To summarize what you have learned so far in this module, here is the simplified Bernouilli's equation:
TDH = P2 + V22 + (Z2 - Z1) + hf 2 g
Each of the four elements in this equation carries its own units of head or pressure. However, the total dynamic head of the pump must be expressed in "height of slurry".
We have already presented an example on the use of this equation on page 20. Solve the following exercise.
Determine the total dynamic head, TDH, of the slurry pumping system illustrated in the following Worksheet 1. This worksheet contains all the information you need to obtain your answer.
Give your final answer in "meters of slurry". Use this space and the blank page that follows the worksheet for your calculations.
Exercise
WORKSHEET
1
SYSTEM INFORMATION
Slurry SG =
Vol. flow rate of slurry = Pressure gauge reading =
PIPING SYSTEM INFORMATION Pipe nominal diameter =
Length of straight pipe = Valves and fittings:
1.60 205 m3/h
55 kPa
6 inches 50.0 m
• One square-mouth inlet
• One non-return valve
• Two 45º elbows
• One long-radius 90º elbow
The answer follows.
Exercise (continued)
Answer
TDH = 19.0 m slurry Solution
TDH = P2 + V22 + (Z2 - Z1) + hf 2 g
The static pressure equals 3.5 m slurry:
55 kPa x 0.102 m water = 5.6 m water kPa
5.6 m water = 3.5 m slurry 1.60
The velocity head equals 0.5 m slurry:
volumetric flow rate of 205 m3/h and a nominal pipe diameter of 6 inches.
Therefore the velocity head equals:
2.982 = 0.5 m slurry 2 x 9.81 m/sec2
The vertical lift equals 12.8 m slurry:
15.8 - 3.0 = 12.8 m slurry
From Table 2, V2 is approximately 2.98 m/sec based on a slurry
The friction loss equals 2.2 m slurry:
The equivalent length of pipe for the piping system is 67.3 meters:
Straight 6-inch pipe: 50.0 m
One square-mouth inlet: 4.9 m
One non-return valve: 6.1 m
f-factor equals 5.3 m water per 100 meters of pipe:
How did you do in this exercise?
• Well? Good work!
• Not so well? Study the solution carefully to make sure that you understand each step.
Move on to learn more about pumping systems!
Up to now, we have defined the boundaries of a pumping system as the level of slurry in the pump box and the point where the slurry is discharged to atmosphere or enters a set of hydrocyclones.
In fact, it doesn't matter where you decide to set the boundaries on each side of the pump: the total dynamic head of the pump will be the same regardless of the location of the boundaries.
Here is an example.
A slurry pumping system is illustrated in the following Worksheet 1.
In this system, there is one initial boundary (1); however, two terminal boundaries (2 and 3) have been identified so that we can calculate the total dynamic head of the pump in two different ways.
Example
WORKSHEET
1
SYSTEM INFORMATION
Slurry SG =
Vol. flow rate of slurry = Pressure gauge reading =
PIPING SYSTEM INFORMATION Pipe nominal diameter =
Length of straight pipe = Valves and fittings:
1.60 300 m3/h 95 kPa
6 inches 42.5 m
• One square-mouth inlet
• One non-return valve
• Two long-radius 90º elbows
We are going to calculate the TDH for this pump in two ways:
from point "1" to "3" and from "1" to "2".
From point "1" to "3":
Since the slurry is at atmospheric pressure at point "3", P
3 equals zero. Since the velocity of the slurry at point "3" is negligible, V
3 is also zero.
Here is the starting equation:
TDH = 0 + 0 + (17.8 - 4.1) + h
f
Let's solve for h
f. The piping system is as follows.
Straight 6-inch pipe: 42.5 m
f-factor equals 11.6 m water per 100 m pipe:
56.9 m pipe x 11.6 m water = 6.6 m water
From point "1" to "2":
Since there is a pressure gauge at point "2", P2 has a value. And since the slurry velocity at point "2" is significant, V
2 also has a value.
Here is the starting equation:
TDH = 95 kPa + V
2
2 + (10.7 - 4.1) + h
f
2 g
The static pressure is 6.1 m slurry:
95 kPa x 0.102 m water = 9.7 m water kPa
9.7 m water = 6.1 m slurry 1.60
The velocity head equals 1.0 m slurry:
From Table 2, V
2 is 4.46 m/sec based on a slurry volumetric flow rate of 300 m3/h and a nominal pipe diameter of 6 inches.
Therefore the velocity head equals:
4.462 = 1.0 m slurry 2 x 9.81 m/sec2
The vertical lift equals only 6.6 m slurry this time.
The friction loss, h
f, remains at 4.1 m slurry since there are no valves nor fittings between points "2" and "3".
Finally we have:
TDH = 6.1 + 1.0 + 6.6 + 4.1
As you can see, the total dynamic head of this pump is 17.8 m slurry no matter where the system boundaries are located. In fact, you can select any two practical boundary locations to calculate the TDH of a pumping system.
Solve the following exercise.
The pumping system for a closed-grinding circuit is shown in the following worksheet.
Exercise
WORKSHEET
1
SYSTEM INFORMATION
Slurry SG =
Vol. flow rate of slurry = Pressure gauge reading =
PIPING SYSTEM INFORMATION Pipe nominal diameter =
Length of straight pipe = Valves and fittings:
1.90 200 m3/h 100 kPa
5 inches 22.8 m
• One bell-mouth inlet
• One non-return valve
• Two regular 90º elbows
Exercise (continued)
Questions
a) What is the TDH for this pump from points "1" to "2"?
Exercise (continued)
b) What is the TDH of this pumping system from points "1" to "3"?
Hints: • Consider the hydrocyclone as a fitting which causes a pressure drop of 100 kPa. Convert this pressure as necessary.
• Consider the average elevation of the two discharge points of the hydrocyclone to be approximately equal to the elevation at the hydrocyclone inlet
• Consider the average slurry velocity at the
discharges of the hydrocyclones to be approximately (i.e., Z3 = Z2).
equal to that at the inlet (i.e., V
3 = V
2).
Answer
TDH for both questions (a) and (b) is 18.1 m slurry.
Solutions
a) Here is the starting equation:
TDH = 100 kPa + V 2
2 + (12.0 - 2.5) + h
f 2 g
The static pressure is 5.37 m slurry:
100 kPa x 0.102 m water = 10.2 m water kPa
10.2 m water = 5.4 m slurry 1.90
The velocity head equals 0.9 m slurry:
rate of 200 m3/h and a nominal pipe diameter of 5 inches.
Therefore the velocity head equals:
4.292 = 0.9 m slurry 2 x 9.81 m/sec2
The vertical lift equals 9.5 m slurry.
From Table 2, V
2 is 4.29 m/sec based on a slurry volumetric flow
The friction loss, hf , equals 2.3 m slurry:
f-factor equals 13.6 m water per 100 m pipe:
32.5 m pipe x 13.6 m water = 4.4 m water
b) Since the pressure at point "3" is atmospheric, P3 equals zero.
We have stated that V
3 is equal to V
This equals the value of hf in (a), i.e., 2.3 m slurry plus the equivalent height of slurry created by the added "fitting" which is the hydrocyclone between points "2" and "3". This fitting creates a pressure drop of 100 kPa (the pressure gauge reading at the hydrocyclone inlet):
100 kPa x 0.102 m water = 10.2 m water kPa
10.2 m water = 5.4 m slurry 1.90
The total hf is:
2.3 + 5.4 = 7.7 m slurry
Finally we have:
TDH = 0 + 0.9 + 9.5 + 7.7 TDH = 18.1 m slurry
Answer (continued)
The friction loss, h
f, from points "1" to "3" equals 7.70 m slurry.
Note
For multiple-diameter piping systems, you can consider the various pipe diameters in your calculations:
• To determine the value of V2, use the diameter of the pipe at the terminal boundary (point "2").
• To determine the total value of h
f, calculate h
f for each pipe diameter, including valves and fittings. Then simply add them up.
Those steps are only necessary if the presence of various pipe diameters is significant in relation to the entire piping system. For example, you don't have to do this if one out of twenty meters of pipe is six inches instead of five inches in diameter.
You have just learned how to determine the total dynamic head, TDH, for a particular pumping system which is transporting a specific volumetric flow rate of slurry. This defines one point on the "system capacity versus head" curve which is presented in the following section.
The total dynamic head of a system (and therefore of the pump in the system), corresponds to a specific volumetric flow rate of slurry, Q.
This TDH versus Q relationship represents a single operating point for this system. Look at Figure 3.
Figure 3. The operating point of the system.
In this figure, point "a" represents the total dynamic head of the system, 20.0 m slurry, at a slurry volumetric flow rate of 630 m3/h.