# Calculating Average Power and Energy Consumption for a Heater

• PhysicsTest
In summary: I agree with your final values but your working has mistakes. The ones I spotted are:Sorry I was little careless, the equation should be ##1500 = \frac{\int_0^{\frac 1 {60}} v^2(t) dt} {\frac{1} {60R}} ## where i replaced ##i(t) = \frac{v(t)} R ## R is the resistance. I hope it is correct now.It is poor practice to give a value without the relevant unit. should have after the '9.6'.Yes, the resistance is ##9.6\Omega##Energy absorbed = ##\int_0^Tp(t) dt## is correct only if p(t) and
PhysicsTest
Homework Statement
An electric resistance space heater rated at 1500 W for a voltage source of v(t)= 120 * (2)^(0.5) sin (2*pi*60t) 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
Relevant Equations
p(t) = v(t)*i(t)
The v(t) = 120##\sqrt(2)\sin(2\pi 60t)##. Let the resistance of the heater be "R", then
the rated power = 1500W. The rated power is the average power
1500 = ##\frac{120^2 *5 * 60} {12*60 * R} ## ->eq1
R = 4 Ohm.
a. Instantaneous power is
##P = \frac{v^2} R ## -> eq2
##P = \frac{{(120\sqrt2})^2} {R} ##
##P = 7200 W ##
b. The average power should be the rated power which is 1500W.
c. The energy is E = P*t Joules
##E = \frac{5*60 *(120\sqrt2)^2} {4} J = 2.16*10^6J##
Are the solutions correct? My main doubt is the rated power the same as average power?

PhysicsTest said:
= ##\frac{120^2 *5 * 60} {12*60 * R} ##

PhysicsTest
haruspex said:
Average Power = ##\frac{\int_0^Tv(t)*i(t)dt} T## --> eq3
T = 12mins = 12*60sec= 720sec
T = 5 mins = 5*60 = 300sec, T=7mins=7*60=420sec
##\frac{\int_0^{720}120\sqrt2 sin(2\pi60t) * 120\sqrt2 sin(2\pi60t)dt} {720R} ##
##\frac{\int_0^{300}120\sqrt2 sin(2\pi60t)*120\sqrt2sin(2\pi60t)dt} {720R} + \frac{\int_{300}^{720} 0dt} {720R} ## ->eq4
The eq4 simplifies to since it is resistor
##P_{Avg} = \frac{(120\sqrt2)^2 * 300} {720R} ##
##1500 = \frac{28800*300} {720R} ##
##R = \frac{28800 * 300} {720*1500} ## ->eq5
##R = 8 ## --> eq6
a. ##P_{Inst} = 3600W##

PhysicsTest said:
Average Power = ##\frac{\int_0^Tv(t)*i(t)dt} T## --> eq3
T = 12mins = 12*60sec= 720sec
T = 5 mins = 5*60 = 300sec, T=7mins=7*60=420sec
##\frac{\int_0^{720}120\sqrt2 sin(2\pi60t) * 120\sqrt2 sin(2\pi60t)dt} {720R} ##
##\frac{\int_0^{300}120\sqrt2 sin(2\pi60t)*120\sqrt2sin(2\pi60t)dt} {720R} + \frac{\int_{300}^{720} 0dt} {720R} ## ->eq4
The eq4 simplifies to since it is resistor
##P_{Avg} = \frac{(120\sqrt2)^2 * 300} {720R} ##
##1500 = \frac{28800*300} {720R} ##
##R = \frac{28800 * 300} {720*1500} ## ->eq5
##R = 8 ## --> eq6
a. ##P_{Inst} = 3600W##
I see you are taking the 1500W as the average over the 12 minute cycle. That's not how I read the question. If that were correct, the answer to b) would be trivial.
I believe the 1500W is the average power while on, so you only need to consider a single 1/60 second cycle.

haruspex said:
I believe the 1500W is the average power while on, so you only need to consider a single 1/60 second cycle
I am not sure if i have understood this correctly, I assume you have asked me to calculate for 1 min during the ON cycle of 5 mins(=300sec)?
##1500 = \frac{\int_0^{60}(120\sqrt2)^2\sin^2(2\pi60t)dt} {300R} ##
##1500 = \frac{(120\sqrt2)^2*60} {300R} ## ->eq1
Can you please confirm, if it is correct?

PhysicsTest said:
I am not sure if i have understood this correctly, I assume you have asked me to calculate for 1 min during the ON cycle of 5 mins(=300sec)?
##1500 = \frac{\int_0^{60}(120\sqrt2)^2\sin^2(2\pi60t)dt} {300R} ##
##1500 = \frac{(120\sqrt2)^2*60} {300R} ## ->eq1
Can you please confirm, if it is correct?
The power source is 60Hz. To find the average, integrating over a single cycle (1/60 of a second) will do. 1 minute is fine too, provided you divide the integral by the same period. But you seem to have divided by 5 minutes,

PhysicsTest
@PhysicsTest. Are you allowed to use the concept of root-mean-square (RMS) values? If so, do you know ##P_{rms} = \frac {P_{peak}}{2}## for a resistance with a sinusoidal supply? (The 'rated' value of 1500W given in the question is ##P_{rms}##.)

Parts b) and c) are short, introductory-level, physics problems requiring only elementary arithmetic.

So, unless you are required to do so, no calculus is required.

PhysicsTest
Ok, thank you for the support, i understand better now.
##1500 = \frac{\int_0^{\frac 1 {60}} v(t)i(t) dt} {\frac{1} {60R}}##
##1500 = \frac{(120\sqrt2)^2} {2R} ##
##R = \frac{28800} {3000} = 9.6 ##
a. Max. Instantaneous power = ##1500*2W = \frac{(120\sqrt2)^2} {9.6} = 3000W##
b. Avg Power = ##\frac{1500*5} {12} = 625 W##
c. Energy absorbed = ##\int_0^Tp(t) dt##
##E = 625*12*60 J = 4.5*10^5 J ##
I hope i have done correct.
Steve4Physics said:
@PhysicsTest. Are you allowed to use the concept of root-mean-square (RMS) values? If so, do you know ##P_{rms} = \frac {P_{peak}}{2}## for a resistance with a sinusoidal supply? (The 'rated' value of 1500W given in the question is ##P_{rms}##.)
I know the RMS concept but the chapter till that section does not cover the RMS value.

PhysicsTest said:
Ok, thank you for the support, i understand better now.
##1500 = \frac{\int_0^{\frac 1 {60}} v(t)i(t) dt} {\frac{1} {60R}}##
##1500 = \frac{(120\sqrt2)^2} {2R} ##
##R = \frac{28800} {3000} = 9.6 ##
a. Max. Instantaneous power = ##1500*2W = \frac{(120\sqrt2)^2} {9.6} = 3000W##
b. Avg Power = ##\frac{1500*5} {12} = 625 W##
c. Energy absorbed = ##\int_0^Tp(t) dt##
##E = 625*12*60 J = 4.5*10^5 J ##
I hope i have done correct.

I know the RMS concept but the chapter till that section does not cover the RMS value.
I agree with your final values but your working has mistakes. The ones I spotted are:

##1500 = \frac{\int_0^{\frac 1 {60}} v(t)i(t) dt} {\frac{1} {60R}}##
should be
##1500 = \frac{\int_0^{\frac 1 {60}} v(t)i(t) dt} {\frac{1} {60}}##
'
##1500 = \frac{(120\sqrt2)^2} {2R}## is correct but the working showing how you got it (from the incorrect starting-equation!) is missing.

It is poor practice to give a value without the relevant unit. ##R = \frac{28800} {3000} = 9.6## should have ##\Omega## after the '9.6'.

Energy absorbed = ##\int_0^Tp(t) dt## is correct only if p(t) and T are clearly defined/understood– which they are not. And then you didn't use the equation! Also ‘absorbed’ is the wrong word. I would have written:
Energy converted to heat in 12 minutes = ##P_{average} t = 625 * (12*60)##

Steve4Physics said:
I agree with your final values but your working has mistakes. The ones I spotted are:
##1500 = \frac{\int_0^{\frac 1 {60}} v(t)i(t) dt} {\frac{1} {60R}}##
should be
##1500 = \frac{\int_0^{\frac 1 {60}} v(t)i(t) dt} {\frac{1} {60}}##
##1500 = \frac{(120\sqrt2)^2} {2R}## is correct but the working showing how you got it (from the incorrect starting-equation!) is missing.
Sorry I was little careless, the equation should be
##1500 = \frac{\int_0^{\frac 1 {60}} v^2(t) dt} {\frac{1} {60R}} ## where i replaced
##i(t) = \frac{v(t)} R ## R is the resistance. I hope it is correct now.
Steve4Physics said:
It is poor practice to give a value without the relevant unit. should have after the '9.6'.
Yes, the resistance is ##9.6\Omega##
Steve4Physics said:
Energy absorbed = ##\int_0^Tp(t) dt## is correct only if p(t) and T are clearly defined/understood– which they are not. And then you didn't use the equation! Also ‘absorbed’ is the wrong word. I would have written:
Energy converted to heat in 12 minutes = ##P_{average} t = 625 * (12*60)##
The idea was to multiply average power calculated for the entire 12min with time of 12min.
##\int_0^Tp(t) dt##
##\int_0^{300} p1(t) dt + \int_{300}^{720}p2(t)dt ##
##p1(t) = v^2(t)##
##p2(t) = 0##
##E = \frac{\int_0^{300}v^2(t) dt} {R} ## I simplified the sine terms and brought the factor of 1/2,
##E=\frac{\int_0^{300} (120\sqrt2)^2} {2R}dt##
##E = 1500* 300 = 4.5*10^5 J##

PhysicsTest said:
Sorry I was little careless, the equation should be
##1500 = \frac{\int_0^{\frac 1 {60}} v^2(t) dt} {\frac{1} {60R}} ## where i replaced
##i(t) = \frac{v(t)} R ## R is the resistance. I hope it is correct now.
It's still wrong. It should be:
##1500 = \frac{\int_0^{\frac 1 {60}} v^2(t) dt} {R(\frac{1} {60})} ##

PhysicsTest said:
The idea was to multiply average power calculated for the entire 12min with time of 12min.
Then why bother writing out the integrals? Just do what you said and multiply the average power (625W) by 720s (12min):
E = 625 * 12*60 = 4.5*10^5 J

(Of course,since the heater operates normally (1500W) for 5 minutes out of the 12 minutes, you get the same answer by multiplying 1500W by 300s (5 minutes).)

Using the integrals is totally unnecessary and increases the risk of errors. The purpose of the question (parts b) and c)) is to get you to use average values so that you don't need any integrations. There should be no working which incudes integrals in your answers to parts b) and c).

PhysicsTest

## What is average power over the cycle?

Average power over the cycle is a measure of the rate at which work is done or energy is transferred over a complete cycle of a periodic process. It is calculated by dividing the total work or energy by the time it takes to complete one cycle.

## Why is average power over the cycle important?

Average power over the cycle is important because it gives us an understanding of the overall efficiency and performance of a system. It allows us to compare different systems and determine which one is more efficient in terms of energy usage.

## How is average power over the cycle different from instantaneous power?

Instantaneous power is the power at a specific moment in time, while average power over the cycle takes into account the entire cycle. Instantaneous power can fluctuate, but average power over the cycle gives a more accurate representation of the overall power output.

## What factors can affect average power over the cycle?

The main factors that can affect average power over the cycle are the duration of the cycle, the amount of work or energy transferred, and any inefficiencies in the system. Other factors such as temperature, friction, and resistance can also impact the average power over the cycle.

## How can average power over the cycle be increased?

To increase average power over the cycle, one can either increase the amount of work or energy transferred per cycle or decrease the duration of the cycle. Improving the efficiency of the system can also lead to an increase in average power over the cycle.

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