Energy Efficient Process Heating

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Energy Efficient Process Heating






Web Seminar


By Kelly Kissock Ph.D., P.E.

University of Dayton Industrial Assessment Center


August 24, 2006


2

Heat Supply and Demand

Furnace Heat
Input

Heat in

Flue Gases

Combustion





Stoichiometric Combustion



Flame Temperature



Available Heat

4

Perfect (Stoichiometric) Combustion

When

fuel

reacts

with

exactly

the

right

amount

of

air,

all

of

the

carbon

and

hydrogen

atoms

combine

with

all

of

the

oxygen

to

form

carbon

dioxide

and

water

vapor
.

This

is

called

“stoiciometric”

combustion
.

CH
4

+ 2O
2



CO
2
+2H
2
O

5

Excess Air

Most

burners

operated

with

more

than

stoichiometric

air

to

guarantee

that

every

fuel

molecule

finds

an

oxygen

molecule
.


The

quantity

of

air

in

“excess”

to

stoichiometeric

air

is

called

excess

air
.


The

quantity

of

excess

air

can

be

measured

by

measuring

the

amount

of

oxygen

in

the

combustion

gasses
.


Excess Air: CH
4

+ 4O
2



CO
2
+2H
2
O + 2O
2

Stoichiometric: CH
4

+ 2O
2



CO
2
+2H
2
O

6

Excess Air

Combustion Products Analysis for a Typical Natural Gas

7

Excess Air

Excess

air

dilutes

the

products

of

combustion,

resulting

in
:


1
)

a

lower

flame

temperature


2
)

less

heat

transfer

from

the

combustion

gasses

to

the

load


3
)

more

heat

carried

away

in

the

exhaust

gasses
.



CH
4

+ 4O
2



CO
2
+2H
2
O +
2O
2

8

Flame (Combustion) Temperature


Flame (Combustion) Temp. (F)



Condition




4,450



NG, With 100% O
2


3,750


NG, 900 F


0% Excess Air


3,545


NG, 900 F
-

20% Excess Air


3,460


NG, 70 F
-

O% Excess Air


3,225


NG, 70 F
-

15 % Excess Air


2,750


NG, 7.0:1 Air/Fuel Ratio




Flame temperature affects heat transfer and
temperature distribution within the heating
system (furnace, oven etc.).

9

Effect of Oxygen Enhancement

Flame Temperature for Natural Gas

10

Calculating Combustion Temperature




Combustion chamber


Stoichiometric combustion equation (for natural gas):


CH
4

+ 2 (O
2

+ 3.76 N
2
)


CO
2

+ 2 H
2
O +7.52 N
2



Air/Fuel ratio for stoichiometric combustion:


AFs = 2 (O
2

+ 3.76 N
2
) / CH
4

= 2 (32 + 3.76 28) / 16 = 17.2


Combustion temperature (T
c
) from energy balance:


T
c

= T
ca

+ LHV / [{1 + (1 + EA) AFs} Cp
g
]

11

Available Heat (Combustion Efficiency)

Available Heat =



Available Heat =


Fraction of energy not


lost in exhaust gasses

supplied
energy

fuel

Total
system
by

absorbed
Energy
12


Calculating Available Heat

















Process heating system


Percent available heat (
h
) from energy balance on system:



h

= [{1 + (1 + EA) AFs} Cp
g

(T
c



T
ex
)] / HHV





m
ng

+ m
ca


T
ca

combustion
chamber

m
ex

T
ex




{Q
out
}


13

Air Flow

-

Usually the largest loss in process heat.


-

Air flow heat loss:


Boilers (250 F


350 F): 20%


Aluminum furnace (1,400 F): 50%


Glass melter (2,500 F): 70%

14

Types of Air Flow


Combustion Air



needed to burn fuel


Ventilation Air



for moisture and volitile removal


Infiltration Air



undesirable; minimized by proper design and
maintenance

15

Managing Combustion Air: Minimize Excess Air


Optimum excess air for energy efficiency and pollution prevention:
10% (yields 2% O
2

in comb gasses)


Combustion temperature increases:

T
c

= T
ca

+ LHV / [{1 + (1 +
ECA
) AFs} Cp
g
]


Percent available heat (combustion efficiency) increases:

h

= [{1 + (1 + ECA) AFs} Cp
g

(
T
c



T
ex
)] / HHV


Example:


Aluminum melt furnace


1,465 F exhaust gas temperature


Operates with 95% excess combustion air


Reducing excess air increases percent available heat from 39% to 60%


Energy use decreases by
35%




16

Managing Combustion Air: Preheating






























Recuperator schematic


A recuperator transfers heat from exhaust gasses to inlet combustion air.


Recuperator effectiveness

(
ε) relationship:

ε = Q / [m
ca

Cp
a

(T
ex1



T
ca1
)] =
(T
ca2



T
ca1
) / (T
ex1



T
ca1
)






T
comb air 1

T
comb air 2

T
exhaust 1

T
exhaust 2

Q

17

Managing Combustion Air: Preheating


Combustion temperature increases:

T
c

=
T
ca

+ LHV / [{1 + (1 + ECA) AFs} Cp
g
]


Percent available heat (combustion efficiency) increases:

h

= [{1 + (1 + ECA) AFs} Cp
g

(
T
c



T
ex
)] / HHV


Example:


Aluminum melt furnace


2,500 F combustion temperature


1,465 exhaust gas temperature


40% effective recuperator increases comb. air temperature to 615 F


Increases combustion temperature to 3,010 F


Energy use decreases by
34%




18

Managing Combustion Air: Use Exhaust Air

Exhaust gasses from ovens with high ventilation rates contain high
O
2

content, and can be redirected back to the burner as
combustion air.


Effective combustion temperature increases:

T
c,eff

=
T
ca

+ LHV / [{1 + (1 + EA) AFs} Cp
g
]


Percent available heat (combustion efficiency) increases:

h

= [{1 + (1 + EA) AFs} Cp
g

(
T
c,eff



T
ex
)] / HHV


Example:


Curing oven at 250 F


18% O
2

in exhaust (about 660% excess air)


Using exhaust for combustion increases percent available heat from 64% to 67%


Energy use decreases by
4%




19

Managing Ventilation Air:


Find Requirement

Industrial ovens must never exceed 25% of lower
explosive limit (LEL)





National Fire Protection Agency Standard 86


This is achieved by:


using 10,000 cu. ft. of ventilation air per gallon of
cured paint in continuous process.


using 380 cfm of ventilation air per gallon of cured
substance in batch process.





20

Managing Ventilation Air: Minimize

Ventilation air can be turned down manually through dampers to meet
process demand or can be controlled with LEL sensors.


Excess air decreases:

T
c,eff

= T
ca

+ LHV / [{1 + (1 +
EA
) AFs} Cp
g
]


Percent available heat (combustion efficiency) increases:

h

= [{1 + (1 + EA) AFs} Cp
g

(
T
c,eff



T
ex
)] / HHV


Example:


Curing oven with 141 F exhaust gasses


3,700% excess air measured


3,470 cfm ventilation air


Ventilation air could be reduced to 45 cfm


Percent available heat would increase from 43% to 82%


Energy use would decrease by
47%




21

Managing Vent Air: Using Thermal Oxidizer Discharge Air


Thermal oxidizers burn off volatile organic compounds in oven exhaust.


Discharge air (usually around 300 F) can be redirected to oven.


Effective combustion temperature increases:

T
c,eff

=
T
a

+ LHV / [{1 + (1 + EA) AFs} Cp
g
]


Percent available heat (combustion efficiency) increases:

h

= [{1 + (1 + EA) AFs} Cp
g

(
T
c,eff



T
ex
)] / HHV


Example:


Curing oven at 200 F internal temperature


75% of air entering oven is ventilation air


Percent available heat would increase from 77% to 90%


Energy use would decrease by
14%




22

Managing Infiltration

Ovens and furnaces are typically under negative
pressure.


Outside air will infiltrate through cracks, open doors,
loose cracks, etc. through differential pressure and
buoyancy effects.





Vertical oven opening

23

Managing Infiltration: Move Opening to Floor

Due to buoyancy effects, little cool air will infiltrate to a warm oven
through its floor.


Energy lost through infiltration (Q
inf
):

Q
inf

= V
infil

A ρ
a

Cp
a

(T
exfil



T
inf
)



Q
inf
could be reduced by

as much as 80%.


Example:


Second story curing oven at 435 F temperature


Door area of 100 sq. ft.


Infiltration was measured to be 2,900 cfm


Energy use would decrease by
40%




Horizontal oven opening

24

Managing Infiltration: Lower Openings

Lowering openings decreases buoyancy effects between cool and warm air.


New infiltration velocity (V
inf2
) from Bernoulli’s Equation.

V
inf2

= V
inf1



Energy saved (Q
sav
):

Q
sav

= A Cp
a

[V
inf1

ρ
a1
(T
oven,1



T
inf
)


V
inf2

ρ
a2
(T
oven,2



T
inf
)]



Example:


Curing oven at 450 F temperature


Door area of 8.5 sq. ft.


Infiltration was measured to be 2,250 cfm


Infiltration would reduce to about 2,000 cfm


Exfiltration temperature would decrease


Energy use would decrease by
28%




Oven opening location
before and after retrofit

25

Summary of Managing Air Flow

Savings opportunities for combustion air:


Minimize combustion air (35% savings)


Preheat combustion air (34% savings)


Use exhaust as combustion air (4% savings)


Savings opportunities for ventilation air:


Minimizing ventilation air (47% savings)


Using thermal oxidizer discharge air for ventilation (14% savings)


Savings opportunities for infiltration:


Move oven opening to floor (40% savings)


Lower oven openings (28% savings)


Savings calculations can be assisted by:


PHAST (www1.eere.energy.gov/industry/bestpractices/software.html)


HeatSim (www.engr.udayton.edu/udiac)

26

Heat Loss

Heat is lost through system walls by
conduction, then convection and radiation.





Heat is lost from heated open tanks by
convection, radiation, and evaporation.



27

Insulating Hot Surfaces


Heat lost (Q) from a surface:


Q

= h A (T
s



T
a
) +
s

A
e

(T
s
4



T
a
4
)




e

is
~
0.9 for dark surface,
~0.1 for shiny surface


Q = A (T
f



T
s
) /
R
shell



convection component

radiation component

conduction

28

Insulating Hot Surfaces

Calculating convection coefficient (h)

Laminar air if: L
3

D
T < 63

Turbulent air if: L
3

D
T > 63


L = (length x width)
1/2

for flat surfaces, L = diameter for cylindrical objects




Horizontal:

h
lam

=
0.27 (
D
T / L)
0.25
; h
tur

= 0.22 (
D
T)
0.33




Vertical:

h
lam

= 0.29 (
D
T / L)
0.25
; h
tur

= 0.19 (
D
T)
0.33




Horizontal:

h
lam

= 0.27 (
D
T / L)
0.25
; h
tur

= 0.18 (
D
T)
0.33





Vertical:

h
lam

= 0.29 (
D
T / L)
0.25
; h
tur

= 0.19 (
D
T)
0.33



Relations taken from ASHRAE Fundamentals


Cylindrical Surfaces{

Flat Surfaces{

29

Covering Heated Tanks

Insulation floats can cover heated tanks to reduce convection
and radiation heat transfer, and virtually eliminate evaporation.


Floats cover up to 79% of

liquid surface area.


Energy balance on float:

h
fs

(
T
fs



T
a
) + ε σ [
T
fs
4



T
a
4
] =



= (T
w



T
fs
) / R
float



HeatSim iterates values of float surface temperature until
equation is balanced. Convection coefficient depends on surface
temperature.



30

Covering Heated Tanks

Convection heat loss (Q
conv
):

Q
conv

= h A (T
w



T
a
)


(h dependent on both water and air
temperature)



Radiation heat loss (Q
rad
):

Q
rad

=
e

s

A (T
w
4



T
a
4
)


Evaporation heat loss (Q
conv
):

Q
evap

= m
w

h
fg

(both values dependent on water and air
temperature)


Total heat loss (Q
tot
):

Q
tot

= Q
conv

+ Q
rad

+ Q
evap







31

Reducing Thermal Mass

Continuous Systems

Energy lost to conveyor (Q
cvr
) traveling at
velocity (V):


Q
cvr

= V
m

Cp
cvr

(T
cvr2



T
cvr1
)



Example:


Brazing oven at 1,900 F


Stainless steel conveyor at velocity 0.7 ft/min


Conveyor weighs 5 lbs/ft


Conveyor only loaded 30% of time


Conveyor is slowed to 0.3 ft/min when unloaded


18,000 Btu/hr, or
40%
, of conveyor energy saved



32

Reducing Thermal Mass

Batch Systems

Thermal mass in oven must heat to temperature
during every batch cycle.



Thermal resistance (R
1
):

R
1

= 1 / (h A) + dx / (2 k A)


R
2

= R
3

= R
4

= R
5

= R
6

= R
7

= R
8

= dx / (k A)



Finite difference equations:

E
in



E
out

=
D
E
store



(T
N



T) / R
N



(T


T
S
) / R
S

= [dx A


Cp (T’


T)] / dt


T’ = dt (T
N



T) / (R
N

dx A


Cp)




dt (T


T
S
) / (R
S

dx A


Cp) + T





Firebrick mass

33

Reducing Thermal Mass

Batch Systems

Example


heat treat oven:


Heat treat oven with 100 sq. ft. floor


8
-
inch thick firebrick layer


Oven is raised to 1,700 F, remains at temperature for 4 hours


Temperature decreases by 50 F every hour until temperature reaches 400 F


Savings measure



reduce firebrick thickness:


Firebricks laid flat so new thickness is 4 inches


Energy savings is 235,000 Btu per cycle












8
-
in (20.3 cm)
thick

4
-
in (10.2 cm)
thick

T1

484 F (251 C)

477 F (247 C)

T2

591 F (310 C)

569 F (298 C)

T3

674 F (357 C)

631 F
(333 C)

T4

736 F (391 C)

662 F (350 C)

T5

782 F (417 C)

T6

813 F (434 C)

T7

832 F (444 C)

T8

841 F (449 C)


Temperature profile for 8
-
inch and 4
-
inch
firebrick after cycle

34

Heat Loss Summary

Shell loss depends on insulation type and thickness. Insulate
surfaces over 150 F.


Heat loss from heated open tanks dominated by evaporation.
Cover tanks to minimize evaporation


Minimize thermal mass of interior structure in batch processes with
short cycle times


Minimize thermal mass of conveyor in continuous processes.


Savings calculations assisted by:




PHAST
(www1.eere.energy.gov/industry/bestpractices/software.html)




HeatSim (www.engr.udayton.edu/udiac)



35

Summary

Energy efficiency improved by:


Minimizing combustion air


Minimizing ventilation air


Minimizing infiltration air


Insulating surfaces over 150 F


Covering open heated tanks


Reducing mass of structure in batch processes


Reducing mass of conveyor in continuous processes


More information see our web site:


www.engr.udayton.edu/udiac

36

Thank you!