3.3. PUMPS AND GAS-MOVING EQUIPMENT

3.3A. Introduction

In order to make a fluid flow from one point to another in a closed conduit or pipe, it is necessary to have a driving force. Sometimes this force is supplied by gravity, where differences in elevation occur. Usually, the energy or driving force is supplied by a mechanical device such as a pump or blower, which increases the mechanical energy of the fluid. This energy may be used to increase the velocity (move the fluid), the pressure, or the elevation of the fluid, as seen in the mechanical-energy-balance equation (2.7-28), which relates ν, p, ρ, and work. The most common methods of adding energy are by positive displacement or centrifugal action.

Generally, the word "pump" designates a machine or device for moving an incompressible liquid. Fans, blowers, and compressors are devices for moving gas (usually air). Fans discharge large volumes of gas at low pressures on the order of several hundred mm of water. Blowers and compressors discharge gases at higher pressures. In pumps and fans the density of the fluid does not change appreciably, and incompressible flow can be assumed. Compressible flow theory is used for blowers and compressors.

3.3B. Pumps

1. Power and work required

Using the total mechanical-energy-balance equation (2.7-28) on a pump and piping system, the actual or theoretical mechanical energy WS J/kg added to the fluid by the pump can be calculated. Example 2.7-5 shows such a case. If η is the fractional efficiency and Wp the shaft work delivered to the pump, Eq. (2.7-30) gives

Equation 3.3-1


The actual or brake power of a pump is as follows:

Equation 3.3-2


where Wp is J/kg, m is the flow rate in kg/s, and 1000 is the conversion factor W/kW. In English units, WS is in ft · lbf/lbm and m in lbm/s. The theoretical or fluid power is

Equation 3.3-3


The mechanical energy WS in J/kg added to the fluid is often expressed as the developed head H of the pump in m of fluid being pumped, where

Equation 3.3-4


To calculate the power of a fan where the pressure difference is on the order of a few hundred mm of water, a linear average density of the gas between the inlet and outlet of the fan is used to calculate WS and brake kW or horsepower.

2. Electric motor efficiency

Since most pumps are driven by electric motors, the efficiency of the electric motor must be taken into account to determine the total electric power to the motor. Typical efficiencies ηe of electric motors are as follows: 77% for -kW motors, 82% for 2 kW, 85% for 5 kW, 88% for 20 kW, 90% for 50 kW, 91% for 100 kW, and 93% for 500 kW and larger. Hence, the total electric power input equals the brake power divided by the electric motor drive efficiency ηe:

Equation 3.3-5


3. Suction lift of pumps (NPSH)

The power calculated by Eq. (3.3-4) depends on the differences in pressures and not on the actual pressures being above or below atmospheric pressure. However, the lower limit of the absolute pressure in the suction (inlet) line to the pump is fixed by the vapor pressure of the liquid at the temperature of the liquid in the suction line. If the pressure on the liquid in the suction line drops below the vapor pressure, some of the liquid flashes into vapor (cavitation). Then no liquid can be drawn into the pump, and vibration can occur.

To avoid flashes of vapor or cavitation, the pressure at the inlet of the pump must be greater than this vapor pressure and exceed it by a value termed the net positive suction head required, or (NPSH)R. Pump manufacturers measure these values experimentally and include them with the pumps furnished.

For water below 100°C at 1750 rpm and centrifugal pumps, typical values of (NPSH)R are as follows (P4): For pressures 3500 kPa (500 psig) or below: up to 200 gpm, 1.5 m (5 ft); 500 gpm, 2.1 m (7 ft); 1000 gpm, 3 m (10 ft); 2000 gpm, 5.5 m (18 ft). At pressures of 7000 kPa (1000 psig) the values are doubled at 200 gpm or less and at 2000 gpm increased by 20%. At an rpm of 3550, the (NPSH)R increases by about a factor of 2.2.

To calculate the net positive suction head that will be available (NPSH)A at the pump suction for the system shown in Fig. 3.3-1, Eq. (3.3-6) can be used:

Equation 3.3-6


Figure 3.3-1. Diagram for (NPSH)A available in a pumping system.


where (NPSH)A is in m (ft), ρ is density of liquid in kg/m3 (lbm/ft3), p1 is pressure above liquid surface in N/m2 (lbf/ft2), pvp is vapor pressure of fluid at the given temperature in N/m2 (lbf/ft2), z1 is height of open surface of liquid above pump centerline in m (ft), ΣF is friction loss in suction line to pump from Eq. (2.10-18) in J/kg (ft lbf/lbm), and ν2/2 is velocity head in J/kg (ν2/2gc is ft lbf/lbm). Note that in Eq. (3.3-6) for SI units, the (NPSH)A in m is multiplied by g to give J/kg.

For cold water and the case where all the terms in Eq. (3.3-6) are small except p1, at atmospheric pressure the (NPSH)A is 10.3 m (33.9 ft). However, a practical limit is about 7.5 m (24.6 ft). The available NPSH for a given pump should be at least 1 m (3 ft) more than that required by the manufacturer.

EXAMPLE 3.3-1. (NPSH)A Available for Pump

Water at 50°C is in an open tank at atmospheric pressure. The pump is 3.0 m above the open tank level. The velocity in the pipe is 0.9 m/s. The friction head loss in the pipe has been calculated as 1.0 m. The required (NPSH)R for this pump is 2.0 m. Calculate the available (NPSH)A.

Solution: From Appendix A.2-3, ρ = 988.07 kg/m3. Also, the vapor pressure pvp = 12.349 kPa from A.2-9 and p1 = 1.01325 × 105 Pa. Substituting into Eq. (3.3-6) and noting that z1 is negative and that the friction loss head of 1.0 m is multiplied by g,


Solving, (NPSH)A = 5.14m


Hence, the available (NPSH)A of 5.14 m is sufficiently greater than that required of 2.0 m.

4. Centrifugal pumps

Process industries commonly use centrifugal pumps. They are available in sizes of about 0.004 to 380 m3/min (1 to 100000 gal/min) and for discharge pressures from a few m of head to 5000 kPa or so. A centrifugal pump in its simplest form consists of an impeller rotating inside a casing. Figure 3.3-2 shows a schematic diagram of a simple centrifugal pump.

Figure 3.3-2. Simple centrifugal pump.


The liquid enters the pump axially at point 1 in the suction line and then enters the rotating eye of the impeller, where it spreads out radially. On spreading radially it enters the channels between the vanes at point 2 and flows through these channels to point 3 at the periphery of the impeller. From here it is collected in the volute chamber 4 and flows out the pump discharge at 5. The rotation of the impeller imparts a high-velocity head to the fluid, which is changed to a pressure head as the liquid passes into the volute chamber and out the discharge. Some pumps are also made as two-stage or even multistage pumps.

Many complicating factors determine the actual efficiency and performance characteristics of a pump. Hence, the actual experimental performance of the pump is usually employed. The performance is usually expressed by the pump manufacturer in terms of curves called characteristic curves, which are usually for water. The head H in m produced will be the same for any liquid of the same viscosity. The pressure produced, p = Hρ g, will be in proportion to the density. Viscosities of less than 0.05 Pa · s (50 cp) have little effect on the head produced. The brake kW varies directly as the density.

Pump efficiencies typical of centrifugal pumps at rated capacities are as follows: 50% for 0.075 m3/min (20 gal/min), 62% for 0.19 m3/min (50 gal/min), 68% for 0.38 m3/min (100 gal/min), 75% for 0.76 m3/min (200 gal/min), 82% for 1.89 m3/min (500 gal/min), and 85% for 3.8 m3/min (1000 gal/min).

As rough approximations, the following relationships, called affinity laws, can be used for a given pump. The capacity q1 in m3/s is directly proportional to the rpm N1 or

Equation 3.3-7


The head H1 is proportional to or

Equation 3.3-8


The power consumed W1 is proportional to the product of H1q1, or

Equation 3.3-9


A given pump can be modified when needed for a different capacity by changing the impeller size. Then the affinity laws for a constant-rpm N are as follows: The capacity q is proportional to the diameter D, the head H is proportional to D2, and the brake horsepower W is proportional to D3.

In most pumps, the speed is generally not varied. Characteristic curves for a typical single-stage centrifugal pump operating at a constant speed are given in Fig. 3.3-3 Most pumps are usually rated on the basis of head and capacity at the point of peak efficiency. The efficiency reaches a peak at about 50 gal/min flow rate. As the discharge rate in gal/min increases, the developed head drops. The brake hp increases, as expected, with flow rate.

Figure 3.3-3. Characteristic curves for a single-stage centrifugal pump with water. (From W. L. Badger and J. T. Banchero, Introduction to Chemical Engineering. New York: McGraw-Hill Book Company, 1955. With permission.)


EXAMPLE 3.3-2. Calculation of Brake Horsepower of a Pump

In order to see how the brake-hp curve is determined, calculate the brake hp at 40 gal/min flow rate for the pump in Fig. 3.3-3.

Solution: At 40 gal/min, the efficiency η from the curve is about 60% and the head H is 38.5 ft. A flow rate of 40 gal/min of water with a density of 62.4 lb mass/ft3 is


The work WS is as follows, from Eq. (3.3-4):


The brake hp from Eq. (3.3-2) is


This value agrees with the value on the curve in Fig. 3.3-3.


5. Positive-displacement pumps

In this class of pumps, a definite volume of liquid is drawn into a chamber and then forced out of the chamber at a higher pressure. There are two main types of positive-displacement pumps. In the reciprocating pump the chamber is a stationary cylinder, and liquid is drawn into the cylinder by withdrawal of a piston in the cylinder. Then the liquid is forced out by the piston on the return stroke. In the rotary pump the chamber moves from inlet to discharge and back again. In a gear rotary pump two intermeshing gears rotate, and liquid is trapped in the spaces between the teeth and forced out the discharge.

Reciprocating and rotary pumps can be used to very high pressures, whereas centrifugal pumps are limited in their head and are used for lower pressures. Centrifugal pumps deliver liquid at uniform pressure without shocks or pulsations and can handle liquids with large amounts of suspended solids. In general, in chemical and biological processing plants, centrifugal pumps are primarily used.

Equations (3.3-1) through (3.3-5) hold for calculation of the power of positive-displacement pumps. At a constant speed, the flow capacity will remain constant with different liquids. In general, the discharge rate will be directly dependent upon the speed. The power increases directly as the head, and the discharge rate remains nearly constant as the head increases.

Pump efficiencies η of reciprocating pumps used to calculate brake horsepower are as follows: 55% at 2.2 kW (3 hp), 70% at 7.5 kW (10 hp), 77% at 14.9 kW (20 hp), 85% at 37 kW (50 hp), and 90% at 373 kW (500 hp).

3.3C. Gas-Moving Machinery

Gas-moving machinery comprises mechanical devices used for compressing and moving gases. They are often classified or considered from the standpoint of the pressure heads produced and include fans for low pressures, blowers for intermediate pressures, and compressors for high pressures.

1. Fans

The commonest method for moving small volumes of gas at low pressures is by means of a fan. Large fans are usually centrifugal and their operating principle is similar to that of centrifugal pumps. The discharge heads are low, from about 0.1 m to 1.5 m H2O. However, in some cases much of the added energy of the fan is converted to velocity energy and a small amount to pressure head.

In a centrifugal fan, the centrifugal force produced by the rotor causes a compression of the gas, called the static pressure head. Also, since the velocity of the gas is increased, a velocity head is produced. Both the static-pressure-head increase and velocity-head increase must be included in estimating efficiency and power. Operating efficiencies are in the range 40–70%. The operating pressure of a fan is generally given as inches of water gage and is the sum of the velocity head and the static pressure of the gas leaving the fan. Incompressible flow theory can be used to calculate the power of fans.

When the rpm or speed of centrifugal fans varies, the performance equations are similar to Eqs. (3.3-7)-(3.3-9) for centrifugal pumps.

EXAMPLE 3.3-3. Brake-kW Power of a Centrifugal Fan

It is desired to use 28.32 m3/min of air (metered at a pressure of 101.3 kPa and 294.1 K) in a process. This amount of air, which is at rest, enters the fan suction at a pressure of 741.7 mm Hg and a temperature of 366.3 K and is discharged at a pressure of 769.6 mm Hg and a velocity of 45.7 m/s. A centrifugal fan having a fan efficiency of 60% is to be used. Calculate the brake-kW power needed.

Solution: Incompressible flow can be assumed, since the pressure drop is only (27.9/741.7)100, or 3.8% of the upstream pressure. The average density of the flowing gas can be used in the mechanical-energy-balance equation.

The density at the suction, point 1, is


(The molecular weight of 28.97 for air, the volume of 22.414 m3/kg mol at 101.3 kPa, and 273.2 K were obtained from Appendix A.1.) The density at the discharge, point 2, is


The average density of the gas is


The mass flow rate of the gas is


The developed pressure head is


The developed velocity head for ν1 = 0 is


Writing the mechanical-energy-balance equation (2.7-28),


Setting z1 = 0, z2 = 0, v1 = 0, and Σ F = 0, and solving for WS,


Substituting into Eq. (3.3-2),



2. Blowers and compressors

For handling gas volumes at higher pressure rises than fans, several distinct types of equipment are used. Turboblowers or centrifugal compressors are widely used to move large volumes of gas for pressure rises from about 5 kPa to several thousand kPa. The principles of operation for a turboblower are the same as for a centrifugal pump. The turboblower resembles the centrifugal pump in appearance, the main difference being that the gas in the blower is compressible. The head of the turboblower, as in a centrifugal pump, is independent of the fluid handled. Multistage turboblowers are often used to go to the higher pressures.

Rotary blowers and compressors are machines of the positive-displacement type and are essentially constant-volume flow-rate machines with variable discharge pressure. Changing the speed will change the volume flow rate. Details of construction of the various types (P1) vary considerably, and pressures up to about 1000 kPa can be obtained, depending on the type.

Reciprocating compressers which are of the positive displacement type using pistons are available for higher pressures. Multistage machines are also available for pressures up to 10 000 kPa or more.

3.3D. Equations for Compression of Gases

In blowers and compressors, pressure changes are large and compressible flow occurs. Since the density changes markedly, the mechanical-energy-balance equation must be written in differential form and then integrated to obtain the work of compression. In compression of gases the static-head terms, velocity-head terms, and friction terms are dropped and only the work term dW and the dp/ρ term remain in the differential form of the mechanical-energy equation; or,

Equation 3.3-10


Integration between the suction pressure p1 and discharge pressure p2 gives the work of compression:

Equation 3.3-11


1. Isothermal compression

To integrate Eq. (3.3-11) for a perfect gas, either isothermal or adiabatic compression is assumed. For isothermal compression, where the gas is cooled on compression, p/ρ is a constant equal to RT/M, where R = 8314.3 J/kg mol · K in SI units and 1545.3 ft · lbf/lb mol · °R in English units. Then,

Equation 3.3-12


Solving for ρ in Eq. (3.3-12) and substituting it in Eq. (3.3-11), the work for isothermal compression is

Equation 3.3-13


Also, T1 = T2, since the process is isothermal.

2. Adiabatic compression

For adiabatic compression, the fluid follows an isentropic path and

Equation 3.3-14


where γ = cp/cν, the ratio of heat capacities. By combining Eqs. (3.3-11) and (3.3-14) and integrating,

Equation 3.3-15


The adiabatic temperatures are related by

Equation 3.3-16


To calculate the brake power when the efficiency is η,

Equation 3.3-17


where m = kg gas/s and WS = J/kg.

The values of γ are approximately 1.40 for air, 1.31 for methane, 1.29 for SO2, 1.20 for ethane, and 1.40 for N2 (P1). For a given compression ratio, the work for isothermal compression in Eq. (3.3-13) is less than the work for adiabatic compression in Eq. (3.3-15). Hence, cooling is sometimes used in compressors.

EXAMPLE 3.3-4. Compression of Methane

A single-stage compressor is to compress 7.56 × 10-3 kg mol/s of methane gas at 26.7°C and 137.9 kPa abs to 551.6 kPa abs.

  1. Calculate the power required if the mechanical efficiency is 80% and the compression is adiabatic.

  2. Repeat, but for isothermal compression.

Solution: For part (a), p1 = 137.9 kPa, p2 = 551.6 kPa, M = 16.0 kg mass/kg mol, and T1 = 273.2 + 26.7 = 299.9 K. The mass flow rate per sec is


Substituting into Eq. (3.3-15) for γ = 1.31 for methane and p2/p1 = 551.6/137.9 = 4.0/1,


Using Eq. (3.3-17),


For part (b), using Eq. (3.3-13) for isothermal compression,


Hence, isothermal compression uses 15.8% less power.


3. Polytropic compression

In large compressors neither isothermal nor adiabatic compression is achieved. This polytropic path is

Equation 3.3-18


For isothermal compression n = 1.0 and for adiabatic, n = γ. The value of n is found by measuring the pressure p1 and density ρ1 at the inlet and p2 and ρ2 at the discharge and substituting these values into Eq. (3.3-18).

4. Multistage compression ratios

Water cooling is used between each stage in multistage compressors to reduce the outlet temperature to near the inlet temperature for minimum power requirement. The compression ratios should be the same for each stage so that the total power is a minimum. This gives the same power in each stage. Hence, for n stages, the compression ratio (pb/pa) for each stage is

Equation 3.3-19


where p1 is the inlet pressure and pn the outlet pressure from n stages. For two stages, the compression ratio is .

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