Lecture 5- he pressure drop and pumping power

19
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

Transcript of Lecture 5- he pressure drop and pumping power

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

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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

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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

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• 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

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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

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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.

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Pressure drop in bends and fittings

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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

.

.

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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

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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

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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