Jones D.S.J., Pujado P.R. Handbook of Petroleum Processing

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SUPPORT SYSTEMS COMMON TO MOST REFINERIES 603

Corrected for emissivity h

= 1.18 × .95 = 1.123 Btu/hr.sqft.

◦

‘b’ =(h

+ h

) × t

=(4.21 + 1.12) × 55

=293 Btu/hr.sqft

‘a’ −‘b’ =244 −293

=−49 Btu/hr.sqft

2nd trial. Assume wall temperature is 110

◦

Carrying out the same calculation procedure as for trial 1:

‘a’ − ‘b’ =+172 Btu/hr.sqft

3rd trial. Assume wall temperature is 115

◦

Again carrying out the calculation procedure as for trial 1:

‘a’ − ‘b’ =+35

The results of the above trials are plotted linearly below:

Final trial. At wall temperature of 117

◦

=8.31 Btu/hr.sqrt.

◦

‘a’ =8.31 × (150 −117) = 274 Btu/hr.sqrt

(corrected) =4.18 Btu/hr.sqft.

◦

(corrected) =1.12 Btu/hr.sqft

◦

‘b’ =(4.18 + 1.12) × (117 − 65)

◦

=275.7 Btu/hr.sqft

‘a’and ‘ b’are close enough call total heat loss 275.7 Btu/hr.sqft.

604 CHAPTER 13

Surface area of wall = circumference × height.

= π D × 180 ft

= 33929 sqft

Total heat loss through wall = 275.7 ×33929

= 9.35 mm Btu/hr

2.0 Calculating heat loss through roof.

Trial 1. Assume roof temperature is 116

◦

=8.38 Btu/hr.sqft.

◦

‘a’ =8.38 ×(150 − 116)

◦

=284.9 Btu/hr.sqft.

(corrected) =(.470 × 2.0) × 1.04 × 6.29

(Note: the number read from Figure 13.9 is multiplied by 2.0 in this case as the roof

is an upward facing plate.)

=6.35 Btu/hr.sqft.

◦

(corrected) =1.165 × 0.95

=1.11 Btu/hr.sqft.

◦

‘b’ =(6.35 + 1.11) × (116 − 65)

◦

=380 Btu/hr.sqft.

‘a’ − ‘b’ =−95 Btu/hr.sqft

Trial 2. Assume a wail temperature of 110

◦

‘a’ − ‘b’in this case =+23 which is within acceptable limits.

The heat loss is taken as an average of ‘a’and ‘ b’

= 338 Btu/hr.sqft.

Total heat loss from the roof = area of roof × 338

= 2827 sqft × 338

= 0.956 mm Btu/hr

3.0 Calculating the heat loss through the ﬂoor.

Assume the ground temperature is 50

◦

F and the heat transfer coefﬁcient is 1.5 Btu/

hr.sqft.

◦

Then heat loss = 1.5 × (150 − 50) × 2827 sqft

= 0.424 mm Btu/hr

SUPPORT SYSTEMS COMMON TO MOST REFINERIES 605

4.0 Total heat loss from the tank.

Heat loss from the Walls =9.350 mm Btu/hr

Heat loss from the Roof =0.956 mm Btu/hr

Heat loss from the Floor =0.424 mm Btu/hr

Total Heat Loss =10.730 mm Btu/hr

5.0 Calculating the tank heater coil surface area required.

The heating medium is saturated 125 psig steam.

Temperature of the steam = 354

◦

Steam side calculations.

Approx resistance of steam, h

=0.001 hr.sqft.

◦

F/Btu

Fouling factor on steam side, r

=0.0005 hr.sqft.

◦

F/Btu

Tube metal resistance r

=0.0005 hr.sqft.

◦

F/Btu

Outside fouling factor r

=0.005 hr.sqft.

◦

F/Btu

Heat transfer coefﬁcient for the steam side =

0.001 + 0.0005 + 0.0005 + 0.005

= 143 Btu/hr.sqft.

◦

Oil side heat transfer coefﬁcient is obtained from Figure 13.11.

1st trial. Assume a tube wall temperature of 310

◦

For steam side ‘a’ = 143 × (354 −310)

=6292 Btu/hr.sqft.

For oil side h

=31 Btu/hr.sqft.

◦

F (Figure 13.11)

‘b’ = 31 ×(310 − 150)

=4960 Btu/hr.sqft

‘a’ −‘b’ =+1332

2nd trial. Assume a tube wall temperature of 320

◦

‘a’in this case calculated to be 4862

‘b’w as calculated to be 5355

‘a’ −‘b’ =−493

Plotted on a linear curve the tube wall temperature to give ‘a’ = ‘b’was 317

◦

606 CHAPTER 13

Final trial. At a tube wall temperature of 317

◦

Steam side =143 × (354 − 317)

‘a’ =5291 Btu/hr.sqft.

Oil side h

=31.2 (From Figure 13.11)

‘b’ =31.2 × (317 −150)

=5210 Btu/hr.sqft

‘a’ −‘b’ =+81 which is acceptable.

Make rate of heat transfer =

5291 + 5210

= 5251 Btu/hr.sqft

Then surface area of coil required =

10.730 mm Btu/hr

5251 Btu/hr.sqft

= 2043 sq ft

Appendix 13.2: Example calculation for sizing a relief value

A vessel containing naphtha C

–C

range is uninsulated and is not ﬁreproofed. The

vessel is vertical and has a skirt 15



in length. Dimensions of the vessel are I/D 6





T- T 2 0



0 liquid height to HLL = 16



0. Calculate the valve size for ﬁre condition relief.

Set pressure is 120 psig.

Latent heat of naphtha at 200

◦

F is 136 Btu/lb = H

Q = 21000 FA

0.82

A = Wetted area and is calculated as follows:

Liquid height above grade = 15 + 16 ft

= 31 ft

Therefore wetted surface of vessel need only be taken to 25 ft above grade which is

25−15 = 10 ft of vessel height.

Wetted surface = π D×10 for walls

= 188.5 sq.ft

plus 28.3 sq.ft for bottom

= 216.8 sq.ft

Q = 21000 × 1.0 × (216.8)

0.82

= 1.729 Btu/hr × 10

Q/H

1.729×10

136

= 12713 lbs/hr = W

A =

√

CKP

√

SUPPORT SYSTEMS COMMON TO MOST REFINERIES 607

where

A = effective discharge area in sq ins.

W = ﬂow through valve in lbs/hr = 12713

∗

= absolute temp of inlet vapor = 460 + 200 = 660

◦

Z = 0.98 (NC

)

C = 356.06 (based on CA/CV = 1.4)

K = 0.65 (from Figure 13.40)

= 0.9 M = 100 (use C

)

= set pressure of valve = 134.7 psia.

∗

Bubble point of C

to C

and say 10 psig

A =

12713 × 25.7 × 0.99

356 × 0.65 × 134.7 × 0.9 × 1

= 1.153 inch

nearest oriﬁce size = ‘J ’at 1 .287 inch

Appendix 13.3: Control valve sizing

Process ﬂow coefﬁcient (C

) and valve sizing

Process ﬂow coefﬁcient C

is deﬁned as the water ﬂow in GPM through a given

restriction for 1 psi pressure drop. These C

s can be determined by the following

equations:

= Q



P

for liquid



P − P

for steam

1360



P − P

for gases

where

= Flow coefﬁcient

= liquid ﬂow in GPM at conditions

P = pressure drop across valve, psi

= speciﬁc gravity of liquid at conditions

= steam rate in lbs/hr

= pressure downstream of valve psia

608 CHAPTER 13

= gas ﬂow in SCFH (60

◦

F, 14.7 psia).

T = temperature of gas in

◦

F + 460)

S = mol weight of gas divided by 29

= compressibility factor at downstream conditions

The following are some special considerations that may have to be made in determin-

ing.

Process C

values

Pressure drop

For compressible ﬂuids the maximum usable pressure drop in equations (b) and (c)

is the critical value. As a rule of thumb and for design purposes this value is 50% of

the absolute upstream pressure. (The valve can take more than the critical pressure

drop, but any pressure drop over the critical takes the form of exit losses).

Flashing liquids

In the absence of accurate information, it is recommended that for ﬂashing service

the valve body be speciﬁed as one nominal size larger than the valve port.

Two-phase ﬂow

If two-phase ﬂow exists upstream of the control valve experience has shown that for

ﬂuids below their critical point a sufﬁciently accurate process C

value can be arrived

at by adding the process C

values for the gas and liquid portions of the stream.

The calculation is based on the quantities of gas and liquid at upstream conditions.

The valve body is speciﬁed to be one nominal size larger than the port to allow for

expansion.

Valve rangeability

The rangeability of a control valve is the ratio of the ﬂow coefﬁcient at the maximum

ﬂow rate to the ﬂow coefﬁcient at the minimum ﬂow rate. (R = C

Max/C

Min).

Valve rangeability is actually a criteria which is used to judge whether a given valve

will be in a controlling position throughout its required range of operation (neither

wide open nor fully closed). In practice the selection of the actual valve to be installed

is the responsibility of the instrument engineer. As the process engineer is usually the

person responsible for the correct operation of the process itself however he must be

satisﬁed that the item selected meets the control criteria required. He must therefore

satisfy himself that the valve will control over the range of the process ﬂow.

Control valves are usually limited to a rangeability of 10 : 1. If R is greater than 10 : 1

then a dual valve installation should be considered in order to assure good control at

the maximum and minimum ﬂow conditions.

SUPPORT SYSTEMS COMMON TO MOST REFINERIES 609

In some applications, particularly on compressor or blower suction, butterﬂy valves

have been speciﬁed to be line size without considering that as a result the valve may

operate almost closed for long periods of time. Under this condition there have been

cases of erosion resulting from this. It is recommended therefore that butterﬂy valves

be sized so that they will not operate below 10% open for any appreciable period of

time and not arbitrarily be made line size.

Valve ﬂow coefﬁcient (C



)

In order to ensure that the valve is in a controlling position at the maximum ﬂow rate,

the valve C



is the maximum process C

value determined above, divided by 0.8.

The reasons for using this factor are that:

(1) It is not desirable to have the valve fully open at maximum ﬂow since it is not

then in a controlling position.

(2) The valves supplied by a single manufacturer often vary as much as 10–20% in

(3) Allowance must be made for pressure drop, ﬂow rate, etc., values which differ

from design.

Control valve sizing

Control valve sizes are determined by the manufacturers from the process data sub-

mitted to them. However there are available some simple equations to give a good

estimate of the required valve sizes to meet a process duty. Three of these are given

below:

S (inches) =



Valve C



1/2

Double seated control valves sizes may be estimated by:

S (inches) =



Valve C



1/2

Butterﬂy valve sizes may be estimated by:

S (inches) =



Valve C



1/2

The constants (9, 12 or 20) in these equations can vary as much as 25% depending

on the valve manufacturer.

A control valve should be no larger than the line size. A control valve size that is

calculated to be greater than line size should be carefully checked together with the

calculation used for determining line size. Ideally a control valve size should be one

size smaller than line size.

610 CHAPTER 13

Once the valve size is estimated and the valve C

known, then the percent opening

of the valve at minimum ﬂow and maximum ﬂow can be obtained by dividing the

respective process C

values conditions by the valve C

. This information is normally

required to check the percent opening of a butterﬂy at minimum ﬂow. It is not normally

necessary to calculate it for any other type of valve.

Valve action on air failure

In the analysis of the design and operation of any process or utility system the question

always arises on the action of control valves in the system on instrument air failure.

Should the control valve fail open or closed is the judgement decision of the process

engineer after evaluating all aspects of safety and damage in each event. For example,

control valves on ﬁred heater tube inlets should always fail open to prevent damage to

the tubes through low or no ﬂow through them when they are hot. On the other hand

control valves controlling fuel to the heaters should fail closed on air failure to avoid

over heating of the heater during the air failure.

The failed action of the valve is established by introducing the motive air to either

above the diaphragm for a failed open requirement or below the diaphragm for a failed

shut situation. The air failure to the valve above the diaphragm allows the spring to

pull up the plugs from the valve seats. Air failure to valves below the diaphragm

forces the spring to seat the valves in the closed position.

Chapter 14

Environmental control and engineering

in petroleum reﬁning

D.S.J. Jones

Introduction

Certainly during the last three decades of the 1900s control of contaminants from

the reﬁning of crude oil has become a signiﬁcant consideration in the design and

operation of the oil reﬁnery. Chapter 2 of this book provides the changes that have

taken place in the speciﬁcation of motive fuels to meet legislative environmental

control. These products have been singled out because they impact more promi-

nently on the daily life style and health of the general public, particularly in the

more advanced industrial countries. Similar changes and legislative control have

been adopted on the levels of efﬂuents leaving the reﬁnery plants in the water and

air. These changes have become more restrictive as greater knowledge of their ef-

fect on the environment emerges. This chapter then deals with some of the measures

adopted in process design and operation of processes to meet these environmental

protection measures. These are described and discussed in the following parts of the

chapter:

Part 1. Aqueous wastes

Part 2. Emission to the atmosphere

Part 3. Noise abatement

The aqueous and atmospheric contaminants that most reﬁneries emit are shown in

Table 14.1. The table also shows the process plant source of these more common

contaminants.

14.1 Aqueous wastes

This part is divided into a description of the pollutants in aqueous wastes, the pro-

cesses that generate the wastes, and the treatment of the various aqueous waste

streams.

611

612 CHAPTER 14

Table 14.1. Sources of waste water and air contaminants in oil reﬁning

Process Waste water Air

Atmospheric and vacuum

distillation

Sour water (NH

and H

Desalter water

Spent caustic

Process area waste water

(pump glands, area drains, etc.)

Furnace

Flue gases—SO

Thermal cracking

delayed coking

Sour water

Decking water (Oil)

Process Area Waste Water

Furnace

Flue gases

Fluid cat cracking

unsat gas plant

Sour water (NH

Phenols)

Spent caustic

Process area waste water

Furnace

Flue gases

,CO

Particulates.

Hydrocracker

hydrogen plant

Sour water (inc Phenols)

Process area waste water

Furnace

Flue Gases

Sat gas plant

alkylation

Spent caustic

Process area waste water

Nil

Naphtha hydrotreater

cat reformer

Sour water

Process area waste water

Furnace

Flue Gases

Sulfur plant Nil Incinerator

Flue Gas—SO

Hydrocarbons—

Flare

Tankage area Tank dike area drains

Non contaminated rain runoff

Tank Vents

Hydrocarbons

Pollutants in aqueous waste streams

The most undesirable pollutants in aqueous waste streams are:

Those that deplete the dissolved oxygen content of the waterways into which they

discharge

Those contaminants that are toxic to all forms of life, such as arsenic, cyanide,

mercury and the like

Those contaminants that impart undesirable tastes and odors to streams and other

waterways into which they discharge

A description and discussion on these contaminants are expressed in the following

paragraphs.

The effect of contaminants on the oxygen balance of natural waterways

Natural waterways have a complex and delicate oxygen balance. The water contains

an amount of dissolved oxygen which has an equilibrium of between 14 ppm in winter