## anonymous one year ago What power of an engine is required to pump 2450 N of water per second from a well 50 m deep to the surface? (a) 2.45*10^3 watts (b) 2.45*10^4 watts (c) 2.45*10^5 watts (d) none of these

1. anonymous

@Michele_Laino please can you solve it?

2. anonymous

okay so work is Fd so the answer is 2450 N times 50 m

3. anonymous

wait... per second

4. anonymous

5. anonymous

I think you're right because a watt is a N-m/s, so the units work out

6. Michele_Laino

we have to consider the manometric head H, which is about 50 m, so the requested power P is: $\Large P = F\left( {H + h} \right) = 2450\left( {50 + 50} \right) = ...?$ being h the deepness of the well

7. Michele_Laino

sorry H is equal to the ratio between the pressure difference and the specific weight of the water and H+h is the manometric head or manometric prevalence

8. anonymous

9. Michele_Laino

since the formula for computing the power, is: $\large P = \frac{{FH}}{t}$ where H is the total prevalence, whose formula is: $H = h + \frac{{\Delta p}}{\gamma }$ \Delta p is the difference between the pressures

10. Michele_Laino

|dw:1434218860993:dw|

11. Michele_Laino

|dw:1434218975568:dw|

12. Michele_Laino

so: $\frac{{\Delta p}}{\gamma } = 5 \times 10.33 = 51.65\;meters$

13. Michele_Laino

please keep in mind that 10.33 meters is the maximum aspiration height for a pump

14. Michele_Laino

and: 1 athmosphere is equivalent to 10.33 meters of water

15. anonymous

so you think that 2.45*10^5 would be the correct option?

16. Michele_Laino

yes!

17. anonymous

are you confirmed liano?

18. IrishBoy123

. i'd go with d) but i'd like the OP to post the answer when that becomes apparent

19. Michele_Laino

yes! I confirm my answer above!

20. anonymous

i asked this to my friend he say to me that you can apply formula $P=mgh/t$ since W=mgh $P=W/t$

21. anonymous

so after this answer is none of these,the (d) option

22. anonymous

i am so much confused about the answer,but i think what you are telling is also correct!