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Solved exercise · level 3 of 5

Efficiency of a goods lift

Skills: Calculate

Statement

Determine the efficiency of a goods-lift installation that raises 1000 kg by 5 m in 50 s while the drive motor delivers a useful power of 1401 W.

m = 1000 kg · h = 5 m · t = 50 s · P = 1401 W

Parametric exercise: the values change with every draw; these are one example.

Guided problem in 4 steps: unknown, formula, units, data, result.

1 hint from Bastaya available in the lab
Solution

Useful power = m·g·h / t = 981 W; η = 981 / 1401 = 0.7.

Course sheet of the chapter : Power, energy, efficiency

This chapter is covered by the paths: General electricity, ISL-oriented exercises

Other exercises of the chapter

Simulated time00:00:00
Q = I·t-
W = P·t-

Circuit diagram CEI 60617

How does it work?

Each big hole is a circuit node: you push one leg of a component into it. The 4 small holes around it are connected to it: chips and leads plug in there.

  1. Place a component Drag it from the palette: it snaps onto the grid. R rotates it.
  2. Bridge neighbouring holes Pick the jumper chip (or C) and drag on the board: bridged holes become a single node.
  3. Draw a lead For long links and the power supply: from one big hole to another, or to a terminal.
  4. Close the loop The circuit is solved live (Ohm, Kirchhoff) and the diagram follows.

Shortcuts

Del
delete
R
rotate
C
chip brush
Esc
cancel
wheel
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fit the board
drag background
pan

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