## Dumb_as_a_Duck 3 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.

1. JamesJ

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. Dumb_as_a_Duck

given - [g = 9.8ms ^{-2} \], $density of water = 1000 kg m ^{-3}$ and $1000L = 1 m^{3}$

3. Dumb_as_a_Duck

(that was a part of the question)

4. Dumb_as_a_Duck

how do you convert the given litres of water to kg?

5. JamesJ

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.

6. Dumb_as_a_Duck

how did you get that/derive that 1 litre of water has a mas of 1 kg from the relation given?

7. eashmore

He got it from density. Note that 1000 litres are equal to 1 m^3.

8. Dumb_as_a_Duck

and then? (im not seeing the relation or the beauty here, i'm sorry.)

9. eashmore

No problem. Let's use dimensional analysis. $1000 \left [\rm kg \over m^3 \right ] \cdot \left [ \rm 1 m^3 \over 1000L \right ] = 1 \left [\rm kg \over L \right ]$

10. Dumb_as_a_Duck

WAIITTTT

11. eashmore

Waiting.

12. Dumb_as_a_Duck

i didnt get a word of that. the thing you wrote under dimensional analysis, i mean. t

13. Dumb_as_a_Duck

if you were gonna explain that, then continue.

14. Dumb_as_a_Duck

i havent learnt about dimensional analysis either.

15. eashmore

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

16. eashmore

Therefore, we are left with $1 \left [ \rm kg \over L \right ]$

17. JamesJ

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.

18. Dumb_as_a_Duck

Oh! I see :)

19. Dumb_as_a_Duck

that makes the rest of the problem obvious. thanx! :)

20. eashmore

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$PE = \rho V gh$

21. eashmore

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.

22. Dumb_as_a_Duck

isn't P=W/t? (Power = Work/Time)?

23. JamesJ

Yes ... eashmore made a small mistake

24. eashmore

Whoops! Sorry about that. I was just testing you. :-P

25. Dumb_as_a_Duck

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).

26. eashmore

Looks good. Assuming you typed it into your calculator right.

27. Dumb_as_a_Duck

I calculated it myself lol. I'll double check with a calculator.

28. Dumb_as_a_Duck

Thanks a bunch! :)

29. eashmore

Whoops! I got that it should be 22.05 kW.

30. Dumb_as_a_Duck

uhoh! did i make a conversion mistake?

31. eashmore

Nope. Everything is good up until you calculate power. Double check that guy and you should be good.

32. Dumb_as_a_Duck

I got your answer (22.05 kW) on the calculator, but that's not what i got when I quadruple-checked my calculation.

33. Dumb_as_a_Duck

><

34. eashmore

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

35. Dumb_as_a_Duck

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

36. Dumb_as_a_Duck

well, maybe i should just listen to the math gods in silence and not be an atheist. :P

37. eashmore

You should get the same answer with both ways that you have described. Maybe you are misinterpreting the results you get.

38. Dumb_as_a_Duck

I found my mistake! I should have divided 1323 by 600, not just 6...i forgot about that :P *sheepish look*.

39. Dumb_as_a_Duck

I salute you for sticking with me til the end. :D *high five* :P

40. eashmore

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.

41. supercrazy92

I salute both of you.. Great work!

42. Dumb_as_a_Duck

LOL! :D