A lift is reported to operate at 75% efficiency. If the lift can raise a load through a height of 30m using a 230V, 15A supply, calculate how long it will take to raise a load of 1000Kg.
g=9.8ms2g = 9.8\frac{m}{s^2}g=9.8s2m
h=30mh = 30mh=30m
m=1000kgm =1000kgm=1000kg
η=75%\eta=75 \%η=75%
I=15AI = 15AI=15A
U=230VU= 230VU=230V
ΔEP=mgh=9.8∗30∗1000=294000 J\Delta EP = mgh = 9.8*30*1000= 294000\ JΔEP=mgh=9.8∗30∗1000=294000 J
W=ΔEP=294000 JW=\Delta EP= 294000\ JW=ΔEP=294000 J
η=PoutPinp∗100%\eta =\frac{P_{out}}{P_{inp}}*100\%η=PinpPout∗100%
Pinp=IU=230∗15=3450 WP_{inp}= IU= 230*15= 3450\ WPinp=IU=230∗15=3450 W
Pout=Pinpη100%=2587.5 WP_{out}=\frac{P_{inp}\eta}{100\%}=2587.5\ WPout=100%Pinpη=2587.5 W
W=PouttW =P_{out}tW=Poutt
t=WPout=2940002587.5=113.62 st =\frac{W}{P_{out}}=\frac{294000}{2587.5}=113.62\ st=PoutW=2587.5294000=113.62 s
Answer: t=113.62 s\text{Answer: }t = 113.62\ sAnswer: t=113.62 s
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