Construction of the head-flow and power-flow plots
These plots can be generated at any scale with any fluid. Measuring the flow, the fluid elevations difference, the power, and the density of the fluid or dispersion are required. It is easiest in the lab, because everything is small and it is easy to make changes [2]. Flow can be measured with flow meters, calibrated weirs or by simply collecting and weighing water during a known time interval on small-scale. Head can be measured by simply measuring the elevation difference of the fluid surface between the pumper box and the separation troughs after the settler. This is easier done on pilot-scale and needs more care on full-scale installations. The friction factor in the pipes must be negligible for this simple method to work. If they are not, your system is not operating at peak performance (see discussion of hydraulic efficiency later on). The power can be measured using torque cells and meters on small-scale. On large-scale plants the power should be measured at the motor. This is often displayed on the variable drive controls. Subtract from this value the product of the motor efficiency, gearbox efficiency (about 98%) and the maximum motor power listed on the motor nameplate. This isn’t perfect, but the impeller power will be close.
Data compiled in this way should yield overlapping plots if everything is geometrically similar. If they are not, here are some things to check.
- The false bottom that has the orifice can bow upward due to the enormous suction generated by the pumper if the plate is not designed strong enough. This can narrow the measured gap between the pumper blades and the false bottom. This increases the suction and increases the head. In most cases this will also increase the flow and the power consumption.
- Line losses are significant. Piping diameters are too small. Friction losses are not negligible. Valves in the piping are not fully open. This restricts the flow and increases the head. This is a waste of hydraulic efficiency.
- Density of the fluid or average density of the dispersion was not determined correctly.
- Auxiliary impellers are not of the same design on both scales. Some impeller designs add to the head requirements. Optimized auxiliary impellers subtract from the overall head requirements.
- Power readings on large scale are inaccurate because the system is operating far from its maximum load and the motor/gearbox efficiency compensation mentioned above is not adequate. Ask for motor efficiency curves for motors when they are purchased and use the actual motor efficiency at partial motor loadings.
Examples
on how to use these graphs
Example
1 (pilot-scale): A 50/50 solution of organic and aqueous has a dispersion density of 1000 kg/m3. A 6” R300 is rotating at 400 RPM. DO/D=0.33. The measured flow rate is 2.35 Liters/second (37.2 GPM). What are Nq, Np, and Nh, and the power and head that is developed?
First determine Nq from Equation 1. Since Nq is dimensionless, keep track of the dimensions of Q, N,
and D. Using SI-units, Q=0.00235 m3/s, N=400/60=6.67 s-1, and D=0.152 m, Nq=0.1004. From Figure 3, Np(Nq=0.1)=1.3. Using Equation 2, PPumper=31.3 Watts. From Figure 5, Nh(Nq=0.1)=0.45. Using Equation 3, the developed head is H=0.23 m (9.2”). The tip speed is 3.2 m/s
(626 FPM). If D/T=0.5, then T=0.305m (12”). Assuming a square box and the liquid depth, Z=0.8T, the
volume of the pump stage above the orifice plate is V=22.7 liters (6 gallons). Thus, P/V=1.38 kW/m
3 (6.9 Hp/1000 gallons). The mean residence time in the pumper stage is qRes
= V/QPumper=9.7 seconds.
Example
2 (full-scale): A 72” R300 is running at the same Nq-value and a tip speed maximum of 5.1 m/s (1000 FPM). How much would it pump, how much power would it consume, and how much head would it develop?
The tip speed requirement dictates that the maximum pumper speed is N=0.884 s-1 (53
RPM). For Nq=0.1, Equation 1 shows Q=0.5108 m3/s (8096 GPM). From Figure 3, Np(Nq=0.1)=1.3
and Equation 2, P=18.38 kW (24.6 Hp). Assuming a 95% efficient motor and a motor load of 80% of
maximum, the motor and gearbox must be designed to handle 24.2 kW (32.4 Hp). Based on Figure 5,
Nh(Nq=0.1)=0.45. From Equation 3, H=0.5966 m (23.5”). Once again, if D/T=0.5, then T=3.66m (144”), and if Z/T=0.8, then the pumper volume is V=39.2 m3=39,200 L (10356 gallons). P/V=0.469 kW/m3 (2.34 Hp/1000 gallons). The mean residence time in the pumper stage is qRes
= V/QPumper=76 seconds.
Discussion of Examples 1 and 2: Example 2 is almost an exact geometrical scale-up of Example 1,
with the exception that the large-scale pumper could not exceed a tip speed criterion. The
differences highlight why pilot plants cannot look and operate like full-scale SX plants. Whereas the
residence time on small-scale is just under 10s, the residence time on large-scale is greater than
a minute. Depending on the organic ligand used, 10s might just be too short for mass transfer. Since the volume of each auxiliary stage is usually only slightly larger than the pumper stage, the total
small-scale residence time may be only 40 seconds with 2 auxiliary stages. The large scale plant,
probably only needs one auxiliary stage so that the total residence time is about 3 minutes. Furthermore,
the P/V on small-scale is almost 3 times greater than full-scale. As I will show later, the small-scale unit may be entraining air, whereas the large-scale is probably seeing phase separation.
The design of an SX plant depends on many questions that determine the location of the operating point on the head-flow and power-flow lines. What is the expected head of the SX-circuit? What is the design flow rate or metal production rate? What is the residence time required for efficient mass transfer? Are there plans to increase production in the future? What are the maximum and minimum P/V conditions? These questions need to be contemplated.
Example 3:Suppose the developed head of H=23.5” in Example 2 is considered not enough?
There are only three ways to increase the head: increase Nh, increase the tip speed, and/or change
to a pumper design that has a higher head-flow curve. To increase Nh for this R300, means that Nq must be decreased. If the tip speed
is kept constant (N and D are unchanged), reducing Nq to 0.075 will increase the Nh to 0.65,
thus increasing the head by 44% or to 33.9”. The trade off is that there will only be 75% as much
flow or only 6072 GPM and the residence time will increase to 105 seconds. The power will also go down
by a ratio of 1.1/1.3.
Example
4 :Suppose the head in Example 3 is desired, but the flow rate in Example 2 must be maintained?
To increase the head and keep the flow constant at the same tip speed, N and D need to be changed.
This means a new impeller and probably a new gearbox/motor. Since Q is proportional to ND3 (Equation 1), TS is proportional to ND, and Q must increase by 1/0.75=1.333, the new diameter must be 1.3330.5 times the previous diameter or 72*1.155=83”. The impeller speed will be reduced to 46 RPM
to maintain the tip speed requirement. Np is now reduced to about 1.15, but since the power is proportionalto D5, the new power will be 21.6 kW=29 Hp. Following
Example 2, the motor and gearbox would have to be designed to handle 28 kW or 38 Hp.
Example 5:Here is one last consideration for the four examples above. Suppose the plant decides to increase production by 50%. What needs to be done to the existing pumper installation of Example 4?
50% more metal production means 50% more flow through the SX circuit. The flow in Example 4 is Q=0.511
m3/s (8096 GPM). The flow for the new production demand is now Q=0.767 m3/s
(12144 GPM). According to Mike Nees of Bechtel, the developed head will be approximately constant at
the operating conditions, when only changing N [7]. Increasing N will increase the flow rate, so tip
speed will increase. This means that Nh must decrease. On Figure 5, a decrease in Nh means an increase
in Nq and an increase in Np. An iterative approach to the answer begins: try a new speed, determine the decrease in Nh, determine the new Nq and see if these conditions result in the higher flow rate.
Iteration 1: Increase the pumper speed from 46 RPM to 53 RPM. H=0.8611 m (33.9”). TS=5.85 m/s. From Equation 3, Nh=0.49. Using Figure 5, Nq=0.09 and using Equation 1, Q=0.745 m3/s (11808 GPM). This is a flow increase of 46% and almost the desired increase. Iteration 2: Trying 54 RPM results in TS=5.96 m/s (1173 fpm), Nh=0.475, Nq=0.095, and Q=0.801 m3/s (12698 GPM). Figure 2 indicates that Np has increased to about 1.25 and P=37.9 kW (50.8 Hp). If a 60 Hp motor/gearbox had been installed originally, this throughput increase would have been accomplished by merely increasing the pumper speed. If the original gearbox/motor were not designed for this service, a retrofit or a new, more power efficient pumper would be required. P/V would increase from 0.469 kW/m3 (2.34 Hp/1000 gallons) in Example 2 to 0.97 kW/m3 (4.83 Hp/1000 gallons). The mean residence time decreases to 51 seconds.
These examples show how important it is to do some careful planning prior to installation and how to use the power-flow and hear-flow plots to design a pumper stage.
Hydraulic Efficiency
|
