Chemical Engineering, January 1999
Shortcut Sizing for Air Cooled Heat Exchanger

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An air-cooled exchanger is used to cool fluids with ambient air. 1 inch OD tube is the most popular diameter and the most common fins are 1/2 inch or 5/8 inch high. Tube configuration used in this guideline : 1 inch OD tube, 5/8 inch fin height, 10 fins per inch and 2.5 inch triangular pitch.

APSF – External area in ft²/ft² of bundle face area.

Rows APSF Face Velocity, ft/min
3 80.4 700
4 107.2 660
5 134.0 625
6 160.8 600

APF – Total external area/ ft of fintube in ft²/ft ~ 5.58.

Face Velocity – Typical air face velocities (VFace ) used in design are tabulated above, these value result in optimum cost of exchanger.

Obtain Process Parameters
Obtain process duty (Q), hot process side inlet (T1) and outlet (T2) temperature. Select an overall heat transfer coefficient (U) from literature based on type of fluids. Select an air inlet temperature (t1) that is not exceeded during a certain percentage of time over the year (e.g. 95% of the time).

Calculate Air Density
Density at air inlet temperature and site elevation

ρo = 14.696 x 29 /(10.7316 x (t1 + 459.67))
ρAir / ρo = exp(-29 x z/ (1545 x (t1 + 459.67)))

where,
t1: Air Inlet Temperature, °F
z : Elevation above sea level, feet

Calculate MTD
Assume an Air outlet temperature (t2) and calculate LMTD.

LMTD = ((T1-t2) -(T2-t1))/ ln((T1-t2)/(T2-t1))
R = (T1 - T2)/(t2 - t1)
S = (t2 - t1)/(T1 - t1)

LMTD Correction factor (F) is estimated based on following graphs

Cross Flow Single Pass

LMTD Correction Factor Cross Flow Single Pass

Cross Flow Two Pass

LMTD Correction Factor Cross Flow Multi Pass

Calculate Air Flowrate
Finned area is estimated using following equation :

AFinned = Q / (U * F * LMTD)

Bundle Face area is estimated using following equation :

AFace = AFinned / APSF

Air Flow is estimated using following relation :

VAir = AFace * VFace

Air Mass flowrate is estimated :

MAir = VAir * ρAir

Air temperature rise is calculated :

ΔT = Q / (MAir * CpAir )

Revised air outlet temperature is calculated :

t2 = t1 + ΔT

This temperature is again used in above steps to re-estimate air outlet temperature. Above steps are iterated till there is no change in air outlet temperature.

Air Cooler Dimension
Air cooler width is calculated :

W = AFace / LTube

Number of Tubes are calculated :

NTube = AFinned / (APF * LTube)

Number of Tubes per Row are calculated :

Nr Tube = NTube / No of Rows

where,
LTube : Length of Tube

Air Side Pressure Drop
Air Side pressure drop is calculated as following :

ΔP Total = ΔPStatic + ΔPVelocity

Static Pressure Drop
Pressure drop across tube bundle.

ΔP Static = FP * No of Rows / DR
FP = 6*10-8 * ( GFace )1.825

where,
DR : ρAir / ρAir at seal level and 70°F
GFace : Air face mass velocity in lb/h.ft² face area
ΔPStatic : Static pressure drop in inch of H2O

Velocity Pressure Drop
Typically 2 fans are used in air cooler. Fan area per fan (FAPF) is calculated as following :

FAPF = 0.4 * AFace / No. of Fans

Fan Diameter is calculated :

D = (4 * FAPF / π )0.5
ΔPVelocity = (ACFM / (4005 * (π* D2/4)) )2* DR

where,
ACFM : Air flowrate in Actual Cubic Feet per Minute
ΔPVelocity : Velocity pressure drop in inch of H2O

Power Calculation
Break power for fan is calculated as following :

BHP = ΔP Total * ACFM / 6356 / ηFan

Motor power is calculated as following :

Power = BHP / ηMotor

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References

  1. Process Heat Transfer: Principles and Applications,2007, Robert W Serth
  2. Handbook of Chemical Engineering Calculations, Nicholas P Chopey
  3. Rules of Thumb for Chemical Engineers, Carl R Branan
  4. GPSA, Engineering Databook, 12th Edition FPS
Double Pipe Heat Exchanger Design

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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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Heat Exchanger Analysis

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Heat Exchanger Analysis based on Effectiveness (ε) – NTU method is done when inlet temperatures are known and outlet temperatures are to be determined.

Obtain Process Parameters

Get process stream mass flowrate (M), specific heat (Cp) and inlet temperature (T). Obtain the heat transfer area (A) and overall heat transfer coefficient (U) for the given dimensions of heat exchanger.

Calculate heat capacities and obtain the minimum heat capacity.

CH = MH * CpH
CC = MC * CpC
CMin = Minimum (CH , CC)
CR = CMin / CMax

where MH, MC are hot and cold fluid mass flowrate; CpH, CpC are hot and cold fluid specific heat.

Calculate NTU and QMax

Number of transfer units ( NTU ) is calculated using following equation :

NTU = U.A/ CMin

Maximum heat transfer rate ( QMax ) is calculated using following equation :

QMax = CMin.(THot In - TCold In)

Determine Effectiveness

Based on NTU and CR (Ratio of heat capacities) determine heat exchanger effectiveness (ε) from Effectiveness – NTU curves available in literature.

ε – NTU Curve for Cross Flow exchanger Both stream unmixed

Effectiveness NTU Curve for Cross Flow Both stream unmixed

Calculate Outlet Temperature

Heat exchanger duty is calculated as:

Q = ε * Q Max

Outlet temperature are estimated as following :

THot Out = THot In - Q /( MH.CpH)
TCold Out = TCold In + Q /( MC.CpC)

Example
Hot exhaust gases 1.5 kg/s, enter a finned-tube, cross-flow heat exchanger at 250°C and is used to heat pressurized water at a flow rate of 1 kg/s available at 35°C. Exhaust gas specific heat is 1000 J/kg.K and water specific heat is 4197 J/kg.K. Overall heat transfer coefficient is 100 W/m².K and area is 40 m². What is rate of heat transfer by exchanger and gas and water outlet temperatures ?

Calculate Heat Capacities

CH = 1.5 kg/s * 1 kJ/kg.°K 
   = 1.5 kW/ °K
CC = 1.0 kg/s * 4.197 kJ/kg.°K 
   = 4.197 kW/ °K
CR = 0.36

Determine NTU and Q Max

NTU = 2.67
QMax = 322.5 kW

Determine Effectiveness
Based on ε – NTU curves for cross flow heat exchanger, determine ε

ε = 0.845

Calculate Outlet Temperature
Heat duty is calculated as following –

Q = 272.36 kW
THot Out = 68.43 ° C
TCold Out = 99.89 ° C

Spreadsheet

All above calculation for some common exchanger geometries have been provided in below spreadsheet.

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Heat Loss from Insulated Pipe

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Heat loss/gain takes place from a pipe carrying hotter/ colder fluid than ambient temperature. Insulation reduces the heat loss to surroundings. Heat loss depends upon number of factors like insulation thickness, ambient temperature, wind speed etc. This article shows how to calculate heat loss from an insulated pipe and a bare pipe to surroundings.

Example
A 3″ Carbon steel pipe is carrying hot oil at 180°C and insulated with 50 mm thick calcium silicate insulation. Insulation is cladded with a sheet with surface emissivity of 0.9. Ambient temperature is 28°C and wind velocity is 3.5 m/s. Calculate surface temperature and heat loss from insulated and bare pipe.

Insulated pipe heat loss example

Overall heat transfer coefficient of an insulated pipe is defined as following.

Insulated pipe heat transfer coefficient

where, kPIPE, kINSULATION are thermal conductivity of pipe and insulation. hin is heat transfer coefficient for fluid flowing in pipe and hair is heat transfer coefficient due to air flowing outside the pipe. The first two terms of denominator in above equation are generally smaller compared to remaining terms and can be neglected. For this example first term due to pipe fluid is ignored.

Air Side Heat Transfer Coefficient, hAIR

Air side heat transfer is due to combined effect of convection and radiation. Assume a temperature at cladding surface t_surface and steel pipe surface t_interface. Calculate average air film temperature as following.

 t_average = ( t_surface + t_ambient )/ 2

Estimate thermodynamic properties of air like thermal conductivity (k), viscosity (μ), expansion coefficient (β = 1/t_average), air density (ρ), kinematic viscosity (ν), specific heat (Cp) and thermal diffusivity (α) at average air film temperature. These properties are available in literature in form of tables, these can be fitted into a polynomial form using excel’s LINEST function. Reynolds’s number (Re), Prandtl number (Pr) and Rayleigh number (Ra) are calculated based on above properties.

h_radiation

Heat transfer coefficient due to radiation is calculated using following relation.

 h_radiation = σ ε (t_surface4 - t_ambient4)/ (t_surface - t_ambient)

where σ is Stefan Boltzmann coefficient and ε is emissivity for cladded surface.

h_convection

Convective heat transfer coefficient comprises of forced and free convection. Forced convection can be modeled based on correlation by Churchill and Bernstein.

Forced convection correlation by Churchill and Bernstein

 h_forced = Nu.k_air / D3

Free convection is calculated based on correlation by Churchill and Chu.

Free convection correlation by Churchill and Chu

 h_free = Nu.k_air / D3

Combined heat transfer coefficient due to forced and free convection is calculated using following relation.

 Nu_combined = ( Nu_forced 4 + Nu_free 4) 0.25
 h_convection = Nu_combined.k_air / D3

Air side heat transfer coefficient is calculated as following.

 h_air = h_radiation + h_convection

Overall Heat Transfer Coefficient, U

Thermal conductivity for insulation material and pipe is available in literature and depends upon temperature. It can be fitted into a polynomial equation using LINEST function in excel. Heat transfer resistance due to pipe and insulation is calculated using following relation.

 r_pipe = D3.ln(D2/D1) / 2.k_pipe
 r_insulation = D3.ln(D3/D2) / 2.k_insulation

Overall heat transfer coefficient is calculated as.

 r_overall =  r_pipe + r_insulation + 1/h_air
 U = 1/r_overall

Heat flowing through insulation is estimated.

 Q = (t_operating - t_ambient)/r_overall

A revised estimate for interface and surface temperature is made.

 t_interface = t_operating - Q.r_pipe
 t_surface   = t_interface - Q.r_insulation

Above steps are repeated with these new estimates till there is negligible difference in temperature.

Heat loss per unit length of pipe is estimated as following.

 HeatLoss = πD3 Q

Bare Pipe

For heat loss from bare pipe all above steps are repeated with resistance due to insulation not considered.

 r_pipe = D2.ln(D2/D1) / 2.k_pipe
 r_overall =  r_pipe + 1/h_air

For this example surface temperature and heat loss are as following.

example for insulated pipe

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