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An engine can pump 30000 litres of water to a vertical height of 45 m in 10 minutes. Calculate work done by the machine and its power.
 2 years ago
 2 years ago
An engine can pump 30000 litres of water to a vertical height of 45 m in 10 minutes. Calculate work done by the machine and its power.
 2 years ago
 2 years ago

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JamesJBest ResponseYou've already chosen the best response.0
The work done is the equal to the change in gravitational potential energy of the water. Calculate that, using PE = mgh. Then Power = (Work)/(Time)
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
given  [g = 9.8ms ^{2} \], \[density of water = 1000 kg m ^{3}\] and \[1000L = 1 m^{3}\]
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
(that was a part of the question)
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
how do you convert the given litres of water to kg?
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
This is the beauty of SI units and you can deduce this from the relation you wrote down above: 1 litre of water has a mass of 1 kg.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
how did you get that/derive that 1 litre of water has a mas of 1 kg from the relation given?
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
He got it from density. Note that 1000 litres are equal to 1 m^3.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
and then? (im not seeing the relation or the beauty here, i'm sorry.)
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
No problem. Let's use dimensional analysis. \[1000 \left [\rm kg \over m^3 \right ] \cdot \left [ \rm 1 m^3 \over 1000L \right ] = 1 \left [\rm kg \over L \right ]\]
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
i didnt get a word of that. the thing you wrote under dimensional analysis, i mean. t
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
if you were gonna explain that, then continue.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
i havent learnt about dimensional analysis either.
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
It is a fact that 1 m^3 equals 1000 L. Can you accept that? Dimensional analysis is just a method of converting units. So the above relation can be written as \[\rm 1 m^3 \over 1000 L\] If we multiply this by the density that is given\[1000 \left [\rm kg \over m^3 \right ] \cdot \left [\rm 1 m^3 \over 1000 L \right ]\]Notice that \(m^3\) cancel out, and that 1000/1000 = 1
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
Therefore, we are left with \[1 \left [ \rm kg \over L \right ]\]
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
Or put another way: if 1 cubic meter of water is 1000 kg, and 1 cubic meter of water holds 1000 liters, it must be that 1 liter has a mass of 1 kg.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
that makes the rest of the problem obvious. thanx! :)
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
Great! Now, we know that, as James said, \[PE = mgh\]and density is defined as\[\rho = {m \over V}\]We can write mass in terms of density as\[m = \rho V\]Therefore, the potential energy can be written as\[PE = \rho V gh\]
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
We know that\[W = \Delta PE\] in this case, and that\[P = W \cdot t\]Therefore, \[P = \rho V ght\]where \(\rho\) is the density of water, \(V\) is the volume of water pumped, \(g\) is the acceleration due to gravity, \(h\) is the height that the pump raises the water, and \(t\) is the time that it takes to pump the given volume of water. Let me know what you get.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
isn't P=W/t? (Power = Work/Time)?
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
Yes ... eashmore made a small mistake
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
Whoops! Sorry about that. I was just testing you. :P
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
oh. is my solution here right? (Given) height= 45 m time = 10 minutes = 600 seconds density of water = 1000 kg m^3 volume of water = 30,000 litres of water = 30 m^3 (as 1000 L = 1 m^3) (solution) mass of water = density of water * volume of water = 30 * 1000 = 30,000 kg. P.E. = mgh = 30,000 * 9.8 * 45 = 1323 * 10^4 Joule Work done= P.E. = 1323 * 10^4 Joule Power = W/t = 1323 * 10^4/600 = 2205 * 10^3 watts. (or 2205 kW).
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
Looks good. Assuming you typed it into your calculator right.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
I calculated it myself lol. I'll double check with a calculator.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
Thanks a bunch! :)
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
Whoops! I got that it should be 22.05 kW.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
uhoh! did i make a conversion mistake?
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
Nope. Everything is good up until you calculate power. Double check that guy and you should be good.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
I got your answer (22.05 kW) on the calculator, but that's not what i got when I quadruplechecked my calculation.
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
I hate when this happens. The math gods of WolframAlpha say that it is 22.05 kW. http://www.wolframalpha.com/input/?i=30000*45*9.8%2F600+W+in+kW
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
i mean, i got 22.05 kW when i first multiplied 1323 * 10^4 and THEN divided by 600....but when i divided 1323 by 600 AND then multiplied by 10^4, its the other wrong answer. lol i suck at math too. :P
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
well, maybe i should just listen to the math gods in silence and not be an atheist. :P
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
You should get the same answer with both ways that you have described. Maybe you are misinterpreting the results you get.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
I found my mistake! I should have divided 1323 by 600, not just 6...i forgot about that :P *sheepish look*.
 2 years ago

Dumb_as_a_DuckBest ResponseYou've already chosen the best response.1
I salute you for sticking with me til the end. :D *high five* :P
 2 years ago

eashmoreBest ResponseYou've already chosen the best response.4
No problem. That is why we are here. I salute you for doing the same. Most users simply post all their homework questions, go eat dinner, and return to copy answers. Thank you for being engaged and striving to learn. Glad I could help.
 2 years ago

supercrazy92Best ResponseYou've already chosen the best response.0
I salute both of you.. Great work!
 2 years ago
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