Pumps and compressors Pumps and compressors Sub-chapters

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Pumps and compressors
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Pumps and compressors

• Sub-chapters
• 9.1. Positive-displacement pumps
• 9.2. Centrifugal pumps
• 9.3. Positive-displacement compressors
• 9.4. Rotary compressors
• 9.5. Compressor efficiency

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• In Chapters 4, 5, and 6, we have written energy
  balance equations which
• involve a dWa,o term (see Sec. 4.8 for a
  definition of dWa.o).
• For steady-flow this term generally represents
  the action of a pump, fan, blower, compressor
  turbine, etc.
• This chapter discusses the fluid mechanics of the
  devices which actually perform that dWa,o

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POSITIVE-DISPLACEMENT PUMPS

• Pumps work on liquid and compressors work on a
  gas.
• Most mechanical pumps are one of these
• Positive-displacement
• Centrifugal
• Special designs with characteristics intermediate
  between the two
• In addition, there are nonmechanical pumps (i.e.,
  electromagnetic, ion, diffusion,jet, etc.), which
  are not considered here.

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• Positive-displacement (PD) pumps work by allowing
  a fluid to flow into enclosed cavity from a
  low-pressure source, trapping the fluid, and then
  forcing it out into a high-pressure receiver by
  decreasing the volume of the cavity.
• Examples are the fuel and oil pumps on
• most automobiles, the pumps on most hydraulic
  systems, and the hearts
• most animals.
• Figure 9.1 shows the cross-sectional view of a
  simple PD pump

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• The operating cycle of such a pump is as
• follows, starting with the piston at the top
• The piston starts downward, creating a slight
  vacuum in the cylinder.
• The pressure of the fluid in the inlet line is
  high enough relative to this vacuum to force open
  the left-hand valve, whose spring has been
  designed to let the valve open under this slight
  pressure difference.
• Fluid flows in during the entire downward
  movement of the piston.

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• The piston reaches the bottom of its stroke and
  starts upward. This raises the pressure in the
  cylinder gt the pressure in the inlet line, so the
  inlet valve is pulled shut by its spring.
• When the pressure in the cylinder gt the pressure
  in the outlet line, the outlet valve is forced
  open.
• The piston pushes the fluid out into the outlet
  line.
• The piston starts downward again the spring
  closes the outlet valve, because the pressure in
  the cylinder has fallen, and the cycle begins
  again

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• Suppose that we test such a pump, using a pump
  test stand, as shown in Fig. 9.2.
• With the pump discharge pressure regulator (a
  control valve) we can regulate the discharge
  pressure and, using the bucket, scale, and clock,
  determine the flow rate corresponding to that
  pressure.
• For a given speed of the pump's motor, the
  results for various discharge pressures are shown
  in Fig 9.3.

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• From Fig. 9.3 we see that PD pumps are
  practically constant-volumetric flow-rate devices
  (at a fixed motor speed) and that they can
  generate large pressures. The danger that these
  large pressures will break something is so severe
  that these pumps must always have some kind of
  safety valve to relieve the pressure if a line is
  accidentally blocked.
• For a perfect PD pump and an absolutely
  incompressible fluid, the volumetric flow rate
  equals the volume swept out per unit time by the
  piston,
• Volumetric flow rate piston area x piston
  travel x cycles/time (9.1)

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• For an actual pump the flow rate will be slightly
  less because of various fluid leakages.
• If we write Bernoulli's equation (Eq. 5.7) from
  the inlet of this pump to the outlet and solve
  for the work input to the pump, we find
• (9.2)
• The 1st term on the right represents the "useful"
  work done by the pump increasing the pressure,
  elevation, or velocity of the
• The 2nd represents the "useless" work done in
  heating either the fluid the surroundings.

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• The normal definition of pump efficiency is
  (9.3)
• This gives ? in terms of a unit mass of fluid
  passing through the pump.
• It is often convenient to multiply the top and
  bottom of this equation by the mass flow rate ,
  which makes the denominator exactly equal to the
  power supplied to the pump
• (9.4)

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• Example 9.1. A pump is pumping 50gal/min of water
  from a pressure of 30psia to a pressure of
  100psia. The changes in elevation and velocity
  are negligible. The motor which drives the pump
  is supplying 2.80 hp. What is the efficiency of
  the pump?
• The mass flow rate through the pump is

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• so, from Eq. 9.4,
• From this calculation we see that the numerator
  in Eq. 9.4 has the dimension of horsepower. This
  numerator is often referred to as the hydraulic
  horsepower of the pump.

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• ? becomes equal to the hydraulic horsepower
  divided by the total horsepower supplied to the
  pump.
• For large PD pumps, ? can be as high as 0.90 for
  small pumps it is less.
• One may show (Prob. 9.4) that for the pump in
  this example the energy which was converted to
  friction heating and thereby heated the fluid
  would cause a negligible temperature rise. The
  same is not true of gas compressors, as discussed
  in Sec. 9.3.

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• If we connect our PD pump to a sump, as shown in
  Fig. 9.4, and start the motor, what will happen?
  A PD pump is generally operable as a vacuum pump.
  Therefore, the pump will create a vacuum in the
  inlet line. This will make the fluid rise in the
  inlet line.
• If we write the head form of Bernoulli's
  equation, Eq. 5.11, between the free surface of
  the fluid (point 1) and the inside of the pump
  cylinder, there is no pump work over this
  section so

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• . (9.5)
• If, as shown in Fig. 9.4, the fluid tank is open
  to atmosphere, then P1 Patm. The maximum value
  of h possible corresponds to P2 0. If there is
  no friction and the velocity at 2 is negligible,
  then
• . (9.6)
• For water under normal Patm and Troom, hmax ? 34
  ft 10 m. This height is called the suction lift.

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• The actual suction lift obtainable with a PD
• pump lt that shown Eq. 9.6 because
• There is always some line friction, some friction
  effect through the pump inlet valve, and some
  inlet velocity.
• The pressure on the liquid cannot be reduced to
  zero with causing the liquid to boil. All liquids
  have some finite vapor pressure. For water at
  room temperature, it is about 0.3 psia or 0.02
  atm. If the pressure lt 0.02 atm, the liquid will
  boil.

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• Example 9.2.
• We wish to pump 200 gal/ min of water at 150oF
  from a sump. We have a PD pump which can reduce
  absolute pressure in its cylinder to 1 psia. We
  have an F/g (for the pipe only) of 4 ft. The
  friction effect in the valve may be considered
  the same as that of a sudden expansion (see
  Sec.5.5) with the inlet velocity equal to the
  fluid flow velocity through the valve, which is
  10 ft/s. The atmospheric pressure at this
  location gt 14.5 psia.
• What is the maximum elevation above the water
  level in the sump at which we can place the pump
  inlet?

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• Answer
• The lowest pressure we can allow in the cylinder
  P is 3.72 psia, the vapor pressure of water at
  150oF. If the pressure lt 3.72 psia, the water
  would boil, interrupting the flow. The density of
  water at 150oF 61.3 lbm/ft3.
• Thus
• 19.8 ft 6.04 m

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CENTRIFUGAL PUMPS

• A centrifugal pump raises the pressure of a
  liquid by giving it a high kinetic energy and
  then converting that kinetic energy to injection
  work. The water pump on most automobiles is a
  typical centrifugal pump.
• As shown in Fig. 9.5. it consists of an impeller
  (i.e., a wheel with blades) and some form of
  housing with a central inlet and a peripheral
  outlet.
• The fluid flows in the central inlet into the
  "eye" of the impeller, is spun outward by the
  rotating impeller, and flows out through the
  peripheral outlet.

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• To analyze such a pump on a very simplified
  basis, use Bernoulli's equation (Eq. 5.7) between
  the inlet pipe (point 1) and the outer tips of
  the impeller blades (point 2)
• The elevation change and V1 ? negligible. We
  assume that the friction losses also are
  negligible.
• Although P2 gt P1, this term is small compared
  with the change in kinetic energy
• (9.7)

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• This equation indicates that the pump work, which
  enters through the rotating shaft, principally
  increases the kinetic energy of the fluid as it
  flows across the impeller from the eye to the
  tips of the blades.
• The tangential velocity at any point is
• tangential velocity radius x angular
• velocity (9.8)
• The angular velocity (2? rpm) is constant for the
  whole impeller, but the radius ? from 0 at the
  eye to a significant value at the tip of the
  blades. Max tangential vel. V2

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• Use Bernoulli's equation from the tip of the
  blades (point 2) to the outlet pipe (point 3).
• The change in elevation is negligible, and
  friction is neglected. No work on the fluid
  between points 2 and 3. V3 ? negligible.
  Thus (9.9)
• This equation indicates that the section of the
  pump from the tip of the rotor blades to the
  outlet pipe converts the kinetic energy of the
  fluid to increased pressure

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• Thus, the centrifugal pump may be considered a
  two-stage device
• 1. The impeller increases the kinetic energy of
  the fluid at practically constant pressure.
• 2. The diffuser converts this kinetic energy to
  increased pressure.
• Equations 9.7 and 9.9 suggest that for a given
  pump size and speed, ?P/(?g) should be constant.
• Figure 9.6 shows the results of such a test for a
  large, high-efficiency pump. As predicted by the
  simple model, for low flow rates, ?P/(?g) ?
  f(flowrate)

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• Example 9.3. A typical centrifugal pump runs at
  1800rpm (mainly because of the convenient
  availability of 1800rpm electric motors). If the
  fluid being pumped is water, what is the maximum
  pressure difference across the pump for impeller
  diameters of 1, 3, and 10 in?
• Using Eq 9.9
• For impeller 1 in

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• 0.41 psi 2.86 kPa
• For 3-in dia impeller
• For 10-in dia impeller

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• This example illustrates the fact that
  centrifugal pumps supply small ?P with small
  impellers and high ?P with large impellers.
• When a large pressure rise is required, we can
  obtain it by hooking several centrifugal pumps in
  series (head to tail).
• The normal practice is to put several impellers
  on a common shaft and to design a casing so that
  the outlet from one impeller is fed through a
  diffuser directly to the inlet of the next
  impeller. This is particularly true of deep-well
  centrifugal irrigation pump.

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• The performance curve of a real pump,
• shown on Fig. 9.6, indicates some the
• limitations of our simple model
• As the flow rate is large, the ?P decreases,
  which is not predicted by the model.
• The model would indicate ? 100 for all flow
  rates, whereas the actual ? varies over a wide
  range, peaking near 90 at the design operating
  range.

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• As in Example 9.3, we let the pump turn at 1800
  rpm and have an impeller of 10-in diameter. If
  the pump is full of water, the pump has a ?P of
  42 psi so there is no problem with the suction
  lift.
• When we start the pump, the fluid around the
  impeller is air. Therefore, from Eq. 9.9 we find
  ?P (V1 is kept constant due to tip design) is

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• If this pump is discharging into the atm and the
  sump is open to atm, then the pump, when full of
  air, can lift the water only a height of
• To get a centrifugal pump going, one must replace
  the air in the system with liquid. This is called
  priming.
• Numerous schemes for performing this function are
  available

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• Centrifugal pumps are often used to pump boiling
  liquids, e.g., at the bottom of many distillation
  columns.
• Between inlet pipe (point 1) and the point on the
  blades of impeller where the pump starts to
  increase pressure (point 1a)
• . (9.10)
• P1a falls and boiling may occur. V1a?, P1a?
• In this case the pump must be located far enough
  below the boiling surface so that the ?P due to
  gravity from the boiling surface to the pump eye
  gt -?P in the pump eye.

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• This elevation is shown as h in Fig. 9.7. This
  distance required below the boiling surface is
  called the net positive suction head (NPSH).

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The pressure in Fig. 9.8 is the pressure measured
at the pump inlet. If there is significant
frictional -?P between the vessel and the pump,
then h in Fig. 9.7 must be increased to overcome
this additional -?P.
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POSITIVE-DISPLACEMENT COMPRESSORS

• A compressor has Poutlet/Pinlet gtgt 1.
• If ?P is small, the device is called a blower or
  fan. Blowers and fans work practically the way as
  centrifugal pumps, and their behavior can be
  readily predicted by equations developed for
  centrifugal pumps.
• To compress a gas to a final (absolute) pressure
  gt 1.1 times its inlet pressure requires a
  compressor, and the change in density of the gas
  must be taken into account.

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• A PD compressor has the same general form as a PD
  pump. The operating sequence is the same as that
  described in Sec. 9.1. The differences are in the
  size and speed of the various parts (recall Eq
  9.9).
• The pressure-volume history of the gas in the
  cylinder of such a compressor is shown
• in Fig. 9-9.
• Cycle ABCD

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• The work of any single-piston process is given by
• . (9.11)
• The work done by the compressor on the gas is the
  gross work done on the gas (under curve CDA)
  minus the work done by the gas on the piston as
  the flowed in (area under curve BC) thus, the
  net work is the area enclosed by curve ABCD. This
  is the work done on the gas.
• Compressors are often used to compress gases
  which can be reasonably well represented by the
  perfect-gas law PV nRT.

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• If a compressor works slowly enough and has good
  cooling facilities, then the gas in the cylinder
  will be at practically a constant temp throughout
  the entire compression process. Then we may
  substitute nRT/P for V in Eq. 9.12 and integrate
• . (9.13)

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• However, in most compressors the piston moves too
  rapidly for the gas to be cooled much by the
  cylinder walls.
• If so, the gas will undergo what is practically a
  reversible, adiabatic process, i.e., an
  isentropic process.
• .PVk P1V1k constant adiabatic (9.14)
• Inserting this in Eq. 9.11
• .
• . (9.15)

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• Example 9.4. A 100 efficient compressor is
  required to compress from 1 to 3 atm. The inlet
  temperature is 68oF. Calculate the work
  pound-mole for an isothermal compressor and an
  adiabatic compressor.
• For an isothermal compressor,
• For an adiabatic compressor

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• Equations 9.13 and 9.15 in Example 9.4 indicate
  that the required work for an adiabatic
  compressor is always gt that for an isothermal
  compressor with the same inlet and outlet
  pressures.
• Therefore, it is advantageous to try to make real
  compressors as nearly isothermal as possible.
• One way to do this is to cool the cylinders of
  the compressor, and this is generally done with
  cooling jackets or cooling fins on compressors.
• Another way is by staging and intercooling see
  Example 9.5

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• Example 9.5. Air is to be compressed from 1 to 10
  atm. The inlet temp is 68oF. What is the work per
  mole for (a) an isothermal compressor, (b) an
  adiabatic compressor, and (c) a two-stage
  adiabatic compressor in which the gas is
  compressed adiabatically to 3 atm, then cooled to
  68oF, and then compressed from 3 to 10 atm?
• For an isothermal compressor,

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• For an adiabatic compressor,
• For a two-stage adiabatic compressor,
• 2862 Btu/lbmol6.66 kJ/mol.

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• This example illustrates the advantage of staging
  and intercooling.
• With an infinite number of stages with
  intercooling, an adiabatic compressor would have
  the same performance as an isothermal compressor
  (Prob. 9.21).
• Thus, the behavior of an isothermal compressor
  represents the best performance obtainable by
  staging.
• The optimum number of stages is found by an
  economic balance between the extra cost of each
  additional stage and the improved performance as
  the number of stages is increased.
• Large PD compressors typically have stage
  Pout/Pin of 3-5 and ? of 75- 85 (Sec. 9.5)

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ROTARY COMPRESSORS

• The PD compressor has been a common industrial
  tool. But, it is a complicated, heavy, expensive,
  low-flow-rate device.
• The need to supercharge aircraft reciprocating
  engines and the development of turbojet
  gas-turbine engines demanded the development of
  lightweight, efficient, low-cost, high-flow-rate
  compressors.
• The resulting compressors, which were developed
  for aircraft service, are now being applied
  industrially in high-capacity applications, for
  example, in ammonia plants.

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• The two types of compressor developed for
  aircraft engines are centrifugal and axial-flow.
• The centrifugal compressor is a centrifugal pump
  with very high-speed (for example, 20,000rpm) and
  large-diameter rotor. To give high-pressure
  ratios, centrifugal compressors are normally
  staged with intercooling the pressure rise per
  stage is small.
• Axial-flow compressors pass the gas between
  numerous rows of blades arranged in an annulus
  (Fig. 9.10). The gas is successively accelerated
  by a moving row of rotor blades and then slowed
  by a stator blade which converts the kinetic
  energy imparted by the rotor blades to pressure

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• The advantages of the axial-flow compressor over
  centrifugal compressors are the small
  cross-sectional area perpendicular to gas flow,
  which makes it easy to build into a streamlined
  airplane, and the lower velocities, which lead to
  lower friction losses and slightly higher ?.
• Centrifugal and axial-flow compressors generally
  handle very large volume of gases in small
  equipment, so the heat transfer from the gases is
  negligible.
• Their performance is well described by the
  equations for adiabatic compressors (see Eq.
  9.15).

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• Chap. 5 showed that for any steady-flow
  compressor in which changes in potential and
  kinetic energy are negligible
• . (4.40)

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COMPRESSOR EFFICIENCIES

• .
• (9.17)
• In the case of an adiabatic compressor, the best
  possible device is a reversible, adiabatic
  compressor for which the inlet and outlet
  entropies are the same. It is commonly called an
  isentropic compressor.
• (9.18)

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• Consider the steady-flow, adiabatic compressor
  shown in Fig. 9.11. The balance for this process
  (taking the compressor as the system and assuming
  that changes in kinetic and potential energies
  are negligible)
• . (9.19)
• the real compressor has a higher outlet entropy,
  temperature, and enthalpy than the outlet stream
  from a reversible compressor would (2s in Fig.
  9.11).

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• Comparing the real compressor with a reversible
  one having the same outlet pressure.
• (9.20)

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• Example 9.6. An adiabatic compressor is
  compressing air from 20oC and 1.4atm. The airflow
  rate is 100kg/h, and the power required to drive
  the compressor is 5.3 kW. What are the efficiency
  of the compressor and the temperature of the
  outlet air? What would the outlet air temperature
  be if the compressor were 100 percent efficient?
• Air may be assumed to be a perfect gas with Cp
  29.3 J/(mol.K) and k 1.40. The work of an
  isentropic compressor doing the same job is given
  by Eq. 9.15

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• .
• .
• .
• .
• .

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• .
• For an isentropic compressor,
• .
• .
• This example illustrates the fact that the effect
  of the compressor inefficiency is to raise the
  outlet temperature of the compressor