Calculating power/energy/energy conversion efficency.. my attempt attached

Discussion in 'Homework Help' started by Alice24, Apr 22, 2011.

  1. Alice24

    Thread Starter Active Member

    Apr 22, 2011
    Hi guys. I'm a mechatronics student and we just started electronics in our second semester. I need some help with basic questions please^^

    An electrical engine requires a flow of 5 A with the useful power of 1kW is connected to a voltage source of 220 V and has been operating for 3 hours. Calculate
    1) Invested power
    2) Invested energy
    3) The energy conversion efficency of the engine
    4) The price of activating the engine, if the price of the electrical energy is 0.42 shekels for 1kWh.

    My attempt at the answers attached
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    You have (1) & (3) correct.

    The input power is more than the useful (mechanical output) power because of inefficiency in the engine. About 9.1% of the input power is lost as heat - the rest goes into doing useful mechanical work.

    Input power is 1100 Watts which equates to 1100 Joules per second.

    Over 3 hours there are 3*3600 seconds=10,800 seconds

    So total Energy in 3 hours =10,800*1100=1,188,000 Joules = 1.188MJ

    The total kilowatt-hours = 1.1 (kW) * 3 (hours)=3.3kWh

    Cost = 0.42 * 3.3=1.386 Shekels