Acentrifugal pump is pumping oil with ?= 860 kg/m3at a rate of 0.1 m3/s.The inlet and outlet diameters of the pipe are 8 cm and 12 cm,respectively. If the pressure rise of oil in the pump is measuredto be 260 kPa and the pumpefficiency is 78percent, determinethe input power required to operate thepump. Take the kineticenergy correction factor to be 1.05. Task (4)A centrifugal pump is pumping oil with p = 860 kg/m³ at a rate of 0.1 m³/s.The inlet and outlet diameters of the pipe are 8 cm and 12 cm, respectively.If the pressure rise of oil in the pump is measured to be 260 kPa and thepump efficiency is 78 percent, determine the input power required tooperate the pump. Take the kinetic energy correction factor to be 1.05.Pump8 cm0.1 m³/s12 cmAP= 240 kPa

Given: Density of the pumping oil:→ρ=860 kg/m3 Flow rate→Q=0.1 m3/s Inlet diameter→Di=8 Cm=0.08 m So…

Answered: A centrifugal pump is pumping oil with…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Acentrifugal pump is pumping oil with ?= 860 kg/m3at a rate of 0.1 m3/s.The inlet and outlet diameters of the pipe are 8 cm and 12 cm,respectively. If the pressure rise of oil in the pump is measuredto be 260 kPa and the pumpefficiency is 78percent, determinethe input power required to operate thepump. Take the kineticenergy correction factor to be 1.05.

Task (4)
A centrifugal pump is pumping oil with p = 860 kg/m³ at a rate of 0.1 m³/s.
The inlet and outlet diameters of the pipe are 8 cm and 12 cm, respectively.
If the pressure rise of oil in the pump is measured to be 260 kPa and the
pump efficiency is 78 percent, determine the input power required to
operate the pump. Take the kinetic energy correction factor to be 1.05.
Pump
8 cm
0.1 m³/s
12 cm
AP= 240 kPa

Expert Solution

Step 1

Given:

Density of the pumping oil: $\to ρ = 860 k g / m^{3}$

Flow rate $\to Q = 0.1 m^{3} / s$

Inlet diameter $\to D_{i} = 8 C m = 0.08 m$

So inlet area $\to A_{i} = \frac{π}{4} \times {D_{i}}^{2} = \frac{π}{4} \times {0.08}^{2} = 0.005 m^{2}$

Hence inlet velocity $\to V_{i} = \frac{Q}{A_{i}} = \frac{0.1}{0.005} = 20 m / s$

Outlet diameter $\to D_{o} = 12 C m = 0.12 m$

So outlet area $\to A_{o} = \frac{π}{4} \times {D_{o}}^{2} = \frac{π}{4} \times {0.12}^{2} = 0.011 m^{2}$

Hence outlet velocity $\to V_{o} = \frac{Q}{A_{o}} = \frac{0.1}{0.011} = 9.09 m / s$

Pressure rise in the pump $\to P_{o} - P_{i} = 260 K P a$

Acceleration due to gravity $\to g = 9.81 m / s^{2}$

Efficiency of the pump $\to η = 78 % = 0.78$

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