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873 Series Quick Ship Contactor Control Panel

873 Series Quick Ship Contactor Control Panel

The 873 Series Quick Ship Contactor Temperature Control panel provides fast shipment, compact packaging, and is pre-engineered for resistive load applications. The panel is completely assembled, pre-wired, tested, and ready for trouble free installation and operation. Designed for use in industrial environments and outdoor installations. 4

STANDARD CONSTRUCTION FEATURES:

  • NEMA 4X Fiberglass Enclosure
  • 16”H x 14”W x 8”D Enclosure for 1 Circuit (48A max)
  • 20”H x 16”W x 8”D Enclosure for 2 Circuits (48-96A)
  • Single or Three Phase Loads
  • 50 Amp Contactor for Resistive Loads (per circuit)
  • 120 VAC Control Transformer
  • UL Listed4
  • Wiring Diagram Permanently Affixed to the Inside of the Cover
  • Terminals for Customer Supplied Remote Interlock

OPTIONAL FEATURES:

□ Digital Indicating Thermostat (°F only) 32-999°F range
□ Digital 1/16th DIN Process Controller with RS485 Communications available (°F or °C) 0-1600°F range
□ Disconnect Switch
□ Power Fusing
□ Enclosure Heater
□ Pilot Light for Indication of Power “ON”
□ Digital Indicating Limit Control with manual reset (°F only) 32-999°F range
□1/16th DIN Digital Indicating Limit Control with manual reset (°F or °C) 0-1600°F range

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How to Prevent Electric Tank Heat Explosions

Being aware of potential dangers can help prevent electric tank heat explosions at industrial facilities.

Electric tank heaters present several possible dangers in process heating applications. Recent incidents include one at a manufacturing plant and another at an oil-and-gas processing facility.

The manufacturing facility incident occurred on the hydraulic oil storage tank similar to thousands in use all over the world. Because of the commonality of these systems, and the prolific use of electric tank immersion heaters, it is important to be aware of the potential danger associated with this setup.

The other explosion occurred at an oil-and-gas processing facility outside of Pittsburgh. There, a fire broke out at a complex of three storage tanks, each about 25′ tall and 8′ wide.

At the facility, the natural gas was pressurized within the tanks and moved along a pipeline to a processing plant, leaving behind a “briny fluid” made up of liquid hydrocarbon waste called condensate. The tank’s electric heaters are designed to keep the liquid mixture from freezing.

The local fire chief explained that the tank designed to hold the liquid apparently was empty or nearly empty but contained gas vapors. When the liquid mixture is low, the heating element is designed to shut off. But in this case, the electric heater continued to operate, eventually warming the vapors to combustible levels, setting off the explosion.

Seeing such disasters, it becomes apparent that any electric tank heaters should be reviewed for design and installation issues that can lead to explosions so facilities can take action to avoid an incident.

How would someone know that they might have a problem? In this article, several key electric tank heater design and installation issues – that, if left unchecked, could lead to fires or explosions – will be reviewed.

Know the Risks of Electric Bayonet Immersion Heaters.

The electric heaters to be concerned about typically extend into a tank through the sidewall. They might be 12 to 36″ long, or longer. Typically, they are in the form of a bundle of elements such as a coil, wire or other shape that resists the electric current, causing the bundle to give off heat. It is the same concept used in an electric stove or toaster.

In many cases, these elements work acceptably if they are covered with a fluid. If they are uncovered and the watt density is high enough, however, they can become an ignition source.

Consider Watt Density and Protection.

Watt density — defined as the number of watts per square inch of heating surface or element — is what the designer planned for the unit selected. It makes sense that if you squeeze a whole lot of watts through a small heater, the heater elements can get very hot. The free air temperature of some of these elements can be more than 1,000°F (537°C). Remember, the flashpoint of many hydraulic oils is only 400 to 600°F (204 to 315°C).

It is not just oil tanks that are an area of concern. Any tank that can accumulate flammable elements must be considered. For instance, a wastewater tank that drains machining fluids may primarily contain water. But just a small amount of oil mixed in can cause a problem. The oil can be thermally cracked or broken down under certain conditions, filling the vapor space of the tank with hydrocarbon fumes. Then, when the level changes in the tank and air is brought in, a flammable mixture can be created.

Consider also the case of a wastewater tank that is part of a food process or sewage system. Whenever biological activity consumes organic materials, methane can be released. If the elements become exposed and there are no protections in place, you can create an ignition source directly in the tank.

Understand the Fire Triangle.

The fire triangle concept is that when you bring fuel, oxygen and an ignition source together, you have all of the elements needed for a fire. When a flammable mixture is confined and ignited, you have everything you need for an explosion. The flammable materials try to expand to many times their volume in fractions of a second. When they cannot escape whatever vent is provided in a timely manner, the vessel or tank is pressurized and usually comes apart at some weak point. Often for cylindrical tanks, the weakest spot is at the top of the container. If an explosion does occur, the flying metal and resulting fireball can be deadly and destructive.

Strategies for Industrial Plant Safety

All users of electric immersion heaters must verify that they have minimized risks related to electric tank heaters wherever they are installed. Methods to minimize risks include the following.

Verify Watt Densities and Maximum Temperatures.

Many options are available when buying electric heaters. You can purchase heaters that have features to minimize the maximum temperature, or you can specify lower watt densities.

Use Operating and High Temperature Limit Controllers.

Operating temperature controllers are essentially like thermostats in your home. They seek to provide the right temperature on a routine basis. High temperature limit controllers are special, separately installed controls that have a manual reset feature once exceeded. In this case, the device shuts the unit off until a person manually hits a button or takes some action. This feature forces someone to investigate and hopefully understand that some unusual event occurred and that an unsafe condition was reached. If a high temperature limit is installed, the setting has to be well below whatever flashpoint could be reached.

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How to Calculate KW Requirements for Typical Heater Applications

Tank Heating Calculations

When selecting a heater for tank heating application you must first determine whether the application requires that the temperature be maintained or if the temperature needs to be raised.  Below are the calculations for each application.  You can also visit our website and utilize our online calculator; look for the free calculator link near the top of the page.

Maintain Temperature

To calculate the KW required to maintain the temperature of a tank you will need to determine the tanks surface area, process temperature to be maintained, minimum ambient temperature and the R-value of the insulation.

Surface area:

Round tank –

A (ft²) = (2 x p x r x h) + (2 x p x r²)

p = 3.14

r = radius (ft)

h = height (ft)

 

Rectangular tank –

A (ft²) = 2 x [(l x w) + (l x h) + h x w)]

l = length (ft)

w = width (ft)

h = height (ft)

 

After determining the tanks surface area the maintain KW can be calculated as follows:

KW = (A x (1/R) x ΔT(°F) x SF)/3412

A = surface area

R = R-value of the insulation

  • Use 0.5 as the R-value of an uninsulated steel tank
  • See the chart below for typical examples
  • R-value = thickness (in.)/k-factor

ΔT = difference between the process set point temperature and lowest ambient temperature

SF = safety factor, recommended 1.2

3412 = conversion of BTU to KW

 

Table 1

Insulation Type R-Value/inch of thickness
Fiberglass R-3
Mineral fiber R-3.7
Calcium silicate R-2
Open-cell polyurethane foam R-3.6
Closed-cell polyurethane foam R-6
Polyisocyanurate spray foam R-6

 

Example:

A 42’ diameter x 40’ tall crude oil tank with R-6 insulation needs to be maintained at 75°F given a minimum ambient temperature of 10°F.

A = (2 x 3.14 x 21 x 40) + (2 x 3.14 x 21²)

A = 8044.68 ft²

KW = (8044.68 x 1/6 x 65 x 1.2)/3412

KW = 30.65

 

Raise Temperature

The KW calculation to raise the temperature of a material in a tank (heat-up) starts with the same information required in the maintain application.  Additionally we’ll need the weight of the material to be heated, the specific heat of the material and the time required to heat the material from its start temperature to its end temperature.  The KW calculation to raise the temperature is as follows:

KWtotal = KWheat-up + KWmaintain

KWheat-up = [(M x Cp x ΔT x SF)/3412]/t

M = weight of the material in pounds

Cp = Specific Heat, see examples in the chart

ΔT = difference between the process set point (end) temperature and the start temperature

SF = safety factor, recommended 1.2

3412 = conversion of BTU to KW

t = time in hours

KWmaintain = (A x (1/R) x ΔT(°F) x SF)/3412

A = surface area

R = R-value of the insulation

  • Use 0.5 as the R-value of an uninsulated steel tank

ΔT = difference between the process set point temperature and lowest ambient temperature

SF = safety factor, recommended 1.2

3412 = conversion of BTU to KW

 

Example:

A 4’ x 6’ x 12’ tank with 1800 gallons of water needs to be heated from 60°F to 95°F in 3 hours.  The tank has R-4 insulation and the minimum ambient temperature is 0°F.

To begin we need to convert the gallons of water to pounds:

lbs = G x D1

G = gallons

D1 = lbs per gallon from the chart below

lbs = 1800 x 8.34

lbs = 15,012

If the volume of the tank is stated in cubic feet (ft³) the formula looks like this:

lbs = C x D2

C = Cubic feet of material

D2 = lbs per ft³ from the chart below

Table 2

Material D1

lbs/gallon

D2

lbs/ft³

Specific Heat
water 8.34 62.4 1
#1 fuel oil 6.8 50.5 0.47
#2 fuel oil 7.2 53.9 0.44
#3,4 fuel oil 7.5 55.7 0.425
#5,6 fuel oil 7.9 58.9 0.41
Bunker C 8.15 61 0.5
SAE 10-50 weight oil 7.4 55.4 0.43
ethylene glycol 9.4 70 0.55
50% ethylene glycol/water 8.8 65.8 0.76
air 0.073 0.24
nitrogen 0.073 0.25

 

 

KWheat-up = [(15,012 x 1 x 35 x 1.2)/3412]/3

KWheat-up = 61.6

plus

KWmaintain = (288 x 1/4 x 95 x 1.2)/3412

KWmaintain = 2.4

KWtotal = 64

 

Calculations for heating air in a duct

Once the volume of air in standard cubic feet per minute (SCFM) and the required temperature rise in °F (ΔT) are known, the required kilowatt rating (KW) of the heater can be determined from the following formula:

KW = (SCFM x ΔT)/3193

Note that CFM is given at standard conditions (SCFM): 80° F and normal atmospheric pressure of 15 psi. The CFM at a higher pressure (P) and inlet air temperature (T) may be calculated as follows:

SCFM = ACFM x (P/15) x [540/(T+460)]

 

Example:

A drying oven, operating at 25 psia (10 psi gauge pressure), recirculates 3000 CFM of air per minute through a heater which raises its temperature from 350 to 400° F.

To select an appropriate heater:

Step 1: Convert 3000 CFM at 25 psia and 350° F to CFM at standard conditions using the above formula:

3000 x (25/15) x [540/(350°F+460)] = 3333 SCFM

Step 2: Calculate the required KW:

[3333 SCFM x (400°F-350°F)]/3193 = 52 KW

 

Calculations for circulation heater applications

When calculating the power required to heat a material flowing through a circulation heater, the KW equation shown below can be applied.  This equation is based on the criteria that there is no vaporization occurring in the heater. The KW equation incorporates a 20% safety factor, allowing for heat losses of the jacket and piping, variation in voltage and wattage tolerance of the elements.

KW = (M x ΔT x x Cp x S.F.)/3412

 

Where:

KW = power in kilowatts

M = flow rate in lbs/hr

ΔT = temperature rise in °F (The difference between the minimum inlet temperature and maximum outlet temperature.)

Cp = specific heat in BTU/lb °F

S.F. = safety factor, 1.2

3412 = conversion of BTU to KWH

Water heating example:

We have 8GPM of water with an inlet temperature of 65°F and outlet temperature of 95°F.  First, convert the flow rate to lbs/hr.

8 gal x 1 ft³ x 60 min = 64.17 ft³/hr
min 7.48 Gal 1 hr

 

Convert to lbs/hr, obtain the density and specific heat from table 2 above.

64.17 ft³/hr x 62.4 lbs/ft³ = 4004 lbs/hr

Now calculate the KW:

KW = 4004 lbs/hr x (95-65)°F x 1 BTU/lb °F x 1.2
3412
KW = 42

 

Gas heating example:

Air is flowing at 187 CFM and 5 PSIG pressure.  It needs to be heated from an inlet temperature of 90°F to an outlet temperature of 250°F.  First, convert the flow rate to SCFM using the formula given earlier.

187 x (20/15) x [540/(90°F+460)] = 243.7 SCFM

Convert to lbs/hr, again referring to table 2 for the density and specific heat.

243.7 SFCM x 60 min x 0.073 lbs = 1067.4 lbs/hr
1 hr ft³

 

Now calculate the KW:

KW = 1067.4 lbs/hr x (250-90)°F x 0.24 BTU/lb °F x 1.2
3412
KW = 14.4