Calculating Average Power and Energy Consumption for a Heater

[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?

 

[haruspex]

= $\frac{120^2 *5 * 60} {12*60 * R}$

How do you get that? Please post your steps.

 

[PhysicsTest]

How do you get that? Please post your steps.

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. $P_{Inst} = 3600W$

 

[haruspex]

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.

 

[PhysicsTest]

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?

 

[haruspex]

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,

 

[Steve4Physics]

@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.

@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.

 

[Steve4Physics]

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 = $P_{average} t = 625 * (12*60)$

 

[PhysicsTest]

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.

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 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,
$E=\frac{\int_0^{300} (120\sqrt2)^2} {2R}dt$
$E = 1500* 300 = 4.5*10^5 J$

 

[Steve4Physics]

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:
$1500 = \frac{\int_0^{\frac 1 {60}} v^2(t) dt} {R(\frac{1} {60})}$

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).