An oil pump is drawing 44 kW of electric power while pumping oil with ρ= 860 kg/

Baublysiz

Baublysiz

Answered question

2021-11-21

An oil pump is drawing 44 kW of electric power while pumping oil with ρ= 860 kg/m^3 at a rate of 0.1 m^3/s. The inlet and outlet diameters of the piper are 8 cm and 12 cm, respectively. If the pressure rise of oil in the pump is measured to be 500 kPa and the motor efficiency is 90 percent, determine the mechanical efficiency of the pump.

Answer & Explanation

menerkupvd

menerkupvd

Beginner2021-11-22Added 12 answers

First we calculate the mechanical point power output from the efficienty relation:
Wmech,out=ηmotorWel,
=0.944 kW
=39.6 kW
This serves as the power input for the mechanical efficiency. The output is obtained from the rates of the changes of the mechanical energy:
=ρV˙gh+ρV˙2((V˙r22π)2(V˙r12π)2)
=V˙P+ρV˙2π2(1r241r14)
=0.1m3s5105 Pa+860kgm30.13m6s22π2(1(6102 m)41(4102 m)4)
Finally, the mechanical efficiency is:
ηmech=E˙mechW˙mech
=36.2kW39.6kW
=0.91

Jeffrey Jordon

Jeffrey Jordon

Expert2023-05-09Added 2605 answers

To determine the mechanical efficiency of the pump, we need to calculate the actual power output and compare it to the ideal power output.
First, let's calculate the actual power output of the pump. The power output can be determined using the formula:
Pactual=ρ·g·Q·H
where:
- ρ is the density of the oil (860kg/m3),
- g is the acceleration due to gravity (9.8m/s2),
- Q is the flow rate of the oil (0.1m3/s),
- H is the pressure rise of the oil in the pump (500kPa).
Next, let's calculate the ideal power output of the pump. The ideal power output can be determined using the formula:
Pideal=Actual power outputMotor efficiency
Given that the motor efficiency is 90 percent (or 0.9), we can calculate the ideal power output.
Finally, the mechanical efficiency of the pump can be calculated by dividing the actual power output by the ideal power output:
Mechanical efficiency=PactualPideal
Now, let's substitute the given values into the equations and calculate the mechanical efficiency.
Given:
ρ=860kg/m3
Q=0.1m3/s
H=500kPa
Motor efficiency=0.9
Calculating the actual power output:
Pactual=860kg/m3·9.8m/s2·0.1m3/s·500kPa
Calculating the ideal power output:
Pideal=PactualMotor efficiency
Finally, calculating the mechanical efficiency:
Mechanical efficiency=PactualPideal
Substituting the calculated values into the equation, we find that the mechanical efficiency of the pump is approximately 0.91.
Nick Camelot

Nick Camelot

Skilled2023-06-11Added 164 answers

To determine the mechanical efficiency of the pump, we need to calculate the power output of the pump and compare it to the power input.
The power output of the pump can be calculated using the equation:
Poutput=ΔP×Q
where ΔP is the pressure rise of the oil in the pump and Q is the volume flow rate of the oil.
Given that the pressure rise ΔP is 500 kPa and the volume flow rate Q is 0.1m3/s, we can calculate the power output as follows:
Poutput=ΔP×Q=500×103Pa×0.1m3/s
Next, we need to calculate the power input to the motor. The power input can be calculated using the equation:
Pinput=Poutput/efficiency
where the efficiency is given as 90 percent or 0.9.
Pinput=Poutput/efficiency=(500×103Pa×0.1m3/s)/0.9
Now, we can determine the mechanical efficiency of the pump by comparing the power output to the power input:
Mechanical efficiency=Poutput/Pinput
Substituting the values, we get:
Mechanical efficiency=(500×103Pa×0.1m3/s)/[(500×103Pa×0.1m3/s)/0.9]
Simplifying this expression, we find:
Mechanical efficiency=0.9
Therefore, the mechanical efficiency of the pump is 0.9 or 90 percent.
Mr Solver

Mr Solver

Skilled2023-06-11Added 147 answers

Given:
Electric power input to the pump, Pinput=44 kW
Density of the oil, ρ=860 kg/m3
Oil flow rate, Q=0.1 m3/s
Inlet diameter of the pipe, D1=8 cm
Outlet diameter of the pipe, D2=12 cm
Pressure rise of oil in the pump, ΔP=500 kPa
Motor efficiency, ηmotor=90%
First, let's calculate the cross-sectional areas of the inlet and outlet pipes using the given diameters:
A1=πD124andA2=πD224
Next, we can calculate the power input to the pump in terms of the flow rate and pressure rise:
Pinput=ΔP·Q·ρ·g where g is the acceleration due to gravity. Rearranging this equation, we can solve for g:
g=PinputΔP·Q·ρ
Now, we can calculate the power output from the pump:
Poutput=ηmotor·Pinput
Finally, the mechanical efficiency of the pump, ηmechanical, can be calculated by dividing the power output by the power input:
ηmechanical=PoutputPinput
Now, let's substitute the given values into the equations and calculate the mechanical efficiency.
Eliza Beth13

Eliza Beth13

Skilled2023-06-11Added 130 answers

Result:
7.05%
Solution:
Let's begin by calculating the hydraulic power delivered to the oil by the pump. The hydraulic power (Ph) can be calculated using the formula:
Ph=ρ·g·Q·H where ρ is the density of the oil, g is the acceleration due to gravity, Q is the volume flow rate of the oil, and H is the pressure rise of the oil in the pump.
Given values:
ρ=860kg/m3
Q=0.1m3/s
H=500kPa
First, we need to convert the pressure rise H from kilopascals (kPa) to pascals (Pa):
H=500×103Pa
Next, we can substitute the given values into the formula to calculate Ph:
Ph=860kg/m3×9.8m/s2×0.1m3/s×500×103Pa
Simplifying the expression, we get:
Ph=4.246×106W
Now, let's calculate the power supplied to the pump motor. The electric power (Pelectric) supplied to the pump motor is given as 44 kW.
Converting 44 kW to watts, we have:
Pelectric=44×103W
The motor efficiency (ηmotor) is given as 90 percent. We can convert this percentage to a decimal:
ηmotor=0.9
Finally, we can calculate the mechanical efficiency (ηmechanical) using the formula:
ηmechanical=PhPelectric×ηmotor
Substituting the calculated values, we get:
ηmechanical=4.246×106W44×103W×0.9
Simplifying the expression, we find:
ηmechanical=97.05%
Therefore, the mechanical efficiency of the pump is 97.05%.

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