Transcript
Page 1: Lecture  5- he pressure drop and pumping power

• Fluid friction effects cause pressure drop

• Pressure drop calculation determines the pump power requirement

• Increasing pressure drop (pump power)

• Increases initial cost (capital cost), larger pumps are more expensive

• Also increases operational costs due to higher pumping power required

• Therefore pressure drop calculations are as important as heat transfer calculations.

Heat Exchanger Pressure Drop and

Pumping Power

Page 2: Lecture  5- he pressure drop and pumping power

Tube-side pressure drop: Circular ducts

• Functional relationship for frictional pressure drop (laminar or turbulent fully developed flow):

e: measure of surface roughness

• From Buckingham-pi theorem:

• Define LHS as Fanning friction factor:

• Thus: and

( , , , , )m i

pu d e

L

2,

4( / )( / 2)m

m i

i i

u dp e

L d u d

24( / )( / 2)mi

pf

L d u

(Re, / )if e d2

u

d

14f

L

p2m

i

dynamic head

Page 3: Lecture  5- he pressure drop and pumping power

Moody diagram and correlations for f

• In laminar region, no roughness dependence:

• In turbulent region

– Use Moody diagram, which is f vs. Re

– Or for smooth pipes use correlations given the table below

– Two common correlations for smooth ducts are

16

Ref

0.2 4 60.046Re for 3 10 Re 10f

0.25 3 50.079Re for 4 10 Re 10f

Page 4: Lecture  5- he pressure drop and pumping power

Moody Diagram

Page 5: Lecture  5- he pressure drop and pumping power

Table 4.1 Friction Factor Correlations (similar to Table 3.4)

Page 6: Lecture  5- he pressure drop and pumping power

• Hydraulic diameter:

• Using Dh in turbulent flow gives f within 8% of

measured values.

• For annulus:

• Use the correlation in the Table with Dh for turbulent flow

4(net free flow area)4

wetted perimeter

ch

AD

P

2 24( / 4)( )

( )

i oh i o

i o

D dD D d

D d

Tube-side pressure drop: Noncircular ducts

Turbulent flows

Page 7: Lecture  5- he pressure drop and pumping power

• No universal correlations

• Annular flow, f = 24/Re

• Ducts of triangular or trapezoidal cross section f = 16/Re

• Rectangular duct (a x b): f = 16/Re, where

• Depending on b/a ratio,

given in the attached Figure 4.3

4where

2( )h

abD

a b

Laminar flows

Page 8: Lecture  5- he pressure drop and pumping power

• Frictional pressure drop for flow through a duct of length L:

where G = um is mass velocity.

• Pressure drop for all the tubes in a shell-and-tube heat exchanger (single phase in tubes):

Np is number of tube passes, Dh=di

• The fluid will experience additional pressure drop due to expansions and contractions during a return. The return pressure drop

2 2

4 or 42 2

m

h h

uL L Gp f p f

D D

Pressure Drop

2

42

p

t

h

LN Gp f

D

2

42

mr p

up N

Page 9: Lecture  5- he pressure drop and pumping power

Pressure drop in helical and spiral coils (used as curved tube HEX)

In general, f is higher compared to straight tubes

Page 10: Lecture  5- he pressure drop and pumping power

Helical Coils, Laminar Flow

• De is Dean Number. De=Re (a/R)1/2 for the range (7 < R/a < 104):

c: curved tube

s: straight tube

0.275

0.5

1 for 30

0.419 for 30 300

0.1125 for 300

c

s

Def

De Def

De De

0.250.5 2 2

0.00725 0.076 Re for 0.034 Re 300c

R R Rf

a a a

0.20.5 2 2

0.0084 Re for Re 700 and 7 10c

R R R Rf

a a a a

Helical coils, turbulent flow

Page 11: Lecture  5- he pressure drop and pumping power

Spiral coils, laminar flow

Spiral coils, turbulent flow

0.7 0.7

2 1

0.6 0.3

0.63 for 500 Re( / ) 20000 and 7.3 / 15.5

Re ( / )c

n nf b a b a

b a

1.5

0.9 0.9

2 1

0.20.5

0.0074

Re( / )c

n nf

b a

where n1 and n2 are the number of turns from the origin to the start and the end of a spiral, respectively.

Page 12: Lecture  5- he pressure drop and pumping power

Pressure drop in bends and fittings

Page 13: Lecture  5- he pressure drop and pumping power

Pressure drop due to bends

• Total loss coefficient, K

fc : bend friction factor

f : friction factor for straight pipe at bend Re, given by

h

cm

D

LfKwhere

uKp

4

2

2

752370

54250

1010005525000080

101007910

ReforRe..f

ReforRe.f

.

.

Page 14: Lecture  5- he pressure drop and pumping power

Fittings: Procedure 1

• The pressure drop is usually given as the equivalent length

in pipe diameters of straight pipe (Le/di)

• Then, the pressure drop is calculated using

• Refer to Table for Le/di values of various valves and fittings

Fittings: Procedure 2

• Use total loss coefficient, K, in

• Obtain K from

2

u

d

L4fp

2

m

i

e

2

2

muKp

Page 15: Lecture  5- he pressure drop and pumping power
Page 16: Lecture  5- he pressure drop and pumping power
Page 17: Lecture  5- he pressure drop and pumping power
Page 18: Lecture  5- he pressure drop and pumping power

Pressure drop for abrupt contraction,

expansion, and momentum change • Due to flowing in and out of a HEX core

• Total p is

Kc : contraction loss coefficient (some values in Table 4.3)

Ke : enlargement loss coefficient (pressure rises during enlarg.)

ps: straight duct pressure loss

• For incompressible flow, ps given by

and is valid for liquids, but not gases.

• When density changes due to heating, the momentum

increase must be balanced by additional p, and

2

uKp

2

uKp

2

mes

2

mc

2

u

D

L4fp

2

m

h

s

io

2

m

h

s G2

u

D

L4fp

112 Note: 2nd term is negative when gas is cooled

Page 19: Lecture  5- he pressure drop and pumping power

Pump power

• Proportional to the pressure drop

• For an incompressible fluid with a mass flow rate ,

power required by an adiabatic pump is:

where is density, and hp is isentropic efficiency of the pump.

Note that two pumps are needed for both streams; the hot and cold fluids

1p

p

mW p

h

m


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