An experimentalelectrical generator collects sunlight with mirrors and to genera

Question

An experimentalelectrical generator collects sunlight with mirrors and to generateheat at a rate of 1.2 megawatts. The generator is mounted on theroof of an environmentally friendly building and is used to operatethe elevator. The elevator has a max operating load of 8000 kg anda maximum velocity of 6 m/s.
a)Determine the power that the generator must supply to operate theelevator at its maximum conditions.
b)What is the efficiency of the experimental generator?
c)Calculate the rate at which the generator exhausts heat to theenvironment.
d)Determine the total heat wasted by this generator in the courseof a single day.(Assume that the generator is operating a maxcapacity all day) An experimentalelectrical generator collects sunlight with mirrors and to generateheat at a rate of 1.2 megawatts. The generator is mounted on theroof of an environmentally friendly building and is used to operatethe elevator. The elevator has a max operating load of 8000 kg anda maximum velocity of 6 m/s.
a)Determine the power that the generator must supply to operate theelevator at its maximum conditions.
b)What is the efficiency of the experimental generator?
c)Calculate the rate at which the generator exhausts heat to theenvironment.
d)Determine the total heat wasted by this generator in the courseof a single day.(Assume that the generator is operating a maxcapacity all day)

Explanation / Answer

The elevator has a max operating load M = 8000kg
Mmaximum velocity v = 6 m/s (a). The power that the generator must supply to operate theelevator at its maximum conditions in one second is                       P = weight * velocity                          = Mg v                         = 8000 kg * 9.8 m / s^2 * 6 m / s                         = 4.704*10^5 watt                        = 0.4704*10^6 watt (b). rate of heat generated P ' = 1.2 M W                                              = 1.2* 10 ^ 6 W So, efficiency   e = P / P '                         = 0.392                         = 39.2 % (c). the rate at which the generator exhausts heat to theenvironment = P ' - P                                                                                                      = 0.7296 * 10 ^ 6 W
(d). the total heat wasted by this generator in the course of asingle day = (P'-P ) * 24 h                                       = ( P'-P ) * 24 * 3600 s                                       = 6.303 * 10 ^ 10 J

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