a filtration system continuously removes water from a swimming pool, passes the water through filters, and then returns it to the pool. both pipes are located near the surface of the water. the flow rate is 15 gallons per minute. the water entering the pump is at 0 psig, and the water leaving the pump is at 10 psig.

Answers

Answer 1

It has been given that:

15 gallons per minute are the water's volumetric flow rate.

The pipe that exits the pump has a 1 inch diameter.

(a) When the diameter of the pipe that leaves the pump is 1 inch, flow work is done by the water as it leaves the pump and enters the pipe will be

W= 7133.2ft.lbf / min

Here, measuring the amount of work flow performed by the water as it exits the pump and enters the pipe is the goal.

The relation can be used to express the work flow that is supposedly complete:

W = P × V

where;

Pressure is P.

Volume is V.

The gauge pressure is included in the specified outlet pressure.

Now, the relation may be used to express the pressure in the pump P:

[tex]P_{absolute}[/tex] = [tex]P_{gauge}[/tex] + [tex]P_{atmospheric}[/tex]

[tex]P_{absolute}[/tex] = 10 psig + 14.7 psig

[tex]P_{absolute}[/tex]  = 24.7 psig

W =  P × V

W =  24.7 psig  ×  15 gal/min

W = (24.7 psig * [tex]\frac{\frac{lb_{f} }{in^{2} } }{psig}[/tex] ) * (15 [tex]\frac{gal}{min}[/tex] * [tex]\frac{0.1337 ft^{3} }{1 gal}[/tex] * [tex]\frac{144 in^{2} }{1 ft^{2} }[/tex] )

W = 7133.2 [tex]\frac{ft. lb_{f} }{min}[/tex]

This means that the water is stated to produce work at a rate of flow of

W= 7133.2 [tex]\frac{ft.lb_{f} }{min}[/tex]

(b) Work done by the water as it leaves the pipe and enters the pool is:

According to the obtained data:

Through a 1.5-inch-diameter aperture, the pool's water is refilled.

when there is 1 atm of pressure.

The water then exerts the following amount of energy when it exits the pipe and enters the pool:

W = P × V

W equals 1 atm x 15 gal/min.

W = 1 atm * (15[tex]\frac{gal}{min}[/tex] * [tex]\frac{0.1337 ft^{3} }{1 gal}[/tex] * [tex]\frac{144 in^{2} }{1 ft^{2} }[/tex])

W = 4245.24 [tex]\frac{ft.lb_{f} }{min}[/tex]

When a result, the water's work rate as it exits the pipe and enters the pool is at W = 4245.24 [tex]\frac{ft.lb_{f} }{min}[/tex]

(c) Due to the fact that the data provided for the flow rate and pressure do not account for changes in time in the space between the pump and the pool, we may say that the system is in a steady state.

However, the main reason why the system is out of equilibrium is that:

As indicated in the inquiry, the pressure coming out of the pipe is different from the pressure of the water at the pool's surface.

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Answers

Answer;

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Explanation:

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