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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

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.

- 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.

- 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

- 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)

- 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

(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)

- 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

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.

- 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,

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.

- 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?

- 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

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)

- 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) ?

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

- 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

range.

- 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

- 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

- 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.

- 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.

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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.

- 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

- 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

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

- 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

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)

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.

- 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).

- Chap. 5 showed that for any steady-flow

compressor in which changes in potential and

kinetic energy are negligible - . (4.40)

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)

- 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)

- 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

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