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Power in kW required to drive Centrifugal pump with discharge of Q m$$^3$$/sec, Head in H mtr, specific weight of fluid as $$\omega N/m^3$$ and overall efficiency of pump as $$\eta$$ shall be
$$\frac{\omega H}{Qx \eta}$$
$$\frac{\omega QH}{\eta}$$
$$\frac{\omega Q}{Hx \eta}$$
$$\frac{\omega Qx \eta}{H}$$
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