# Power consumption costs for a malfunctioning lightbulb

1. Mar 19, 2016

### reed2100

1. The problem statement, all variables and given/known data
Consider a refrigerator whose 40-W light bulb remains on continuously as a result of a malfunctioning of the switch inside the refrigerator. The refrigerator has a coefficient of performance (COP) of 1.3 and the cost of electricity is estimated to be 10 cents per kWh (kilowatt-hour). Assume that the refrigerator is normally opened 20 times a day for an average of 30 s.

1-Determine additional power consumed by the refrigerator.

2-Estimate the additional electric power costs per year due to such malfunctioning per year.

2. Relevant equations
Coefficient of Performance = QL / W = heat transfer to/from cold body to system divided by work cost
Power = W / unit time
3. The attempt at a solution

I'd just like verification as to whether or not I did this correctly, because I don't trust this problem.

The problem asks me to find the additional power consumed by the fridge, so I take this to mean the additional power cost of the malfunctioning lightbulb (not sure why the messed up lightbulb would affect the actual refrigeration much). So the EXTRA power consumed is just "actual power consumed per year - the normal power consumed per year".

First I convert the lightbulb power to KW, 40 W = .04 KW.

With a normally functioning bulb, the total time per day it'd be active is
20 opens/day * 30 sec/open = 600 sec/day = 10 minute/day = .1666 hours / day = 60.83 hours / year
normal active time = 60.83 hour / year

With this malfunctioning bulb, it would be active at all times
actual active time = 24 hour / day * 365 day / year = 8760 hour / year

Extra active time = actual - normal = 8760 - 60.83 = 8699 hours
Extra power = extra time * wattage = .04 KW * 8699 hour = 347.96 KW-Hour

Extra cost
= extra time * cost per KW-hour = 10 cent/KW-Hour * 347.96 KW-Hour = 3479.6 cent = 34.79 \$ per year

So if I'm right, those are the answers to the asked questions. What I'm leery of is the coefficient of performance. Am I right in assuming that was info meant to misdirect or distract me, or did I butcher this problem? Any and all help is greatly appreciated, thanks.

2. Mar 19, 2016

### Staff: Mentor

The light bulb not only adds light to the food compartment, it adds heat.

3. Mar 19, 2016

### reed2100

I was afraid of that, so this problem becomes more complicated. If that's the case then looking back at the COP, 1.3 = QL(per sec) / W-input (per sec)

But that's for the entire fridge I would assume, so W-input(per sec) is not 40 J/sec, it's not a work input of the lightbulb.

I'm having trouble assigning values here...The lightbulb consumes 40 J/sec regardless of the fridge's COP, so the numbers I calculated are still true correct? At least so far as "this is the extra amount of energy used to power the lightbulb nonstop for the year" ?

Now it's a question of "the lightbulb adds heat to the fridge, heat that the fridge requires Work energy to offset and maintain a COP of 1.3", correct?
Is the added heat just equivalent to the power consumed by the bulb? Like, what does it mean when you say "a 40 watt bulb" ? The bulb consumes 40 watts or puts out 40 watts of energy? It has a work input of 40 watts or a heat transfer to its surroundings of 40 watts? I'm pretty confused here.

4. Mar 19, 2016

### Staff: Mentor

A 40W light-bulb is a 40W heater. (Besides its heat losses, the energy in its visible light output is absorbed in the enclosure and converted to heat, so it's essentially a 100% efficient heater.)

5. Mar 19, 2016

### reed2100

So if the bulb adds 40 watts of heat to the fridge, the fridge has to increase its QL by 40 J/sec? And that requires a new work input such that the COP remains 1.3? How exactly do I get to any of these numbers if I wasn't given a normal QL, QH, or W for the fridge?

EDIT - so I played around with the definition of the COP and my (potentially incorrect) understanding that an addition of 40 watts of heat by the bulb necessitates an additional 40 watts of heat drained out by the fridge, so an increase of QL by 40 watts. I picked arbitrary numbers for the work input and found the resulting QL, then increased QL by 40 watts and found the new work input. It seems an increase of QL by this much necessitates an increase in work input by 30.76 watts. So I think I'm getting somewhere.

I would then say the fridge consumed an extra 30.76 watts for (a nonstop year - time of extra work if bulb functioned normally). And that would be the extra power consumed to make up for the bulb's extra heat.

I would then add the extra power consumed by the bulb itself, which I already calculated. Since they're on the same circuit. That would tally up to the total extra power consumption due to this bulb. Does this all sound right?

Last edited: Mar 19, 2016
6. Mar 19, 2016

### Staff: Mentor

Do you need any more data for determining just the extra electricity used?

7. Mar 19, 2016

### reed2100

Sorry about the bad timing but I just edited my earlier reply, I think I got at what you're suggesting here

8. Mar 19, 2016

### Staff: Mentor

That sounds like the right approach, but take care with the units. Q and W have units of Joules.

9. Mar 19, 2016

### reed2100

Thank you! But as for the units, I can use watts instead of joules can't I? That's what our textbook says at least. The COP is a dimensionless ratio so the units would cancel out, so I thought I'd keep it all in watts since the question asks for power consumption anyway. I was always using watts in my calculations here but I couldn't find a way to add a dot above the Q or W.

10. Mar 19, 2016

### Staff: Mentor

If it's continuous process you can work in watts because the duration is common, in which case you refer to power, not "work". But where the fridge door is opened for short periods you had to calculate using energy (Joules).

11. Mar 19, 2016

### reed2100

I'm not clear on this, so you're saying it's not arbitrary how you define a short period here? If I used joules to find the extra work input I'd still get an extra 30.76 joules needed whenever the bulb is on, but that's for 1 second because the bulb gives out 40 joules of heat per second. The fridge is open for 30 seconds at a time which I thought is a lot more.

Basically the way I'm thinking of it is - the fridge instantly starts to operate with an extra work input of 30.76 joules/sec whenever the light is on, that's the answer to question 1 (since it doesn't specify over any period of time)

Then for question 2 - total yearly cost of entire extra power consumption due to heat of bulb and powering the bulb = (1 year - time that a functioning bulb would be on in 1 year) * (power cost to keep bulb on + extra power cost of fridge to counteract bulb's heat)

This would give me units of (hours * watts) which I can convert to kWh, which I have a cost per unit for

12. Mar 19, 2016