Basics of Electric Heaters

Hotwatt, made in USA, is a quality supplier
to oem’s in the industrial, medical,
commercial, packaging, instrumentation,
aviation, transportation and military fields.
Contents
Basics of
Electric Heaters
The simplest definition
of an electric heater is any
device that changes electrical
energy into heat energy. But
from that simple explanation,
electric heaters explode into
a myriad of types, sizes,
applications, and designs
depending upon what’s being
heated, the degree of heating
needed, and the method by
which the heat is applied.
The measure of electrical
energy is called the Joule
after its discoverer, James
Prescott Joule. Through
numerous experiments, Joule
determined that the quantity
(Q) of heat transferred
from electrical energy is
proportional to the square of
the current (I 2 ) multiplied
by the resistance (R) for the
period of time (t) through
which it passes:
Q ∝ I2 × R × t
However, one seldom sees a reference to
Joules used in modern
Instead,
Q ∝ I 2electric
× R × circuits.
t
the controlling factor
P =becomes
I 2 × R that of power
(P):
P = I2 × R
A = πD × h or 6π × 5 = 941/4 in2
You’llS note the only difference between the
formula for determining power and that of
determining Joules is the time component.2
AS =factor
πD ×inhheating
or 6π becomes
× 5 = 941/4
in
The time
readily
×c×
�Fhot when
apparent in any devicemthat
gets
W=
an electric current flows
through
3.412
× T it: its
temperature rises as time passes.
m × c × �F
W=
Types of heaters
3.412 × T
Heaters come in
Wmany
= WL/SFdifferent
× A styles
and configurations, each design optimized
Sponsored by Hotwatt W = WL/SF × A
m ×H
Types of
heaters …1
Transfer of
heat …2
Heat losses …2
Basic heat
equations …3
Specific
applications …3
An application
example …4
for a specific application. The selection of a
heater depends on the material being heated,
the heater style, the sheath material that
surrounds the heating element and protects
it from the material being heated, and the
operating voltage of the heater.
Styles of heaters include strip, ring,
rope and cable, cartridge, tubular, band,
immersion, circulation, process air and duct,
radiant, comfort, flexible, tote, and drum.
The style of heater chosen depends on the
application. The type chosen depends on the
heated object’s shape, size, and mass, as well
as performance requirements.
For example, band heaters work best on
cylindrical object such as tanks and pipes.
Cartridge heaters are typically inserted in
close fitting holes in large blocks of metal
1
Watt density …5
Presented by
March 2013
Type of heater …1
Basic heat equations …3
An application example …4
Watt density …5
Hotwatt manufactures cartridge, air
process, immersion, strip and finned strip,
tubular, band, foil, flexible glasrope®,
crankcase and ceramic heaters.
such as platens and
molds. And immersion
heaters may, as their name
implies, be immersed
directly into the material
being heated. As added
protection, an immersion
heater may use a heat
well, a protective device
that isolates the heater
from the material being
heated but still allows heat
transfer to the material.
Heat wells also allow
immersion heaters to be
replaced without the need
to drain the tank.
Sheath materials can
include steel, Incoloy,
iron, silicone, Kapton,
copper, ceramic, and glass
overbraid among others to handle specific materials. The
choice of sheath material depends upon the material being
heated.
infrared wavelengths, although that’s not the only frequency
used. For example, microwave ovens use wavelengths many
times longer than infrared to heat food while medical
diathermy machines use even lower radio frequencies to
warm parts of the human body.
Electric heaters use all three methods to warm specific
objects. Which of the three methods becomes the best choice
depends upon the application.
Transfer of heat
When a body starts to generate heat, meaning its
temperature rises above that of other nearby objects, it
is called a heat source. As its temperature rises, it starts
to raise the temperature of the materials in its vicinity
using any combination of three different methods. Those
three methods are known as conduction, convection, and
radiation.
Conduction transfers heat energy from one material to
another via direct contact. It is the most direct method of
transferring heat energy, and is usually considered the most
efficient: the highest percentage of heat energy created
transfers to the colder object from the heat source.
Convection also uses physical contact to transfer heat
energy, but the contact entails the use of an intermediary
gas, typically air. In convection heating, the heat source
warms the gas. The warmed gas now weighs less per
volume measure, producing a buoyancy effect. This
buoyancy is easily seen in the flight of hot air balloons. The
hot gas rises, creating convection currents that move the
heated gas into contact with the colder objects, warming
them. The now cooled gas flows back to the heat source
where it is warmed again, and the process repeats.
The third method of transferring heat energy, radiation,
does not rely on any physical contact between the heat
source and the object to be heated. Instead, heat energy
is transmitted through space from the heat source to the
object in the form of electromagnetic radiation: typically as
Sponsored by Hotwatt
Heat losses
As stated previously, an object subjected to electric heating
continues to become hotter as long as current passes through
it until it reaches its burn out temperature and the heat
source is destroyed. Obviously, when used properly modern
electric heaters do not experience this burn out condition.
The answer lies in heat losses, or the transfer of heat away
from the heat source that keeps its temperature down.
The term heat loss is typically reserved for heat energy
transferred to undesired areas, but heat energy is also
transferred to the desired point. All three methods of heat
transfer can lead to heat losses. For example, a contact heater
placed on the bottom of a container heats the contents
of the container. The primary form of heat transfer is via
conduction from the heater to the container and then its
contents, but conduction losses can occur between the
container and any supporting structure. As the container
and heater are both in air, a convection heat loss occurs as air
circulates around the container. In addition, a hand held one
foot away from the side of the container can feel the radiated
heat loss.
Needless to say, the size of these objects also plays a
considerable role. For a container the amount of radiated
2
March 2013
Type of heater …1
Hotwatt can also supply compatible
accessory items to offer our customers
a complete heater system.
heat energy lost isQdetermined
∝ I 2 × R ×by
t the outer surface
area. Assume the container is a cylinder 6-in.
diameter and 5-in. tall. The surface area of the
container (AS) equals the circumference of the
container (calculatedP as
2
= Ieither
× R π times the diameter
D, or 2π times the radius R) multiplied by its height
(h):
Transfer of heat …2
AS = πD × h or 6π × 5 = 941/4 in2
Just enlarging the diameter of the container by one
inch means the surface area of the container increases
by 15.7 in2, an almost 17% increase in surface
area and thus heat loss.mWhen
determining the size
× c × �F
W = needed to reach a specific
of the heating element
3.412 × T
temperature, all of these heat losses must be taken
into effect.
Q ∝ I2 × R × t
Basic heat equations
W = WL/SF × A
It would appear at first that calculating all of
the heat transfers and losses in a design would be
a daunting task. Fortunately
Q ∝ I 2 a×number
R × t of equations were
developed that help simplify this task. First the equations
m ×H
were divided intoWthree
= tasks referred to as Equations A,
3.412 ×
B, or C: the wattage needed
to Theat a material to a specific
temperature in a given amount
of time (Equation A);
P = I 2 ×the
R losses at operating
the wattage needed to overcome
2
temperature (Equation
special
calculation needed
Q B);
∝ I and
× Ra ×
t
WL/SF
× A(Equation C.)
×
c
×
�F
m
to reach a melting or vaporizing
point
W=
+
Equation A 3.412
calculates
× Tthe amount
2 of wattage2 (W) needed
= πD × h ofora 6π
× 5 =a specific
941/4 inamount in °F
to raise the Atemperature
material
S
(ΔF) in a given number Pof=hours
I 2 × R(T), you first need to know
the mass (m) of the material being heated and its specific
heat
m × c × �F2 WL/SF × A
c ×value
�F1 (c): m × Hf
m×
+
+m × c × �F
+
W=
3.412 × T
2
3.412 × T
3.412 × W
T =
AS = πD × h or3.412
6π ××5T = 941/4 in2
cylindrical objects as well as the top surface losses from a
horizontal surface laid flat. Curve B shows the average heat
loss from the
and×Qbottom
ofR ×
a×horizontal
laid
AStop
= πD
h∝or
6π
5t = 941/4 surface
in2
I2 ×
flat while Curve
C shows the heat loss from the bottom
only of a horizontal surface laid flat.
The final equation, Equation C, determines the
additional wattage necessary for
melting or vaporizing a
2
P =mI ×
×c ×
R �F
material. Whenever a W
material
changes
state, whether from
=
3.412
a solid to liquid, or liquid to
gas, ×
it T
requires an additional
influx of energy to initiate the change.
The heat needed to melt a solid material is known
as
2
AS =of
πD
× h (Hf
or 6π
× 5 the
= 941/4
the latent heat
fusion
) while
latentinheat of
W = WL/SF
A
vaporization (Hv) determines
the×energy
needed to change
a substance from a liquid to a vapor. The equation for
both is identical, with the value of Hf or Hv substituted as
necessary for the value H: m × c × �F
W=
m ××HT
3.412
W=
3.412 × T
)
(
Sponsored by Hotwatt
3
)
March 2013
Watt density …5
The mass and specific heat of some materials may be
found on our Web site at: www.hotwatt.com/table1.htm for
metallic solids, table2.htm
solids
W = for
WL/SF
× A other than metals, and
320 in.3 ×table3.htm
2 × 0.284for
= certain
181.7 lb
×m0.12
BTU/lb/°F
× (400-70)°F
Again, m is the weight of the material in lb, while T is
liquids
gases.
×and
c×
�F
= 2,110 W
W
=
the
heat-up time in hours.
The values
Equation3.412
B calculates
the
amount
of
wattage
(W)
needed
3.412
×
T
W= W
× A for the latent heat of
× 1hr
fusion or vaporization can beL/SF
found in the same look-up
to maintain the temperature of a material based upon its
WL/SF × A
c × �F
m × A.
tables used in Equation
known wattage loss per square foot (WL/SF) multiplied by
W=
+
3 area (A) in square feet:
m ×H
3.412 × T
2
= 270 in.the
× 2 × 0.286 = 154.5
W =lb × 0.12 BTU/lb/°F × (400-70)°F = 1,800 W
3.412
×
T
Specific applications
W = WL/SF × A
3.412 × 1hr
m ×H
To heat a liquid requires
W = additional energy due to losses
3.412
× T by adding half of
The wattage loss per square foot is found for specific
by evaporation. This is usually covered
materials in tables located at www.hotwatt.com/heatloss.htm.
themwattage
to
maintain
the m
material’s
WL/SF × A
× c × �Ftemperature
m×
Hf
× c × �Fneeded
2
1 +
5.6 lb × 0.55
BTU/lb/°F
× Metal
(400-70)°FW (Equation
+ heating period
+(Equation
×A
WL/SF
A sample table is shown
Uninsulated
B)
to
that
of
its
initial
==300
W
× �F from
m × c(Losses
3.412 × T
2
3.412 × T
3.412 × T
m×
= some materials
+H have
Surfaces). NoteWthat
different
A):
3.412
× 1hr
W =×T
3.412
2 three
3.412
× T A represents the
WL/SF × A
curves identified as A, B, or
C. Curve
m × c × �F
W
=
+
heat losses from vertical surfaces of tanks, pipes, and other
3.412 × T
2
An application example …4
(
P = I2 × R
WL/SF × A= 320 in.3 × 2 × 0.284 = 181.7 lb × 0.12 BTU/lb/°F × (400-70)°F
c ×A�F2 (10×8×4)
m × c × �F1 10 × 4m××4H+
f 8 × 4W
×m4××
W = WL/SF × A
Hotwatt is a quality supplier to oem’s
in the industrial, medical, commercial,
Q ∝ I × R × t m ×H
packaging, instrumentation,
aviation,
W=
3.412 ×fields.
T
transportation and military
Type of heater …1
2
P = I2 × R
WL/SF × A
m × c × �F
+
To heat a solid to a liquid state temperature requires
3.412three
× T different2calculations. First, use the solid calculation
W=
(Equation A) from the initial temperature of the material to its melting point. Then add in the wattage needed to
AS = πDC).
× hFinally,
or 6πuse
× 5the=equation
941/4 in2 for a liquid material from its melting point to the
transition from solid to liquid (Equation
final temperature:
W=
m × Hf
m × c × �F1
+
+
3.412 × T
3.412 × T
W
× c × �F
+
( m3.412
×T
L/SF
2
×A
2
)
Transfer of heat …2
m × c × �F
For example, you want to raise the temperature
W = of solid lead from a room temperature of 70°F to a molten 950°F.
× Tlead is 620°F. That makes the first change in temperature,
From Table 1 on our Web site, the melting point3.412
of solid
∆F1, equal to 620° - 70° or 550°F. The second change in temperature, ∆F 2, equals 950° - 620° or 330°F.
(10×8×4) = 320 in.3 × 2 × 0.284 = 181.7 lb × 0.12 BTU/lb/°F × (400-70)°F
An application example
Basic heat equations …3
W=
Watt density …5
= 2,110 W
W =3.412
WL/SF ×
×A
A steel mold is being used to form polyethylene
parts.1hr
Every hour 90 ounces of nylon is introduced to the mold. The
mold itself measures 10 × 8 × 4 in. The mold is attached between two stainless steel platens, each measuring 15 × 12 ×
1.5 in. thick. The
platens are=insulated
press=mechanism
with BTU/lb/°F
0.5 in. thick
(15×12×1.5)
270 in.3 from
× 2 ×the
0.286
154.5 lb × 0.12
× insulation.
(400-70)°FOperating temperature
1,800 W
of the mold is 400°F and must reach this temperature in 1 hour from an ambient temperature of=70°F.
3.412
×
1hr
From Table 1: Properties of Metals
m ×H
W=
Specific heat of steel: 0.12 BTU/lb/°F
3.412 × T
Specific heat of stainless steel: 0.12 BTU/lb/°F
5.6 lb × 0.55 BTU/lb/°F × (400-70)°F
Converting cubic inches into pounds (density lb/in2)
= 300 W
3.412 × 1hr
WL/SF × A
m × c × �F
+
3.412 × T
2
From Table 2: Properties of Non-Metallic Solids
10 × 4 × 4 + 8 × 4 × 4
= 2 ft2 × 350 W/ft2/hr
Specific heat of polyethelyne: 0.55 BTU/lb/°F
144 in.2/ft2
W
m × c × �F
+
(
3.412 × T
1.5 × 15 × 4 + 1.5 × 12 × 4
× c Heat
× �F1 Lossesm × Hf
m for
From Graph 1: Guide
+
+
W=
Heat losses curves: A + 3.412
B @ 400°F
×T
3.412 × T
2
×A
2
)
= 1.1 ft2 × 350 W/ft2/hr = 385 W
144 in.2/ft2
Equation A: Wattage required for heat-up
L/SF
= 700 W
To heat the mold:
15 ×
× 2 - BTU/lb/°F
(10 × 8 × 2)× (400-70)°F
(10×8×4) = 320 in.3 × 2 × 0.284 = 181.7
lb12
× 0.12
= 1.3 ft2 × 250 W/ft2/hr = 350 W
= 2,110 W
144 in.2/ft2
3.412 × 1hr
To heat the platens:
× 12 × 2 × (400-70)°F
2
(15×12×1.5) = 270 in.3 × 2 × 0.286 = 154.5 lb × 0.1215
BTU/lb/°F
= 2.5 ft2 × 100 W/ft
/hr W
= 250 W
= 1,800
144 in.
3.412 × 1hr
To heat the polyethelyne (90 oz / 16 oz/lb = 5.6 lb): (700W + 385W + 350W + 250W) × 0.20 = 340 W
5.6 lb × 0.55 BTU/lb/°F × (400-70)°F
= 300 W
3.412 × 1hr
Compensation Factor: 20%
(2,101 + 1,800 + 300) × 0.20 = 840 W
AS = 𝜋𝜋D × L
Total wattage needed for heat-up =
5,050
W
2
3.1416 × 0.75 × 9 = 21.1 in
10 × 4 × 4 + 8 × 4 × 4
144 in. /ft
2
Sponsored by Hotwatt
2
= 2 ft2 × 350 W/ft2/hr
500W ÷ 21.1 in2 = 23.7W /in.2
1.5 × 15 × 4 + 1.5 × 12 4× 4
2
2
= 700 W
= 1.1 ft2 × 350 W/ft2/hr = 385 W
March 2013
= 1,
Type of heater …1
(10×8×4) = 320 in. × 2 ×W0.284
= WL/SF=×181.7
A lb × 0.12 BTU/lb/°F × (400-70)°F
= 2,110
3.412
× 1hr W
3.412 × 1hr
ESTABLISHED
HOTWATT
IS A LEADER
5.6 lb × 0.55 BTU/lb/°F × (400-70)°F
(15×12×1.5)
= 270 in.3 ×IN
2 ×1952,
0.286 = 154.5
lb × 0.12 BTU/lb/°F
× (400-70)°F
=
= 1,800 W
m × H RESISTANCE HEATING
IN MANUFACTURING
3.412
×
1hr
W=
3.412
× 1hr
3.412
× TOVER
ELEMENTS FOR
60 YEARS.
5.6 lb × 0.55 BTU/lb/°F × (400-70)°F
= 300 W
WL/SF × A
3.412 × 1hr
10 × 4 × 4 + 8 × 4 × 4
m × c × �F
= 2 ft2 × 350 W/ft2/hr
W=
+
2
2
3.412
× at
T operating2 temperature (See graphs: Guide for Heat Losses)
Equation B: Wattage
losses
144 in. /ft
=
Heat loss from mold (vertical surfaces)
1.5 × 15 × 4 + 1.5 × 12 × 4
10 × 4 × 4 + 8 × 4 × 4
= 1.1 ft2 × 350 W/ft2/hr =
2
2
=
2
ft
×
350
W/ft
/hr =
700
W
m × c × �F2 WL/SF × A
2
2
m × Hf
m × c × �F1
144 in. /ft
+
+
+2
W=
144 in.2/ft
3.412 × T
2
3.412 × T
3.412 × T
)
(
144 in. /ft
2
2
Transfer of heat …2
Heat loss from platen (vertical surfaces)
1.5 × 15 × 4 + 1.5 × 12 × 4
15 × 12 × 2 - (10 × 8 × 2) = 1.3 ft2 × 250 W/ft2/hr =
= 1.1 ft2 × 350 W/ft2/hr = 385 W
144 in.2/ft2
10in×the
4 ×mold
4 + would
8 × 4AS×
=4𝜋𝜋D ×theL2
The number of holes
dictate
= 2 ft × 350 W/ft2/hr = 700 W
500W ÷ 21.1 in2 = 23.7W /in.2
2 the
2 wattage by the
number of heaters required. Dividing
1443.1416
in. /ft × 0.75 × 9 = 21.1 in2
number of heaters will equal the wattage rating of each
heater.
Our Web site at www.hotwatt.com/watt.htm lists
1.5 × 15 × 4 + 1.5 × 12 × 4
2
suggested
wattW
densities for use with various materials.
= 1.1 ft2 × 350 W/ft
/hr = 385
Watt density
2
144 in.
/ft2 ÷ 21.1 in2 = 23.7W /in.2 Safe values vary with operating temperature, flow
500W
Watt density is the rated wattage of the heater divided
velocity, and heat transfer rates. In general, the higher the
by the overall area being heated. The watt density permitted
material temperature, the lower the watt density should
for any given application
depends
on
how
well
the
material
be, especially those materials which coke or carbonize,
15 × 12 × 2 - (10 × 8 × 2) = 1.3 ft2 × 250 W/ft
2
350Watt
W densities should be low if a material is
being heated distributes its heat throughout
its volume.
such/hras =
oils.
2
2
/ft
144
in.
Water, light oils, and metals typically have high heat
being heated to a temperature near the change of state
checked. However, no responsibility for possible errors or
omissions can be assumed.
It is the express responsibility of the customer to
determine the suitability of the product for the intended
application and Hotwatt Inc. makes no claims or provides
no guarantee in this respect, either written or applied.
AS = 𝜋𝜋D × L
3.1416 × 0.75 × 9 = 21.1 in2
Then divide the wattage of the heater by the surface area:
Sponsored by Hotwatt
500W ÷ 21.1 in2 = 23.7W /in.2
5
March 2013
An application example …4
distribution rates that permit the use of high watt densities.
to a vapor occurs, such as water to steam at 212°F. The
Heavy oils, syrups, hydraulic fluid, and other materials with
vapor state has much poorer heat transfer capabilities
15 ×densities
12 × 2 to prevent
low heat distribution need lower watt
than
2
2 liquids, and a watt density chosen for a liquid might
= 2.5 ft × 100 W/ft
/hr = 250 W
spot overheating. This can lead to damage
of the heating
be too great for a gas.
144 in.
element, the well, and even the heated material.
As an example, a 10 in. immersion heater is rated at 500
W. There is a 0.5 in. cold area at
each end,
making
its total
(700W
+ 385W
+ 350W
+ 250W) × 0.20 = 340 W
heating length (L) 9 in. The diameter (D) of the heating area
is 0.75 in. To calculate the watt density of this heater, first
All information, data and dimension tables in this
article have been carefully compiled and thoroughly
determine the total surface area (AS) being heated:
Basic heat equations …3
8×4) = 320 in.3 × 2 × 0.284 = 181.7 lb × 0.12 BTU/lb/°F × (400-70)°F
= 2,110 W
Heat loss from platen (horizontal surfaces, uninsulated)
3.412 × 1hr
15 × 12 × 2
15 × 12 × 2 - (10 × 8 × 2) = 1.3 ft2 × 250 W/ft2/hr = 350 W
= 2.5 ft2 × 100 W/ft2/hr =
2
2
144
in.
144 in. /ft
12×1.5) = 270 in.3 × 2 × 0.286 = 154.5 lb × 0.12 BTU/lb/°F × (400-70)°F
= 1,800 W
Heat loss from platen
surface)
3.412 (insulated
× 1hr
(700W + 385W + 350W + 250W) × 0.20 =
15 × 12 × 2
= 2.5 ft2 × 100 W/ft2/hr = 250 W
144 in.
5.6 lb × 0.55 BTU/lb/°F × (400-70)°F
= 300 W
Compensation factor: 20%
3.412
×
1hr
= 𝜋𝜋D
(700 + 385 + 350 + 250) × 0.20 W ×L
(700W + 385W + 350W + 250W) × 0.20 ==A340
340
W
S
Total wattage losses at operating temperature = 2,025 W
3.1416 × 0.75 × 9 = 21.1 in2
Total wattage required for heat-up = 5,050 W
Total wattage required
= 7,075 W