DESIGN AND COST ANALYSIS OF HEAT TRANSFER...

DESIGN AND COST ANALYSIS OF HEAT TRANSFER EQUIPMENTS

Md. Khairul IslamLecturer

Department of Applied Chemistry and Chemical Engineering. University of Rajshahi.

What is design?

• Design includes all the aspects involved in the development of either a new modified or expanded equipment or plant.

• Design of heat transfer equipment includes all the aspects involved in the development of either a new, modified or expanded heat transfer equipment.

4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 2

Heat Transfer Equipment

Usual terminology:

Exchanger: heat exchange between two process streams.

Heater or Cooler: a process stream is heated/cooled by a utility stream.

Vaporiser: a process stream is completely vaporised.

Reboiler: vaporiser associated with a distillation column.

Evaporator: used to concentrate a solution.

Fired heater: heating is done by combustion.


Tube and shell heat exchanger


Shell and Tube Heat Exchangers

The most commonly used heat exchangers.

Advantages:

Large surface area in a small volume.

Good mechanical layout.

Uses well established fabrication methods.

Can be constructed from a wide variety of materials.

Easily cleaned and maintained.

Well established design procedures.


Approach to Heat-Exchanger Design

Design of a heat exchanger is an iterative (trial & error) process. Here is a set of steps for the process:

1. Calculate the required heat transfer rate, Q, in Btu/hr from specified information about fluid flow rates and temperatures.

2. Make an initial estimate of the overall heat transfer coefficient, U, based on the fluids involved.

3. Calculate the log mean temperature difference, ΔTm, from the inlet and outlet temperatures of the two fluids.

4. Calculate the estimated heat transfer area required, using: A = Q/(U ΔTm).

5. Select a preliminary heat exchanger configuration.

6. Make a more detailed estimate of the overall heat transfer coefficient, U, based on the preliminary heat exchanger configuration.


Approach to Heat-Exchanger Design (contd.)

7. Estimate the pressure drop across the heat exchanger. If it is too high, revise the heat exchanger configuration until the pressure drop is acceptable.

8. If the new estimate of U is significantly different than the previous estimate, repeat steps 4 through 7 as many times as necessary until the two estimates are the same to the desired degree of accuracy.

9. Input information needed. In order to start the heat exchanger design process, several items of information are needed as follows:

• The two fluids involved need to be identified.

• The heat capacity of each fluid is needed.

• The required initial and final temperatures for one of the fluids are needed.

• The design value of the initial temperature for the other fluid is needed.


Considerations in selection of heat-transfer equipment

Tube size:

Length is standard, commonly 8, 12 or 16 ft.

Diameter: most common 3/4 or 1 in OD.

(Pipes are used to transport something, and tubes to construct something; hence, tubes are defined by the outside diameter and wall thickness (for construction stability), and pipes are measured by inside diameter to allow a calculation for transportation viz., speed, volumes etc)



• Tube pitch and clearance:

Pitch is the shortest center-to-center distance between adjacent tubes. Commonly 1.25 to 1.5 time the tube diameter.

Clearance is the distance between tubes. It should be larger than 25% of the tube diameter. Minimum 3/16 inch

Triangular or square arrangement of tubes are quite common.PT = do+ C

PT = tube pitch

do = tube outside diameter and C = clearance



• Baffles:

Baffles are usually spaced between 20% and 100% of the ID of the shell.

• Shell:

Up to 24 in nominal size, use standard pipes. In general, a shell thickness of 3/8 in. is used for shell diameters between 12 and 24 in. unless the fluids are extremely corrosive.

Nominal Pipe Size (NPS) is a North American set of standard sizes for pipes. Nominal pipe size (NPS) for diameter based on inches.


Nominal Pipe sizeNPS[inches]

Nominal DiameterDN[mm]

Outside DiameterOD[mm]

1/2 15 21.3

3/4 20 26.7

1 25 33.4

1 1/4 32 42.16


• Passes:

Most usual pass is one. Split flow arrangement are used for pressure drop reduction, when the pressure drop is the controlling factor in the design.

• Fluid location:

Corrosive fluids flow inside the tubes.

Fluid with higher fouling tendency inside the tubes.

High pressure fluid inside the tubes (if everything else the same).

Hot fluid inside the tubes.

Typical velocities:

Liquids: 1-2 m/s in tubes, max 4 m/s to reduce fouling. 0.3 to 1 m/s in shell

Vapors: 50-70 m/s (vacuum), 10-30 m/s (1 bar), 5-10 m/s (high P)



• Cleaning and Maintenance

• Use of Water in Heat Exchangers

• Use of Steam in Heat Exchangers

• Ineffective Surface

• Thermal Strains


Fouling Factor

Fouling is the build up of sediments and debris on the surface area of a heat exchanger that inhibits heat transfer. Fouling will reduce heat transfer, impede fluid flow, and increase the pressure drop across the heat exchanger. The design engineer can include a resistance, called a dirt, scale, or fouling factor, when determining an overall coefficient of heat transfer.


Types of FoulingThere are several types of fouling, each forming depending on the type of

fluid and conditions.Crystallization is one of the most common type of fouling. Certain salts

commonly present in natural waters have a lower solubility in warm water than cold. Ex. Cerium (III) sulphate, lithium selenite etc [Common Solution: reducing the temperature of the heat transfer surface often softens the deposits]

Sedimentation, the depositing of dirt, sand, rust, and other small matter is also common when fresh water is used. [Common Solution: velocity control]

Biological Organic growth material occurs from chemical reactions, and can cause considerable damage when built up. [Common Solution: material selection]

Chemical Reaction Coking appears where hydrocarbon deposits in a high temperature application. [Common Solution: reducing the temperature between the fluid and the heat transfer surface]

Corrosion can destroy surface areas of the heat exchangers, creating costly damage. [Common Solution: material selection]

Freezing Fouling results from overcooling at the heat transfer surface causing solidification of some of the fluid stream components. [Common Solution: reducing the temperature gradient between the fluid and the heat transfer surface]


Mechanisms of Heat Transfer:

1. Conduction:

where Q is the amount of heat, Btu, transferred in time𝝷 , h

k is the thermal conductivity, Btu/[h ft2 (oF/ft)]

A is the area of heat transfer normal to heat flow, ft2

t is the temperature, oF

x is the thickness or length of the conduction path, ft.



2. Convection:

h is the heat transfer coefficient, Btu/[h ft2 oF].

3. Radiation:

where

σ is the Stefan-Boltzmann constant = 0.1713 10-8 Btu/(h ft2 oR4)

ε is the emissivity of surface

A is the exposed area for heat transfer, ft2

T is absolute temperature, oR.



Rate of heat interchange between two bodys,

Or

Where, Fa is a correction factor based on the relative orientation of the two surfaces and the surface chosen for the evaluation of A, and Fe is a correction factor based on the emissivities and absorptivities of the surfaces.



q radiated from pipe= hrA1𝛥tf

where h, is a fictitious heat-transfer coefficient based on the rate at which radiant heat leaves the surface of the pipe.

q radiated from pipe= hrA1(T1-T2)


Comparison

• For conduction

• for convection

• For radiation q= hrA1(T1-T2)


Overall Heat Transfer Coefficient

Definition of the overall heat transfer coefficient, U

q =U A 𝛥t

Or, U= q/A𝛥t

U [=] Btu/(h ft2 oF)


Overall Heat Transfer CoefficientOverall resistance, R=1/UA

Resistances in series:

Overall resistance = Sum of resistances

In this case:

Ud: overall heat transfer coefficient

h, h : outside/inside film heat transfer coefficient

Af , Af outside/inside pipe diameter

k: wall thermal conductivity

hd , hd

outside/inside fouling heat transfer coefficient4/15/2016 Md. Khairul Islam, ACCE, University of Rajshahi 21

Temperature difference The log mean temperature difference (DTm or LMTD) can be calculated from the terminal temperatures of the two streams if the following assumptions are valid:

• All elements of a given fluid stream have the same thermal history in passing through the exchanger.*

• The exchanger operates at steady state.

• The specific heat is constant for each stream (or if either stream undergoes an isothermal phase transition).

• The overall heat-transfer coefficient is constant.

• Heat losses are negligible.


Mean area of heat transfer1. Conduction of heat through a solid of constant cross section

(example, a large flat plate).

Am=Aarith.avg= (A1+A2)/2

2. Conduction of heat through a solid when the cross-sectional area of heat transfer is proportional to the radius (example, a long hollow cylinder).

3. Conduction of heat through a solid when the cross-sectional area of heat transfer is proportional to the square of the radius (example, a hollow sphere).

Am=Ageom.mean=√(A1.A2)


Film Coefficients for Fluids in Pipes and Tubes (No Change in Phase)


Film Coefficients for Fluids Flowing Outside Pipes and Tubes (No change in Phase)

where a, = 0.33 if tubes in tube bank are staggered and 0.26 if tubes are in line

F, = safety factor to account for bypassing effects

G, = shell-side mass velocity across tubes, based on minimum free area between baffles at shell axis? lb/(hXft2> Subscript f refers to properties at average film temperature


Tube-side Pressure Drop

• fi is the friction factor for isothermal flow at the mean temperature

• np is the number of tube passes

• gc is the conversion factor in Newton’s law of motion, (32.17) X (3600)2

ft.lbm/h2lbf

• φi is a correction factor for non-isothermal flow

• φi =1.1( μ i / μw )0.25 for Re < 2100

• φi =1.02( μ i / μw )0.14 for Re > 2100

• Bi is a correction factor for friction due to contraction, expansion and reversal of flow direction = = 1 + (Fe + Fe + Fr)/(2fiG

2L/gcpi2Diφi).

G • is the mass velocity inside the tube.


Shell-side Pressure Drop

Where

f’ = special friction factor for shell-side flow

Nr= number of rows of tubes across which shell fluid flows

B0 = correction factor to account for friction due to reversal in direction of flow, recrossing of tubes, and variation in cross section; when the flow is across unbaffled tubes, B0 can be taken as 1.0


HEAT-TRANSFER EQUIPMENT COSTS

The major factors that can influence the cost for heat transferequipment are indicated in the following list:

1. Heat-transfer area

2. Tube diameter and gauge

3. Tube length

4. Pressure

5. Materials of construction for tubes and shell

6. Degree and type of baffling

7. Supports, auxiliaries, and installation

8. Special features, such as floating heads; removable bundles;multipass, finnedsurfaces; and U bends


Problems and Solutions

Calculate a preliminary estimate of the heat exchanger area needed to cool 55,000 lb/hr of a light oil (specific heat = 0.74 Btu/lb.°F) from 190°F to 140°F using cooling water that is available at 50°F. The cool-ing water can be allowed to heat to 90°F. An initial estimate of the Overall Heat Transfer Coefficient is 120 Btu/hr.ft².°F. Also estimate the required mass flow rate of cooling water.



Solution: required heat transfer rate for the above indicated light oil:

Q = m.Cp. (T2 – T1)

Imperial Units:

Q = (55,000 lb/hr) (0.74 Btu/lb.°F) (190 – 140) °F = 2,035,000 Btu/hr.

Next calculate the log mean temperature difference:

T1 = Inlet tube side fluid temperature (light oil hot side = 190 °F)

t2 = Outlet shell side fluid temperature (water cold side = 90 °F)

T2 = Outlet tube side fluid temperature (light oil cold side = 140 °F)

t1 = Inlet shell side fluid temperature (water cold side = 50 °F)

ΔTm=(190-90)-(140-50)/[ ln (190-90)/(140-50)]


Problems and SolutionsThe preliminary area estimate of the heat exchanger can now be calculated as

A = Q / (U x ΔTm) =

A = [2, 035, 000 Btu/hr]/[ (120Btu/h.ft².°F).(94.9°F)]= 178.7 ft²

The required mass flow rate of water can be calculated from

Q = m.Cp.ΔTm.

Rearranging:

m = Q /Cp. ΔTm

m = [(2,035,000 Btu/hr)]/[(0.74Btu/lb.°F)(94.9°F)]= 28,978 lb/hr



A manufacturer is to design a heat exchanger in which the specification takes some steam at 4 bar g (58 psi g) to heat secondary water from 10°C (50°F) to 60°C (140°F).It is assumed that 1 litre of water has a mass of 1 kg.Mass flow rate = 1.5 l/s (24 gpm) x 1 kg/l = 1.5 kg/s (3.30 lb/s).

The manufacturer uses a Heat Transfer Coefficient 'U' for the heat exchanger of 2500 W/m2°C (440 Btu/hr.ft².°F). Take the Specific Heat of water (cp) as 4.19 kJ/kg°C = 4190 J/ kg°C (1.0 Btu/lb.°F).Determine:

• The design heat load. • The corresponding steam flow rate. • The minimum heating area required. • When the minimum heat load occurs when the inlet water temperature

rises to 30°C (86°F), determine:• The minimum heat load. • The corresponding steam pressure in the heat exchanger. • The corresponding steam flow rate.



A) Find the design heat load using the heat transfer flow rate equation:

Q = m.cp. ΔTm

Where:

Q = Heat transfer flow rate (kJ/s) (kW) – (Btu/s) (HP) m = mass of steam (kg/s) (lb/s)

cp = Specific heat capacity of the secondary fluid (kJ/kg°C) – (Btu/lb.°F) ΔTm = Temperature rise of the secondary fluid (K or °C) -

Q = 1.5 kg/s x 4.19 kJ/kg°C (60 – 10)°C = 314.25 kJ/s = 314.25 kW

Or Q = 3.30 lb/s x 1.0 Btu/lb.°F x (140°F - 50°F) = 297 Btu/s = 420 HP



(B) Find the corresponding steam flow rate:

At 4 bar g, (39°F) the saturation temperature (Ts) is 152°C (305°F), and he = 2108.1 kJ/kg = 2,108,100 J/kg = 906 Btu/lb (from steam tables).

Calculate the required steam flow at the design condition using equation below:

Metric Units:

Steam flowrate (m) =[ kW x 3600 (kg/h)]/ he

Steam flowrate (m) = [314.25 x 3600 (kg/h) ]/2108.1

Steam flowrate (m) = 536.6 kg/h

Imperial Units:

Steam flowrate (m) =[297 x 3600 (lb/h)]/906 Btu/lb

Steam flowrate (m) = 1180 lb/h


Problems and Solutions( C) Use the LMTD (ΔTm) to calculate the minimum amount of heating area to satisfy the design rating:

ΔTm = [ (T2 – T1)]In (Ts – T1)/(Ts – T2)

ΔTm = Logarithmic Mean Temperature Difference (LMTD)

T s = Steam temperature = (152 °C) (305°F)

T 1= Secondary fluid in temperature = (10 °C) (50°F)

T 2= Secondary fluid out temperature = (60°C) (140°F)

In = The mathematical function known as 'natural logarithm'

ΔTm = 115°C= 206°F


A = Q

U .ΔTm

Where:

A = Heating area (m²)Q = Mean heat transfer rate (W)U = Heat transfer coefficient (W/m²°C)

ΔTM = Mean Temperature Difference.

By re-arranging the general heat transfer equation (Q = U x A x ΔTm):


Metric Units:

A = [314,250 W]/[2500 W/m².°C x 115.2°C]

A = 1.09 m²

Imperial Units:

A = [1,072,800 Btu/h]/[440 Btu/hr.ft².°Fx 239°F]

A = 10.2 ft²

(D) Find the minimum heat load, when the inlet water temperature is 30°C:

Q = m.cp. ΔT

Metric Units:

Q = (1.5 kg/s).(4.19 kJ/kg.oC) (60 oC - 30 oC) = 188.5 kJ/s = 188.5 kW

Imperial Units:

Q = 3.30 lb/s x 1.0 Btu/lb.°F x (140°F - 86°F) =178.2 Btu/s = 252 HP



(E) Find the corresponding heat exchanger steam pressure and enthalpy at minimum load:

1) A steam temperature of 115.2°C (239.3°F) corresponds a steam pressure of 0.7 bar g.

2) The Specific Enthalpy of evaporation at 0.7 bar g (he) = 2 215 kJ/kg (see steam tables).

(F) Find the steam flow rate at minimum load:

From (D) the minimum heat load is 188.5 kW = 252 HP

From (E) the he is 2 215 kJ/kg = 952 Btu/lb

Steam flowrate (m) =[kW x 3600kg/h]/he

Steam flowrate (m) = [188.5 kW x 3600kg/h]/2 215 kJ/kg

Steam flowrate (m) = 306.4 kg/h – at minimum load

Imperial Units:

Steam flowrate (m) = 178 Btu/s x 3600 lb/h

952 Btu/lb

Steam flowrate (m) = 673 lb/h - at minimum load.


DESIGN AND COST ANALYSIS OF HEAT TRANSFER...

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