9. Pumps (compressors & turbines) Partly based on Chapter 10 of the De Nevers textbook.


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1 Lecture Notes CHE 31 Fluid Mechanics (Fall 010) 9. Pumps (compressors & turbines) Partly based on Chapter 10 of the De Nevers textbook. Basics (pressure head, efficiency, working point, stability) Pumps and compressors are generators of mechanical energy. They enter as a positive (source) term W ɺ in the Bernoulli equation. 1 p Wɺ + + = v gz e ρ φm (1) The terminology is roughly such that pumps involve liquid flow, and compressors involve gas flow. Turbines extract mechanical energy (e.g. for electric power generation) and are a sink term ( W ɺ < 0 ) in the Bernoulli equation (Eq. 1). In this part of the course we will limit ourselves to pumps and will not go into details regarding compressors and turbines. Figure 1 In Figure 1 the pump is a black box with mass coming in on one side and leaving at the other side. For a constant density fluid the Bernoulli equation om inlet to outlet reads 1 p 1 p Wɺ 1 v + gz + v1 gz1 = e. If the pump inlet and outlet are roughly at the ρ ρ φ m same level ( z 1 z ) and if the pipe leaving the pump has roughly the same diameter as the pipe entering (so that again at constant density a mass balance over the pump implies v1 v ) p p1 p Wɺ we get = e. ρ ρ ρ φm Pump efficiency η is defined as useful work (per unit mass) divided by actual work done p / ρ p φ e φm (per unit mass): η = = 1 (the latter equality being the result of Wɺ / φm Wɺ Wɺ applying the Bernoulli equation). It shows that nonideal efficiency is the result of iction in the pump. It also shows that p φ is the mechanical power (mechanical energy per unit time) being generated. Next to efficiency, pumps are characterized by graphs relating pressure increase versus volumetric flow rate (socalled pump curves). In the world of pumps pressure increase is p usually expressed as pressure head (with unit meters) h with ρ the density of the ρg pumped liquid. Figure 8 is an example of such a graph. The details in the graph will be discussed later.
2 The working point of a pumppipeline system is determined by the pump curve and the characteristics of the pipeline system it is placed in. In Figure (left panel) we see a common situation: a pump that delivers high pressure at relatively low flow rates. The pressure head decreases when the flow rate increases. The pipeline system usually has a no flow p, (e.g. if we pump liquid up om a lower to a higher level noflow p = hρ g with h the difference in levels), and a pressure drop due to iction. The latter is a function of the flow rate; the higher the flow rate, the more iction, and the more pressure drop due to iction. Closing or opening valves iluences the ictional loss and moves the working point over the pump curve (Figure, left). Specific forms of pump curves can lead to unstable behavior. In the left panel of Figure we have a stable situation: If (for some reason) the flow rate decreases, the pressure delivered by the pump increases, which (eventually) will bring back the flow rate to its original value. If the same thing happens with the pump curve in the right panel of Figure, the decreased flow rate will now reduce the pressure delivered by the pump, bringing back the flow rate even more until the whole system comes to a standstill. Figure Types of pumps Positive displacement (PD) pumps Figure 3 Our heart is a PD pump. The basic concept is displayed in Figure 3. A piston moves up and down in a cylinder. In the down stroke the springloaded inlet valve opens due to the vacuum created by the increasing volume above the piston; for the same reason the outlet valve closes.
3 Fluid enters the piston. The upwards stroke opens the outlet valve and closes the inlet valve and pushes the fluid out through the outlet. Also see the online animation at The volumetric flow rate can be written as φ = ApistonLstroke Ncycle. At specific motor speed N cycles/s, piston area A piston, and stroke length L stroke (and constant density liquids) the volumetric flow rate φ is virtually independent of the pressure increase p, see Figure 3 for the pump curve. The line is only slightly negatively inclined: if the pressure increases the valves (may) start leaking, reducing the flow rate. PD pumps have high efficiencies (as high as η = 0.9). PD pumps can act as vacuum pumps, i.e. they can create pressures below atmospheric pressure. If they would be able to create a true vacuum (i.e. p abs = 0, they would be able to patm lift a liquid with density ρ to a level h =. For a liquid as water with ρ =1000 kg/m 3, we ρ g will get h 10 m. An indicator for a pump s performance in creating vacuum is called suction lift; this is the level to which a specific pump can lift a specific liquid. Also see Figure 4. Figure 4 Figure 5 (reprinted om De Nevers, p 367)
4 Next to the basic design as shown in Figure 3 there is a large number of PD pump designs available. As an example I show the sliding vane PD pump in Figure 5. Centrifugal pumps Figure 6 (reprinted om De Nevers, p 37) In a centrifugal pump, liquid is accelerated by the action of an impeller. The mechanical energy gained by acceleration in the pump is (usually) converted into pressure at the pump s outlet (the diffuser ). Here we go through a stepbystep analysis of a centrifugal pump. Applying a Bernoulli equation om the inlet (point 1 in Figure 6) to the eye of the impeller (point ) yields ρ p p1 = ( v1 v ) ρe,1 (no level differences assumed). From point to 3 we take an approach based on solid body rotation : we assume that between and 3 the liquid spins with the same speed as the impeller. The latter has an angular velocity ω (unit: radians per second). Since we assume the liquid also rotates with ω, it feels a centrifugal force. At a radial location r (somewhere in between r and r 3 ) this force per unit volume is ρω r. This dp force is balanced by a radial pressure gradient: = ρω r (see the analogy with hydrostatic dr pressure: there we had a vertical gravity force per unit volume ρ g that was balanced by a hydrostatic pressure gradient dp = ρg ). If we integrate we can relate p 3 to p : dz p3 r3 p p = dp = ρω rdr = ρω r r. From point 3 to point 4 we again apply Bernoulli: ( ) 3 3 p r ρ p4 p3 = ( v4 v3 ) ρe,3 4. If we combine the equations om the stepbystep approach ρω ρ we get p4 p1 = ( r3 r ) + ( v3 v4 + v1 v ) e,1 e,3 4. If we realize that p4 p1 v = ωr and v3 = ωr3 and write the equation in terms of pressure head h = we get ρg 4 1 h r r ( ) p p ω 1 4,1,3 4 ( 3 ) v v e e + = = + ρg g g g ()
5 If the inlet and outlet tube of the centrifugal pump would have the same diameter, and the ( v v ) fluid would have constant density, the term 1 4 g would drop out. Usually, however, the ( v v ) inlet diameter is larger so that v 4 gets bigger than v 1 and the term 1 4 becomes negative, g and more negative when the flow rate increases. This helps in having stable operation of a pumppipeline system: the flow rate versus pump head curve (the pump curve) gets a negative slope which we showed benefited stability. e,1 + e,3 4 Friction losses (the term in Eq. ) are significant in centrifugal g pumps. A reduction of the head by a factor of due to ictional losses is no exception. Figure 7 Cavitation is a problem in centrifugal pumps. Cavitation is boiling of liquid flowing through the pump as a result of acceleration of the liquid. The vapor bubbles can damage pump rotors and pump housing, see Figure 7, left panel. Acceleration of the liquid implies a reduction in pressure (according to the Bernoulli equation) that can cause boiling (also see Figure 7 (right panel) that shows boiling in the low pressure centre of an impeller, and at the highspeed blade tips). Pump manufacturers provide there clients with a required Net Positive Suction Head (NPSH R ) which is the pressure required at the suction side of the pump to avoid cavitation. Pump practicalities In Figure 8 we show a typical pump curves as supplied by a manufacturer of centrifugal pumps. What sense does this graph make, and what terminology is in there? 1. Horizontal axis denotes flow rate; vertical axis pressure increase (in terms of head).. We see four pump curves with letters A, B, C, D; the size with the letters denotes the impeller diameter; the bigger the pump, the more head. 3. The dashdotted negatively sloped curves indicate power demand in horse power (1 HP = 746 W). 4. The drawn lines roughly perpendicular to the pump curves are isoefficiency curves. 5. At the top of the graph it says that the curve is for an impeller speed of 1750 rpm; and that the outlet has a diameter of 1.5 inch and the inlet a diameter of inch (1½ ). 6. The dashed lines show the NPSH R in feet; the higher the flow rate, the higher the velocities and thus the higher the chance of cavitation, and thus the higher NPSH needs to be.
6 Figure 8 (reprinted om De Nevers, p 375) Pipeline systems & pump curve interaction We now revisit one of the pipeline examples om LN06b. There we considered the system as given in Figure 9. Liquid needs to be pumped om a lower level to a higher level vessel. In the example in LN06b the question was what (pump) power would be needed to achieve a certain flow rate. We here make the problem a bit more complicated by adding a pump curve (see Figure p φ working point of this pipeline system is, 9 right panel). The question now is what the ( ) going to be. For this we need to set up a few Bernoulli equations between different points. The data needed to solve the problem are as follows (they slightly differ om the numbers in LN06b): The liquid is oil, µ=0.00 Pa s, ρ=980 kg/m 3. Total pipe length L=50 m, the inner diameter of all pipes D=5 cm, rough pipes with constant f= Levels: H 1 =10 m, H =5 m, H 3 =3 m. Loss factors: two sharp 90 o bends K 90 =0.74 (each); a sudden contraction at the exit of vessel A K A =0.5; an open gate valve K ov =0.13; a sudden expansion at the entrance of vessel B K B =1.0. Pump curve: p = p0 aφ with p0 = 1.55 bar, and a= Pa s m 6 (valid for 0 < φ < 0 litre/s).
7 Figure 9 Solution The working point is the point where the pump curve and the pipeline curve cross. The pump curve we know, the pipeline curve we need to figure out. The pipeline curve is the relation between flow rate and pressure difference over the pump as demanded by the pipeline system. In the figure we define four points: (1) At the surface of vessel A. v1 0 m/s, p 1 = 0 bar (pressure we take as gauge pressures), z1 = H1. () At the suction side of the pump. v and p are unknowns, z = H3. (3) At the pressure side of the pump. v 3 and p 3 are unknowns, z3 = H3. (4) At the surface of vessel B. v4 0 m/s, p 4 = 0 bar, z4 = H. 1 p K A Bernoulli equation between points 1 and : ( H3 H1) g + v + = e = v. If we ρ write this equation in terms of p it may make more sense: 1 1 p = ρ H H g ρv K ρv. The first term on the righthand side is the hydrostatic ( ) 1 3 A pressure. The flow reduces the hydrostatic pressure because part of the potential energy has been converted to kinetic energy ( nd term on the rhs), and because part is lost in iction (3 rd term on the rhs). Bernoulli equation between points 3 and 4: 1 p3 1 L ( H H3 ) g v3 = e = v3 K90 + Kov + KB + 4 f. Again we write this ρ D 1 1 L equation in terms of pressure: p3 = ρ ( H H3 ) g v3 v3 K90 Kov KB 4 f ρ + ρ D. Same structure as the Bernoulli equation between 1 and, but watch out for the signs: ρ H H g being reduced by kinetic energy effects is the same, but hydrostatic pressure ( ) 3 now iction has to be added. Makes sense: to pump oil through the pipe against iction we need pressure. We should at this point realize that since the pipeline has constant diameter, and the oil has constant density the velocity is the same everywhere: v = v3 v.
8 We now subtract the two Bernoulli equations: 3 ( 1) 1 L p p = ρ H H g + v K A K90 Kov KB 4 f ρ This is the pipeline curve D 4φ relating pumppressureincrease and flow rate we are looking for (since v = ). We write π D it as p = α + βφ. Substituting the numbers: α = 1.44 bar, β = Pa s m 6. The working point we find by demanding α + βφ = p aφ so that p0 α φ = = m 3 ρvd /s=1.8 litre/s. As a result v=0.9 m/s and Re = a + β µ indicating turbulent flow. 0
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