Feet of head loss

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. ... 

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. 

Water is pumped from a large reservoir to a point 65 ft higher than the reservoir. How many feet of head must be added by the pump if 8000 lb,yh flows through a 6 in. pipe and the frictional head loss is 2 ft The density of the fluid is 62.4 Ib ft and the pump efficiency is 60%. Assume the kinetic energy correction factor equals 1. 

E = Head loss due to friction in feet of flowing fluid... 

Friction head loss develops as fluids flow through the various pipes, elbows, tees, vessel connections, valves, etc. These losses are expressed as loss of fluid static head in feet of fluid flowing. 

Determine expansion or contraction losses, if any, including tank or vessel entrance or exit losses from Figures 2-12A, 2-15, or 2-16. Convert units to psi, head loss in feet times 0.4331 = psi (for water), or adjust for Sp Gr of other liquids. 

Hf = loss of head in meters per 100 meters (feet per 100 ft) of effective pipe run,... 

Ah = Head loss in feet of flowing fluid Up = Velocity in the pipe leading to and from the exchanger, ft/sec UT = Velocity in the tubes N = Number of tube passes ... 

Lack of available NPSH may also be caused by high frictional loss in the suction piping. If this is the case, a small reduction in flow will not noticeably increase the pressure at the suction of the pump. A properly designed suction line to a centrifugal pump should have a frictional head loss of only a few feet of liquid. However, having a large-diameter suction line, and a relatively small draw-off nozzle, usually will lead to excessive loss of available NPSH. 

Pumping Head—The energy required to raise water to the distribution elevation and overcome friction losses through pipe, valves, fittings and nozzles. It is expressed in feet of liquid the pump must move and is equal to the total friction loss, static head and pressure drop through the distribution system. 

Second, the friction loss hL from pipe entrance to exit may also be expressed as foot-pounds per pound of flowing fluid. This term is more commonly referred to as the head loss in feet of fluid. It is loss, because the hydraulic grade line shown in Fig. 6.1 exhibits the hL loss. 

Compute the friction-head loss using the Hazen-WiUiams formula. The Hazen-Williams formula is hf = (u/1.318C/ ° 63) 85, where h j = friction-head loss per foot of pipe, in feet of water v = water velocity, in ft/s C = a constant depending on the condition and kind of pipe and Rh = hydraulic radius of pipe, in ft. 

Compute the net positive suction head on the pump. The net positive suction head hn on a pump when the liquid supply is above the pump inlet equals pressure on liquid surface + static suction head-friction-head loss in suction piping and pump inlet-vapor pressure of the liquid, all expressed in feet absolute of liquid handled. When the liquid supply is below the pump centerline—i.e., there is a static suction lift—the vertical distance of the lift is subtracted from the pressure on the liquid surface instead of added as in the preceding relation. 

The friction-head loss is 5 ft of water at maximum density. To convert to feet absolute, multiply by the ratio of water densities at 39.2°F and the operating temperature or (5)(62.4/62.0) = 5.03 ft. 

Compute the total pumping head. The total head, expressed in feet of water, equals static head + friction head + required nozzle head = 10 + 35 + 8(0.434) = 48.5 ft of water (145.0 kPa). A pump having a total head of at least 50 ft of water (15.2 m) would be chosen for this spray pond. If future expansion of the pond is anticipated, compute the probable total head required at a future date and choose a pump to deliver that head. Until the pond is expanded, the pump would operate with a throttled discharge. Normal nozzle inlet pressures range from about 6 to 10 lb/in2 (41.4 to 69.0 kPa). Higher pressures should not be used, because there will be excessive spray loss and rapid wear of the nozzles. 

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