Question

10. An engine pumps a fluid of density 1200 kg/m3
through a tube of cross-section area of 4 cm2
through a height of 5 m with a speed of 40 cm/s.
Find the power of engine.​

Answer 1

power of engine = 9.6 watt

first find mass of fluid pumped per sec,

i.e., dm/dt = ρA v

where ρ is density, A is cross sectional area and v is speed of fluid .

so, dm/dt = 1200 kg/m³ × 4 × 10^-4 m² × 40 × 10^-2 m/s

= 1200 × 4 × 40 × 10^-6 kg /s

= 192000 × 10^-6 kg /s

= 0.192 kg/s

mass of water after time t, m = t × dm/dt

= t × 0.192 kg

now, potential energy of water = mgh

= t × 0.192 × 10m/s² × 5m

= 9.6 t

then power of energy = potential energy/time = 9.6t/t = 9.6 watt

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Answer 2

$\bf\large\underline\green{Answer:-}$

first find mass of fluid pumped per sec,

i.e., dm/dt = ρA v

where ρ is density, A is cross sectional area and v is speed of fluid .

so, dm/dt = 1200 kg/m³ × 4 × 10^-4 m² × 40 × 10^-2 m/s

= 1200 × 4 × 40 × 10^-6 kg /s

= 192000 × 10^-6 kg /s

= 0.192 kg/s

mass of water after time t, m = t × dm/dt

= t × 0.192 kg

now, potential energy of water = mgh

= t × 0.192 × 10m/s² × 5m

= 9.6 t