An oil pump is drawing 44 kW of electric power while pumping oil with ρ=860 kg / m^3 at a rate o

An oil pump is drawing 44 kW of electric power while pumping...

Key Concepts

Pressure and Flow Rate Relationship

The relationship between pressure rise and flow rate is essential for assessing the performance of pumps. The product of the pressure increase and the volumetric flow rate gives the hydraulic power delivered to the fluid. This relationship helps in interpreting how changes in these parameters affect the overall energy imparted to the fluid and the efficiency of the pumping process.

Hydraulic Power

Hydraulic power is the rate at which work is done on a fluid by a pump, calculated as the product of the pressure increase imparted to the fluid and the volumetric flow rate. This concept is fundamental in fluid mechanics, as it quantifies the energy transferred to the fluid, which is then available for performing tasks like moving or lifting.

Energy Transfer in Fluid Systems

The study of energy transfer in fluid systems involves understanding how mechanical energy is converted into hydraulic energy in pumps and other fluid transport devices. This concept uses conservation of energy principles to analyze how input energy is partitioned between useful work done on the fluid and the losses incurred during the process.

Mechanical Efficiency

Mechanical efficiency in pump systems is the ratio of the useful hydraulic power delivered to the fluid to the mechanical power supplied to the pump. This efficiency measures how effectively the pump converts the mechanical energy into fluid work, accounting for losses due to friction, leakage, and other detrimental effects in the pumping mechanism.

Motor Efficiency

Motor efficiency refers to the effectiveness with which an electric motor converts electrical energy into mechanical energy. It is defined as the ratio of the motor's mechanical output power to its electrical input power, and it plays a crucial role in determining the overall performance of a system that includes both a motor and a pump.

Transcript

00:01 In this problem, we're determining the mechanical efficiency for an oil pump.

00:07 So what we have here is a pump motor system, and we can consider it either as the two things working together as a whole system, or we can also look at it as the motor and then the pump.

00:21 So we'll have an input of electricity into the motor, lose some power through the motor, give a mechanical input to the pump, again, some energy losses, and then we'll get an output for our pump.

00:36 So what we'll do here is we'll use our definitions of efficiency and mechanical energy to go from our electrical input to our motor, to our power out for our pump.

00:52 And we're given a number of data points here.

00:57 Here so our electrical power into our motor is 44 kilowatts our volumetric flow rate of our oil is 0 .1 cubic meter per second our density of oil is 860 kilograms per cubic meter, the diameter of our output of our pump is 12 centimeter, and the diameter of the input, the inlet is 8 centimeters.

01:55 The pressure change from the inlet to the outlet is 500 kilopascals, and the efficiency of our motor is 0 .9.

02:15 So we're given the efficiency of the motor and we want to find the efficiency of the pump part of the system.

02:24 So looking at the pump, we can define the efficiency aeta pump as our change in power that we get out.

02:40 Over the power going into the and so that power going into the pump w dot derivative of work we can say is the motor it's due to the motor so efficiency of the motor times the motor input work of the motor in so that's the electrical input and so we could say the power going into the pump is the same thing as the power going out of the motor.

03:29 Now we want to figure out what this change, this mechanical energy term is.

03:37 And so our change in energy is going to be due to potential and kinetic.

03:51 So potential and kinetic energy potential mghh with mass rate over time and kinetic mv squared over two and mass varying over time we'll have a v2 and a v1 so we've got we don't quite have everything we need here we're not sure how to write this height or the volumes at the inlet and the outlet.

04:40 But we can figure out ways to convert the information we have.

04:45 So i'll call this equation for mechanical energy equation one.

04:51 And we can say that velocity, small b, would be equal to volumetric flow rate, capital b, divided by area.

05:02 Alright, so in this pipe, it's going to have a cross -sectional area, and it's going to have a flow rate, volumetric flow rate.

05:12 So the volumetric flow rate doesn't change, but as the cross -sectional area changes, the velocity is going to change.

05:20 And we can say that area is pi r squared.

05:27 And so that'll be equation two to find the velocities in the pipe.

05:34 Terms of things we know.

05:36 And we can also say that our m .g .h can be rewritten, we can rewrite mass as density to volumetric flow rate, and we know both of those, gh.

05:56 Okay, so we've got v volumetric flow rate and row gh, and so row gh is going to be our pressure differential...

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