Question

A tank has a volume of 2000m³.It is filled with water in half an hour from a well 50m below that tank. Calculate the power of the engine used to pump the water. If 40% of the power is wasted what would be the actual power.? Given:Density of water-1000kgm³​

Answer 1

Answer:

Here, Volume of water =30 m

3

t=15 min=15×60 s=900 s;Height,h=40 m

Efficiency, η=30%

Density of water =10

3

kg m

−3

∴ Mass of water pumped =Volume×Density

=(30 m

3

)(10

3

kg m

−3

)=3×10

4

kg

P

output

=

t

W

=

t

mgh

=

900 s

(3×10

4

kg)(10 ms

−2

)(40 m)

=

3

4

×10

4

W

Efficiency, η=

P

input

P

output

P

input

=

η

P

output

=

3×

100

30

4×10

4

=

9

4

×10

5

=44.4×10

3

W=44.4 kW.

Answer 2

Answer:

Explanation:

mass of water to be pumped = volume x density

= 2000 x 1000

= 2 x $10^{6}$kg

when the water is completely filled in the tank

PE of water = mgh = 2 x 10^6 x 10 x 50

= $10^{9}$ J

Since it must be filled in half an hour

required power = $10^{9} / 1800$

=$5.5 * 10^{5}$W = energy output

Efficiency η% =energy output/energy input x 100

40 = $5.5 * 10^{5}$/E x 100

E = $5.5 * 10^{7$ /  40

E = 1375 kJ .