Feet of head

The above graphie shows three identieal pumps, eaeh designed to develop 92.4 feet of head. When they pump liquids of different speeifie gravities, the heads remain the same, but the pressures vary in proportion to the speeifie gravity. 

A pump capable of generating 125 feet of head would provide the... 

To express the quantity of energt available in the lit]iiid entering into the pump, the unit of measure for NPSH is feet of head or elevation in the pump suction. The pump has its NPSHr, or Net Positive Suction Head Required. The system, meaning all pipe, tanks and connections on the suction side of the pump has the NPSHa, or the Net Positive Suction Head Available. There should always be more NPSHa in the system dian the NPSHr of the pump. Let s look at them, beginning with what the pump recgiires ... 

Where ATM = the atmospheric pre.ssure at the elevation of the installation expressed in feet of head. 

Pgs = the suction pressure gauge reading taken at the pump centerline and converted into feet of head. 

Hvp = the vapor pressure of the fluid expressed in feet of head. The vapor pressure is tied to the fluid temperature. 

To understand pumps and analyze their problems, its neee,ssary to dominate the formula that changes feet of head (H) into psi. This is explained in Chapter 2, but here is a brief review ... 

The pump companies develop their curves using head in feet (H), because when they make a new pump, they don t know the ultimate service of the pump (they don t know the liquid that the pump will be pumping), but they do know how many feet of elevation the pump can raise that liquid. This is why it s necessary to specify pumps in feet of head and not in psi. Let s begin by exploring the H-Q curve of the pump, using feet of head. 

At point A on the H-Q curve, the pump is pumping Q gpm (gallons per minute), at H feet of head. This point on the curve corresponds to the best efficiency, and it is also seen at approximately the middle of the energ) curve, and al.so on the NPSHr curve where it begins its sharp rise. 

Provide roughly 50-75 feet of head or a little higher. 

Note that psi (pounds per square inch) is pressure on the system and is not expressed as absolute unless the system is under absolute pressure. Feet are expressed as head, not head absolute or gauge (see later example). Note the conversion of psi pressure to feet of head pressure. 

If a pump were required to deliver 50 psig to a system, for water, the feet of head on the pump curve must read,... 

Enter capacity at 125 GPM, follow vertically to 86 feet of head, then to right to Mscosity of 27.8 centistokes, and up to correction factors ... 

Casing Gives direction to the flow from the impeller and converts this velocity energy into pressure energy which is usually measured in feet of head. 

The problem at Big Spring was rather typical. The two parallel, second-stage jets were not working as a team. Jet A in Fig. 16.5 was a real strong worker. Jet B was a loafer. It is rather like running two centrifugal pumps in parallel. Unless both pumps can develop about the same feet of head, the strong pump takes all the flow, and the weak... 

Centrifugal pumps are dynamic machines, which means that they convert velocity into feet of head. 

This idea of the conservation of energy is at the heart of any process plant. In the case of the Chicago Water Treatment Plant, the reduced velocity of the water was converted into feet of head. That is, the elevation of the water in the sump suddenly increased, and blew the manhole covers off the top of the sump. 

The mass of water in our pipeline weighed 160 X106 lb. This much water, moving at 8 ft/s, represents a tremendous amount of energy (about 500 million Btu per hour). If the flow of water is cut in half, then the momentum of the water flowing in the pipeline is also cut in half. This energy cannot simply disappear. It has to go somewhere. The energy is converted to an increase of feet of head in the sump that is, the water level in the sump jumps up, and blows the manhole covers off the top of the sump. 

Let us imagine that the six pumps in Fig. 23.1 have not run for a few days. The water level in the sump and the level of Lake Michigan will be the same. I now start all six pumps at the same time. An hour later, the water level in the sump is 12 ft below the water level in the lake. This 12 ft is called feet of head loss. If the pipeline is 3 mi long, we say we have lost 4 ft of head per mile of pipeline. ... 

The loss of 12 ft of head as the water flows through the pipeline is due to friction that is, 12 feet of head is converted to heat. But why do we have a temporary loss of an extra 3 ft The answer lies in the concept of acceleration. 

But what is the only source of energy that the lake possesses Answer—elevation or potential energy. To accelerate the water in the pipeline to 8 ft/s requires more energy than to keep the water flowing at that same velocity. And this extra energy comes from the 3 ft of elevation difference between the water level in the sump and the water level in the lake. Once the water has reached its steady-state velocity of 8 ft/s, the need for this extra conversion of feet of head to acceleration disappears, and the water level in the sump rises to within 12 ft of the lake s level. 

Volute—converts the velocity imparted to the liquid by the impeller, to feet of head... 

The impeller is the working part of a centrifugal pump. The function of the impeller is to increase the velocity or kinetic energy of the liquid. The liquid flows into the impeller, and leaves the impeller, at the same pressure. The black dot shown at the top of the impeller in Fig. 23.6 is called the vane tip. The pressure at the vane tip is the same as the pump s suction pressure. However, as the high-velocity liquid escapes from the impeller and flows into the volute, its velocity decreases. The volute (which is also called the diffuser) is shaped like a cone. It widens out in the manner illustrated in Fig. 23.7. As the liquid flows into the wider section of the volute, its velocity is reduced, and the lost velocity is converted—well, not into pressure, but into feet of head. 

A centrifugal pump develops the same feet of head, regardless of the density of the liquid pumped, as long as the flow is constant. This statement is valid as long as the viscosity of the liquid is below 40 cp or 200 SSU (Saybolt Seconds Universal). But, as process operators or engineers, we are not interested in feet of head. We are interested only in pressure. Differential pressure is related to differential feet of head as follows ... 

Answer The power demand will go down to 60 Amperes. Amp(ere)s are a form of electrical work. The units of work are foot-pounds. The feet of head developed by the pump is not affected by the specific gravity of the liquid. But the weight of liquid pumped is proportional to the specific gravity. If the specific gravity drops by 40 percent, and the liquid volume (GPM) stays constant, then the pounds lifted by the pump, drops by 40 percent and so does the electrical work. 

Answer The same elevation. Centrifugal pumps develop the same feet of head at a given volumetric flow rate, regardless of the specific gravity of the liquid pumped. This means the ability of the pump to push liquid uphill is the same, even if the density of the liquid changes. 

The feet of head developed by a pump is affected by the volume of liquid pumped. Figure 23.10 is a typical pump curve. As the flow of liquid from a centrifugal pump increases, the feet of head developed by the pump goes down, as does the pump discharge pressure. 

On the steep part of the pump curve, the flow increases slowly, while the head drops off rapidly. Hence, the product of GPM times feet of head remains the same, or goes down. 

To summarize, the specific gravity of the liquid has increased by 10 percent. The feet of head of liquid has decreased by 10 percent. Since ... 

AP remains constant. But the lower required feet of head permits the pump to operate further on its performance curve. This produces more volumetric (GPM) flow. It also requires more work to pump the greater flow. So we have to have sufficient excess amperage on the motor driver, to accommodate this maneuver. But that is the price we pay for expanding the pump s capacity, without increasing the impeller diameter. Of course, the larger-diameter impeller might still require more motor amperage. 

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