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Water Pumped Vertically

Displays of water-and-light at hotels in Las Vegas involve quite complex pumping, piping and lighting systems. The sketch below right depicts the components (reservoir, submerged motor, pump housing and piping) of one such cosmetic water display. For our simple analysis we assume the instant electric power is supplied to the pump water passes through it at a constant volume rate, at a constant number of gallons per second. In addition we assume ideal operation, we assume there is no friction whatsoever.

Two aspects of the pumping event are its transient and and its steady operation. Inspecting the diagram we see that upon starting the pump will commence (at a constant volumetric rate of flow ) to fill the stand-pipe. As the level of water in the pipe rises, the power required by the pump with increase. This trend will continue until the water attains the height, H, whereupon it will begin to spill from the pipe. The operation after the water height is, H, is steady.

Once started and past its mechanical startup transient, water passes through the pump and into the vertical (effluent) pipe attached to the pump exit. It is obvious the power of the pump must increase as the height of water in the effluent pipe increases. The event is "steady" with a time-dependent exiting condition.

Calculate the transient and steady power required of the pump.
♦  We will analyse the pump by use of two distinct system representations. However the system is defined, our mass equation and energy equation apply. Both equations have a "rate" form.

Mass Equation:  The summation signs remind use to include all instances of "system, in or out."

1 (1)1

Terms of the mass equation can be written explicitly for use (as needed):


1 (2)2

Energy Equation:  The energy equation is a bit long to write. We write it out every time because it contains all terms that might apply.

3 (3)3

Equations 1,2 and 3 contain "all of the terms." In any application equation usage amounts to specification of the system perspective, open, closed or unsteady, then specialization by striking out the inapplicable terms.

schematic
Constant volume system is open
with an infinite, “free surface”
entrance and a “finite” exit.

Open System Perspective:  To the right is a sketch of our pumping system as the constant volume of water contained within an imaginary, dotted boundary. Water enters the system across its "in" boundary, at all places on the perimeter of a large circular annulus shown as a "dashed" ellipse. The pump (not shown) drives water from the space to an exit and "up the pipe." The pump supplies power. It is represented schematically as its power input, W-dot. We reduce the mass and energy equation terms "left to right" as they appear in the equation.

Their is a single mass of the system, (m), and it is constant. The system energies KE and PE relate to the motion and elevation of the center of mass of the system. The center of mass does not move, hence the derivatives of KE and of PE are zero. The left-of-equality terms are as follows:

(4) 4

The system has only one entering stream and one leaving stream. Further steps for the mass equation are:

(5)5

Using the reduced mass equation (above) we look at the first two terms (right-of-equality) of the energy equation.

6 (6) 6

The next term is the system work as a consequence of change of volume of the system. We use the mean value theorem on this term:

7 (7) 7

There is but one "shaft work" term and we know the work is "to the system." that is, positive."

8 (8) 8

The term "extrinsic work rate" is non-zero for systems that accelerate or experience elevation change of the center of mass. Our system does neither. And there are no "special" work mechanisms for this system.

9 (9) 9

By these steps we reduce the energy equation from its general form (3) to the equation below which applies to our system.

10 (10) 10
  1. Drop the summations because there is only a single species and a single stream.
  2. There is no "special" manner of work active, Wspecial.
  3. dKE and dPE of the system are zero.
  4. pein = 0 and pein =0.
  5. kein = 0
11 (11) 11
(12) 12
(13) 13
This inlet "construction" proceeds as follows (for simplicity - suppose the water to be flat and still):
  • imagine a very long fishing net. The net has floats along the top and weights along the bottom.
  • the net hangs in the water like a curtain.
  • the net is deployed in a great circle on the surface of the reservoir.
  • the net is somewhat strange since the vertical distance - floats above to weights below is only an inch or so.
  • the water enters the system moving along radial lines pointing toward the center of the circular area.
  • the perimeter of the circle is so large that the entering water velocity is essentially zero, hence kein = 0.
  • the surface of the water is designated at "zero elevation." Hence pein = 0, for the entering water.
14 (14) 14

Now, this goes into the Energy Equation as follows:

15 (15) 15

Closed System Perspective  To analyze the pumping from a closed perspective we divide the large mass of water into two parts. The larger part, II is essentially inactive. The mass equation for the system's two parts is:

16 (16)16

Obviously the level of II diminishes as water moves up the vertical pump. However, the vastness of II is such that the change of potential energy and kinetic energy of II is inconsequential.

17 (17)17
18 (18) 18
19 (19)19
20(20) 20

This PE and KE belong to the "part I" water in time as it proceeds up the pipe. There is no water in the pipe initially. The speed the surface of the water moves upward equals dh(t)/dt which equals the volume rate of the pump divided by the pipe area. The center of mass is located at h(t)/2. The result of this analysis is the same as the previous open system analysis.

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