Question

Designing a building 700 ft high, 50 stories. 6500 lb elevator in building takes 1 min to rise 20 stories. Boss thinks this is too long, so she asks to buy a bigger power supply for the elevator. Specifications for new elevator are the same except it operates at twice the voltage of the old. You estimate that while the elevator runs at max speed, the whole system, including power supply, is 60% efficient. If cost of electricity is $0.06 per kilowatt-hour, will the operating system of the new elevator be more or less expensive than the old?

Answer 1

we have to compute the mechanical power developed by the elevator

Answer 2

yes, i'm having trouble with that without knowing the actual voltage

Answer 3

for the power of the old operating system, I plugged in P = mgh/t and got the power for that that way. but without knowing the voltage, I can't decipher a different equation between the old and the new... if that makes sense

Answer 4

the electrical power is:
\[\frac{{mechanical\;power}}{{0.6}}\]

Answer 5

so we can write:
\[V \times I = \frac{{mechanical\;power}}{{0.6}}\]

Answer 6

is the 0.6 from the 60% efficiency?

Answer 7

yes!

Answer 8

that 60% takes account of efficiency of the electric engine and mechanical friction forces

Answer 9

okay. so i'm thinking by mechanical power you are referring to "work" correct? sorry my professor has never used the phrase mechanical power so I'm just making sure I'm understanding :/

Answer 10

yes! that's right. What I keep in mi mind is the transformation of electrical energy into mechanical energy, and that transformation take place into the electric engine

Answer 11

oops..my*

Answer 12

Got it. So I'm getting (4.1 x 10^6)/I = V

Answer 13

So my final equation for power would still have a complex fraction. Also, i have the equation\[P = (q \Delta V) \div t \] which would still leave me with the q...

Answer 14

idk sorry i'm lost :/

Answer 15

If I double the voltage applied to the electric engine, then the electrical absorbed power is doubled