Lecture Notes CHE 312 Fluid Mechanics (Fall 2010) 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 nfWɺ in the Bernoulli equation.

21

2nf

frm

Wpv gz e

ρ φ ∆ + + = −

ɺ

(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 ( 0nfW <ɺ ) 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 from inlet to outlet reads

2 22 12 2 1 1

1 1

2 2nf

frm

Wp pv gz v gz e

ρ ρ φ+ + − − − = −

ɺ

. If the pump inlet and outlet are roughly at the

same level (1 2z 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 1 2v v≈ )

we get 2 1 nffr

m

Wp p pe

ρ ρ ρ φ∆− ≡ = −

ɺ

.

Pump efficiency η is defined as useful work (per unit mass) divided by actual work done

(per unit mass): /

1/

fr mV

nf m nf nf

epp

W W W

φφρηφ

∆ ⋅∆≡ = = −ɺ ɺ ɺ

(the latter equality being the result of

applying the Bernoulli equation). It shows that non-ideal efficiency is the result of friction in the pump. It also shows that Vp 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 (so-called pump curves). In the world of pumps pressure increase is

usually expressed as pressure head (with unit meters) p

hgρ

∆≡ with ρ the density of the

pumped liquid. Figure 8 is an example of such a graph. The details in the graph will be discussed later.

The working point of a pump-pipeline system is determined by the pump curve and the characteristics of the pipeline system it is placed in. In Figure 2 (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 from a lower to a higher level no-flow- p h gρ∆ = ∆ with h∆ the difference in levels), and a pressure drop due to friction. The latter is a function of the flow rate; the higher the flow rate, the more friction, and the more pressure drop due to friction. Closing or opening valves influences the frictional loss and moves the working point over the pump curve (Figure 2, left).

Specific forms of pump curves can lead to unstable behavior. In the left panel of Figure 2 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 2, 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 2

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 spring-loaded inlet valve opens due to the vacuum created by the increasing volume above the piston; for the same reason the outlet valve closes.

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 http://www.rpi.edu/dept/chem-eng/Biotech-Environ/PUMPS/reciprocating.html

The volumetric flow rate can be written as V piston stroke cycleA L Nφ = . At specific motor speed N

cycles/s, piston area Apiston, and stroke length Lstroke (and constant density liquids) the volumetric flow rate Vφ 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. 0absp = , they would be able to

lift a liquid with density ρ to a level atmph

gρ= . For a liquid as water with ρ =1000 kg/m3, we

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 from De Nevers, p 367)

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 from De Nevers, p 372) 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 step-by-step analysis of a centrifugal pump. Applying a Bernoulli equation from the inlet (point 1 in Figure 6) to the eye of the impeller (point 2) yields

( )2 22 1 1 2 ,1 22 frp p v v e

ρ ρ →− = − − (no level differences assumed). From point 2 to 3 we take an

approach based on ‘solid body rotation’: we assume that between 2 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 r2 and r3) this force per unit volume is 2rρω . This

force is balanced by a radial pressure gradient: 2dpr

drρω= (see the analogy with hydrostatic

pressure: there we had a vertical gravity force per unit volume gρ− that was balanced by a

hydrostatic pressure gradient dp

gdz

ρ= − ). If we integrate we can relate p3 to p2:

( )23 3

2 2 23 2 3 2

2 2 2

p r

p r

p p dp rdr r rρωρω− = = = −∫ ∫ . From point 3 to point 4 we again apply Bernoulli:

( )2 24 3 4 3 ,3 42 frp p v v e

ρ ρ →− = − − . If we combine the equations from the step-by-step approach

we get ( ) ( )2

2 2 2 2 2 24 1 3 2 3 4 1 2 ,1 2 ,3 42 2 fr frp p r r v v v v e e

ρω ρ→ →− = − + − + − − − . If we realize that

2 2 3 3andv r v rω ω= = and write the equation in terms of pressure head 4 1p ph

gρ−= we get

( ) ( )2 221 4 ,1 2 ,3 42 24 1

3 2 2fr fr

v v e ep ph r r

g g g g

ωρ

→ →− + −= = − + −

(2)

If the inlet and outlet tube of the centrifugal pump would have the same diameter, and the

fluid would have constant density, the term ( )2 2

1 4

2

v v

g

− would drop out. Usually, however, the

inlet diameter is larger so that v4 gets bigger than v1 and the term ( )2 2

1 4

2

v v

g

− becomes negative,

and more negative when the flow rate increases. This helps in having stable operation of a pump-pipeline system: the flow rate versus pump head curve (the pump curve) gets a negative slope which we showed benefited stability.

Friction losses (the ,1 2 ,3 4fr fre e

g→ →+

−

term in Eq. 2) are significant in centrifugal

pumps. A reduction of the head by a factor of 2 due to frictional 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 high-speed blade tips). Pump manufacturers provide there clients with a required Net Positive Suction Head (NPSHR) 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). 2. 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 dash-dotted negatively sloped curves indicate power demand in horse power (1 HP = 746 W). 4. The drawn lines roughly perpendicular to the pump curves are iso-efficiency 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 2 inch (1½×2). 6. The dashed lines show the NPSHR 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.

Figure 8 (reprinted from De Nevers, p 375) Pipeline systems & pump curve interaction We now revisit one of the pipeline examples from LN06b. There we considered the system as given in Figure 9. Liquid needs to be pumped from 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 9 right panel). The question now is what the ( ), Vp φ∆ working point of this pipeline system is

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 from the

numbers in LN06b): • The liquid is oil, µ=0.002 Pa·s, ρ=980 kg/m3. • Total pipe length L=50 m, the inner diameter of all pipes D=5 cm, rough pipes with

constant f= 0.006. • Levels: H1=10 m, H2=25 m, H3=3 m. • Loss factors: two sharp 90o bends K90=0.74 (each); a sudden contraction at the exit of

vessel A KA=0.5; an open gate valve Kov=0.13; a sudden expansion at the entrance of vessel B KB=1.0.

• Pump curve: 20 Vp p aφ∆ = ∆ − with 0p∆ = 1.55 bar, and a=1.9·107 Pa·s2·m-6 (valid for

0 Vφ< < 20 litre/s).

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. 1 0v ≈ m/s, 1 0p = bar (pressure we take as gauge

pressures), 1 1z H= .

(2) At the suction side of the pump. 2v and 2p are unknowns, 2 3z H= .

(3) At the pressure side of the pump. 3v and 3p are unknowns, 3 3z H= .

(4) At the surface of vessel B. 4 0v ≈ m/s, 4 0p = bar, 4 2z H= .

Bernoulli equation between points 1 and 2: ( ) 2 223 1 2 2

1

2 2A

fr

p KH H g v e v

ρ− + + = − = − . If we

write this equation in terms of 2p it may make more sense:

( ) 2 22 1 3 2 2

1 1

2 2Ap H H g v K vρ ρ ρ= − − − . The first term on the right-hand side is the hydrostatic

pressure. The flow reduces the hydrostatic pressure because part of the potential energy has been converted to kinetic energy (2nd term on the rhs), and because part is lost in friction (3rd term on the rhs). Bernoulli equation between points 3 and 4:

( ) 2 232 3 3 3 90

1 12 4

2 2fr ov B

p LH H g v e v K K K f

Dρ − − − = − = − + + +

. Again we write this

equation in terms of pressure: ( ) 2 23 2 3 3 3 90

1 12 4

2 2 ov B

Lp H H g v v K K K f

Dρ ρ ρ = − − + + + +

.

Same structure as the Bernoulli equation between 1 and 2, but watch out for the signs: hydrostatic pressure ( )2 3H H gρ − being reduced by kinetic energy effects is the same, but

now friction has to be added. Makes sense: to pump oil through the pipe against friction 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: 2 3v v v= ≡ .

We now subtract the two Bernoulli equations:

( ) 23 2 2 1 90

12 4

2 A ov B

Lp p H H g v K K K K f

Dρ ρ − = − + + + + +

. This is the pipeline curve

relating pump-pressure-increase and flow rate we are looking for (since 2

4 VvD

φπ

= ). We write

it as 2Vp α βφ∆ = + . Substituting the numbers: α = 1.44 bar, β = 3.4·109 Pa·s2·m-6.

The working point we find by demanding 2 20V Vp aα βφ φ+ = ∆ − so that

0V

p

a

αφβ

∆ −=+

=1.8·10-3 m3/s=1.8 litre/s. As a result v=0.92 m/s and RevDρµ

= ≈ 2.2·104

indicating turbulent flow.