Two heater of 180 watt and 1400 watt which heater is having less resistance

By following a few rules-of-thumb you can determine the wattage requirement for your application.

Calculating the wattage requirements to heat a system is a straightforward process as long as all the parameters of heat energy flowing in and out of a system are considered. Heat requirements that must be considered are:

• Initial heat for the startup of the system, usually from ambient temperature to the desired processing temperature.
• Heat losses to the environment due to conduction, convection, and radiation.
• Heating of material being processed during operation.
• Heating of material flowing through the process such as a liquid that will be heated and pumped to be used elsewhere.
• The losses due to phase changes of materials, either during initial heat-up or while processing (melting a solid to a liquid or boiling a liquid to a gas).

Luckily, such precision is usually not required since temperature controllers are commonly employed in most heating applications, meaning that a quick calculation can be used to get you up and running.

Before beginning the calculations, it is important to realize the distinction between energy and power and their relationship to wattage requirements.

• In metric units, power is measured in watts (W) and energy is measured in watt-hours (W x hr).
• In imperial units, power is measured in British Thermal Units per hour (BTU / hr) and energy is measured in BTUs. A BTU is the amount of energy required to heat 1 pound of water by 1°F (specifically from 39° to 40°F).

Think of power as the rate at which energy is used. A 60 watt light bulb uses 60 watt-hours of energy in one hour.

It also should be noted that the difference between the starting temperature and the desired final application temperature is commonly referred to as delta T (ΔT). If a process is started at room temperature, say 72°F, and the application process temperature is 500°F, then ΔT is 500°F - 72°F, or 428°F.

Below are some good guidelines for heating different materials in different situations.

To calculate the wattage requirement to heat steel, use the following equation:

Watts = 0.05 x Lbs of Steel x ΔT (in °F) / Heat-Up Time (in hrs)

Example: To heat 50 lbs of steel by 250°F in 1 hour; .05 x 50 x 250 / 1 = 625 Watts. Using the same example, reaching temperature in 15 minutes (0.25 hrs); .05 x 50 x 250 / .25 = 2,500 Watts. This equation is suitable for mild and stainless steel. If you are heating a different material than steel, you can replace the 0.05 in the equation above with the following coefficients:

• Brass: 0.053
• Aluminum: 0.018
• Copper: 0.056

To calculate the wattage requirement to heat water in a tank, use the following equation:

Watts = 3.1 x Gallons x ΔT (in °F) / Heat-Up Time (in hrs)

Example: To heat 20 gallons of water by 100°F in 30 mins (0.5 hrs); 3.1 x 20 x 100 / 0.5 = 12,400 Watts

To calculate the wattage requirement to heat flowing water, use the following equation:

Watts = 165 x Gallons Per Minute X ΔT (in °F)

Example: To heat water flowing at 2 gallons per minute by 45°F; 165 x 2 x 45 = 14,850 Watts

To calculate the wattage requirement to heat oil in a tank, use the following equation:

Watts = 1.35 x Gallons x ΔT (in °F) / Heat-Up Time (in hrs)

Example: To heat 5 gallons of oil by 300°F in 15 mins (0.25 hrs); 1.35 x 5 x 300 / 0.25 = 8,100 Watts

There are a variety of other equations that can be used to provide good estimates of the power requirement to heat substances in varying situations. The numbers calculated from these formulas do account for typical losses in most applications and will provide a number that is more than adequate to get your process to temperature in the time required.

For more complicated heating applications, the Tutco engineering team can provide support to help with your calculations and provide solutions for your heating needs.

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  All of our electric heater calculators in one place.

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  This interactive calculation tool helps to visualize the relationships among airflow, temperature, and power.

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• Watt density is a useful measure when considering the various types of heating elements available. Tutco-Farnam Custom Products manufactures open coil elements which are generally used to heat a gas flow, such as air or nitrogen.

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and assuming that they all genuinely use 2000W of power

This is the crucial assumption. If both have real power 2000 W, this means 2000 W of heat introduced into the wires and then (by assumption of stationarity of temperature of the heater) this heat leaves the wires and goes into the heater environment.

But equal heat introduced per unit time does not mean equal effect on average temperature or equal efficiency of operation. Bigger heater with greater exchange surface will operate on lower temperatures and will heat up the air more evenly quicker (because of energy losses from the room). This can be more efficient if we want to achieve same temperature everywhere as fast as possible, or less efficient, if we want to heat up only our frozen legs.

Also in practice, even if two heaters have 2000 W written on them, this does not mean that actual consumption is 2000 W all the time. For example, as the heat producing wires get hotter, their ohmic resistance increases, so the effective value of current should decrease and real power should decrease as well. The only way to be sure of the consumption of the heater is to measure it with a wattmeter.

The argument:

If your heating element is rated to dissipate 2300 watts with 230 volts across it, then when it's dissipating 2300 watts the current through it will be:

I = P/E = 2300W/230V = 10 amperes,

and its resistance will be:

R = E/I = 230V/10A = 23 ohms.

In order for the element to dissipate 1400 watts, then, the voltage across it must be decreased to:

E = sqrt (PR) = sqrt (1400W * 23R) ~ 180 volts,

and the current through it decreased to:

I = P/E = 1400W/180V ~ 7.8 amperes

Since the mains voltage is unchangeable at 230 volts, then in order to limit the current through the heating element to 7.8 amperes, a series attenuating element must be introduced into the circuit which will drop the mains' 230 volts down to the 180 volts required by the heater.

That's 50 volts, and it must pass the 7.8 ampere heater current, so its resistance will be:

R = E/I = 50V/7.8A = 6.4 ohms.

The total resistance of the series string will therefore be 29.4 ohms, the sum of the two resistances,.

If we use a capacitor to losslessly drop the voltage to the heater, then we can define the resistance of the string as an impedance,

Z² = R² + Xc², where:

Z is the impedance of the string, R is the heater resistance, and Xc is the reactance of the series capacitor, all in ohms.

Rearranging to solve for Xc we get:

Xc = sqrt (Z² - R²) = sqrt (864.4R - 529R) = 18.3 ohms

and, finally, to get the capacitance,

C = 1/(2pi f Xc) = 1/(6.28 * 50Hz * 18.3R )= 174 microfarads .

Cornell-Dubilier makes a nice line of caps which seem perfect for this application.

Pricey, no doubt, but probably less so - and certainly lighter and smaller - than an inductor or a transformer capable of doing the same job.

The LTspice proof: