An oil pump is drawing 52kW of electric power while pumping oil with ρ=830kg/m^3 at a rate of 0.15m^3/s. The inlet and outlet diameters of the pipe are 10cm and 15cm, respectively. If the pressure rise of oil in the pump is measured to be 380 kpa and the moter efficiency is 89%, determine the mechanical efficiency of the pump. 已知 ρ=830 (kg/m^3) P2-P1=380 (kpa) dV/dt=0.15 (m^3/s) 可得 v=1/ρ=1/830 dm/dt=ρdV/dt=830*0.15 dV/dt=uA (u为速度) 进出流量相同 u1=0.15/(pi(0.1/2)^2)=19.09 u1=0.15/(pi(0.15/2)^2)=8.49 Δe=Δu+ΔPv+Δu^2/2 假设温度不变 Δu=0 Δe=(P1-P2)v+(u1^2-u2^2)/2 =(380*1000/830)+(8.49^2-19.09^2) =311.66 (J) ΔE=dm/dt*Δe=830*0.15*311.66 52*1000*0.89*η=311.66 功*换成J*马达效率*pump效率 η=0.84 --

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