Question

a horse power of engine used to fill the tank of tank size 1m×2m×1m situated at average height 20 m with water from a well of depth 17.3 m in 5 minutes assume 20 percent power is wasted

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Answer 1

Answer:

• The size of the tank = (1 × 2 × 1) m³

volume of the tank = 2 m³

mass of water completely filling the tank = volume × density

= 2m³ × 1000kg/ m³

= 2000 kg

• Tank situated at a height = 20m

Water to be filled in the tank from a well of depth = 17.3 m

Total height through which the water is to be carried (h) = (20 + 17.3) m

= 37.3 m

Useful energy = mass × g × height

= 2000 × 10 × 37.3

= 746000 J

• Time taken to fill the tank with water (t) = 5 mins

= 5 × 60 s

= 300 s

Useful power = energy/ time

= 746000/ 300

= 2486.67 W

Since, 20 % power is wasted, useful power = (100 - 20) = 80 %

=> $\frac{useful power }{total power} * 100 = 80$

=> $\frac{2486.67}{total P} * 100 = 80$

=> total Power = (100 * 2486.67)/ 80

P = 3108. 34 W

CONVERTING POWER FROM WATTS TO HORSE POWER :-

=> 745.69 W = 1 hp

=> 1 W = 1/ 745.69 hp

=> 3108.34 W =  (1/ 745.69) x 3108.34 hp

= 4.16 hp

The horse power of the engine required is 4.16 hp (Ans

Answer 2

Answer:

volume of tank

mass of water

total height

energy

power