transp fluida dasar 3 juni_14

8/17/2019 Transp Fluida Dasar 3 Juni_14

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TRANSPORTASI FLUIDA

Dr. Ir. Ahmad Rifandi, M.Sc. Cert. IV

Teknik Kimia - POLBAN


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PENGERTIAN-PENGERTIAN DASAR

Properties of Fluids

A f lu id is any substance which flows because its particles are not rigidlyattached to one another. This includes liquids, gases and even some

materials which are normally considered solids, such as glass.

Essentially, fluids are materials which have no repeating crystalline

structure.

Buoyancy is defined as the tendency of a body to float or rise whensubmerged in a fluid. We all have had numerous opportunities of observing the

buoyant effects of a liquid. When we go swimming, our bodies are held up

almost entirely by the water.

Compressib i l i ty is the measure of the change in volume a substance undergoes

when a pressure is exerted on the substance. Liquids are generally considered tobe incompressible. For instance, a pressure of 16,400 psig will cause a given

volume of water to decrease by only 5% from its volume at atmospheric pressure.

Gases on the other hand, are very compressible. The volume of a gas can be

readily changed by exerting an external pressure on the gas


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Relationship Between Depth and Pressure

Anyone who dives under the surface of the water notices that the pressure on his

eardrums at a depth of even a few feet is noticeably greater than atmosphericpressure. Careful measurements show that the pressure of a liquid is directly

proportional to the depth, and for a given depth the liquid exerts the same

pressure in all directions

As shown in Figure 1 the pressure at

different levels in the tank varies and

this causes the fluid to leave the tankat varying velocities. Pressure was

defined to be force per unit area. In

the case of this tank, the force is due

to the weight of the water above the

point where the pressure is being

determined


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Volumetr ic Flow Rate

The volumetric flow rate ( V ) of a system is a measure of the volume of luid

passing a point in the system per unit time.The volumetric flow rate can be calculated as the product of the rcosssectional

area (A) for flow and the average flow velocity (v).

Example: A pipe with an inner diameter of 4 inches contains water that

flows at an average velocity of 14 feet per second. Calculate

the volumetric flow rate of water in the pipe.

Solution: Use Equation 3-1 and substitute for the area.


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Mass Flow Rate

Example:

Solution:

The mass flow rate (m˙ ) of a system is a measure of the mass of fluid passing

a point in the system per unit time. The mass flow rate is related to thevolumetric flow rate as shown in Equation 3-2 where r is the density of the fluid.

The water in the pipe of the previous example had a density

of 62.44 lbm/ft3. Calculate the mass flow rate


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Conservation of Mass

In thermodynamics, energy can neither be created nor destroyed, only

changed in form.

The same is true for mass. Conservation of mass is a principle of

engineering that states that all mass flow rates into a control volume are

equal to all mass flow rates out of the control volume plus the rate of

change of mass within the control volume


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Two storage tanks, A and B, containing a petroleum product, discharge rough

pipes each 0.3 m in diameter and 1.5 km long to a junction at D, as shown in

Figure 3.9. From D the liquid is passed through a 0.5 m diameter pipe to a

third storage tank C, 0.75 km away. The surface of the liquid in A is nitially 10m above that in C and the liquid level in B is 6 rn higher than that in A.

Calculate the initial rate of discharge of liquid into tank C assuming the ipes

are of mild steel. The density and viscosity of the liquid are 870 kg/m3 and 0.7

mN s/m2 respectively.


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Taking the roughness of mild steel pipe e as 0.00005 m, e/d varies from 0.0001 to

0.00017.

As a first approximation, R/pu2 may be taken as 0.002 in each pipe, and

substituting this value in equations 4

and 5 then:


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Flow Regimes

All fluid flow is classified into one of two broad categories or regimes.

These two flow regimes are laminar flow and turbulent flow.

Flow Velocity Profiles

Not all fluid particles

travel at the same

velocity within a pipe.The shape of the

velocity curve (the

velocity profile across

any given section of the

pipe) depends upon

whether the flow islaminar or turbulent.


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For practical purposes, if the Reynolds number is less than 2000, the flow

is laminar. If it is greater than 3500, the flow is turbulent. Flows with

Reynolds numbers between 2000 and 3500 are sometimes referred to as

transitional flows


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An ideal fluid is one that is incompressible and has no viscosity. Ideal fluids

do not actually exist, but sometimes it is useful to consider what would

happen to an ideal fluid in a particular fluid flow problem in order to simplify

the problem.

Ideal Fluid

The flow regime (either laminar or turbulent) is determined by evaluating the

Reynolds number of the flow (refer to figure 5). The Reynolds number,based on studies of Osborn Reynolds, is a dimensionless number

comprised of the physical characteristics of the flow. Equation 3-7 is used to

calculate the Reynolds number (NR) for fluid flow.

Reynolds Number


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The conservation of energy principle states that energy can be neither created

nor destroyed. This is equivalent to the First Law of Thermodynamics, whichwas used to develop the general energy equation in the module on

thermodynamics. Equation 3-8 is a statement of the general energy equation

for an open system.

General Energy Equation

Karena cairan tidakmengalami perubahan

bentuk (fasa) maka U = 0)


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Bernoulli’s equation results from the application of the general energy

equation and the first law of thermodynamics to a steady flow system inwhich no work is done on or by the fluid, no heat is transferred to or from

the fluid, and no change occurs in the internal energy (i.e., no temperature

change) of the fluid. Under these conditions, the general energy equation

is simplified to Equation 3-9.

Persamaan Bernoulli Sederhana


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Energi dalam sistem tertutup terdiri atas:

“pressure”, “motion”, dan “position”

Persamaan Bernoulli Sederhana ……lanjutan

pressuremotion

position

Apabila “motion” diubah tetapi “position” tidak berubah maka “pressure”

akan berubah tetapi jumlah energi dalam sistem tetap tidak berubah.Contoh: apabila kita menekan ujung selang plastik ketika kita menyiram

kebun maka tekanan air akan berubah dan air akan memancar lebih kuat

dari ujung selang karet.


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Multiplying all terms in Equation 3-10 by the factor gc/mg r esults in the

form of Bernoulli’s equation shown by Equation 3-11

The Bernoulli equation can be modified to take into account gains and

losses of head. The resulting equation, referred to as the Extended Bernoulli

equation, is very useful in solving most fluid flow problems


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Contoh Soal

Water is pumped from a large reservoir to a point 65 feet higher than the

reservoir. How many feet of head must be added by the pump if 8000lbm/hr flows through a 6-inch pipe and the frictional head loss is 2 feet? The

density of the fluid is 62.4 lbm/ft3, and the cross-sectional area of a 6-inch

pipe is 0.2006 ft2.

Penyelesaian

To use the modified form of Bernoulli’s equation, reference points arechosen at the surface of the reservoir (point 1) and at the outlet of the pipe

(point 2). The pressure at the surface of the reservoir is the same as the

pressure at the exit of the pipe, i.e., atmospheric pressure. The velocity at

point 1 will be essentially zero.

Using the equation for the mass flow rate to determine the velocity at point 2:


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Now we can use the Extended Bernoulli equation to determine the

required pump head.


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The main elements of a pumping system are:

PUMPING SYSTEM

Supply side (suction or inlet side)

Pump (with a driver)

Delivery side (discharge or process)

Energy input = Energy useful + Losses

Efficiency = Energy useful /Energy input

Losses = Mechanical + Volumetric + Hydraulic⇓

bearings leakage (slip) friction

coupling entrance/exit

Rubbing vortices

separationdisc friction

⇓ ⇓

Head

losses


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PUMPING SYSTEM

Work equals force multipled by distance

For a steady motion, the force is balanced by the pressure “p,” acting on area, “A”:

W = (p × A) × L = p × (A × L) = p × V

INPUT POWER, LOSSES, AND EFFICIENCY

Work per unit of time equals power. So, dividing both sides of the equation by “t,”

we get:“Q” is the volume per unit of time, which in

pump language is called “flow,” “capacity,”

or “delivery.”


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INPUT POWER, LOSSES, AND EFFICIENCY

Ideal Power = Fluid Horsepower = FHP = p × Q × constant

since all power goes to “fluid horsepower,” in the ideal world

pressure is measured in ps i , and flow in gpm , (US units)

Power =


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Application of Bernoulli’s Equation to a Venturi


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Application of Bernoulli’s Equation to a Venturi

Therefore the flow velocity at

the throat of the venturi and the

volumetric flow rate are directly

proportional to the square root

of the differential pressure.


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Head Losses

The head loss that occurs in pipes is dependent on the f low veloci ty,

pipe length and d iameter, and a frict ion factor based on th e roughness

of the pipe and the Reyno lds n umber of the f low. The head los s that

occu rs in the components of a flow path can be corre lated to a pip ing

length that would cause an equiv alent head loss .

Head loss is a measure of the reduction in the total head (sum of elevation

head, velocity head and pressure head) of the fluid as it moves through afluid system

Frictional loss is that part of the total head loss that occurs as the fluid flows

through straight pipes. The head loss for fluid flow is directly proportional to

the length of pipe, the square of the fluid velocity, and a term accounting for

fluid friction called the friction factor. The head loss is inversely proportional to

the diameter of the pipe.


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Friction Factor

The friction factor has been determined to depend on the Reynolds number

for the flow and the degree of roughness of the pipe’s inner surface The quantity used to measure the roughness of the pipe is called the relative

roughness, which equals the average height of surface irregularities (e)

divided by the pipe diameter (D).

The value of the friction factor is usually obtained from the Moody Chart (Figure

B-1 of Appendix B). The Moody Chart can be used to determine the friction

factor based on the Reynolds number and the relative roughness.

Determine the friction factor (f) for fluid flow in a pipe that has a Reynolds

number of 40,000 and a relative roughness of 0.01

Using the Moody Chart, a Reynolds number of 40,000 intersects the curve

corresponding to a relative roughness of 0.01 at a friction factor of 0.04.

Contoh Soal

Penyelesaian

M d Ch


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Moody Chart

M d Ch t


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Moody Chart


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The frictional head loss can be calculated using a mathematical relationship

that is known as Darcy’s equation for head loss. The equation takes two

distinct forms. The first form of Darcy’s equation determines the losses in the

system associated with the length of the pipe.

Persamaan Darcy


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A pipe 100 feet long and 20 inches in diameter contains water at 200°F

flowing at a mass flow rate of 700 lbm/sec. The water has a density of 60lbm/ft3 and a viscosity of 1.978 x 10-7 lbf-sec/ft2. The relative roughness

of the pipe is 0.00008. Calculate the head loss for the pipe.

The sequence of steps necessary to solve this problem is first to determine

the flow velocity. Second, using the flow velocity and the fluid properties

given, calculate the Reynolds number. Third, determine the friction factor

from the Reynolds number and the relative roughness. Finally, use Darcy’s

equation to determine the head loss.

Contoh Soal

Penyelesaian


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Use the Moody Chart for a

Reynolds number of 8.4 x 107and a relative roughness of

0.00008.


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Minor Losses

The losses that occur in pipelines due to bends, elbows, joints, valves, etc.

are sometimes called minor losses .

Equivalent Length

Minor losses may be expressed in terms of the equivalent length (Leq) of

pipe that would have the same head loss for the same discharge flow rate.

This relationship can be found by setting the two forms of Darcy’s equation

equal to each other

This yields two relationships that are useful


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A fully-open gate valve is in a

pipe with a diameter of 10inches. What equivalent length

of pipe would cause the same

head loss as the gate valve

From Table 1, we find that the

value of Leq/D for a fully-open

gate valve is 10.

Leq /D = 10

Leq = 10 D

Leq = 10 (10 inches)

= 100 inches

Contoh Soal

Penyelesaian


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Contoh Soal

LC

Hitung ukuran pipa dan spesifikasi pompa yang diperlukan untuk memompa cairan ortho-

dichlorobenzene (ODBC) dengan kecepatan aliran 10.000 kg/j, suhu 20oC dan jenis pipa

carbon steel (density ODBC = 1306 kg/m3 dan viscosity = 0,9 cp

C 201

1 bar

C 203

2 bar

HE 205

2 , 5 m

7 , 5 m

1 , 0 m

3

, 5 m

3 , 0 m

0 , 5 m

20 m

2 m

1 m

4 m


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Contoh Soal

Hitung ukuran pipa dan spesifikasi pompa yang diperlukan untuk memompa cairan ortho-

dichlorobenzene (ODBC) dengan kecepatan aliran 10.000 kg/j, suhu 20oC dan jenis pipa

carbon steel (density ODBC = 1306 kg/m3 dan viscosity = 0,9 cp


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Penyelesaian


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Penyelesaian

atau menggunakan rumus:

Hf = ft


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Konversi satuan:

Viscosity (M/Lθ):

1 pound mass/foot hour = 0.00413 g/cm s

0.000413 kg/m s

1 centipoise = 0.01 poise

= 0.01 g/cm s

= 0.001 kg/m s

= 0.000672 lbm/ft s

= 0.0000209 lbf s/ft2

Pressure (F/L2):

Normal atmospheric pressure

1 atm = 760 millimeters of mercury at 0°C

(density 13.5951 g/cm3)= 29.921 inches of mercury at 32°F

= 14.696 pounds force/square inch

= 33.899 feet of water at 39.1°F

= 1.01325 × 106 dynes/square centimeter

= 1.01325 × 105 Newtons/square meter

= 1.01325 x 105 Pa (pascal)

Density (M/L3)

1 pound mass/cubic foot = 0.01601846 grams/cubic centimeter

= 16.01846 kilogram/cubic meter


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Penyelesaian


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Tabel Leq/D (Coulson and Richardson, Chem Eng. Vol 6)


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Penyelesaian


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Penyelesaian


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Penyelesaian

P = 65,27 psi, Q = 32,93 gpm;

BHP = 2,09 PK= 1,5 kVA


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Penyelesaian


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JENIS POMPA

Pump Classification Table

Classification Type Name Application Volumetric Pumps

(Displacement)

Reciprocating Type

(Piston)

Single Action

Double ActionDiaphragm

Beam

PlungerHand

Rotary Type Gear Type

Lobe Type

Rotary Plunger

Screw Type

Vane TypeImpeller Type

Peristaltic

Oil

Fish Pond Pump

Air Con Unit

Top Drive Rotary

Hazardous

Chemicals

Dynamic Pumps

(Centrifugal)

Radial Flow Pumps

Axial Flow Pumps

Single Impeller

Multiple Impeller

Cylinder

Electrical

SubmersiblesElectrical

Submersibles

Fan Heater

Static Pumps Ejectors Jet Nozzle

Venturi Nozzle

Vacuum Generators


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JENIS POMPA

Reciprocat ing Piston Pump

JENIS POMPA


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JENIS POMPA

Gear Pump

JENIS POMPA


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JENIS POMPA

Centrifugal Pumps

PUMP SELECTION


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PUMP SELECTION

SYSTEM CURVE


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SYSTEM CURVE

Essentially, any flow restriction requires a pressure gradient to overcome it.

These restrictions are valves, orifices, turns, and pipe friction.From the fundamentals of hydraulics based on the Bernoulli equation, a

pressure drop (i.e., hydraulic loss) is proportional to velocity head:

For the flow of liquid through a duct (such as pipe), the velocity is equal to:

which means that pressure loss is proportional to the square of flow:

From the discussion above, we have established that flow and pressure are

the two main parameters for a given application. Other parameters, such as

pump speed, fluid viscosity, specific gravity, and so on, will have an effect on

flow and/or pressure, by modifying the hydraulics of a pumping system inwhich a given pump operates. A mechanism of such changes can be traced

directly to one of the components of losses, namely the hyd raul ic losses .

SYSTEM CURVE


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SYSTEM CURVE

Essentially, any flow restriction requires a pressure gradient to overcome it.

These restrictions are valves, orifices, turns, and pipe friction.

Hydraulic losses, as a function of flow.

PUMP CURVE


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PUMP CURVE

A pump curve shows a relationship between its two main parameters: flow

and pressure

The shape of this curve (see Figure) depends on the particular pump type.

Pump curves, relating pressure and flow. The slope of the centrifugal pump curve is

“mostly” flat or horizontal; the slope of the PD-pump is almost a vertical line.

PUMP CURVE


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PUMP CURVE

Therefore, the pump operating point is an intersection of the pump curve

and a system curve (see Figure 6). In addition to friction, a pump must

also overcome the elevation difference between fluid levels in thedischarge and suction side tanks, a so called static head, that is

independent of flow (see Figure 7). If pressure inside the tanks is not

equal to atmospheric pressure then the static head must be calculated as

equivalent difference between total static pressures (expressed in feet of

head) at the pump discharge and suction, usually referenced to the pump

centerline (see Figure 8).The above discussion assumes that the suction and discharge piping near

the pump flanges are of the same diameter, resulting in the same

velocities. In reality, suction and discharge pipe diameters are different

(typically, a discharge pipe diameter is smaller). This results in difference

between suction and discharge velocities, and their

energies (velocity heads) must be accounted for. Therefore, a total pumphead is the difference between all three components of the discharge and

suction fluid energy per unit mass: static pressure heads, velocity heads,

and elevations

PUMP CURVE


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PUMP CURVE

For example,

Note that the units in Equation 16 are feet of head of water. The conversion

between pressure and head is:

PUMP CURVE


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PUMP CURVE

FIGURE 6Pump operating point — intersection of a pump

and a system curves.

Note : Due to the almost vertical curve slope of

rotary pumps (b), their performance curves

are usually and historically plotted as shown on

(c) (i.e., flow vs. pressure).

PUMP CURVE


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PUMP CURVE

PUMP CURVE


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PUMP CURVE

From our high school days and basic hydraulics, we remember that

the pressure, exerted by a column of water of height, “h,” is

Where γ is a specific weight of the substance, measured in lbf/ft3

.A specific gravity (SG) is defined as a ratio of the specific weight of the

substance to the specific weight of cold water: γo = 62.4 lbf/ft 3 . (SG is

also equal to the ratio of densities, due to a gravitational constant

between the specific weight and density). So,

(To obtain pressure in more often used units of lbf/in2 (psi), divide by 144).

PUMP CURVE


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PUMP CURVE

Clearly, if the system resistance changes, such as an opening or a closing of

the discharge valve, or increased friction due to smaller or longer piping, the

slope of the system curve will change (see Figure 9). The operating point

moves: 1 → 2, as valve becomes “more closed,” or 1 → 3, if it opens more.

CENTRIFUGAL PUMP


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Stuffing Box - A : :

Packing - B : :

Shaft - C : :

Shaft Sleeve - D : :

Vane - E : :Casing - F : :

Eye of Impeller - G : :

Impeller - H : :

Casing wear Ring - I : :

Impeller - J : :

Discharge Nozzle - K : :

Self priming/non self priming


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PUMP IMPELLER

http://www.sparepairs.com.au/category139_1.htm


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http://www.cfturbo.com/en/impeller.htmlhttp://www.sparepairs.com.au/category139_1.htm


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EXTERNAL GEAR PUMP

INTERNAL GEAR PUMP

http://www.answers.com/main/Record2?a=NR&url=http://commons.wikimedia.org/wiki/Image:Gear%20pump%20exploded.png


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INTERNAL GEAR PUMP

LOBE PUMP

http://upload.wikimedia.org/wikipedia/commons/f/f9/Rotary_piston_pump.svghttp://upload.wikimedia.org/wikipedia/commons/f/f9/Rotary_piston_pump.svg


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1. As the lobes come out of mesh, they create expanding volume on the inlet side of the

pump. Liquid flows into the cavity and is trapped by the lobes as they rotate.

2. Liquid travels around the interior of the casing in the pockets between the lobes and

the casing -- it does not pass between the lobes.

3. Finally, the meshing of the lobes forces liquid through the outlet port under pressure.

DOUBLE SCREW PUMP

http://upload.wikimedia.org/wikipedia/commons/f/f9/Rotary_piston_pump.svg


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VANE PUMP

http://upload.wikimedia.org/wikipedia/commons/2/2d/Rotary_vane_pump.svg


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1. body

2. rotor

3. piston valve

4. spring

SINGLE SCREW PUMP

http://upload.wikimedia.org/wikipedia/commons/2/2d/Rotary_vane_pump.svghttp://edis.ifas.ufl.edu/EDISImagePage?imageID=937263795&dlNumber=WI006&tag=IMAGE%20WI:WI00607&credits=http://upload.wikimedia.org/wikipedia/commons/2/2d/Rotary_vane_pump.svg


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PERISTALTIC PUMP

PISTON PUMP

http://prettech.en.alibaba.com/product/50355353/50923100/Pump/Peristaltic_Pump.html


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DIAPHRAGM PUMP

http://www.animatedsoftware.com/pumpglos/wobble.htm


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FLEXIBLE IMPELLER PUMP


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CHAIN PUMP


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ECCENTRIC

CAM PUMP

PUMP CHARACTERISTICS


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System Curves

For a specified impeller diameter and speed, a centrifugal pump has a fixed and

predictable performance curve. The point where the pump operates on its curve is

dependent upon the characteristics of the system In which it is operating, commonlycalled the System Head Curve. ..or, the relationship between flow and hydraulic losses* in

a system. This representation is in a graphic form and, since friction losses vary as a

square of the flow rate, the system curve is parabolic in shape.

By plotting the system head

curve and pump curve

together, it can be

determined:

Where the pump will

operate on its curve.

What changes will occur if

the system head curve or

the pump performance

curve changes.

NO STATIC HEAD - ALL FRICTION

A th l l i th ti d di h th (Fi 1) th i t ti

http://www.gouldspumps.com/cpf_0011.htmlhttp://www.gouldspumps.com/cpf_0011.html


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As the levels in the suction and discharge are the same (Fig. 1), there is no static

head and, therefore, the system curve starts at zero flow and zero head and its

shape is determined solely from pipeline losses. The point of operation is at the

intersection of the system head curve and the pump curve. The flow rate may be

reduced by throttling valve.


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POSITIVE STATIC HEAD

The parabolic shape of the

system curve is againdetermined by the friction

losses through the system

including all bends and valves.

But in this case there is a

positive static head involved.

This static head does not affectthe shape of the system curve

or its "steepness", but it does

dictate the head of the system

curve at zero flow rate.

The operating point is at the

intersection of the system curveand pump curve. Again, the

flow rate can be reduced by

throttling the discharge valve.


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NEGATIVE (GRAVITY) HEAD

In the illustration below, a

certain flow rate will occur by

gravity head alone. But toobtain higher flows, a pump Is

required to overcome the pipe

friction losses in excess of "H" -

the head of the suction above

the level of the discharge. In

other words, the system curveis plotted exactly as for any

other case involving a static

head and friction head, except

the static head is now negative.

The system curve begins at a

negative value and shows thelimited flow rate obtained by

gravity alone. More capacity

requires extra work.

MOSTLY LIFT- LITTLE FRICTION HEAD

The system head curve in the illustration below starts at the static head "H" and


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The system head curve in the illustration below starts at the static head H and

zero flow. Since the friction losses are relatively small (possibly due to the large

diameter pipe), the system curve is "flat". In this case. the pump is required to

overcome the comparatively large static head before it will deliver any flow at all.

*Hydraulic losses in

piping systems are

composed of pipe

friction losses, valves,

elbows and otherfittings, entrance and

exit losse (these to the

entrance and exit to

and from the pipeline

normally at the

beginning and end notthe pump) and losses

from changes in pipe

size by enlargement

or reduction in

diameter.

CHARACTERISTIC CURVE OF PUMP


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How to Read Pump Curves


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STEP 1: The basic pump curves are no different than reading any other

head - flow curve. For a known head value, follow the head over to the

pump curve then drop down to the capacity axis and this will be the flowrate. What you are trying to figure out here is what diameter impeller is

needed to get the required head and capacity.

STEP 2: The next thing to figure out is what motor is needed to drive

this impeller without overloading. To do this use the dashed horsepower

lines. To the right of the horsepower line is overloading and to the left isnon-overloading.

STEP 3: The last thing to determine is at what pump efficiency the pump

will operate. Look at the U-shaped lines and interpolate to get the

efficiency.

Now let's try an example using ZM1570, Performance Data for Models 6650-6671

(5-15 BHp 4" discharge units). For the example we will size a pump for 400 GPM

at 54 feet of total dynamic head.

STEP 1: Locate the point of 400 GPM at 54 feet on the pump curve. This

point is slightly above the 8.31" impeller but well below the 8.63"


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p g y p

impeller so I would go with an 8.38" impeller to hit the duty point.

STEP 2:


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Next, draw a new pump curve that passes through the duty point and is

parallel to the existing pump curves. This will give you a close

representation of the actual performance the pump will deliver. Look tosee where this curve crosses the horsepower line to the right of the

design point.

In this example the pump curve crosses the 10 BHp curve at about 48

feet and crosses the 12.5 BHp curve at about 21 feet. We will not oversize

an impeller on a pump if the overload point on the pump curve is greaterthan the static head for the system.

So for this example, if the static head is greater than 48 feet then we can

use the 10 BHp unit. If the static head is between 21 feet and 48 feet, use

the 12.5 BHp motor. If the static head is less than 21 feet then use the 15

BHp motor.


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STEP 3


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STEP 3:

Now let's figure the pump efficiency we can expect. The design point

is about half way in between the efficiency lines of 60% and 63%. So,

for the design point of 400 GPM at 54 feet, we would expect about61.5% pump efficiency.

As you can tell from the above example, we would consider oversizing

an impeller on a unit and not overload the unit due to engineering the

right pump for the system. If this were the case we would also able to

provide a more competitively priced unit since pricing is based onmotor size (i.e. smaller motors cost less).

The only exception to this rule is a single-phase unit. ZOELLER

COMPANY DOES NOT SELL SINGLE-PHASE UNITS WITH OVERSIZED

IMPELLERS because we feel that this will compromise the life of a

single-phase unit.


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TOTAL DYNAMIC HEAD


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The Total Dynamic Head (TDH) is the sum of the total static head, the total

friction head and the pressure head. The components of the total static head for

a surface water and well wate pumping system are shown

Total Static Head

The total static head is the total vertical distance the pump must lift the water. When pumping from a


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The total static head is the total vertical distance the pump must lift the water. When pumping from a

well, it would be the distance from the pumping water level in the well to the ground surface plus the

vertical distance the water is lifted from the ground surface to the discharge point. When pumping

from an open water surface it would be the total vertical distance from the water surface to the

discharge point.

Pressure Head

Sprinkler and drip irrigation systems require pressure to operate. Center pivot systems require a

certain pressure at the pivot point to distribute the water properly. The pressure head at any point

where a pressure gage is located can be converted from pounds per square inch (PSI) to feet of

head by multiplying by 2.31. For example, 20 PSI is equal to 20 times 2.31 or 46.2 feet of head.

Friction Head

Friction head is the energy loss or pressure decrease due to friction when water flows through

pipe networks. The velocity of the water has a significant effect on friction loss. Loss of head due

to friction occurs when water flows through straight pipe sections, fittings, valves, around

corners, and where pipes increase or decrease in size. Values for these losses can be

calculated or obtained from friction loss tables. The friction head for a piping system is the sumof all the friction losses

Velocity Head

Velocity head is the energy of the water due to its velocity. This is a very small amount of energy

and is usually negligible when computing losses in an irrigation system


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Determining Flow and Head

The pump is installed and running, but how do you know if it is

operating at its design point? There is a simple way to check. Knowing

that a pump will provide a certain flow at a given head, we candetermine the point at which the pump is operating. To determine the

head, a few gage readings will be necessary. Take one reading from

the suction of the pump and one from the discharge after the system is

balanced and with all the control valves wide open. The difference

between the two gage readings will give you the head that the pump is

providing. Remember to convert your gage readings to feet of head.Knowing the head and the impeller size, you can determine the flow of

the pump.

Now that we have the flow and head of the pump, let’s see how close

we are to the design point. Most often, the head will be less than what

we expected, and the flow will be more. Why does this happen? There

are many reasons, but it does no good to blame anyone. Let’s just fixthe problem.

http://www.bellgossett.com/Press/CntrTk1199A.htm


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Solutions

Trimming the impeller is one of best solutions. Before we can trim the impeller,

http://www.bellgossett.com/Press/CntrTk1199A.htm


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we need to determine where the pump is operating. In the pump curve above,

let’s call point “D” the design point, and draw the system curve that corresponds

with that design point. Point “A” is where we actually are, which we determined

from our gage readings. Along with that is our actual system curve. Rememberthat we are concerned with the actual system curve. This shows us how our

system operates, not how it was designed. Operational and design points are

often completely different.

We would like to be on the unmodified actual system curve, but where on that

curve? If our load has not changed and our heat transfer is the same, we want to

be at our design flow. That is“I,” the ideal point.

Trimming the Impeller

But how do we get there? Although it’s off our impeller curve, we can trim our

impeller down to the right size. In this particular case, our ideal impeller size falls

between 10-1/2” and 11-1/2 (actually about 11”). Fortunately, trimming an

impeller is not too difficult or expensive, and in fact it pays for itself very quickly.

Notice from the figure that when we trim our impeller we lose some pump

efficiency, but we’re more concerned about the cost of operating our pump and

that cost has dropped tremendously. In this case we have dropped from 85Hp to

40 Hp-that’s a lot. Even if your electric rates are low and you don’t operate all

year long, there is still the potential for great energy savings.


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PB = Barometric pressure in feet absolute.

VP = Vapor pressure of the liquid at maximum pumping temperature, in feet absolute.

P = Pressure on surface of liquid in closed suction tank, in feet absolute.

Ls = Maximum static suction lift in feet.

LH = Minimum static suction head in feet.

hf = Friction loss in feet in suction pipe at required capacity

NET POSITIVE

SUCTION HEAD


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Net Positive Suction Head Available (NPSHA)The net positi e s ction head a ailable is a f nction of the p mp s ction s stem


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The NPSH available in a

flooded suction system is:

Atmospheric Pressure (- )Vapor Pressure (+) Liquid

Height (-) Friction in the

Suction Line

The NPSH available in a suction

lift system is:Atmospheric Pressure (-) Vapor

Pressure (-) Liquid Ht. (-) Friction

in the Suction Line.

The net positive suction head available is a function of the pump suction system.

The Net Positive Suction Head is the absolute total suction head in feet.


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NPSHA = Atmospheric pressure(converted to head) + static head + surface pressure head -

vapor pressure of your product - loss in the piping, valves and fittings

Given:

Atmospheric pressure = 14 7 psi


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Atmospheric pressure = 14.7 psi

Gage pressure =The tank is at sea level and open to atmospheric pressure.

Liquid level above pump centerline = 5 feet

Piping = a total of 10 feet of 2 inch pipe plus one 90° long radius screwed elbow.Pumping =100 gpm. 68°F. fresh water with a specific gravity of one (1).

Vapor pressure of 68°F. Water = 0.27 psia from the vapor chart.

Specific gravity = 1

NPSHR (net positive suction head required, from the pump curve) = 9 feet

NPSHA = Atmospheric pressure(converted to head) + static head + surface pressure head - vapor pressure of your

product - loss in the piping, valves and fittings

Static head = 5 feet

Atmospheric pressure = pressure x 2.31/sg. = 14.7 x 2.31/1 = 34 feet absolute

Gage pressure = 0

Vapor pressure of 68°F. water converted to head = pressure x 2.31/sg = 0.27 x 2.31/1 = 0.62 feet

Looking at the friction charts:

100 gpm flowing through 2 inch pipe shows a loss of 17.4 feet for each 100 feet of pipe or 17.4/10 = 1.74 feetof head loss in the piping

The K factor for one 2 inch elbow is 0.4 x 1.42 = 0.6 feet

Adding these numbers together, 1.74 + 0.6 = a total of 2.34 feet friction loss in the pipe and fitting.

NPSHA (net positive suction head available) = 34 + 5 + 0 - 0.62 - 2.34 = 36.04 feet

The pump required 9 feet of head at 100 gpm. And we have 36.04 feet so we have plenty to spare.

Given:

Gage pressure = - 20 inches of vacuum

Atmospheic pressure = 14.7 psi

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Liquid level above pump centerline = 5 feet

Piping = a total of 10 feet of 2 inch pipe plus one 90° long radius screwed elbow.

Pumping = 100 gpm. 68°F fresh water with a specific gravity of one (1).

Vapor pressure of 68°F water = 0.27 psia from the vapor chart.NPSHR (net positive suction head required) = 9 feet

Now for the calculations:

NPSHA = Atmospheric pressure(converted to head) + static head + surface pressure head

- vapor pressure of your product - loss in the piping, valves and fittings

Atmospheric pressure = 14.7 psi x 2.31/sg. =34 feet

Static head = 5 feetGage pessure pressure = 20 inches of vacuum converted to head

inches of mercury x 1.133 / specific gravity = feet of liquid

-20 x 1.133 /1 = -22.7 feet of pressure head absolute

Vapor pressure of 68°F water = pressure x 2.31/sg. = 0.27 x 2.31/1 = 0.62 feet

Looking at the friction charts:

100 gpm flowing through 2.5 inch pipe shows a loss of 17.4 feet or each 100 feet ofpipe or 17.4/10 = 1.74 feet loss in the piping

The K factor for one 2 inch elbow is 0.4 x 1.42 = 0.6 feet

Adding these two numbers together: (1.74 + 0.6) = a total of 2.34 feet friction loss in the

pipe and fitting.

NPSHA (net positive suction head available) = 34 + 5 - 22.7 - 0.62 - 2.34 = 13.34 feet.

This is enough to stop cavitation also.

SPECIFIC HEAD

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Where

N = Pump speed RPM

Q = GPM = Pump flow at best efficiency point at impeller inlet

(for double suction impellers divide total pump flow by two).

hsv = NPSHR = Pump NPSH required at best efficiency point.

Suction specific speed (S or NS) is defined as:

CAVITATION


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Suction Cavitation

Suction Cavitation occurs when the pump suction is under a low pressure/high vacuum

condition where the liquid turns into a vapor at the eye of the pump impeller. This vapor is


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Discharge Cavitation

Discharge Cavitation occurs when the pump discharge is extremely high. It normally occursin a pump that is running at less than 10% of its best efficiency point. The high discharge

pressure causes the majority of the fluid to circulate inside the pump instead of being allowed

to flow out the discharge. As the liquid flows around the impeller it must pass through the

small clearance between the impeller and the pump cutwater at extremely high velocity. This

velocity causes a vacuum to develop at the cutwater similar to what occurs in a venturi and

turns the liquid into a vapor. A pump that has been operating under these conditions shows

premature wear of the impeller vane tips and the pump cutwater. In addition due to the high

pressure condition premature failure of the pump mechanical seal and bearings can be

expected and under extreme conditions will break the impeller shaft.

condition where the liquid turns into a vapor at the eye of the pump impeller. This vapor is

carried over to the discharge side of the pump where it no longer sees vacuum and is

compressed back into a liquid by the discharge pressure. This imploding action occurs

violently and attacks the face of the impeller. An impeller that has been operating under asuction cavitation condition has large chunks of material removed from its face causing

premature failure of the pump.

BHP = Flow(GPM) X TDH(FT) x SG /3960xEFFICIENCY(%)

BRAKE HORSE POWER


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BHP = Flow(GPM) X TDH(FT) x SG /3960xEFFICIENCY(%)

Example: BHP = (100 GPM) x (95 Ft) x (1.0) / 3960 x .6

BHP = 4.0

Horsepower at the output shaft of

an engine, turbine, or motor is

termed brake horsepower or shaft

horsepower, depending on what

kind of instrument is used to

measure it. Horsepower of

reciprocating engines, particularly

in the larger sizes, is often

expressed as indicated

horsepower, which is determined

from the pressure in the cylinders.

Brake or shaft horsepower is less

than indicated...

The affinity laws express the mathematical relationship between the several

variables involved in pump performance They apply to all types of centrifugal

THE AFFINITY LAWS


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variables involved in pump performance. They apply to all types of centrifugal

and axial flow pumps.

With impeller diameter D held constant:

With speed N held constant:

Where:

Q = Capacity, GPM

H = Total Head, Feet

BHP = Brake Horsepower

N = Pump Speed, RPM

When the performance (Q1, H1, &

BHP1) is known at some particular

speed (N1) or diameter (D1), theformulas can be used to estimate the

performance (Q2, H2, & BHP2) at some

other speed (N2) or diameter (D2). The

efficiency remains nearly constant for

speed changes and for small changes

in impeller diameter


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The affinity laws listed under 1 above will be used to determine the new performance, with N1 1750

RPM and N2 = 2000 RPM. The first step is to read the capacity, head, and horsepower at several


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points on the 13" dia. curve in Fig. 9 below. For example, one point may be near the best efficiency

point where the capacity is 300 GPM, the head is 160 ft, and the BHP is approx. 20 hp.

This will then be the best efficiency point on the new 2000 RPM curve. By performing the same

calculations for several other points on the 1750 RPM curve, a new curve can be drawn which will

approximate the pump's performance at 2000 RPM, Fig. 9.

Trial and error would be required to solve this problem in reverse. In other words, assume you want

to determine the speed required to make a rating of 343 GPM at a head of 209 ft. You would begin by

selecting a trial speed and applying the affinity laws to convert the desired rating to the corresponding

rating at 1750 RPM. When you arrive at the correct speed, 2000 RPM in this case, the corresponding

1750 RPM rating will fall on the 13" diameter curve.


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PUMP PERFORMANCE CURVE


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TRANSPORTATION OF GAS


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Agar volume bahan dalam bentuk gas sedapat mungkin kecil, gas

disimpan dengan cara dimanpatkan. Bila jumlahnya kecil gas disimpan

dan diangkut dalam tabunjg gas bertekanan, tetapi apabila jumlanyabesar, gas diangkut melalui saluran pipa atau diangkut menggunakan tanki

berukuran besar yang bertekanan tinggi. Untuk membuat gas bertekanan

tinggi, dipergunakan kompresor

Kompresi gas berkaitan dengan kenaikan temperatur karena timbulnya

panas kompresi. Misalnya apabila udara ditekan dari 1 bar menjadi 3 bar,

maka temperaturnya naik dar 20oC menjadi 100oC. Untuk keselamatan

kerja tidak dibenarkan bagi suatu kompresor satu tingkat mencapai

kenaikan suhu melebihi 200oC. Dengan perbandingan tekanan antaratekanan awal dengan tekanan akhir dibatasi maksimal 5 kalinya. Bila

diperlukan tekanan akhir yang tinggi, maka kompresi dilakukan secara

bertingkat dengan menggunakan pendingin pada setiap antara tingkatnya

( intercooler )

KOMPRESI GAS

MEKANIKA GAS


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Dari gambar di atas terlihat bahwa hasil perkalian antara tekanan dan

volume selalu 1

Dengan demikian, maka

p x V = konstan

p1 x V1 = p2 x V2 = konstan

Hukum Boyle-Mariotte

1 bar

2 bar

10 bar

1 L 0,5 L 0,1 L

Gambar 1: Hubungan antara tekanan dan volume gas

MEKANIKA GAS


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Hubungan antara variabel keadaan tekanan, suhu dan volume

dihubungkan satu sama lain dengan persamaan keadaan gas ideal.

p1 x V1 p2 x V2

----------- = -------------- = konstan

T1 T2

Suhu Kelvin

Tekanan bar

Volume m3

Hukum Boyle-Mariotte dan Gay Lussac (persamaan gas ideal)

Konstanata Gas Universal dan Persamaan Umum Gas Ideal

Pada keadaan standar, To = 273 oK, po = 1,013 x 105 N/m2, dan vo = 22,4 l

per mole

po x Vo

---------- = R = konstan dimana R = 8,31 Nm/mole K = 8,31 J/mole K

To

MEKANIKA GAS


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Konstanata Gas Universal dan Persamaan Umum Gas Ideal

Apabila jumlah molekul sama dengan n, maka

po x Vo

---------- = R T

n

P V = n R T Persamaan keadaan umum Gas Ideal

n adalah jumlah molekul = massa/massa molekul

EQUIPMENT FOR GAS TRANSPORT (Walas ChPE)


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Gas handling equipment is used to transfer materials

through pipelines, during which just enough pressure or

head is generated to overcome line friction, or to raise or

lower the pressure to some required operating level in

connected process equipment

Fans are used for the supply of gases at relatively low pressures

(


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Blowers is a term applied to machines that raise the pressure to an

intermediate level, usually to less than 40 psig, but more than

accomplished by fans.

BLOWERS

Compressors are any machines that raise the pressure above the

levels for which fans are used. Thus, in modern terminology they

include blowers.

COMPRESSORS

Jet compressors utilize a high pressure gas to raise other gases at

low pressure to some intermediate value by mixing with them.

JET COMPRESSORS

EQUIPMENT FOR GAS TRANSPORT


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Vacuum pumps produce subatmospheric pressures in process

equipment. Often they are compressors operating in reverse but

other devices also are employed.

VACUUM PUMPS

Steam jet ejectors are used primarily to evacuate equipment but also

as pumps or compressors.

STEAM EJECTORS

Compressor operations can be categorized under three thermodynamic

categories:

Isothermal. When compression takes place at constant temperature. This

situation is strictly hypothetical. A constant-temperature operation can be

approached only when the compressor runs at an infinitely slow speed.

COMPRESSOR OPERATIONS (Nicholas: Hanbook of ChPE)

COMPRESSOR OPERATIONS


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Compressor operations can be categorized under three thermodynamic

categories:

Adiabat ic. When the net heat lost or gained by the unit to or from the

surroundings is zero. Most plant installations approach this operation, and

the adiabatic equations are widely used.

(reversible adiabat ic = isentrop ic)

Polytropic. Sometimes the compression process has certain associatedirreversibilities. The actual operation is therefore approaching adiabatic,

but not quite. This "approximately adiabatic" operation is called polytropic.

Work per Cycle

COMPRESSOR OPERATIONS (Recporocating Compressor)


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Work per Cycle

For adiabatic compression, the adiabatic change is described by thefollowing equations:

where k is the ratio of specific heats and some typical values for common

gases are : 1.67 for monatomic gases (e.g., He, A, etc.); 1.40 for diatomicgases (e.g., H,, CO, NJ; and 1.30 for tri-, tetra- and penta-atomic gases

(e.g., CO,, CH,, etc.

In the case of isothermal compression, for an ideal gas, we may state the

following:



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For polytropic compression (reversible and adiabat ic ), the change may

be described by the following equations :



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For adiabatic compression: (reversible adiabat ic = isentro pic )

Since non-ideal gases do not obey the ideal gas law (i.e., PV = nRT),

corrections for nonideality must be made using an equation of state such as

the Van der Waals or Redlich-Kwong equations. Another method for anonideal gas situation is the use of the compressibility factor Z, where Z

equals PV/nRT.

And form PV = NZRT, we note that T = PV/NZR, and hence, at any initial

state 1, we may write the following:

For polytropic compression:


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Point 1 represents the initial condition of the gas (P 1 and V 1).

Line 1-2 represents the compression of gas to pressure P 2 , volume V 2

Line 2-3 represents the expulsion of the gas at a constant pressure P 2Line 3-4 represents an expansion of the pressure in the cylinder from P 2 to P1Line 4-1 represents the suction stroke of the piston, during which a volume V 1 of

gas is admitted at constant pressure, P1.


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The total work done per cycle


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The work of compression for an ideal gas per cycle under isothermal

conditions:

Under isentropic conditions, the work of compression:

The work done on the gas during each stage of the cycle is

as follo s


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as follows.

If the compression and expansion are taken as isentropic,

the work done per cycle is therefore:


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Now (V1 — V3) is the swept volume, Vs, say; and V3/(V1 — V3) is the

clearance c.

and:

The value of V4 is not known explicitly, but can be calculated interms of V3, the clearance volume. For isentropic conditions:


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The gas is frequently cooled during compression so that the workdone per cycle is less than that given by equation above, and ˠ is

replaced by some smaller quantity k.

The total work done on the fluid per cycle is therefore:

Thus:

Compression Ratio (R)


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Compression ratio pada praktiknya tidak lebih dari 4:1

Total power untuk kompresor akan minimum apabila compression ratio padasetip stage sama


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PUMP SELECTION


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COMPRESSOR SELECTION


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JENIS KOMPRESOR


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transp fluida dasar 3 juni_14

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