Objective

• Heat Exchangers• Learn about different types

• Define Heat Exchanger Effectivness (ε)

• Analyze how geometry affects ε

• Solve examples

Systems: residential

Indoor Air

Outdoor Air

Large building system Chiller

Large building system Chiller

Outdoor air

53oF

43oF

Water to

building

Water from

building

95oF

Air-liquid Tube heat exchanger

Plate heat exchanger

Heat exchangers

Air-air

Some Heat Exchanger Facts

• All of the energy that leaves the hot fluid enters the cold fluid

• If a heat exchanger surface is not below the dew point of the air, you will not get any dehumidification• Water takes time to drain off of the coil

• Heat exchanger effectivness varies greatly

Heat Exchanger Effectivness (ε)

C=mcp

exchangeheatposible Maximum

exchangedHeat

Location B Location A

THout

TCin

TCout

THin

Mass flow rate Specific capacity of fluid

Example:What is the saving with the residential heat recovery system?

Furnace

72ºF

32ºF 72ºF

Outdoor Air

For ε=0.5 and if mass flow rate for outdoor and exhaust air are the same

50% of heating energy for ventilation is recovered!

For ε=1 → free ventilation! (or maybe not)

52ºF Exhaust

Gas

Combustion

products

Fresh Air

Air-Liquid Heat Exchangers

• Fins added to refrigerant tubes

Extended surfaces (fins) from air side

Analysis of Compact Heat Exchangers

• Geometry is very complex

• Assume flat circular-plate fin

Overall Heat Transfer

Q = U0A0Δtm

Overall Heat

Transfer Coefficient

Mean temperature

difference

Δtm for Heat Exchangers

Depends on

flow direction:

• Parallel flow• Counterflow• Crossflow

Ref: Incropera & Dewitt (2002)

Heat Exchanger Analysis - Δtm

ocih

icoh

ocihicohm

tt

tt

ttttt

,,

,,

,,,,

ln

Counterflow

A

B

ABm

tt

ttt

ln

For parallel flow is the same

or

A

B

ABm

tt

ttt

ln

Logarithmic mean temperature difference

Counterflow Heat Exchangers

ocih

icoh

ocihicohm

tt

tt

ttttt

,,

,,

,,,,

ln

icoc

ohih

tt

ttR

,,

,,

icih

icoc

tt

ttP

,,

,,

Important parameters:

RPP

Rttt icocm

11

ln

1,,

ExampleAssume that the residential heat recovery system is counterflow heat exchanger with ε=0.5.

Calculate Δtm for the residential heat recovery system if : mcp,hot= 0.8· mc p,cold

Furnace

72ºF

32ºF 72ºF

Outdoor Air

Exhaust

Combustion

products

Fresh Air

mcp,hot= 0.8· mc p,cold mc p,cold 0.2· mc p,cold

th,i=72 ºF, tc,i=32 ºF

For ε = 0.5 → th,o=52 ºF, tc,o=48 ºF

Δtm,cf=(20-16)/ln(20/16)=17.9 ºF

What about crossflow heat exchangers?

Δtm= F·Δtm,cf

Correction

factor Δt for

counterflow

Derivation of F is in the text book:

………

Example:Calculate the real Δtm for the residential heat recovery cross flow system

(both fluids unmixed):

For: th,i=72 ºF, tc,i=32 ºF , th,o=52 ºF, tc,o=48 ºF

R=1.25, P=0.4 → From diagram → F=0.92 Δtm= Δtm,cf · F =17.9 ·0.92=16.5 ºF

Overall Heat Transfer

Q = U0A0Δtm

ExternalPipecondInternalo RRRRU

110

Need to find this

AP,oAF

Heat Transfer

tP,o

1) (eq. )(

)()(

,,

,,,

ttAAhQ

ttAhttAhQ

P,oFoPoc

F,mFocP,ooPoc

Heat transfer from fin and pipe to air (External):

tF,m

t

)/()( ,, tttt BFmF where is fin efficiency

Heat transfer from hot fluid to pipe (Internal ):

Heat transfer through the wall:

2) (eq. )(, P,ifiPi ttAhQ

3) (eq. )(,

P

P,oP,imPP

x

ttAkQ

Resistance model

Q = U0A0Δtm

From eq. 1, 2, and 3:

• We can often neglect conduction through pipe walls• Sometime more important to add fouling coefficients

ocFA

oPAoc

pmP

po

iiP

o

hhkA

xA

hAA

U,,

,,,

110

1

R Internal R cond-Pipe R External

Example

With given:

Calculate the needed area of heat exchanger A0=?

Btu/hsfF 10h ,sfF/Btu/h 002.0R , Btu/hsfF 10 h ExternalcondInternal

The air to air heat exchanger in the heat recovery system from previous example has flow rate of fresh air of 200 cfm.

Solution: Q = mcp,cold Δtcold = mcp,hot Δthot = U0A0Δtm

From heat exchanger side: Q = U0A0Δtm → A0 = Q/ U0Δtm

U0 = 1/(RInternal+RCond+RFin+RExternal) = (1/10+0.002+0+1/10) = 4.95 Btu/hsfF

Δtm = 16.5 F

From air side: Q = mcp,cold Δtcold =

= 200cfm·60min/h·0.075lb/cf·0.24Btu/lbF·16 = 3456 Btu/h

Then: A0 = 3456 / (4.95·16.5) = 42 sf

For Air-Liquid Heat Exchanger we need Fin Efficiency

• Assume entire fin is at fin base temperature• Maximum possible heat transfer• Perfect fin

• Efficiency is ratio of actual heat transfer to perfect case

• Non-dimensional parameter

)/()( ,, tttt BFmF

tF,m

Fin Theory

pL=L(hc,o /ky)0.5

k – conductivity

of material hc,o – convection

coefficient

Reading Assignment

• Chapter 11- From 11.1-11.7

Final project topics

• Beside 3 introduced in last class:• Duct design • DOAS design • VAV design