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In our university there's a posted sign that encourages students to take the stairs instead of the elevator to save the university electricity during the hot NYC summers.

When you climb the stairs you generate something like 8x your mechanical energy in metabolism which is dissipated as heat and must be cooled by the air conditioning in our cooled stairwells.

What are plausible assumptions about the energy efficiency limits of air conditioning and mechanical lifting of an elevator? What (fundamental) limits are there that prevent 100% efficiency in the elevator case? Are there other important factors for this analysis?

Is it plausible that the university is wrong? Should we take the elevator instead of the stairs to save the university energy? Do the humans of an energy-efficient future look like Wall-E?

The source of this question is a bathroom physics problem posted above university physics department toilets near grad offices. Problems are submitted by department members. To the best of my knowledge, none of these were ever homework questions. This question is adapted by my toilet memory from a problem that was provided by Andrei Gruzinov.

GPhys
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2 Answers2

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I wondered the same thing and so did some empirical testing with my elevator.

TLDR;

The elevator uses a huge amount of energy to move just the elevator.

For 5 floor ascent using my single piston hydraulic lift elevator on the day I tested...

Elevator: ~550KWs

Stairs: 92KWs / (22 calories)

If the elevator is going up anyway, then you should hop on since your additional weight has almost no effect on the total power used.

If you are considering taking the elevator alone, you should take the stairs.

Also keep in mind that most the energy used by the elevator is turned into heat and that heat is generated inside the envelope of the building, so must be rejected by the AC system.

More explanation and data here...

https://wp.josh.com/2013/05/29/elevator-power-usage-should-i-take-the-stairs/

bigjosh
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Accepting that climbing stairs generates heat equal to 7 times the mechanical energy (one part goes into you lifting yourself), how much energy does it take to remove that heat?

The 2nd Law of Thermodynamics says the best you can do is

$W = Q \frac {T_h - T_c}{T_c} \sim Q \frac{\Delta T}{T}$

Where $\Delta T$ is the difference between inside and outside temperature, and T is the outside temperature. For ease of calculation, taking T around room temperature of 280K, you get that the air conditioner is better (uses less energy, i,e, W/Q is better than 1/7) if it’s working into a temperature difference less than $\Delta T = 40\rm{K}$.

Large-building air-conditioners routinely even better than this for electrical power usage by using e.g. evaporative chillers to do most of the thermodynamics.

Unless it’s really really hot outside, take the stairs.

Bob Jacobsen
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