Coker A.K. Fortran Programs for Chemical Process Design, Analysis, and Simulation

Подождите немного. Документ загружается.

Heat Transfer 637

AF a =

annulus pressure drop, ft

AFr = pressure drop due to reversal of flow in the annulus, ft

G- mass velocity, lb/ft2hr

G a - mass velocity in annulus, lb/ft2hr

Gp - mass velocity of inner pipe, lb/ft2hr

g = acceleration due to gravity, 32.2 ft/sec 2 (4.18 x 108 ft/hr 2)

H = fin height, inch

h~, h o - heat transfer film coefficient for inside fluid and outside fluid,

respectively, Btu/hr.ft.2~

h~o - value of h~ when referred to the pipe outside diameter, Btu/hr.

ft.2~

hie = heat transfer film coefficient corrected for fouling, Btu/hr.ft.2~

hie d = corrected film heat transfer rate, Btu/hr.ft.2~

ID = inside diameter, inch or ft

JH = heat transfer factor, dimensionless

k = fluid thermal conductivity, Btu/hr.ft.2~

K t = thermal conductivity of tube material, Btu/hr.ft.2~

L- pipe length, ft

LMTD = log mean temperature difference, ~

NFA = net free cross-sectional area, shell side,

ft 2

N = number of fins

OD = outside diameter, inch or ft

Pr = Prandtl number, dimensionless

AP = pressure drop, psi

Q = exchanger duty, Btu/hr

R D = fouling resistance, hrft2~

Re = Reynolds number, dimensionless

R w - wall resistance

s = specific gravity

Tl, T 2 --

temperature of hot fluid, inlet and outlet respectively, ~

tl, t 2 =

temperature of cold fluid, inlet and outlet respectively, ~

U = overall heat transfer coefficient, Btu/hr.ft.2~

V = fluid velocity, ft/sec

W = mass flow rate of hot fluid, lb/hr

w = mass flow rate of cold fluid, lb/hr

X = parameter used in fin efficiency

p = fluid density, lb/ft 3

bt, bt,v = fluid viscosity (flowing and at wall), cP

0 = fin thickness (normally 0.035 inch)

rt = 3.1415927

638

Fortran Programs for Chemical Process Design

a =

annulus

p = pipe

Subscripts

AIR COOLER DESIGN

Air coolers are widely used in the chemical process industries where

there are shortages of cooling water and regulatory controls on thermal

pollution. An economic comparison is often made between air and

water cooling to determine which is advantageous. Air coolers can be

used for cooling high pressure streams where the fluid is always inside

the tubes. They minimize heat exchanger maintenance costs because

descaling of the water-contact surface is eliminated. In addition, there is

some savings in first cost because no additional surface area is required

to allow for water side fouling between descaling. However, the

exchangers are larger because the gas film coefficient is smaller. Air

coolers eliminate problems of temperature rise and chemical pollution

of water resources.

Air cooled heat exchangers have rectangular bundles containing sev-

eral rows of tubes, horizontally aligned and vertically offset. Air flows

vertically upward across the tube bank. The flow can be induced by

fans above the bundle or forced by fans below the bundle. The heat

transfer is countercurrent, because the hot fluid enters at the top of the

bundle and flows downward through successive passes. The cost of an

air cooler depends on the length of the tubes and the number of the

tube rows.

A forced draft air cooler design has a power consumption advantage,

if the temperature rise of the air is high. It also allows a more conve-

nient and economical mounting arrangement when a number of bundles

and services are to be combined in a single unit. In addition, it develops

a high turbulence, and consequently, higher heat transfer coefficients.

Conversely, an induced draft design provides a more even distribution

of air across the bundles, and for a given bundle elevation, affords more

space for location of additional plant equipment beneath the unit. Induced

draft units are less likely to recirculate hot exhaust air, because the exit

air velocity is two or three times that of a forced draft unit. These units

give higher pressure drops and are more expensive. Figure 8-14 illus-

trates forced and induced air cooled exchanger arrangements.

Standard air coolers are in widths of 8, 10, 12, 16, or 20-ft lengths of

4-40ft, and stacks of 3-6 rows of tubes. Tubes are 0.75-1.0 inch OD

FAN~ ' FLUID

PLENUM ~ INLET

HEADER

ADER

SUPPORT

DRIVER I I FLUI[T

i i oo

_ __ __ I NDUCED DRAFT _~L

HEADER

TUBES

FLUID

I~LET

CHAMBER J~~AN

SUPPORT

FAN

DRIVER

FORCED DRAFT

HEADER

x 7

FLUID

OUTLET

Figure 8-14. Forced and induced draft air cooled exchangers.

639

640

Fortran Programs for Chemical Process Design

with 7-11 fins/inch each 0.5-0.625 in height and provide a total surface

15-20

times greater than the bare surface of the tube. Fans are 4-12 ft/

diameter; developed pressures are 0.5-1.5 inch water, and require power

inputs of 2-5 hp/MBtu/hr or about 7.5 hp/100

ft 2

of exchanger cross

section. Spacings are 1.8 times the width of the cooler. Face velocities

are about 10ft/sec at a depth of three rows and 8ft/sec at a depth of six

rows. The rating methods for air cooled exchangers are outlined by Cook

[16] and Ganapathy [17]. Here, the design of air coolers is based upon

correlations, tables, and graphs presented by Smith [18] and Brown [19]

and fitted to a series of equations developed by Blackwell [20].

The Equations

(1) The number of tube rows, R, is:

R-C~

+ C2 lnIT2-t2

U (8-92)

where C l = 3.1679

C 2 = 3.7948

(2) The area ratio is:

= C 3 (R) c4 (8-93)

where C 3 = 1.2557

C 4 =

1.0031

(3) The face velocity of air, FV, ft/min.

= C5(C6) R

(8-94)

where C 5 = 720.8542

C 6 = 0.9530

Typical face velocities (FVs) used for design are in Table 8-15. These

values result in air cooled heat exchangers that approach an optimum

cost [21 ]. This takes into account the purchase cost, the cost of installa-

tion, and the cost of power to drive the fans.

Table 8-16 lists an estimate of the outlet air temperature, based on

90 ~ to 95~ design ambient air temperature.

(4) Estimated air outlet temperature, t~, ~

Table 8-15

Design Face Velocities for Air-Cooled Exchangers

FACE VELOCITY, ft/min (m/sec.)

Number of

tube rows

8 fins/in.

(315 fins/m)

2.375 in.

(0.0603 m)

pitch

10 fins/in.

(394 fins/m)

2.375 in.

(0.0635 m)

pitch

10 fins/in.

(394 fins/m )

2.5 in.

(0.0635 m)

pitch

650 (3.30)

615(3.12)

585 (2.97)

560 (2.84)

625 (3.18)

600 (3.05)

575 (2.92)

550 (2.79)

700 (3.56)

600 (3.35)

625 (3.18)

600 (3.05)

Source: Chopey and Hicks [21].

Table 8-16

Estimated Outlet Air Temperature for Air-Cooled Exchangers

OUTLET AIR TEMPERATURE ~

PROCESS INLET

TEMPERATURE ~

175

150

125

100

U=50

U=100

U=150

lO0

Source: Chopey and Hicks [21].

641

642

Fortran Programs for Chemical Process Design

tJ - 0"005u[(T~ + T2)- t I + t2 2 2

(5) The effective log mean temperature difference, ~

(8-95)

LMTD -

(T,- t2)- (T 2 -t,)

In T~-t2}

(8-5)

(6) The heat duty Q, Btu/hr, is"

Q- wcAt

where w- fluid flow rate, lb/hr

c - specific heat capacity, Btu/lb~

At- temperature difference, ~

(7) The bare tube surface area based on the tube, OD, ft 2

(8-96)

A - Q (8-97)

(U)(LMTD)

(8) The face area of bundle FA, ft 2

FA - (8-98)

C3(R) c~

o F

(9) The calculated outlet air temperature, tj,

'= +t

t, (1.08)(FA)(FV) 2 (8-99)

(10) The air flow over tubes F, std.ftVmin

F - (FA)(FV) (8-100)

(1 l) The ratio of the bare tube surface area based on the tube OD to

the fan horsepower

Bhp

= + C (R) (8-101)

where

C 7 -

7.4212

C 8 - 12.5342

Heat Transfer 643

(12) The ratio of air cooler weight to the face area of bundle is"

= C 9 + C,0 (R) (8-102)

where

C 9 -

36.4

C~0- 9.35

(13) The tube bundle width, W, ft

W = (8-103)

(14) The area available, A v,

ft 2

A v - rc.N~.Nt.D o.L, ft 2

(

8-104)

where N~- number of rows

N t -

number of tubes per row

= (tube bundle width)/(tube spacing) - W/s

L- tube length, ft

D o - tube outside diameter, ft

If the area available, A v, is less than the area required, A, increase the

bundle width, and calculate the new outlet air temperature. Re-calculate

LMTD, the new required area, and the new available area.

(15) The air-side heat transfer coefficient, h a, Btu/hrftZ~ The air side

coefficient is determined on the basis of the outside surface of a bare

tube. h, is expressed as:

h a - 8(FV) ~ for 10 fins per inch (8-105)

h a - 6.75(FV) ~ for 8 fins per inch (8-106)

(16) The tube wall heat transfer coefficient h w, Btu/hrftZ~

2. k

h w = (8-107)

D O -D i

where k- thermal conductivity, Btu/hrft~

D o - outside diameter of tube, ft

D i - inside diameter of tube, ft

(17) The overall heat transfer coefficient U, Btu/hrft2~ is

1 1 1 1 1

= t + + ~ (8-108)

h a

hi(Di/Do) h w h

644

Fortran Programs for Chemical Process Design

where h i = inside of tube heat transfer coefficient

h~ = fouling coefficient

The Air-Side Pressure Drop,

AP a

(inch HzO )

The following equations are used to calculate the pressure drop on

the air side [21 ].

AP a = 0.0047N~(FV/100) ~s for 10 fins per inch, 2.375-inch spacing

(8-109)

AP~ = 0.0044N~(FV/100) ~8 for 8 fins per inch, 2.375-inch spacing

(8-110)

AP~ = 0.0037Nr(FV/100) ~8 for 10 fins per inch, 2.5 inch spacing

(8-111)

where

N r =

the number of tube rows

FV = face velocity, ft/min

AP~ = inch H20

Nomenclature

A = bare-tube surface area, ft 2

A v = available area, ft 2

Bhp = fan horsepower

c = specific heat of fluid, Btu/lb~

C~, C2, C3 = constants

C a, C 5, C 6 = constants

C 7, C 8, C 9 = constants

C~o = constant

D i = inside diameter of tube, ft

D o = outside diameter of tube, ft

F = air flow over tubes, stdft3/min

FA = face area of bundle,

ft 2

FV = face velocity of air, ft/min

h a = air-side heat transfer coefficient, Btu/hrftZ~

h i = inside of tube heat transfer coefficient, Btu/hrftZ~

h~ = fouling coefficient, Btu/hrftZ~

h w = tube wall heat transfer coefficient, Btu/hrft2~

k = thermal conductivity, Btu/hrft~

Heat Transfer 645

L = pipe length, ft

LMTD = log mean temperature difference, ~

N r = number of rows

N~ = number of tubes per row

APa = air-side pressure drop, inch

H20

Q = exchanger duty, Btu/hr

R = number of tube rows

T~ = outlet process fluid temperature, ~

T 2 -

inlet process fluid temperature, ~

t, = estimated air outlet temperature, ~

t~ = calculated outlet air temperature, ~

t 2 -

inlet air temperature, ~

At = temperature difference, ~

U = overall heat transfer coefficient (based on bare-tube O.D.),

Btu/hrftZ~

W = tube bundle width, ft

w = fluid flow rate, lb/hr

Wt = air cooler weight, lb

= 3.1415926

HEAT TRACER REQUIREMENTS FOR PIPELINES

AND HEAT LOSS FROM INSULATED PIPELINES

Pipelines conveying viscous fluids are maintained at an elevated tem-

perature by means of heat tracing. Pipelines containing vapors may also

be heat traced to prevent components from condensing out. The heat

loss from pipes is often reduced by thermal insulation. The thickness of

insulation depends on an economic analysis involving both capital cost

and the cost of heat loss from the insulated line. Water lines are com-

monly insulated to avoid freezing. However, heat is inevitably lost from

insulated lines, and if the cooling resulting from this heat loss cannot be

tolerated, heat tracing the lines becomes necessary. There are cases where

blockages occur in pipelines, which cannot be unblocked by flushing

with a solvent or by blowing in steam or air, unless they are heat traced.

Many types of fluid heating media are used for heat tracing, such as

steam, hot oil, or Dowtherm. Electric heating tape may be used. Steam

tracing is selected for about 60 percent of pipe footage in chemical pro-

cess plants Lam and Sandberg [22] have provided a comprehensive

checklist and criteria between steam and electric tracing. Table 8-17

lists a typical checklist for a petroleum refinery plant.

Table 8-17

Checklist to Decide Between Steam and Electric Heat Tracing

PRE-EXISTING PROJECT CONSTRAINTS

Heat tracing recommended in company (Yes/no)

engineering practices

Plant preference

PROJECT DESIGN PHASE CRITERIA

Total feet of traced pipe

Cost of electricity (S/kWh) and steam energy

($/1,000 lb)

Estimated annual maintenance cost per foot of

tracing (S/ft.)

Number of heat-tracing circuits required

Needed accuracy of temperature control (degrees)

Temperature control cost per circuit for needed

accuracy ($)

Cost of monitoring one circuit in distributed

control system ($)

Design time per circuit (h)

Design tools available for engineers (Good, Fair,

Poor, None)

Capital cost of steam capacity, condensate

return, etc. ($)

CalSital cost of required electrical power ($)

Other

PROJECT INSTALLATION CRITERIA

Labor cost to install tracing and accessories ($/ft)

Electric

Yes

4,700

$0.067

$1.67

+50 ~

No extra

NA*

0.5

Good

Low

$30.00

Training time per plant laborer (h)

Labor and overhead costs (S/h)

Total installed heat-tracing costs ($/ft)

Other

PROJECT OPERATION CRITERIA

Annual maintenance required (h/ft.)

Annual cost for replacement parts (S/ft.)

Total annual maintenance cost ($)

Total annual operating cost ($)

Other

$23.50

$62.00

0.03

$~.oo

$6,182

$7,179

Steam

Yes

4,700

Consider free

$1.33

+50 ~

No extra

NA*

1.5

Good

Low

$40.00

$23.5O

$74.00

0.05

$0.80

$6,807

*NA = not applicable, as the plant does not have a distributed control system capable of moni-

toring heat tracing

Reproduced by permission of the American Institute of Chemical Engineers, ~ 1992, A.I. ChE. All

rights reserved.

646