1500w electric space heater problem

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SUMMARY

The discussion focuses on calculating the power and energy output of a 1500W electric resistance space heater operating on a 120V AC source. The maximum instantaneous power is established as 3000W, while the average power over a 12-minute cycle is calculated to be 625W, based on the duty cycle of 5 minutes on and 7 minutes off. Additionally, the total energy converted to heat in each cycle can be expressed in both Joules and kilowatt-hours (kWh), emphasizing the importance of using RMS values for accurate calculations.

PREREQUISITES
  • Understanding of electric power calculations, including RMS and peak values.
  • Familiarity with duty cycle concepts in electrical engineering.
  • Knowledge of basic electrical formulas, including energy conversion (E=Pt).
  • Ability to perform integration for average power calculations over time.
NEXT STEPS
  • Study the principles of RMS voltage and current in AC circuits.
  • Learn about duty cycle calculations and their impact on average power.
  • Explore energy conversion formulas in electrical systems, particularly Joules and kWh.
  • Investigate the effects of varying load conditions on electric heaters.
USEFUL FOR

Electrical engineering students, HVAC technicians, and anyone involved in the design or analysis of electric heating systems will benefit from this discussion.

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Homework Statement


*An electric resistance space heater rated at 1500W for a voltage source of v(t)=120√(2) sin(2pi60t)V has a thermostatically controlled switch. The
heater periodically switches on for 5 min and off for 7 min. Determine
(a) the maximum instantaneous power, (b) the average power over the
12-min cycle, and (c) the electric energy converted to heat in each 12-min
cycle.

Homework Equations


none.

The Attempt at a Solution


1. the maximum instantaneous power is twice of the rating right? 3000Watts?

2. and average power is the integral of instantaneous over time, divided by the time interval? but can't quite figure out what values to use.
3. for energy we use e=vit? am i supposed to multiply just the amplitudes of both voltage and current with time in seconds?
 
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The power rating for an electrical appliance will be the RMS value so the peak input power (V2/R) will be twice as much.

For part 2) I would use the RMS value multiplied by the mark/space ratio eg

1500W * 5/12 = 635W

For part 3) you have a choice. You can either specify the number of Joules converted (remembering that a Watt is a Joule per second) or you can convert "635W for 12 min" to kWH. Both answer the question but I'd do both to be safe.
 
thanks cwatters.
much appreciated..
 

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