Heat transfer and pipe flow

Transcription

Introduction
Mechanisms of heat transfer
Heat exchangers
Pumps
Heat transfer and pipe ow
Professor Eric S Fraga
Room 2.05 Engineering Front Building
Department of Chemical Engineering
UCL
c 2009
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Introduction
Objectives
Heat exchangers
Pumps
What is heat transfer?
How is heat transferred?
What equipment is used to transfer heat between two uids
and how does it work?
What is the power required for pumping a uid?
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Introduction
Heat exchangers
Reading material
Pumps
Coulson, J.M and Richardson J.F. Chemical Engineering Vol.
1, Pergamon Press.
Pitts, D. and Sissom, L. Heat Transfer, Schaum's Outlines,
McGraw-Hill.
Cengel, Y.A. Heat transfer a practical approach,
McGraw-Hill.
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Introduction
Heat exchangers
Heat transfer
Pumps
Heat transfer is concerned with
temperature: represents the thermal energy, or heat, that is
available, and
ow of heat: represents the movement of thermal energy from
one place to another.
Heat transfer is thermal energy in transit due to a
!temperature dierence!.
Temperature dierence is the driving force that causes heat to
be transferred.
Heat transfer plays a major role in the design of process
equipment.
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Introduction
Heat exchangers
How is heat transferred?
Pumps
Heat may be transferred in three dierent ways:
conduction transfer of heat as a result of molecular motion
and the subsequent transfer of kinetic energy.
Conduction is predominant in solid materials and
in static uids.
convection the ow of heat as a result of macroscopic
movement of matter from a hot to a cool region
radiation transfer of energy in the form of rays or waves or
particles (α, β , γ )
We will consider only conduction and convection.
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Introduction
Heat exchangers
Pumps
Conduction through a wall
Consider a wall of thickness x and surface area A that has a
uniform temperature T1 on one side and T2 on the other:
The heat transfer, q (W ), through the wall is
k
q = A∆T
x
(1)
where k (in mWK ) is the thermal conductivity of the material
which gives a measure of the ability of the material to conduct
heat and ∆T ≡ T1 − T2 .
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Introduction
Heat exchangers
Thermal conductivity
Materials with a large thermal conductivity,
Pumps
k , are good
thermal conductors. These will transfer large amounts of heat
over time, e.g. copper.
Conversely, materials with low thermal conductivities are poor
thermal conductors. These will transfer small amounts of heat
over time, e.g. concrete.
The ratio
k
x
is called the heat transfer coecient.
The inverse of the heat transfer coecient,
resistance.
x
k,
is the thermal
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Introduction
Heat exchangers
Composite wall example
Pumps
Applying eq. 1 to each section of the
composite wall:
⇒
k
q = i A(Ti − Ti +1 ) i = 1, . . . , 3
xi
1
q = P xi A∆T
ki
so overall thermal resistance is sum
of the individual resistances
(analogous to electrical circuits).
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Introduction
Heat exchangers
Heat transfer through convection
Pumps
Example: heating a pot of water.
At rst, when the water is cold and
still, it behaves as a solid and heat
is transferred by conduction
through the bottom of the pot.
Bubbles, when created, will transfer
heat from the bottom to the top
by convection due to buoyancy.
Cooler, more dense water at the
top will sink to the bottom.
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Introduction
Heat exchangers
Types of convection
Pumps
Natural convection occurs when the motion of uid is due to
buoyancy eects. Example, the cooling of a heated pipe:
Forced convection is when the uid motion is produced by an
external agent.
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Introduction
Heat exchangers
Pumps
Convective heat transfer
Heat transfer by convection, q , is
q = hA∆T
(2)
where
A (m2 ) is the characteristic area of contact.
∆T
(
K ) is the temperature dierence between the solid and
the uid.
h ( mW2 K ) is the convective heat transfer coecient and is a
property of the system, not a property of the uid as is the
thermal conductivity,
k.
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Introduction
Heat exchangers
Convective heat transfer coecient
Pumps
The value of h depends on the surface geometry, the
properties of the uid and the uid motion regime:
Type of convection Material h mW2 K
Natural
Gases
2-25
Liquids
50-1000
Forced
Gases
25-250
Liquids 100-20000
Liquids transfer greater amounts of heat than gases, which are
good thermal insulators and forced convection gives greater
heat transfer than natural convection for both gases and
liquids.
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Introduction
Heat exchangers
Pumps
What aects convective heat
transfer?
h is a function of the properties of the system and depends on:
geometry of the system, i.e. a characteristic length
physical properties of the uid: i.e. viscosity,
heat capacity,
uid regime, i.e. a characteristic velocity
To determine
µ,
cp , and thermal conductivity, k .
L.
density,
ρ,
v.
h is therefore a complex task and we rely on
experiments.
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Introduction
Heat exchangers
Prandtl and Nusselt numbers
Pumps
The properties of the uids and the dierent forms of heat
transfer can be described by some dimensionless numbers:
Prandtl the ratio between uid ability to store heat and to
transfer heat through conduction, independent of
the system geometry:
Pr =
µ cp
k
Nusselt ratio between heat transfer through convection
and conduction:
Nu =
hL
k
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Introduction
Heat exchangers
Pumps
Reynolds and Grashof numbers
Other numbers describe uid ow properties:
Reynolds The ratio between inertial and viscous forces in
the uid which identies the uid ow regime:
Re =
ρvL
µ
(3)
High values correspond to turbulent ow regime,
and therefore to high convection h.t.c.
Grashof replaces Re when uid motion is driven by
thermal expansion of the uid:
Gr =
β∆T ρ2 gL3
µ2
where β is the volumetric thermal expansion
coecient.
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Introduction
Heat exchangers
Flow patterns and temperature
proles
Pumps
Consider the ow of air over an innite at plate.
Ts
Tf is the air ow
Ts > Tf .
is the temperature of the plate, while
temperature away from the plate, with
Regardless of the type of convection, we analyse the ow
pattern and the temperature proles at the wall and away
from it.
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Introduction
Flow layers
Heat exchangers
Pumps
Adjacent to the wall, a laminar sub-layer forms with no uid
mixing. Heat transfer across the sub-layer is by conduction
only with large resistance to heat transfer and large
temperature change.
Away from the wall, turbulent ow with large eddies and high
uid mixing. Heat transfer is by convection with low
resistance to heat transfer and small temperature change.
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Introduction
Heat exchangers
Pumps
Overall heat transfer coecient
Given a at wall of uniform, homogeneous material having
constant thermal conductivity, k , exposed to uids h, at
temperature Th , and c , at Tc , on either side:
with convective heat transfer coecients hh and hc , wall
thermal conductivity, k , and wall thickness, x . What is the
overall heat transfer coecient, U ( mW2 K ), for the combined
conductive-convective heat transfer:
q = UA∆T
(4)
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Introduction
Heat exchangers
Assumptions
In steady state, the same amount of heat
Pumps
q must pass through
each section.
Heat transfer is by convection across the hot and cold lm
and by conduction through the solid wall.
Th and
Tc , are unaected by the heat transfer and are known.
The uid temperatures suciently far from the wall,
The surface temperatures
The heat transfer area,
T1 and T2 are unknown.
A, is the same for all sections.
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Introduction
Heat exchangers
Overall heat transfer
q = hh A(Th − T1 )
k
q = A(T1 − T2 )
x
q = hc A(T2 − Tc )
Pumps
hot lm
solid wall
⇓
A(T − T )
q = 1 h x c1 = UA∆T
cold lm
(solve)
hh + k + hc
Note: U can be calculated in a manner similar to resistances
in electrical networks:
1 X 1
≡
(5)
U
i
Ui
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Introduction
Heat exchangers
Example: double-paned glass
window
Pumps
Consider a 0.8m-high and 1.5m-wide double-pane glass
window consisting of two 4 mm thick layers of glass
W ) separated by a 10 mm wide stagnant air space
kg = 0.78 mK
W
(ka = 0.026
mK ).
(
Determine the steady rate of heat transfer,
q , through the
double-pane window and the temperature of its inner surface,
T1 , for a day during which the room is maintained at 20 ◦ C
while the temperature outside is
−10 ◦ C.
Take the convection heat transfer coecients on the inner and
outer surfaces of the window to be
h2
= 40 mW2 K .
h1 = 10 mW2 K
and
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Introduction
Heat exchangers
Solution: Diagram
Pumps
Assume that the heat transfer through the window is steady
state since the surface temperatures remain constant.
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Introduction
Heat exchangers
Solution: overall thermal resistance
From eq. 5,
where
1
U
≡
1
h1
L1
kg
L2
k2
L3
kg
1
h2
1
h1
=
=
=
=
=
+
Pumps
L1 L2 L1 1
+
+
+
kg k2 kg h2
1
m2 K
= 0.100
W
W
10 m2 K
0.004m
m2 K
=
0
.
00513
W
0.78 mWK
0.01m
m2 K
= 0.385
W
W
0.026 m K
0.004m
m2 K
=
0
.
00513
W
0.78 mWK
1
m2 K
= 0.025
W
W
40 m2 K
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Introduction
Heat exchangers
Solution: heat transferred
Pumps
m2 K
= 0.52026
U
W
W
U = 1.92 2
mK
A = 0.8 m × 1.5 m = 1.2 m2
q = UA(Ti − To )
W
= 1.92 2 × 1.2m2 × 30 K
mK
= 69.12W
1
⇒
and
∴
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Introduction
Heat exchangers
Solution: inner surface temperature
Pumps
q = h1 A(Ti − T1 )
⇓
T1 = Ti −
q
h1 A
= (20 + 273) K −
= 287.24 K
= 14.24◦ C
69.12W
10 mW2 K × 1.2 m2
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Introduction
Heat exchangers
Non-uniform heat transfer area
Pumps
In some cases, the area for transfer applicable to each media
could dier.
For example, the radial ow of heat through a thick pipe wall
or cylinder.
The area of transfer in these cases is a function of position.
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Introduction
Heat exchangers
Example: curved wall
Pumps
Consider a curved solid wall with constant thermal
conductivity k exposed to a convective hot outer uid and a
convective cold inner uid:
To
and
Ti
ho
and
hi
are the temperatures at
ro
and
ri
respectively.
are the convective heat transfer coecients in the
outer and inner lm, and impact of the thickness of these
lms is assumed to be negligible.
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Introduction
Derivation
Start with eq. 1,
Heat exchangers
Pumps
q = kx A∆T , and let x ≡ r .
Consider heat transfer over a small part of the pipe and the
corresponding change in temperature:
dT
=
q
dr
kA
Integrate over the pipe wall:
Z To
⇒
∴
Z ro
ro q
q
dT =
dr =
dr
ri kA
ri 2π rLk
Ti
q
To − Ti =
(log ro − log ri )
2π Lk
k
q=
2π L (To − Ti )
log(ro /ri )
Z
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Introduction
Heat exchangers
Combined system
Pumps
Assuming thin lms on either side of the pipe, we can write
three equations:
q = ho 2πro L(Th − To )
k
2π L(To − Ti )
q=
log(ro /ri )
q = hi 2πri L(Ti − Tc )
which have three unknowns, To , Ti and q , so we can solve as
usual.
Note: the heat transfer coecient will often be given with
respect to a specic reference area.
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Introduction
Heat exchangers
Engineers and heat transfer
Pumps
In practice, engineers often have to design equipment to eect
heat transfer, say to achieve a specic temperature change in
a uid stream of known mass ow rate,
Such equipment will typically be in the form of a heat
exchanger and the engineer will need to
determine the surface area to transfer heat at a given rate for
given uid temperatures and ow rates.
predict the outlet temperatures of hot and cold uid streams
for a specied heat transfer.
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Introduction
Heat exchangers
Heat exchangers
Pumps
A heat exchanger is any device that eects transfer of thermal
energy between two uids that are at dierent temperature.
The two uids do not come in direct contact but are
separated by a solid surface or tube wall.
Common heat exchangers include:
Shell-and-tube (single pass or multi-pass)
Flat-plate
Finned tubes
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Introduction
Heat exchangers
Double pipe heat exchanger
Pumps
The simplest form of an heat exchanger consists of two
concentric cylindrical tubes, the double pipe heat exchanger:
Heat transfer involves convection in each uid and conduction
through the wall separating the two uids.
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Introduction
Heat exchangers
Parallel or co-current ow
Pumps
The uids both ow in the same direction:
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Introduction
Heat exchangers
Counter-current ow
Pumps
The uids ow in opposite directions:
In both cases, the uids are forced to ow using pumps or fans.
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Introduction
Heat exchangers
Shell and tube single pass
Pumps
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Introduction
Heat exchangers
Pumps
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Introduction
Heat exchangers
Multi-pass shell and tube
Pumps
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Introduction
Heat exchangers
Multi-pass ow arrangement
Pumps
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Introduction
Heat exchangers
Pumps
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Introduction
Heat exchangers
Plate heat exchangers
Pumps
Plate heat exchangers are built up from individual plates
separated by gaskets, assembled in a pack and clamped in a
frame.
They are applied in the energy recovery section of many
processes because of low initial cost, high eciency and low
maintenance costs.
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Introduction
Plates
Heat exchangers
Pumps
Thin sheet material, resulting in economic
units, particularly when expensive material
is involved.
Plates are especially corrugated to
promote turbulence also at low
Re ,
resulting in:
very high heat transfer coecients
reduces fouling
facilitates chemical cleaning
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Introduction
Heat exchangers
Finned tube heat exchangers
Pumps
Finned tubes exchangers are
employed in large air/liquid heat
exchanger systems to give greater
heat transfer area because gases,
which are good thermal insulators,
transfer smaller amounts of heat
than liquids.
There are various types of nned
tubes, depending on the
application.
Applied in various systems:
large air conditioning systems
radiators for large truck.
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Introduction
Heat exchangers
Heat transfer in a heat exchanger
Pumps
Heat exchangers operate for long periods of time with no
change in the operating conditions, thus they can be modelled
as steady-ow devices:
1
The overall heat transfer coecient,
U , is constant
throughout the exchanger.
2
The mass ow rate of each uid remains constant.
3
The specic heats of the uids are constant.
4
The temperature of the two uids are constant over a specic
cross-section.
5
The outer surface is perfectly insulated, so that any heat
transfer occurs between the two uids only.
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Introduction
Heat exchangers
Pumps
Energy balance
Under these assumptions, it follows that the rate of heat
transfer from the hot uid to be equal to the rate of heat
transfer to the cold one.
The basic design equations for heat exchangers are therefore
the energy balance for each uid:
q = −ṁh cph (Tho − Thi )
q = ṁc cpc (Tco − Tci )
where
(Energy given up by hot uid)
(Energy gained by cold uid)
ṁ (kg/s) is the mass ow rate and cp
specic heat.
But what is the driving force,
(
kJ
kg K )
is the
∆T ?
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Introduction
Heat exchangers
Pumps
Temperature prole: co-current ow
The temperature dierence
∆T
is
large at the inlet but decreases
exponentially towards the outlet.
Temperature of the hot uid
decreases and the temperature of
the cold uid increases along the
heat exchanger.
The outlet temperature of the cold
uid can never exceed that of the
hot uid, no matter how long the
heat exchanger.
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Introduction
Heat exchangers
Pumps
Temperature prole: counter-current
ow
The hot and cold uids enter the
heat exchanger from opposite ends.
uid may exceed the outlet
temperature of the hot uid,
temperature cross.
uid can never exceed the inlet
temperature of the hot uid.
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Introduction
Design
Recall eq. 4,
Heat exchangers
Pumps
q = UA∆T .
Need to calculate
∆T
for either
co-current or counter-current
exchange.
However, the temperature
dierence varies across the range of
operation.
We introduce the log mean
temperature dierence (LMTD),
∆TLM ,
as an expression of
∆T .
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Introduction
Heat exchangers
Pumps
Log mean temperature dierence
derivation I
At any point along the
exchanger:
δ q = −ṁh cph δ Th
δ q = ṁc cpc δ Tc
δ q = U (Th − Tc )δ A
(6)
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Introduction
LMTD II
Heat exchangers
Pumps
Rearrange for δ T in each equation and nd the dierence:
δ Th = −
δq
ṁh cph
δq
δ Tc =
ṁc cpc
⇒
δ Th − δ Tc = −δ q
1
ṁh cph
+
1
ṁc cpc
δ(Th − Tc ) = −U δ A(Th − Tc )
δ(Th − Tc )
1
⇒
= −U δ A
+
(Th − Tc )
ṁh cph
+(6) :
1
+
ṁh cph
1
1
ṁc cpc
ṁc cpc
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Introduction
LMTD III
Heat exchangers
Pumps
Let the δ terms be dierentials and integrate along the length
of the exchanger:
log (Tho − Tco ) − log (Thi − Tci ) = −UA
1
ṁh cph
+
1
ṁc cpc
which, when combined with overall energy balance on each
uid (solve for ṁcp term and substitute):
log (Tho − Tco ) − log (Thi − Tci ) = −UA
⇓
log ∆T2 − log ∆T1 = UA
(Thi − Tho ) + (Tco − Tci )
q
∆T2 − ∆T1
q
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Introduction
Heat exchangers
∴
q = UA
∆T2 − ∆T1
log ∆T2 − log ∆T1
or
q = UA
LMTD IV
The term in the box is
Pumps
∆T2 − ∆T1
T2
log ∆
∆T1
∆TLM ,
the log mean temperature
dierence (LMTD).
For counter-current exchangers, the same result is obtained
although for dierent
∆T1
and
∆T2 .
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Introduction
Heat exchangers
Correction factor for complex heat
exchangers
Determination of the average temperature dierence,
Pumps
∆TLM ,
is dicult for complex heat exchangers.
It is practice to introduce a correction factor,
Ft .
The heat
transfer rate is then given by:
q = UA Ft ∆TLM
where
∆TLM
is that for the counter ow double-pipe heat
exchangers with the same uid inlet and outlet temperatures
as in the more complex design.
Ft
values for several congurations are given in reference
books (cf. Perry or Kern).
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Introduction
Heat exchangers
Example: area comparison
Pumps
Suppose we wish to exchange heat between a hot stream, with
Thi = 200 ◦ C and Tho = 150 ◦ C, and a cold stream, with
Tci = 80 ◦ C and Tco = 120 ◦ C.
Assuming the same heat transfer coecient,
and the same amount of heat,
U , in both cases
q , to be exchanged, what is the
ratio of area required for co-current exchange to area required
for counter-current exchange?
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Introduction
Solution
co-current
⇒
Heat exchangers
Pumps
∆T1 = 200 ◦ C − 80 ◦ C = 120 ◦ C
∆T2 = 150 ◦ C − 120 ◦ C = 30 ◦ C
30 ◦ C − 120 ◦ C
∆Tco-current =
≈ 65 ◦ C
◦
30 C
log 120
◦C
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Introduction
Solution
co-current
Heat exchangers
∆T1 = 200 ◦ C − 80 ◦ C = 120 ◦ C
⇒
∆T2 = 150 ◦ C − 120 ◦ C = 30 ◦ C
30 ◦ C − 120 ◦ C
∆Tco-current =
≈ 65 ◦ C
◦
counter-current
∆T1 = 200 ◦ C − 120 ◦ C = 80 ◦ C
⇒
Pumps
30 C
log 120
◦C
∆T2 = 150 ◦ C − 80 ◦ C = 70 ◦ C
70 ◦ C − 80 ◦ C
∆Tcounter-current =
≈ 75 ◦ C
◦
C
log 70
80 ◦ C
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Introduction
Solution
co-current
Heat exchangers
∆T1 = 200 ◦ C − 80 ◦ C = 120 ◦ C
⇒
∆T2 = 150 ◦ C − 120 ◦ C = 30 ◦ C
30 ◦ C − 120 ◦ C
∆Tco-current =
≈ 65 ◦ C
◦
counter-current
∆T1 = 200 ◦ C − 120 ◦ C = 80 ◦ C
⇒
∴
Pumps
30 C
log 120
◦C
∆T2 = 150 ◦ C − 80 ◦ C = 70 ◦ C
70 ◦ C − 80 ◦ C
∆Tcounter-current =
≈ 75 ◦ C
◦
C
log 70
80 ◦ C
q
U ∆Tco-current
Aco-current
=
q
Acounter-current U ∆T
counter-current
=
∆Tcounter-current
∆Tco-current
≈ 1.15
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Introduction
Heat exchangers
Pumps
Example: Cooling a hot process uid
in a counter-ow heat exchanger
A counter-ow double-pipe heat exchanger is used to cool a
hot process uid using water. The process uid ows at 18 kgs
and is cooled from 105 ◦ C to 45 ◦ C. The water ows
counter-currently to the process uid, entering at 25 ◦ C and
leaving at 50 ◦ C.
1
Assuming no heat losses, calculate the required ow-rate for
the cooling water. The specic heat for water is 4.2
that of the process uid is
2
kJ
3.4
kg K .
kJ
kg K
and
Neglecting the tube wall curvature, calculate the required area
for heat exchange. Under these conditions, the process uid
W
m2 K , the cooling
W
water side heat transfer coecient is 1200 2 . The tube wall
mK
W
thickness is 3 mm and the thermal conductivity is 220
mK .
side lm heat transfer coecient is 2500
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Introduction
Heat exchangers
Solution: Diagram
Pumps
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Introduction
Heat exchangers
Pumps
Solution: 1. required water ow rate
q = −Gf cpf (Tho − Thi ) = 3672kW (heat from hot stream)
= Gw cpw (Tco − Tci )
(energy balance)
⇓
Gw
= 34.97
kg
s
(water ow rate required)
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Introduction
Heat exchangers
Solution: 2. heat exchanger area
q = UA∆TLM
∆T2 − ∆T1
∆TLM =
∴
= 34.6 K
x 1
+
(thermal resistance)
U hf k hw
W
U = 802 2
(overall htc)
mK
q
= 132.3m2
A=
U ∆TLM
1
⇒
T2
log ∆
∆T1
Pumps
=
1
+
58 / 104
Introduction
Fouling
Heat exchangers
Pumps
The performance of a heat exchanger depends upon surfaces
being clean but deposits form over time.
The layer of deposits presents additional resistance to heat
transfer and must be accounted for by a fouling factor,
Rf
(cf.
Perry, Kern).
1
U
= Rconv,hot
uid
+ Rcond,wall +
Rf
+ Rconv,cold
uid
Deposits can occur by the precipitation of solid deposits (e.g.
calcium in a kettle), corrosion or chemical fouling due to
chemical reactions, and the growth of algae, biological fouling.
Can apply water treatment, coatings and chemical treatments.
Periodic cleaning of exchangers and the resulting down time
are additional penalties associated with fouling.
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Introduction
Heat exchangers
Pumping of liquid uids
Pumps
Pumps are the devices used to add energy to a uid in order
to maintain ow.
Fluids ow in the direction of decreasing pressure.
Pumps direct a uid from one vessel to another or through a
long pipeline.
The energy added to the uid compensates for the mechanical
energy losses due to friction and provides an increase in the
velocity, the pressure, or the height of the uid.
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Introduction
Valves
Heat exchangers
Pumps
Valves are used to control the ow rate:
On/O valves: gate (30% of all on/o valves) and ball
Throttling valves: globe (50%), needle and diaphragm
Check valves: allow ow only in one direction
Automatic valves
Relief valves (for safety): spring loaded, bursting disk.
61 / 104
Introduction
Heat exchangers
Pumping devices for liquids
Pumps
Liquids used in the chemical industry dier considerably in
physical and chemical properties so a variety of pump types
exists.
Most pumps fall into one of two major classications:
Positive-displacement pumps
Centrifugal pumps
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Introduction
Heat exchangers
Positive-displacement pumps
Pumps
Reciprocating pumps involve a back-and-forth motion of a
piston in a cylinder.
Rotary pumps depend upon a rotating motion.
The ow from these pumps is pulsating. The higher the speed
of the pump, the higher the ow rate delivered.
Deliver a controlled amount of liquid for each stroke or
revolution.
Used when nearly constant delivery rates are required.
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Introduction
Examples
Heat exchangers
Pumps
Reciprocating pumps:
Diaphragm pump
Piston pump
Rotary pumps:
Gear pump
Lobe pump
Peristaltic pump
Screw pump
For details on other types of pump refer to Coulson &
Richardson.
64 / 104
Introduction
Heat exchangers
The Piston pump
Pumps
In these pumps, the motion of a rotor is converted into a
back-and-forward motion of a piston.
The rate of liquid delivery is a function of the volume swept
out by the piston in the cylinder and the number of strokes
the piston make per unit time.
For each stroke of the piston, a xed volume of liquid is
discharged from the pump.
65 / 104
Introduction
Heat exchangers
Piston pumps
Pumps
Piston pumps may be single-acting, with the liquid admitted
only to the portion of the cylinder in front of the piston.
When the piston moves towards the disk it creates a partial
vacuum in the chamber. This allows atmospheric pressure to
push the uid from below.
On the return half of the cycle the top check valve opens and
the bottom one closes. The water is forced up through the
pipe.
66 / 104
Introduction
Heat exchangers
Double-acting piston pumps
Pumps
Piston pumps can also be double-acting, in which case the
feed is admitted to both sides of the piston.
As the piston moves downwards, it forces the uid out
through the bottom right valve and creates at the same time a
partial vacuum in the upper chamber, pushing the uid
through the top left valve.
When the piston is moving upwards it forces the uid out
through the top right valve, while creates at the same time a
partial vacuum in the lower chamber, pushing the uid in
through the bottom left valve.
67 / 104
Introduction
Heat exchangers
Piston pump applications
Pumps
These pumps can deliver the highest pressure of any other
pump.
Piston reciprocating pumps have long been used in many
applications, including pumping of oil, feed water and mud.
However, their capacities are relatively small compared to
centrifugal pumps.
Not used with liquid containing abrasive material as it can
damage the machined surfaces of the cylinder and piston.
68 / 104
Introduction
Heat exchangers
Diaphragm pump
Pumps
In one section a piston operates in a cylinder
in which an inert liquid is displaced.
The movement of the uid is transmitted by
means of the exible diaphragm.
They have been developed to handle corrosive
liquids or suspensions with abrasive solids.
They are used for example to move gasoline
from the gas tank to the carburetor.
They are not used for high pressure
applications.
69 / 104
Introduction
Heat exchangers
Rotary pumps: the Gear pump
Pumps
Has been developed to deal with
viscous uids.
Two gears operate within a casing.
Small packages of uids are carried
between the teeth and the casing
from the low pressure inlet side to
the high pressure delivery side.
70 / 104
Introduction
Heat exchangers
The 3 lobe pump
Pumps
Works on the same principle of the
gear pump but the gear teeth are
replaced by two or three lobes.
A small clearance between the
lobes can be maintained and wear
is reduced.
71 / 104
Introduction
Heat exchangers
The Peristaltic pump
Pumps
Delivers ow precisely controlled by the
speed of a rotor.
An elastic tube is compressed in stages by
a rotor.
As the rollers rotate, they atten the tube
against the track at the points of contact.
These ats move the uid along the tube.
They are mainly used in labs.
They are particularly good in handling
biological uids, where all forms of
contacts must be avoided.
72 / 104
Introduction
Heat exchangers
The screw pump
Pumps
One of the oldest pumps: its usage goes back 2000 years. It
consist of a helical screw. Suitable for very viscous uids (e.g.
polymers) and also for sewage uids. The uid is sheared in
the screw channel and so is raised to the delivery side.
73 / 104
Introduction
Heat exchangers
Pumps
Figure: Example of a screw pump used in a waste water treatment
plant.
74 / 104
Introduction
Heat exchangers
Centrifugal pumps
Pumps that cause the pumped uid to
rotate
Pumps
are called
centrifugal pumps.
Centrifugal pumps are the most widely applied in the chemical
and petroleum industry.
They are applied for large capacity applications.
They pump liquids with very wide-ranging properties and
suspensions with high solids content.
They cannot handle highly viscous uids.
75 / 104
Introduction
Heat exchangers
Pumps
A Centrifugal pump has an impeller enclosed by a casing or
volute. The impeller consists of a series of curved vanes. The
greater the number of vanes, the greater is the control over
the direction of motion of the uid.
76 / 104
Introduction
Heat exchangers
Pumps
The uid is fed to the centre of the
rotating impeller and is thrown
from the impeller vanes into the
casing by centrifugal force.
As a result, the liquid acquires a
high kinetic energy. This velocity
energy is converted in pressure as
the uid leaves the impeller and
enters the casing.
77 / 104
Introduction
Heat exchangers
Turbine pump
Pumps
In the turbine pump, the liquid
ows from the impeller into a series
of xed vanes, called diusers.
These give more gradual change in
direction to the uid and more
ecient conversion of kinetic
energy into pressure energy.
78 / 104
Introduction
Heat exchangers
Summary of types
Pumps
79 / 104
Introduction
Pumps
Pump characteristics
Centrifugal
Head
Heat exchangers
High, single stage
up to 600 ft
Multistage
up to 6000 psi
Capacity Low (100 gal/min)
to very high
(200,000 gal/min)
Liquids Clear or dirty,
non viscous
Reciprocating
(piston)
Highest available
100,000 psi
Rotary
(gear or screw)
Intermediate
up to 600 psi
Intermediate
(500 gal/min)
Low (1 gal/min)
to intermediate
(500 gal/min)
High viscosity,
non abrasive
Clean
no solids
80 / 104
Introduction
Pump
Heat exchangers
Summary of characteristics
Pressure
delivered
Piston
very high
Diaphragm not high
Rotating
not high
Centrifugal not high
Flow
delivered
pulsating
pulsating
pulsating
continuous
Pumps
Capacity
Fluid
not high
not high
not high
very high
non-corrosive
corrosive
highly viscous
not viscous
81 / 104
Introduction
Heat exchangers
Operating characteristics of
centrifugal pumps
Pumps
Centrifugal pumps are the most widely applied in the chemical
and petroleum industry.
Centrifugal pumps operate at constant speed and the capacity
depends upon the total head,
H , the design and the suction
conditions.
Pumps usually achieve maximum eciency at one particular
ow rate.
Operating characteristics are described through use of
characteristic curves.
82 / 104
Introduction
Heat exchangers
Characteristic curves
Pumps
For a pump at a particular speed, the characteristic curves
show the inter-relation between:
Total head and capacity, the H − Q curve.
Power input and pump capacity, the P − Q curve.
Pump eciency and capacity, the η − Q curve.
The duty point shows the optimum conditions for operation.
This is the point where the head curve cuts the ordinate
through the point of maximum eciency.
83 / 104
Introduction
Heat exchangers
Characteristic curve graph
Pumps
For a pump having the characteristics shown above: maximum
eciency would occur at a capacity of 2500 gal/min and a
total head of 80ft.
84 / 104
Introduction
Heat exchangers
Variable speeds
Pumps
When a pump is capable of being operated at variable speeds,
then, at higher speed of rotation gives higher capacity and
requires more horsepower to supply the increased supply of
liquid.
85 / 104
Introduction
Heat exchangers
Pump design
Pumps
The energy required for a pump will depend on
the height through which the uid raises,
the pressure required at the delivery point,
the length and diameter of the pipe,
the rate of ow, and
the physical properties of the uid, density and viscosity.
86 / 104
Introduction
Heat exchangers
General scenario for pumping
Pumps
87 / 104
Introduction
Heat exchangers
Energy balance
Pumps
The work done on a uid by a pump is expressed as head, H
(m), and is given by the mechanical energy balance (per unit
mass):
p u2
H=∆
+
+ z + ∆hfT
ρg
2g
P2 − P1 u22 − u12
+
+ (z2 − z1 ) + ∆hfT
(7)
=
ρg
2g
where ∆hfT represents the total friction loss in the system and
is the sum of the losses in the whole pipe length and other
losses due to ttings:
∆hft = ∆hf + ∆htting loss
where hf is the frictional head loss.
88 / 104
Introduction
Heat exchangers
Frictional losses
Pumps
∆hf is expressed as a frictional pressure drop:
∆hf = 2cf
1 u2
dg
=
∆pf
ρg
where the Fanning friction factor, cf , is given as a function of
the Reynolds number, Re (eq. 3):
16
Laminar ow cf = Re
Turbulent ow cf = 0.079Re −0.25
Rough pipes use Moody diagram to estimate the value
89 / 104
Introduction
Heat exchangers
Pumps
Fitting losses
Loss due to ttings, such as bends and valves, can generate
large-scale turbulence in which energy is dissipated as heat.
For turbulent ow, these losses are proportional to the square
of the uid velocity and can be expressed as the frictional loss
due to an
equivalent length of straight pipe, le ,
estimated
as a multiple of the pipe diameter:
le = nd
so that
∆htting
loss
= 2cf
le u 2
d g
90 / 104
Introduction
Heat exchangers
Pumps
Total friction losses
Total friction loss is therefore equal to
l + le u 2
∆hfT = 2cf
d g
(8)
which in terms of pressure losses is
∆pfT = ρg ∆hfT = 2cf
l + le 2
ρu
d
91 / 104
Introduction
Heat exchangers
Power requirements
Pumps
Thus, from eq. 7, the head, H , that must be supplied by the
pump is:
P2 − P1 u22 − u12
l + le u 2
H=
+
+ (z2 − z1 ) + 2cf
ρg
2g
d g
The power required by the pump to deliver H is given by:
Power = ∆Pp Q = H ρg Q
where Q is the volumetric owrate of the uid that the pump
moves and is known as the capacity of the pump.
92 / 104
Introduction
Example
Heat exchangers
Pumps
2.32 mh of water is pumped in a 35 mm internal diameter pipe
through a distance of 125 m in a horizontal direction and then
up through a vertical height of 12 m. The friction loss in the
90◦ square elbow may be taken as equivalent to 60 pipe
diameters. Also in the line there is a control valve fully open
and frictional losses may be taken equivalent to 200 pipe
diameters.
Calculate the total head ∆hfT to be developed to overcome
the total frictional losses in the pipeline. You may assume that
for this pipe f = 0.079Re −0.25 . Assume the water to ow in
turbulent regime through the pipe. Density and viscosity of
water in the pipe are 1000 mkg3 and 0.65 mN
m2 s respectively.
3
93 / 104
Introduction
Heat exchangers
Pumps
Solution: data collection
Quantity
Volumetric Flow rate
Density of water
Viscosity of water
Value
Q
2.32
ρ
1000
µ
0.65
0.65 × 10−3
Internal pipe diameter
d
35
0.035
Horizontal pipe length
lh
125
Vertical pipe length
lv
12
◦
90 square elbow friction loss
60d
Control valve friction loss
200d
Friction coecient
cf 0.079Re −0.25
Units
m3
h
kg
m3
mN s
m2
N
m2 s
mm
m
m
m
m
m
94 / 104
Introduction
Solution
Heat exchangers
Pumps
The total friction losses, eq. 8, in the pipeline for turbulent
ow regime is expressed as:
∆hfT =
2cf (l + le )u 2
gd
with l = lh + lv = 125m + 12m = 137m and le from the
denition of the equivalent pipe length:
le = 200d + 60d = 200x 0.035m + 60x 0.035m = 9.1m
The velocity of the water in the pipe is given by the volumetric
ow rate divided by the pipe cross-section area:
u=
Q
π
=
d 2
2
1h
2.32 mh × 3600
s
3
π×
= 0.67
0.035m 2
2
m
s
95 / 104
Introduction
Solution
Heat exchangers
Pumps
The friction coecient depends on Re , the Reynolds number
(eq. 3):
0.035m × 0.67 ms × 1000 mkg3
Re =
= 36076
=
µ
0.65 × 10−3 mN2 s
cf = 0.079Re −0.25 = 0.079 × 36076−0.25 = 0.0057
2cf (l + le )u 2
du ρ
⇒
⇒ ∆hfT =
=
gd
2 × 0.0057 × (137 + 9.1)m × (0.67 ms )2
= 2.17m
9.81 sm2 × 0.035m
⇒ ∆pfT = ∆hfT ρg = 2.17m × 9.81
m
kg
×
1000
s2
m3
= 21287
N
m2
96 / 104
Introduction
Heat exchangers
Cavitation in centrifugal pumps
Pumps
Cavitation is the formation of cavities or bubbles in a pumped
uid when the pressure on the uid falls below the vapour
pressure of the liquid,
Pv
(or
p∗ ).
When a centrifugal pump is operated at high capacity, low
pressure may develop at the impeller eye or vane tips and so
vapourisation may occur.
Cavitation leads to:
A reduction in pump capacity.
A reduction in the head of the pump.
A noise that can be heard when the pump is running.
Mechanical damage that can be seen on the pump impeller
and volute.
To avoid cavitation, the pressure at the pump inlet must
exceed the vapour pressure of the liquid.
97 / 104
Introduction
Heat exchangers
Pump congurations
Pumps
Negative suction head
Positive suction head
Pump is placed above the
reservoir of uid to be pumped,
drawing uid up with suction:
Pump is placed below the
reservoir of uid to be pumped
and is fed by gravity action:
98 / 104
Introduction
Heat exchangers
Net positive suction head (NPSH)
Pumps
To avoid cavitation, the pressure at the pump inlet must
exceed the vapour pressure of the liquid. There are two cases:
NPSH is the amount by which the pressure at the suction
point of the pump must exceed the vapour pressure of the
liquid and is expressed as a head of the liquid to be pumped.
For any pump, the manufacturers specify the minimum value
of the NPSH required at the impeller eye to avoid cavitation.
99 / 104
Introduction
Heat exchangers
Typical conguration
Pumps
P1 , pressure at the tank
liquid surface.
h1 , height of the liquid
surface above the pump
centre-line at the suction
inlet.
u1 , liquid velocity on the
surface.
u2 , velocity at the pump
inlet.
hf , total piping friction loss
between (1) and (2).
100 / 10
Introduction
Heat exchangers
Pumps
Available NPSH
Start with energy balance (per unit mass):
P1
ρg
u12
+
2g
P2 u22
+ h1 − hf =
+
+ h2
(9)
ρg
2g
If the reference plane is taken at h2 , and the liquid velocity in
the reservoir, u1 , is negligible compared with u2 :
P1
ρg
+ h1 − hf =
P2
ρg
+
u22
2g
(10)
The available NPSH is the dierence between the static head
and the head corresponding to the vapour pressure of the
liquid at the suction inlet.
NPSHA =
P2
ρg
u2
+ 2
2g
−
Pv
ρg
=
P1 − Pv
ρg
+ h1 − hf
(11)
101 / 10
Introduction
Heat exchangers
Pumps
Required NPSH - I
From eq. 10, the total head at the suction inlet is:
P2
ρg
=
P1
ρg
+ h1 − hf −
u22
2g
(12)
Cavitation usually occurs at the impeller eye where the
pressure will be less than at the suction inlet by
∆P = φ
u32
2g
(13)
Where φ is a pressure drop coecient characteristic of pump
geometry and u3 is the uid velocity at the eye.
102 / 10
Introduction
Heat exchangers
Required NPSH II
Pumps
Cavitation is probable if the total head at the impeller eye (rhs
of eq. 12 minus eq. 13) is equal to or less than the vapour
pressure:
P1
ρg
+ h1 − hf −
Limiting case therefore is
P1 − Pv
ρg
u22
2g
−φ
+ h1 − hf =
u32
2g
u22
2g
≤
Pv
ρg
+φ
u32
2g
where the right hand side is the NPSH required at the impeller
eye:
NPSHR =
u22
2g
+φ
u32
2g
103 / 10
Introduction
Heat exchangers
Avoiding cavitation
Pumps
In order to avoid cavitation, the NPSHA available has to be
greater than the NPSHR required at impeller eye:
NPSHA
> NPSHR
The value of the required NPSHR for the particular pump
being used may be obtained from the pump manufacturer.
If NPSHA is too low, then cavitation can be avoided by
increasing
h1 .
This is why pumping uids that are close to saturated
conditions require that the vessel upstream be elevated . . .
. . . or, more generally, that the pump be lowered.
104 / 10

Heat transfer and pipe flow

Transcription

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