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INTRODUCTION

Fuel Oil:

Fuel oil is a fraction obtained from petroleum distillation, either as a

distillate or a residue. Broadly speaking, fuel oil is an liquid petroleum product that is

burned in a furnace or boiler for the generation of heat or used in an engine for the

generation of power, except oils having a flash point of approximately +40 °C (104 °F)

and oils burned in cotton or wool-wick burners. In this sense, diesel is a type of fuel oil.

Fuel oil is made of long hydrocarbon chains, particularly alkanes, cycloalkanes and

aromatics. The term fuel oil is also used in a stricter sense to refer only to the heaviest

commercial fuel that can be obtained from crude oil, heavier than gasoline and naphtha.

Fuel oil in the United States is classified into six classes, according to its

boiling temperature, composition and purpose. The boiling point, ranging from 175 to

600 °C, and carbon chain length, 20 to 70 atoms, of the fuel increases with number.

Viscosity also increases with fuel oil number and the heaviest oil has to be heated to get it

to flow. Price usually decreases as the fuel number increases.

Some commonly used fuel oils are

High Speed Diesel

Light Furnace Oil

Heavy Furnace Oil

Light Diesel Oil

Properties of Fuel Oil:

Some of the important properties of fuel oil are

1. Viscosity: Viscosity is a measure of the resistance of a fluid to being deformed by

either shear stress or extensional stress. It is commonly perceived as "thickness",

or resistance to flow. Viscosity describes a fluid's internal resistance to flow and

may be thought of as a measure of fluid friction.

2. Flash Point: The flash point of a flammable liquid is the lowest temperature at

which it can form an ignitable mixture in air. At this temperature the vapor may

cease to burn when the source of ignition is removed. The flash point is often used

as one descriptive characteristic of liquid fuel, but it is also used to describe

liquids that are not used intentionally as fuels.

3. Pour Point: The pour point of a liquid is the lowest temperature at which it will

pour or flow under prescribed conditions. It is a rough indication of the lowest

temperature at which oil is ready to pump. Also, the pour point can be defined as

the minimum temperature of a liquid, particularly a lubricant, after which, on

decreasing the temperature, the liquid ceases to flow.

4. Density: The density of fuel oil is an important factor in the design of the fuel oil

handling system. Density greatly depends on the temperature of the fuel oil and in

turn will determine the viscosity and the power required for pumping it.

Uses of Fuel Oil:

1. Fuels are used in house hold purpose, Industrial purpose and transportation

purpose etc.

2. Electricity power can also produced by diesel generators but it is very pollutant

and it is not economical to use the power continuously.

3. In power plants Fuel oils are very useful to produce steam by heating up the water

to convert into Electric power

4. Heavy fuel oils continue to be used in the boiler "lighting up" facility in every

coal-fired power plant. It is analogous to lighting kindling to start a fire - without

performing this simple function it is difficult to begin the large-scale combustion

process.

Fuel Oil Handling System:

A fuel oil handling system is a system which

is used to supply the fuel oil to the boiler/furnace/plant. The basic functions which a fuel

oil handling system must perform in order to achieve this are

1. Filtration

2. Heating

3. Pumping

4. Storage and

5. Transfer of the fuel oil.

A schematic diagram of a general fuel oil handling system is as shown in the

given figure

Fig.1 Fuel Oil Handling System

Applications of Fuel Oil Handling System:

1. Used in thermal power plants for storage and supply of fuel to boilers.

2. Used in manufacturing industries for the firing of oil furnaces.

3. Used at fuel processing and petroleum refining industries.

4. Used for handling of various types of oils in food processing, transportation and

petroleum industries.

5. It is generally used wherever heavy fuel oil is used.

Components of Fuel Oil Handling System:

The main components of a fuel oil handling system as illustrated in

the figure are

1. Skids

2. Storage Tanks

3. Industrial Heaters

4. Piping

5. Day Tank

The fuel oil handling system is designed according to the specifications of the customer

as well as the standards of the American Petroleum Institute.

The various components listed above will now be discussed in brief

Skids:

A skid is basically an assembly of various components such as

Frame: The frame is constructed using steel beams and fabricated using arc

welding. All the main components of the skid such as pumps, piping, etc are

supported by the frame.

Pumps: Generally three types of pumps are used for pumping of fuel oil according

to the discharge and pressure requirements. They are

1. Centrifugal Pump: For high discharge and high pressure heads.

2. Screw Pump: For low discharge and high pressures.

3. Gear pump: For medium discharges and pressures.

Here we use mainly Gear pump to pump the oil from Supplier tank to

Storage tank and from Storage tank to Boiler.

GEAR PUMP:

A gear pump uses the meshing of gears to pump fluid by

displacement. They are one of the most common types of pumps for hydraulic fluid

power applications. Gear pumps however are also widely used in chemical installations

to pump fluid with a certain viscosity. Gear pumps are fixed displacement, meaning they

pump a constant amount of fluid for each revolution. Some gear pumps are designed to

function as either a motor or pump.

Theory of operation

As the gears rotate they separate on the intake side of the pump,

creating a void and suction which is filled by fluid. The fluid is carried by the gears to the

discharge side of the pump, where the meshing of the gears displaces the fluid. The

mechanical clearances are small—on the order of a thousandth of an inch (micrometers).

The tight clearances, along with the speed of rotation, effectively prevent the fluid from

leaking backwards.

The rigid design of the gears and housing allow for very high pressures and the ability to

pump highly viscous fluids.

http://en.wikipedia.org/wiki/Viscosity

http://en.wikipedia.org/wiki/Fluid

http://en.wikipedia.org/wiki/Hydraulic_motor

http://en.wikipedia.org/wiki/Hydraulic_machinery


http://en.wikipedia.org/wiki/Pump

Many variations exist, including; helical and herringbone gear sets (instead of spur

gears), lobe shaped rotors similar to Roots Blowers (commonly used as superchargers),

and mechanical designs that allow the stacking of pumps. The most common variations

are shown below (the drive gear is shown blue and the idler is shown purple).

Gear pump design for hydraulic

power applications.

Suction and pressure ports need to interface where the gears mesh (shown as dim gray

lines in the internal pump images). Some internal gear pumps have an additional, crescent

shaped seal (shown above, right).

Piping: All the piping of the skid is as per IS 2062 material standard and is

designed for the required discharge and pressure.

Valves: Various types of valves are used in the skids such as flow control valves,

pressure relief valves, non return valves, etc.

Meters: Different types of measuring instruments such as a thermometer, pressure

gauge, flow meter and viscometer are used in order to measure the temperature,

pressure, rate of flow and viscosity of the fluid.

Heaters: Sometimes an In-line heater is used on the skid if the fuel oil is very

viscous and the surrounding temperature does not allow for the smooth flow of

the oil.



http://en.wikipedia.org/wiki/Superchargers

http://en.wikipedia.org/wiki/Herringbone_gear

http://en.wikipedia.org/wiki/File:Gear_pump.png

Filters : Filters are used in the skids to filter the fuel oil as per the requirement.

They are made of MS Fabricated basket with flange connection. Filters are

generally of two types. They are

1. Simplex Filter

2. Duplex Filter (combination of two simplex filters connected with two

3-way interconnecting ball valves).

Fig.2 Duplex Filter

Simplex and duplex filters contain a stainless steel mesh backed with MS

perforated sheet. They are provided on the unloading skid with 500microns of

filtration and on the transfer skid with 250microns of filtration.

Fig.3 Front View of a Skid

Fig.4 Rear View of a Skid

Pressure Gauge

Frame

Docking ports

Gear Pumps

Duplex Filter

Skids are used to transfer fuel oil from one point to another.

They are divided into two types based on their function. They are

1. Unloading Skid: which is used to unload fuel from the tanker and supply it

to the storage tank. This has a filtration of about 500microns in order to

remove the sediments from the fuel oil before being pumped to the storage

tank.

2. Transfer Skid: This is used to transfer fuel oil from the storage tank to the

day tank. It has a filtration of around 250microns before being pumped to

the day tank and then to the furnace.

Design Considerations: The skid is designed based on the specifications of the customer

such as the number of unloading ports, rate of discharge of the fuel oil, number of pumps,

etc.

Storage Tanks:

These are basically large tanks which are used to store fuel oil in

bulk for later use in the plant. The capacity of these tanks usually varies from a few kl

(kilo liters) to sometimes up to 4000kl depending upon the customer’s specifications.

There are generally two types of storage tanks

1. Horizontal Storage Tank

2. Vertical Storage Tank

Components of a Storage Tank: The following are the basic components of a storage

tank. They are

Air vent: This vent is provided in order to relieve the build up of internal pressure

inside the storage tank due to the heating of the fuel oil.

Drain vent: It is provided in order to drain the storage tank during regular

maintenance and repair work.

Inlet nozzle: The fuel oil is pumped into the storage through the inlet nozzle

which is usually located radially near the top of the tank.

Outlet nozzle: This nozzle is also positioned radially but near the bottom of the

storage tank.

Manhole: It is basically an opening in the storage tank through which the repair

and maintenance worker can climb into the tank and is usually located at the top

of the tank on its roof.

Level indicator: The level indicator is basically an arrow and scale type of a

measuring device which is used to determine the level of fluid in the storage tank.

Spare nozzle: A couple of spare nozzles are provided on the storage tank in case

the primary nozzles do not work or need maintenance and repair work.

Heaters: Industrial heaters of the immersion or outflow type are used in the

storage tank in order to heat the fuel oil to its required temperature.

Fig.5 Vertical Cylindrical Storage Tank

Design Considerations: The storage tanks are designed as per the standards of the

American Petroleum Institute (API 650) or Indian Standards (IS 803) and the given

specifications of the customer. The design is based upon the estimated or required

capacity of the tank as well as the space conditions.

Industrial Heaters:

There are Electrical heaters and Steam traced heaters which

are used to heat the fuel oil in order to decrease its viscosity and to increase its flow. We

are mainly concentrated on the Steam traced heaters which use steam for heating and are

specified according to the Surface area of the heater. There are three types of heaters.

They are

1. Floor coil Heater: This type of heater is used inside the storage tank and is

submerged in the fuel oil, below the outlet nozzle of the tank. These are usually

high capacity heaters.

2. Outflow Heater: This type of heater is used in the outlet nozzle of the tank to

heat the fuel oil uniformly when it leaves the storage tank. This type of heater

differs from the floor coil type in the way that these are usually low capacity

heaters and is used if the fuel oil is not so viscous and can be easily made to flow

with the addition of only a small amount of heat.

Fig.8 Outflow Heaters

Fig.9 Outflow heater

3. In-line Heater: These type of heaters are used in-line with the pipeline and are

used for pre-heating the fuel oil at regular intervals in the course of the piping.

Inlet to Fuel Oil

Outlet

Inlet Nozzle

Outlet Nozzle

Storage Tank

Outflow heater

Fuel Oil InFuel Oil Out

Fig.10 In-line Heater

Design Considerations: The industrial heaters are designed based upon the type of fuel

oil, its viscosity, its flash point and pour point, the flow rate required and the required

temperature.

Day Tanks:

These are basically smaller capacity tanks which are used for supplying

the fuel oil to the furnace/boiler/plant according to the running capacity of the plant per

day. The capacity of the day tank usually lies between a few kl (kilo liters).

Fig.11 Day Tank

Heater ShellHeating Elements

Design Considerations: These tanks are designed as per the standards of the American

Petroleum Institute and the given specifications of the customer. Heaters of smaller

capacities are provided in order to heat the fuel oil.

Other Equipment Used:

3 way ball valve

2 way ball valve

Non return valves

L-port valve

Pressure relief valve

Flow meters

Pressure gauges

Viscosity meter

Temperature gauges

Insulation (wool blanket covered with Aluminum sheet)

Heating jackets

Level indicators

Level switches

Thermostats

Control panels

DESIGN CALCULATIONS OF FUEL HANDLING SYSTEM

`The fuel handling system is the system is used to supply the fuel from the

supplier tank to the storage tank and from storage tank to the boiler. The components

which we have to design in the fuel handling system are

a. Unloading Skid :

Pipe for guide the fuel from supplier tank to the storage tank

Strainer for filtering the oil

Pump for pressurized the oil from supplier tank to the storage tank

Storage tank to store the oil for further use

Floor coil heater to raise the temperature of the oil up to its pour

point temperature

b. Transport Skid :

Pipe for guide the fuel from the storage tank to the Boiler

Strainer for filtering the oil

Pump for pressurized the oil from Storage tank to the Boiler

c. Total steam requirement :

Steam requirement for across the Fuel Handling System to raise

the temperature of the fuel to its Pour point temperature.

Design of the above mentioned components are calculated by using the

inputs given by the customer according to his plant requirement. Our main intension is to

Design the components involve in the fuel handling system up to the satisfaction of the

customer requirement.

Input data given by the customer to design the Fuel handling system to

there plant are :

UNLOADING SKID :

I. PIPE AND PUMP

Required Discharge (Q) = 10 m3 / h

Velocity of flow at the suction side of the Gear pump (V1) = 0.9 m / sec

Vapour head of the oil (Vh) = 0.2 mwc

Level of the supplier tank outlet w.r.t 0.0 m = 1.3 m

Level of the pump suction w.r.t 0.0 m = 0.6 m

Suction head = (1.3 – 0.6) = 0.7mwc

Level of Pump discharge w.r.t 0.0m = 0.6 m

Level of Tank inlet w.r.t 0.0m = 11.9 m

Static head = (11.9 – 0.6) = 11.3

Design Viscosity (K) at 50ºC = 180 centi strokes

Velocity of flow at the delivery side of the Gear pump (V2) = 1.5 m / sec

II. STORAGE TANK

No. of Storage tank = 1

Diameter of the Storage tank (D) = 10 m

Height of the Storage tank (H) = 8.5 m

Net Design Liquid Height (He) = 8 m

Effective Capacity, (Q1) = 628 m3

Design Specific Gravity (G) = 1

Design Density of liquid (ρ) = 1000 kg / m3

Design Temperature, (T) = 120 0 c

Corrosion Allowance for Roof Plate, (C) = 1 mm

Corrosion Allowance for Bottom & Shell plate, (C) = 2 mm

Material of Construction = IS 2062 Gr. B

Maximum Allowable working Stress, = 165 MPa

Width of Shell Plates, (Ws) = 1.5

III. FLOOR COIL HEATER

Operating Pressure = 4.5 ata

Pressure at inlet of Floor coil heater = 4.5 ata

Satruated Temp. at 4.5 ata = 138 0 c

Latent heat at inlet pressure of 4.5 ata (hf) = 512.89 Kcal / kg.

Type of Oil = HFO

Maxi Density, (D1) = 875 Kg / m3

Specific heat, (Cp) = 0.46 Kcal / kg 0 c

Initial Temperature, (T1) = 12 0 c

Final Temperature (T2) = 55 0 c

TRANSFER SKID :

IV. PIPE AND PUMP




Length of the pipe = 12 m



According to the plot plan and the input data given by the customer we

have to draw a “PROCESS AND INSTRUMENTATION DIAGRAM” and send to the

customer for corrections. After refering of our “PROCESS AND INSTRUMENTATION

DIAGRAM” if there is any correction they will correct on the “P &ID” and send to us to

Design according to that corrections.

The “P & ID” according to the above information

After corrections we have start design the components according to the input data given

by the customer.

UNLOADING SKID :

Pipe Design:

All the piping used in the fuel oil handling system is made of material SA

106 Grade B. The required data to design the pipe is :

Required Discharge (Q) = 0.002 m3 / sec


Q = A x V

A = Q / V

= 0.002 / 0.9

= 0.003

d = [(0.003 x 4) / 3.12] 1/2

= 0.061 m

Inner diameter of the pipe = 0.061m

Length of the pipe = According to the plot plan.

PIPE THICKNESS

We have calculate the pipe thickness according to the internal pressure of the

steam which flow under the pipe.

T = ( P x D) / [2((S x E) + (P x Y))] + CA

Where

Pressure (P) = 3.5 Kg / cm2

Diameter(D) = 6.1 cm

Allowable Stress (S) = 1680 Kg / cm2

Joint efficiency = 0.7

Wall thickness factor (y) = 0.04

T = ( P x D) / [2((S x E) + (P x Y))] + CA

= (3.5 x 6.1) / [ 2((1680 x 0.7) + (3.5 x 0.04))] + CA

= 0.029 cm

Thickness of the Pipe is equal to 0.00029m

FILTER CALCULATIONS

The volume of the basket = 10 x diameter of the pipe (As per

Customer)

= 10 x 0.061

= 0.61 m

The filtration Capacity is 1000 microns

1 micron = 10-6 m

1000 micron = 1000 x 10-6

= 0.001 m

= 1 mm

PUMP SELECTION:

Usually, Gear pumps are used for pumping fuel oil in most of the fuel oil

handling systems since they require lesser maintenance than other type of pumps and also

they are used for medium pressures and discharges.

P = Q x [(NPSH) + (DH)] / (270 x η)

P = power of pump in (HP)

Q = discharge of pump in (m3/sec)

NPSH = Net Positive Suction Head (m)

DH = Discharge Head (m)

η = efficiency of pump=0.5

Net Positive Suction Head :

The head of the fuel at the suction valve of the pump is known as the Net

Positive Suction Head.

NPSH = (Atm. head + Suction head) - (Friction head + Vapour head)

Required Input data :

Atm. Head = 10.3 mwc


Suction head = (1.3 – 0.6) = 0.7mwc


Now we have to find the Friction head loss by using the “DARCESE FORMULA”

H = (4fLV2) / (2gd)

Where,

H = head loss in m = ?

f = friction factor = ?

L = Overall length of the pipe and fittings = ?

V = Velocity of fluid in m/sec = 0.9m/sec

G = acceleration due to gravity = 9.8m/sec2

d = dia. of pipe in m = 0.061m

Overall length of the pipe and fittings before suction valve of the pump(L)

= [(length of the suction pipe) + (2 x length of the 900bend)

+ (length of the non return value) + ( length of the 45 0

bend) + (length of the ball valve)]

According to Standards of Pipe Fittings :

45º Elbow (Le/D=15.75)


Tee (Le/D=45)

Gate Valve (Le/D=7.65)

Globe Valve (Le/D=270)

Ball Valve (Le/D=2.25)

Plug valve (Le/D=7.65)

Non return Valve (Le/D=90)

Overall length (L) = 30 + (2 x 1.23) + (5.49) + (0.96) + (0.13)

= 40 m

Friction Coefficient (f) = 16 / Re (When Re <= 2100, if Re > 2100 we have

to find the “f” by modi diagram)

Reynolds number = VD / Design viscosity

= (0.9 x 0.061) / (180 x 10 -6 )

= 305

Friction Coefficient (f) = 16 / Re = 16 / 305 = 0.05

By substituting the values of Overall length and Friction Coefficient in the

Friction head loss formula we can find the Friction head.

H = (4fLV2) / (2gd)

Friction head loss ( H ) = (4 x 0.05 x 40 x 0.92) / (2 x 9.81 x 0.061)

= 5.4m


= (10.3 + 0.7 ) - (5.4 + 0.2 )

= 5.4 mwc

DELIVERY HEAD = (STATIC HEAD LOSS) + (FRICTION HEAD LOSS)

Required input data :

Level of Pump discharge w.r.t 0.0m = 0.6 m

Level of Tank inlet w.r.t 0.0m = 11.9 m

Static head = (11.9 – 0.6) = 11.3

For Friction Head Loss

H = (4fLV2) / (2gd)

Where,








= [(length of the delivery pipe) + (2 x length of the

900bend) + (length of the ball valve)]

Overall length (L) = 40 + (2 x 0.83) + (0.092)

= 42 m



Reynolds number = VD / kinematic viscosity

= (1.5 x 0.041) / (180 x 10 -6 )

= 341




H = (4fLV2) / (2gd)


= 18.79m

DELIVERY HEAD = (STATIC HEAD LOSS) + (FRICTION HEAD LOSS)

= (11.3 + 18.79)

= 30 m

P = Q x [(NPSH) + (DH)] / (270 x η)

= 10 x [5.4 + 30] / (270 x 0.5)

= 2.62 H.P

We require 2.62 Horse Power motor to pump the oil from the supplier tank to the storage

tank.

FLOOR COIL HEATER

The Floor Coil Heater is used to raise and maintain the pour point

temperature of the oil, which stored in the storage tank for further use. This heater is

arranged at the bottom of the Storage tank which transfer heat from the steam to the oil.

Floor Coil Heater is just like a shell and tube type Heat Exchanger that the steam is flow

through the tubes and the oil which we want to boil is stored in the shell.

To design the Floor Coil Heater we have to calculate

The heat load taken by the oil for initial heating

The heat loss through Tank shell, Bottom & Roof during initial heating

The heat loss through Tank shell, Bottom & Roof during maintaining heating

The heat load taken by the storage tank material

To calculate above heat loads and heat losses the required input data is

Inner Diameter of the Storage tank (d) = 10 m

Outer Diameter of the Storage tank without insulation (d1) = 10.016 m

Outer Diameter of the Storage tank with insulation (d2) = 10.09 m

Height of the Storage tank (h) = 8.5 m


Operating Pressure (P) = 4.5 ata

Pressure at inlet of Floor coil heater = 4.5 ata

Satruated Temp. at 4.5 ata (T) = 138 0 c

Latent heat at inlet pressure of 4.5 ata (hf) = 512.89 Kcal / kg.

Type of Oil = HFO

Maxi Density, (D1) = 980 Kg / m3

Specific heat, (Cp) = 0.44 Kcal / kg 0 c

Initial Temperature, (T1) = 12 0 c

Final Temperature (T2) = 55 0 c

Average Temperature (Tavg) = (55 +12) / 2 = 33.50c

Heating time (Ht) = 72 hours (Assumed)

THE HEAT LOAD TAKEN BY THE OIL DURING INITIAL HEATING (Q1)

The heat load taken by the oil during initial heating (Q1) = [m Cp(T2 - T1)] / Ht

Mass(m) = Volume x density

= [(3.14 / 4) x d1 x h] x 980

= 615752 kg

Q1 = [m Cp(T2 - T1)] / Ht

= [ 615752 x 0.44 x (55 – 12)] / (72)

= 161805 Kcal / hr.

THE HEAT LOSS THROUGH SHELL, ROOF AND BOTTOM PLATES

DURING INITIAL HEATING :

HEAT LOSS THROUGH SHELL OF THE TANK :

The heat loss through the shell during initial heating (Q2)

Q2 = (Tavg – T1) / [(d1 / 2k) x (ln(d2 / d1)) + (d1 / fd2)]

Film Coefficient (f ) = (f) convection + (f) radiation

(f) convection = 1.957x(T3-T1)^0.25 x (2.857xV+1)^0.5

(f) radiation = 5.755 x 10^-8 x E x (T34-T14)/(T3-T1)

Where

Outside surface temp. of insulation, T3 (Consider Max. Outside Temp. As 50ºC) = 323

Ambient Air Temp, T1 = 273 k

Average wind speed, V = 50 m / sec

By substituting all the above values in the (f) convection and (f) radiation and by

summing them we get Film Coefficient value as given below

Film Coefficient = 17.46 kcal / hr.m2 oc

Q2 = (Tavg – T1) / [(d1 / 2k) x (ln(d2 / d1)) + (d1 / fd2)]

= (33.5 – 12) / [( 10.016 / 0.088) x (ln(10.9 / 10.016) ) + (10.016 / (17.46 x 10.09))

= 24.03 Kcal / hr m2

Surface area (A) = (3.14 / 4) x d x h

= (3.14 / 4) x 10.016 x 8.5

= 67 m2

Heat loss through tank shell = Q2 x A

= 24.03 X 67

= 1606 Kcal / hr.

HEAT LOSS THROUGH ROOF PLATE :

Rate of heat loss through Roof plate(Q3) = (Tavg – T1) / [(L /k) + (1 / f)]

= 21.5 / [(0.04 / 0.044) + (1 / 17.46)]

= 22.25 Kcal / hr m2

Surface area (A) = (3.14 / 4) x d2

= (3.14 / 4) x 10.112

= 80.27 m2

Heat loss through Roof plate = Q3 x A

= 22.23 x 80.27

= 1806 Kcal / hr.

HEAT LOSS THROUGH BOTTOM PLATE :

Heat loss through Bottom plate(Q4) = KA (Tavg – T1)

= 0.86 x [(3.14 / 4) x (10)2] (33.5)

= 1417 Kcal / hr.

THE HEAT LOSS THROUGH SHELL, ROOF AND BOTTOM PLATES

DURING MAINTAINING HEAT :

HEAT LOSS THROUGH SHELL OF THE TANK :

Rate of heat loss through the shell during maintaining heat (Q5)

Q5 = (T2 – T1) / [(d1 / 2k) x (ln(d2 / d1)) + (d1 / fd2)]

Film Coefficient (f ) = (f) convection + (f) radiation

(f) convection = 1.957x(T3-T1)^0.25 x (2.857xV+1)^0.5

(f) radiation = 5.755 x 10^-8 x E x (T34-T14)/(T3-T1)

Where

Outside surface temp. of insulation, T3 (Consider Max. Outside Temp. As 50ºC) = 323

Ambient Air Temp, T1 = 273 k

Average wind speed, V = 50 m / sec

By substituting all the above values in the (f) convection and (f) radiation and by

summing them we get Film Coefficient value as given below

Film Coefficient = 17.46 kcal / hr.m2 oc

Q5 = (T2 – T1) / [(d1 / 2k) x (ln(d2 / d1)) + (d1 / fd2)]

= (55 – 12) / [( 10.016 / 0.088) x (ln(10.9 / 10.016) ) + (10.016 / (17.46 x 10.09))

= 48.31 Kcal / hr m2

Surface area (A) = (3.14 / 4) x d x h

= (3.14 / 4) x 10.016 x 8.5

= 67 m2

Heat loss through tank shell = Q5 x A

= 48.31 X 67

= 3236 Kcal / hr.

HEAT LOSS THROUGH ROOF PLATE :

Rate of heat loss through Roof plate = (T2 – T1) / [(L /k) + (1 / f)]

= 43 / [(0.04 / 0.044) + (1 / 17.46)]

= 44.51 Kcal / hr m2

Surface area (A) = (3.14 / 4) x d2

= (3.14 / 4) x 10.112

= 80.27 m2

Heat loss through Roof plate (Q6) = Q3 x A

= 44.51 x 80.27

= 3572 Kcal / hr.

HEAT LOSS THROUGH BOTTOM PLATE :

Heat loss through Bottom plate(Q7) = KA (Tavg – T1)

= 0.86 x [(3.14 / 4) x (10)2] (43)

= 2902 Kcal / hr.

Heat load taken by the material of the tank = m Cp (T1 – T2) / Ht


= [(3.14 / 4) x d2 x h] x 7850

= 5240569 kg

Q8 = [m Cp(T2 - T1)] / Ht

= [ 5240569 x 0.52 x (55 – 12)] / (72)

= 1721381 Kcal / hr.

TOTAL HEAT LOAD PER HOUR

Initial Heat load consider 10 % Marjin (Q1) = (Q1 + Q2 + Q3 + Q4 + Q8) x 1.1

= 2076816 Kcal / hr.

Maintain heat load consider10 % Marjin(Q2)= (Q5 + Q6 + Q7 ) x 1.1

= 9710 Kcal / hr.

HEAT TRANSFER AREA OF THE FLOOR COIL HEATER

Heat transfer area required for floor coil heater(a) = (Q1) / {K x [T sat - (T1 + T2) / 2])

= (2076816) / {75 x 114.42)

= 242 m2

LENGTH OF THE FLOOR COIL HEATER

Actual Length of Floor Coil pipe required = Heat transfer area / Area of pipe per

meter length

Floor Coil Pipe considered is 50NB Seamless Carbon Steel to ASTM A-106, Gr.B, Sch-

40

Surface area of 50NB Pipe per meter, a4 = 0.189 m2 / m

Length of Pipe required (L) = 242 / 0.189

= 1280 m

DESIGN OF THE STORAGE TANK

The main components which involve in the construction of the storage

tank that which we are going to design are

Shell Courses

Rafters

Center column

SHELL COURSES :

Shell courses are the structural steel plates conforming to the material standard IS 2062

which are rolled and welded to form a cylindrical shape according to our requirement and

are generally available in the market in standard dimensions, i.e. 1.25x6.3m and

1.5x6.3m of variable thicknesses. Therefore, according to the required dimensions of the

tank, plates of optimum dimensions have to be chosen in order to minimize the waste.

RAFTER :

Rafter is a structural beam which is used to support the roof of the tank. It is generally in

“I” section and one end of the section is supported by the Center column and other is

supported by the shell courses.

CENTER COLUMN:

Center column is a structural Column which is used to support the Rafter.

Requied input data for design

No. of Storage tank = 1

Diameter of the Storage tank (D) = 10 m

Height of the Storage tank (H) = 8.5 m


Effective Capacity, (Q1) = 628 m3

Design Specific Gravity (G) = 1

Design Density of liquid (ρ) = 1000 kg / m3

Design Temperature, (T) = 120 0 c

Corrosion Allowance for Roof Plate, (C) = 1 mm

Corrosion Allowance for Bottom & Shell plate, (C) = 2 mm


Maximum Allowable working Stress, (Sd) = 165 MPa

Maximum Allowable Hydrostatic Stress, (St) = 175.7 MPa

Width of Shell Plates, (Ws) = 1.5


Density of material (ρ1) = 7850 kg / m3

SHELL COURSES

No of shell courses required to construct the require storage tank = H / Ws

= 8.5 / 1.5

= 6 Nos.

Thickness of the Courses

Thickness of 1st Course :

Thickness of the 1st course = (4.9xGxDx(Hc1-0.3)/Sd) + C

Where

Hc1 = Ht - ((n-1) x Ws)

Because it is 1st course n = 1

Therefore

Hc1 = 8.5 m

Thickness of the 1st course according to Maximum Allowable working stress = 4.44 mm

Thickness of the 1st course = (4.9xGxDx(Hc1-0.3)/St) + C

Where

Hc1 = Ht - ((n-1) x Ws)

Because it is 1st course n = 1

Therefore

Hc1 = 8.5 m

Thickness of the 1st course according to Maximum Hydrostatic stress = 3.87 mm

Thickness of 2nd Course :

Thickness of the 2nd course = (4.9xGxDx(Hc1-0.3)/Sd) + C

Where

Hc2 = Ht - ((n-1) x Ws)

Because it is 2nd course n = 2

Therefore

Hc2 = 7 m

Thickness of the 2nd course according to Maximum Allowable working stress = 3.99 mm

Thickness of the 2nd course according to Maximum Hydrostatic stress = 3.53 mm

Thickness of 3rd Course :

Thickness of the 3rd course = (4.9xGxDx(Hc1-0.3)/Sd) + C

Where

Hc3 = Ht - ((n-1) x Ws)

Because it is 3rd course n = 3

Therefore

Hc3 = 5.5 m

Thickness of the 3rd course according to Maximum Allowable working stress = 3.33 mm

Thickness of the 3rd course according to Maximum Hydrostatic stress = 3.18 mm

Thickness of 4th Course :

Thickness of the 4th course = (4.9xGxDx(Hc1-0.3)/Sd) + C

Where

Hc4 = Ht - ((n-1) x Ws)

Because it is 4th course n = 4

Therefore

Hc4 = 4 m

Thickness of the 4th course according to Maximum Allowable working stress = 3.1 mm

Thickness of the 4th course according to Maximum Hydrostatic stress = 2.84 mm



Where

Hc5 = Ht - ((n-1) x Ws)


Therefore

Hc5 = 2.25 m





Where

Hc6 = Ht - ((n-1) x Ws)


Therefore

Hc6 = 1 m



As per Cl.3.6.1.1, of API 650, Refer page 3-6:

For < 15m nominal tank diameter, nominal plate thickness is 8 mm. The thickness of all

the shell courses are less than 5mm, So we can take the shell thickness as

Thickness of the shell of the storage tank = 8 mm

THICKNESS OF BOTTOM PLATE

As per Clause No.5.5.3,Table 5-1, Page 5-10 of API 650, 2007; Bottom plate thickness

required is 6mm exclusive of Corrosion Allowance. Hence Thickness of bottom plate

provided is 8 mm.

THICKNESS OF ROOF PLATE

As per Clause No. 5.10.2.2, Page 5-60 of API 650, 2007; Roof plate shall have a

minimum nominal thickness of 4 mm, Also Corrosion Allowance (2mm) shall be added

for the plates of supported roofs, Hence the thickness of roof plate provided is 6 m.

By the above calculations we came to known that the

No of Shell courses required = 6

Thickness of shell courses = 8 mm

Thickness of Bottom plate = 8 mm

Thickness of Roof plate = 6 mm

DESIGN OF RAFTER

Design of Rafter for roof with slope 1:16.

Slant length of the roof (l) = [ DO2 + (DO /16)2] 1/2

DO = outer radius with insulation of tank

= [5.0082 + 0.3132] ½ = = 5.02 m

Thickness of the roof plate (t ) = 6mm

Weight of the roof (w) = Volume x Density

= [(3.14 / 4) x (2 x l)2 x t] x 7850

= 3728 kg

Roof Area of Tank, (Ar) = Π x (D/2) x l = 78.7 m2

Dead Weight of the roof per area(Wd) = w / Ar

= 3728 / 78.7

= 47.31 Kg /m2

Live load on the roof (Assumed)(Wl) = 150 Kg / m2

As per Cl. 5.10.4.4, Rafters shall be spaced so that, in the outer ring, their centers shall

not be more than 0.6Π m i.e 1.884 m, apart measured along the circumference of the

tank.

No. of rafters required = (3.14 x D) / 2

= 17

We should always take even number rafters. So the No. of rafters(n) = 18 Nos

Max. Distance between two adjacent Rafter, (Xmax) = (3.14*D*100)/18

= 174.5 cm

Min. Distance between two adjacent Rafter, (XMIN) = (3.14*d) / 18

Where d = dia of the crown plate = 63

= 11cm

Length of the rafter (Lr) = (Slant length of the roof) – (dia of crown)

= 5.008 - 0.63

= 437 cm

Average width of Roof Area supported by individual Rafter, b = (Xmax + Xmin)/2

= (174.5 + 11) / 2

= 92.75 cm

Dead weight on each rafter(Wd) = (Wd / 104) x b

= 47.31 x 92.75

= 0.43 Kg / cm

Live weight on each rafter(Wl) = (Wl / 104) x b

= (150 / 104) x 92.75

= 1.39 Kg / cm

Total weight on each rafter (Wr) = (Wd) + (Wl)

= 0.43 +1.39

= 1.82 Kg / cm

Loading on Rafter member is considered as Simply Supported beam with uniform

loading,

Therefore

Max. Bending Moment, (Mmax) = (Wr x Lr2)/8

= (1.82 x 190969) / 8

= 43445.45 Kg-cm

Section Modulus (Z) = Mmax / Sd

= 43445.45 / 1682.54

= 25.82 cm3

So ISMC 150 member is Considered as a member for the above Rafter.

Deflection Checking for Rafter

Uniform load on Rafter, Wr = 1.82 Kg / cm

Maximum length of Rafter, L = 437 cm

Moment of Inertia for Rafter i.e. ISMC 150, I = 779.4 cm4

Youngs Modulus of Elasticity, E = 200000 MPa

= 2039432 Kg/cm2

Max. Vertical Deflection for Rafter, δ = (5 x WrL4)/(384E x I)

= (5 x 1.82 x 4374) / (384 x

2039432 x 779)

= 0.54 cm

Limiting Vertical Deflection for Rafter, δl = L/325

= 437 / 325

= 1.34 cm

Max. Deflection is under permissible limit, Hence member considered for Rafter i.e.

ISMC 150 is Safe Under loading and deflection.

DESIGN OF CENTER COLUMN

Considering Center Column configuration as a composite section of ISMC of ISMC 250

&200

Slenderness Ratio is the ration of Length of the Column to the least radius of gyration.

When Slenderness ratio is Less then the column strength is high and if the Slederness

ratio is high then the column strength is less.

As per above configuration least radius of gyration, Rm = 10.26 cm

Length of Centre Column considered, Lcc = 1232.81 cm

Induced Slenderness Ratio, λ = Lcc/Rm

= 1232.81 / 10.26

= 120.16

Induced Slenderness ratio is less than Max. Permissible, Hence satisfactory

Allowable Stress in compression, σac = 652.62 Kg / cm2

Induced Compressive stress, σic = Total compressive load(Wc)/Cross Sectional

Area of centre column(Ac)

So we have to find Total compressive load and Cross sectional Area of center

column(Ac)

Total load of each Rafter = (Length of rafter x Weight on rafter) / 2

= ( 437 x 1.82 ) / 2

= 398 Kg

Total Number of Rafter = 18 No.s

Total Compressive Load, Wc = (Total load of each Rafter x Total

Number of Rafter)

= ( 398 x 18 )

= 7164 Kg

Cross sectional area of Center column = 50.51 cm2 ( According to Stand

Induced Compressive stress, σic = Total compressive load(Wc)/Cross

Sectional Area of centre column(Ac)

= ( 7164 / 50.51)

= 142 Kg / cm2

Since σic < σac, Centre column Provided is a composite section of ISMC 200 & ISMC

250 is Satisfactory.

TRANSFER SKID :

PIPE AND PUMP




Length of the pipe = 12 m



TRANSFER SKID :

Pipe Design:

All the piping used in the fuel oil handling system is made of material SA

106 Grade B. The required data to design the pipe is :

Required Discharge (Q) = 0.0002 m3 / sec


Q = A x V

A = Q / V

= 0.0002 / 0.9

= 0.0002

d = [(0.0002 x 4) / 3.12] 1/2

= 0.016 m

Inner diameter of the pipe = 0.016m

Length of the pipe = According to the plot plan.

PIPE THICKNESS

We have calculate the pipe thickness according to the internal pressure of the

steam which flow under the pipe.

T = ( P x D) / [2((S x E) + (P x Y))] + CA

Where

Pressure (P) = 3.5 Kg / cm2

Diameter(D) = 1.6 cm

Allowable Stress (S) = 1680 Kg / cm2

Joint efficiency = 0.7

Wall thickness factor (y) = 0.04

T = ( P x D) / [2((S x E) + (P x Y))] + CA

= (3.5 x 1.6) / [ 2((1680 x 0.7) + (3.5 x 0.04))] + CA

= 0.022 cm

Thickness of the Pipe is equal to 0.00022m

FILTER CALCULATIONS

The volume of the basket = 10 x diameter of the pipe (As per

Customer)

= 10 x 0.016

= 0.16 m

The filtration Capacity is 500 microns

1 micron = 10-6 m

500 micron = 500 x 10-6

= 0.0005 m

= 0.5 mm

PUMP SELECTION :

Usually, Gear pumps are used for pumping fuel oil in most of the fuel oil

handling systems since they require lesser maintenance than other type of pumps and also

they are used for medium pressures and discharges.

P = Q x [(NPSH) + (DH)] / (270 x η)

P = power of pump in (HP)

Q = discharge of pump in (m3/sec)

NPSH = Net Positive Suction Head (m)

DH = Discharge Head (m)

η = efficiency of pump=0.5

Net Positive Suction Head :

The head of the fuel at the suction valve of the pump is known as the Net

Positive Suction Head.

NPSH = (Atm. head) - (Friction head + Vapour head)

Required Input data :

Atm. Head = 10.3 mwc



Now we have to find the Friction head loss by using the “DARCESE FORMULA”

H = (4fLV2) / (2gd)

Where,








= [(length of the suction pipe)

According to Standards of Pipe Fittings :



Tee (Le/D=45)

Gate Valve (Le/D=7.65)

Globe Valve (Le/D=270)

Ball Valve (Le/D=2.25)

Plug valve (Le/D=7.65)

Non return Valve (Le/D=90)

Overall length (L) = 2 m



Reynolds number = VD / Design viscosity

= (0.9 x 0.016) / (180 x 10 -6 )

= 80




H = (4fLV2) / (2gd)


= 4.12 m


= (10.3 ) - (4.12 + 0.2 )

= 5.97 mwc

DELIVERY HEAD = (FRICTION HEAD LOSS)

For Friction Head Loss

H = (4fLV2) / (2gd)

Where,






d = dia. of pipe in m

= 0.015m


= [(length of the delivery pipe)

Overall length (L) = 12 m



Reynolds number = VD / kinematic viscosity

= (1.5 x 0.016) / (180 x 10 -6 )

= 133




H = (4fLV2) / (2gd)


= 55m

DELIVERY HEAD = (FRICTION HEAD LOSS)

= 55 m

P = Q x [(NPSH) + (DH)] / (270 x η)

= 1 x [5.9 + 55] / (270 x 0.5)

= 0.5 H.P

We require 0.5 Horse Power motor to pump the oil from the Storage tank to the Boiler

furnance.

TOTAL STEAM REQUIREMENT

THE HEAT LOAD TAKEN BY THE OIL IN THE SUPPLIER TANK (Q1)

The heat load taken by the oil (Q1) = [m Cp(T2 - T1)] / Ht

Assume Capacity of tanker as 12m3


= 12 x 980

= 11760kg

Q1 = [m Cp(T2 - T1)] / Ht

= [ 11760 x 0.44 x (55 – 12)] / (72)

= 3090 Kcal / hr.

Heat load taken by the material of the tank = m Cp (T1 – T2) / Ht


= 12 x 7850

= 94200 kg

Q2 = [m Cp(T2 - T1)] / Ht

= [ 94200 x 0.52 x (55 – 12)] / (72)

= 29254 Kcal / hr.

Heat loss from the tank Q3 = U x (T2 – T1) x A

= 72 x (55 – 12) x 6

= 18576kcal / hr.

Steam Requirement for Supplier tank(s1) = (Q1 + Q2 + Q3) x 1.1 / Latent heat

= ( 3090 + 29254 + 18576) x 1.1 /

512.89

= 109 Kg / hr

Steam Requirement for Storage tank(s2) = (Q1 + Q2 + Q3 + Q4 + Q5 +

Q6 + Q7+ Q8) x 1.1 / Latent heat

= (2076816 + 9710) / 512

= 4075 Kg / hr.

Heat loss through the pipe Q4 = (T2 – T1) / [(d1 / 2k) x (ln(d2 / d1))

+ (d1 / fd2)]

= (55 -12) / [ (0.061 / 2 x 0.044) x (ln

0.1 / 0.061)) + (0.061 / 17.46 x 0.1 )

= 113 Kcal / hr. m2

Surface area (A) = 3.14 x 0.1 x 134

= 42.076 m2

Rate of heat loss through the pipe(Q4) = Q4 x A

= 113 x 42.07

= 4754 Kcal / hr.

Steam requirement for pipe (s3) = 4754 / 512

= 9.28 Kg / hr.

TOTAL STEAM REQUIREMENT = s1 +s2 + s3 + s4

= 109 + 4075 + 9.28

= 4193 kg / hr

fohs

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