Pumps and compressors Pumps and compressors Sub-chapters - PowerPoint PPT Presentation


PPT – Pumps and compressors Pumps and compressors Sub-chapters PowerPoint presentation | free to download - id: 3cf2ae-OWNmY


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation

Pumps and compressors Pumps and compressors Sub-chapters


Pumps and compressors Pumps and compressors Sub-chapters 9.1. Positive-displacement pumps 9.2. Centrifugal pumps 9.3. Positive-displacement compressors 9.4. – PowerPoint PPT presentation

Number of Views:3260
Avg rating:5.0/5.0
Slides: 63
Provided by: staffUiA
Learn more at: http://staff.ui.ac.id


Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: Pumps and compressors Pumps and compressors Sub-chapters

Pumps and compressors
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

  • 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

  • Pumps work on liquid and compressors work on a
  • 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.

  • 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

(No Transcript)
  • 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.

  • 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
  • The piston pushes the fluid out into the outlet
  • The piston starts downward again the spring
    closes the outlet valve, because the pressure in
    the cylinder has fallen, and the cycle begins

  • 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
  • For a given speed of the pump's motor, the
    results for various discharge pressures are shown
    in Fig 9.3.

(No Transcript)
(No Transcript)
  • 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
  • Volumetric flow rate piston area x piston
    travel x cycles/time (9.1)

  • 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.

  • The normal definition of pump efficiency is
  • 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)

  • 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

  • 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.

  • ? becomes equal to the hydraulic horsepower
    divided by the total horsepower supplied to the
  • 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.

  • 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

  • . (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,
  • . (9.6)
  • For water under normal Patm and Troom, hmax ? 34
    ft 10 m. This height is called the suction lift.

(No Transcript)
  • 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

  • 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

  • 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

  • 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
  • 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.

(No Transcript)
  • 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
  • Although P2 gt P1, this term is small compared
    with the change in kinetic energy
  • (9.7)

  • 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

  • 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

  • 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) ?

(No Transcript)
  • 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

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

  • 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.

  • 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

  • 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
  • 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

  • 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
  • Numerous schemes for performing this function are

  • Centrifugal pumps are often used to pump boiling
    liquids, e.g., at the bottom of many distillation
  • 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.

  • 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).

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.
(No Transcript)
  • 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.

  • 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
  • The pressure-volume history of the gas in the
    cylinder of such a compressor is shown
  • in Fig. 9-9.
  • Cycle ABCD

  • 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.

  • 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)

  • 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)

  • 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

  • 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
  • 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

  • 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,

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

  • 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
  • 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)

  • 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
  • The resulting compressors, which were developed
    for aircraft service, are now being applied
    industrially in high-capacity applications, for
    example, in ammonia plants.

  • 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

(No Transcript)
  • 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
  • Their performance is well described by the
    equations for adiabatic compressors (see Eq.

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

  • .
  • (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)

  • 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.

(No Transcript)
  • Comparing the real compressor with a reversible
    one having the same outlet pressure.
  • (9.20)

  • 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

  • .
  • .
  • .
  • .
  • .

  • .
  • 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
About PowerShow.com