SALMO 49 : CANTATA,
V. BIBLIOGRAFIA
3.5 Centrifugal pump laws
(b) By projecting a vertical line upwards from the point of design flow (1.5 l/s) on your graph it intersects the pump characteristic at the final operating point of the pump on the system. Refer again to Figure 3.2. The final location of the system characteristic can now be sketched on your graph by hand. Two horizontal lines can now be drawn back to the pressure axis of your graph and the required regulating pd read from the pressure scale. This should come to
72.5-50=22.5 kPa
(c) From your graph, which should now look similar to Figure 3.2, the final operating point of the pump on the system after regulation is 1.5 l/s at 72.5 kPa.
Note: The vertical line from the design flow on your graph and in Figure 3.2 intersects two points: the original system characteristic at a pressure of 50 kPa, and the pump curve at 72.5 kPa. The first intersection is at the design pressure drop.
Example 3.2
A pump having the following characteristic is to be employed in a system whose design conditions are 3.5 l/s at 24 kPa. Determine whether or not two identical pumps operating together are required and state whether a series or parallel arrangement should be adopted. Assess the pressure reduction required to achieve design flow.
Pump characteristic:
P 0 10 20 30 40 50 60 kPa
Q 3.0 2.85 2.6 2.28 1.85 1.3 01/s
Solution
For the series arrangement the pressure developed is doubled. Refer to Figure 3.3:
2P 0 20 40 60 80 100 120 kPa
Table 3.1
For the parallel arrangement the flow rate is doubled. Refer to Figure 3.4:
2Q 6.0 5.7 5.2 4.56 3.7 2.6 0 l/s
By analysing the options before plotting the pump characteristics it is clear that:
1. a single pump is not big enough, as the maximum flow of 3.0 l/s is achieved at zero pressure;
2. two pumps simultaneously operating in series do not increase the flow above 3.0 l/s;
3. two pumps operating simultaneously in parallel achieve 3.7 l/s at 40 kPa, which is in excess of the design conditions and is therefore worthy of investigation.
The solution therefore is obtained from the two pumps operating simultaneously in parallel.
After obtaining further values for the system design condition of 3.5 l/s at 24 kPa using K=P1/(Q1)2=47, the system and parallel pump characteristics are plotted, and from the graph the regulation required to achieve design flow is
42-24=18 kPa
It is recommended that you now plot the characteristics and check the above solution. What is the final operating point of the pump on the system? You should have 3.5 l/s at 42 kPa.
Example 3.3
Consider the system shown in Figure 3.9(a). Design flow rate is 2.5 l/s and the system and pump characteristics are given in the data below.
Analyse the pump pressure distribution around the system and determine whether or not there will be discharge at the open vent.
If the regulating valve is relocated to the pump discharge will the system operate satisfactorily?
Data
Hydraulic resistances around the system:
Section A–B B–C C–D D–E E–F F–A
Resistance (kPa) 2 2 8 2 2 3
Pump characteristic:
P 50 48 43.5 32.5 23.5 0 kPa
Q 0 1 2 3 3.5 4 l/s
Solution
The sum of the hydraulic resistances around the system identifies the net pump pressure required, as the single circuit forms the index run. Thus the system design conditions are 2.5 l/s at 19 kPa. Now look at the pump characteristic data and you will see that the system design conditions fall between the values in bold type, so the proposed pump is clearly big enough.
From the system design conditions other values can be calculated as shown in Example 3.1, and the system and pump characteristics can now be plotted and system regulation considered as shown in Figure 3.2.
The operating point of the pump on the system is 3.2 l/s at 30 kPa. By regulating back to the design flow of 2.5 l/s the pump pressure developed is 39 kPa and the regulation required is 20 kPa.
Knowing the pump pressure under operating conditions we can now plot the pump pressure effects around the system as shown in Figure 3.9b.
Pump pressure at the pump outlet will be positive, and the pump pressure effects around the system will therefore be positive up to the neutral point at the cold feed entry, where it must change to a negative pump effect.
Remember, at the neutral point pump pressure changes from positive to negative or vice versa. From the data the hydraulic resistance from point E to point F is 2 kPa; this will have a negative value and thus at the pump inlet the pump pressure is -2 kPa. The pressure rise through the pump is 39 kPa: thus the discharge pump pressure will be +37 kPa. By using the other section resistances from the data the remaining pump pressure effects can now be plotted around the system.
Make sure you agree with the analysis in Figure 3.9b.
Now we need to consider what happens at the open vent. If water does not flow in the open vent pipe the residual pump pressure effect at point A will be Figure 3.9 Example 3.3.
the same at the original water level in the open vent pipe, namely +34 kPa. This is equivalent to a rise in water level of h=P/ρg metres. Thus
Clearly water will flow in the open vent pipe as it is only 1.6 m above the water level and some of the residual pump pressure here will be absorbed.
However, it would be foolhardy to leave the system as it is: the likelihood of flow in the vent pipe is overwhelming, as the height of the pipe above the water level in the F&E tank is only 1.6 m. If water flow takes place, circulation to the system will be reduced and the tank room will fill with vapour.
How would you rectify the situation? There are two or three solutions.
One of them is indicated in the question, for if the regulating valve is relocated to the pump discharge, section F–A now has a hydraulic resistance of 23 kPa and the pump effect at point A and hence at the water level in the open vent is reduced to 14 kPa, from which
As the height of the vent above the water level in the tank is 1.6 m the system should operate satisfactorily.
Can you suggest other solutions to the problem of pumping over?
Example 3.4
A diagram of a system is shown in elevation in Figure 3.10(a). The design flow rate is 2.5 l/s and the system and pump characteristics are given in the data.
(a) Determine the operating point of the pump on the system.
(b) What is the pressure loss required across the regulating valve to achieve design flow?
(c) Analyse the pump pressure distribution around the system and determine whether or not air will be drawn in through the open vent at point A.
Data
System hydraulic resistances:
Section A–B B–C C–D D–E E–F F–A
Resistance (kPa) 2 2 8 2 2 3
Pump characteristic:
P 50 48 43.5 32.5 23.5 0 kPa
Q 0 1 2 3 3.5 4 l/s
Solution
You will see that the system and pump characteristics are similar to those used in Example 3.3. The solution to parts (a) and (b) are therefore similar:
(a) The operating point of the pump on the system is 3.2 l/s at 30 kPa.
(b) The regulating pressure drop is 20 kPa to achieve design flow of 2.5 l/s.
The pump duty is now 2.5 l/s at 39 kPa.
(c) Here the similarity ends, as the location of the cold feed is now at the pump discharge instead of the pump inlet. This requires another analysis of the pump effects around the system.
As the pump pressure changes from positive to negative at the neutral point F, pump discharge pressure will be +2 kPa, which from the data is the hydraulic resistance in pipe section E–F. For a pressure rise through the pump of 39 kPa the inlet pump pressure must be -37 kPa. The pump pressure effects can now be plotted anticlockwise from pump inlet E around the system or clockwise from the cold feed entry at point F, using the hydraulic resistances for each pipe section from the data.
The results are shown in Figure 3.10(b).
Using a similar argument as in Example 3.3 the negative pump effect at point A is transferred to the original water level in the open vent pipe under no-flow conditions.
We are reminded that the reason for this is that under no flow conditions there is no loss in pump pressure. Thus the negative pump effect in the vent is 23 kPa, which Figure 3.10 Example 3.4.
is equivalent to
As the F&E tank is only 2 m above point A, air will be drawn into the system at this point, and furthermore the pump will not operate in a stable manner.
Example 3.5
The diagram in Figure 3.11(a) shows a simple LTHW heating system in elevation. From the data determine:
(a) the operating point of the pump on the system;
(b) the presure loss required across the regulating valve to achieve design flow;
(c) the antiflash margin at point E in the system if water temperature at this point is 86 °C.
Data
System characteristic:
Section A–B B–C C–D D–E E–F F–A
Hydraulic resistance (kPa) 5 10 10 20 20 5 System design flow: 4.0 l/s
Pump characteristic:
P 120 114 105 95 83 67 45 0 kPa
Q 0 1 2 3 4 5 6 7 l/s
Steam tables data:
Absolute pressure (kPa) 110 100 80 70 60
Saturation temperature (°C) 102 100 93 90 86 Take atmospheric pressure as 100 kPa.
Solution
By summing up the hydraulic resistances in the index circuit, index pressure drop is 70 kPa at 4.0 l/s. Before proceeding further, look at the pump characteristic in the data, and you will see that the nearest duty is 83 kPa at 4.0 l/s. Clearly this pump should be suitable for the system. It also offers a flow rate that matches the system design flow. It is therefore necessary only to plot the pump and system characteristics to solve part (a) of the question.
(a) Adopting the appropriately adapted pump law (P2=K×Q22), a series of
Figure 3.11 Example 3.5.
system pressure and flow readings are obtained and the system characteristic plotted along with the pump characteristic.
You should undertake this piece of work now.
From the graph, which should be similar to Figure 3.2, the operating point of the pump on the system is 4.25 l/s at 80 kPa.
(b) By drawing a vertical line on your graph through the design flow of 4.0 l/s, the required pressure drop across the regulating valve from the graph and from the system design conditions is 83-70=13 kPa.
(c) This solution requires an analysis of the pump pressure distribution around the system, which is shown in Figure 3.11(b).
The initial step commences at the neutral point, where the pump pressure effect passes through zero. Pump discharge pressure must therefore be +18 kPa to overcome the hydraulic resistance of pipe section A–B and that required of the regulating valve. For a pressure rise through the pump of 83 kPa, the inlet pump pressure must be -65 kPa. With this knowledge the pump pressure distribution around the system can now be found with the aid of the system section resistances.
At point E the algebraic sum of pump and static pressure is (+10+-40) kPa=-30 kPa gauge.
Clearly this is a case where subatmospheric pressures exist in this high-level pipe when the pump runs.
As absolute pressure Pabs=gauge Pg+atmospheric Pat, then absolute pressure will be -30 +100=70 kPa.
From the data in the question relating to the steam tables, the saturation temperature of water at 70 kPa abs is 90 °C. As the water temperature at point E is 86 °C, the antiflash margin is 90-86= 4 K.
This margin is too small; a minimum margin of 10 K is required, as water becomes unstable at 5 °C below its boiling point.
Therefore unless the configuration of the pump and cold feed are changed the maximum water temperature sustainable at point E will be 90-10=80 °C.
Note: Although the examples used here relate to vented systems so that pumping over and air ingress can be identified, the same procedure can be adopted for pressurized systems. In such cases the neutral point is where the cold feed and expansion pipe from the pressurization unit enter the system.
For pressurized systems, potential problems at the open vent do not arise.
However, the effects of negative pump pressure around the system should be investigated to ensure that cavitation does not occur when the system is run up to operating temperature with the pump running.
The ideal location for the pump is in the flow as shown in Figure 3.7(b).
Do you agree?
This completes the work on pump and system. You now have the necessary skills to analyse the effects of different configurations of cold feed and pump locations in closed systems and to select the best position for the pump at the design stage.
You are able to offer solutions to some of the problems associated with existing installations, such as continual air in the system, erratic operation of the pump, considerable noise in high-level pipes or from within the boiler (cavitation), discharge from the open vent, and air in high-level pipes at the beginning of the heating season.
Finally, you are now in a position to recommend a suitable pump type for a given application.
3.6 Chapter closure