Heat Exchanger Training 02.20.15

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JGC America’s Vision Process Training “Quick” Heat Exchanger Calculation By Sharon Wenger February 26, 2015 1

Transcript of Heat Exchanger Training 02.20.15

Page 1: Heat Exchanger Training 02.20.15

JGC America’s Vision

Process Training

“Quick” Heat Exchanger Calculation

By Sharon Wenger

February 26, 2015

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Overview

Introduction–Types of Exchangers–Basic Heat Transfer

• Pure Counter-current Flow• Now Count-current Flow

–Heat Transfer Calculation Formulas• General• Sensible Heating or Cooling of Fluids• Steam Condensing• Heating of Cooling a Solid• Terminology and Definitions

–Working Problem from Mozambique LNG• Mozambique LNG PJ FEED - DeC2 OVHD Condenser (251-E-1006)

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Types of Heat Exchangers– Shell & Tube – Hairpin – Plate & Frame – Brazed Plate – Welded Plate – Finned Tube TEMA Heat

Exchangers

Plate and Frame Heat Exchanger

Casketed Plate & Frame Heat Exchanger

Brazed Plate Heat Exchanger

Hairpin Heat Exchanger 

Twisted Tube Heat Exchangers

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Terminology and Definitions

TerminologyDimensionless Numbers

– Nusselt No (Nu) = Film coefficient * diameter / thermal conductivity – Prandtl No (Pr) = Heat capacity * viscosity / thermal conductivity– Reynolds No (Re) = Diameter * density * flow rate / viscosity

Biot NumberIs a comparison of the internal thermal resistance to the external resistance of a body. If small, then the body temp will be uniform during heating or cooling

– Bi = h * Lc / k– Lc = V/As = the characteristic length– h is the film coefficient– k is the thermal conductivity– V is volume– As is the surface area

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Terminology and Definitions

TerminologyFourier Number

– For transient heat transfer– Ratio of heat transferred by conduction to rate of energy stored– Fo = alpha * t / Lc2– alpha is the thermal diffusivity– t is time– Lc is the characteristic length– As is the surface area

Laminar and Turbulent– Shellside– Turbulent at Re of 1000 to 2000 and higher– Tubeside– Laminar has Re < 2000– Turbulent has Re > 10000

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Main heat transfer equationBasic Heat Transfer Equation Q = UA ΔTm

where, – Q = rate of heat transfer – U = mean overall heat transfer coefficient – A = heat transfer surface area – ΔTm = logarithmic mean temperature difference

Q = m Cp T – Q = rate of heat transfer, Btu/hr– m = flow rate, pounds/hr– Cp = heat capacity, Btu/pound/F– T = temperature change, F

• Q = mphase_change – Q = rate of heat transfer, Btu/hr– mphase_change = mass that changes phase, pounds/hr– = heat of vaporization, Btu/pound

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Main heat transfer equationBasic Heat Transfer

Non Counter-Current Flow Flow

Pure Counter-current

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Basic Heat Transfer

FT is the LMTD correction factor. The value of FT is depends upon the stream inlet and outlet temperatures and the exchanger geometry

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Heat Transfer Calculation Formulas -Most of the formula listed below are “rules of thumb” for quick estimation purposes and are limited in their application. The estimation is under standard temperatures and pressures.

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Heat Transfer Calculation Formulas -

– = GPM x Density 8.022 = GPM × 501.375 × Specific GravitySpecific Gravity = Density/62.4 Psia = Psig+14.7

– = Evaporative Cooling Tower Tons x 15000– SCFM of air = [ACFM x (psig + 14.7) x 528] / [(Temp + 460) x 14.7]– SCFM of air = Lbs/ Hr of air / 4.5 (at atmospheric temperature and pressure)

General

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Heat Transfer Calculation Formulas

Sensible Heating or Cooling of Fluids

– Btu/hr = Lbs./ Hr x Specific Heat x Specific Gravity X Temp Rise– Btu/hr = GPM x Temp Rise x K

water = 500 K 30% glycol = 470 K 40% glycol = 450 K 50% glycol = 433K hydraulic oil = (210 – 243) K

– For Air, Btu/hr = 1.085 x SCFM x Temp Rise

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Heat Transfer Calculation Formulas

Steam Condensing

 – Btu/hr = Lbs./ Hr x Latent Heat

Heating or cooling a solid 

– Btu/hr for Solids = Lbs./Hr x Specific Heat x Delta-T

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Approach of a Temperature Difference (ATD)What is Approach of a Temperature Difference (ATD)

That is to transfer heat from the refrigerant coolant temperature difference between the tow, and that is the approach of a temperature difference, Shown in Fig. below. It must be large enough to provide a flow of heat, necessary to achieve the required system capacity.

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Information Required to Design a Heat Exchanger

We are often asked what information we require to design a heat exchanger. It depends on the project phases

1) Pre-FEED2) FEED3) EPC detailed design.

In case 1) and 2) we only have minimum amount of information, can we prepare a quick budget design? Yes!In case 3) we have all the detail information to design for a particular application, by using program such as HTRI will help us optimize a design result.

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Information Required to Design a Heat Exchanger

Fluid Properties1. Fluid Composition and Percentage 2. Specific Heat3. Viscosity, cp 4. Specific Gravity or Density5. Thermal Conductivity6. Latent Heat, (if phase change)7. Operating Pressure and Temperature

Support Information 8. Allowable Pressure Drop9. Fouling Factor10. Design Pressure11. Design Temperature

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Working Problem from Mozambique LNG

Do we have to have all the required information to design a heat exchanger? Not necessary. That is where we use the quick calculation formulas to design it

Example - Working Problem from Mozambique LNG• Mozambique LNG PJ FEED - DeC2 OVHD Condenser (251-E-1006)

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Mozambique LNG FEED – Deethanizer OVHD Condenser (251-E-1006)

Steps to solve the problem1.Gather all the Information Required to

Design a Heat Exchanger2.PFD3.HMB4.Find the streams in and out to the

Exchanger

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Calculated

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Approach of a Temperature Difference (ATD)

Streams 112 113 374 375

Temperature [C] -0.14 -16.3 -22.3 -19.3Pressure [bara] 26.7 26.7 2.5 2.5Mass Flow [kg/h] 3980 3980 4500 4500

Calculate condenser supply temperature to achieve condensing temperature of -16.23 °C.

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Calculate Heat Exchanger Duty Required

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Streams 1 2 3 4

Temperature [C] tc1 tc2 Th1 Th2

  -0.14 -16.3 -22.3 -19.3

Pressure [bara] 26.7 26.7 2.5 2.5

Mass Flow [kg/h] 3980 3980 4500 4500

LMTD (°C) 11.35      

UA (Kw/°C) 33 - 40 From Go-By  

For UA = 33   Q (Kw) = 375

For UA = 40   Q (Kw) = 454

         

Result UniSim Results

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Single Phase Fluid Heat

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Two Phase Heat @ Isothermal Conditions

Q = mphase_change

• Q = rate of heat transfer, Btu/hr• mphase_change = mass that changes phase, pounds/hr• = heat of vaporization, Btu/pound

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Information Required to Design a Heat Exchanger• Streams

Stream No. 374 375

Pressure bar_a 2.50 2.50Temperature oC -19.32 -19.32

VAPOR PROPERTIES

Flow kgmole/hr 25.30 102.16kg/hr 1,116 4,505MMSCF60F/day 5.080E-01 2.05

Volumetric Flow m3/hr 198.11 799.99Enthalpy kJ/kgmole -107,494 -107,494Entropy kJ/kgmole*K -289.614 -289.614Molecular Weight kg/kgmole 44.097 44.097Density kg/m3 5.631 5.631Heat Capacity kJ/kg*K 1.604 1.604Viscosity cpoise 6.744E-03 6.744E-03Thermal Conductivity W/m*K 1.365E-02 1.365E-02Specific Heat Ratio 1.198 1.198Gas Compressibility 9.286E-01 9.286E-01LIQUID PROPERTIES

Flow kgmole/hr 76.86 0.00kg/hr 3,389

Volumetric Flow m3/hr 6.12 0.00Enthalpy kJ/kgmole -125,135Entropy kJ/kgmole*K -359.113 0.000Molecular Weight kg/kgmole 44.097 0.000Density kg/m3 553.422 0.000Heat Capacity kJ/kg*K 2.364 0.000Viscosity cpoise 1.512E-01 0.000E+00Thermal Conductivity W/m*K 1.183E-01 0.000E+00Specific Heat Ratio 1.567

TOTAL PROPERTIES

Flow kgmole/hr 102.16 102.16kg/hr 4,505 4,505MMSCF60F/day 2.05 2.05

Enthalpy kW -3,427 -3,051Entropy kJ/hr*K -34,930 -29,588Molecular Weight kg/kgmole 44.097 44.097

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Heat Transfer Calculation Formulas