Explanations & Calculations
- Those given conditions are valid only at the pump outlet where the water is ejected at the speed of 30ms−1 and beyond that these calculations may not seem to be effective.
- The power, as it is defined is = ΔtimeΔwork
- And the work done at the outlet is W=21mv2 and lets assume that this work is done at a small duration of t(s). Then the power needed is
P=tW=21(tm)v2
- That term within the parentheses is the rate of mass flow which can be deduced to be Avρ where A is the are of the pump.
- Then,
P=21(Avρ)v2=21Av3ρ=21(π(0.01m)2×(30ms−1)3×1000kgm−3)=4241.15W≈4.2kW
Comments