Lecture 5- he pressure drop and pumping power
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Transcript of 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
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
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
Moody Diagram
Table 4.1 Friction Factor Correlations (similar to Table 3.4)
• 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
• 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
• 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
Pressure drop in helical and spiral coils (used as curved tube HEX)
In general, f is higher compared to straight tubes
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
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.
Pressure drop in bends and fittings
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
.
.
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
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
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