The industrial blower sucks in air through the axial opening A with a velocity v1 and discharges

The industrial blower sucks in air through the axial opening...

Key Concepts

Power Consumption and Efficiency

This concept relates to how much mechanical power is drawn by devices like motors and fans, and how efficiently this power is converted into useful work such as moving air. Evaluating efficiency is crucial since additional losses (like friction and turbulence) often require that extra power be supplied beyond the ideal based on energy conversion calculations.

Kinetic Energy of Fluids

The kinetic energy imparted to a fluid is a function of its velocity and mass. When a fan or pump operates, mechanical energy is converted into kinetic energy, which accelerates the fluid. Understanding this energy conversion is vital for calculating the power required to achieve a specific flow velocity in fluid handling applications.

Dynamic Pressure

In fluid dynamics, dynamic pressure reflects the kinetic energy per unit volume of the fluid and is proportional to the square of the velocity. This idea is key when analyzing systems where a velocity change occurs, as it represents the pressure drop associated with the conversion between pressure energy and kinetic energy. It plays a critical role in determining the driving forces in engineered fluid systems.

Continuity Equation

This fundamental principle in fluid mechanics states that the mass flow rate must remain constant from one cross-section of a flow system to another. In cases of constant density fluids such as air under many engineering conditions, the product of cross-sectional area and velocity is conserved. This concept is essential when relating the inlet and outlet flow properties in any system involving fluid transport.

Volumetric Flow Rate

This is the volume of fluid passing through a given surface per unit time. It is a critical parameter in determining the performance of equipment such as fans and pumps, and it ties directly to the velocity and cross-sectional area through which the fluid flows. Accurately characterizing this rate is fundamental to the design and analysis of fluid systems.

Transcript

00:01 So here on this problem we are given this industrial blower as you can see in this figure and it sucks air through the axial opening at a with velocity v1 and discharge the air at velocity v2 in the atmospheric pressure at the atmospheric pressure now the diameter of the section at point b we are given to be so at this section basically this diameter d we are given to be 150 mm now the the flow rate for the air queue we are given to be 16 meter cube per minute and the rpm for the motor in we are given to be as 3450 revolution per minute and the power consumed by the blower at no load condition is given to be 0 .3 .3 to kilowatt this is the power of the motor at no load condition now we have to calculate the power consumed by the pump when it deliver this much of air now let the motor is rotating in this direction like this and the moment would be m this is the moment now we know that the resultant moment of all external forces about any fixed point let it be this point or off the system it equals to the time rate of angular momentum that is we can write that sigma m0 is equals to sigma hg dot that is the rate of change of angular momentum so in this case m would be equals to m dash v2 d2 minus zero now here m dash is the master rate for the air and mass for the air and mass for it for air m dash can be calculated as this is small m so it would be equals to density times q in density we are for the air we can take it to be 1 .2 06 kg per meter cube and the q we are given as 16 meter cube per minute and per minute we can convert the units into meter queue per second so we're divided with 60 so we'll get this value to be 0 .322 kg per second now next we can calculate this velocity v2 so velocity v2 will be equals to q over a where a is the cross -sectional area of this dimension of this cross -section so q again we have to convert it into meter cube per minute 60 over 16 over 60 so this would be meter cube per second divide by the area no area we have pi by 4 d square so we have to take the diameter in the meter so it would be 0 .15 meter and it's square so we'll get this value to be 15 .906 sorry 15 .09 meter per second now we have a v2 we have a d2 this will be this distance would be d2 so we have everything we can plug it here to calculate m so m will be equals to m dash we have 0 .322 times v2 we have 15 .09 times d2 we have 200 m m means is 0 .2 meter so from here we'll get the value of a moment to be nearly 0 .0...

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