I recently bought a new refrigerator for my kitchen. The feet are adjustable but I've been lazy.
Whenever the motor runs and the feet aren't all touching the floor there is a loud buzzing noise. As soon as I move the fridge around to level it up the buzzing stops. One day I'll sort it out properly maybe.
Question
Clearly the fridge keeps its contents at a roughly even temperature in my warm kitchen and that takes energy.
Without buying a multi-meter or doing a long term experiment, can I get a quick idea of how much energy is being wasted by the noise? Maybe the resonance of the noise somehow makes things more efficient and is a good thing?
Suppose I borrowed a decibel meter. Could I work it out from that?
Best Answer
The answer by niels nielson is much more useful than my answer. But just in case you really do want a rough estimate of how much power is emitted as sound...
According to [1] (references are listed at the end), a sound level meter is a hand-held instrument with a microphone that measures the sound pressure level (SPL). I'll assume that this is the kind of decibel meter that you'll borrow. To convert SPL to power, we need to account for three things:
The relationship between sound pressure and intensity
The variation of intensity with distance from the source (the refrigerator)
The fact that the sound is not being emitting equally in all directions
For this analysis, let's account for the first two things and ignore the third one. This should be good enough for an order-of-magnitude estimate. According to [2], the relationship between sound pressure $p$ and sound intensity $I$ is $$ I = \frac{p^2}{Z_0} \tag{1} $$ where $Z_0$ is the accoustic impedance, which is $$ Z_0 = 400\ \frac{\text{Newton}\cdot\text{second}}{\text{meter}^3}. \tag{2} $$ Also,
The intensity $I$ has units of Watts per square meter
The sound pressure $p$ has units of Newtons per square meter (Pascals)
The sound level meter might register the sound pressure level as a number of decibels. According to [3], the SPL in decibels (dB) is defined by $$ \text{SPL in dB} = 10\log_{10}\left(\frac{p^2}{p_0^2}\right) \tag{3} $$ where $p$ is the root mean square sound pressure and $p_0$ is the reference sound pressure $$ p_0 = 2\times 10^{-5} \text{ Pascal}. \tag{4} $$ Equations (1)-(4) give the following relationship between the SPL in dB, which the meter measures, and the intensity $I$, which is closer to what we want: $$ I = \frac{10^{\text{SPL}/10}\times p_0^2}{Z_0} = 10^{\text{SPL}/10}\times 10^{-12}\ \frac{\text{Watts}}{\text{meter}^2}, \tag{5} $$ where SPL is the sound pressure level in decibels. If the sound were being emitted isotropically (the same in all directions), then we could get the total emitted power from the intensity $I$ times the area of the sphere: $$ P = 4\pi R^2 I \tag{6} $$ where $R$ is the distance from the source (the radius of the sphere). Equations (5)-(6) are the desired relationships, assuming that the sound is emitted equally in all directions without any reflection — which might be a semi-realistic assumption if the refrigerator is suspended in mid-air in a large anechoic chamber, but probably much less realistic in a typical kitchen.
As an example, consder the sound emitted by a vacuum cleaner. According to [2], the SPL is about $70$ dB at a distance of $1$ meter. (Not all vacuum cleaners are identical, of course. This is only an estimate.) According to equation (5), the corresponding intensity is $$ I = 10^{70/10}\times 10^{-12}\ \frac{\text{Watts}}{\text{meter}^2} = 10^{-5} \frac{\text{Watts}}{\text{meter}^2}. \tag{7} $$ This is at a distance of $1$ meter, so if we assume that the sound is emitted isotropically at this intensity, the total power emitted as sound is $$ 4\pi R^2 I = 4\pi \,(1\text{ meter})^2\times 10^{-5} \frac{\text{Watts}}{\text{meter}^2} \approx 10^{-4}\text{ Watts}. \tag{8} $$ Every increase in the SPL of $10$ dB corresponds to a factor of $10$ increase in the power. So if we replace the vacuum cleaner with a chainsaw at the same distance, which corresponds to $110$ dB SPL according to [2], the total power emitted as sound (with the same isotropic assumption) would be $1$ Watt. For comparison to the more-useful answer by niels nielson, one horsepower is roughly 746 Watts.
Hopefully your refrigerator is not quite as noisy as a chainsaw.
References:
[1] https://en.wikipedia.org/wiki/Sound_level_meter
[2] http://www.sengpielaudio.com/TableOfSoundPressureLevels.htm
[3] https://en.wikipedia.org/wiki/Sound_pressure