ENERGY CONVERSIONMME 9617A

Eric Savorywww.eng.uwo.ca/people/esavory/

mme9617a.htm

Lecture 8 – Basics of heat exchangers

Department of Mechanical and Material EngineeringUniversity of Western Ontario

Heat Exchangers

The most common types of energy conversion systems (e.g. internal combustion engines, gas/steam turbines, boilers) consist of three parts:

1. a combustion process generating heat and kinetic energy (K.E.)

2. a device for converting K.E. to mechanical (useful) energy

3. heat exchangers to recuperate the heat either for heating purposes or to increase

efficiency.

The different applications of heat exchangers require different designs (geometries):

Heat Exchangers are classified according to their function and geometry:

Function:1. Recuperative: two fluids separated by a solid

wall (this is the most common type)2. Evaporative: enthalpy of evaporation of one

fluid is used to heat or cool the other fluid (condensers/evaporators and boilers)

3. Regenerative: use a third material which stores/releases heat

Geometry: 1. Double Tube 2. Shell and Tube 3. Cross-flow Heat Exchangers 4. Compact Heat Exchangers

Underlying calculation approach

The heat transfer rate for most heat exchangers can be calculated using the LMTD-method (Log Mean Temperature Difference), if the inlet (T1) and outlet (T2) temperatures are known:

TAUQ

FT/Tln

TTT

12

12

U = Overall heat transfer coefficient [ W/m2-oC ]A = Effective heat transfer surface area [ m2 ]F = Geometry correction factor = Log mean temperature differenceT

Otherwise, the Effectiveness () – Number of Transfer Units (NTU) method may be used:

minmax CmAU

NTUQ

Q

General Formulation for Heat Exchanger Analysis (LMTD-method)

Most heat exchangers are characterized relative to a double-pipe heat exchanger (H = Hot, C = Cold):

T2

T1

CiCoCHoHiH TTCmTTCmTAUQ

We now want to derive the expression for LMTD for a counter-flow double-pipe heat exchanger.

This will be done by considering the first law (for counter flow):

First globally:

Then locally: Apply the first law between points 1 and 2 (for counter-flow)

Heat lost by hot side = Heat gained by cold side

For counter-flow:

By using the notation 1 and 2, as shown on the graphs, this definition is valid for both Counter-current and Co-flow (parallel) double-pipe heat exchangers.

CoHi

CiHo

CoHiCiHo

CoHi

CiHo

CiCoHiHo

TTTT

ln

TTTTAU

TTTT

ln

TTTTAUQ

1

2

12

TT

ln

TTAUQ

ξ-NTU (Effectiveness – Number of Transfer Units) Method

maxQQ

transferheat.maxlTheoreticatransferheatActual

If the inlet or outlet temperatures are not given, the LMTD-method becomes cumbersome to use. It is thus advisable to use the Effectiveness-NTU method. The method can be formulated fromthe following definitions:

Effectiveness:

minCmAU

NTU

Minimum thermal capacity max. temp. difference

In general:

Actual heat transfer is given by

Theoretical maximum heat transfer by:

Hence, we obtain the effectiveness as:

CiCoCHoHiH TTCmTTCmQ

CiHimin TTCmQ

CiHimin

CiCoC

CiHimin

HoHiH

TTCmTTCm

TTCmTTCm

For a counter-flow heat exchanger:

Let

and

Which, on using the definition for LMTD, leads to an expression for the effectiveness as:

minH CmCm

C

H

max

min

Cm

Cm

Cm

CmR

R1NTU

R1NTU

eR1e1

minC CmCm

H

C

max

min

Cm

Cm

Cm

CmR

R1NTU

R1NTU

eR1e1

If, instead

then

We end up with the same effectiveness:

Counter flow Parallel flow

HC CmCm HC CmCm

Similar expressions are used for other types of geometry.

For example, for a parallel double-pipe heat exchanger, the effectiveness is:

R1e1 R1NTU

Next we shall look at some applications of these concepts.

Typical thermal design problems

• Problem #1– Given the entrance temperature of the two

streams, given one exit temperature;– Find heat transfer area, A.

• Problem #2– Given entrance temperature of the two

streams, given the heat transfer area, A;– Find the exit temperatures of the two

streams.

Objective: Calculation procedure and advantages / disadvantages of:

Double pipe Shell and tube Cross flow heat exchangers

1. Double Pipe Heat Exchangers:

Double Pipe Heat Exchangers

Arrangements:

Advantages:

- low pressure loss- small applications (simple, cheap to build)- counter flow: high effectiveness; parallel flow: quick (short) fetches.

Disadvantage:

- requires large surface area (footprint on floor) if large heat transfer rates are needed.

2. Shell-and-Tube Heat Exchangers:

Advantages:

- ideal for large scale applications- commonly used in petrochemical industry where dangerous substances are present (protective shell)- compact design or double tube heat exchanger.

Disadvantages:

- very bulky (heavy construction), baffles are used to increase mixing- subject to water hammer and corrosion (behind baffles)- high pressure loses (recirculation behind baffles)

Heat transfer calculations:

Using counter flow, double pipe heat exchanger definition for the temperatures

TAUQ

FT/Tln

TTT

12

12

CoHi1CiHo2 TTTTTT

Heat exchanger correction factor plot for one shell pass and an even number of tube passes

= +

Heat exchanger correction factor plot for two shell passes and twice an even number of tube passes

For n-shell passes with an even number of tubes:

Again, for boiling or evaporation R 0so that = 1 – e-NTU

Cross flow and compact heat exchangers

Overview: Cross-flow and compact heat exchangers are used where space is limited. These aim to maximize the heat transfer surface area.

Cross-flow Heat Exchangers: Commonly used in gas (air) heating applications. The heat transfer is influenced by whether the fluids are unmixed (i.e. confined in a channel) or mixed (i.e. not confined, hence free to contact several different heat transfer surfaces).

e.g.: both fluids unmixed: air-conditioning devices e.g.: both fluids mixed: boilers

In a cross-flow heat exchanger the direction of fluids are perpendicular to each other. The required surface area, Across for this heat exchanger is usually calculated by using tables. It is between the required surface area for counter-flow (Acounter) and parallel-flow (Aparallel) i.e. Acounter< Across <Aparallel

Cross-Flowsmay be mixedor unmixed

Advantage: large surface area-good for transferring heat to gases

Disadvantages: heavy, high pressure losses

Both fluids unmixed

Both fluids unmixed

One fluid unmixed

TAUQ

FT/Tln

TTT

12

12

Cross-flow heat exchangers have the same analysis equations as before:

with F as the correction factor (see graphs). The -NTU method may also be used

Heat exchanger correction factor plot for single pass, cross-flow with one fluid mixed

Heat exchanger correction factor plot for single pass, cross-flow with both fluids unmixed

Compact heat exchangers: These are cross-flow heat exchangers characterized by very large heat transfer area per unit volume. In fact, the contact area is so large that much of the flow behaves as duct or channel flow.

For this reason, the heat-transfer is dominated by wall effects and the characteristics cannot be evaluated as for the other types.

For these heat exchangers, the heat transfer rate is directly related to pressure loss.

Advantages:- very small- ideal for transferring heat to / from fluids with very low conductivity or where the heat transfer must be done in very small spaces (e.g. electronic component cooling, cryogenic cooling, domestic furnaces). Disadvantages:- high manufacturing costs- very heavy- extremely high pressure losses.

Examples of compact heat exchangers

To solve problems involving design and selection (sizing) of compact heat exchangers it is first required to find the effective pressure (static) loss. This loss can be shown, based on fundamental heat transfer principles, to be directly related to the heat transfer rate based on Colburn’s analogy:

f – friction factor, St – Stanton number,Pr – Prandtl number and jH = Colburn factor

32

H PrSt8f

j

These calculations can be quite involved and so most design or sizing applications use data in tables and graphs.

All material properties are calculated at the bulk average temperature, i.e. at (T1+T2)/2, if T1 = inlet, T2 = exit

CpPrnumberandtlPr

2max

H

U

DdxdP

ffactorFriction

GDRe H

m

Reynolds number at the smallest diameter:

DH = hydraulic diameter at smallest cross-section = 4 Ac / PAc = smallest cross-sectional areaP = perimeter (circumference) of tubeµ = dynamic viscosity = thermal diffusivityG = maximum mass flow rate flux = mass flow rate

cAm

G

AAc

CpGh

St

= ratio of open area to total frontal area (A)h = heat transfer coefficientCp = specific heat capacity

1

m

c1

222

1

VV

AA

f1VV

12GV

p

p = pressure loss through heat exchangerVm = (V2 + V1) / 2

Overall heat transfer coefficient UA is computed from:

hc Ah1

Ah1

AU1

(h A)h = hot fluid(h A)c = cold fluidA = effective heat transfer area

Then the heat transfer Q is:

TAUQ

Heat transfer and friction factor for a finned flat tubeheat exchanger

Heat transfer and friction factor for a finned circulator-tube

heat exchanger (details on next slide)

Summary

Summary of effectiveness equations

Heat exchanger Effectiveness:type:

Heat exchanger Effectiveness:type:

= +

Heat exchanger Effectiveness:type:

Example questions

Example 1 – Finned flat tube heat exchanger

Air at 1 atm and 300 K enters a finned flat tube heat exchanger (as in graph in an earlier slide) with a velocity of 15 m/s. Calculate the heat transfer coefficient (h).

Note at this temperature the air properties (found from tables) are:

= 1.1774 kg/m3

= 1.983 x 10-5 kg/msCp = 1.0057 kJ/KgoCPr = 0.708

Example 2 – Shell and tube heat exchanger

Hot oil at 100oC is used to heat air in a shell and tube heat exchanger. The oil makes 6 tube passes and the air makes 1 shell pass. 2.0 kg/s of air (specific heat of 1009 J/kgoC) is to be heated from 20 to 80oC. The specific heat of the oil is 2100 J/kgoC and its flow rate is 3.0 kg/s. Calculate the area required for the heat exchanger for U = 200 W/m2oC.

Example 3 – Finned-tube (both fluids unmixed) cross-flow heat exchanger

A finned-tube exchanger is used to heat 2.36 m3/s of air (specific heat of 1006 J/kgoC) at 1 atm from 15.55 to 29.44oC. Hot water enters the tubes at 82.22oC and the air flows across the tubes, producing an average overall heat transfer coefficient of 227 W/m2oC. The total surface area of the exchanger is 9.29m2. Calculate the heat transfer rate (kW) and the exit water temperature.

Note: We don’t know whether the air or the water is the minimum thermal capacity fluid. So try with the air as the minimum fluid first and see if the -NTU equations give a possible solution. If not then we have to use water as the minimum and iterate to a solution.