equation simply becomes
Note that TDHfullsd has been changed to TDHsetup. This equation demonstrates that the above setup of the unit is a convenient arrangement for determining the TDH. As shown in this equation, TDH can simply be calculated using the pressure gage reading and a measured velocity at point 2.
As depicted in Figure 4.8, the performance of this particular pump has been characterized in terms of total developed head on the ordinate and discharge on the abscissa. The other characteristics are the parameters rpm, power, and efficiency. To illustrate how this chart was developed, consider one curve: when the curve for the 1,170 rpm was developed, the setup in Figure 4.7 was adjusted to 1,170 rpm and the discharge was varied from 0, the shut-off flow, up to the abscissa value depicted in Figure 4.8. The reading of the pressure gage was then taken. This reading converted to head, along with the velocity head obtained from the flow meter reading and the cross-sectional of the discharge pipe, gives the TDH. This was repeated for the other rpm's as well as for the powers consumed (which are indicated in kW).
The relationship of discharge versus total developed head at the shut-off flow cannot be developed for the positive displacement pump operating under a cylinder, without breaking the cylinder head or the cylinder, itself. For pumps operating under a cylinder, the element that pushes the fluid is either the piston or the plunger. This piston or plunger moves inside the cylinder and pushes the fluid inside into the cylinder head located at the end of the forward travel. This pushing creates a tremendous amount of pressure that can rupture the cylinder head or the cylinder body itself, if the piston or plunger has not given way first. In the case of centrifugal pumps, this situation would not be a problem since the fluid will just be churned inside the impeller casing, and testing at shut-off flow is possible.
The activity inside the pump volute incurs several losses: first is the backflow of the flow that had already been acted upon but is slipping back into the suction eye of the impeller or, in general, toward the suction side of the pump. Because energy had already been expended on this flow but failed to exit into the discharge, this backflow represents a loss. The other loss is the turbulence induced as the impeller acts on the flow and swirls it around. Turbulence is a loss of energy. As the impeller rotates, its tips and sides shear off the fluid; this also causes what is called disk friction and is a loss of energy. All these losses cause the inefficiency of the pump; hlp is these losses.
During the testing, the power to the motor or prime mover driving the pump is recorded. Multiplying this input power by the prime mover efficiency produces the brake or shaft power. In the figure, the powers are measured in terms of kilowatts (or kW). Call the head corresponding to the brake power as hbrake. The brake efficiency of a pump is defined as the ratio of TDH to the brake input power to the pump. Therefore, brake efficiency n is n = ™ (="4+^) (4.17)
Y 2g as far as the arrangement in Figure 4.7 is concerned. With this equation substituted into Equation (4.17), the efficiency during a trial run can be determined. This new equation for efficiency is n = -L V+ p) (4.19)
hbrakeV Y 2g)
As shown in the figure, along a certain curve there are several values of efficiencies determined. Among these efficiencies is one that is the highest of all. This particular value of the efficiency corresponds to the best operating performance of the pump; hence, this point is called the best operating efficiency. For example, for the characteristic curve determined at a brake kilowatt input of 40 kW, the best operating efficiency is approximately 67%. This corresponds to a TDH of approximately 16 m and approximately a discharge of 0.18 m /s. To operate this pump, its discharge should be set at 0.18 m /s to take advantage of the best operating efficiency. In practice, the operating performance is normally set anywhere from 60 to 120% of the best operating efficiency.
Note that the brake power has been given in terms of its head equivalent hbrake. To obtain hbrake from a given brake power expressed in horsepower, hp, use the equivalent that hp = 745.7 N- m/s. If Q is the rate of flow in m3/s, and yis the specific weight in N/m3, then hbrake in meters is hbrake = 74lr (4.20)
Example 4.2 Pump characteristics curves are developed in accordance with the setup of Figure 4.7. The pressure at the outlet of the pump is found to be 196
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